ADVANCE PRAISE FOR INTRODUCTION TO OPERATIONS RESEARCH, SEVENTH EDITION
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ADVANCE PRAISE FOR INTRODUCTION TO OPERATIONS RESEARCH, SEVENTH EDITION

Reviewers seem to agree that this is clearly the best edition yet. Here is a sampling of comments: “The new edition seems to contain the most current information available.” “The new edition of Hillier/Lieberman is very well done and greatly enhances this classic text.” “The authors have done an admirable job of rewriting and reorganizing to reflect modern management practices and the latest software developments.” “It is a complete package.” “Hillier/Lieberman has recaptured any advantage it may have lost (to other competitors) in the past.” “The changes in this new edition make Hillier/Lieberman the preeminent book for operations research and I would highly recommend it.”

INTRODUCTION TO OPERATIONS RESEARCH

McGraw-Hill Series in Industrial Engineering and Management Science CONSULTING EDITORS Kenneth E. Case, Department of Industrial Engineering and Management, Oklahoma State University Philip M. Wolfe, Department of Industrial and Management Systems Engineering, Arizona State University Barnes Statistical Analysis for Engineers and Scientists: A Computer-Based Approach Bedworth, Henderson, and Wolfe Computer-Integrated Design and Manufacturing Blank and Tarquin Engineering Economy Ebeling Reliability and Maintainability Engineering Grant and Leavenworth Statistical Quality Control Harrell, Ghosh, and Bowden Simulation Using PROMODEL Hillier and Lieberman Introduction to Operations Research Gryna Quality Planning and Analysis: From Product Development through Use Kelton, Sadowski, and Sadowski Simulation with ARENA Khalil Management of Technology Kolarik Creating Quality: Concepts, Systems, Strategies, and Tools Creating Quality: Process Design for Results Law and Kelton Simulation Modeling and Analysis Nash and Sofer Linear and Nonlinear Programming Nelson Stochastic Modeling: Analysis and Simulation Niebel and Freivalds Methods, Standards, and Work Design Pegden Introduction to Simulation Using SIMAN Riggs, Bedworth, and Randhawa Engineering Economics Sipper and Bulfin Production: Planning, Control, and Integration Steiner Engineering Economics Principles

INTRODUCTION TO OPERATIONS RESEARCH Seventh Edition

FREDERICK S. HILLIER, Stanford University

GERALD J. LIEBERMAN, Late of Stanford University

Cases developed by Karl Schmedders and Molly Stephens Tutorial software developed by Mark Hillier and Michael O’Sullivan

Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Bangkok Bogotá Caracas Lisbon London Madrid Mexico City Milan New Delhi Seoul Singapore Sydney Taipei Toronto

McGraw-Hill Higher Education A Division of The McGraw-Hill Companies

INTRODUCTION TO OPERATIONS RESEARCH Published by McGraw-Hill, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY, 10020. Copyright © 2001, 1995, 1990, 1986, 1980, 1974, 1967, by The McGraw-Hill Companies, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGrawHill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 ISBN

DOC/DOC

0 9 8 7 6 5 4 3 2 1 0

0072321695

Vice president/Editor-in-chief: Kevin Kane Publisher: Thomas Casson Executive editor: Eric M. Munson Developmental editor: Maja Lorkovic Marketing manager: John Wannemacher Project manager: Christine A. Vaughan Manager, new book production: Melonie Salvati Coordinator, freelance design: Gino Cieslik Supplement coordinator: Cathy Tepper Media technology producer: Judi David Cover design: Gino Cieslik Cover Illustration: Paul Turnbaugh Compositor: York Graphic Services, Inc. Typeface: 10/12 Times Printer: R. R. Donnelley & Sons Company Library of Congress Cataloging-in-Publication Data Hillier, Frederick S. Introduction to operations research/Frederick S. Hillier, Gerald J. Lieberman; cases developed by Karl Schmedders and Molly Stephens; tutorial software developed by Mark Hillier and Michael O’Sullivan.—7th ed. p. cm. ISBN 0-07-232169-5 1. Operations research. I. Lieberman, Gerald J. II. Title. T57.6. H53 2001 658.4034—dc21 00-025683 www.mhhe.com

ABOUT THE AUTHORS

Frederick S. Hillier was born and raised in Aberdeen, Washington, where he was an award winner in statewide high school contests in essay writing, mathematics, debate, and music. As an undergraduate at Stanford University he ranked first in his engineering class of over 300 students. He also won the McKinsey Prize for technical writing, won the Outstanding Sophomore Debater award, played in the Stanford Woodwind Quintet, and won the Hamilton Award for combining excellence in engineering with notable achievements in the humanities and social sciences. Upon his graduation with a B.S. degree in Industrial Engineering, he was awarded three national fellowships (National Science Foundation, Tau Beta Pi, and Danforth) for graduate study at Stanford with specialization in operations research. After receiving his Ph.D. degree, he joined the faculty of Stanford University, and also received visiting appointments at Cornell University, Carnegie-Mellon University, the Technical University of Denmark, the University of Canterbury (New Zealand), and the University of Cambridge (England). After 35 years on the Stanford faculty, he took early retirement from his faculty responsibilities in 1996 in order to focus full time on textbook writing, and so now is Professor Emeritus of Operations Research at Stanford. Dr. Hillier’s research has extended into a variety of areas, including integer programming, queueing theory and its application, statistical quality control, and the application of operations research to the design of production systems and to capital budgeting. He has published widely, and his seminal papers have been selected for republication in books of selected readings at least ten times. He was the first-prize winner of a research contest on “Capital Budgeting of Interrelated Projects” sponsored by The Institute of Management Sciences (TIMS) and the U.S. Office of Naval Research. He and Dr. Lieberman also received the honorable mention award for the 1995 Lanchester Prize (best English-language publication of any kind in the field of operations research), which was awarded by the Institute of Operations Research and the Management Sciences (INFORMS) for the 6th edition of this book. Dr. Hillier has held many leadership positions with the professional societies in his field. For example, he has served as Treasurer of the Operations Research Society of America (ORSA), Vice President for Meetings of TIMS, Co-General Chairman of the 1989 TIMS International Meeting in Osaka, Japan, Chair of the TIMS Publications Committee, Chair of the ORSA Search Committee for Editor of Operations Research, Chair of the ORSA Resources Planning Committee, Chair of the ORSA/TIMS Combined Meetings Committee, and Chair of the John von Neumann Theory Prize Selection Committee for INFORMS. vii

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ABOUT THE AUTHORS

He currently is serving as the Series Editor for the International Series in Operations Research and Management Science being published by Kluwer Academic Publishers. In addition to Introduction to Operations Research and the two companion volumes, Introduction to Mathematical Programming and Introduction to Stochastic Models in Operations Research, his books are The Evaluation of Risky Interrelated Investments (NorthHolland, 1969), Queueing Tables and Graphs (Elsevier North-Holland, 1981, co-authored by O. S. Yu, with D. M. Avis, L. D. Fossett, F. D. Lo, and M. I. Reiman), and Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets (Irwin/McGraw-Hill, co-authored by M. S. Hillier and G. J. Lieberman). The late Gerald J. Lieberman sadly passed away shortly before the completion of this edition. He had been Professor Emeritus of Operations Research and Statistics at Stanford University, where he was the founding chair of the Department of Operations Research. He was both an engineer (having received an undergraduate degree in mechanical engineering from Cooper Union) and an operations research statistician (with an A.M. from Columbia University in mathematical statistics, and a Ph.D. from Stanford University in statistics). Dr. Lieberman was one of Stanford’s most eminent leaders in recent decades. After chairing the Department of Operations Research, he served as Associate Dean of the School of Humanities and Sciences, Vice Provost and Dean of Research, Vice Provost and Dean of Graduate Studies, Chair of the Faculty Senate, member of the University Advisory Board, and Chair of the Centennial Celebration Committee. He also served as Provost or Acting Provost under three different Stanford presidents. Throughout these years of university leadership, he also remained active professionally. His research was in the stochastic areas of operations research, often at the interface of applied probability and statistics. He published extensively in the areas of reliability and quality control, and in the modeling of complex systems, including their optimal design, when resources are limited. Highly respected as a senior statesman of the field of operations research, Dr. Lieberman served in numerous leadership roles, including as the elected President of The Institute of Management Sciences. His professional honors included being elected to the National Academy of Engineering, receiving the Shewhart Medal of the American Society for Quality Control, receiving the Cuthbertson Award for exceptional service to Stanford University, and serving as a fellow at the Center for Advanced Study in the Behavioral Sciences. In addition, the Institute of Operations Research and the Management Sciences (INFORMS) awarded him and Dr. Hillier the honorable mention award for the 1995 Lanchester Prize for the 6th edition of this book. In 1996, INFORMS also awarded him the prestigious Kimball Medal for his exceptional contributions to the field of operations research and management science. In addition to Introduction to Operations Research and the two companion volumes, Introduction to Mathematical Programming and Introduction to Stochastic Models in Operations Research, his books are Handbook of Industrial Statistics (Prentice-Hall, 1955, co-authored by A. H. Bowker), Tables of the Non-Central t-Distribution (Stanford University Press, 1957, co-authored by G. J. Resnikoff), Tables of the Hypergeometric Probability Distribution (Stanford University Press, 1961, co-authored by D. Owen), Engineering Statistics, Second Edition (Prentice-Hall, 1972, co-authored by A. H. Bowker), and Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets (Irwin/McGraw-Hill, 2000, co-authored by F. S. Hillier and M. S. Hillier).

ABOUT THE CASE WRITERS

Karl Schmedders is assistant professor in the Department of Managerial Economics and Decision Sciences at the Kellogg Graduate School of Management (Northwestern University), where he teaches quantitative methods for managerial decision making. His research interests include applications of operations research in economic theory, general equilibrium theory with incomplete markets, asset pricing, and computational economics. Dr. Schmedders received his doctorate in operations research from Stanford University, where he taught both undergraduate and graduate classes in operations research. Among the classes taught was a case studies course in operations research, and he subsequently was invited to speak at a conference sponsored by the Institute of Operations Research and the Management Sciences (INFORMS) about his successful experience with this course. He received several teaching awards at Stanford, including the university’s prestigious Walter J. Gores Teaching Award. Molly Stephens is currently pursuing a J.D. degree with a concentration in technology and law. She graduated from Stanford University with a B.S. in Industrial Engineering and an M.S. in Operations Research. A champion debater in both high school and college, and president of the Stanford Debating Society, Ms. Stephens taught public speaking in Stanford’s School of Engineering and served as a teaching assistant for a case studies course in operations research. As a teaching assistant, she analyzed operations research problems encountered in the real world and the transformation of these problems into classroom case studies. Her research was rewarded when she won an undergraduate research grant from Stanford to continue her work and was invited to speak at an INFORMS conference to present her conclusions regarding successful classroom case studies. Following graduation, Ms. Stephens worked at Andersen Consulting as a systems integrator, experiencing real cases from the inside, before resuming her graduate studies.

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DEDICATION

To the memory of our parents and To the memory of one of the true giants of our field, Jerry Lieberman, whose recent passing prevented him from seeing the publication of this edition

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It now is 33 years since the first edition of this book was published in 1967. We have been humbled by having had both the privilege and the responsibility of introducing so many students around the world to our field over such a long span of time. With each new edition, we have worked toward the goal of meeting the changing needs of new generations of students by helping to define the modern approach to teaching the current status of operations research effectively at the introductory level. Over 33 years, much has changed in both the field and the pedagogical needs of the students being introduced to the field. These changes have been reflected in the substantial revisions of successive editions of this book. We believe that this is true for the current 7th edition as well. The enthusiastic response to our first six editions has been most gratifying. It was a particular pleasure to have the 6th edition receive honorable mention for the 1995 INFORMS Lanchester Prize (the prize awarded for the year’s most outstanding Englishlanguage publication of any kind in the field of operations research), including receiving the following citation. “This is the latest edition of the textbook that has introduced approximately one-half million students to the methods and models of Operations Research. While adding material on a variety of new topics, the sixth edition maintains the high standard of clarity and expositional excellence for which the authors have long been known. In honoring this work, the prize committee noted the enormous cumulative impact that the Hillier-Lieberman text has had on the development of our field, not only in the United States but also around the world through its many foreign-language editions.” As we enter a new millennium, the particular challenge for this new edition was to revise a book with deep roots in the 20th century so thoroughly that it would become fully suited for the 21st century. We made a special effort to meet this challenge, especially in regard to the software and pedagogy in the book.

A WEALTH OF SOFTWARE OPTIONS The new CD-ROM that accompanies the book provides an exciting array of software options that reflect current practice. One option is to use the increasingly popular spreadsheet approach with Excel and its Solver. Using spreadsheets as a key medium of instruction clearly is one new wave in xxiii

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the teaching of operations research. The new Sec. 3.6 describes and illustrates how to use Excel and its Solver to formulate and solve linear programming models on a spreadsheet. Similar discussions and examples also are included in several subsequent chapters for other kinds of models. In addition, the CD-ROM provides an Excel file for many of the chapters that displays the spreadsheet formulation and solution for the relevant examples in the chapter. Several of the Excel files also include a number of Excel templates for solving the models in the chapter. Another key resource is a collection of Excel add-ins on the CD-ROM (Premium Solver, TreePlan, SensIt, and RiskSim) that are integrated into the corresponding chapters. In addition, Sec. 22.6 describes how some simulations can be performed efficiently on spreadsheets by using another popular Excel add-in (@RISK) that can be downloaded temporarily from a website. Practitioners of operations research now usually use a modeling language to formulate and manage models of the very large size commonly encountered in practice. A modeling language system also will support one or more sophisticated software packages that can be called to solve a model once it has been formulated appropriately. The new Sec. 3.7 discusses the application of modeling languages and illustrates it with one modeling language (MPL) that is relatively amenable to student use. The student version of MPL is provided on the CD-ROM, along with an extensive MPL tutorial. Accompanying MPL as its primary solver is the student version of the renowned state-of-the-art software package, CPLEX. The student version of CONOPT also is provided as the solver for nonlinear programming. We are extremely pleased to be able to provide such powerful and popular software to students using this book. To further assist students, many of the chapters include an MPL/CPLEX file (or MPL/CPLEX/CONOPT file in the case of the nonlinear programming chapter) on the CD-ROM that shows how MPL and CPLEX would formulate and solve the relevant examples in the chapter. These files also illustrate how MPL and CPLEX can be integrated with spreadsheets. As described in the appendix to Chaps. 3 and 4, a third attractive option is to employ the student version of the popular and student-friendly software package LINDO and its modeling language companion LINGO. Both packages can be downloaded free from the LINDO Systems website. Associated tutorial material is included on the CD-ROM, along with a LINDO/LINGO file for many of the chapters showing how LINDO and LINGO would formulate and solve the relevant examples in the chapter. Once again, integration with spreadsheets also is illustrated. Complementing all these options on the CD-ROM is an updated version of the tutorial software that many instructors have found so useful for their students with the 5th and 6th editions. A program called OR Tutor provides 16 demonstration examples from the 6th edition, but now with an attractive new design based on JavaScript. These demos vividly demonstrate the evolution of an algorithm in ways that cannot be duplicated on the printed page. Most of the interactive routines from the 6th edition also are included on the CD-ROM, but again with an attractive new design. This design features a spreadsheet format based on VisualBasic. Each of the interactive routines enables the student to interactively execute one of the algorithms of operations research, making the needed decision at each step while the computer does the needed arithmetic. By enabling the student to focus on concepts rather than mindless number crunching when doing homework to learn an algorithm, we have found that these interactive routines make the learning process far more efficient and effective as well as more stimulating. In addition to these

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routines, the CD-ROM includes a few of the automatic routines from the 6th edition (again redesigned with VisualBasic) for those cases that are not covered by the software options described above. We were very fortunate to have the services of Michael O’Sullivan, a talented programmer and an advanced Ph.D. student in operations research at Stanford, to do all this updating of the software that had been developed by Mark S. Hillier for the 5th and 6th editions. Microsoft Project is introduced in Chap. 10 as a useful tool for project management. This software package also is included on the CD-ROM.

NEW EMPHASES Today’s students in introductory operations research courses tend to be very interested in learning more about the relevance of the material being covered, including how it is actually being used in practice. Therefore, without diluting any of the features of the 6th edition, the focus of the revision for this edition has been on increasing the motivation and excitement of the students by making the book considerably more “real world” oriented and accessible. The new emphasis on the kinds of software that practitioners use is one thrust in this direction. Other major new features are outlined below. Twenty-five elaborate new cases, embedded in a realistic setting and employing a stimulating storytelling approach, have been added at the end of the problem sections. All but one of these cases were developed jointly by two talented case writers, Karl Schmedders (a faculty member at the Kellogg Graduate School of Management at Northwestern University) and Molly Stephens (recently an operations research consultant with Andersen Consulting). We also have further fleshed out six cases that were in the 6th edition. The cases generally require relatively challenging and comprehensive analyses with substantial use of the computer. Therefore, they are suitable for student projects, working either individually or in teams, and can then lead to class discussion of the analysis. A complementary new feature is that many new problems embedded in a realistic setting have been added to the problem section of many chapters. Some of the current problems also have been fleshed out in a more interesting way. This edition also places much more emphasis on providing perspective in terms of what is actually happening in the practice of operations research. What kinds of applications are occurring? What sizes of problems are being solved? Which models and techniques are being used most widely? What are their shortcomings and what new developments are beginning to address these shortcomings? These kinds of questions are being addressed to convey the relevance of the techniques under discussion. Eight new sections (Secs. 10.7, 12.2, 15.6, 18.5, 19.8, 20.1, 20.10, and 22.2) are fully devoted to discussing the practice of operations research in such ways, along with briefer mentions elsewhere. The new emphases described above benefited greatly from our work in developing our recent new textbook with Mark S. Hillier (Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Irwin/McGraw-Hill, 2000). That book has a very different orientation from this one. It is aimed directly at business students rather than students who may be in engineering and the mathematical sciences, and it provides almost no coverage of the mathematics and algorithms of operations research. Nevertheless, its applied orientation enabled us to adapt some excellent material developed for that book to provide a more well-rounded coverage in this edition.

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OTHER FEATURES In addition to all the new software and new emphases just described, this edition received a considerable number of other enhancements as well. The previous section on project planning and control with PERT/CPM has been replaced by a complete new chapter (Chap. 10) with an applied orientation. Using the activity-on-node (AON) convention, this chapter provides an extensive modern treatment of the topic in a very accessible way. Other new topics not yet mentioned include the SOB mnemonic device for determining the form of constraints in the dual problem (in Sec. 6.4), 100 percent rules for simultaneous changes when conducting sensitivity analysis (in Sec. 6.7), sensitivity analysis with Bayes’ decision rule (in Sec. 15.2), a probability tree diagram for calculating posterior probabilities (in Sec. 15.3), a single-server variation of the nonpreemptive priorities model where the service for different priority classes of customers now have different mean service rates (in Sec. 17.8), a new simpler analysis of a stochastic continuous-review inventory model (Sec. 19.5), the mean absolute deviation as a measure of performance for forecasting methods (in Sec. 20.7), and the elements of a major simulation study (Sec. 22.5). We also have added much supplementary text material on the book’s new website, www.mhhe.com/hillier. Some of these supplements are password protected, but are available to all instructors who adopt this textbook. For the most part, this material appeared in previous editions of this book and then was subsequently deleted (for space reasons), to the disappointment of some instructors. Some also appeared in our Introduction to Mathematical Programming textbook. As delineated in the table of contents, this supplementary material includes a chapter on additional special types of linear programming problems, a review or primer chapter on probability theory, and a chapter on reliability, along with supplements to a few chapters in the book. In addition to providing this supplementary text material, the website will give updates about the book, including an errata, as the need arises. We made two changes in the order of the chapters. The decision analysis chapter has been moved forward to Chap. 15 in front of the stochastic chapters. The game theory chapter has been moved backward to Chap. 14 to place it next to the related decision analysis chapter. We believe that these changes provide a better transition from topics that are mainly deterministic to those that are mainly stochastic. Every chapter has received significant revision and updating, ranging from modest refining to extensive rewriting. Chapters receiving a particularly major revision and reorganization included Chaps. 15 (Decision Analysis), 19 (Inventory Theory), 20 (Forecasting), and 22 (Simulation). Many sections in the linear programming and mathematical programming chapters also received major revisions and updating. The overall thrust of all the revision efforts has been to build upon the strengths of previous editions while thoroughly updating and clarifying the material in a contemporary setting to fully meet the needs of today’s students. We think that the net effect has been to make this edition even more of a “student’s book”—clear, interesting, and well-organized with lots of helpful examples and illustrations, good motivation and perspective, easy-to-find important material, and enjoyable homework, without too much notation, terminology, and dense mathematics. We believe

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and trust that the numerous instructors who have used previous editions will agree that this is the best edition yet. This feeling has been reinforced by the generally enthusiastic reviews of drafts of this edition. The prerequisites for a course using this book can be relatively modest. As with previous editions, the mathematics has been kept at a relatively elementary level. Most of Chaps. 1 to 14 (introduction, linear programming, and mathematical programming) require no mathematics beyond high school algebra. Calculus is used only in Chaps. 13 (Nonlinear Programming) and in one example in Chap. 11 (Dynamic Programming). Matrix notation is used in Chap. 5 (The Theory of the Simplex Method), Chap. 6 (Duality Theory and Sensitivity Analysis), Sec. 7.4 (An Interior-Point Algorithm), and Chap. 13, but the only background needed for this is presented in Appendix 4. For Chaps. 15 to 22 (probabilistic models), a previous introduction to probability theory is assumed, and calculus is used in a few places. In general terms, the mathematical maturity that a student achieves through taking an elementary calculus course is useful throughout Chaps. 15 to 22 and for the more advanced material in the preceding chapters. The content of the book is aimed largely at the upper-division undergraduate level (including well-prepared sophomores) and at first-year (master’s level) graduate students. Because of the book’s great flexibility, there are many ways to package the material into a course. Chapters 1 and 2 give an introduction to the subject of operations research. Chapters 3 to 14 (on linear programming and on mathematical programming) may essentially be covered independently of Chaps. 15 to 22 (on probabilistic models), and vice versa. Furthermore, the individual chapters among Chaps. 3 to 14 are almost independent, except that they all use basic material presented in Chap. 3 and perhaps in Chap. 4. Chapter 6 and Sec. 7.2 also draw upon Chap. 5. Sections 7.1 and 7.2 use parts of Chap. 6. Section 9.6 assumes an acquaintance with the problem formulations in Secs. 8.1 and 8.3, while prior exposure to Secs. 7.3 and 8.2 is helpful (but not essential) in Sec. 9.7. Within Chaps. 15 to 22, there is considerable flexibility of coverage, although some integration of the material is available. An elementary survey course covering linear programming, mathematical programming, and some probabilistic models can be presented in a quarter (40 hours) or semester by selectively drawing from material throughout the book. For example, a good survey of the field can be obtained from Chaps. 1, 2, 3, 4, 15, 17, 19, 20, and 22, along with parts of Chaps. 9, 11, 12, and 13. A more extensive elementary survey course can be completed in two quarters (60 to 80 hours) by excluding just a few chapters, for example, Chaps. 7, 14, and 21. Chapters 1 to 8 (and perhaps part of Chap. 9) form an excellent basis for a (one-quarter) course in linear programming. The material in Chaps. 9 to 14 covers topics for another (one-quarter) course in other deterministic models. Finally, the material in Chaps. 15 to 22 covers the probabilistic (stochastic) models of operations research suitable for presentation in a (one-quarter) course. In fact, these latter three courses (the material in the entire text) can be viewed as a basic one-year sequence in the techniques of operations research, forming the core of a master’s degree program. Each course outlined has been presented at either the undergraduate or the graduate level at Stanford University, and this text has been used in the manner suggested. To assist the instructor who will be covering only a portion of the chapters and who prefers a slimmer book containing only those chapters, all the material (including the supplementary text material on the book’s website) has been placed in McGraw-Hill’s PRIMIS

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system. This system enables an instructor to pick and choose precisely which material to include in a self-designed book, and then to order copies for the students at an economical price. For example, this enables instructors who previously used our Introduction to Mathematical Programming or Introduction to Stochastic Models in Operations Research textbooks to obtain updated versions of the same material from the PRIMIS system. For this reason, we will not be publishing new separate editions of these other books. Again, as in previous editions, we thank our wives, Ann and Helen, for their encouragement and support during the long process of preparing this 7th edition. Our children, David, John, and Mark Hillier, Janet Lieberman Argyres, and Joanne, Michael, and Diana Lieberman, have literally grown up with the book and our periodic hibernations to prepare a new edition. Now, most of them have used the book as a text in their own college courses, given considerable advice, and even (in the case of Mark Hillier) become a software collaborator. It is a joy to see them and (we trust) the book reach maturity together. And now I must add a very sad note. My close friend and co-author, Jerry Lieberman, passed away on May 18, 1999, while this edition was in preparation, so I am writing this preface on behalf of both of us. Jerry was one of the great leaders of our field and he had a profound influence on my life. More than a third of a century ago, we embarked on a mission together to attempt to develop a path-breaking book for teaching operations research at the introductory level. Ever since, we have striven to meet and extend the same high standards for each new edition. Having worked so closely with Jerry for so many years, I believe I understand well how he would want the book to evolve to meet the needs of each new generation of students. As the substantially younger co-author, I am grateful that I am able to carry on our joint mission to continue to update and improve the book, both with this edition and with future editions as well. It is the least I can do to honor Jerry. I welcome your comments, suggestions, and errata to help me improve the book in the future.

ACKNOWLEDGMENTS We are indebted to an excellent group of reviewers who provided sage advice throughout the revision process. This group included Jeffery Cochran, Arizona State University; Yahya Fathi, North Carolina State University; Yasser Hosni and Charles Reilly, University of Central Florida; Cerry Klein, University of Missouri—Columbia; Robert Lipset, Ohio University; Mark Parker, United States Air Force Academy; Christopher Rump, State University of New York at Buffalo; and Ahmad Seifoddini, California Polytechnic State University—San Luis Obispo. We also received helpful advice from Judith Liebman, Siegfried Schaible, David Sloan, and Arthur F. Veinott, Jr., as well as many instructors who sent us letters or e-mail messages. In addition, we also thank many dozens of Stanford students and many students at other universities who gave us helpful written suggestions. This edition was very much of a team effort. Our case writers, Karl Schmedders and Molly Stephens (both graduates of our department), made a vital contribution. One of our department’s current Ph.D. students, Roberto Szechtman, did an excellent job in preparing the solutions manual. Another Ph.D. student, Michael O’Sullivan, was very skillful in updating the software that Mark Hillier had developed for the 5th and 6th editions. Mark

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(who was born the same year as the first edition and now is a tenured faculty member in the Management Science Department at the University of Washington) helped to oversee this updating and also provided both the spreadsheets and the Excel files (including many Excel templates) for this edition. Linus Schrage of the University of Chicago and LINDO Systems (and who took an introductory operations research course from me 37 years ago) supervised the development of LINGO/LINDO files for the various chapters as well as providing tutorial material for the CD-ROM. Another long-time friend, Bjarni Kristjansson (who heads Maximal Software), did the same thing for the MPL/CPLEX files and MPL tutorial material, as well as arranging to provide student versions of MPL, CPLEX, CONOPT, and OptiMax 2000 for the CD-ROM. One of our department’s Ph.D. graduates, Irv Lustig, was the ILOG project manager for providing CPLEX. Linus, Bjarni, and Irv all were helpful in checking material going into this edition regarding their software. Ann Hillier devoted numerous long days and nights to sitting with a Macintosh, doing word processing and constructing many figures and tables, in addition to endless cutting and pasting, photocopying, and FedExing of material. Helen Lieberman also carried a heavy burden in supporting Jerry. They all were vital members of the team. The inside back cover lists the various companies and individuals who have provided software for the CD-ROM. We greatly appreciate their key contributions. It was a real pleasure working with McGraw-Hill’s thoroughly professional editorial and production staff, including Eric Munson (executive editor), Maja Lorkovic (developmental editor), and Christine Vaughan (project manager). Frederick S. Hillier Stanford University ([email protected])

January 2000

TABLE OF CONTENTS

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CHAPTER 1 Introduction

1

1.1 The Origins of Operations Research 1 1.2 The Nature of Operations Research 2 1.3 The Impact of Operations Research 3 1.4 Algorithms and OR Courseware 5 Problems 6 CHAPTER 2 Overview of the Operations Research Modeling Approach 2.1 Defining the Problem and Gathering Data 2.2 Formulating a Mathematical Model 10 2.3 Deriving Solutions from the Model 14 2.4 Testing the Model 16 2.5 Preparing to Apply the Model 18 2.6 Implementation 20 2.7 Conclusions 21 Selected References 22 Problems 22 CHAPTER 3 Introduction to Linear Programming

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3.1 Prototype Example 25 3.2 The Linear Programming Model 31 3.3 Assumptions of Linear Programming 36 3.4 Additional Examples 44 3.5 Some Case Studies 61 3.6 Displaying and Solving Linear Programming Models on a Spreadsheet 3.7 Formulating Very Large Linear Programming Models 73 3.8 Conclusions 79 Appendix 3.1 The LINGO Modeling Language 79

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Selected References 89 Learning Aids for This Chapter in Your OR Courseware 90 Problems 90 Case 3.1 Auto Assembly 103 Case 3.2 Cutting Cafeteria Costs 104 Case 3.3 Staffing a Call Center 106 CHAPTER 4 Solving Linear Programming Problems: The Simplex Method

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4.1 The Essence of the Simplex Method 109 4.2 Setting Up the Simplex Method 114 4.3 The Algebra of the Simplex Method 118 4.4 The Simplex Method in Tabular Form 123 4.5 Tie Breaking in the Simplex Method 128 4.6 Adapting to Other Model Forms 132 4.7 Postoptimality Analysis 152 4.8 Computer Implementation 160 4.9 The Interior-Point Approach to Solving Linear Programming Problems 4.10 Conclusions 168 Appendix 4.1 An Introduction to Using LINDO 169 Selected References 171 Learning Aids for This Chapter in Your OR Courseware 172 Problems 172 Case 4.1 Fabrics and Fall Fashions 182 Case 4.2 New Frontiers 185 Case 4.3 Assigning Students to Schools 188 CHAPTER 5 The Theory of the Simplex Method

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5.1 Foundations of the Simplex Method 190 5.2 The Revised Simplex Method 202 5.3 A Fundamental Insight 212 5.4 Conclusions 220 Selected References 220 Learning Aids for This Chapter in Your OR Courseware 221 Problems 221 CHAPTER 6 Duality Theory and Sensitivity Analysis 6.1 6.2 6.3 6.4 6.5 6.6

230

The Essence of Duality Theory 231 Economic Interpretation of Duality 239 Primal-Dual Relationships 242 Adapting to Other Primal Forms 247 The Role of Duality Theory in Sensitivity Analysis The Essence of Sensitivity Analysis 254

252

163

TABLE OF CONTENTS

xv

6.7 Applying Sensitivity Analysis 262 6.8 Conclusions 284 Selected References 284 Learning Aids for This Chapter in Your OR Courseware 285 Problems 285 Case 6.1 Controlling Air Pollution 302 Case 6.2 Farm Management 304 Case 6.3 Assigning Students to Schools (Revisited) 307 CHAPTER 7 Other Algorithms for Linear Programming

309

7.1 The Dual Simplex Method 309 7.2 Parametric Linear Programming 312 7.3 The Upper Bound Technique 317 7.4 An Interior-Point Algorithm 320 7.5 Linear Goal Programming and Its Solution Procedures 332 7.6 Conclusions 339 Selected References 340 Learning Aids for This Chapter in Your OR Courseware 340 Problems 341 Case 7.1 A Cure for Cuba 347 CHAPTER 8 The Transportation and Assignment Problems

350

8.1 The Transportation Problem 351 8.2 A Streamlined Simplex Method for the Transportation Problem 8.3 The Assignment Problem 381 8.4 Conclusions 391 Selected References 391 Learning Aids for This Chapter in Your OR Courseware 392 Problems 392 Case 8.1 Shipping Wood to Market 401 Case 8.2 Project Pickings 402 CHAPTER 9 Network Optimization Models

405

9.1 Prototype Example 406 9.2 The Terminology of Networks 407 9.3 The Shortest-Path Problem 411 9.4 The Minimum Spanning Tree Problem 415 9.5 The Maximum Flow Problem 420 9.6 The Minimum Cost Flow Problem 429 9.7 The Network Simplex Method 438 9.8 Conclusions 448 Selected References 449

365

xvi

TABLE OF CONTENTS

Learning Aids for This Chapter in Your OR Courseware 449 Problems 450 Case 9.1 Aiding Allies 458 Case 9.2 Money in Motion 464 CHAPTER 10 Project Management with PERT/CPM

468

10.1 A Prototype Example—The Reliable Construction Co. Project 10.2 Using a Network to Visually Display a Project 470 10.3 Scheduling a Project with PERT/CPM 475 10.4 Dealing with Uncertain Activity Durations 485 10.5 Considering Time-Cost Trade-Offs 492 10.6 Scheduling and Controlling Project Costs 502 10.7 An Evaluation of PERT/CPM 508 10.8 Conclusions 512 Selected References 513 Learning Aids for This Chapter in Your OR Courseware 514 Problems 514 Case 10.1 Steps to Success 524 Case 10.2 “School’s out forever . . .” 527 CHAPTER 11 Dynamic Programming

469

533

11.1 A Prototype Example for Dynamic Programming 533 11.2 Characteristics of Dynamic Programming Problems 538 11.3 Deterministic Dynamic Programming 541 11.4 Probabilistic Dynamic Programming 562 11.5 Conclusions 568 Selected References 568 Learning Aids for This Chapter in Your OR Courseware 568 Problems 569 CHAPTER 12 Integer Programming

576

12.1 Prototype Example 577 12.2 Some BIP Applications 580 12.3 Innovative Uses of Binary Variables in Model Formulation 585 12.4 Some Formulation Examples 591 12.5 Some Perspectives on Solving Integer Programming Problems 600 12.6 The Branch-and-Bound Technique and Its Application to Binary Integer Programming 604 12.7 A Branch-and-Bound Algorithm for Mixed Integer Programming 616 12.8 Other Developments in Solving BIP Problems 622 12.9 Conclusions 630 Selected References 631

TABLE OF CONTENTS

xvii

Learning Aids for This Chapter in Your OR Courseware Problems 632 Case 12.1 Capacity Concerns 642 Case 12.2 Assigning Art 645 Case 12.3 Stocking Sets 649 Case 12.4 Assigning Students to Schools (Revisited Again) CHAPTER 13 Nonlinear Programming

631

653

654

13.1 Sample Applications 655 13.2 Graphical Illustration of Nonlinear Programming Problems 659 13.3 Types of Nonlinear Programming Problems 664 13.4 One-Variable Unconstrained Optimization 670 13.5 Multivariable Unconstrained Optimization 673 13.6 The Karush-Kuhn-Tucker (KKT) Conditions for Constrained Optimization 13.7 Quadratic Programming 683 13.8 Separable Programming 690 13.9 Convex Programming 697 13.10 Nonconvex Programming 702 13.11 Conclusions 706 Selected References 706 Learning Aids for This Chapter in Your OR Courseware 707 Problems 708 Case 13.1 Savvy Stock Selection 720 CHAPTER 14 Game Theory 726 14.1 The Formulation of Two-Person, Zero-Sum Games 726 14.2 Solving Simple Games—A Prototype Example 728 14.3 Games with Mixed Strategies 733 14.4 Graphical Solution Procedure 735 14.5 Solving by Linear Programming 738 14.6 Extensions 741 14.7 Conclusions 742 Selected References 743 Learning Aids for This Chapter in Your OR Courseware 743 Problems 743 CHAPTER 15 Decision Analysis 15.1 15.2 15.3 15.4 15.5

749

A Prototype Example 750 Decision Making without Experimentation 751 Decision Making with Experimentation 758 Decision Trees 764 Utility Theory 770

679

xviii

TABLE OF CONTENTS

15.6 The Practical Application of Decision Analysis 778 15.7 Conclusions 781 Selected References 781 Learning Aids for This Chapter in Your OR Courseware 782 Problems 782 Case 15.1 Brainy Business 795 Case 15.2 Smart Steering Support 798 CHAPTER 16 Markov Chains

802

16.1 Stochastic Processes 802 16.2 Markov Chains 803 16.3 Chapman-Kolmogorov Equations 808 16.4 Classification of States of a Markov Chain 810 16.5 Long-Run Properties of Markov Chains 812 16.6 First Passage Times 818 16.7 Absorbing States 820 16.8 Continuous Time Markov Chains 822 Selected References 827 Learning Aids for This Chapter in Your OR Courseware 828 Problems 828 CHAPTER 17 Queueing Theory 834 17.1 Prototype Example 835 17.2 Basic Structure of Queueing Models 835 17.3 Examples of Real Queueing Systems 840 17.4 The Role of the Exponential Distribution 841 17.5 The Birth-and-Death Process 848 17.6 Queueing Models Based on the Birth-and-Death Process 852 17.7 Queueing Models Involving Nonexponential Distributions 871 17.8 Priority-Discipline Queueing Models 879 17.9 Queueing Networks 885 17.10 Conclusions 889 Selected References 890 Learning Aids for This Chapter in Your OR Courseware 890 Problems 891 Case 17.1 Reducing In-Process Inventory 905 CHAPTER 18 The Application of Queueing Theory 18.1 Examples 907 18.2 Decision Making 909 18.3 Formulation of Waiting-Cost Functions

907

912

TABLE OF CONTENTS

18.4 Decision Models 917 18.5 Some Award-Winning Applications of Queueing Theory 18.6 Conclusions 926 Selected References 926 Learning Aids for This Chapter in Your OR Courseware 926 Problems 927 Case 18.1 Queueing Quandary 932

xix

923

CHAPTER 19 Inventory Theory 935 19.1 Examples 936 19.2 Components of Inventory Models 938 19.3 Deterministic Continuous-Review Models 941 19.4 A Deterministic Periodic-Review Model 951 19.5 A Stochastic Continuous-Review Model 956 19.6 A Stochastic Single-Period Model for Perishable Products 19.7 Stochastic Periodic-Review Models 975 19.8 Larger Inventory Systems in Practice 983 19.9 Conclusions 987 Selected References 987 Learning Aids for This Chapter in Your OR Courseware 987 Problems 988 Case 19.1 Brushing Up on Inventory Control 1000 Case 19.2 TNT: Tackling Newsboy’s Teachings 1002 Case 19.3 Jettisoning Surplus Stock 1004

961

CHAPTER 20 Forecasting 1009 20.1 Some Applications of Forecasting 1010 20.2 Judgmental Forecasting Methods 1013 20.3 Time Series 1014 20.4 Forecasting Methods for a Constant-Level Model 1016 20.5 Incorporating Seasonal Effects into Forecasting Methods 1018 20.6 An Exponential Smoothing Method for a Linear Trend Model 1021 20.7 Forecasting Errors 1025 20.8 Box-Jenkins Method 1026 20.9 Causal Forecasting with Linear Regression 1028 20.10 Forecasting in Practice 1036 20.11 Conclusions 1038 Selected References 1038 Learning Aids for This Chapter in Your OR Courseware 1038 Problems 1039 Case 20.1 Finagling the Forecasts 1048

xx

TABLE OF CONTENTS

CHAPTER 21 Markov Decision Processes

1053

21.1 A Prototype Example 1053 21.2 A Model for Markov Decision Processes 1056 21.3 Linear Programming and Optimal Policies 1059 21.4 Policy Improvement Algorithm for Finding Optimal Policies 21.5 Discounted Cost Criterion 1069 21.6 Conclusions Selected References 1077 Learning Aids for This Chapter in Your OR Courseware 1078 Problems 1078

1064

CHAPTER 22 Simulation 1084 22.1 The Essence of Simulation 1084 22.2 Some Common Types of Applications of Simulation 1097 22.3 Generation of Random Numbers 1101 22.4 Generation of Random Observations from a Probability Distribution 22.5 Outline of a Major Simulation Study 1110 22.6 Performing Simulations on Spreadsheets 1115 22.7 Variance-Reducing Techniques 1126 22.8 Regenerative Method of Statistical Analysis 1131 22.9 Conclusions 1138 Selected References 1140 Learning Aids for This Chapter in Your OR Courseware 1140 Problems 1141 Case 22.1 Planning Planers 1151 Case 22.2 Pricing under Pressure 1153 APPENDIXES 1. Documentation for the OR Courseware 1156 2. Convexity 1159 3. Classical Optimization Methods 1165 4. Matrices and Matrix Operations 1169 5. Tables 1174 PARTIAL ANSWERS TO SELECTED PROBLEMS INDEXES Author Index 1195 Subject Index 1199

1176

1105

1 Introduction

1.1

THE ORIGINS OF OPERATIONS RESEARCH Since the advent of the industrial revolution, the world has seen a remarkable growth in the size and complexity of organizations. The artisans’ small shops of an earlier era have evolved into the billion-dollar corporations of today. An integral part of this revolutionary change has been a tremendous increase in the division of labor and segmentation of management responsibilities in these organizations. The results have been spectacular. However, along with its blessings, this increasing specialization has created new problems, problems that are still occurring in many organizations. One problem is a tendency for the many components of an organization to grow into relatively autonomous empires with their own goals and value systems, thereby losing sight of how their activities and objectives mesh with those of the overall organization. What is best for one component frequently is detrimental to another, so the components may end up working at cross purposes. A related problem is that as the complexity and specialization in an organization increase, it becomes more and more difficult to allocate the available resources to the various activities in a way that is most effective for the organization as a whole. These kinds of problems and the need to find a better way to solve them provided the environment for the emergence of operations research (commonly referred to as OR). The roots of OR can be traced back many decades, when early attempts were made to use a scientific approach in the management of organizations. However, the beginning of the activity called operations research has generally been attributed to the military services early in World War II. Because of the war effort, there was an urgent need to allocate scarce resources to the various military operations and to the activities within each operation in an effective manner. Therefore, the British and then the U.S. military management called upon a large number of scientists to apply a scientific approach to dealing with this and other strategic and tactical problems. In effect, they were asked to do research on (military) operations. These teams of scientists were the first OR teams. By developing effective methods of using the new tool of radar, these teams were instrumental in winning the Air Battle of Britain. Through their research on how to better manage convoy and antisubmarine operations, they also played a major role in winning the Battle of the North Atlantic. Similar efforts assisted the Island Campaign in the Pacific. When the war ended, the success of OR in the war effort spurred interest in applying OR outside the military as well. As the industrial boom following the war was run1

2

1 INTRODUCTION

ning its course, the problems caused by the increasing complexity and specialization in organizations were again coming to the forefront. It was becoming apparent to a growing number of people, including business consultants who had served on or with the OR teams during the war, that these were basically the same problems that had been faced by the military but in a different context. By the early 1950s, these individuals had introduced the use of OR to a variety of organizations in business, industry, and government. The rapid spread of OR soon followed. At least two other factors that played a key role in the rapid growth of OR during this period can be identified. One was the substantial progress that was made early in improving the techniques of OR. After the war, many of the scientists who had participated on OR teams or who had heard about this work were motivated to pursue research relevant to the field; important advancements in the state of the art resulted. A prime example is the simplex method for solving linear programming problems, developed by George Dantzig in 1947. Many of the standard tools of OR, such as linear programming, dynamic programming, queueing theory, and inventory theory, were relatively well developed before the end of the 1950s. A second factor that gave great impetus to the growth of the field was the onslaught of the computer revolution. A large amount of computation is usually required to deal most effectively with the complex problems typically considered by OR. Doing this by hand would often be out of the question. Therefore, the development of electronic digital computers, with their ability to perform arithmetic calculations thousands or even millions of times faster than a human being can, was a tremendous boon to OR. A further boost came in the 1980s with the development of increasingly powerful personal computers accompanied by good software packages for doing OR. This brought the use of OR within the easy reach of much larger numbers of people. Today, literally millions of individuals have ready access to OR software. Consequently, a whole range of computers from mainframes to laptops now are being routinely used to solve OR problems.

1.2

THE NATURE OF OPERATIONS RESEARCH As its name implies, operations research involves “research on operations.” Thus, operations research is applied to problems that concern how to conduct and coordinate the operations (i.e., the activities) within an organization. The nature of the organization is essentially immaterial, and, in fact, OR has been applied extensively in such diverse areas as manufacturing, transportation, construction, telecommunications, financial planning, health care, the military, and public services, to name just a few. Therefore, the breadth of application is unusually wide. The research part of the name means that operations research uses an approach that resembles the way research is conducted in established scientific fields. To a considerable extent, the scientific method is used to investigate the problem of concern. (In fact, the term management science sometimes is used as a synonym for operations research.) In particular, the process begins by carefully observing and formulating the problem, including gathering all relevant data. The next step is to construct a scientific (typically mathematical) model that attempts to abstract the essence of the real problem. It is then hypothesized that this model is a sufficiently precise representation of the essential features of the situation that the conclusions (solutions) obtained from the model are also

1.3 THE IMPACT OF OPERATIONS RESEARCH

3

valid for the real problem. Next, suitable experiments are conducted to test this hypothesis, modify it as needed, and eventually verify some form of the hypothesis. (This step is frequently referred to as model validation.) Thus, in a certain sense, operations research involves creative scientific research into the fundamental properties of operations. However, there is more to it than this. Specifically, OR is also concerned with the practical management of the organization. Therefore, to be successful, OR must also provide positive, understandable conclusions to the decision maker(s) when they are needed. Still another characteristic of OR is its broad viewpoint. As implied in the preceding section, OR adopts an organizational point of view. Thus, it attempts to resolve the conflicts of interest among the components of the organization in a way that is best for the organization as a whole. This does not imply that the study of each problem must give explicit consideration to all aspects of the organization; rather, the objectives being sought must be consistent with those of the overall organization. An additional characteristic is that OR frequently attempts to find a best solution (referred to as an optimal solution) for the problem under consideration. (We say a best instead of the best solution because there may be multiple solutions tied as best.) Rather than simply improving the status quo, the goal is to identify a best possible course of action. Although it must be interpreted carefully in terms of the practical needs of management, this “search for optimality” is an important theme in OR. All these characteristics lead quite naturally to still another one. It is evident that no single individual should be expected to be an expert on all the many aspects of OR work or the problems typically considered; this would require a group of individuals having diverse backgrounds and skills. Therefore, when a full-fledged OR study of a new problem is undertaken, it is usually necessary to use a team approach. Such an OR team typically needs to include individuals who collectively are highly trained in mathematics, statistics and probability theory, economics, business administration, computer science, engineering and the physical sciences, the behavioral sciences, and the special techniques of OR. The team also needs to have the necessary experience and variety of skills to give appropriate consideration to the many ramifications of the problem throughout the organization.

1.3

THE IMPACT OF OPERATIONS RESEARCH Operations research has had an impressive impact on improving the efficiency of numerous organizations around the world. In the process, OR has made a significant contribution to increasing the productivity of the economies of various countries. There now are a few dozen member countries in the International Federation of Operational Research Societies (IFORS), with each country having a national OR society. Both Europe and Asia have federations of OR societies to coordinate holding international conferences and publishing international journals in those continents. It appears that the impact of OR will continue to grow. For example, according to the U.S. Bureau of Labor Statistics, OR currently is one of the fastest-growing career areas for U.S. college graduates. To give you a better notion of the wide applicability of OR, we list some actual awardwinning applications in Table 1.1. Note the diversity of organizations and applications in the first two columns. The curious reader can find a complete article describing each application in the January–February issue of Interfaces for the year cited in the third col-

4

1 INTRODUCTION

TABLE 1.1 Some applications of operations research Organization

Nature of Application

The Netherlands Rijkswaterstaat

Develop national water management policy, including mix of new facilities, operating procedures, and pricing. Optimize production operations in chemical plants to meet production targets with minimum cost. Schedule shift work at reservation offices and airports to meet customer needs with minimum cost. Optimize refinery operations and the supply, distribution, and marketing of products. Optimally schedule and deploy police patrol officers with a computerized system. Optimally blend available ingredients into gasoline products to meet quality and sales requirements. Integrate a national network of spare parts inventories to improve service support.

Monsanto Corp.

United Airlines

Citgo Petroleum Corp. San Francisco Police Department Texaco, Inc.

IBM

Yellow Freight System, Inc. New Haven Health Department AT&T

Delta Airlines

Digital Equipment Corp. China

South African defense force Proctor and Gamble

Taco Bell

Hewlett-Packard

Optimize the design of a national trucking network and the routing of shipments. Design an effective needle exchange program to combat the spread of HIV/AIDS. Develop a PC-based system to guide business customers in designing their call centers. Maximize the profit from assigning airplane types to over 2500 domestic flights. Restructure the global supply chain of suppliers, plants, distribution centers, potential sites, and market areas. Optimally select and schedule massive projects for meeting the country’s future energy needs. Optimally redesign the size and shape of the defense force and its weapons systems. Redesign the North American production and distribution system to reduce costs and improve speed to market. Optimally schedule employees to provide desired customer service at a minimum cost. Redesign the sizes and locations of buffers in a printer production line to meet production goals.

Year of Publication*

Related Chapters†

1985

2–8, 13, 22

$15 million

1985

2, 12

$2 million

1986

2–9, 12, 17, 18, 20

$6 million

1987

2–9, 20

$70 million

1989

2–4, 12, 20

$11 million

1989

2, 13

$30 million

1990

2, 19, 22

1992 1993

2, 9, 13, 20, 22 2

$20 million $250 million less inventory $17.3 million

1993

17, 18, 22

33% less HIV/AIDS $750 million

1994

12

$100 million

1995

12

$800 million

1995

12

$425 million

1997

12

$1.1 billion

1997

8

$200 million

1998

12, 20, 22

$13 million

1998

17, 18

$280 million more revenue

*Pertains to a January–February issue of Interfaces in which a complete article can be found describing the application. † Refers to chapters in this book that describe the kinds of OR techniques used in the application.

Annual Savings

1.4 ALGORITHMS AND OR COURSEWARE

5

umn of the table. The fourth column lists the chapters in this book that describe the kinds of OR techniques that were used in the application. (Note that many of the applications combine a variety of techniques.) The last column indicates that these applications typically resulted in annual savings in the millions (or even tens of millions) of dollars. Furthermore, additional benefits not recorded in the table (e.g., improved service to customers and better managerial control) sometimes were considered to be even more important than these financial benefits. (You will have an opportunity to investigate these less tangible benefits further in Probs. 1.3-1 and 1.3-2.) Although most routine OR studies provide considerably more modest benefits than these award-winning applications, the figures in the rightmost column of Table 1.1 do accurately reflect the dramatic impact that large, well-designed OR studies occasionally can have. We will briefly describe some of these applications in the next chapter, and then we present two in greater detail as case studies in Sec. 3.5.

1.4

ALGORITHMS AND OR COURSEWARE An important part of this book is the presentation of the major algorithms (systematic solution procedures) of OR for solving certain types of problems. Some of these algorithms are amazingly efficient and are routinely used on problems involving hundreds or thousands of variables. You will be introduced to how these algorithms work and what makes them so efficient. You then will use these algorithms to solve a variety of problems on a computer. The CD-ROM called OR Courseware that accompanies the book will be a key tool for doing all this. One special feature in your OR Courseware is a program called OR Tutor. This program is intended to be your personal tutor to help you learn the algorithms. It consists of many demonstration examples that display and explain the algorithms in action. These “demos” supplement the examples in the book. In addition, your OR Courseware includes many interactive routines for executing the algorithms interactively in a convenient spreadsheet format. The computer does all the routine calculations while you focus on learning and executing the logic of the algorithm. You should find these interactive routines a very efficient and enlightening way of doing many of your homework problems. In practice, the algorithms normally are executed by commercial software packages. We feel that it is important to acquaint students with the nature of these packages that they will be using after graduation. Therefore, your OR Courseware includes a wealth of material to introduce you to three particularly popular software packages described below. Together, these packages will enable you to solve nearly all the OR models encountered in this book very efficiently. We have added our own automatic routines to the OR Courseware only in a few cases where these packages are not applicable. A very popular approach now is to use today’s premier spreadsheet package, Microsoft Excel, to formulate small OR models in a spreadsheet format. The Excel Solver then is used to solve the models. Your OR Courseware includes a separate Excel file for nearly every chapter in this book. Each time a chapter presents an example that can be solved using Excel, the complete spreadsheet formulation and solution is given in that chapter’s Excel file. For many of the models in the book, an Excel template also is pro-

6

1 INTRODUCTION

vided that already includes all the equations necessary to solve the model. Some Excel add-ins also are included on the CD-ROM. After many years, LINDO (and its companion modeling language LINGO) continues to be a dominant OR software package. Student versions of LINDO and LINGO now can be downloaded free from the Web. As for Excel, each time an example can be solved with this package, all the details are given in a LINGO/LINDO file for that chapter in your OR Courseware. CPLEX is an elite state-of-the-art software package that is widely used for solving large and challenging OR problems. When dealing with such problems, it is common to also use a modeling system to efficiently formulate the mathematical model and enter it into the computer. MPL is a user-friendly modeling system that uses CPLEX as its main solver. A student version of MPL and CPLEX is available free by downloading it from the Web. For your convenience, we also have included this student version in your OR Courseware. Once again, all the examples that can be solved with this package are detailed in MPL/CPLEX files for the corresponding chapters in your OR Courseware. We will further describe these three software packages and how to use them later (especially near the end of Chaps. 3 and 4). Appendix 1 also provides documentation for the OR Courseware, including OR Tutor. To alert you to relevant material in OR Courseware, the end of each chapter from Chap. 3 onward has a list entitled Learning Aids for This Chapter in Your OR Courseware. As explained at the beginning of the problem section for each of these chapters, symbols also are placed to the left of each problem number or part where any of this material (including demonstration examples and interactive routines) can be helpful.

PROBLEMS 1.3-1. Select one of the applications of operations research listed in Table 1.1. Read the article describing the application in the January–February issue of Interfaces for the year indicated in the third column. Write a two-page summary of the application and the benefits (including nonfinancial benefits) it provided.

1.3-2. Select three of the applications of operations research listed in Table 1.1. Read the articles describing the applications in the January–February issue of Interfaces for the years indicated in the third column. For each one, write a one-page summary of the application and the benefits (including nonfinancial benefits) it provided.

2 Overview of the Operations Research Modeling Approach The bulk of this book is devoted to the mathematical methods of operations research (OR). This is quite appropriate because these quantitative techniques form the main part of what is known about OR. However, it does not imply that practical OR studies are primarily mathematical exercises. As a matter of fact, the mathematical analysis often represents only a relatively small part of the total effort required. The purpose of this chapter is to place things into better perspective by describing all the major phases of a typical OR study. One way of summarizing the usual (overlapping) phases of an OR study is the following: 1. Define the problem of interest and gather relevant data. 2. Formulate a mathematical model to represent the problem. 3. Develop a computer-based procedure for deriving solutions to the problem from the model. 4. Test the model and refine it as needed. 5. Prepare for the ongoing application of the model as prescribed by management. 6. Implement. Each of these phases will be discussed in turn in the following sections. Most of the award-winning OR studies introduced in Table 1.1 provide excellent examples of how to execute these phases well. We will intersperse snippets from these examples throughout the chapter, with references to invite your further reading.

2.1

DEFINING THE PROBLEM AND GATHERING DATA In contrast to textbook examples, most practical problems encountered by OR teams are initially described to them in a vague, imprecise way. Therefore, the first order of business is to study the relevant system and develop a well-defined statement of the problem to be considered. This includes determining such things as the appropriate objectives, constraints on what can be done, interrelationships between the area to be studied and other areas of the organization, possible alternative courses of action, time limits for making a decision, and so on. This process of problem definition is a crucial one because it greatly affects how relevant the conclusions of the study will be. It is difficult to extract a “right” answer from the “wrong” problem! 7

8

2

OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

The first thing to recognize is that an OR team is normally working in an advisory capacity. The team members are not just given a problem and told to solve it however they see fit. Instead, they are advising management (often one key decision maker). The team performs a detailed technical analysis of the problem and then presents recommendations to management. Frequently, the report to management will identify a number of alternatives that are particularly attractive under different assumptions or over a different range of values of some policy parameter that can be evaluated only by management (e.g., the tradeoff between cost and benefits). Management evaluates the study and its recommendations, takes into account a variety of intangible factors, and makes the final decision based on its best judgment. Consequently, it is vital for the OR team to get on the same wavelength as management, including identifying the “right” problem from management’s viewpoint, and to build the support of management for the course that the study is taking. Ascertaining the appropriate objectives is a very important aspect of problem definition. To do this, it is necessary first to identify the member (or members) of management who actually will be making the decisions concerning the system under study and then to probe into this individual’s thinking regarding the pertinent objectives. (Involving the decision maker from the outset also is essential to build her or his support for the implementation of the study.) By its nature, OR is concerned with the welfare of the entire organization rather than that of only certain of its components. An OR study seeks solutions that are optimal for the overall organization rather than suboptimal solutions that are best for only one component. Therefore, the objectives that are formulated ideally should be those of the entire organization. However, this is not always convenient. Many problems primarily concern only a portion of the organization, so the analysis would become unwieldy if the stated objectives were too general and if explicit consideration were given to all side effects on the rest of the organization. Instead, the objectives used in the study should be as specific as they can be while still encompassing the main goals of the decision maker and maintaining a reasonable degree of consistency with the higher-level objectives of the organization. For profit-making organizations, one possible approach to circumventing the problem of suboptimization is to use long-run profit maximization (considering the time value of money) as the sole objective. The adjective long-run indicates that this objective provides the flexibility to consider activities that do not translate into profits immediately (e.g., research and development projects) but need to do so eventually in order to be worthwhile. This approach has considerable merit. This objective is specific enough to be used conveniently, and yet it seems to be broad enough to encompass the basic goal of profitmaking organizations. In fact, some people believe that all other legitimate objectives can be translated into this one. However, in actual practice, many profit-making organizations do not use this approach. A number of studies of U.S. corporations have found that management tends to adopt the goal of satisfactory profits, combined with other objectives, instead of focusing on long-run profit maximization. Typically, some of these other objectives might be to maintain stable profits, increase (or maintain) one’s share of the market, provide for product diversification, maintain stable prices, improve worker morale, maintain family control of the business, and increase company prestige. Fulfilling these objectives might achieve long-run profit maximization, but the relationship may be sufficiently obscure that it may not be convenient to incorporate them all into this one objective.

2.1 DEFINING THE PROBLEM AND GATHERING DATA

9

Furthermore, there are additional considerations involving social responsibilities that are distinct from the profit motive. The five parties generally affected by a business firm located in a single country are (1) the owners (stockholders, etc.), who desire profits (dividends, stock appreciation, and so on); (2) the employees, who desire steady employment at reasonable wages; (3) the customers, who desire a reliable product at a reasonable price; (4) the suppliers, who desire integrity and a reasonable selling price for their goods; and (5) the government and hence the nation, which desire payment of fair taxes and consideration of the national interest. All five parties make essential contributions to the firm, and the firm should not be viewed as the exclusive servant of any one party for the exploitation of others. By the same token, international corporations acquire additional obligations to follow socially responsible practices. Therefore, while granting that management’s prime responsibility is to make profits (which ultimately benefits all five parties), we note that its broader social responsibilities also must be recognized. OR teams typically spend a surprisingly large amount of time gathering relevant data about the problem. Much data usually are needed both to gain an accurate understanding of the problem and to provide the needed input for the mathematical model being formulated in the next phase of study. Frequently, much of the needed data will not be available when the study begins, either because the information never has been kept or because what was kept is outdated or in the wrong form. Therefore, it often is necessary to install a new computer-based management information system to collect the necessary data on an ongoing basis and in the needed form. The OR team normally needs to enlist the assistance of various other key individuals in the organization to track down all the vital data. Even with this effort, much of the data may be quite “soft,” i.e., rough estimates based only on educated guesses. Typically, an OR team will spend considerable time trying to improve the precision of the data and then will make do with the best that can be obtained. Examples. An OR study done for the San Francisco Police Department1 resulted in the development of a computerized system for optimally scheduling and deploying police patrol officers. The new system provided annual savings of $11 million, an annual $3 million increase in traffic citation revenues, and a 20 percent improvement in response times. In assessing the appropriate objectives for this study, three fundamental objectives were identified: 1. Maintain a high level of citizen safety. 2. Maintain a high level of officer morale. 3. Minimize the cost of operations. To satisfy the first objective, the police department and city government jointly established a desired level of protection. The mathematical model then imposed the requirement that this level of protection be achieved. Similarly, the model imposed the requirement of balancing the workload equitably among officers in order to work toward the second objective. Finally, the third objective was incorporated by adopting the long-term goal of minimizing the number of officers needed to meet the first two objectives. 1

P. E. Taylor and S. J. Huxley, “A Break from Tradition for the San Francisco Police: Patrol Officer Scheduling Using an Optimization-Based Decision Support System,” Interfaces, 19(1): 4–24, Jan.–Feb. 1989. See especially pp. 4–11.

10

2

OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

The Health Department of New Haven, Connecticut used an OR team1 to design an effective needle exchange program to combat the spread of the virus that causes AIDS (HIV), and succeeded in reducing the HIV infection rate among program clients by 33 percent. The key part of this study was an innovative data collection program to obtain the needed input for mathematical models of HIV transmission. This program involved complete tracking of each needle (and syringe), including the identity, location, and date for each person receiving the needle and each person returning the needle during an exchange, as well as testing whether the returned needle was HIVpositive or HIV-negative. An OR study done for the Citgo Petroleum Corporation2 optimized both refinery operations and the supply, distribution, and marketing of its products, thereby achieving a profit improvement of approximately $70 million per year. Data collection also played a key role in this study. The OR team held data requirement meetings with top Citgo management to ensure the eventual and continual quality of data. A state-of-the-art management database system was developed and installed on a mainframe computer. In cases where needed data did not exist, LOTUS 1-2-3 screens were created to help operations personnel input the data, and then the data from the personal computers (PCs) were uploaded to the mainframe computer. Before data was inputted to the mathematical model, a preloader program was used to check for data errors and inconsistencies. Initially, the preloader generated a paper log of error messages 1 inch thick! Eventually, the number of error and warning messages (indicating bad or questionable numbers) was reduced to less than 10 for each new run. We will describe the overall Citgo study in much more detail in Sec. 3.5.

2.2

FORMULATING A MATHEMATICAL MODEL After the decision maker’s problem is defined, the next phase is to reformulate this problem in a form that is convenient for analysis. The conventional OR approach for doing this is to construct a mathematical model that represents the essence of the problem. Before discussing how to formulate such a model, we first explore the nature of models in general and of mathematical models in particular. Models, or idealized representations, are an integral part of everyday life. Common examples include model airplanes, portraits, globes, and so on. Similarly, models play an important role in science and business, as illustrated by models of the atom, models of genetic structure, mathematical equations describing physical laws of motion or chemical reactions, graphs, organizational charts, and industrial accounting systems. Such models are invaluable for abstracting the essence of the subject of inquiry, showing interrelationships, and facilitating analysis. 1

E. H. Kaplan and E. O’Keefe, “Let the Needles Do the Talking! Evaluating the New Haven Needle Exchange,” Interfaces, 23(1): 7–26, Jan.–Feb. 1993. See especially pp. 12–14. 2 D. Klingman, N. Phillips, D. Steiger, R. Wirth, and W. Young, “The Challenges and Success Factors in Implementing an Integrated Products Planning System for Citgo,” Interfaces, 16(3): 1–19, May–June 1986. See especially pp. 11–14. Also see D. Klingman, N. Phillips, D. Steiger, and W. Young, “The Successful Deployment of Management Science throughout Citgo Petroleum Corporation,” Interfaces, 17(1): 4–25, Jan.–Feb. 1987. See especially pp. 13–15. This application will be described further in Sec. 3.5.

2.2 FORMULATING A MATHEMATICAL MODEL

11

Mathematical models are also idealized representations, but they are expressed in terms of mathematical symbols and expressions. Such laws of physics as F ma and E mc2 are familiar examples. Similarly, the mathematical model of a business problem is the system of equations and related mathematical expressions that describe the essence of the problem. Thus, if there are n related quantifiable decisions to be made, they are represented as decision variables (say, x1, x2, . . . , xn) whose respective values are to be determined. The appropriate measure of performance (e.g., profit) is then expressed as a mathematical function of these decision variables (for example, P 3x1 2x2 + 5xn). This function is called the objective function. Any restrictions on the values that can be assigned to these decision variables are also expressed mathematically, typically by means of inequalities or equations (for example, x1 3x1x2 2x2 10). Such mathematical expressions for the restrictions often are called constraints. The constants (namely, the coefficients and right-hand sides) in the constraints and the objective function are called the parameters of the model. The mathematical model might then say that the problem is to choose the values of the decision variables so as to maximize the objective function, subject to the specified constraints. Such a model, and minor variations of it, typifies the models used in OR. Determining the appropriate values to assign to the parameters of the model (one value per parameter) is both a critical and a challenging part of the model-building process. In contrast to textbook problems where the numbers are given to you, determining parameter values for real problems requires gathering relevant data. As discussed in the preceding section, gathering accurate data frequently is difficult. Therefore, the value assigned to a parameter often is, of necessity, only a rough estimate. Because of the uncertainty about the true value of the parameter, it is important to analyze how the solution derived from the model would change (if at all) if the value assigned to the parameter were changed to other plausible values. This process is referred to as sensitivity analysis, as discussed further in the next section (and much of Chap. 6). Although we refer to “the” mathematical model of a business problem, real problems normally don’t have just a single “right” model. Section 2.4 will describe how the process of testing a model typically leads to a succession of models that provide better and better representations of the problem. It is even possible that two or more completely different types of models may be developed to help analyze the same problem. You will see numerous examples of mathematical models throughout the remainder of this book. One particularly important type that is studied in the next several chapters is the linear programming model, where the mathematical functions appearing in both the objective function and the constraints are all linear functions. In the next chapter, specific linear programming models are constructed to fit such diverse problems as determining (1) the mix of products that maximizes profit, (2) the design of radiation therapy that effectively attacks a tumor while minimizing the damage to nearby healthy tissue, (3) the allocation of acreage to crops that maximizes total net return, and (4) the combination of pollution abatement methods that achieves air quality standards at minimum cost. Mathematical models have many advantages over a verbal description of the problem. One advantage is that a mathematical model describes a problem much more concisely. This tends to make the overall structure of the problem more comprehensible, and it helps to reveal important cause-and-effect relationships. In this way, it indicates more clearly what additional data are relevant to the analysis. It also facilitates dealing with the problem in its

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entirety and considering all its interrelationships simultaneously. Finally, a mathematical model forms a bridge to the use of high-powered mathematical techniques and computers to analyze the problem. Indeed, packaged software for both personal computers and mainframe computers has become widely available for solving many mathematical models. However, there are pitfalls to be avoided when you use mathematical models. Such a model is necessarily an abstract idealization of the problem, so approximations and simplifying assumptions generally are required if the model is to be tractable (capable of being solved). Therefore, care must be taken to ensure that the model remains a valid representation of the problem. The proper criterion for judging the validity of a model is whether the model predicts the relative effects of the alternative courses of action with sufficient accuracy to permit a sound decision. Consequently, it is not necessary to include unimportant details or factors that have approximately the same effect for all the alternative courses of action considered. It is not even necessary that the absolute magnitude of the measure of performance be approximately correct for the various alternatives, provided that their relative values (i.e., the differences between their values) are sufficiently precise. Thus, all that is required is that there be a high correlation between the prediction by the model and what would actually happen in the real world. To ascertain whether this requirement is satisfied, it is important to do considerable testing and consequent modifying of the model, which will be the subject of Sec. 2.4. Although this testing phase is placed later in the chapter, much of this model validation work actually is conducted during the modelbuilding phase of the study to help guide the construction of the mathematical model. In developing the model, a good approach is to begin with a very simple version and then move in evolutionary fashion toward more elaborate models that more nearly reflect the complexity of the real problem. This process of model enrichment continues only as long as the model remains tractable. The basic trade-off under constant consideration is between the precision and the tractability of the model. (See Selected Reference 6 for a detailed description of this process.) A crucial step in formulating an OR model is the construction of the objective function. This requires developing a quantitative measure of performance relative to each of the decision maker’s ultimate objectives that were identified while the problem was being defined. If there are multiple objectives, their respective measures commonly are then transformed and combined into a composite measure, called the overall measure of performance. This overall measure might be something tangible (e.g., profit) corresponding to a higher goal of the organization, or it might be abstract (e.g., utility). In the latter case, the task of developing this measure tends to be a complex one requiring a careful comparison of the objectives and their relative importance. After the overall measure of performance is developed, the objective function is then obtained by expressing this measure as a mathematical function of the decision variables. Alternatively, there also are methods for explicitly considering multiple objectives simultaneously, and one of these (goal programming) is discussed in Chap. 7. Examples. An OR study done for Monsanto Corp.1 was concerned with optimizing production operations in Monsanto’s chemical plants to minimize the cost of meeting the target for the amount of a certain chemical product (maleic anhydride) to be produced in a given 1

R. F. Boykin, “Optimizing Chemical Production at Monsanto,” Interfaces, 15(1): 88–95, Jan.–Feb. 1985. See especially pp. 92–93.

2.2 FORMULATING A MATHEMATICAL MODEL

13

month. The decisions to be made are the dial setting for each of the catalytic reactors used to produce this product, where the setting determines both the amount produced and the cost of operating the reactor. The form of the resulting mathematical model is as follows: Choose the values of the decision variables Rij (i 1, 2, . . . , r; j 1, 2, . . . , s) so as to r

s

cij Rij , i1 j1

Minimize subject to r

s

pijRij T

i1 j1 s

Rij 1, j1

for i 1, 2, . . . , r Rij 0 or 1,

0

if reactor i is operated at setting j otherwise cij cost for reactor i at setting j pij production of reactor i at setting j T production target r number of reactors s number of settings (including off position)

where Rij

1

The objective function for this model is cijRij. The constraints are given in the three lines below the objective function. The parameters are cij, pij, and T. For Monsanto’s application, this model has over 1,000 decision variables Rij (that is, rs 1,000). Its use led to annual savings of approximately $2 million. The Netherlands government agency responsible for water control and public works, the Rijkswaterstaat, commissioned a major OR study1 to guide the development of a new national water management policy. The new policy saved hundreds of millions of dollars in investment expenditures and reduced agricultural damage by about $15 million per year, while decreasing thermal and algae pollution. Rather than formulating one mathematical model, this OR study developed a comprehensive, integrated system of 50 models! Furthermore, for some of the models, both simple and complex versions were developed. The simple version was used to gain basic insights, including trade-off analyses. The complex version then was used in the final rounds of the analysis or whenever greater accuracy or more detailed outputs were desired. The overall OR study directly involved over 125 person-years of effort (more than one-third in data gathering), created several dozen computer programs, and structured an enormous amount of data. 1

B. F. Goeller and the PAWN team: “Planning the Netherlands’ Water Resources,” Interfaces, 15(1): 3–33, Jan.–Feb. 1985. See especially pp. 7–18.

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2.3

OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

DERIVING SOLUTIONS FROM THE MODEL After a mathematical model is formulated for the problem under consideration, the next phase in an OR study is to develop a procedure (usually a computer-based procedure) for deriving solutions to the problem from this model. You might think that this must be the major part of the study, but actually it is not in most cases. Sometimes, in fact, it is a relatively simple step, in which one of the standard algorithms (systematic solution procedures) of OR is applied on a computer by using one of a number of readily available software packages. For experienced OR practitioners, finding a solution is the fun part, whereas the real work comes in the preceding and following steps, including the postoptimality analysis discussed later in this section. Since much of this book is devoted to the subject of how to obtain solutions for various important types of mathematical models, little needs to be said about it here. However, we do need to discuss the nature of such solutions. A common theme in OR is the search for an optimal, or best, solution. Indeed, many procedures have been developed, and are presented in this book, for finding such solutions for certain kinds of problems. However, it needs to be recognized that these solutions are optimal only with respect to the model being used. Since the model necessarily is an idealized rather than an exact representation of the real problem, there cannot be any utopian guarantee that the optimal solution for the model will prove to be the best possible solution that could have been implemented for the real problem. There just are too many imponderables and uncertainties associated with real problems. However, if the model is well formulated and tested, the resulting solution should tend to be a good approximation to an ideal course of action for the real problem. Therefore, rather than be deluded into demanding the impossible, you should make the test of the practical success of an OR study hinge on whether it provides a better guide for action than can be obtained by other means. Eminent management scientist and Nobel Laureate in economics Herbert Simon points out that satisficing is much more prevalent than optimizing in actual practice. In coining the term satisficing as a combination of the words satisfactory and optimizing, Simon is describing the tendency of managers to seek a solution that is “good enough” for the problem at hand. Rather than trying to develop an overall measure of performance to optimally reconcile conflicts between various desirable objectives (including well-established criteria for judging the performance of different segments of the organization), a more pragmatic approach may be used. Goals may be set to establish minimum satisfactory levels of performance in various areas, based perhaps on past levels of performance or on what the competition is achieving. If a solution is found that enables all these goals to be met, it is likely to be adopted without further ado. Such is the nature of satisficing. The distinction between optimizing and satisficing reflects the difference between theory and the realities frequently faced in trying to implement that theory in practice. In the words of one of England’s OR leaders, Samuel Eilon, “Optimizing is the science of the ultimate; satisficing is the art of the feasible.”1 OR teams attempt to bring as much of the “science of the ultimate” as possible to the decision-making process. However, the successful team does so in full recognition of the 1

S. Eilon, “Goals and Constraints in Decision-making,” Operational Research Quarterly, 23: 3–15, 1972—address given at the 1971 annual conference of the Canadian Operational Research Society.

2.3 DERIVING SOLUTIONS FROM THE MODEL

15

overriding need of the decision maker to obtain a satisfactory guide for action in a reasonable period of time. Therefore, the goal of an OR study should be to conduct the study in an optimal manner, regardless of whether this involves finding an optimal solution for the model. Thus, in addition to pursuing the science of the ultimate, the team should also consider the cost of the study and the disadvantages of delaying its completion, and then attempt to maximize the net benefits resulting from the study. In recognition of this concept, OR teams occasionally use only heuristic procedures (i.e., intuitively designed procedures that do not guarantee an optimal solution) to find a good suboptimal solution. This is most often the case when the time or cost required to find an optimal solution for an adequate model of the problem would be very large. In recent years, great progress has been made in developing efficient and effective heuristic procedures (including so-called metaheuristics), so their use is continuing to grow. The discussion thus far has implied that an OR study seeks to find only one solution, which may or may not be required to be optimal. In fact, this usually is not the case. An optimal solution for the original model may be far from ideal for the real problem, so additional analysis is needed. Therefore, postoptimality analysis (analysis done after finding an optimal solution) is a very important part of most OR studies. This analysis also is sometimes referred to as what-if analysis because it involves addressing some questions about what would happen to the optimal solution if different assumptions are made about future conditions. These questions often are raised by the managers who will be making the ultimate decisions rather than by the OR team. The advent of powerful spreadsheet software now has frequently given spreadsheets a central role in conducting postoptimality analysis. One of the great strengths of a spreadsheet is the ease with which it can be used interactively by anyone, including managers, to see what happens to the optimal solution when changes are made to the model. This process of experimenting with changes in the model also can be very helpful in providing understanding of the behavior of the model and increasing confidence in its validity. In part, postoptimality analysis involves conducting sensitivity analysis to determine which parameters of the model are most critical (the “sensitive parameters”) in determining the solution. A common definition of sensitive parameter (used throughout this book) is the following. For a mathematical model with specified values for all its parameters, the model’s sensitive parameters are the parameters whose value cannot be changed without changing the optimal solution.

Identifying the sensitive parameters is important, because this identifies the parameters whose value must be assigned with special care to avoid distorting the output of the model. The value assigned to a parameter commonly is just an estimate of some quantity (e.g., unit profit) whose exact value will become known only after the solution has been implemented. Therefore, after the sensitive parameters are identified, special attention is given to estimating each one more closely, or at least its range of likely values. One then seeks a solution that remains a particularly good one for all the various combinations of likely values of the sensitive parameters. If the solution is implemented on an ongoing basis, any later change in the value of a sensitive parameter immediately signals a need to change the solution.

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In some cases, certain parameters of the model represent policy decisions (e.g., resource allocations). If so, there frequently is some flexibility in the values assigned to these parameters. Perhaps some can be increased by decreasing others. Postoptimality analysis includes the investigation of such trade-offs. In conjunction with the study phase discussed in the next section (testing the model), postoptimality analysis also involves obtaining a sequence of solutions that comprises a series of improving approximations to the ideal course of action. Thus, the apparent weaknesses in the initial solution are used to suggest improvements in the model, its input data, and perhaps the solution procedure. A new solution is then obtained, and the cycle is repeated. This process continues until the improvements in the succeeding solutions become too small to warrant continuation. Even then, a number of alternative solutions (perhaps solutions that are optimal for one of several plausible versions of the model and its input data) may be presented to management for the final selection. As suggested in Sec. 2.1, this presentation of alternative solutions would normally be done whenever the final choice among these alternatives should be based on considerations that are best left to the judgment of management. Example. Consider again the Rijkswaterstaat OR study of national water management policy for the Netherlands, introduced at the end of the preceding section. This study did not conclude by recommending just a single solution. Instead, a number of attractive alternatives were identified, analyzed, and compared. The final choice was left to the Dutch political process, culminating with approval by Parliament. Sensitivity analysis played a major role in this study. For example, certain parameters of the models represented environmental standards. Sensitivity analysis included assessing the impact on water management problems if the values of these parameters were changed from the current environmental standards to other reasonable values. Sensitivity analysis also was used to assess the impact of changing the assumptions of the models, e.g., the assumption on the effect of future international treaties on the amount of pollution entering the Netherlands. A variety of scenarios (e.g., an extremely dry year and an extremely wet year) also were analyzed, with appropriate probabilities assigned.

2.4

TESTING THE MODEL Developing a large mathematical model is analogous in some ways to developing a large computer program. When the first version of the computer program is completed, it inevitably contains many bugs. The program must be thoroughly tested to try to find and correct as many bugs as possible. Eventually, after a long succession of improved programs, the programmer (or programming team) concludes that the current program now is generally giving reasonably valid results. Although some minor bugs undoubtedly remain hidden in the program (and may never be detected), the major bugs have been sufficiently eliminated that the program now can be reliably used. Similarly, the first version of a large mathematical model inevitably contains many flaws. Some relevant factors or interrelationships undoubtedly have not been incorporated into the model, and some parameters undoubtedly have not been estimated correctly. This is inevitable, given the difficulty of communicating and understanding all the aspects and

2.4 TESTING THE MODEL

17

subtleties of a complex operational problem as well as the difficulty of collecting reliable data. Therefore, before you use the model, it must be thoroughly tested to try to identify and correct as many flaws as possible. Eventually, after a long succession of improved models, the OR team concludes that the current model now is giving reasonably valid results. Although some minor flaws undoubtedly remain hidden in the model (and may never be detected), the major flaws have been sufficiently eliminated that the model now can be reliably used. This process of testing and improving a model to increase its validity is commonly referred to as model validation. It is difficult to describe how model validation is done, because the process depends greatly on the nature of the problem being considered and the model being used. However, we make a few general comments, and then we give some examples. (See Selected Reference 2 for a detailed discussion.) Since the OR team may spend months developing all the detailed pieces of the model, it is easy to “lose the forest for the trees.” Therefore, after the details (“the trees”) of the initial version of the model are completed, a good way to begin model validation is to take a fresh look at the overall model (“the forest”) to check for obvious errors or oversights. The group doing this review preferably should include at least one individual who did not participate in the formulation of the model. Reexamining the definition of the problem and comparing it with the model may help to reveal mistakes. It is also useful to make sure that all the mathematical expressions are dimensionally consistent in the units used. Additional insight into the validity of the model can sometimes be obtained by varying the values of the parameters and/or the decision variables and checking to see whether the output from the model behaves in a plausible manner. This is often especially revealing when the parameters or variables are assigned extreme values near their maxima or minima. A more systematic approach to testing the model is to use a retrospective test. When it is applicable, this test involves using historical data to reconstruct the past and then determining how well the model and the resulting solution would have performed if they had been used. Comparing the effectiveness of this hypothetical performance with what actually happened then indicates whether using this model tends to yield a significant improvement over current practice. It may also indicate areas where the model has shortcomings and requires modifications. Furthermore, by using alternative solutions from the model and estimating their hypothetical historical performances, considerable evidence can be gathered regarding how well the model predicts the relative effects of alternative courses of actions. On the other hand, a disadvantage of retrospective testing is that it uses the same data that guided the formulation of the model. The crucial question is whether the past is truly representative of the future. If it is not, then the model might perform quite differently in the future than it would have in the past. To circumvent this disadvantage of retrospective testing, it is sometimes useful to continue the status quo temporarily. This provides new data that were not available when the model was constructed. These data are then used in the same ways as those described here to evaluate the model. Documenting the process used for model validation is important. This helps to increase confidence in the model for subsequent users. Furthermore, if concerns arise in the

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future about the model, this documentation will be helpful in diagnosing where problems may lie. Examples. Consider once again the Rijkswaterstaat OR study of national water management policy for the Netherlands, discussed at the end of Secs. 2.2 and 2.3. The process of model validation in this case had three main parts. First, the OR team checked the general behavior of the models by checking whether the results from each model moved in reasonable ways when changes were made in the values of the model parameters. Second, retrospective testing was done. Third, a careful technical review of the models, methodology, and results was conducted by individuals unaffiliated with the project, including Dutch experts. This process led to a number of important new insights and improvements in the models. Many new insights also were gleaned during the model validation phase of the OR study for the Citgo Petroleum Corp., discussed at the end of Sec. 2.1. In this case, the model of refinery operations was tested by collecting the actual inputs and outputs of the refinery for a series of months, using these inputs to fix the model inputs, and then comparing the model outputs with the actual refinery outputs. The process of properly calibrating and recalibrating the model was a lengthy one, but ultimately led to routine use of the model to provide critical decision information. As already mentioned in Sec. 2.1, the validation and correction of input data for the models also played an important role in this study. Our next example concerns an OR study done for IBM1 to integrate its national network of spare-parts inventories to improve service support for IBM’s customers. This study resulted in a new inventory system that improved customer service while reducing the value of IBM’s inventories by over $250 million and saving an additional $20 million per year through improved operational efficiency. A particularly interesting aspect of the model validation phase of this study was the way that future users of the inventory system were incorporated into the testing process. Because these future users (IBM managers in functional areas responsible for implementation of the inventory system) were skeptical about the system being developed, representatives were appointed to a user team to serve as advisers to the OR team. After a preliminary version of the new system had been developed (based on a multiechelon inventory model), a preimplementation test of the system was conducted. Extensive feedback from the user team led to major improvements in the proposed system.

2.5

PREPARING TO APPLY THE MODEL What happens after the testing phase has been completed and an acceptable model has been developed? If the model is to be used repeatedly, the next step is to install a welldocumented system for applying the model as prescribed by management. This system will include the model, solution procedure (including postoptimality analysis), and oper1

M. Cohen, P. V. Kamesam, P. Kleindorfer, H. Lee, and A. Tekerian, “Optimizer: IBM’s Multi-Echelon Inventory System for Managing Service Logistics,” Interfaces, 20(1): 65–82, Jan.–Feb. 1990. See especially pp. 73–76. This application will be described further in Sec. 19.8.

2.5 PREPARING TO APPLY THE MODEL

19

ating procedures for implementation. Then, even as personnel changes, the system can be called on at regular intervals to provide a specific numerical solution. This system usually is computer-based. In fact, a considerable number of computer programs often need to be used and integrated. Databases and management information systems may provide up-to-date input for the model each time it is used, in which case interface programs are needed. After a solution procedure (another program) is applied to the model, additional computer programs may trigger the implementation of the results automatically. In other cases, an interactive computer-based system called a decision support system is installed to help managers use data and models to support (rather than replace) their decision making as needed. Another program may generate managerial reports (in the language of management) that interpret the output of the model and its implications for application. In major OR studies, several months (or longer) may be required to develop, test, and install this computer system. Part of this effort involves developing and implementing a process for maintaining the system throughout its future use. As conditions change over time, this process should modify the computer system (including the model) accordingly. Examples. The IBM OR study introduced at the end of Sec. 2.4 provides a good example of a particularly large computer system for applying a model. The system developed, called Optimizer, provides optimal control of service levels and spare-parts inventories throughout IBM’s U.S. parts distribution network, which includes two central automated warehouses, dozens of field distribution centers and parts stations, and many thousands of outside locations. The parts inventory maintained in this network is valued in the billions of dollars. Optimizer consists of four major modules. A forecasting system module contains a few programs for estimating the failure rates of individual types of parts. A data delivery system module consists of approximately 100 programs that process over 15 gigabytes of data to provide the input for the model. A decision system module then solves the model on a weekly basis to optimize control of the inventories. The fourth module includes six programs that integrate Optimizer into IBM’s Parts Inventory Management System (PIMS). PIMS is a sophisticated information and control system that contains millions of lines of code. Our next example also involves a large computer system for applying a model to control operations over a national network. This system, called SYSNET, was developed as the result of an OR study done for Yellow Freight System, Inc.1 Yellow Freight annually handles over 15 million shipments by motor carrier over a network of 630 terminals throughout the United States. SYSNET is used to optimize both the routing of shipments and the design of the network. Because SYSNET requires extensive information about freight flows and forecasts, transportation and handling costs, and so on, a major part of the OR study involved integrating SYSNET into the corporate management information system. This integration enabled periodic updating of all the input for the model. The implementation of SYSNET resulted in annual savings of approximately $17.3 million as well as improved service to customers. 1

J. W. Braklow, W. W. Graham, S. M. Hassler, K. E. Peck, and W. B. Powell, “Interactive Optimization Improves Service and Performance for Yellow Freight System,” Interfaces, 22(1): 147–172, Jan.–Feb. 1992. See especially p. 163.

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Our next example illustrates a decision support system. A system of this type was developed for Texaco1 to help plan and schedule its blending operations at its various refineries. Called OMEGA (Optimization Method for the Estimation of Gasoline Attributes), it is an interactive system based on a nonlinear optimization model that is implemented on both personal computers and larger computers. Input data can be entered either manually or by interfacing with refinery databases. The user has considerable flexibility in choosing an objective function and constraints to fit the current situation as well as in asking a series of what-if questions (i.e., questions about what would happen if the assumed conditions change). OMEGA is maintained centrally by Texaco’s information technology department, which enables constant updating to reflect new government regulations, other business changes, and changes in refinery operations. The implementation of OMEGA is credited with annual savings of more than $30 million as well as improved planning, quality control, and marketing information.

2.6

IMPLEMENTATION After a system is developed for applying the model, the last phase of an OR study is to implement this system as prescribed by management. This phase is a critical one because it is here, and only here, that the benefits of the study are reaped. Therefore, it is important for the OR team to participate in launching this phase, both to make sure that model solutions are accurately translated to an operating procedure and to rectify any flaws in the solutions that are then uncovered. The success of the implementation phase depends a great deal upon the support of both top management and operating management. The OR team is much more likely to gain this support if it has kept management well informed and encouraged management’s active guidance throughout the course of the study. Good communications help to ensure that the study accomplishes what management wanted and so deserves implementation. They also give management a greater sense of ownership of the study, which encourages their support for implementation. The implementation phase involves several steps. First, the OR team gives operating management a careful explanation of the new system to be adopted and how it relates to operating realities. Next, these two parties share the responsibility for developing the procedures required to put this system into operation. Operating management then sees that a detailed indoctrination is given to the personnel involved, and the new course of action is initiated. If successful, the new system may be used for years to come. With this in mind, the OR team monitors the initial experience with the course of action taken and seeks to identify any modifications that should be made in the future. Throughout the entire period during which the new system is being used, it is important to continue to obtain feedback on how well the system is working and whether the assumptions of the model continue to be satisfied. When significant deviations from the original assumptions occur, the model should be revisited to determine if any modifications should be made in the system. The postoptimality analysis done earlier (as described in Sec. 2.3) can be helpful in guiding this review process. 1

C. W. DeWitt, L. S. Lasdon, A. D. Waren, D. A. Brenner, and S. A. Melhem, “OMEGA: An Improved Gasoline Blending System for Texaco,” Interfaces, 19(1): 85–101, Jan.–Feb. 1989. See especially pp. 93–95.

2.7 CONCLUSIONS

21

Upon culmination of a study, it is appropriate for the OR team to document its methodology clearly and accurately enough so that the work is reproducible. Replicability should be part of the professional ethical code of the operations researcher. This condition is especially crucial when controversial public policy issues are being studied. Examples. This last point about documenting an OR study is illustrated by the Rijkswaterstaat study of national water management policy for the Netherlands discussed at the end of Secs. 2.2, 2.3, and 2.4. Management wanted unusually thorough and extensive documentation, both to support the new policy and to use in training new analysts or in performing new studies. Requiring several years to complete, this documentation aggregated 4000 single-spaced pages and 21 volumes! Our next example concerns the IBM OR study discussed at the end of Secs. 2.4 and 2.5. Careful planning was required to implement the complex Optimizer system for controlling IBM’s national network of spare-parts inventories. Three factors proved to be especially important in achieving a successful implementation. As discussed in Sec. 2.4, the first was the inclusion of a user team (consisting of operational managers) as advisers to the OR team throughout the study. By the time of the implementation phase, these operational managers had a strong sense of ownership and so had become ardent supporters for installing Optimizer in their functional areas. A second success factor was a very extensive user acceptance test whereby users could identify problem areas that needed rectifying prior to full implementation. The third key was that the new system was phased in gradually, with careful testing at each phase, so the major bugs could be eliminated before the system went live nationally. Our final example concerns Yellow Freight’s SYSNET system for routing shipments over a national network, as described at the end of the preceding section. In this case, there were four key elements to the implementation process. The first was selling the concept to upper management. This was successfully done through validating the accuracy of the cost model and then holding interactive sessions for upper management that demonstrated the effectiveness of the system. The second element was the development of an implementation strategy for gradually phasing in the new system while identifying and eliminating its flaws. The third involved working closely with operational managers to install the system properly, provide the needed support tools, train the personnel who will use the system, and convince them of the usefulness of the system. The final key element was the provision of management incentives and enforcement for the effective implementation of the system.

2.7

CONCLUSIONS Although the remainder of this book focuses primarily on constructing and solving mathematical models, in this chapter we have tried to emphasize that this constitutes only a portion of the overall process involved in conducting a typical OR study. The other phases described here also are very important to the success of the study. Try to keep in perspective the role of the model and the solution procedure in the overall process as you move through the subsequent chapters. Then, after gaining a deeper understanding of mathematical models, we suggest that you plan to return to review this chapter again in order to further sharpen this perspective.

22

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OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

OR is closely intertwined with the use of computers. In the early years, these generally were mainframe computers, but now personal computers and workstations are being widely used to solve OR models. In concluding this discussion of the major phases of an OR study, it should be emphasized that there are many exceptions to the “rules” prescribed in this chapter. By its very nature, OR requires considerable ingenuity and innovation, so it is impossible to write down any standard procedure that should always be followed by OR teams. Rather, the preceding description may be viewed as a model that roughly represents how successful OR studies are conducted.

SELECTED REFERENCES 1. Fortuin, L., P. van Beek, and L. van Wassenhove (eds.): OR at wORk: Practical Experiences of Operational Research, Taylor & Francis, Bristol, PA, 1996. 2. Gass, S. I.: “Decision-Aiding Models: Validation, Assessment, and Related Issues for Policy Analysis,” Operations Research, 31: 603–631, 1983. 3. Gass, S. I.: “Model World: Danger, Beware the User as Modeler,” Interfaces, 20(3): 60–64, May–June 1990. 4. Hall, R. W.: “What’s So Scientific about MS/OR?” Interfaces, 15(2): 40–45, March–April 1985. 5. Miser, H. J.: “The Easy Chair: Observation and Experimentation,” Interfaces, 19(5): 23–30, Sept.–Oct. 1989. 6. Morris, W. T.: “On the Art of Modeling,” Management Science, 13: B707–717, 1967. 7. Murthy, D. N. P., N. W. Page, and E. Y. Rodin: Mathematical Modeling: A Tool for Problem Solving in Engineering, Physical, Biological and Social Sciences, Pergamon Press, Oxford, England, 1990. 8. Simon, H. A.: “Prediction and Prescription in Systems Modeling,” Operations Research, 38: 7–14, 1990. 9. Tilanus, C. B., O. B. DeGans, and J. K. Lenstra (eds.): Quantitative Methods in Management: Case Studies of Failures and Successes, Wiley, New York, 1986. 10. Williams, H. P.: Model Building in Mathematical Programming, 3d ed., Wiley, New York, 1990.

PROBLEMS 2.1-1. Read the article footnoted in Sec. 2.1 that describes an OR study done for the San Francisco Police Department. (a) Summarize the background that led to undertaking this study. (b) Define part of the problem being addressed by identifying the six directives for the scheduling system to be developed. (c) Describe how the needed data were gathered. (d) List the various tangible and intangible benefits that resulted from the study. 2.1-2. Read the article footnoted in Sec. 2.1 that describes an OR study done for the Health Department of New Haven, Connecticut. (a) Summarize the background that led to undertaking this study.

(b) Outline the system developed to track and test each needle and syringe in order to gather the needed data. (c) Summarize the initial results from this tracking and testing system. (d) Describe the impact and potential impact of this study on public policy. 2.2-1. Read the article footnoted in Sec. 2.2 that describes an OR study done for the Rijkswaterstaat of the Netherlands. (Focus especially on pp. 3–20 and 30–32.) (a) Summarize the background that led to undertaking this study. (b) Summarize the purpose of each of the five mathematical models described on pp. 10–18.

CHAPTER 2 PROBLEMS

(c) Summarize the “impact measures” (measures of performance) for comparing policies that are described on pp. 6–7 of this article. (d) List the various tangible and intangible benefits that resulted from the study. 2.2-2. Read Selected Reference 4. (a) Identify the author’s example of a model in the natural sciences and of a model in OR. (b) Describe the author’s viewpoint about how basic precepts of using models to do research in the natural sciences can also be used to guide research on operations (OR). 2.3-1. Refer to Selected Reference 4. (a) Describe the author’s viewpoint about whether the sole goal in using a model should be to find its optimal solution. (b) Summarize the author’s viewpoint about the complementary roles of modeling, evaluating information from the model, and then applying the decision maker’s judgment when deciding on a course of action. 2.4-1. Refer to pp. 18–20 of the article footnoted in Sec. 2.2 that describes an OR study done for the Rijkswaterstaat of the Netherlands. Describe an important lesson that was gained from model validation in this study. 2.4-2. Read Selected Reference 5. Summarize the author’s viewpoint about the roles of observation and experimentation in the model validation process.

23

(c) OMEGA is constantly being updated and extended to reflect changes in the operating environment. Briefly describe the various kinds of changes involved. (d) Summarize how OMEGA is used. (e) List the various tangible and intangible benefits that resulted from the study. 2.5-2. Refer to the article footnoted in Sec. 2.5 that describes an OR study done for Yellow Freight System, Inc. (a) Referring to pp. 147–149 of this article, summarize the background that led to undertaking this study. (b) Referring to p. 150, briefly describe the computer system SYSNET that was developed as a result of this study. Also summarize the applications of SYSNET. (c) Referring to pp. 162–163, describe why the interactive aspects of SYSNET proved important. (d) Referring to p. 163, summarize the outputs from SYSNET. (e) Referring to pp. 168–172, summarize the various benefits that have resulted from using SYSNET. 2.6-1. Refer to pp. 163–167 of the article footnoted in Sec. 2.5 that describes an OR study done for Yellow Freight System, Inc., and the resulting computer system SYSNET. (a) Briefly describe how the OR team gained the support of upper management for implementing SYSNET. (b) Briefly describe the implementation strategy that was developed. (c) Briefly describe the field implementation. (d) Briefly describe how management incentives and enforcement were used in implementing SYSNET.

2.4-3. Read pp. 603–617 of Selected Reference 2. (a) What does the author say about whether a model can be completely validated? (b) Summarize the distinctions made between model validity, data validity, logical/mathematical validity, predictive validity, operational validity, and dynamic validity. (c) Describe the role of sensitivity analysis in testing the operational validity of a model. (d) What does the author say about whether there is a validation methodology that is appropriate for all models? (e) Cite the page in the article that lists basic validation steps.

2.6-2. Read the article footnoted in Sec. 2.4 that describes an OR study done for IBM and the resulting computer system Optimizer. (a) Summarize the background that led to undertaking this study. (b) List the complicating factors that the OR team members faced when they started developing a model and a solution algorithm. (c) Briefly describe the preimplementation test of Optimizer. (d) Briefly describe the field implementation test. (e) Briefly describe national implementation. (f) List the various tangible and intangible benefits that resulted from the study.

2.5-1. Read the article footnoted in Sec. 2.5 that describes an OR study done for Texaco. (a) Summarize the background that led to undertaking this study. (b) Briefly describe the user interface with the decision support system OMEGA that was developed as a result of this study.

2.7-1. Read Selected Reference 3. The author describes 13 detailed phases of any OR study that develops and applies a computer-based model, whereas this chapter describes six broader phases. For each of these broader phases, list the detailed phases that fall partially or primarily within the broader phase.

3 Introduction to Linear Programming The development of linear programming has been ranked among the most important scientific advances of the mid-20th century, and we must agree with this assessment. Its impact since just 1950 has been extraordinary. Today it is a standard tool that has saved many thousands or millions of dollars for most companies or businesses of even moderate size in the various industrialized countries of the world; and its use in other sectors of society has been spreading rapidly. A major proportion of all scientific computation on computers is devoted to the use of linear programming. Dozens of textbooks have been written about linear programming, and published articles describing important applications now number in the hundreds. What is the nature of this remarkable tool, and what kinds of problems does it address? You will gain insight into this topic as you work through subsequent examples. However, a verbal summary may help provide perspective. Briefly, the most common type of application involves the general problem of allocating limited resources among competing activities in a best possible (i.e., optimal) way. More precisely, this problem involves selecting the level of certain activities that compete for scarce resources that are necessary to perform those activities. The choice of activity levels then dictates how much of each resource will be consumed by each activity. The variety of situations to which this description applies is diverse, indeed, ranging from the allocation of production facilities to products to the allocation of national resources to domestic needs, from portfolio selection to the selection of shipping patterns, from agricultural planning to the design of radiation therapy, and so on. However, the one common ingredient in each of these situations is the necessity for allocating resources to activities by choosing the levels of those activities. Linear programming uses a mathematical model to describe the problem of concern. The adjective linear means that all the mathematical functions in this model are required to be linear functions. The word programming does not refer here to computer programming; rather, it is essentially a synonym for planning. Thus, linear programming involves the planning of activities to obtain an optimal result, i.e., a result that reaches the specified goal best (according to the mathematical model) among all feasible alternatives. Although allocating resources to activities is the most common type of application, linear programming has numerous other important applications as well. In fact, any problem whose mathematical model fits the very general format for the linear programming model is a linear programming problem. Furthermore, a remarkably efficient solution pro24

3.1 PROTOTYPE EXAMPLE

25

cedure, called the simplex method, is available for solving linear programming problems of even enormous size. These are some of the reasons for the tremendous impact of linear programming in recent decades. Because of its great importance, we devote this and the next six chapters specifically to linear programming. After this chapter introduces the general features of linear programming, Chaps. 4 and 5 focus on the simplex method. Chapter 6 discusses the further analysis of linear programming problems after the simplex method has been initially applied. Chapter 7 presents several widely used extensions of the simplex method and introduces an interior-point algorithm that sometimes can be used to solve even larger linear programming problems than the simplex method can handle. Chapters 8 and 9 consider some special types of linear programming problems whose importance warrants individual study. You also can look forward to seeing applications of linear programming to other areas of operations research (OR) in several later chapters. We begin this chapter by developing a miniature prototype example of a linear programming problem. This example is small enough to be solved graphically in a straightforward way. The following two sections present the general linear programming model and its basic assumptions. Sections 3.4 and 3.5 give some additional examples of linear programming applications, including three case studies. Section 3.6 describes how linear programming models of modest size can be conveniently displayed and solved on a spreadsheet. However, some linear programming problems encountered in practice require truly massive models. Section 3.7 illustrates how a massive model can arise and how it can still be formulated successfully with the help of a special modeling language such as MPL (described in this section) or LINGO (described in the appendix to this chapter).

3.1

PROTOTYPE EXAMPLE The WYNDOR GLASS CO. produces high-quality glass products, including windows and glass doors. It has three plants. Aluminum frames and hardware are made in Plant 1, wood frames are made in Plant 2, and Plant 3 produces the glass and assembles the products. Because of declining earnings, top management has decided to revamp the company’s product line. Unprofitable products are being discontinued, releasing production capacity to launch two new products having large sales potential: Product 1: An 8-foot glass door with aluminum framing Product 2: A 4 6 foot double-hung wood-framed window Product 1 requires some of the production capacity in Plants 1 and 3, but none in Plant 2. Product 2 needs only Plants 2 and 3. The marketing division has concluded that the company could sell as much of either product as could be produced by these plants. However, because both products would be competing for the same production capacity in Plant 3, it is not clear which mix of the two products would be most profitable. Therefore, an OR team has been formed to study this question. The OR team began by having discussions with upper management to identify management’s objectives for the study. These discussions led to developing the following definition of the problem: Determine what the production rates should be for the two products in order to maximize their total profit, subject to the restrictions imposed by the limited production capacities

26

3

INTRODUCTION TO LINEAR PROGRAMMING

available in the three plants. (Each product will be produced in batches of 20, so the production rate is defined as the number of batches produced per week.) Any combination of production rates that satisfies these restrictions is permitted, including producing none of one product and as much as possible of the other.

The OR team also identified the data that needed to be gathered: 1. Number of hours of production time available per week in each plant for these new products. (Most of the time in these plants already is committed to current products, so the available capacity for the new products is quite limited.) 2. Number of hours of production time used in each plant for each batch produced of each new product. 3. Profit per batch produced of each new product. (Profit per batch produced was chosen as an appropriate measure after the team concluded that the incremental profit from each additional batch produced would be roughly constant regardless of the total number of batches produced. Because no substantial costs will be incurred to initiate the production and marketing of these new products, the total profit from each one is approximately this profit per batch produced times the number of batches produced.) Obtaining reasonable estimates of these quantities required enlisting the help of key personnel in various units of the company. Staff in the manufacturing division provided the data in the first category above. Developing estimates for the second category of data required some analysis by the manufacturing engineers involved in designing the production processes for the new products. By analyzing cost data from these same engineers and the marketing division, along with a pricing decision from the marketing division, the accounting department developed estimates for the third category. Table 3.1 summarizes the data gathered. The OR team immediately recognized that this was a linear programming problem of the classic product mix type, and the team next undertook the formulation of the corresponding mathematical model. Formulation as a Linear Programming Problem To formulate the mathematical (linear programming) model for this problem, let x1 number of batches of product 1 produced per week x2 number of batches of product 2 produced per week Z total profit per week (in thousands of dollars) from producing these two products Thus, x1 and x2 are the decision variables for the model. Using the bottom row of Table 3.1, we obtain Z 3x1 5x2. The objective is to choose the values of x1 and x2 so as to maximize Z 3x1 5x2, subject to the restrictions imposed on their values by the limited production capacities available in the three plants. Table 3.1 indicates that each batch of product 1 produced per week uses 1 hour of production time per week in Plant 1, whereas only 4 hours per week are available. This restriction is expressed mathematically by the inequality x1 4. Similarly, Plant 2 imposes the restriction that 2x2 12. The number of hours of production

3.1 PROTOTYPE EXAMPLE

27

TABLE 3.1 Data for the Wyndor Glass Co. problem Production Time per Batch, Hours Product Plant

1

2

Production Time Available per Week, Hours

1 2 3

1 0 3

0 2 2

4 12 18

Profit per batch

$3,000

$5,000

time used per week in Plant 3 by choosing x1 and x2 as the new products’ production rates would be 3x1 2x2. Therefore, the mathematical statement of the Plant 3 restriction is 3x1 2x2 18. Finally, since production rates cannot be negative, it is necessary to restrict the decision variables to be nonnegative: x1 0 and x2 0. To summarize, in the mathematical language of linear programming, the problem is to choose values of x1 and x2 so as to Maximize

Z 3x1 5x2,

subject to the restrictions 3x1 2x2 4 3x1 2x2 12 3x1 2x2 18 and x1 0,

x2 0.

(Notice how the layout of the coefficients of x1 and x2 in this linear programming model essentially duplicates the information summarized in Table 3.1.) Graphical Solution This very small problem has only two decision variables and therefore only two dimensions, so a graphical procedure can be used to solve it. This procedure involves constructing a two-dimensional graph with x1 and x2 as the axes. The first step is to identify the values of (x1, x2) that are permitted by the restrictions. This is done by drawing each line that borders the range of permissible values for one restriction. To begin, note that the nonnegativity restrictions x1 0 and x2 0 require (x1, x2) to lie on the positive side of the axes (including actually on either axis), i.e., in the first quadrant. Next, observe that the restriction x1 4 means that (x1, x2) cannot lie to the right of the line x1 4. These results are shown in Fig. 3.1, where the shaded area contains the only values of (x1, x2) that are still allowed. In a similar fashion, the restriction 2x2 12 (or, equivalently, x2 6) implies that the line 2x2 12 should be added to the boundary of the permissible region. The final restriction, 3x1 2x2 18, requires plotting the points (x1, x2) such that 3x1 2x2 18

28

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INTRODUCTION TO LINEAR PROGRAMMING

x2

5 4 3 2 FIGURE 3.1 Shaded area shows values of (x1, x2) allowed by x1 0, x2 0, x1 4.

1 0

1

2

3

4

5

6

7

x1

(another line) to complete the boundary. (Note that the points such that 3x1 2x2 18 are those that lie either underneath or on the line 3x1 2x2 18, so this is the limiting line above which points do not satisfy the inequality.) The resulting region of permissible values of (x1, x2), called the feasible region, is shown in Fig. 3.2. (The demo called Graphical Method in your OR Tutor provides a more detailed example of constructing a feasible region.)

FIGURE 3.2 Shaded area shows the set of permissible values of (x1, x2), called the feasible region.

x2 10 3x1 2x2 18 8 x1 4 2x2 12

6

4 Feasible region 2

0

2

4

6

8

x1

3.1 PROTOTYPE EXAMPLE

29

The final step is to pick out the point in this feasible region that maximizes the value of Z 3x1 5x2. To discover how to perform this step efficiently, begin by trial and error. Try, for example, Z 10 3x1 5x2 to see if there are in the permissible region any values of (x1, x2) that yield a value of Z as large as 10. By drawing the line 3x1 5x2 10 (see Fig. 3.3), you can see that there are many points on this line that lie within the region. Having gained perspective by trying this arbitrarily chosen value of Z 10, you should next try a larger arbitrary value of Z, say, Z 20 3x1 5x2. Again, Fig. 3.3 reveals that a segment of the line 3x1 5x2 20 lies within the region, so that the maximum permissible value of Z must be at least 20. Now notice in Fig. 3.3 that the two lines just constructed are parallel. This is no coincidence, since any line constructed in this way has the form Z 3x1 5x2 for the chosen value of Z, which implies that 5x2 3x1 Z or, equivalently, 3 1 x2 x1 Z 5 5 This last equation, called the slope-intercept form of the objective function, demonstrates that the slope of the line is 35 (since each unit increase in x1 changes x2 by 35), whereas the intercept of the line with the x2 axis is 15 Z (since x2 15 Z when x1 0). The fact that the slope is fixed at 35 means that all lines constructed in this way are parallel. Again, comparing the 10 3x1 5x2 and 20 3x1 5x2 lines in Fig. 3.3, we note that the line giving a larger value of Z (Z 20) is farther up and away from the origin than the other line (Z 10). This fact also is implied by the slope-intercept form of the objective function, which indicates that the intercept with the x1 axis ( 15 Z) increases when the value chosen for Z is increased.

FIGURE 3.3 The value of (x1, x2) that maximizes 3x1 5x2 is (2, 6).

x2

8 Z 36 3x1 5x2 6

Z 20 3x1 5x2

(2, 6)

4

Z 10 3x1 5x2 2

0

2

4

6

8

10

x1

30

3 INTRODUCTION TO LINEAR PROGRAMMING

These observations imply that our trial-and-error procedure for constructing lines in Fig. 3.3 involves nothing more than drawing a family of parallel lines containing at least one point in the feasible region and selecting the line that corresponds to the largest value of Z. Figure 3.3 shows that this line passes through the point (2, 6), indicating that the optimal solution is x1 2 and x2 6. The equation of this line is 3x1 5x2 3(2) 5(6) 36 Z, indicating that the optimal value of Z is Z 36. The point (2, 6) lies at the intersection of the two lines 2x2 12 and 3x1 2x2 18, shown in Fig. 3.2, so that this point can be calculated algebraically as the simultaneous solution of these two equations. Having seen the trial-and-error procedure for finding the optimal point (2, 6), you now can streamline this approach for other problems. Rather than draw several parallel lines, it is sufficient to form a single line with a ruler to establish the slope. Then move the ruler with fixed slope through the feasible region in the direction of improving Z. (When the objective is to minimize Z, move the ruler in the direction that decreases Z.) Stop moving the ruler at the last instant that it still passes through a point in this region. This point is the desired optimal solution. This procedure often is referred to as the graphical method for linear programming. It can be used to solve any linear programming problem with two decision variables. With considerable difficulty, it is possible to extend the method to three decision variables but not more than three. (The next chapter will focus on the simplex method for solving larger problems.) Conclusions The OR team used this approach to find that the optimal solution is x1 2, x2 6, with Z 36. This solution indicates that the Wyndor Glass Co. should produce products 1 and 2 at the rate of 2 batches per week and 6 batches per week, respectively, with a resulting total profit of $36,000 per week. No other mix of the two products would be so profitable—according to the model. However, we emphasized in Chap. 2 that well-conducted OR studies do not simply find one solution for the initial model formulated and then stop. All six phases described in Chap. 2 are important, including thorough testing of the model (see Sec. 2.4) and postoptimality analysis (see Sec. 2.3). In full recognition of these practical realities, the OR team now is ready to evaluate the validity of the model more critically (to be continued in Sec. 3.3) and to perform sensitivity analysis on the effect of the estimates in Table 3.1 being different because of inaccurate estimation, changes of circumstances, etc. (to be continued in Sec. 6.7). Continuing the Learning Process with Your OR Courseware This is the first of many points in the book where you may find it helpful to use your OR Courseware in the CD-ROM that accompanies this book. A key part of this courseware is a program called OR Tutor. This program includes a complete demonstration example of the graphical method introduced in this section. Like the many other demonstration examples accompanying other sections of the book, this computer demonstration highlights concepts that are difficult to convey on the printed page. You may refer to Appendix 1 for documentation of the software. When you formulate a linear programming model with more than two decision variables (so the graphical method cannot be used), the simplex method described in Chap. 4

3.2 THE LINEAR PROGRAMMING MODEL

31

enables you to still find an optimal solution immediately. Doing so also is helpful for model validation, since finding a nonsensical optimal solution signals that you have made a mistake in formulating the model. We mentioned in Sec. 1.4 that your OR Courseware introduces you to three particularly popular commercial software packages—the Excel Solver, LINGO/LINDO, and MPL/CPLEX—for solving a variety of OR models. All three packages include the simplex method for solving linear programming models. Section 3.6 describes how to use Excel to formulate and solve linear programming models in a spreadsheet format. Descriptions of the other packages are provided in Sec. 3.7 (MPL and LINGO), Appendix 3.1 (LINGO), Sec. 4.8 (CPLEX and LINDO), and Appendix 4.1 (LINDO). In addition, your OR Courseware includes a file for each of the three packages showing how it can be used to solve each of the examples in this chapter.

3.2

THE LINEAR PROGRAMMING MODEL The Wyndor Glass Co. problem is intended to illustrate a typical linear programming problem (miniature version). However, linear programming is too versatile to be completely characterized by a single example. In this section we discuss the general characteristics of linear programming problems, including the various legitimate forms of the mathematical model for linear programming. Let us begin with some basic terminology and notation. The first column of Table 3.2 summarizes the components of the Wyndor Glass Co. problem. The second column then introduces more general terms for these same components that will fit many linear programming problems. The key terms are resources and activities, where m denotes the number of different kinds of resources that can be used and n denotes the number of activities being considered. Some typical resources are money and particular kinds of machines, equipment, vehicles, and personnel. Examples of activities include investing in particular projects, advertising in particular media, and shipping goods from a particular source to a particular destination. In any application of linear programming, all the activities may be of one general kind (such as any one of these three examples), and then the individual activities would be particular alternatives within this general category. As described in the introduction to this chapter, the most common type of application of linear programming involves allocating resources to activities. The amount available of each resource is limited, so a careful allocation of resources to activities must be made. Determining this allocation involves choosing the levels of the activities that achieve the best possible value of the overall measure of performance. TABLE 3.2 Common terminology for linear programming Prototype Example

General Problem

Production capacities of plants 3 plants

Resources m resources

Production of products 2 products Production rate of product j, xj

Activities n activities Level of activity j, xj

Profit Z

Overall measure of performance Z

32

3

INTRODUCTION TO LINEAR PROGRAMMING

Certain symbols are commonly used to denote the various components of a linear programming model. These symbols are listed below, along with their interpretation for the general problem of allocating resources to activities. Z value of overall measure of performance. xj level of activity j (for j 1, 2, . . . , n). cj increase in Z that would result from each unit increase in level of activity j. bi amount of resource i that is available for allocation to activities (for i 1, 2, . . . , m). aij amount of resource i consumed by each unit of activity j. The model poses the problem in terms of making decisions about the levels of the activities, so x1, x2, . . . , xn are called the decision variables. As summarized in Table 3.3, the values of cj, bi, and aij (for i 1, 2, . . . , m and j 1, 2, . . . , n) are the input constants for the model. The cj, bi, and aij are also referred to as the parameters of the model. Notice the correspondence between Table 3.3 and Table 3.1. A Standard Form of the Model Proceeding as for the Wyndor Glass Co. problem, we can now formulate the mathematical model for this general problem of allocating resources to activities. In particular, this model is to select the values for x1, x2, . . . , xn so as to Maximize

Z c1x1 c2x2 cnxn,

subject to the restrictions a11x1 a12x2 a1nxn b1 a21x1 a22x2 a2nxn b2 am1x1 am2x2 amnxn bm, TABLE 3.3 Data needed for a linear programming model involving the allocation of resources to activities Resource Usage per Unit of Activity Activity Resource

1

2

...

n

1 2 . . . m

a11 a21

a12 a22

... ...

a1n a2n

...

...

...

...

am1

am2

...

amn

c1

c2

...

cn

Contribution to Z per unit of activity

Amount of Resource Available b1 b2 . . . bm

3.2 THE LINEAR PROGRAMMING MODEL

33

and x1 0,

x2 0,

...,

xn 0.

We call this our standard form1 for the linear programming problem. Any situation whose mathematical formulation fits this model is a linear programming problem. Notice that the model for the Wyndor Glass Co. problem fits our standard form, with m 3 and n 2. Common terminology for the linear programming model can now be summarized. The function being maximized, c1x1 c2x2 cn xn, is called the objective function. The restrictions normally are referred to as constraints. The first m constraints (those with a function of all the variables ai1x1 ai2x2 ain xn on the left-hand side) are sometimes called functional constraints (or structural constraints). Similarly, the xj 0 restrictions are called nonnegativity constraints (or nonnegativity conditions). Other Forms We now hasten to add that the preceding model does not actually fit the natural form of some linear programming problems. The other legitimate forms are the following: 1. Minimizing rather than maximizing the objective function: Minimize

Z c1x1 c2 x2 cn xn.

2. Some functional constraints with a greater-than-or-equal-to inequality: ai1x1 ai2x2 ain xn bi

for some values of i.

3. Some functional constraints in equation form: ai1x1 ai2x2 ain xn bi

for some values of i.

4. Deleting the nonnegativity constraints for some decision variables: xj unrestricted in sign

for some values of j.

Any problem that mixes some of or all these forms with the remaining parts of the preceding model is still a linear programming problem. Our interpretation of the words allocating limited resources among competing activities may no longer apply very well, if at all; but regardless of the interpretation or context, all that is required is that the mathematical statement of the problem fit the allowable forms. Terminology for Solutions of the Model You may be used to having the term solution mean the final answer to a problem, but the convention in linear programming (and its extensions) is quite different. Here, any specification of values for the decision variables (x1, x2, . . . , xn) is called a solution, regardless of whether it is a desirable or even an allowable choice. Different types of solutions are then identified by using an appropriate adjective. 1

This is called our standard form rather than the standard form because some textbooks adopt other forms.

34

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INTRODUCTION TO LINEAR PROGRAMMING

A feasible solution is a solution for which all the constraints are satisfied. An infeasible solution is a solution for which at least one constraint is violated. In the example, the points (2, 3) and (4, 1) in Fig. 3.2 are feasible solutions, while the points (1, 3) and (4, 4) are infeasible solutions. The feasible region is the collection of all feasible solutions. The feasible region in the example is the entire shaded area in Fig. 3.2. It is possible for a problem to have no feasible solutions. This would have happened in the example if the new products had been required to return a net profit of at least $50,000 per week to justify discontinuing part of the current product line. The corresponding constraint, 3x1 5x2 50, would eliminate the entire feasible region, so no mix of new products would be superior to the status quo. This case is illustrated in Fig. 3.4. Given that there are feasible solutions, the goal of linear programming is to find a best feasible solution, as measured by the value of the objective function in the model. An optimal solution is a feasible solution that has the most favorable value of the objective function. The most favorable value is the largest value if the objective function is to be maximized, whereas it is the smallest value if the objective function is to be minimized. Most problems will have just one optimal solution. However, it is possible to have more than one. This would occur in the example if the profit per batch produced of product 2 were changed to $2,000. This changes the objective function to Z 3x1 2x2, so that all the points

FIGURE 3.4 The Wyndor Glass Co. problem would have no feasible solutions if the constraint 3x1 5x2 50 were added to the problem.

x2 Maximize Z 3x1 5x2, 4 x1 subject to 2x2 12 3x1 2x2 18 3x1 5x2 50 x1 0, and x2 0

10 3x1 5x2 50 8

6 2x2 12 4

3x1 2x2 18 x1 0

2

x1 4 x2 0

0

2

4

6

8

10

x1

3.2 THE LINEAR PROGRAMMING MODEL

35

FIGURE 3.5 The Wyndor Glass Co. problem would have multiple optimal solutions if the objective function were changed to Z 3x1 2x2.

on the line segment connecting (2, 6) and (4, 3) would be optimal. This case is illustrated in Fig. 3.5. As in this case, any problem having multiple optimal solutions will have an infinite number of them, each with the same optimal value of the objective function. Another possibility is that a problem has no optimal solutions. This occurs only if (1) it has no feasible solutions or (2) the constraints do not prevent improving the value of the objective function (Z) indefinitely in the favorable direction (positive or negative). The latter case is referred to as having an unbounded Z. To illustrate, this case would result if the last two functional constraints were mistakenly deleted in the example, as illustrated in Fig. 3.6. We next introduce a special type of feasible solution that plays the key role when the simplex method searches for an optimal solution. A corner-point feasible (CPF) solution is a solution that lies at a corner of the feasible region. Figure 3.7 highlights the five CPF solutions for the example. Sections 4.1 and 5.1 will delve into the various useful properties of CPF solutions for problems of any size, including the following relationship with optimal solutions. Relationship between optimal solutions and CPF solutions: Consider any linear programming problem with feasible solutions and a bounded feasible region. The problem must possess CPF solutions and at least one optimal solution. Furthermore, the best CPF solution must be an optimal solution. Thus, if a problem has exactly one optimal solution, it must be a CPF solution. If the problem has multiple optimal solutions, at least two must be CPF solutions.

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(4, ), Z

x2 10

(4, 10), Z 62

8

(4, 8), Z 52

6

FIGURE 3.6 The Wyndor Glass Co. problem would have no optimal solutions if the only functional constraint were x1 4, because x2 then could be increased indefinitely in the feasible region without ever reaching the maximum value of Z 3x1 5x2.

(4, 6), Z 42 Feasible region

4

(4, 4), Z 32

2

(4, 2), Z 22

0

Maximize Z 3x1 5x2, subject to x1 4 x1 0, and x2 0

2

4

6

8

10

x1

The example has exactly one optimal solution, (x1, x2) (2, 6), which is a CPF solution. (Think about how the graphical method leads to the one optimal solution being a CPF solution.) When the example is modified to yield multiple optimal solutions, as shown in Fig. 3.5, two of these optimal solutions—(2, 6) and (4, 3)—are CPF solutions.

3.3

ASSUMPTIONS OF LINEAR PROGRAMMING All the assumptions of linear programming actually are implicit in the model formulation given in Sec. 3.2. However, it is good to highlight these assumptions so you can more easily evaluate how well linear programming applies to any given problem. Furthermore, we still need to see why the OR team for the Wyndor Glass Co. concluded that a linear programming formulation provided a satisfactory representation of the problem. Proportionality Proportionality is an assumption about both the objective function and the functional constraints, as summarized below. Proportionality assumption: The contribution of each activity to the value of the objective function Z is proportional to the level of the activity xj, as represented by the cj xj term in the objective function. Similarly, the contribution of each activity to the left-hand side of each functional constraint is proportional to the level of the activity xj, as represented by the aij xj term in the constraint.

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

FIGURE 3.7 The five dots are the five CPF solutions for the Wyndor Glass Co. problem.

37

x2 (0, 6)

(2, 6)

Feasible region

(4, 3)

(0, 0)

(4, 0)

x1

Consequently, this assumption rules out any exponent other than 1 for any variable in any term of any function (whether the objective function or the function on the left-hand side of a functional constraint) in a linear programming model.1 To illustrate this assumption, consider the first term (3x1) in the objective function (Z 3x1 5x2) for the Wyndor Glass Co. problem. This term represents the profit generated per week (in thousands of dollars) by producing product 1 at the rate of x1 batches per week. The proportionality satisfied column of Table 3.4 shows the case that was assumed in Sec. 3.1, namely, that this profit is indeed proportional to x1 so that 3x1 is the appropriate term for the objective function. By contrast, the next three columns show different hypothetical cases where the proportionality assumption would be violated. Refer first to the Case 1 column in Table 3.4. This case would arise if there were start-up costs associated with initiating the production of product 1. For example, there 1 When the function includes any cross-product terms, proportionality should be interpreted to mean that changes in the function value are proportional to changes in each variable (xj) individually, given any fixed values for all the other variables. Therefore, a cross-product term satisfies proportionality as long as each variable in the term has an exponent of 1. (However, any cross-product term violates the additivity assumption, discussed next.)

TABLE 3.4 Examples of satisfying or violating proportionality Profit from Product 1 ($000 per Week) Proportionality Violated x1

Proportionality Satisfied

Case 1

Case 2

Case 3

0 1 2 3 4

0 3 6 9 12

0 2 5 8 11

0 3 7 12 18

0 3 5 6 6

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INTRODUCTION TO LINEAR PROGRAMMING

might be costs involved with setting up the production facilities. There might also be costs associated with arranging the distribution of the new product. Because these are one-time costs, they would need to be amortized on a per-week basis to be commensurable with Z (profit in thousands of dollars per week). Suppose that this amortization were done and that the total start-up cost amounted to reducing Z by 1, but that the profit without considering the start-up cost would be 3x1. This would mean that the contribution from product 1 to Z should be 3x1 1 for x1 0, whereas the contribution would be 3x1 0 when x1 0 (no start-up cost). This profit function,1 which is given by the solid curve in Fig. 3.8, certainly is not proportional to x1. At first glance, it might appear that Case 2 in Table 3.4 is quite similar to Case 1. However, Case 2 actually arises in a very different way. There no longer is a start-up cost, and the profit from the first unit of product 1 per week is indeed 3, as originally assumed. However, there now is an increasing marginal return; i.e., the slope of the profit function for product 1 (see the solid curve in Fig. 3.9) keeps increasing as x1 is increased. This violation of proportionality might occur because of economies of scale that can sometimes be achieved at higher levels of production, e.g., through the use of more efficient highvolume machinery, longer production runs, quantity discounts for large purchases of raw materials, and the learning-curve effect whereby workers become more efficient as they gain experience with a particular mode of production. As the incremental cost goes down, the incremental profit will go up (assuming constant marginal revenue). If the contribution from product 1 to Z were 3x1 1 for all x1 0, including x1 0, then the fixed constant, 1, could be deleted from the objective function without changing the optimal solution and proportionality would be restored. However, this “fix” does not work here because the 1 constant does not apply when x1 0.

1

FIGURE 3.8 The solid curve violates the proportionality assumption because of the start-up cost that is incurred when x1 is increased from 0. The values at the dots are given by the Case 1 column of Table 3.4.

Contribution of x1 to Z 12

9 Satisfies proportionality assumption 6

Violates proportionality assumption

3

0 Start-up cost 3

1

2

3

4

x1

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

39

Contribution of x1 to Z 18

15 12 9 FIGURE 3.9 The solid curve violates the proportionality assumption because its slope (the marginal return from product 1) keeps increasing as x1 is increased. The values at the dots are given by the Case 2 column of Table 3.4.

Violates proportionality assumption Satisfies proportionality assumption

6 3

0

1

2

3

4

x1

Referring again to Table 3.4, the reverse of Case 2 is Case 3, where there is a decreasing marginal return. In this case, the slope of the profit function for product 1 (given by the solid curve in Fig. 3.10) keeps decreasing as x1 is increased. This violation of proportionality might occur because the marketing costs need to go up more than proportionally to attain increases in the level of sales. For example, it might be possible to sell product 1 at the rate of 1 per week (x1 1) with no advertising, whereas attaining sales to sustain a production rate of x1 2 might require a moderate amount of advertising, x1 3 might necessitate an extensive advertising campaign, and x1 4 might require also lowering the price. All three cases are hypothetical examples of ways in which the proportionality assumption could be violated. What is the actual situation? The actual profit from produc-

FIGURE 3.10 The solid curve violates the proportionality assumption because its slope (the marginal return from product 1) keeps decreasing as x1 is increased. The values at the dots are given by the Case 3 column in Table 3.4.

Contribution of x1 to Z 12 9

Satisfies proportionality assumption

6 Violates proportionality assumption

3

0

1

2

3

4

x1

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INTRODUCTION TO LINEAR PROGRAMMING

ing product 1 (or any other product) is derived from the sales revenue minus various direct and indirect costs. Inevitably, some of these cost components are not strictly proportional to the production rate, perhaps for one of the reasons illustrated above. However, the real question is whether, after all the components of profit have been accumulated, proportionality is a reasonable approximation for practical modeling purposes. For the Wyndor Glass Co. problem, the OR team checked both the objective function and the functional constraints. The conclusion was that proportionality could indeed be assumed without serious distortion. For other problems, what happens when the proportionality assumption does not hold even as a reasonable approximation? In most cases, this means you must use nonlinear programming instead (presented in Chap. 13). However, we do point out in Sec. 13.8 that a certain important kind of nonproportionality can still be handled by linear programming by reformulating the problem appropriately. Furthermore, if the assumption is violated only because of start-up costs, there is an extension of linear programming (mixed integer programming) that can be used, as discussed in Sec. 12.3 (the fixed-charge problem). Additivity Although the proportionality assumption rules out exponents other than 1, it does not prohibit cross-product terms (terms involving the product of two or more variables). The additivity assumption does rule out this latter possibility, as summarized below. Additivity assumption: Every function in a linear programming model (whether the objective function or the function on the left-hand side of a functional constraint) is the sum of the individual contributions of the respective activities. To make this definition more concrete and clarify why we need to worry about this assumption, let us look at some examples. Table 3.5 shows some possible cases for the objective function for the Wyndor Glass Co. problem. In each case, the individual contributions from the products are just as assumed in Sec. 3.1, namely, 3x1 for product 1 and 5x2 for product 2. The difference lies in the last row, which gives the function value for Z when the two products are produced jointly. The additivity satisfied column shows the case where this function value is obtained simply by adding the first two rows (3 5 8), so that Z 3x1 5x2 as previously assumed. By contrast, the next two columns show hypothetical cases where the additivity assumption would be violated (but not the proportionality assumption). TABLE 3.5 Examples of satisfying or violating additivity for the objective function Value of Z Additivity Violated (x1, x2)

Additivity Satisfied

Case 1

Case 2

(1, 0) (0, 1)

3 5

3 5

3 5

(1, 1)

8

9

7

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

41

Referring to the Case 1 column of Table 3.5, this case corresponds to an objective function of Z 3x1 5x2 x1x2, so that Z 3 5 1 9 for (x1, x2) (1, 1), thereby violating the additivity assumption that Z 3 5. (The proportionality assumption still is satisfied since after the value of one variable is fixed, the increment in Z from the other variable is proportional to the value of that variable.) This case would arise if the two products were complementary in some way that increases profit. For example, suppose that a major advertising campaign would be required to market either new product produced by itself, but that the same single campaign can effectively promote both products if the decision is made to produce both. Because a major cost is saved for the second product, their joint profit is somewhat more than the sum of their individual profits when each is produced by itself. Case 2 in Table 3.5 also violates the additivity assumption because of the extra term in the corresponding objective function, Z 3x1 5x2 x1x2, so that Z 3 5 1 7 for (x1, x2) (1, 1). As the reverse of the first case, Case 2 would arise if the two products were competitive in some way that decreased their joint profit. For example, suppose that both products need to use the same machinery and equipment. If either product were produced by itself, this machinery and equipment would be dedicated to this one use. However, producing both products would require switching the production processes back and forth, with substantial time and cost involved in temporarily shutting down the production of one product and setting up for the other. Because of this major extra cost, their joint profit is somewhat less than the sum of their individual profits when each is produced by itself. The same kinds of interaction between activities can affect the additivity of the constraint functions. For example, consider the third functional constraint of the Wyndor Glass Co. problem: 3x1 2x2 18. (This is the only constraint involving both products.) This constraint concerns the production capacity of Plant 3, where 18 hours of production time per week is available for the two new products, and the function on the left-hand side (3x1 2x2) represents the number of hours of production time per week that would be used by these products. The additivity satisfied column of Table 3.6 shows this case as is, whereas the next two columns display cases where the function has an extra crossproduct term that violates additivity. For all three columns, the individual contributions from the products toward using the capacity of Plant 3 are just as assumed previously, namely, 3x1 for product 1 and 2x2 for product 2, or 3(2) 6 for x1 2 and 2(3) 6 for

TABLE 3.6 Examples of satisfying or violating additivity for a functional constraint Amount of Resource Used Additivity Violated (x1, x2)

Additivity Satisfied

Case 3

Case 4

(2, 0) (0, 3)

6 6

6 6

6 6

(2, 3)

12

15

10.8

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INTRODUCTION TO LINEAR PROGRAMMING

x2 3. As was true for Table 3.5, the difference lies in the last row, which now gives the total function value for production time used when the two products are produced jointly. For Case 3 (see Table 3.6), the production time used by the two products is given by the function 3x1 2x2 0.5x1x2, so the total function value is 6 6 3 15 when (x1, x2) (2, 3), which violates the additivity assumption that the value is just 6 6 12. This case can arise in exactly the same way as described for Case 2 in Table 3.5; namely, extra time is wasted switching the production processes back and forth between the two products. The extra cross-product term (0.5x1x2) would give the production time wasted in this way. (Note that wasting time switching between products leads to a positive crossproduct term here, where the total function is measuring production time used, whereas it led to a negative cross-product term for Case 2 because the total function there measures profit.) For Case 4 in Table 3.6, the function for production time used is 3x1 2x2 0.1x 21x2, so the function value for (x1, x2) (2, 3) is 6 6 1.2 10.8. This case could arise in the following way. As in Case 3, suppose that the two products require the same type of machinery and equipment. But suppose now that the time required to switch from one product to the other would be relatively small. Because each product goes through a sequence of production operations, individual production facilities normally dedicated to that product would incur occasional idle periods. During these otherwise idle periods, these facilities can be used by the other product. Consequently, the total production time used (including idle periods) when the two products are produced jointly would be less than the sum of the production times used by the individual products when each is produced by itself. After analyzing the possible kinds of interaction between the two products illustrated by these four cases, the OR team concluded that none played a major role in the actual Wyndor Glass Co. problem. Therefore, the additivity assumption was adopted as a reasonable approximation. For other problems, if additivity is not a reasonable assumption, so that some of or all the mathematical functions of the model need to be nonlinear (because of the crossproduct terms), you definitely enter the realm of nonlinear programming (Chap. 13). Divisibility Our next assumption concerns the values allowed for the decision variables. Divisibility assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values. Since each decision variable represents the level of some activity, it is being assumed that the activities can be run at fractional levels. For the Wyndor Glass Co. problem, the decision variables represent production rates (the number of batches of a product produced per week). Since these production rates can have any fractional values within the feasible region, the divisibility assumption does hold. In certain situations, the divisibility assumption does not hold because some of or all the decision variables must be restricted to integer values. Mathematical models with this restriction are called integer programming models, and they are discussed in Chap. 12.

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

43

Certainty Our last assumption concerns the parameters of the model, namely, the coefficients in the objective function cj, the coefficients in the functional constraints aij, and the right-hand sides of the functional constraints bi. Certainty assumption: The value assigned to each parameter of a linear programming model is assumed to be a known constant. In real applications, the certainty assumption is seldom satisfied precisely. Linear programming models usually are formulated to select some future course of action. Therefore, the parameter values used would be based on a prediction of future conditions, which inevitably introduces some degree of uncertainty. For this reason it is usually important to conduct sensitivity analysis after a solution is found that is optimal under the assumed parameter values. As discussed in Sec. 2.3, one purpose is to identify the sensitive parameters (those whose value cannot be changed without changing the optimal solution), since any later change in the value of a sensitive parameter immediately signals a need to change the solution being used. Sensitivity analysis plays an important role in the analysis of the Wyndor Glass Co. problem, as you will see in Sec. 6.7. However, it is necessary to acquire some more background before we finish that story. Occasionally, the degree of uncertainty in the parameters is too great to be amenable to sensitivity analysis. In this case, it is necessary to treat the parameters explicitly as random variables. Formulations of this kind have been developed, as discussed in Secs. 23.6 and 23.7 on the book’s web site, www.mhhe.com/hillier. The Assumptions in Perspective We emphasized in Sec. 2.2 that a mathematical model is intended to be only an idealized representation of the real problem. Approximations and simplifying assumptions generally are required in order for the model to be tractable. Adding too much detail and precision can make the model too unwieldy for useful analysis of the problem. All that is really needed is that there be a reasonably high correlation between the prediction of the model and what would actually happen in the real problem. This advice certainly is applicable to linear programming. It is very common in real applications of linear programming that almost none of the four assumptions hold completely. Except perhaps for the divisibility assumption, minor disparities are to be expected. This is especially true for the certainty assumption, so sensitivity analysis normally is a must to compensate for the violation of this assumption. However, it is important for the OR team to examine the four assumptions for the problem under study and to analyze just how large the disparities are. If any of the assumptions are violated in a major way, then a number of useful alternative models are available, as presented in later chapters of the book. A disadvantage of these other models is that the algorithms available for solving them are not nearly as powerful as those for linear programming, but this gap has been closing in some cases. For some applications, the powerful linear programming approach is used for the initial analysis, and then a more complicated model is used to refine this analysis.

44

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As you work through the examples in the next section, you will find it good practice to analyze how well each of the four assumptions of linear programming applies.

3.4

ADDITIONAL EXAMPLES The Wyndor Glass Co. problem is a prototype example of linear programming in several respects: It involves allocating limited resources among competing activities, its model fits our standard form, and its context is the traditional one of improved business planning. However, the applicability of linear programming is much wider. In this section we begin broadening our horizons. As you study the following examples, note that it is their underlying mathematical model rather than their context that characterizes them as linear programming problems. Then give some thought to how the same mathematical model could arise in many other contexts by merely changing the names of the activities and so forth. These examples are scaled-down versions of actual applications (including two that are included in the case studies presented in the next section). Design of Radiation Therapy

FIGURE 3.11 Cross section of Mary’s tumor (viewed from above), nearby critical tissues, and the radiation beams being used. Beam 2 1 3

2

3

Beam 1 1. Bladder and tumor 2. Rectum, coccyx, etc. 3. Femur, part of pelvis, etc.

MARY has just been diagnosed as having a cancer at a fairly advanced stage. Specifically, she has a large malignant tumor in the bladder area (a “whole bladder lesion”). Mary is to receive the most advanced medical care available to give her every possible chance for survival. This care will include extensive radiation therapy. Radiation therapy involves using an external beam treatment machine to pass ionizing radiation through the patient’s body, damaging both cancerous and healthy tissues. Normally, several beams are precisely administered from different angles in a twodimensional plane. Due to attenuation, each beam delivers more radiation to the tissue near the entry point than to the tissue near the exit point. Scatter also causes some delivery of radiation to tissue outside the direct path of the beam. Because tumor cells are typically microscopically interspersed among healthy cells, the radiation dosage throughout the tumor region must be large enough to kill the malignant cells, which are slightly more radiosensitive, yet small enough to spare the healthy cells. At the same time, the aggregate dose to critical tissues must not exceed established tolerance levels, in order to prevent complications that can be more serious than the disease itself. For the same reason, the total dose to the entire healthy anatomy must be minimized. Because of the need to carefully balance all these factors, the design of radiation therapy is a very delicate process. The goal of the design is to select the combination of beams to be used, and the intensity of each one, to generate the best possible dose distribution. (The dose strength at any point in the body is measured in units called kilorads.) Once the treatment design has been developed, it is administered in many installments, spread over several weeks. In Mary’s case, the size and location of her tumor make the design of her treatment an even more delicate process than usual. Figure 3.11 shows a diagram of a cross section of the tumor viewed from above, as well as nearby critical tissues to avoid. These tissues include critical organs (e.g., the rectum) as well as bony structures (e.g., the femurs and pelvis) that will attenuate the radiation. Also shown are the entry point and direction for the only two beams that can be used with any modicum of safety in this case. (Actually,

3.4 ADDITIONAL EXAMPLES

45

we are simplifying the example at this point, because normally dozens of possible beams must be considered.) For any proposed beam of given intensity, the analysis of what the resulting radiation absorption by various parts of the body would be requires a complicated process. In brief, based on careful anatomical analysis, the energy distribution within the twodimensional cross section of the tissue can be plotted on an isodose map, where the contour lines represent the dose strength as a percentage of the dose strength at the entry point. A fine grid then is placed over the isodose map. By summing the radiation absorbed in the squares containing each type of tissue, the average dose that is absorbed by the tumor, healthy anatomy, and critical tissues can be calculated. With more than one beam (administered sequentially), the radiation absorption is additive. After thorough analysis of this type, the medical team has carefully estimated the data needed to design Mary’s treatment, as summarized in Table 3.7. The first column lists the areas of the body that must be considered, and then the next two columns give the fraction of the radiation dose at the entry point for each beam that is absorbed by the respective areas on average. For example, if the dose level at the entry point for beam 1 is 1 kilorad, then an average of 0.4 kilorad will be absorbed by the entire healthy anatomy in the two-dimensional plane, an average of 0.3 kilorad will be absorbed by nearby critical tissues, an average of 0.5 kilorad will be absorbed by the various parts of the tumor, and 0.6 kilorad will be absorbed by the center of the tumor. The last column gives the restrictions on the total dosage from both beams that is absorbed on average by the respective areas of the body. In particular, the average dosage absorption for the healthy anatomy must be as small as possible, the critical tissues must not exceed 2.7 kilorads, the average over the entire tumor must equal 6 kilorads, and the center of the tumor must be at least 6 kilorads. Formulation as a Linear Programming Problem. The two decision variables x1 and x2 represent the dose (in kilorads) at the entry point for beam 1 and beam 2, respectively. Because the total dosage reaching the healthy anatomy is to be minimized, let Z denote this quantity. The data from Table 3.7 can then be used directly to formulate the following linear programming model.1 1

Actually, Table 3.7 simplifies the real situation, so the real model would be somewhat more complicated than this one and would have dozens of variables and constraints. For details about the general situation, see D. Sonderman and P. G. Abrahamson, “Radiotherapy Treatment Design Using Mathematical Programming Models,” Operations Research, 33:705–725, 1985, and its ref. 1.

TABLE 3.7 Data for the design of Mary’s radiation therapy Fraction of Entry Dose Absorbed by Area (Average) Area Healthy anatomy Critical tissues Tumor region Center of tumor

Beam 1

Beam 2

Restriction on Total Average Dosage, Kilorads

0.4 0.3 0.5 0.6

0.5 0.1 0.5 0.4

Minimize 2.7 6 6

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INTRODUCTION TO LINEAR PROGRAMMING

Minimize

Z 0.4x1 0.5x2,

subject to 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6 and x1 0,

x2 0.

Notice the differences between this model and the one in Sec. 3.1 for the Wyndor Glass Co. problem. The latter model involved maximizing Z, and all the functional constraints were in form. This new model does not fit this same standard form, but it does incorporate three other legitimate forms described in Sec. 3.2, namely, minimizing Z, functional constraints in form, and functional constraints in form. However, both models have only two variables, so this new problem also can be solved by the graphical method illustrated in Sec. 3.1. Figure 3.12 shows the graphical solution. The feasible region consists of just the dark line segment between (6, 6) and (7.5, 4.5), because the points on this segment are the only ones that simultaneously satisfy all the constraints. (Note that the equality constraint limits the feasible region to the line containing this line segment, and then the other two functional constraints determine the two endpoints of the line segment.) The dashed line is the objective function line that passes through the optimal solution (x1, x2) (7.5, 4.5) with Z 5.25. This solution is optimal rather than the point (6, 6) because decreasing Z (for positive values of Z) pushes the objective function line toward the origin (where Z 0). And Z 5.25 for (7.5, 4.5) is less than Z 5.4 for (6, 6). Thus, the optimal design is to use a total dose at the entry point of 7.5 kilorads for beam 1 and 4.5 kilorads for beam 2. Regional Planning The SOUTHERN CONFEDERATION OF KIBBUTZIM is a group of three kibbutzim (communal farming communities) in Israel. Overall planning for this group is done in its Coordinating Technical Office. This office currently is planning agricultural production for the coming year. The agricultural output of each kibbutz is limited by both the amount of available irrigable land and the quantity of water allocated for irrigation by the Water Commissioner (a national government official). These data are given in Table 3.8.

TABLE 3.8 Resource data for the Southern Confederation of Kibbutzim Kibbutz

Usable Land (Acres)

Water Allocation (Acre Feet)

1 2 3

400 600 300

600 800 375

3.4 ADDITIONAL EXAMPLES

47

x2 15

0.6x1 0.4x2 6

10

(6, 6)

5 (7.5, 4.5) Z 5.25 0.4x1 0.5x2

0.3x1 0.1x2 2.7 FIGURE 3.12 Graphical solution for the design of Mary’s radiation therapy.

0.5x1 0.5x2 6 0

5

10

x1

The crops suited for this region include sugar beets, cotton, and sorghum, and these are the three being considered for the upcoming season. These crops differ primarily in their expected net return per acre and their consumption of water. In addition, the Ministry of Agriculture has set a maximum quota for the total acreage that can be devoted to each of these crops by the Southern Confederation of Kibbutzim, as shown in Table 3.9. TABLE 3.9 Crop data for the Southern Confederation of Kibbutzim Crop Sugar beets Cotton Sorghum

Maximum Quota (Acres)

Water Consumption (Acre Feet/Acre)

Net Return ($/Acre)

600 500 325

3 2 1

1,000 750 250

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Because of the limited water available for irrigation, the Southern Confederation of Kibbutzim will not be able to use all its irrigable land for planting crops in the upcoming season. To ensure equity between the three kibbutzim, it has been agreed that every kibbutz will plant the same proportion of its available irrigable land. For example, if kibbutz 1 plants 200 of its available 400 acres, then kibbutz 2 must plant 300 of its 600 acres, while kibbutz 3 plants 150 acres of its 300 acres. However, any combination of the crops may be grown at any of the kibbutzim. The job facing the Coordinating Technical Office is to plan how many acres to devote to each crop at the respective kibbutzim while satisfying the given restrictions. The objective is to maximize the total net return to the Southern Confederation of Kibbutzim as a whole. Formulation as a Linear Programming Problem. The quantities to be decided upon are the number of acres to devote to each of the three crops at each of the three kibbutzim. The decision variables xj ( j 1, 2, . . . , 9) represent these nine quantities, as shown in Table 3.10. Since the measure of effectiveness Z is the total net return, the resulting linear programming model for this problem is Maximize

Z 1,000(x1 x2 x3) 750(x4 x5 x6) 250(x7 x8 x9),

subject to the following constraints: 1. Usable land for each kibbutz: x1 x4 x7 400 x2 x5 x8 600 x3 x6 x9 300 2. Water allocation for each kibbutz: 3x1 2x4 x7 600 3x2 2x5 x8 800 3x3 2x6 x9 375 3. Total acreage for each crop: x1 x2 x3 600 x4 x5 x6 500 x7 x8 x9 325 TABLE 3.10 Decision variables for the Southern Confederation of Kibbutzim problem Allocation (Acres) Kibbutz Crop

1

2

3

Sugar beets Cotton Sorghum

x1 x4 x7

x2 x5 x8

x3 x6 x9

3.4 ADDITIONAL EXAMPLES

49

4. Equal proportion of land planted: x1 x4 x7 x2 x5 x8 400 600 x2 x5 x8 x3 x6 x9 600 300 x3 x6 x9 x1 x4 x7 300 400 5. Nonnegativity: xj 0,

for j 1, 2, . . . , 9.

This completes the model, except that the equality constraints are not yet in an appropriate form for a linear programming model because some of the variables are on the righthand side. Hence, their final form1 is 3(x1 x4 x7) 2(x2 x5 x8) 0 (x2 x5 x8) 2(x3 x6 x9) 0 4(x3 x6 x9) 3(x1 x4 x7) 0 The Coordinating Technical Office formulated this model and then applied the simplex method (developed in the next chapter) to find an optimal solution

1 (x1, x2, x3, x4, x5, x6, x7, x8, x9) 133, 100, 25, 100, 250, 150, 0, 0, 0 , 3 as shown in Table 3.11. The resulting optimal value of the objective function is Z 633,33313, that is, a total net return of $633,333.33.

1

Actually, any one of these equations is redundant and can be deleted if desired. Also, because of these equations, any two of the usable land constraints also could be deleted because they automatically would be satisfied when both the remaining usable land constraint and these equations are satisfied. However, no harm is done (except a little more computational effort) by including unnecessary constraints, so you don’t need to worry about identifying and deleting them in models you formulate.

TABLE 3.11 Optimal solution for the Southern Confederation of Kibbutzim problem Best Allocation (Acres) Kibbutz Crop Sugar beets Cotton Sorghum

1

2

3

13313 100 0

100 250 0

25 150 0

50

3

INTRODUCTION TO LINEAR PROGRAMMING

Controlling Air Pollution The NORI & LEETS CO., one of the major producers of steel in its part of the world, is located in the city of Steeltown and is the only large employer there. Steeltown has grown and prospered along with the company, which now employs nearly 50,000 residents. Therefore, the attitude of the townspeople always has been, “What’s good for Nori & Leets is good for the town.” However, this attitude is now changing; uncontrolled air pollution from the company’s furnaces is ruining the appearance of the city and endangering the health of its residents. A recent stockholders’ revolt resulted in the election of a new enlightened board of directors for the company. These directors are determined to follow socially responsible policies, and they have been discussing with Steeltown city officials and citizens’ groups what to do about the air pollution problem. Together they have worked out stringent air quality standards for the Steeltown airshed. The three main types of pollutants in this airshed are particulate matter, sulfur oxides, and hydrocarbons. The new standards require that the company reduce its annual emission of these pollutants by the amounts shown in Table 3.12. The board of directors has instructed management to have the engineering staff determine how to achieve these reductions in the most economical way. The steelworks has two primary sources of pollution, namely, the blast furnaces for making pig iron and the open-hearth furnaces for changing iron into steel. In both cases the engineers have decided that the most effective types of abatement methods are (1) increasing the height of the smokestacks,1 (2) using filter devices (including gas traps) in the smokestacks, and (3) including cleaner, high-grade materials among the fuels for the furnaces. Each of these methods has a technological limit on how heavily it can be used (e.g., a maximum feasible increase in the height of the smokestacks), but there also is considerable flexibility for using the method at a fraction of its technological limit. Table 3.13 shows how much emission (in millions of pounds per year) can be eliminated from each type of furnace by fully using any abatement method to its technological limit. For purposes of analysis, it is assumed that each method also can be used less fully to achieve any fraction of the emission-rate reductions shown in this table. Furthermore, the fractions can be different for blast furnaces and for open-hearth furnaces. For either type of furnace, the emission reduction achieved by each method is not substantially affected by whether the other methods also are used. 1

Subsequent to this study, this particular abatement method has become a controversial one. Because its effect is to reduce ground-level pollution by spreading emissions over a greater distance, environmental groups contend that this creates more acid rain by keeping sulfur oxides in the air longer. Consequently, the U.S. Environmental Protection Agency adopted new rules in 1985 to remove incentives for using tall smokestacks.

TABLE 3.12 Clean air standards for the Nori & Leets Co. Pollutant Particulates Sulfur oxides Hydrocarbons

Required Reduction in Annual Emission Rate (Million Pounds) 60 150 125

3.4 ADDITIONAL EXAMPLES

51

TABLE 3.13 Reduction in emission rate (in millions of pounds per year) from the maximum feasible use of an abatement method for Nori & Leets Co. Taller Smokestacks

Pollutant

Filters

Better Fuels

Blast Open-Hearth Blast Open-Hearth Blast Open-Hearth Furnaces Furnaces Furnaces Furnaces Furnaces Furnaces

Particulates Sulfur oxides Hydrocarbons

12 35 37

9 42 53

25 18 28

20 31 24

17 56 29

13 49 20

After these data were developed, it became clear that no single method by itself could achieve all the required reductions. On the other hand, combining all three methods at full capacity on both types of furnaces (which would be prohibitively expensive if the company’s products are to remain competitively priced) is much more than adequate. Therefore, the engineers concluded that they would have to use some combination of the methods, perhaps with fractional capacities, based upon the relative costs. Furthermore, because of the differences between the blast and the open-hearth furnaces, the two types probably should not use the same combination. An analysis was conducted to estimate the total annual cost that would be incurred by each abatement method. A method’s annual cost includes increased operating and maintenance expenses as well as reduced revenue due to any loss in the efficiency of the production process caused by using the method. The other major cost is the start-up cost (the initial capital outlay) required to install the method. To make this one-time cost commensurable with the ongoing annual costs, the time value of money was used to calculate the annual expenditure (over the expected life of the method) that would be equivalent in value to this start-up cost. This analysis led to the total annual cost estimates (in millions of dollars) given in Table 3.14 for using the methods at their full abatement capacities. It also was determined that the cost of a method being used at a lower level is roughly proportional to the fraction of the abatement capacity given in Table 3.13 that is achieved. Thus, for any given fraction achieved, the total annual cost would be roughly that fraction of the corresponding quantity in Table 3.14. The stage now was set to develop the general framework of the company’s plan for pollution abatement. This plan specifies which types of abatement methods will be used and at what fractions of their abatement capacities for (1) the blast furnaces and (2) the open-hearth furnaces. Because of the combinatorial nature of the problem of finding a TABLE 3.14 Total annual cost from the maximum feasible use of an abatement method for Nori & Leets Co. ($ millions) Abatement Method Taller smokestacks Filters Better fuels

Blast Furnaces

Open-Hearth Furnaces

8 7 11

10 6 9

52

3

INTRODUCTION TO LINEAR PROGRAMMING

plan that satisfies the requirements with the smallest possible cost, an OR team was formed to solve the problem. The team adopted a linear programming approach, formulating the model summarized next. Formulation as a Linear Programming Problem. This problem has six decision variables xj, j 1, 2, . . . , 6, each representing the use of one of the three abatement methods for one of the two types of furnaces, expressed as a fraction of the abatement capacity (so xj cannot exceed 1). The ordering of these variables is shown in Table 3.15. Because the objective is to minimize total cost while satisfying the emission reduction requirements, the data in Tables 3.12, 3.13, and 3.14 yield the following model: Minimize

Z 8x1 10x2 7x3 6x4 11x5 9x6,

subject to the following constraints: 1. Emission reduction: 12x1 9x2 25x3 20x4 17x5 13x6 60 35x1 42x2 18x3 31x4 56x5 49x6 150 37x1 53x2 28x3 24x4 29x5 20x6 125 2. Technological limit: xj 1,

for j 1, 2, . . . , 6

xj 0,

for j 1, 2, . . . , 6.

3. Nonnegativity: The OR team used this model1 to find a minimum-cost plan (x1, x2, x3, x4, x5, x6) (1, 0.623, 0.343, 1, 0.048, 1), with Z 32.16 (total annual cost of $32.16 million). Sensitivity analysis then was conducted to explore the effect of making possible adjustments in the air standards given in Table 3.12, as well as to check on the effect of any inaccuracies in the cost data given in Table 3.14. (This story is continued in Case 6.1 at the end of Chap. 6.) Next came detailed planning and managerial review. Soon after, this program for controlling air pollution was fully implemented by the company, and the citizens of Steeltown breathed deep (cleaner) sighs of relief. 1

An equivalent formulation can express each decision variable in natural units for its abatement method; for example, x1 and x2 could represent the number of feet that the heights of the smokestacks are increased.

TABLE 3.15 Decision variables (fraction of the maximum feasible use of an abatement method) for Nori & Leets Co. Abatement Method Taller smokestacks Filters Better fuels

Blast Furnaces

Open-Hearth Furnaces

x1 x3 x5

x2 x4 x6

3.4 ADDITIONAL EXAMPLES

53

Reclaiming Solid Wastes The SAVE-IT COMPANY operates a reclamation center that collects four types of solid waste materials and treats them so that they can be amalgamated into a salable product. (Treating and amalgamating are separate processes.) Three different grades of this product can be made (see the first column of Table 3.16), depending upon the mix of the materials used. Although there is some flexibility in the mix for each grade, quality standards may specify the minimum or maximum amount allowed for the proportion of a material in the product grade. (This proportion is the weight of the material expressed as a percentage of the total weight for the product grade.) For each of the two higher grades, a fixed percentage is specified for one of the materials. These specifications are given in Table 3.16 along with the cost of amalgamation and the selling price for each grade. The reclamation center collects its solid waste materials from regular sources and so is normally able to maintain a steady rate for treating them. Table 3.17 gives the quantities available for collection and treatment each week, as well as the cost of treatment, for each type of material. The Save-It Co. is solely owned by Green Earth, an organization devoted to dealing with environmental issues, so Save-It’s profits are used to help support Green Earth’s activities. Green Earth has raised contributions and grants, amounting to $30,000 per week, to be used exclusively to cover the entire treatment cost for the solid waste materials. The board of directors of Green Earth has instructed the management of Save-It to divide this money among the materials in such a way that at least half of the amount available of each material is actually collected and treated. These additional restrictions are listed in Table 3.17. Within the restrictions specified in Tables 3.16 and 3.17, management wants to determine the amount of each product grade to produce and the exact mix of materials to be used for each grade. The objective is to maximize the net weekly profit (total sales income minus total amalgamation cost), exclusive of the fixed treatment cost of $30,000 per week that is being covered by gifts and grants. Formulation as a Linear Programming Problem. Before attempting to construct a linear programming model, we must give careful consideration to the proper definition of the decision variables. Although this definition is often obvious, it sometimes becomes TABLE 3.16 Product data for Save-It Co. Grade

Specification 1: 2: 3: 4:

Amalgamation Cost per Pound ($)

Selling Price per Pound ($)

A

Material Material Material Material

Not more than 30% of total Not less than 40% of total Not more than 50% of total Exactly 20% of total

3.00

8.50

B

Material 1: Not more than 50% of total Material 2: Not less than 10% of total Material 4: Exactly 10% of total

2.50

7.00

C

Material 1: Not more than 70% of total

2.00

5.50

54

3

INTRODUCTION TO LINEAR PROGRAMMING

TABLE 3.17 Solid waste materials data for the Save-It Co. Material

Pounds per Week Available

Treatment Cost per Pound ($)

1 2 3 4

3,000 2,000 4,000 1,000

3.00 6.00 4.00 5.00

Additional Restrictions 1. For each material, at least half of the pounds per week available should be collected and treated. 2. $30,000 per week should be used to treat these materials.

the crux of the entire formulation. After clearly identifying what information is really desired and the most convenient form for conveying this information by means of decision variables, we can develop the objective function and the constraints on the values of these decision variables. In this particular problem, the decisions to be made are well defined, but the appropriate means of conveying this information may require some thought. (Try it and see if you first obtain the following inappropriate choice of decision variables.) Because one set of decisions is the amount of each product grade to produce, it would seem natural to define one set of decision variables accordingly. Proceeding tentatively along this line, we define yi number of pounds of product grade i produced per week

(i A, B, C).

The other set of decisions is the mix of materials for each product grade. This mix is identified by the proportion of each material in the product grade, which would suggest defining the other set of decision variables as zij proportion of material j in product grade i

(i A, B, C; j 1, 2, 3, 4).

However, Table 3.17 gives both the treatment cost and the availability of the materials by quantity (pounds) rather than proportion, so it is this quantity information that needs to be recorded in some of the constraints. For material j ( j 1, 2, 3, 4), Number of pounds of material j used per week zAj yA zBj yB zCj yC. For example, since Table 3.17 indicates that 3,000 pounds of material 1 is available per week, one constraint in the model would be zA1 yA zB1 yB zC1 yC 3,000. Unfortunately, this is not a legitimate linear programming constraint. The expression on the left-hand side is not a linear function because it involves products of variables. Therefore, a linear programming model cannot be constructed with these decision variables. Fortunately, there is another way of defining the decision variables that will fit the linear programming format. (Do you see how to do it?) It is accomplished by merely replacing each product of the old decision variables by a single variable! In other words, define (for i A, B, C; j 1, 2, 3, 4) xij zij yi xij number of pounds of material j allocated to product grade i per week,

3.4 ADDITIONAL EXAMPLES

55

and then we let the xij be the decision variables. Combining the xij in different ways yields the following quantities needed in the model (for i A, B, C; j 1, 2, 3, 4). xi1 xi2 xi3 xi4 number of pounds of product grade i produced per week. xAj xBj xCj number of pounds of material j used per week. xij proportion of material j in product grade i. xi1 xi2 xi3 xi4 The fact that this last expression is a nonlinear function does not cause a complication. For example, consider the first specification for product grade A in Table 3.16 (the proportion of material 1 should not exceed 30 percent). This restriction gives the nonlinear constraint xA1 0.3. xA1 xA2 xA3 xA4 However, multiplying through both sides of this inequality by the denominator yields an equivalent constraint xA1 0.3(xA1 xA2 xA3 xA4), so 0.7xA1 0.3xA2 0.3xA3 0.3xA4 0, which is a legitimate linear programming constraint. With this adjustment, the three quantities given above lead directly to all the functional constraints of the model. The objective function is based on management’s objective of maximizing net weekly profit (total sales income minus total amalgamation cost) from the three product grades. Thus, for each product grade, the profit per pound is obtained by subtracting the amalgamation cost given in the third column of Table 3.16 from the selling price in the fourth column. These differences provide the coefficients for the objective function. Therefore, the complete linear programming model is Maximize

Z 5.5(xA1 xA2 xA3 xA4) 4.5(xB1 xB2 xB3 xB4) 3.5(xC1 xC2 xC3 xC4),

subject to the following constraints: 1. Mixture specifications (second column of Table 3.16): xA1 0.3(xA1 xA2 xA3 xA4) xA2 0.4(xA1 xA2 xA3 xA4) xA3 0.5(xA1 xA2 xA3 xA4) xA4 0.2(xA1 xA2 xA3 xA4)

(grade (grade (grade (grade

xB1 0.5(xB1 xB2 xB3 xB4) xB2 0.1(xB1 xB2 xB3 xB4) xB4 0.1(xB1 xB2 xB3 xB4)

(grade B, material 1) (grade B, material 2) (grade B, material 4).

xC1 0.7(xC1 xC2 xC3 xC4)

(grade C, material 1).

A, A, A, A,

material material material material

1) 2) 3) 4).

56

3

INTRODUCTION TO LINEAR PROGRAMMING

2. Availability of materials (second column of Table 3.17): xA1 xB1 xC1 xA2 xB2 xC2 xA3 xB3 xC3 xA4 xB4 xC4

3,000 2,000 4,000 1,000

(material (material (material (material

1) 2) 3) 4).

3. Restrictions on amounts treated (right side of Table 3.17): xA1 xB1 xC1 1,500 xA2 xB2 xC2 1,000 xA3 xB3 xC3 2,000 xA4 xB4 xC4 500

(material (material (material (material

1) 2) 3) 4).

4. Restriction on treatment cost (right side of Table 3.17): 3(xA1 xB1 xC1) 6(xA2 xB2 xC2) 4(xA3 xB3 xC3) 5(xA4 xB4 xC4) 30,000. 5. Nonnegativity constraints: xA1 0,

xA2 0,

...,

xC4 0.

This formulation completes the model, except that the constraints for the mixture specifications need to be rewritten in the proper form for a linear programming model by bringing all variables to the left-hand side and combining terms, as follows: Mixture specifications: 0.7xA1 0.3xA2 0.3xA3 0.3xA4 0 0.4xA1 0.6xA2 0.4xA3 0.4xA4 0 0.5xA1 0.5xA2 0.5xA3 0.5xA4 0 0.2xA1 0.2xA2 0.2xA3 0.8xA4 0

(grade A, material 1) (grade A, material 2) (grade A, material 3)

0.5xB1 0.5xB2 0.5xB3 0.5xB4 0 0.1xB1 0.9xB2 0.1xB3 0.1xB4 0 0.1xB1 0.1xB2 0.1xB3 0.9xB4 0

(grade B, material 1) (grade B, material 2) (grade B, material 4).

0.3xC1 0.7xC2 0.7xC3 0.7xC4 0

(grade C, material 1).

(grade A, material 4).

An optimal solution for this model is shown in Table 3.18, and then these xij values are used to calculate the other quantities of interest given in the table. The resulting optimal value of the objective function is Z 35,108.90 (a total weekly profit of $35,108.90). The Save-It Co. problem is an example of a blending problem. The objective for a blending problem is to find the best blend of ingredients into final products to meet certain specifications. Some of the earliest applications of linear programming were for gasoline blending, where petroleum ingredients were blended to obtain various grades of gasoline. The award-winning OR study at Texaco discussed at the end of

3.4 ADDITIONAL EXAMPLES

57

TABLE 3.18 Optimal solution for the Save-It Co. problem Pounds Used per Week Material Grade

1

A

412.3 (19.2%) 2587.7 (50%) 0

B C Total

2 859.6 (40%) 517.5 (10%) 0

3000

1377

Number of Pounds Produced per Week

3

4

447.4 (20.8%) 1552.6 (30%) 0

429.8 (20%) 517.5 (10%) 0

2000

947

2149 5175 0

Sec. 2.5 dealt with gasoline blending (although Texaco used a nonlinear programming model). Other blending problems involve such final products as steel, fertilizer, and animal feed. Personnel Scheduling UNION AIRWAYS is adding more flights to and from its hub airport, and so it needs to hire additional customer service agents. However, it is not clear just how many more should be hired. Management recognizes the need for cost control while also consistently providing a satisfactory level of service to customers. Therefore, an OR team is studying how to schedule the agents to provide satisfactory service with the smallest personnel cost. Based on the new schedule of flights, an analysis has been made of the minimum number of customer service agents that need to be on duty at different times of the day to provide a satisfactory level of service. The rightmost column of Table 3.19 shows the number of agents needed for the time periods given in the first column. The other entries TABLE 3.19 Data for the Union Airways personnel scheduling problem Time Periods Covered Shift Time Period

1

6:00 A.M. to 8:00 A.M. 8:00 A.M. to 10:00 A.M. 10:00 A.M. to noon Noon to 2:00 P.M. 2:00 P.M. to 4:00 P.M. 4:00 P.M. to 6:00 P.M. 6:00 P.M. to 8:00 P.M. 8:00 P.M. to 10:00 P.M. 10:00 P.M. to midnight Midnight to 6:00 A.M.

✔ ✔ ✔ ✔

Daily cost per agent

$170

2 ✔ ✔ ✔ ✔

$160

3

✔ ✔ ✔ ✔

$175

4

✔ ✔ ✔ ✔ $180

5

Minimum Number of Agents Needed

✔ ✔

48 79 65 87 64 73 82 43 52 15

$195

58

3

INTRODUCTION TO LINEAR PROGRAMMING

in this table reflect one of the provisions in the company’s current contract with the union that represents the customer service agents. The provision is that each agent work an 8-hour shift 5 days per week, and the authorized shifts are Shift Shift Shift Shift Shift

1: 2: 3: 4: 5:

6:00 A.M. to 2:00 P.M. 8:00 A.M. to 4:00 P.M. Noon to 8:00 P.M. 4:00 P.M. to midnight 10:00 P.M. to 6:00 A.M.

Checkmarks in the main body of Table 3.19 show the hours covered by the respective shifts. Because some shifts are less desirable than others, the wages specified in the contract differ by shift. For each shift, the daily compensation (including benefits) for each agent is shown in the bottom row. The problem is to determine how many agents should be assigned to the respective shifts each day to minimize the total personnel cost for agents, based on this bottom row, while meeting (or surpassing) the service requirements given in the rightmost column. Formulation as a Linear Programming Problem. Linear programming problems always involve finding the best mix of activity levels. The key to formulating this particular problem is to recognize the nature of the activities. Activities correspond to shifts, where the level of each activity is the number of agents assigned to that shift. Thus, this problem involves finding the best mix of shift sizes. Since the decision variables always are the levels of the activities, the five decision variables here are xj number of agents assigned to shift j,

for j 1, 2, 3, 4, 5.

The main restrictions on the values of these decision variables are that the number of agents working during each time period must satisfy the minimum requirement given in the rightmost column of Table 3.19. For example, for 2:00 P.M. to 4:00 P.M., the total number of agents assigned to the shifts that cover this time period (shifts 2 and 3) must be at least 64, so x2 x3 64 is the functional constraint for this time period. Because the objective is to minimize the total cost of the agents assigned to the five shifts, the coefficients in the objective function are given by the last row of Table 3.19. Therefore, the complete linear programming model is Minimize

Z 170x1 160x2 175x3 180x4 195x5,

subject to x1 x1 x2 x1 x2 x1 x2 x3 x2 x3 x3 x4

48 79 65 87 64 73

(6–8 A.M.) (8–10 A.M.) (10 A.M. to noon) (Noon–2 P.M.) (2–4 P.M.) (4–6 P.M.)

3.4 ADDITIONAL EXAMPLES

59

x3 x4 82 x4 43 x4 x5 52 x5 15

(6–8 P.M.) (8–10 P.M.) (10 P.M.–midnight) (Midnight–6 A.M.)

and xj 0,

for j 1, 2, 3, 4, 5.

With a keen eye, you might have noticed that the third constraint, x1 x2 65, actually is not necessary because the second constraint, x1 x2 79, ensures that x1 x2 will be larger than 65. Thus, x1 x2 65 is a redundant constraint that can be deleted. Similarly, the sixth constraint, x3 x4 73, also is a redundant constraint because the seventh constraint is x3 x4 82. (In fact, three of the nonnegativity constraints—x1 0, x4 0, x5 0—also are redundant constraints because of the first, eighth, and tenth functional constraints: x1 48, x4 43, and x5 15. However, no computational advantage is gained by deleting these three nonnegativity constraints.) The optimal solution for this model is (x1, x2, x3, x4, x5) (48, 31, 39, 43, 15). This yields Z 30,610, that is, a total daily personnel cost of $30,610. This problem is an example where the divisibility assumption of linear programming actually is not satisfied. The number of agents assigned to each shift needs to be an integer. Strictly speaking, the model should have an additional constraint for each decision variable specifying that the variable must have an integer value. Adding these constraints would convert the linear programming model to an integer programming model (the topic of Chap. 12). Without these constraints, the optimal solution given above turned out to have integer values anyway, so no harm was done by not including the constraints. (The form of the functional constraints made this outcome a likely one.) If some of the variables had turned out to be noninteger, the easiest approach would have been to round up to integer values. (Rounding up is feasible for this example because all the functional constraints are in form with nonnegative coefficients.) Rounding up does not ensure obtaining an optimal solution for the integer programming model, but the error introduced by rounding up such large numbers would be negligible for most practical situations. Alternatively, integer programming techniques described in Chap. 12 could be used to solve exactly for an optimal solution with integer values. Section 3.5 includes a case study of how United Airlines used linear programming to develop a personnel scheduling system on a vastly larger scale than this example. Distributing Goods through a Distribution Network The Problem. The DISTRIBUTION UNLIMITED CO. will be producing the same new product at two different factories, and then the product must be shipped to two warehouses, where either factory can supply either warehouse. The distribution network available for shipping this product is shown in Fig. 3.13, where F1 and F2 are the two factories, W1 and W2 are the two warehouses, and DC is a distribution center. The amounts to be shipped from F1 and F2 are shown to their left, and the amounts to be received at W1 and W2 are shown to their right. Each arrow represents a feasible shipping lane. Thus, F1 can ship directly to W1 and has three possible routes (F1 DC W2, F1 F2

60

3

INTRODUCTION TO LINEAR PROGRAMMING

DC W2, and F1 W1 W2) for shipping to W2. Factory F2 has just one route to W2 (F2 DC W2) and one to W1 (F2 DC W2 W1). The cost per unit shipped through each shipping lane is shown next to the arrow. Also shown next to F1 F2 and DC W2 are the maximum amounts that can be shipped through these lanes. The other lanes have sufficient shipping capacity to handle everything these factories can send. The decision to be made concerns how much to ship through each shipping lane. The objective is to minimize the total shipping cost. Formulation as a Linear Programming Problem. With seven shipping lanes, we need seven decision variables (xF1-F2, xF1-DC, xF1-W1, xF2-DC, xDC-W2, xW1-W2, xW2-W1) to represent the amounts shipped through the respective lanes. There are several restrictions on the values of these variables. In addition to the usual nonnegativity constraints, there are two upper-bound constraints, xF1-F2 10 and xDC-W2 80, imposed by the limited shipping capacities for the two lanes, F1 F2 and DC W2. All the other restrictions arise from five net flow constraints, one for each of the five locations. These constraints have the following form. Net flow constraint for each location: Amount shipped out amount shipped in required amount. As indicated in Fig. 3.13, these required amounts are 50 for F1, 40 for F2, 30 for W1, and 60 for W2.

50 units produced

$900/unit

F1

W1

30 units needed

$4 00 /u ni t

DC

ni

00 /u t

its

ni

un

/u

$300/unit

$1

00

$200/unit

80

t

$200/unit 10 units max.

m

$3

FIGURE 3.13 The distribution network for Distribution Unlimited Co.

ax .

40 units produced

F2

W2

60 units needed

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61

What is the required amount for DC? All the units produced at the factories are ultimately needed at the warehouses, so any units shipped from the factories to the distribution center should be forwarded to the warehouses. Therefore, the total amount shipped from the distribution center to the warehouses should equal the total amount shipped from the factories to the distribution center. In other words, the difference of these two shipping amounts (the required amount for the net flow constraint) should be zero. Since the objective is to minimize the total shipping cost, the coefficients for the objective function come directly from the unit shipping costs given in Fig. 3.13. Therefore, by using money units of hundreds of dollars in this objective function, the complete linear programming model is Z 2xF1-F2 4xF1-DC 9xF1-W1 3xF2-DC xDC-W2 3xW1-W2 2xW2-W1,

Minimize

subject to the following constraints: 1. Net flow constraints: xF1-F2 xF1-DC xF1-W1 xF1-F2 xF2-DC xF1-DC xF2-DC xDC-W2 xF1-W1

xW1-W2 xDC-W2 xW1-W2

50 (factory 1) 40 (factory 2) 0 (distribution center) xW2-W1 30 (warehouse 1) xW2-W1 60 (warehouse 2)

2. Upper-bound constraints: xF1-F2 10,

xDC-W2 80

3. Nonnegativity constraints: xF1-F2 0,

xF1-DC 0, xF1-W1 0, xF2-DC 0, xW1-W2 0, xW2-W1 0.

xDC-W2 0,

You will see this problem again in Sec. 9.6, where we focus on linear programming problems of this type (called the minimum cost flow problem). In Sec. 9.7, we will solve for its optimal solution: xF1-F2 0, xW1-W2 0,

xF1-DC 40, xW2-W1 20.

xF1-W1 10,

xF2-DC 40,

xDC-W2 80,

The resulting total shipping cost is $49,000. You also will see a case study involving a much larger problem of this same type at the end of the next section.

3.5

SOME CASE STUDIES To give you a better perspective about the great impact linear programming can have, we now present three case studies of real applications. Each of these is a classic application, initiated in the early 1980s, that has come to be regarded as a standard of excellence for future applications of linear programming. The first one will bear some strong similarities to the Wyndor Glass Co. problem, but on a realistic scale. Similarly, the second and

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third are realistic versions of the last two examples presented in the preceding section (the Union Airways and Distribution Unlimited examples). Choosing the Product Mix at Ponderosa Industrial1 Until its sale in 1988, PONDEROSA INDUSTRIAL was a plywood manufacturer based in Anhuac, Chihuahua, that supplied 25 percent of the plywood in Mexico. Like any plywood manufacturer, Ponderosa’s many products were differentiated by thickness and by the quality of the wood. The plywood market in Mexico is competitive, so the market establishes the prices of the products. The prices can fluctuate considerably from month to month, and there may be great differences between the products in their price movements from even one month to the next. As a result, each product’s contribution to Ponderosa’s total profit was continually varying, and in different ways for different products. Because of its pronounced effect on profits, a critical issue facing management was the choice of product mix—how much to produce of each product—on a monthly basis. This choice was a very complex one, since it had to take into account the current amounts available of various resources needed to produce the products. The most important resources were logs in four quality categories and production capacities for both the pressing operation and the polishing operation. Beginning in 1980, linear programming was used on a monthly basis to guide the product-mix decision. The linear programming model had an objective of maximizing the total profit from all products. The model’s constraints included the various resource constraints as well as other relevant restrictions such as the minimum amount of a product that must be provided to regular customers and the maximum amount that can be sold. (To aid planning for the procurement of raw materials, the model also considered the impact of the product-mix decision for the upcoming month on production in the following month.) The model had 90 decision variables and 45 functional constraints. This model was used each month to find the product mix for the upcoming month that would be optimal if the estimated values of the various parameters of the model prove to be accurate. However, since some of the parameter values could change quickly (e.g., the unit profits of the products), sensitivity analysis was done to determine the effect if the estimated values turned out to be inaccurate. The results indicated when adjustments in the product mix should be made (if time permitted) as unanticipated market changes occurred that affected the price (and so the unit profit) of certain products. One key decision each month concerned the number of logs in each of the four quality categories to purchase. The amounts available for the upcoming month’s production actually were parameters of the model. Therefore, after the purchase decision was made and then the corresponding optimal product mix was determined, postoptimality analysis was conducted to investigate the effect of adjusting the purchase decision. For example, it is very easy with linear programming to check what the impact on total profit would be if a quick purchase were to be made of additional logs in a certain quality category to enable increasing production for the upcoming month. Ponderosa’s linear programming system was interactive, so management received an immediate response to its “what-if questions” about the impact of encountering parame1

A. Roy, E. E. DeFalomir, and L. Lasdon: “An Optimization-Based Decision Support System for a Product Mix Problem,” Interfaces, 12(2):26–33, April 1982.

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ter values that differ from those in the original model. What if a quick purchase of logs of a certain kind were made? What if product prices were to fluctuate in a certain way? A variety of such scenarios can be investigated. Management effectively used this power to reach better decisions than the “optimal” product mix from the original model. The impact of linear programming at Ponderosa was reported to be “tremendous.” It led to a dramatic shift in the types of plywood products emphasized by the company. The improved product-mix decisions were credited with increasing the overall profitability of the company by 20 percent. Other contributions of linear programming included better utilization of raw material, capital equipment, and personnel. Two factors helped make this application of linear programming so successful. One factor is that a natural language financial planning system was interfaced with the codes for finding an optimal solution for the linear programming model. Using natural language rather than mathematical symbols to display the components of the linear programming model and its output made the process understandable and meaningful for the managers making the product-mix decisions. Reporting to management in the language of managers is necessary for the successful application of linear programming. The other factor was that the linear programming system used was interactive. As mentioned earlier, after an optimal solution was obtained for one version of the model, this feature enabled managers to ask a variety of “what-if” questions and receive immediate responses. Better decisions frequently were reached by exploring other plausible scenarios, and this process also gave managers more confidence that their decision would perform well under most foreseeable circumstances. In any application, this ability to respond quickly to management’s needs and queries through postoptimality analysis (whether interactive or not) is a vital part of a linear programming study. Personnel Scheduling at United Airlines1 Despite unprecedented industry competition in 1983 and 1984, UNITED AIRLINES managed to achieve substantial growth with service to 48 new airports. In 1984, it became the only airline with service to cities in all 50 states. Its 1984 operating profit reached $564 million, with revenues of $6.2 billion, an increase of 6 percent over 1983, while costs grew by less than 2 percent. Cost control is essential to competing successfully in the airline industry. In 1982, upper management of United Airlines initiated an OR study of its personnel scheduling as part of the cost control measures associated with the airline’s 1983–1984 expansion. The goal was to schedule personnel at the airline’s reservations offices and airports so as to minimize the cost of providing the necessary service to customers. At the time, United Airlines employed over 4,000 reservations sales representatives and support personnel at its 11 reservations offices and about 1,000 customer service agents at its 10 largest airports. Some were part-time, working shifts from 2 to 8 hours; most were full-time, working 8- or 10-hour-shifts. Shifts start at several different times. Each reservations office was open (by telephone) 24 hours a day, as was each of the major airports. However, the number of employees needed at each location to provide the re1

T. J. Holloran and J. E. Bryn, “United Airlines Station Manpower Planning System,” Interfaces, 16(1): 39–50, Jan.–Feb. 1986.

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quired level of service varied greatly during the 24-hour day, and might fluctuate considerably from one half-hour to the next. Trying to design the work schedules for all the employees at a given location to meet these service requirements most efficiently is a nightmare of combinatorial considerations. Once an employee begins working, he or she will be there continuously for the entire shift (2 to 10 hours, depending on the employee), except for either a meal break or short rest breaks every 2 hours. Given the minimum number of employees needed on duty for each half-hour interval over a 24-hour day (where these requirements change from day to day over a 7-day week), how many employees of each shift length should begin work at what start time over each 24-hour day of a 7-day week? Fortunately, linear programming thrives on such combinatorial nightmares. Actually, several OR techniques described in this book were used in the computerized planning system developed to attack this problem. Both forecasting (Chap. 20) and queuing theory (Chaps. 17 and 18) were used to determine the minimum number of employees needed on duty for each half-hour interval. Integer programming (Chap. 12) was used to determine the times of day at which shifts would be allowed to start. However, the core of the planning system was linear programming, which did all the actual scheduling to provide the needed service with the smallest possible labor cost. A complete work schedule was developed for the first full week of a month, and then it was reused for the remainder of the month. This process was repeated each month to reflect changing conditions. Although the details about the linear programming model have not been published, it is clear that the basic approach used is the one illustrated by the Union Airways example of personnel scheduling in Sec. 3.4. The objective function being minimized represents the total personnel cost for the location being scheduled. The main functional constraints require that the number of employees on duty during each time period will not fall below minimum acceptable levels. However, the Union Airways example has only five decision variables. By contrast, the United Airlines model for some locations has over 20,000 decision variables! The difference is that a real application must consider myriad important details that can be ignored in a textbook example. For example, the United Airlines model takes into account such things as the meal and break assignment times for each employee scheduled, differences in shift lengths for different employees, and days off over a weekly schedule, among other scheduling details. This application of linear programming was reported to have had “an overwhelming impact not only on United management and members of the manpower planning group, but also for many who had never before heard of management science (OR) or mathematical modeling.” It earned rave reviews from upper management, operating managers, and affected employees alike. For example, one manager described the scheduling system as Magical, . . . just as the [customer] lines begin to build, someone shows up for work, and just as you begin to think you’re overstaffed, people start going home.1

In more tangible terms, this application was credited with saving United Airlines more than $6 million annually in just direct salary and benefit costs. Other benefits included improved customer service and reduced need for support staff. 1

Holloran and Bryn, “United Airlines Station Manpower Planning System,” p. 49.

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After some updating in the early 1990s, the system is providing similar benefits today. One factor that helped make this application of linear programming so successful was “the support of operational managers and their staffs.” This was a lesson learned by experience, because the OR team initially failed to establish a good line of communication with the operating managers, who then resisted the team’s initial recommendations. The team leaders described their mistake as follows: The cardinal rule for earning the trust and respect of operating managers and support staffs—”getting them involved in the development process”—had been violated.1

The team then worked much more closely with the operating managers—with outstanding results. Planning Supply, Distribution, and Marketing at Citgo Petroleum Corporation2 CITGO PETROLEUM CORPORATION specializes in refining and marketing petroleum. In the mid-1980s, it had annual sales of several billion dollars, ranking it among the 150 largest industrial companies in the United States. After several years of financial losses, Citgo was acquired in 1983 by Southland Corporation, the owner of the 7-Eleven convenience store chain (whose sales include 2 billion gallons of quality motor fuels annually). To turn Citgo’s financial losses around, Southland created a task force composed of Southland personnel, Citgo personnel, and outside consultants. An eminent OR consultant was appointed director of the task force to report directly to both the president of Citgo and the chairman of the board of Southland. During 1984 and 1985, this task force applied various OR techniques (as well as information systems technologies) throughout the corporation. It was reported that these OR applications “have changed the way Citgo does business and resulted in approximately $70 million per year profit improvement.”3 The two most important applications were both linear programming systems that provided management with powerful planning support. One, called the refinery LP system, led to great improvements in refinery yield, substantial reductions in the cost of labor, and other important cost savings. This system contributed approximately $50 million to profit improvement in 1985. (See the end of Sec. 2.4 for discussion of the key role that model validation played in the development of this system.) However, we will focus here on the other linear programming system, called the supply, distribution, and marketing modeling system (or just the SDM system), that Citgo is continuing to use. The SDM system is particularly interesting because it is based on a special kind of linear programming model that uses networks, just like the model for the Distribution Unlimited example presented at the end of Sec. 3.4. The model for the SDM system provides a representation of Citgo’s entire marketing and distribution network. At the time the task force conducted its OR study, Citgo owned or leased 36 product storage terminals which were supplied through five distribution center terminals via a dis1

Ibid, p. 47. See the references cited in footnote 2 on p. 10. 3 See p. 4 of the second reference cited in footnote 2 on p. 10. 2

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tribution network of pipelines, tankers, and barges. Citgo also sold product from over 350 exchange terminals that were shared with other petroleum marketers. To supply its customers, product might be acquired by Citgo from its refinery in Lake Charles, LA, or from spot purchases on one of five major spot markets, product exchanges, and trades with other industry refiners. These product acquisition decisions were made daily. However, the time from such a decision until the product reached the intended customers could be as long as 11 weeks. Therefore, the linear programming model used an 11-week planning horizon. The SDM system is used to coordinate the supply, distribution, and marketing of each of Citgo’s major products (originally four grades of motor fuel and No. 2 fuel oil) throughout the United States. Management uses the system to make decisions such as where to sell, what price to charge, where to buy or trade, how much to buy or trade, how much to hold in inventory, and how much to ship by each mode of transportation. Linear programming guides these decisions and when to implement them so as to minimize total cost or maximize total profit. The SDM system also is used in “what-if” sessions, where management asks what-if questions about scenarios that differ from those assumed in the original model. The linear programming model in the SDM system has the same form as the model for the Distribution Unlimited example presented at the end of Sec. 3.4. In fact, both models fit an important special kind of linear programming problem, called the minimum cost flow problem, that will be discussed in Sec. 9.6. The main functional constraints for such models are equality constraints, where each one prescribes what the net flow of goods out of a specific location must be. The Distribution Unlimited model has just seven decision variables and five equality constraints. By contrast, the Citgo model for each major product has about 15,000 decision variables and 3,000 equality constraints! At the end of Sec. 2.1, we described the important role that data collection and data verification played in developing the Citgo models. With such huge models, a massive amount of data must be gathered to determine all the parameter values. A state-of-the-art management database system was developed for this purpose. Before using the data, a preloader program was used to check for data errors and inconsistencies. The importance of doing so was brought forcefully home to the task force when, as mentioned in Sec. 2.1, the initial run of the preloader program generated a paper log of error messages an inch thick! It was clear that the data collection process needed to be thoroughly debugged to help ensure the validity of the models. The SDM linear programming system has greatly improved the efficiency of Citgo’s supply, distribution, and marketing operations, enabling a huge reduction in product inventory with no drop in service levels. During its first year, the value of petroleum products held in inventory was reduced by $116.5 million. This huge reduction in capital tied up in carrying inventory resulted in saving about $14 million annually in interest expenses for borrowed capital, adding $14 million to Citgo’s annual profits. Improvements in coordination, pricing, and purchasing decisions have been estimated to add at least another $2.5 million to annual profits. Many indirect benefits also are attributed to this application of linear programming, including improved data, better pricing strategies, and elimination of unnecessary product terminals, as well as improved communication and coordination between supply, distribution, marketing, and refinery groups.

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Some of the factors that contributed to the success of this OR study are the same as for the two preceding case studies. Like Ponderosa Industrial, one factor was developing output reports in the language of managers to really meet their needs. These output reports are designed to be easy for managers to understand and use, and they address the issues that are important to management. Also like Ponderosa, another factor was enabling management to respond quickly to the dynamics of the industry by using the linear programming system extensively in “what-if” sessions. As in so many applications of linear programming, postoptimality analysis proved more important than the initial optimal solution obtained for one version of the model. Much as in the United Airlines application, another factor was the enthusiastic support of operational managers during the development and implementation of this linear programming system. However, the most important factor was the unlimited support provided to the task force by top management, ranging right up to the chief executive officer and the chairman of the board of Citgo’s parent company, Southland Corporation. As mentioned earlier, the director of the task force (an eminent OR consultant) reported directly to both the president of Citgo and the chairman of the board of Southland. This backing by top management included strong organizational and financial support. The organizational support took a variety of forms. One example was the creation and staffing of the position of senior vice-president of operations coordination to evaluate and coordinate recommendations based on the models which spanned organizational boundaries. When discussing both this linear programming system and other OR applications implemented by the task force, team members described the financial support of top management as follows: The total cost of the systems implemented, $20 million to $30 million, was the greatest obstacle to this project. However, because of the information explosion in the petroleum industry, top management realized that numerous information systems were essential to gather, store, and analyze data. The incremental cost of adding management science (OR) technologies to these computers and systems was small, in fact very small in light of the enormous benefits they provided.1

3.6

DISPLAYING AND SOLVING LINEAR PROGRAMMING MODELS ON A SPREADSHEET Spreadsheet software, such as Excel, is a popular tool for analyzing and solving small linear programming problems. The main features of a linear programming model, including all its parameters, can be easily entered onto a spreadsheet. However, spreadsheet software can do much more than just display data. If we include some additional information, the spreadsheet can be used to quickly analyze potential solutions. For example, a potential solution can be checked to see if it is feasible and what Z value (profit or cost) it achieves. Much of the power of the spreadsheet lies in its ability to immediately see the results of any changes made in the solution. In addition, the Excel Solver can quickly apply the simplex method to find an optimal solution for the model. 1

Ibid, p. 21.

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To illustrate this process, we now return to the Wyndor example introduced in Sec. 3.1. Displaying the Model on a Spreadsheet After expressing profits in units of thousands of dollars, Table 3.1 in Sec. 3.1 gives all the parameters of the model for the Wyndor problem. Figure 3.14 shows the necessary additions to this table for an Excel spreadsheet. In particular, a row is added (row 9, labeled “Solution”) to store the values of the decision variables. Next, a column is added (column E, labeled “Totals”). For each functional constraint, the number in column E is the numerical value of the left-hand side of that constraint. Recall that the left-hand side represents the actual amount of the resource used, given the values of the decision variables in row 9. For example, for the Plant 3 constraint in row 7, the amount of this resource used (in hours of production time per week) is Production time used in Plant 3 3x1 2x2. In the language of Excel, the equivalent equation for the number in cell E7 is E7 C7*C9 D7*D9. Notice that this equation involves the sum of two products. There is a function in Excel, called SUMPRODUCT, that will sum up the product of each of the individual terms in two different ranges of cells. For instance, SUMPRODUCT(C7:D7,C9:D9) takes each of the individual terms in the range C7:D7, multiplies them by the corresponding term in the range C9:D9, and then sums up these individual products, just as shown in the above equation. Although optional with such short equations, this function is especially handy as a shortcut for entering longer linear programming equations. Next, signs are entered in cells F5, F6, and F7 to indicate the form of the functional constraints. (When using a trial-and-error approach, the spreadsheet still will allow you to enter infeasible trial solutions that violate the signs, but these signs serve as a reminder to reject such trial solutions if no changes are made in the numbers in column G.)

FIGURE 3.14 The spreadsheet for the Wyndor problem before using the Excel Solver, so the values of the decision variables and the objective function are just entered as zeros.

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Finally, the value of the objective function is entered in cell E8. Much like the other values in column E, it is the sum of products. The equation for cell E8 is SUMPRODUCT(C8:D8,C9:D9). The lower right-hand side of Fig. 3.14 shows all the formulas that need to be entered in the “Totals” column (column E) for the Wyndor problem. Once the model is entered in this spreadsheet format, it is easy to analyze any potential solution. When values for the decision variables are entered in the spreadsheet, the “Totals” column immediately calculates the total amount of each resource used, as well as the total profit. Hence, by comparing column E with column G, it can be seen immediately whether the potential solution is feasible. If so, cell E8 shows how much profit it would generate. One approach to trying to solve a linear programming model would be trial and error, using the spreadsheet to analyze a variety of solutions. However, you will see next how Excel also can be used to quickly find an optimal solution. Using the Excel Solver to Solve the Model Excel includes a tool called Solver that uses the simplex method to find an optimal solution. (A more powerful version of Solver, called Premium Solver, also is available in your OR Courseware.) Before using Solver, all the following components of the model need to be included on the spreadsheet: 1. Each decision variable 2. The objective function and its value 3. Each functional constraint The spreadsheet layout shown in Fig. 3.14 includes all these components. The parameters for the functional constraints are in rows 5, 6, and 7, and the coefficients for the objective function are in row 8. The values of the decision variables are in cells C9 and D9, and the value of the objective function is in cell E8. Since we don’t know what the values of the decision variables should be, they are just entered as zeros. The Solver will then change these to the optimal values after solving the problem. The Solver can be started by choosing “Solver” in the Tools menu. The Solver dialogue box is shown in Fig. 3.15. The “Target Cell” is the cell containing the value of the objective function, while the “Changing Cells” are the cells containing the values of the decision variables. Before the Solver can apply the simplex method, it needs to know exactly where each component of the model is located on the spreadsheet. You can either type in the cell addresses or click on them. Since the target cell is cell E8 and the changing cells are in the range C9:D9, these addresses are entered into the Solver dialogue box as shown in Fig. 3.15. (Excel then automatically enters the dollar signs shown in the figure to fix these addresses.) Since the goal is to maximize the objective function, “Max” also has been selected. Next, the addresses for the functional constraints need to be added. This is done by clicking on the “Add . . .” button on the Solver dialogue box. This brings up the “Add Constraint” dialogue box shown in Fig. 3.16. The location of the values of the left-hand sides and the right-hand sides of the functional constraints are specified in this dialogue box. The cells E5 through E7 all need to be less than or equal to the corresponding cells in G5 through G7. There also is a menu to choose between , , or , so has been chosen for these constraints. (This choice is needed even though signs were pre-

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FIGURE 3.15 The Solver dialogue box after specifying which cells in Fig. 3.14 contain the values of the objective function and the decision variables, plus indicating that the objective function is to be maximized.

viously entered in column F of the spreadsheet because Solver uses only the functional constraints that are specified with the Add Constraint dialogue box.) If there were more functional constraints to add, you would click on Add to bring up a new Add Constraint dialogue box. However, since there are no more in this example, the next step is to click on OK to go back to the Solver dialogue box. The Solver dialogue box now summarizes the complete model (see Fig. 3.17) in terms of the spreadsheet in Fig. 3.14. However, before asking Solver to solve the model, one more step should be taken. Clicking on the Options . . . button brings up the dialogue box shown in Fig. 3.18. This box allows you to specify a number of options about how the problem will be solved. The most important of these are the Assume Linear Model option and the Assume Non-Negative option. Be sure that both options are checked as shown in the figure. This tells Solver that the problem is a linear programming problem with nonnegativity constraints for all the decision variables, and that the simplex method

FIGURE 3.16 The Add Constraint dialogue box after specifying that cells E5, E6, and E7 in Fig. 3.14 are required to be less than or equal to cells G5, G6, and G7, respectively.

3.6 DISPLAYING AND SOLVING LINEAR PROGRAMMING MODELS ON A SPREADSHEET

FIGURE 3.17 The Solver dialogue box after specifying the entire model in terms of the spreadsheet.

FIGURE 3.18 The Solver Options dialogue box after checking the Assume Linear Model and Assume Non-Negative options to indicate that we are dealing with a linear programming model with nonnegativity constraints that needs to be solved by the simplex method.

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FIGURE 3.19 The Solver Results dialogue box that indicates that an optimal solution has been found.

should be used to solve the problem.1 Regarding the other options, accepting the default values shown in the figure usually is fine for small problems. Clicking on the OK button then returns you to the Solver dialogue box. Now you are ready to click on Solve in the Solver dialogue box, which will cause the Solver to execute the simplex method in the background. After a few seconds (for a small problem), Solver will then indicate the results. Typically, it will indicate that it has found an optimal solution, as specified in the Solver Results dialogue box shown in Fig. 3.19. If the model has no feasible solutions or no optimal solution, the dialogue box will indicate that instead by stating that “Solver could not find a feasible solution” or that “the Set Cell values do not converge.” The dialogue box also presents the option of generating various reports. One of these (the Sensitivity Report) will be discussed in detail in Sec. 4.7. After solving the model, the Solver replaces the original value of the decision variables in the spreadsheet with the optimal values, as shown in Fig. 3.20. The spreadsheet also indicates the value of the objective function, as well as the amount of each resource that is being used. 1

In older versions of Excel prior to Excel 97, the Assume Non-Negative option is not available, so nonnegativity constraints have to be added with the Add Constraint dialogue box.

FIGURE 3.20 The spreadsheet obtained after solving the Wyndor problem.

3.7 FORMULATING VERY LARGE LINEAR PROGRAMMING MODELS

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73

FORMULATING VERY LARGE LINEAR PROGRAMMING MODELS Linear programming models come in many different sizes. For the examples in Secs. 3.1 and 3.4, the model sizes range from three functional constraints and two decision variables (for the Wyndor and radiation therapy problems) up to 17 functional constraints and 12 decision variables (for the Save-It Company problem). The latter case may seem like a rather large model. After all, it does take a substantial amount of time just to write down a model of this size. However, by contrast, the models for the classic case studies presented in Sec. 3.5 are much, much larger. For example, the models in the Citgo case study typically have about 3,000 functional constraints and 15,000 decision variables. The Citgo model sizes are not at all unusual. Linear programming models in practice commonly have hundreds or thousands of functional constraints. In fact, there have been some recently reported cases of a few hundred thousand constraints. The number of decision variables frequently is even larger than the number of functional constraints, and occasionally will range into the millions. Formulating such monstrously large models can be a daunting task. Even a “mediumsized” model with a thousand functional constraints and a thousand decision variables has over a million parameters (including the million coefficients in these constraints). It simply is not practical to write out the algebraic formulation, or even to fill in the parameters on a spreadsheet, for such a model. So how are these very large models formulated in practice? It requires the use of a modeling language. Modeling Languages A mathematical programming modeling language is software that has been specifically designed for efficiently formulating large linear programming models (and related models). Even with thousands of functional constraints, they typically are of a relatively few types where the constraints of the same type follow the same pattern. Similarly, the decision variables will fall into a small number of categories. Therefore, using large blocks of data in databases, a modeling language will simultaneously formulate all the constraints of the same type by simultaneously dealing with the variables of each type. We will illustrate this process soon. In addition to efficiently formulating large models, a modeling language will expedite a number of model management tasks, including accessing data, transforming data into model parameters, modifying the model whenever desired, and analyzing solutions from the model. It also may produce summary reports in the vernacular of the decision makers, as well as document the model’s contents. Several excellent modeling languages have been developed over the last couple of decades. These include AMPL, MPL, GAMS, and LINGO. The student version of one of these, MPL (short for mathematical programming language), is provided for you on the CD-ROM along with extensive tutorial material. The latest student version also can be downloaded from the website, maximal-usa.com. MPL is a product of Maximal Software, Inc. A new feature is extensive support for Excel in MPL. This includes both importing and exporting Excel ranges from MPL. Full support also is provided for the Excel VBA macro language through OptiMax 2000. (The student version of OptiMax 2000 is on the CD-ROM as well.) This product allows the user to

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fully integrate MPL models into Excel and solve with any of the powerful solvers that MPL supports, including CPLEX (described in Sec. 4.8). LINGO is a product of LINDO Systems, Inc. The latest student version of LINGO is available by downloading it from the website, www.lindo.com. LINDO Systems also provides a completely spreadsheet-oriented optimizer called What’sBest, also available on this website. The CD-ROM includes MPL, LINGO, and What’sBest formulations for essentially every example in this book to which these modeling languages can be applied. Now let us look at a simplified example that illustrates how a very large linear programming model can arise. An Example of a Problem with a Huge Model Management of the WORLDWIDE CORPORATION needs to address a product-mix problem, but one that is vastly more complex than the Wyndor product-mix problem introduced in Sec. 3.1. This corporation has 10 plants in various parts of the world. Each of these plants produces the same 10 products and then sells them within its region. The demand (sales potential) for each of these products from each plant is known for each of the next 10 months. Although the amount of a product sold by a plant in a given month cannot exceed the demand, the amount produced can be larger, where the excess amount would be stored in inventory (at some unit cost per month) for sale in a later month. Each unit of each product takes the same amount of space in inventory, and each plant has some upper limit on the total number of units that can be stored (the inventory capacity). Each plant has the same 10 production processes (we’ll refer to them as machines), each of which can be used to produce any of the 10 products. Both the production cost per unit of a product and the production rate of the product (number of units produced per day devoted to that product) depend on the combination of plant and machine involved (but not the month). The number of working days ( production days available) varies somewhat from month to month. Since some plants and machines can produce a particular product either less expensively or at a faster rate than other plants and machines, it is sometimes worthwhile to ship some units of the product from one plant to another for sale by the latter plant. For each combination of a plant being shipped from (the fromplant) and a plant being shipped to (the toplant), there is a certain cost per unit shipped of any product, where this unit shipping cost is the same for all the products. Management now needs to determine how much of each product should be produced by each machine in each plant during each month, as well as how much each plant should sell of each product in each month and how much each plant should ship of each product in each month to each of the other plants. Considering the worldwide price for each product, the objective is to find the feasible plan that maximizes the total profit (total sales revenue minus the sum of the total production costs, inventory costs, and shipping costs). The Structure of the Resulting Model Because of the inventory costs and the limited inventory capacities, it is necessary to keep track of the amount of each product kept in inventory in each plant during each month. Consequently, the linear programming model has four types of decision variables: pro-

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duction quantities, inventory quantities, sales quantities, and shipping quantities. With 10 plants, 10 machines, 10 products, and 10 months, this gives a total of 21,000 decision variables, as outlined below. Decision Variables. 10,000 production variables: one for each combination of a plant, machine, product, and month 1,000 inventory variables: one for each combination of a plant, product, and month 1,000 sales variables: one for each combination of a plant, product, and month 9,000 shipping variables: one for each combination of a product, month, plant (the fromplant), and another plant (the toplant) Multiplying each of these decision variables by the corresponding unit cost or unit revenue, and then summing over each type, the following objective function can be calculated: Objective Function. Maximize

profit total sales revenue total cost,

where Total cost total production cost total inventory cost total shipping cost. When maximizing this objective function, the 21,000 decision variables need to satisfy nonnegativity constraints as well as four types of functional constraints—production capacity constraints, plant balance constraints (equality constraints that provide appropriate values to the inventory variables), maximum inventory constraints, and maximum sales constraints. As enumerated below, there are a total of 3,100 functional constraints, but all the constraints of each type follow the same pattern. Functional Constraints. 1,000 production capacity constraints (one for each combination of a plant, machine, and month): Production days used production days available, where the left-hand side is the sum of 10 fractions, one for each product, where each fraction is that product’s production quantity (a decision variable) divided by the product’s production rate (a given constant). 1,000 plant balance constraints (one for each combination of a plant, product, and month): Amount produced inventory last month amount shipped in sales current inventory amount shipped out, where the amount produced is the sum of the decision variables representing the production quantities at the machines, the amount shipped in is the sum of the decision variables representing the shipping quantities in from the other plants, and the amount shipped out is the sum of the decision variables representing the shipping quantities out to the other plants.

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100 maximum inventory constraints (one for each combination of a plant and month): Total inventory inventory capacity, where the left-hand side is the sum of the decision variables representing the inventory quantities for the individual products. 1,000 maximum sales constraints (one for each combination of a plant, product, and month): Sales demand. Now let us see how the MPL modeling language, a product of Maximal Software, Inc., can formulate this huge model very compactly.

Formulation of the Model in MPL The modeler begins by assigning a title to the model and listing an index for each of the entities of the problem, as illustrated below. TITLE Production_Planning; INDEX product month plant fromplant toplant machine

: (A1, A2, A3, A4, A5, : (Jan, Feb, Mar, Apr, : (p1, p2, p3, p4, p5, : plant; : plant; : (m1, m2, m3, m4, m5,

A6, A7, A8, A9, A10); May, Jun, Jul, Aug, Sep, Oct); p6, p7, p8, p9, p10);

m6, m7, m8, m9, m10);

Except for the months, the entries on the right-hand side are arbitrary labels for the respective products, plants, and machines, where these same labels are used in the data files. Note that a colon is placed after the name of each entry and a semicolon is placed at the end of each statement (but a statement is allowed to extend over more than one line). A big job with any large model is collecting and organizing the various types of data into data files. In this case, eight data files are needed to hold the product prices, demands, production costs, production rates, production days available, inventory costs, inventory capacities, and shipping costs. Numbering these data files as 1, 2, 3, . . . , 8, the next step is to give a brief suggestive name to each one and to identify (inside square brackets) the index or indexes over which the data in the file run, as shown below. DATA Price [product] Demand [plant, product, month] ProdCost [plant, machine, product] ProdRate [plant, machine, product] ProdDaysAvail [month] InvtCost [product] InvtCapacity [plant] ShipCost [fromplant, toplant]

: DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE

1; 2; 3; 4; 5; 6; 7; 8;

Next, the modeler gives a short name to each type of decision variable. Following the name, inside square brackets, is the index or indexes over which the subscripts run.

3.7 FORMULATING VERY LARGE LINEAR PROGRAMMING MODELS

VARIABLES Produce [plant, machine, product, month] Inventory [plant, product, month] Sales [plant, product, month] Ship [product, month, fromplant, toplant] WHERE (fromplant toplant);

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Prod; Invt; Sale;

In the case of the decision variables with names longer than four letters, the arrows on the right point to four-letter abbreviations to fit the size limitations of many solvers. The last line indicates that the fromplant subscript and toplant subscript are not allowed to have the same value. There is one more step before writing down the model. To make the model easier to read, it is useful first to introduce macros to represent the summations in the objective function. MACROS Total Revenue : SUM (plant, product, month: Price*Sales); TotalProdCost : SUM (plant, machine, product, month: ProdCost*Produce); TotalInvtCost : SUM (plant, product, month: InvtCost*Inventory); TotalShipCost : SUM (product, month, fromplant, toplant: ShipCost*Ship); TotalCost : TotalProdCost TotalInvtCost TotalShipCost;

The first four macros use the MPL keyword SUM to execute the summation involved. Following each SUM keyword (inside the parentheses) is, first, the index or indexes over which the summation runs. Next (after the colon) is the vector product of a data vector (one of the data files) times a variable vector (one of the four types of decision variables). Now this model with 3,100 functional constraints and 21,000 decision variables can be written down in the following compact form. MODEL MAX Profit TotalRevenue TotalCost; SUBJECT TO ProdCapacity [plant, machine, month] PCap; SUM (product: Produce/ProdRate) ProdDaysAvail; PlantBal [plant, product, month] PBal; SUM (machine: Produce) Inventory [month 1] SUM (fromplant: Ship[fromplant, toplant: plant]) Sales Inventory SUM (toplant: Ship[from plant: plant, toplant]); MaxInventory [plant, month] MaxI: SUM (product: Inventory) InvtCapacity; BOUNDS Sales Demand; END

For each of the four types of constraints, the first line gives the name for this type. There is one constraint of this type for each combination of values for the indexes inside

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the square brackets following the name. To the right of the brackets, the arrow points to a four-letter abbreviation of the name that a solver can use. Below the first line, the general form of constraints of this type is shown by using the SUM operator. For each production capacity constraint, each term in the summation consists of a decision variable (the production quantity of that product on that machine in that plant during that month) divided by the corresponding production rate, which gives the number of production days being used. Summing over the products then gives the total number of production days being used on that machine in that plant during that month, so this number must not exceed the number of production days available. The purpose of the plant balance constraint for each plant, product, and month is to give the correct value to the current inventory variable, given the values of all the other decision variables including the inventory level for the preceding month. Each of the SUM operators in these constraints involves simply a sum of decision variables rather than a vector product. This is the case also for the SUM operator in the maximum inventory constraints. By contrast, the left-hand side of the maximum sales constraints is just a single decision variable for each of the 1,000 combinations of a plant, product, and month. (Separating these upper-bound constraints on individual variables from the regular functional constraints is advantageous because of the computational efficiencies that can be obtained by using the upper bound technique described in Sec. 7.3.) No lower-bound constraints are shown here because MPL automatically assumes that all 21,000 decision variables have nonnegativity constraints unless nonzero lower bounds are specified. For each of the 3,100 functional constraints, note that the left-hand side is a linear function of the decision variables and the right-hand side is a constant taken from the appropriate data file. Since the objective function also is a linear function of the decision variables, this model is a legitimate linear programming model. To solve the model, MPL supports various leading solvers (software packages for solving linear programming models and related models) that can be installed into MPL. As discussed in Sec. 4.8, CPLEX is a particularly prominent and powerful solver. The version of MPL in your OR Courseware already has installed the student version of CPLEX, which uses the simplex method to solve linear programming models. Therefore, to solve such a model formulated with MPL, all you have to do is choose Solve CPLEX from the Run menu or press the Run Solve button in the Toolbar. You then can display the solution file in a view window by pressing the View button at the bottom of the Status Window. This brief introduction to MPL illustrates the ease with which modelers can use modeling languages to formulate huge linear programming models in a clear, concise way. To assist you in using MPL, an MPL Tutorial is included on the CD-ROM. This tutorial goes through all the details of formulating smaller versions of the production planning example considered here. You also can see elsewhere on the CD-ROM how all the other linear programming examples in this chapter and subsequent chapters would be formulated with MPL and solved by CPLEX. The LINGO Modeling Language LINGO is another popular modeling language that is featured in this book. The company that produces LINGO, LINDO Systems, also produces a widely used solver called LINDO as well as a spreadsheet solver, What’sBest. All three share a common set of solvers based

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on the simplex method and, in more advanced versions, on the kind of algorithmic techniques introduced in Secs. 4.9 and 7.4 as well. (We will discuss LINDO further in Sec. 4.8 and Appendix 4.1.) As mentioned earlier, the student version of LINGO is available to you through downloading from the website, www.lindo.com. Like MPL, LINGO enables a modeler to efficiently formulate a huge linear programming model in a clear, concise way. It also can be used for a wide variety of other models. LINGO uses sets as its fundamental building block. For example, in the Worldwide Corp. production planning problem, the sets of interest include the collections of products, plants, machines, and months. Each member of a set may have one or more attributes associated with it, such as the price of a product, the inventory capacity of a plant, the production rate of a machine, and the number of production days available in a month. These attributes provide data for the model. Some set attributes, such as production quantities and shipping quantities, can be decision variables for the model. As with MPL, the SUM operator is commonly used to write the objective function and each constraint type in a compact form. After completing the formulation, the model can be solved by selecting the Solve command from the LINGO menu or pressing the Solve button on the toolbar. An appendix to this chapter describes LINGO further and illustrates its use on a couple of small examples. A supplement on the CD-ROM shows how LINGO can be used to formulate the model for the Worldwide Corp. production planning example. A LINGO tutorial on the CD-ROM provides the details needed for doing basic modeling with this modeling language. The LINGO formulations and solutions for the various examples in both this chapter and many other chapters also are included on the CD-ROM.

3.8

CONCLUSIONS Linear programming is a powerful technique for dealing with the problem of allocating limited resources among competing activities as well as other problems having a similar mathematical formulation. It has become a standard tool of great importance for numerous business and industrial organizations. Furthermore, almost any social organization is concerned with allocating resources in some context, and there is a growing recognition of the extremely wide applicability of this technique. However, not all problems of allocating limited resources can be formulated to fit a linear programming model, even as a reasonable approximation. When one or more of the assumptions of linear programming is violated seriously, it may then be possible to apply another mathematical programming model instead, e.g., the models of integer programming (Chap. 12) or nonlinear programming (Chap. 13).

APPENDIX 3.1 THE LINGO MODELING LANGUAGE LINGO is a mathematical modeling language designed particularly for formulating and solving a wide variety of optimization problems, including linear programming, integer programming (Chap. 12), and nonlinear programming (Chap. 13) problems. Extensive details and a downloadable student version can be found at www.lindo.com.

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Simple problems are entered into LINGO in a fairly natural fashion. To illustrate, consider the following linear programming problem. Maximize

Z 20x 31y,

subject to 2x 5y 16 4x 3y 6

FIGURE A3.1 Screen shots showing the LINGO formulation and the LINGO solution report for a simple linear programming problem.

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and x 0,

y 0.

The screen shot in the top half of Fig. A3.1 shows how this problem would be formulated with LINGO. The first line of this formulation is just a comment describing the model. Note that the comment is preceded by an exclamation point and ended by a semicolon. This is a requirement for all comments in a LINGO formulation. The second line gives the objective function (without bothering to include the Z variable) and indicates that it is to be maximized. Note that each multiplication needs to be indicated by an asterisk. The objective function is ended by a semicolon, as is each of the functional constraints on the next two lines. The nonnegativity constraints are not shown in this formulation because these constraints are automatically assumed by LINGO. (If some variable x did not have a nonnegativity constraint, you would need to add @FREE(x); at the end of the formulation.) Variables can be shown as either lowercase or uppercase, since LINGO is case-insensitive. For example, a variable x1 can be typed in as either x1 or X1. Similarly, words can be either lowercase or uppercase (or a combination). For clarity, we will use uppercase for all reserved words that have a predefined meaning in LINGO. Notice the menu bar at the top of the LINGO window in Fig. A3.1. The ‘File’ and ‘Edit’ menu items behave in a standard Windows fashion. To solve a model once it has been entered, click on the ‘bullseye’ icon. (If you are using a platform other than a Windows-based PC, instead type the GO command at the colon prompt and press the enter key.) Before attempting to solve the model, LINGO will first check whether your model has any syntax errors and, if so, will indicate where they occur. Assuming no such errors, a solver will begin solving the problem, during which time a solver status window will appear on the screen. (For linear programming models, the solver used is LINDO, which will be described in some detail in the appendix to the next chapter.) When the solver finishes, a Solution Report will appear on the screen. The bottom half of Fig. A3.1 shows the solution report for our example. The Value column gives the optimal values of the decision variables. The first entry in the Slack or Surplus column shows the corresponding value of the objective function. The next two entries indicate the difference between the two sides of the respective constraints. The Reduced Cost and Dual Price columns provide some sensitivity analysis information for the problem. After discussing postoptimality analysis (including sensitivity analysis) in Sec. 4.7, we will explain what reduced costs and dual prices are while describing LINDO in Appendix 4.1. These quantities provide only a portion of the useful sensitivity analysis information. To generate a full sensitivity analysis report (such as shown in Appendix 4.1 for LINDO), the Range command in the LINGO menu would need to be chosen next. Just as was illustrated with MPL in Sec. 3.7, LINGO is designed mainly for efficiently formulating very large models by simultaneously dealing with all constraints or variables of the same type. We soon will use the following example to illustrate how LINGO does this.

Example. Consider a production-mix problem where we are concerned with what mix of four products we should produce during the upcoming week. For each product, each unit produced requires a known amount of production time on each of three machines. Each machine has a certain number of hours of production time available per week. Each product provides a certain profit per unit produced. Table A3.1 shows three types of data: machine-related data, product-related data, and data related to combinations of a machine and product. The objective is to determine how much to produce of each product so that total profit is maximized while not exceeding the limited production capacity of each machine.

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TABLE A3.1 Data needed for the product-mix example Production Time per Unit, Hours Product Machine

P01

P02

P03

P04

Production Time Available per Week, Hours

Roll Cut Weld

1.7 1.1 1.6

2.1 2.5 1.3

1.4 1.7 1.6

2.4 2.6 0.8

28 34 21

Profit per unit

26

35

25

37

In standard algebraic form, the structure of the linear programming model for this problem is to choose the nonnegative production levels (number of units produced during the upcoming week) for the four products so as to 4

Maximize

cj x j , j1

subject to 4

aij x j bj j1

for i 1, 2, 3;

where xj production level for product P0j cj unit profit for product P0j aij production time on machine i per unit of product P0j bi production time available per week on machine i. This model is small enough, with just 4 decision variables and 3 functional constraints, that it could be written out completely, term by term, but it would be tedious. In some similar applications, there might instead be hundreds of decision variables and functional constraints, so writing out a term-by-term version of this model each week would not be practical. LINGO provides a much more efficient and compact formulation, comparable to the above summary of the model, as we will see next.

Formulation of the Model in LINGO This model has a repetitive nature. All the decision variables are of the same type and all the functional constraints are of the same type. LINGO uses sets to describe this repetitive nature.1 The simple sets of interest in this case are 1. The set of machines, {Roll, Cut, Weld}. 2. The set of products, {P01, P02, P03, P04}. 1

Order is implied in LINGO sets so, strictly speaking, they are not truly sets in the usual mathematical sense.

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The attributes of interest for the members of these sets are 1. Attribute for each machine: Number of hours of production time available per week. 2. Attributes for each product: Profit per unit produced; Number of units produced per week. Thus, the first two types of attributes are input data that will become parameters of the model, whereas the last type (number of units produced per week of the respective products) provides the decision variables for the model. Let us abbreviate these attributes as follows. machine: ProdHoursAvail product: Profit, Produce. One other key type of information is the number of hours of production time that each unit of each product would use on each of the machines. This number can be viewed as an attribute for the members of the set of all combinations of a product and a machine. Since this set is derived from the two simple sets, it is referred to as a derived set. Let us abbreviate the attribute for members of this set as follows. MaPr (machine, product): ProdHoursUsed A LINGO formulation typically has three sections. 1. A SETS section that specifies the sets and their attributes. You can think of it as describing the structure of the data. 2. A DATA section that either provides the data to be used or indicates where it is to be obtained. 3. A section that provides the mathematical model itself. We begin by showing the first two sections for the example below. ! LINGO3h; ! Product mix example; ! Notice: the SETS section says nothing about the number or names of the machines or products. That information is determined completely by supplied data; SETS: ! The simple sets; Machine: ProdHoursAvail; Product: Profit, Produce; ! A derived set; MaPr (Machine, Product): ProdHoursUsed; ENDSETS DATA: ! Get the names of the machines; Machine Roll Cut Weld; ! Hours available on each machine; ProdHoursAvail 28 34 21; ! Get the names of the products; Product P01 P02 P03 P04; ! Profit contribution per unit; Profit 26 35 25 37; ! Hours needed per ProdHoursUsed 1.7 1.1 1.6 ENDDATA

unit 2.1 2.5 1.3

of product; 1.4 2.4 ! Roll; 1.7 2.6 ! Cut; 1.6 0.8; ! Weld;

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Before presenting the mathematical model itself, we need to introduce two key set looping functions that enable applying an operation to all members of a set by using a single statement. One is the @SUM function, which computes the sum of an expression over all members of a set. The general form of @SUM is @SUM( set: expression). For every member of the set, the expression is computed, and then they are all added up. For example, @SUM( Product(j): Profit(j)*Produce(j))

sums the expression following the colon—the unit profit of a product times the production rate of the product—over all members of the set preceding the colon. In particular, since this set is the set of products {Product( j) for j 1, 2, 3, 4}, the sum is over the index j. Therefore, this specific @SUM function provides the objective function, 4

cj xj, j1 given earlier for the model. The second key set looping function is the @FOR function. This function is used to generate constraints over members of a set. The general form is @FOR( set: constraint). For example, @FOR(Machine(i): @SUM( Product(i): ProdHoursUsed(i, j)*Produce (j)) ProdHoursAvail (i, j); );

says to generate the constraint following the colon for each member of the set preceding the colon. (The “less than or equal to” symbol, , is not on the standard keyboard, so LINGO treats the standard keyboard symbols as equivalent to .) This set is the set of machines {Machine (i) for i 1, 2, 3}, so this function loops over the index i. For each i, the constraint following the colon was expressed algebraically earlier as 4

aij x j bj. j1 Therefore, after the third section of the LINGO formulation (the mathematical model itself) is added, we obtain the complete formulation shown below: ! LINGO3h; ! Product mix example; SETS: !The simple sets; Machine: ProdHoursAvail; Product: Profit, Produce; !A derived set; MaPr( Machine, Product): ProdHoursUsed; ENDSETS DATA: !Get the names of the machines; Machine Roll Cut Weld; ! Hours available on each machine; ProdHoursAvail 28 34 21;

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! Get the names of the products; Product P01 P02 P03 P04; ! Profit contribution per unit; Profit 26 35 25 37; ! Hours needed per unit of product; ProdHoursUsed 1.7 2.1 1.4 2.4 ! Roll; 1.1 2.5 1.7 2.6 ! Cut; 1.6 1.3 1.6 0.8; ! Weld; ENDDATA ! Maximize total profit contribution; MAX @SUM( Product(i): Profit(i) * Produce(i)); ! For each machine i; @FOR( Machine( i): ! Hours used must be hours available; @SUM( Product( j): ProdHoursUsed( i, j) * Produce( j)) ProdHoursAvail; );

The model is solved by pressing the ‘bullseye’ button on the LINGO command bar. Pressing the ‘x ’ button on the command bar produces a report that looks in part as follows: Variable PRODUCE( P01) PRODUCE( P02) PRODUCE( P03) PRODUCE( P04) Row 1 2 3 4

Value 0.0000000 10.00000 5.000000 0.0000000

Slack or Surplus 475.0000 0.0000000 0.5000000 0.0000000

Reduced Cost 3.577921 0.0000000 0.0000000 1.441558 Dual Price 1.000000 15.25974 0.0000000 2.272727

Thus, we should produce 10 units of product P02 and 5 units of product P03, where Row 1 gives the resulting total profit of 475. Notice that this solution exactly uses the available capacity on the first and third machines (since Rows 2 and 4 give a Slack or Surplus of 0) and leaves the second machine with 0.5 hour of idleness. (We will discuss reduced costs and dual prices in Appendix 4.1 in conjunction with LINDO.) The rows section of this report is slightly ambiguous in that you need to remember that Row 1 in the model concerns the objective function and the subsequent rows involve the constraints on machine capacities. This association can be made more clear in the report by giving names to each constraint in the model. This is done by enclosing the name in [ ], placed just in front of the constraint. See the following modified fragment of the model. [Totprof] MAX @SUM( Product: Profit * Produce); ! For each machine i; @FOR( Machine( i): ! Hours used must be hours available; [Capc] @SUM( Product( j): ProdHoursUsed( i, j) * Produce( j)) ProdHoursAvail; );

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The solution report now contains these row names. Row TOTPROF CAPC( ROLL) CAPC( CUT) CAPC( WELD)

Slack or Surplus 475.0000 0.0000000 0.5000000 0.0000000

Dual Price 1.000000 15.25974 0.0000000 2.272727

An important feature of a LINGO model like this one is that it is completely “scalable” in products and machines. In other words, if you wanted to solve another version of this product-mix problem with a different number of machines and products, you would only have to enter the new data in the DATA section. You would not need to change the SETS section or any of the equations. This conversion could be done by clerical personnel without any understanding of the model equations.

Importing and Exporting Spreadsheet Data with LINGO The above example was completely self-contained in the sense that all the data were directly incorporated into the LINGO formulation. In some other applications, a large body of data will be stored in some source and will need to be entered into the model from that source. One popular place for storing data is spreadsheets. LINGO has a simple function, @OLE(), for retrieving and placing data from and into spreadsheets. To illustrate, let us suppose the data for our product-mix problem were originally entered into a spreadsheet as shown in Fig. A3.2. For the moment we are interested only in the shaded cells

FIGURE A3.2 Screen shot showing data for the product-mix example entered in a spreadsheet.

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in columns A-B and E-H. The data in these cells completely describe our little product-mix example. We want to avoid retyping these data into our LINGO model. Suppose that this spreadsheet is stored in the file d:\dirfred7\wbest03i.xls. The only part of the LINGO model that needs to be changed is the DATA section as shown below. DATA: ! Get the names of the machines; Machine @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Hours available on each machine; ProdHoursAvail @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Get the names of the products; Product @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Profit contribution per unit; Profit @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Hours needed per unit of product; ProdHoursUsed @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Send the solution values back; @OLE( ‘d:\dirfred7\wbest03i.xls’) Produce; ENDDATA

The @OLE() function acts as your “plumbing contractor.” It lets the data flow from the spreadsheet to LINGO and back to the spreadsheet. So-called Object Linking and Embedding (OLE) is a feature of the Windows operating system. LINGO exploits this feature to make a link between the LINGO model and a spreadsheet. The first five uses of @OLE() above illustrate that this function can be used on the right of an assignment statement to retrieve data from a spreadsheet. The last use above illustrates that this function can be placed on the left of an assignment statement to place solution results into the spreadsheet instead. Notice from Fig. A3.2 that the optimal solution has been placed back into the spreadsheet in cells E6:H6. One simple but hidden step that had to be done beforehand in the spreadsheet was to define range names for the various collections of cells containing the data. Range names can be defined in Excel by using the mouse and the Insert, Name, Define menu item. For example, the set of cells A9:A11 was given the range name of Machine. Similarly, the set of cells E4:H4 was given the range name Product.

Importing and Exporting from a Database with LINGO Another common repository for data in a large firm is in a database. In a manner similar to @OLE(), LINGO has a connection function, @ODBC(), for transferring data from and to a database. This function is based around the Open DataBase Connectivity (ODBC) standard for communicating with SQL (Structured Query Language) databases. Most popular databases, such as Oracle, Paradox, DB/2, MS Access, and SQL Server, support the ODBC convention. Let us illustrate the ODBC connection for our little product-mix example. Suppose that all the data describing our problem are stored in a database called acces03j. The modification required in the LINGO model is almost trivial. Only the DATA section needs to be changed, as illustrated in the following fragment from the LINGO model. DATA: ! Get the names of the machines and available hours; Machine, ProdHoursAvail @ODBC( ‘acces03j’); ! Get the names of the products and profits; Product, Profit @ODBC( ‘acces03j’);

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! Hours needed per unit of product; ProdHoursUsed @ODBC( ‘acces03j’); ! Send the solution values back; @ODBC( ‘acces03j’) Produce; ENDDATA

Notice that, similar to the spreadsheet-based model, the size of the model in terms of the number of variables and constraints is determined completely by what is found in the database. The LINGO model automatically adjusts to what is found in the database. Now let us show what is in the database considered above. It contains three related tables. We give these tables names to match those in the LINGO model, namely, ‘Machine,’ to hold machinerelated data, ‘Product,’ to hold product-related data, and ‘MaPr,’ to hold data related to combinations of machines and products. Here is what the tables look like on the screen:

Machine Machine

ProdHoursAvail

Roll Cut Weld

28 34 21

Product Product

Profit

P01 P02 P03 P04

26 35 25 37

Produce

MaPr Machine Roll Roll Roll Roll Cut Cut Cut Cut Weld Weld Weld Weld

Product

ProdHoursUsed

P01 P02 P03 P04 P01 P02 P03 P04 P01 P02 P03 P04

1.7 2.1 1.4 2.4 1.1 2.5 1.7 2.6 1.6 1.3 1.6 0.8

SELECTED REFERENCES

89

Notice that the ‘Produce’ column has been left blank in the Product table. Once we solve the model, the ‘Produce’ amounts get inserted into the database and the Product table looks as follows: Product Product

Profit

Produce

P01 P02 P03 P04

26 35 25 37

0 10 5 0

There is one complication in using ODBC in Windows 95. The user must “register” the database with the Windows ODBC administrator. One does this by accessing (with mouse clicks) the My Computer/Control Panel/ODBC32 window. Once there, the user must give a name to the database (which may differ from the actual name of the file in which the data tables reside) and specify the directory in which the database file resides. It is this registered name that should be used in the LINGO model. Because the database has been registered, you did not see a directory specification in the @ODBC( ‘acces03j’) in the LINGO model. The ODBC manager knows the location of the database just from its name.

More about LINGO Only some of the capabilities of LINGO have been illustrated in this appendix. More details can be found in the documentation that accompanies LINGO when it is downloaded. LINGO is available in a variety of sizes. The smallest version is the demo version that can be downloaded from www.lindo.com. It is designed for textbook-sized problems (currently a maximum of 150 functional constraints and 300 decision variables). However, the largest version (called the extended version) is limited only by the storage space available. Tens of thousands of functional constraints and hundreds of thousands of decision variables are not unusual. If you would like to see how LINGO can formulate a huge model like the production planning example introduced in Sec. 3.7, a supplement to this appendix on the book’s website, www.mhhe.com/hillier, shows the LINGO formulation of this example. By reducing the number of products, plants, machines, and months, the supplement also introduces actual data into the formulation and then shows the complete solution. The supplement goes on to discuss and illustrate the debugging and verification of this large model. The supplement also describes further how to retrieve data from external files (including spreadsheets) and how to insert results in existing files. In addition to this supplement, the CD-ROM includes both a LINGO tutorial and LINGO/LINDO files with numerous examples of LINGO formulations.

SELECTED REFERENCES 1. Anderson, D. R., D. J. Sweeney, and T. A. Williams: An Introduction to Management Science, 9th ed., West, St. Paul, MN, 2000, chaps. 2, 4. 2. Gass, S.: An Illustrated Guide to Linear Programming, Dover Publications, New York, 1990. 3. Hillier, F. S., M. S. Hillier, and G. J. Lieberman: Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Irwin/McGraw-Hill, Burr Ridge, IL, 2000, chaps. 2, 3.

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3 INTRODUCTION TO LINEAR PROGRAMMING

4. LINGO User’s Guide, LINDO Systems, Inc., Chicago, IL, e-mail: [email protected], 1999. 5. MPL Modeling System (Release 4.0) manual, Maximal Software, Inc., Arlington, VA, e-mail: [email protected], 1998. 6. Williams, H. P.: Model Building in Mathematical Programming, 3d ed., Wiley, New York, 1990.

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE A Demonstration Example in OR Tutor: Graphical Method

An Excel Add-In: Premium Solver

“Ch. 3—Intro to LP” Files for Solving the Examples: Excel File LINGO/LINDO File MPL/CPLEX File

Supplement to Appendix 3.1: More about LINGO (appears on the book’s website, www.mhhe.com/hillier).

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The demonstration example listed above may be helpful. C: Use the computer to solve the problem by applying the simplex method. The available software options for doing this include the Excel Solver or Premium Solver (Sec. 3.6), MPL/CPLEX (Sec. 3.7), LINGO (Appendix 3.1), and LINDO (Appendix 4.1), but follow any instructions given by your instructor regarding the option to use. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 3.1-1.* For each of the following constraints, draw a separate graph to show the nonnegative solutions that satisfy this constraint. (a) x1 3x2 6 (b) 4x1 3x2 12 (c) 4x1 x2 8

(d) Now combine these constraints into a single graph to show the feasible region for the entire set of functional constraints plus nonnegativity constraints. 3.1-2. Consider the following objective function for a linear programming model: D

Maximize Z 2x1 3x2 (a) Draw a graph that shows the corresponding objective function lines for Z 6, Z 12, and Z 18. (b) Find the slope-intercept form of the equation for each of these three objective function lines. Compare the slope for these three lines. Also compare the intercept with the x2 axis.

D

3.1-3. Consider the following equation of a line: 20x1 40x2 400 (a) Find the slope-intercept form of this equation.

CHAPTER 3 PROBLEMS

(b) Use this form to identify the slope and the intercept with the x2 axis for this line. (c) Use the information from part (b) to draw a graph of this line. D

3.1-4.* Use the graphical method to solve the problem: Z 2x1 x2,

Maximize subject to x2 2x1 5x2 x1 x2 3x1 x2

10 60 18 44

and x1 0, D

x2 0.

3.1-5. Use the graphical method to solve the problem: Maximize

Z 10x1 20x2,

subject to x1 2x2 15 x1 x2 12 5x1 3x2 45 and x1 0,

x2 0.

3.1-6. The Whitt Window Company is a company with only three employees which makes two different kinds of hand-crafted windows: a wood-framed and an aluminum-framed window. They earn $60 profit for each wood-framed window and $30 profit for each aluminum-framed window. Doug makes the wood frames, and can make 6 per day. Linda makes the aluminum frames, and can make 4 per day. Bob forms and cuts the glass, and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass. The company wishes to determine how many windows of each type to produce per day to maximize total profit. (a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources. (b) Formulate a linear programming model for this problem. D (c) Use the graphical model to solve this model. (d) A new competitor in town has started making wood-framed windows as well. This may force the company to lower the price they charge and so lower the profit made for each woodframed window. How would the optimal solution change (if at

91

all) if the profit per wood-framed window decreases from $60 to $40? From $60 to $20? (e) Doug is considering lowering his working hours, which would decrease the number of wood frames he makes per day. How would the optimal solution change if he makes only 5 wood frames per day? 3.1-7. The Apex Television Company has to decide on the number of 27- and 20-inch sets to be produced at one of its factories. Market research indicates that at most 40 of the 27-inch sets and 10 of the 20-inch sets can be sold per month. The maximum number of work-hours available is 500 per month. A 27-inch set requires 20 work-hours and a 20-inch set requires 10 work-hours. Each 27-inch set sold produces a profit of $120 and each 20-inch set produces a profit of $80. A wholesaler has agreed to purchase all the television sets produced if the numbers do not exceed the maxima indicated by the market research. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. 3.1-8. The WorldLight Company produces two light fixtures (products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce so as to maximize profit. For each unit of product 1, 1 unit of frame parts and 2 units of electrical components are required. For each unit of product 2, 3 units of frame parts and 2 units of electrical components are required. The company has 200 units of frame parts and 300 units of electrical components. Each unit of product 1 gives a profit of $1, and each unit of product 2, up to 60 units, gives a profit of $2. Any excess over 60 units of product 2 brings no profit, so such an excess has been ruled out. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. What is the resulting total profit? 3.1-9. The Primo Insurance Company is introducing two new product lines: special risk insurance and mortgages. The expected profit is $5 per unit on special risk insurance and $2 per unit on mortgages. Management wishes to establish sales quotas for the new product lines to maximize total expected profit. The work requirements are as follows: Work-Hours per Unit Department

Special Risk

Mortgage

Work-Hours Available

Underwriting Administration Claims

3 0 2

2 1 0

2400 800 1200

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INTRODUCTION TO LINEAR PROGRAMMING

(a) Formulate a linear programming model for this problem. (b) Use the graphical method to solve this model. (c) Verify the exact value of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations. D

3.1-10. Weenies and Buns is a food processing plant which manufactures hot dogs and hot dog buns. They grind their own flour for the hot dog buns at a maximum rate of 200 pounds per week. Each hot dog bun requires 0.1 pound of flour. They currently have a contract with Pigland, Inc., which specifies that a delivery of 800 pounds of pork product is delivered every Monday. Each hot dog requires 14 pound of pork product. All the other ingredients in the hot dogs and hot dog buns are in plentiful supply. Finally, the labor force at Weenies and Buns consists of 5 employees working full time (40 hours per week each). Each hot dog requires 3 minutes of labor, and each hot dog bun requires 2 minutes of labor. Each hot dog yields a profit of $0.20, and each bun yields a profit of $0.10. Weenies and Buns would like to know how many hot dogs and how many hot dog buns they should produce each week so as to achieve the highest possible profit. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. 3.1-11.* The Omega Manufacturing Company has discontinued the production of a certain unprofitable product line. This act created considerable excess production capacity. Management is considering devoting this excess capacity to one or more of three products; call them products 1, 2, and 3. The available capacity on the machines that might limit output is summarized in the following table:

Machine Type

Available Time (Machine Hours per Week)

Milling machine Lathe Grinder

500 350 150

The sales department indicates that the sales potential for products 1 and 2 exceeds the maximum production rate and that the sales potential for product 3 is 20 units per week. The unit profit would be $50, $20, and $25, respectively, on products 1, 2, and 3. The objective is to determine how much of each product Omega should produce to maximize profit. (a) Formulate a linear programming model for this problem. C (b) Use a computer to solve this model by the simplex method. 3.1-12. Consider the following problem, where the value of c1 has not yet been ascertained. D

Z c1x1 x2,

Maximize subject to x1 x2 6 x1 2x2 10 and x1 0,

x2 0.

Use graphical analysis to determine the optimal solution(s) for (x1, x2) for the various possible values of c1( c1 ). 3.1-13. Consider the following problem, where the value of k has not yet been ascertained.

D

Maximize

Z x1 2x2,

subject to x1 x2 2 x2 3 kx1 x2 2k 3,

where k 0

and x1 0,

x2 0.

The solution currently being used is x1 2, x2 3. Use graphical analysis to determine the values of k such that this solution actually is optimal. 3.1-14. Consider the following problem, where the values of c1 and c2 have not yet been ascertained. D

The number of machine hours required for each unit of the respective products is Productivity coefficient (in machine hours per unit) Machine Type

Product 1

Product 2

Product 3

Milling machine Lathe Grinder

9 5 3

3 4 0

5 0 2

Maximize

Z c1x1 c2x2,

subject to 2x1 x2 11 x1 2x2 2 and x1 0,

x2 0.

Use graphical analysis to determine the optimal solution(s) for (x1, x2) for the various possible values of c1 and c2. (Hint: Sepa-

CHAPTER 3 PROBLEMS

rate the cases where c2 0, c2 0, and c2 0. For the latter two cases, focus on the ratio of c1 to c2.) 3.2-1. The following table summarizes the key facts about two products, A and B, and the resources, Q, R, and S, required to produce them. Resource Usage per Unit Produced Resource

Product A Product B

Q R S

2 1 3

1 2 3

Profit per unit

3

2

Amount of Resource Available 2 2 4

All the assumptions of linear programming hold. (a) Formulate a linear programming model for this problem. D (b) Solve this model graphically. (c) Verify the exact value of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations. 3.2-2. The shaded area in the following graph represents the feasible region of a linear programming problem whose objective function is to be maximized. x2

(3, 3) (6, 3)

93

3.2-3.* This is your lucky day. You have just won a $10,000 prize. You are setting aside $4,000 for taxes and partying expenses, but you have decided to invest the other $6,000. Upon hearing this news, two different friends have offered you an opportunity to become a partner in two different entrepreneurial ventures, one planned by each friend. In both cases, this investment would involve expending some of your time next summer as well as putting up cash. Becoming a full partner in the first friend’s venture would require an investment of $5,000 and 400 hours, and your estimated profit (ignoring the value of your time) would be $4,500. The corresponding figures for the second friend’s venture are $4,000 and 500 hours, with an estimated profit to you of $4,500. However, both friends are flexible and would allow you to come in at any fraction of a full partnership you would like. If you choose a fraction of a full partnership, all the above figures given for a full partnership (money investment, time investment, and your profit) would be multiplied by this same fraction. Because you were looking for an interesting summer job anyway (maximum of 600 hours), you have decided to participate in one or both friends’ ventures in whichever combination would maximize your total estimated profit. You now need to solve the problem of finding the best combination. (a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources. (b) Formulate a linear programming model for this problem. D (c) Use the graphical method to solve this model. What is your total estimated profit? 3.2-4. Use the graphical method to find all optimal solutions for the following model: D

(0, 2)

Maximize

Z 500x1 300x2,

subject to

(0, 0) (6, 0)

x1

Label each of the following statements as True or False, and then justify your answer based on the graphical method. In each case, give an example of an objective function that illustrates your answer. (a) If (3, 3) produces a larger value of the objective function than (0, 2) and (6, 3), then (3, 3) must be an optimal solution. (b) If (3, 3) is an optimal solution and multiple optimal solutions exist, then either (0, 2) or (6, 3) must also be an optimal solution. (c) The point (0, 0) cannot be an optimal solution.

15x1 5x2 300 10x1 6x2 240 8x1 12x2 450 and x1 0,

x2 0.

3.2-5. Use the graphical method to demonstrate that the following model has no feasible solutions. D

Maximize

Z 5x1 7x2,

subject to 2x1 x2 1 x1 2x2 1

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INTRODUCTION TO LINEAR PROGRAMMING

and x1 0,

(a) Design of radiation therapy (Mary). (b) Regional planning (Southern Confederation of Kibbutzim). (c) Controlling air pollution (Nori & Leets Co.).

x2 0.

D 3.2-6. Suppose that the following constraints have been provided for a linear programming model.

x1 3x2 30 3x1 x2 30 and x1 0,

x2 0.

(a) Demonstrate that the feasible region is unbounded. (b) If the objective is to maximize Z x1 x2, does the model have an optimal solution? If so, find it. If not, explain why not. (c) Repeat part (b) when the objective is to maximize Z x1 x2. (d) For objective functions where this model has no optimal solution, does this mean that there are no good solutions according to the model? Explain. What probably went wrong when formulating the model?

3.4-2. For each of the four assumptions of linear programming discussed in Sec. 3.3, write a one-paragraph analysis of how well it applies to each of the following examples given in Sec. 3.4. (a) Reclaiming solid wastes (Save-It Co.). (b) Personnel scheduling (Union Airways). (c) Distributing goods through a distribution network (Distribution Unlimited Co.). D

3.4-3. Use the graphical method to solve this problem: Z 15x1 20x2,

Maximize subject to

x1 2x2 10 2x1 3x2 6 x1 x2 6

3.3-1. Reconsider Prob. 3.2-3. Indicate why each of the four assumptions of linear programming (Sec. 3.3) appears to be reasonably satisfied for this problem. Is one assumption more doubtful than the others? If so, what should be done to take this into account?

and

3.3-2. Consider a problem with two decision variables, x1 and x2, which represent the levels of activities 1 and 2, respectively. For each variable, the permissible values are 0, 1, and 2, where the feasible combinations of these values for the two variables are determined from a variety of constraints. The objective is to maximize a certain measure of performance denoted by Z. The values of Z for the possibly feasible values of (x1, x2) are estimated to be those given in the following table:

subject to

x1 0, D

x2 0.

3.4-4. Use the graphical method to solve this problem: Minimize

Z 3x1 2x2,

x1 2x2 12 2x1 3x2 12 2x1 x2 8 and x1 0,

x2 0.

3.4-5. Consider the following problem, where the value of c1 has not yet been ascertained. D

x2 x1

0

1

2

0 1 2

0 3 6

4 8 12

8 13 18

Z c1x1 2x2,

Maximize subject to 4x1 x2 12 x1 x2 2 and

Based on this information, indicate whether this problem completely satisfies each of the four assumptions of linear programming. Justify your answers. 3.4-1.* For each of the four assumptions of linear programming discussed in Sec. 3.3, write a one-paragraph analysis of how well you feel it applies to each of the following examples given in Sec. 3.4:

x1 0,

x2 0.

Use graphical analysis to determine the optimal solution(s) for (x1, x2) for the various possible values of c1. D

3.4-6. Consider the following model: Minimize

Z 40x1 50x2,

CHAPTER 3 PROBLEMS

subject to

Each pig requires at least 8,000 calories per day and at least 700 units of vitamins. A further constraint is that no more than one-third of the diet (by weight) can consist of Feed Type A, since it contains an ingredient which is toxic if consumed in too large a quantity. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. What is the resulting daily cost per pig?

2x1 3x2 30 x1 x2 12 2x1 x2 20 and x1 0,

x2 0.

(a) Use the graphical method to solve this model. (b) How does the optimal solution change if the objective function is changed to Z 40x1 70x2? (c) How does the optimal solution change if the third functional constraint is changed to 2x1 x2 15? 3.4-7. Ralph Edmund loves steaks and potatoes. Therefore, he has decided to go on a steady diet of only these two foods (plus some liquids and vitamin supplements) for all his meals. Ralph realizes that this isn’t the healthiest diet, so he wants to make sure that he eats the right quantities of the two foods to satisfy some key nutritional requirements. He has obtained the following nutritional and cost information: Grams of Ingredient per Serving Steak

Potatoes

Daily Requirement (Grams)

Carbohydrates Protein Fat

5 20 15

15 5 2

50 40 60

Cost per serving

$4

$2

Ingredient

95

Ralph wishes to determine the number of daily servings (may be fractional) of steak and potatoes that will meet these requirements at a minimum cost. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. C (c) Use a computer to solve this model by the simplex method. 3.4-8. Dwight is an elementary school teacher who also raises pigs for supplemental income. He is trying to decide what to feed his pigs. He is considering using a combination of pig feeds available from local suppliers. He would like to feed the pigs at minimum cost while also making sure each pig receives an adequate supply of calories and vitamins. The cost, calorie content, and vitamin content of each feed is given in the table below.

3.4-9. Web Mercantile sells many household products through an on-line catalog. The company needs substantial warehouse space for storing its goods. Plans now are being made for leasing warehouse storage space over the next 5 months. Just how much space will be required in each of these months is known. However, since these space requirements are quite different, it may be most economical to lease only the amount needed each month on a monthby-month basis. On the other hand, the additional cost for leasing space for additional months is much less than for the first month, so it may be less expensive to lease the maximum amount needed for the entire 5 months. Another option is the intermediate approach of changing the total amount of space leased (by adding a new lease and/or having an old lease expire) at least once but not every month. The space requirement and the leasing costs for the various leasing periods are as follows:

Month

Required Space (Sq. Ft.)

Leasing Period (Months)

Cost per Sq. Ft. Leased

1 2 3 4 5

30,000 20,000 40,000 10,000 50,000

1 2 3 4 5

$ 65 $100 $135 $160 $190

The objective is to minimize the total leasing cost for meeting the space requirements. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-10. Larry Edison is the director of the Computer Center for Buckly College. He now needs to schedule the staffing of the center. It is open from 8 A.M. until midnight. Larry has monitored the usage of the center at various times of the day, and determined that the following number of computer consultants are required:

Time of Day Contents Calories (per pound) Vitamins (per pound) Cost (per pound)

Feed Type A

Feed Type B

800 140 units $0.40

1,000 70 units $0.80

8 A.M.–noon Noon–4 P.M. 4 P.M.–8 P.M. 8 P.M.–midnight

Minimum Number of Consultants Required to Be on Duty 4 8 10 6

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Two types of computer consultants can be hired: full-time and part-time. The full-time consultants work for 8 consecutive hours in any of the following shifts: morning (8 A.M.–4 P.M.), afternoon (noon–8 P.M.), and evening (4 P.M.–midnight). Full-time consultants are paid $14 per hour. Part-time consultants can be hired to work any of the four shifts listed in the above table. Part-time consultants are paid $12 per hour. An additional requirement is that during every time period, there must be at least 2 full-time consultants on duty for every parttime consultant on duty. Larry would like to determine how many full-time and how many part-time workers should work each shift to meet the above requirements at the minimum possible cost. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-11.* The Medequip Company produces precision medical diagnostic equipment at two factories. Three medical centers have placed orders for this month’s production output. The table to the right shows what the cost would be for shipping each unit from each factory to each of these customers. Also shown are the number of units that will be produced at each factory and the number of units ordered by each customer. (Go to the next column.)

40 tons produced

$2,000/ton

M1

30 tons max.

Unit Shipping Cost To From

Customer 1 Customer 2 Customer 3

Factory 1 Factory 2

$600 $400

$800 $900

$700 $600

Order size

300 units

200 units

400 units

3.4-12. The Fagersta Steelworks currently is working two mines to obtain its iron ore. This iron ore is shipped to either of two storage facilities. When needed, it then is shipped on to the company’s steel plant. The diagram below depicts this distribution network, where M1 and M2 are the two mines, S1 and S2 are the two storage facilities, and P is the steel plant. The diagram also shows the monthly amounts produced at the mines and needed at the plant, as well as the shipping cost and the maximum amount that can be shipped per month through each shipping lane. (Go to the left column below the diagram.)

S1

$1

70

n /to x. a

m ns

to

00

,7

30

$40

0/to n sm ax.

ton

$1 ,6 to 00/t ns o m n ax . $1,100/ton 50 tons max.

Management now wants to determine the most economical plan for shipping the iron ore from the mines through the distribution network to the steel plant. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

100 tons needed

n 0/to $80 max. s ton 70

50 M2

400 units 500 units

A decision now needs to be made about the shipping plan for how many units to ship from each factory to each customer. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

P

60 tons produced

Output

S2

3.4-13.* Al Ferris has $60,000 that he wishes to invest now in order to use the accumulation for purchasing a retirement annuity in 5 years. After consulting with his financial adviser, he has been offered four types of fixed-income investments, which we will label as investments A, B, C, D.

CHAPTER 3 PROBLEMS

97

Investments A and B are available at the beginning of each of the next 5 years (call them years 1 to 5). Each dollar invested in A at the beginning of a year returns $1.40 (a profit of $0.40) 2 years later (in time for immediate reinvestment). Each dollar invested in B at the beginning of a year returns $1.70 three years later. Investments C and D will each be available at one time in the future. Each dollar invested in C at the beginning of year 2 returns $1.90 at the end of year 5. Each dollar invested in D at the beginning of year 5 returns $1.30 at the end of year 5. Al wishes to know which investment plan maximizes the amount of money that can be accumulated by the beginning of year 6. (a) All the functional constraints for this problem can be expressed as equality constraints. To do this, let At, Bt, Ct, and Dt be the amount invested in investment A, B, C, and D, respectively, at the beginning of year t for each t where the investment is available and will mature by the end of year 5. Also let Rt be the number of available dollars not invested at the beginning of year t (and so available for investment in a later year). Thus, the amount invested at the beginning of year t plus Rt must equal the number of dollars available for investment at that time. Write such an equation in terms of the relevant variables above for the beginning of each of the 5 years to obtain the five functional constraints for this problem. (b) Formulate a complete linear programming model for this problem. C (c) Solve this model by the simplex model. 3.4-14. The Metalco Company desires to blend a new alloy of 40 percent tin, 35 percent zinc, and 25 percent lead from several available alloys having the following properties: Alloy Property

1

2

3

4

5

Percentage of tin Percentage of zinc Percentage of lead

60 10 30

25 15 60

45 45 10

20 50 30

50 40 10

Cost ($/lb)

22

20

25

24

27

The objective is to determine the proportions of these alloys that should be blended to produce the new alloy at a minimum cost. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-15. The Weigelt Corporation has three branch plants with excess production capacity. Fortunately, the corporation has a new product ready to begin production, and all three plants have this capability, so some of the excess capacity can be used in this way. This product can be made in three sizes—large, medium, and

small—that yield a net unit profit of $420, $360, and $300, respectively. Plants 1, 2, and 3 have the excess capacity to produce 750, 900, and 450 units per day of this product, respectively, regardless of the size or combination of sizes involved. The amount of available in-process storage space also imposes a limitation on the production rates of the new product. Plants 1, 2, and 3 have 13,000, 12,000, and 5,000 square feet, respectively, of in-process storage space available for a day’s production of this product. Each unit of the large, medium, and small sizes produced per day requires 20, 15, and 12 square feet, respectively. Sales forecasts indicate that if available, 900, 1,200, and 750 units of the large, medium, and small sizes, respectively, would be sold per day. At each plant, some employees will need to be laid off unless most of the plant’s excess production capacity can be used to produce the new product. To avoid layoffs if possible, management has decided that the plants should use the same percentage of their excess capacity to produce the new product. Management wishes to know how much of each of the sizes should be produced by each of the plants to maximize profit. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-16* A cargo plane has three compartments for storing cargo: front, center, and back. These compartments have capacity limits on both weight and space, as summarized below:

Compartment Front Center Back

Weight Capacity (Tons)

Space Capacity (Cubic Feet)

12 18 10

7,000 9,000 5,000

Furthermore, the weight of the cargo in the respective compartments must be the same proportion of that compartment’s weight capacity to maintain the balance of the airplane. The following four cargoes have been offered for shipment on an upcoming flight as space is available:

Cargo

Weight (Tons)

Volume (Cubic Feet/Ton)

Profit ($/Ton)

1 2 3 4

20 16 25 13

500 700 600 400

320 400 360 290

Any portion of these cargoes can be accepted. The objective is to determine how much (if any) of each cargo should be accepted and

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how to distribute each among the compartments to maximize the total profit for the flight. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method to find one of its multiple optimal solutions. 3.4-17. Comfortable Hands is a company which features a product line of winter gloves for the entire family—men, women, and children. They are trying to decide what mix of these three types of gloves to produce. Comfortable Hands’ manufacturing labor force is unionized. Each full-time employee works a 40-hour week. In addition, by union contract, the number of full-time employees can never drop below 20. Nonunion part-time workers can also be hired with the following union-imposed restrictions: (1) each part-time worker works 20 hours per week, and (2) there must be at least 2 full-time employees for each part-time employee. All three types of gloves are made out of the same 100 percent genuine cowhide leather. Comfortable Hands has a long-term contract with a supplier of the leather, and receives a 5,000 square feet shipment of the material each week. The material requirements and labor requirements, along with the gross profit per glove sold (not considering labor costs) is given in the following table.

Glove Men’s Women’s Children’s

Material Required Labor Required Gross Profit (Square Feet) (Minutes) (per Pair) 2 1.5 1

30 45 40

$8 $10 $6

Maximum Hours of Availability Operators

Wage Rate

Mon.

Tue.

Wed.

Thurs.

Fri.

K. C. D. H. H. B. S. C. K. S. N. K.

$10.00/hour $10.10/hour $ 9.90/hour $ 9.80/hour $10.80/hour $11.30/hour

6 0 4 5 3 0

0 6 8 5 0 0

6 0 4 5 3 0

0 6 0 0 8 6

6 0 4 5 0 2

There are six operators (four undergraduate students and two graduate students). They all have different wage rates because of differences in their experience with computers and in their programming ability. The above table shows their wage rates, along with the maximum number of hours that each can work each day. Each operator is guaranteed a certain minimum number of hours per week that will maintain an adequate knowledge of the operation. This level is set arbitrarily at 8 hours per week for the undergraduate students (K. C., D. H., H. B., and S. C.) and 7 hours per week for the graduate students (K. S. and N. K.). The computer facility is to be open for operation from 8 A.M. to 10 P.M. Monday through Friday with exactly one operator on duty during these hours. On Saturdays and Sundays, the computer is to be operated by other staff. Because of a tight budget, Beryl has to minimize cost. She wishes to determine the number of hours she should assign to each operator on each day. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

Each full-time employee earns $13 per hour, while each parttime employee earns $10 per hour. Management wishes to know what mix of each of the three types of gloves to produce per week, as well as how many full-time and how many part-time workers to employ. They would like to maximize their net profit—their gross profit from sales minus their labor costs. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

3.4-19. Slim-Down Manufacturing makes a line of nutritionally complete, weight-reduction beverages. One of their products is a strawberry shake which is designed to be a complete meal. The strawberry shake consists of several ingredients. Some information about each of these ingredients is given below.

3.4-18. Oxbridge University maintains a powerful mainframe computer for research use by its faculty, Ph.D. students, and research associates. During all working hours, an operator must be available to operate and maintain the computer, as well as to perform some programming services. Beryl Ingram, the director of the computer facility, oversees the operation. It is now the beginning of the fall semester, and Beryl is confronted with the problem of assigning different working hours to her operators. Because all the operators are currently enrolled in the university, they are available to work only a limited number of hours each day, as shown in the following table.

Ingredient Strawberry flavoring Cream Vitamin supplement Artificial sweetener Thickening agent

Calories Total Vitamin from Fat Calories Content Thickeners Cost (per (per (mg/ (mg/ (¢/ tbsp) tbsp) tbsp) tbsp) tbsp)

1 75

50 100

20 0

3 8

10 8

0

0

50

1

25

0

120

0

2

15

30

80

2

25

6

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The nutritional requirements are as follows. The beverage must total between 380 and 420 calories (inclusive). No more than 20 percent of the total calories should come from fat. There must be at least 50 milligrams (mg) of vitamin content. For taste reasons, there must be at least 2 tablespoons (tbsp) of strawberry flavoring for each tablespoon of artificial sweetener. Finally, to maintain proper thickness, there must be exactly 15 mg of thickeners in the beverage. Management would like to select the quantity of each ingredient for the beverage which would minimize cost while meeting the above requirements. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-20. Joyce and Marvin run a day care for preschoolers. They are trying to decide what to feed the children for lunches. They would like to keep their costs down, but also need to meet the nutritional requirements of the children. They have already decided to go with peanut butter and jelly sandwiches, and some combination of graham crackers, milk, and orange juice. The nutritional content of each food choice and its cost are given in the table below.

Food Item Bread (1 slice) Peanut butter (1 tbsp) Strawberry jelly (1 tbsp) Graham cracker (1 cracker) Milk (1 cup) Juice (1 cup)

Calories Total Vitamin C Protein Cost from Fat Calories (mg) (g) (¢) 10

70

0

3

5

75

100

0

4

4

0

50

3

0

7

20 70 0

60 150 100

0 2 120

1 8 1

8 15 35

The nutritional requirements are as follows. Each child should receive between 400 and 600 calories. No more than 30 percent of the total calories should come from fat. Each child should consume at least 60 milligrams (mg) of vitamin C and 12 grams (g) of protein. Furthermore, for practical reasons, each child needs exactly 2 slices of bread (to make the sandwich), at least twice as much peanut butter as jelly, and at least 1 cup of liquid (milk and/or juice). Joyce and Marvin would like to select the food choices for each child which minimize cost while meeting the above requirements. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.5-1. Read the article footnoted in Sec. 3.5 that describes the first case study presented in that section: “Choosing the Product Mix at Ponderosa Industrial.”

(a) Describe the two factors which, according to the article, often hinder the use of optimization models by managers. (b) Section 3.5 indicates without elaboration that using linear programming at Ponderosa “led to a dramatic shift in the types of plywood products emphasized by the company.” Identify this shift. (c) With the success of this application, management then was eager to use optimization for other problems as well. Identify these other problems. (d) Photocopy the two pages of appendixes that give the mathematical formulation of the problem and the structure of the linear programming model. 3.5-2. Read the article footnoted in Sec. 3.5 that describes the second case study presented in that section: “Personnel Scheduling at United Airlines.” (a) Describe how United Airlines prepared shift schedules at airports and reservations offices prior to this OR study. (b) When this study began, the problem definition phase defined five specific project requirements. Identify these project requirements. (c) At the end of the presentation of the corresponding example in Sec. 3.4 (personnel scheduling at Union Airways), we pointed out that the divisibility assumption does not hold for this kind of application. An integer solution is needed, but linear programming may provide an optimal solution that is noninteger. How does United Airlines deal with this problem? (d) Describe the flexibility built into the scheduling system to satisfy the group culture at each office. Why was this flexibility needed? (e) Briefly describe the tangible and intangible benefits that resulted from the study. 3.5-3. Read the 1986 article footnoted in Sec. 2.1 that describes the third case study presented in Sec. 3.5: “Planning Supply, Distribution, and Marketing at Citgo Petroleum Corporation.” (a) What happened during the years preceding this OR study that made it vastly more important to control the amount of capital tied up in inventory? (b) What geographical area is spanned by Citgo’s distribution network of pipelines, tankers, and barges? Where do they market their products? (c) What time periods are included in the model? (d) Which computer did Citgo use to solve the model? What were typical run times? (e) Who are the four types of model users? How does each one use the model? (f) List the major types of reports generated by the SDM system. (g) What were the major implementation challenges for this study? (h) List the direct and indirect benefits that were realized from this study.

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3.6-1.* You are given the following data for a linear programming problem where the objective is to maximize the profit from allocating three resources to two nonnegative activities. Resource Usage per Unit of Each Activity Resource

Activity 1

Activity 2

Amount of Resource Available

1 2 3

2 3 2

1 3 4

10 20 20

Contribution per unit

$20

$30

(c) Make three guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, to find the value of the objective function. Which feasible guess has the best objective function value? (d) Use the Excel Solver to solve the model by the simplex method. 3.6-3. You are given the following data for a linear programming problem where the objective is to minimize the cost of conducting two nonnegative activities so as to achieve three benefits that do not fall below their minimum levels. Benefit Contribution per Unit of Each Activity Benefit

Contribution per unit profit per unit of the activity.

(a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. (c) Display the model on an Excel spreadsheet. (d) Use the spreadsheet to check the following solutions: (x1, x2) (2, 2), (3, 3), (2, 4), (4, 2), (3, 4), (4, 3). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? C (e) Use the Excel Solver to solve the model by the simplex method. 3.6-2. Ed Butler is the production manager for the Bilco Corporation, which produces three types of spare parts for automobiles. The manufacture of each part requires processing on each of two machines, with the following processing times (in hours): Part Machine

A

B

C

1 2

0.02 0.05

0.03 0.02

0.05 0.04

Each machine is available 40 hours per month. Each part manufactured will yield a unit profit as follows:

Activity 1

1 2 3 Unit cost

Activity 2

5 2 7

3 2 9

$60

$50

Minimum Acceptable Level 60 30 126

(a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. (c) Display the model on an Excel spreadsheet. (d) Use the spreadsheet to check the following solutions: (x1, x2) (7, 7), (7, 8), (8, 7), (8, 8), (8, 9), (9, 8). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? C (e) Use the Excel Solver to solve this model by the simplex method. 3.6-4.* Fred Jonasson manages a family-owned farm. To supplement several food products grown on the farm, Fred also raises pigs for market. He now wishes to determine the quantities of the available types of feed (corn, tankage, and alfalfa) that should be given to each pig. Since pigs will eat any mix of these feed types, the objective is to determine which mix will meet certain nutritional requirements at a minimum cost. The number of units of each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table, along with the daily nutritional requirements and feed costs:

Part

Profit

A

B

C

$50

$40

$30

Ed wants to determine the mix of spare parts to produce in order to maximize total profit. (a) Formulate a linear programming model for this problem. (b) Display the model on an Excel spreadsheet.

Nutritional Ingredient

Kilogram Kilogram Kilogram Minimum of of of Daily Corn Tankage Alfalfa Requirement

Carbohydrates Protein Vitamins

90 30 10

20 80 20

40 60 60

Cost (¢)

84

72

60

200 180 150

CHAPTER 3 PROBLEMS

(a) Formulate a linear programming model for this problem. (b) Display the model on an Excel spreadsheet. (c) Use the spreadsheet to check if (x1, x2, x3) (1, 2, 2) is a feasible solution and, if so, what the daily cost would be for this diet. How many units of each nutritional ingredient would this diet provide daily? (d) Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the daily cost for your solution? C (e) Use the Excel Solver to solve the model by the simplex method. 3.6-5. Maureen Laird is the chief financial officer for the Alva Electric Co., a major public utility in the midwest. The company has scheduled the construction of new hydroelectric plants 5, 10, and 20 years from now to meet the needs of the growing population in the region served by the company. To cover at least the construction costs, Maureen needs to invest some of the company’s money now to meet these future cash-flow needs. Maureen may purchase only three kinds of financial assets, each of which costs $1 million per unit. Fractional units may be purchased. The assets produce income 5, 10, and 20 years from now, and that income is needed to cover at least minimum cash-flow requirements in those years. (Any excess income above the minimum requirement for each time period will be used to increase dividend payments to shareholders rather than saving it to help meet the minimum cash-flow requirement in the next time period.) The following table shows both the amount of income generated by each unit of each asset and the minimum amount of income needed for each of the future time periods when a new hydroelectric plant will be constructed.

101

(d) Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the total amount invested for your solution? C (e) Use the Excel Solver to solve the model by the simplex method. 3.7-1. The Philbrick Company has two plants on opposite sides of the United States. Each of these plants produces the same two products and then sells them to wholesalers within its half of the country. The orders from wholesalers have already been received for the next 2 months (February and March), where the number of units requested are shown below. (The company is not obligated to completely fill these orders but will do so if it can without decreasing its profits.) Plant 1

Plant 2

Product

February

March

February

March

1 2

3,600 4,500

6,300 5,400

4,900 5,100

4,200 6,000

Each plant has 20 production days available in February and 23 production days available in March to produce and ship these products. Inventories are depleted at the end of January, but each plant has enough inventory capacity to hold 1,000 units total of the two products if an excess amount is produced in February for sale in March. In either plant, the cost of holding inventory in this way is $3 per unit of product 1 and $4 per unit of product 2. Each plant has the same two production processes, each of which can be used to produce either of the two products. The production cost per unit produced of each product is shown below for each process in each plant.

Income per Unit of Asset Year 5 10 20

Asset 1

Asset 2

Asset 3

$2 million $1 million $0.5 million $0.5 million $0.5 million $1 million 0 $1.5 million $2 million

Minimum Cash Flow Required $400 million $100 million $300 million

Maureen wishes to determine the mix of investments in these assets that will cover the cash-flow requirements while minimizing the total amount invested. (a) Formulate a linear programming model for this problem. (b) Display the model on a spreadsheet. (c) Use the spreadsheet to check the possibility of purchasing 100 units of Asset 1, 100 units of Asset 2, and 200 units of Asset 3. How much cash flow would this mix of investments generate 5, 10, and 20 years from now? What would be the total amount invested?

Plant 1

Plant 2

Product

Process 1

Process 2

Process 1

Process 2

1 2

$62 $78

$59 $85

$61 $89

$65 $86

The production rate for each product (number of units produced per day devoted to that product) also is given below for each process in each plant. Plant 1

Plant 2

Product

Process 1

Process 2

Process 1

Process 2

1 2

100 120

140 150

130 160

110 130

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The net sales revenue (selling price minus normal shipping costs) the company receives when a plant sells the products to its own customers (the wholesalers in its half of the country) is $83 per unit of product 1 and $112 per unit of product 2. However, it also is possible (and occasionally desirable) for a plant to make a shipment to the other half of the country to help fill the sales of the other plant. When this happens, an extra shipping cost of $9 per unit of product 1 and $7 per unit of product 2 is incurred. Management now needs to determine how much of each product should be produced by each production process in each plant during each month, as well as how much each plant should sell of each product in each month and how much each plant should ship of each product in each month to the other plant’s customers. The objective is to determine which feasible plan would maximize the total profit (total net sales revenue minus the sum of the production costs, inventory costs, and extra shipping costs). (a) Formulate a complete linear programming model in algebraic form that shows the individual constraints and decision variables for this problem. C (b) Formulate this same model on an Excel spreadsheet instead. Then use the Excel Solver to solve the model. C (c) Use MPL to formulate this model in a compact form. Then use the MPL solver CPLEX to solve the model. C (d) Use LINGO to formulate this model in a compact form. Then use the LINGO solver to solve the model. 3.7-2. Reconsider Prob. 3.1-11. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-3. Reconsider Prob. 3.4-11. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-4. Reconsider Prob. 3.4-15. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-5. Reconsider Prob. 3.4-18. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-6. Reconsider Prob. 3.6-4. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-7. Reconsider Prob. 3.6-5. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-8. A large paper manufacturing company, the Quality Paper Corporation, has 10 paper mills from which it needs to supply 1,000 customers. It uses three alternative types of machines and four types of raw materials to make five different types of paper. Therefore, the company needs to develop a detailed production distribution plan on a monthly basis, with an objective of minimizing the total cost of producing and distributing the paper during the month. Specifically, it is necessary to determine jointly the amount of each type of paper to be made at each paper mill on each type of machine and the amount of each type of paper to be shipped from each paper mill to each customer. The relevant data can be expressed symbolically as follows: Djk number of units of paper type k demanded by customer j, rklm number of units of raw material m needed to produce 1 unit of paper type k on machine type l, Rim number of units of raw material m available at paper mill i, ckl number of capacity units of machine type l that will produce 1 unit of paper type k, Cil number of capacity units of machine type l available at paper mill i, Pikl production cost for each unit of paper type k produced on machine type l at paper mill i, Tijk transportation cost for each unit of paper type k shipped from paper mill i to customer j. (a) Using these symbols, formulate a linear programming model for this problem by hand. (b) How many functional constraints and decision variables does this model have? C (c) Use MPL to formulate this problem. C (d) Use LINGO to formulate this problem.

CASE 3.1

CASE 3.1

AUTO ASSEMBLY

103

AUTO ASSEMBLY Automobile Alliance, a large automobile manufacturing company, organizes the vehicles it manufactures into three families: a family of trucks, a family of small cars, and a family of midsized and luxury cars. One plant outside Detroit, MI, assembles two models from the family of midsized and luxury cars. The first model, the Family Thrillseeker, is a four-door sedan with vinyl seats, plastic interior, standard features, and excellent gas mileage. It is marketed as a smart buy for middle-class families with tight budgets, and each Family Thrillseeker sold generates a modest profit of $3,600 for the company. The second model, the Classy Cruiser, is a two-door luxury sedan with leather seats, wooden interior, custom features, and navigational capabilities. It is marketed as a privilege of affluence for upper-middle-class families, and each Classy Cruiser sold generates a healthy profit of $5,400 for the company. Rachel Rosencrantz, the manager of the assembly plant, is currently deciding the production schedule for the next month. Specifically, she must decide how many Family Thrillseekers and how many Classy Cruisers to assemble in the plant to maximize profit for the company. She knows that the plant possesses a capacity of 48,000 laborhours during the month. She also knows that it takes 6 labor-hours to assemble one Family Thrillseeker and 10.5 labor-hours to assemble one Classy Cruiser. Because the plant is simply an assembly plant, the parts required to assemble the two models are not produced at the plant. They are instead shipped from other plants around the Michigan area to the assembly plant. For example, tires, steering wheels, windows, seats, and doors all arrive from various supplier plants. For the next month, Rachel knows that she will be able to obtain only 20,000 doors (10,000 left-hand doors and 10,000 right-hand doors) from the door supplier. A recent labor strike forced the shutdown of that particular supplier plant for several days, and that plant will not be able to meet its production schedule for the next month. Both the Family Thrillseeker and the Classy Cruiser use the same door part. In addition, a recent company forecast of the monthly demands for different automobile models suggests that the demand for the Classy Cruiser is limited to 3,500 cars. There is no limit on the demand for the Family Thrillseeker within the capacity limits of the assembly plant. (a) Formulate and solve a linear programming problem to determine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled.

Before she makes her final production decisions, Rachel plans to explore the following questions independently except where otherwise indicated. (b) The marketing department knows that it can pursue a targeted $500,000 advertising campaign that will raise the demand for the Classy Cruiser next month by 20 percent. Should the campaign be undertaken? (c) Rachel knows that she can increase next month’s plant capacity by using overtime labor. She can increase the plant’s labor-hour capacity by 25 percent. With the new assembly plant capacity, how many Family Thrillseekers and how many Classy Cruisers should be assembled? (d) Rachel knows that overtime labor does not come without an extra cost. What is the maximum amount she should be willing to pay for all overtime labor beyond the cost of this labor at regular time rates? Express your answer as a lump sum.

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(e) Rachel explores the option of using both the targeted advertising campaign and the overtime labor-hours. The advertising campaign raises the demand for the Classy Cruiser by 20 percent, and the overtime labor increases the plant’s labor-hour capacity by 25 percent. How many Family Thrillseekers and how many Classy Cruisers should be assembled using the advertising campaign and overtime labor-hours if the profit from each Classy Cruiser sold continues to be 50 percent more than for each Family Thrillseeker sold? (f) Knowing that the advertising campaign costs $500,000 and the maximum usage of overtime labor-hours costs $1,600,000 beyond regular time rates, is the solution found in part (e) a wise decision compared to the solution found in part (a)? (g) Automobile Alliance has determined that dealerships are actually heavily discounting the price of the Family Thrillseekers to move them off the lot. Because of a profit-sharing agreement with its dealers, the company is therefore not making a profit of $3,600 on the Family Thrillseeker but is instead making a profit of $2,800. Determine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled given this new discounted price. (h) The company has discovered quality problems with the Family Thrillseeker by randomly testing Thrillseekers at the end of the assembly line. Inspectors have discovered that in over 60 percent of the cases, two of the four doors on a Thrillseeker do not seal properly. Because the percentage of defective Thrillseekers determined by the random testing is so high, the floor supervisor has decided to perform quality control tests on every Thrillseeker at the end of the line. Because of the added tests, the time it takes to assemble one Family Thrillseeker has increased from 6 to 7.5 hours. Determine the number of units of each model that should be assembled given the new assembly time for the Family Thrillseeker. (i) The board of directors of Automobile Alliance wishes to capture a larger share of the luxury sedan market and therefore would like to meet the full demand for Classy Cruisers. They ask Rachel to determine by how much the profit of her assembly plant would decrease as compared to the profit found in part (a). They then ask her to meet the full demand for Classy Cruisers if the decrease in profit is not more than $2,000,000. (j) Rachel now makes her final decision by combining all the new considerations described in parts ( f ), (g), and (h). What are her final decisions on whether to undertake the advertising campaign, whether to use overtime labor, the number of Family Thrillseekers to assemble, and the number of Classy Cruisers to assemble?

CASE 3.2

CUTTING CAFETERIA COSTS A cafeteria at All-State University has one special dish it serves like clockwork every Thursday at noon. This supposedly tasty dish is a casserole that contains sautéed onions, boiled sliced potatoes, green beans, and cream of mushroom soup. Unfortunately, students fail to see the special quality of this dish, and they loathingly refer to it as the Killer Casserole. The students reluctantly eat the casserole, however, because the cafeteria provides only a limited selection of dishes for Thursday’s lunch (namely, the casserole). Maria Gonzalez, the cafeteria manager, is looking to cut costs for the coming year, and she believes that one sure way to cut costs is to buy less expensive and perhaps lower-quality ingredients. Because the casserole is a weekly staple of the cafeteria menu, she concludes that if she can cut costs on the ingredients purchased for the casserole, she can significantly reduce overall cafeteria operating costs. She therefore de-

CASE 3.2

CUTTING CAFETERIA COSTS

105

cides to invest time in determining how to minimize the costs of the casserole while maintaining nutritional and taste requirements. Maria focuses on reducing the costs of the two main ingredients in the casserole, the potatoes and green beans. These two ingredients are responsible for the greatest costs, nutritional content, and taste of the dish. Maria buys the potatoes and green beans from a wholesaler each week. Potatoes cost $0.40 per pound, and green beans cost $1.00 per pound. All-State University has established nutritional requirements that each main dish of the cafeteria must meet. Specifically, the total amount of the dish prepared for all the students for one meal must contain 180 grams (g) of protein, 80 milligrams (mg) of iron, and 1,050 mg of vitamin C. (There are 453.6 g in 1 lb and 1,000 mg in 1 g.) For simplicity when planning, Maria assumes that only the potatoes and green beans contribute to the nutritional content of the casserole. Because Maria works at a cutting-edge technological university, she has been exposed to the numerous resources on the World Wide Web. She decides to surf the Web to find the nutritional content of potatoes and green beans. Her research yields the following nutritional information about the two ingredients:

Protein Iron Vitamin C

Potatoes

Green Beans

1.5 g per 100 g 0.3 mg per 100 g 12 mg per 100 g

5.67 g per 10 ounces 3.402 mg per 10 ounces 28.35 mg per 10 ounces

(There are 28.35 g in 1 ounce.)

Edson Branner, the cafeteria cook who is surprisingly concerned about taste, informs Maria that an edible casserole must contain at least a six to five ratio in the weight of potatoes to green beans. Given the number of students who eat in the cafeteria, Maria knows that she must purchase enough potatoes and green beans to prepare a minimum of 10 kilograms (kg) of casserole each week. (There are 1,000 g in 1 kg.) Again for simplicity in planning, she assumes that only the potatoes and green beans determine the amount of casserole that can be prepared. Maria does not establish an upper limit on the amount of casserole to prepare, since she knows all leftovers can be served for many days thereafter or can be used creatively in preparing other dishes. (a) Determine the amount of potatoes and green beans Maria should purchase each week for the casserole to minimize the ingredient costs while meeting nutritional, taste, and demand requirements.

Before she makes her final decision, Maria plans to explore the following questions independently except where otherwise indicated. (b) Maria is not very concerned about the taste of the casserole; she is only concerned about meeting nutritional requirements and cutting costs. She therefore forces Edson to change the recipe to allow for only at least a one to two ratio in the weight of potatoes to green beans. Given the new recipe, determine the amount of potatoes and green beans Maria should purchase each week.

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(c) Maria decides to lower the iron requirement to 65 mg since she determines that the other ingredients, such as the onions and cream of mushroom soup, also provide iron. Determine the amount of potatoes and green beans Maria should purchase each week given this new iron requirement. (d) Maria learns that the wholesaler has a surplus of green beans and is therefore selling the green beans for a lower price of $0.50 per lb. Using the same iron requirement from part (c) and the new price of green beans, determine the amount of potatoes and green beans Maria should purchase each week. (e) Maria decides that she wants to purchase lima beans instead of green beans since lima beans are less expensive and provide a greater amount of protein and iron than green beans. Maria again wields her absolute power and forces Edson to change the recipe to include lima beans instead of green beans. Maria knows she can purchase lima beans for $0.60 per lb from the wholesaler. She also knows that lima beans contain 22.68 g of protein per 10 ounces of lima beans, 6.804 mg of iron per 10 ounces of lima beans, and no vitamin C. Using the new cost and nutritional content of lima beans, determine the amount of potatoes and lima beans Maria should purchase each week to minimize the ingredient costs while meeting nutritional, taste, and demand requirements. The nutritional requirements include the reduced iron requirement from part (c). (f) Will Edson be happy with the solution in part (e)? Why or why not? (g) An All-State student task force meets during Body Awareness Week and determines that AllState University’s nutritional requirements for iron are too lax and that those for vitamin C are too stringent. The task force urges the university to adopt a policy that requires each serving of an entrée to contain at least 120 mg of iron and at least 500 mg of vitamin C. Using potatoes and lima beans as the ingredients for the dish and using the new nutritional requirements, determine the amount of potatoes and lima beans Maria should purchase each week.

CASE 3.3

STAFFING A CALL CENTER1 California Children’s Hospital has been receiving numerous customer complaints because of its confusing, decentralized appointment and registration process. When customers want to make appointments or register child patients, they must contact the clinic or department they plan to visit. Several problems exist with this current strategy. Parents do not always know the most appropriate clinic or department they must visit to address their children’s ailments. They therefore spend a significant amount of time on the phone being transferred from clinic to clinic until they reach the most appropriate clinic for their needs. The hospital also does not publish the phone numbers of all clinic and departments, and parents must therefore invest a large amount of time in detective work to track down the correct phone number. Finally, the various clinics and departments do not communicate with each other. For example, when a doctor schedules a referral with a colleague located in another department or clinic, that department or clinic almost never receives word of the referral. The parent must contact the correct department or clinic and provide the needed referral information. 1

This case is based on an actual project completed by a team of master’s students in the Department of Engineering-Economic Systems and Operations Research at Stanford University.

CASE 3.3

STAFFING A CALL CENTER

107

In efforts to reengineer and improve its appointment and registration process, the children’s hospital has decided to centralize the process by establishing one call center devoted exclusively to appointments and registration. The hospital is currently in the middle of the planning stages for the call center. Lenny Davis, the hospital manager, plans to operate the call center from 7 A.M. to 9 P.M. during the weekdays. Several months ago, the hospital hired an ambitious management consulting firm, Creative Chaos Consultants, to forecast the number of calls the call center would receive each hour of the day. Since all appointment and registration-related calls would be received by the call center, the consultants decided that they could forecast the calls at the call center by totaling the number of appointment and registration-related calls received by all clinics and departments. The team members visited all the clinics and departments, where they diligently recorded every call relating to appointments and registration. They then totaled these calls and altered the totals to account for calls missed during data collection. They also altered totals to account for repeat calls that occurred when the same parent called the hospital many times because of the confusion surrounding the decentralized process. Creative Chaos Consultants determined the average number of calls the call center should expect during each hour of a weekday. The following table provides the forecasts. Work Shift 7 9 11 1 3 5 7

A.M.–9 A.M. A.M.–11 A.M. A.M.–1 P.M. P.M.–3 P.M. P.M.–5 P.M. P.M.–7 P.M. P.M.–9 P.M.

Average Number of Calls 40 85 70 95 80 35 10

calls calls calls calls calls calls calls

per per per per per per per

hour hour hour hour hour hour hour

After the consultants submitted these forecasts, Lenny became interested in the percentage of calls from Spanish speakers since the hospital services many Spanish patients. Lenny knows that he has to hire some operators who speak Spanish to handle these calls. The consultants performed further data collection and determined that on average, 20 percent of the calls were from Spanish speakers. Given these call forecasts, Lenny must now decide how to staff the call center during each 2 hour shift of a weekday. During the forecasting project, Creative Chaos Consultants closely observed the operators working at the individual clinics and departments and determined the number of calls operators process per hour. The consultants informed Lenny that an operator is able to process an average of six calls per hour. Lenny also knows that he has both full-time and part-time workers available to staff the call center. A full-time employee works 8 hours per day, but because of paperwork that must also be completed, the employee spends only 4 hours per day on the phone. To balance the schedule, the employee alternates the 2-hour shifts between answering phones and completing paperwork. Full-time employees can start their day either by answering phones or by completing paperwork on the first shift. The full-time em-

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ployees speak either Spanish or English, but none of them are bilingual. Both Spanish-speaking and English-speaking employees are paid $10 per hour for work before 5 P.M. and $12 per hour for work after 5 P.M. The full-time employees can begin work at the beginning of the 7 A.M. to 9 A.M. shift, 9 A.M. to 11 A.M. shift, 11 A.M. to 1 P.M. shift, or 1 P.M. to 3 P.M. shift. The part-time employees work for 4 hours, only answer calls, and only speak English. They can start work at the beginning of the 3 P.M. to 5 P.M. shift or the 5 P.M. to 7 P.M. shift, and like the full-time employees, they are paid $10 per hour for work before 5 P.M. and $12 per hour for work after 5 P.M. For the following analysis consider only the labor cost for the time employees spend answering phones. The cost for paperwork time is charged to other cost centers. (a) How many Spanish-speaking operators and how many English-speaking operators does the hospital need to staff the call center during each 2-hour shift of the day in order to answer all calls? Please provide an integer number since half a human operator makes no sense. (b) Lenny needs to determine how many full-time employees who speak Spanish, full-time employees who speak English, and part-time employees he should hire to begin on each shift. Creative Chaos Consultants advise him that linear programming can be used to do this in such a way as to minimize operating costs while answering all calls. Formulate a linear programming model of this problem. (c) Obtain an optimal solution for the linear programming model formulated in part (b) to guide Lenny’s decision. (d) Because many full-time workers do not want to work late into the evening, Lenny can find only one qualified English-speaking operator willing to begin work at 1 P.M. Given this new constraint, how many full-time English-speaking operators, full-time Spanish-speaking operators, and part-time operators should Lenny hire for each shift to minimize operating costs while answering all calls? (e) Lenny now has decided to investigate the option of hiring bilingual operators instead of monolingual operators. If all the operators are bilingual, how many operators should be working during each 2-hour shift to answer all phone calls? As in part (a), please provide an integer answer. (f) If all employees are bilingual, how many full-time and part-time employees should Lenny hire to begin on each shift to minimize operating costs while answering all calls? As in part (b), formulate a linear programming model to guide Lenny’s decision. (g) What is the maximum percentage increase in the hourly wage rate that Lenny can pay bilingual employees over monolingual employees without increasing the total operating costs? (h) What other features of the call center should Lenny explore to improve service or minimize operating costs?

4 Solving Linear Programming Problems: The Simplex Method We now are ready to begin studying the simplex method, a general procedure for solving linear programming problems. Developed by George Dantzig in 1947, it has proved to be a remarkably efficient method that is used routinely to solve huge problems on today’s computers. Except for its use on tiny problems, this method is always executed on a computer, and sophisticated software packages are widely available. Extensions and variations of the simplex method also are used to perform postoptimality analysis (including sensitivity analysis) on the model. This chapter describes and illustrates the main features of the simplex method. The first section introduces its general nature, including its geometric interpretation. The following three sections then develop the procedure for solving any linear programming model that is in our standard form (maximization, all functional constraints in form, and nonnegativity constraints on all variables) and has only nonnegative right-hand sides bi in the functional constraints. Certain details on resolving ties are deferred to Sec. 4.5, and Sec. 4.6 describes how to adapt the simplex method to other model forms. Next we discuss postoptimality analysis (Sec. 4.7), and describe the computer implementation of the simplex method (Sec. 4.8). Section 4.9 then introduces an alternative to the simplex method (the interior-point approach) for solving large linear programming problems.

4.1

THE ESSENCE OF THE SIMPLEX METHOD The simplex method is an algebraic procedure. However, its underlying concepts are geometric. Understanding these geometric concepts provides a strong intuitive feeling for how the simplex method operates and what makes it so efficient. Therefore, before delving into algebraic details, we focus in this section on the big picture from a geometric viewpoint. To illustrate the general geometric concepts, we shall use the Wyndor Glass Co. example presented in Sec. 3.1. (Sections 4.2 and 4.3 use the algebra of the simplex method to solve this same example.) Section 5.1 will elaborate further on these geometric concepts for larger problems. To refresh your memory, the model and graph for this example are repeated in Fig. 4.1. The five constraint boundaries and their points of intersection are highlighted in this figure because they are the keys to the analysis. Here, each constraint boundary is a line that forms the boundary of what is permitted by the corresponding constraint. The points 109

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4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

x2

Maximize Z 3x1 5x2, subject to 4 x1 2x2 12 3x1 2x2 18 and x1 0, x2 0

x1 0 (0, 9) 3x1 2x2 18

(0, 6)

(2, 6)

(4, 6)

2x2 12

x1 4

Feasible region

FIGURE 4.1 Constraint boundaries and corner-point solutions for the Wyndor Glass Co. problem.

(4, 3)

x2 0

(0, 0) (4, 0)

(6, 0)

x1

of intersection are the corner-point solutions of the problem. The five that lie on the corners of the feasible region—(0, 0), (0, 6), (2, 6), (4, 3), and (4, 0)—are the corner-point feasible solutions (CPF solutions). [The other three—(0, 9), (4, 6), and (6, 0)—are called corner-point infeasible solutions.] In this example, each corner-point solution lies at the intersection of two constraint boundaries. (For a linear programming problem with n decision variables, each of its corner-point solutions lies at the intersection of n constraint boundaries.1) Certain pairs of the CPF solutions in Fig. 4.1 share a constraint boundary, and other pairs do not. It will be important to distinguish between these cases by using the following general definitions. For any linear programming problem with n decision variables, two CPF solutions are adjacent to each other if they share n 1 constraint boundaries. The two adjacent CPF solutions are connected by a line segment that lies on these same shared constraint boundaries. Such a line segment is referred to as an edge of the feasible region.

Since n 2 in the example, two of its CPF solutions are adjacent if they share one constraint boundary; for example, (0, 0) and (0, 6) are adjacent because they share the x1 0 constraint boundary. The feasible region in Fig. 4.1 has five edges, consisting of the five line segments forming the boundary of this region. Note that two edges emanate from each CPF solution. Thus, each CPF solution has two adjacent CPF solutions (each lying at the other end of one of the two edges), as enumerated in Table 4.1. (In each row 1

Although a corner-point solution is defined in terms of n constraint boundaries whose intersection gives this solution, it also is possible that one or more additional constraint boundaries pass through this same point.

4.1 THE ESSENCE OF THE SIMPLEX METHOD

111

TABLE 4.1 Adjacent CPF solutions for each CPF solution of the Wyndor Glass Co. problem CPF Solution (0, (0, (2, (4, (4,

0) 6) 6) 3) 0)

Its Adjacent CPF Solutions (0, (2, (4, (4, (0,

6) 6) 3) 0) 0)

and and and and and

(4, (0, (0, (2, (4,

0) 0) 6) 6) 3)

of this table, the CPF solution in the first column is adjacent to each of the two CPF solutions in the second column, but the two CPF solutions in the second column are not adjacent to each other.) One reason for our interest in adjacent CPF solutions is the following general property about such solutions, which provides a very useful way of checking whether a CPF solution is an optimal solution. Optimality test: Consider any linear programming problem that possesses at least one optimal solution. If a CPF solution has no adjacent CPF solutions that are better (as measured by Z), then it must be an optimal solution. Thus, for the example, (2, 6) must be optimal simply because its Z 36 is larger than Z 30 for (0, 6) and Z 27 for (4, 3). (We will delve further into why this property holds in Sec. 5.1.) This optimality test is the one used by the simplex method for determining when an optimal solution has been reached. Now we are ready to apply the simplex method to the example. Solving the Example Here is an outline of what the simplex method does (from a geometric viewpoint) to solve the Wyndor Glass Co. problem. At each step, first the conclusion is stated and then the reason is given in parentheses. (Refer to Fig. 4.1 for a visualization.) Initialization: Choose (0, 0) as the initial CPF solution to examine. (This is a convenient choice because no calculations are required to identify this CPF solution.) Optimality Test: Conclude that (0, 0) is not an optimal solution. (Adjacent CPF solutions are better.) Iteration 1: Move to a better adjacent CPF solution, (0, 6), by performing the following three steps. 1. Considering the two edges of the feasible region that emanate from (0, 0), choose to move along the edge that leads up the x2 axis. (With an objective function of Z 3x1 5x2, moving up the x2 axis increases Z at a faster rate than moving along the x1 axis.) 2. Stop at the first new constraint boundary: 2x2 12. [Moving farther in the direction selected in step 1 leaves the feasible region; e.g., moving to the second new constraint boundary hit when moving in that direction gives (0, 9), which is a corner-point infeasible solution.]

112

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

3. Solve for the intersection of the new set of constraint boundaries: (0, 6). (The equations for these constraint boundaries, x1 0 and 2x2 12, immediately yield this solution.) Optimality Test: Conclude that (0, 6) is not an optimal solution. (An adjacent CPF solution is better.) Iteration 2: Move to a better adjacent CPF solution, (2, 6), by performing the following three steps. 1. Considering the two edges of the feasible region that emanate from (0, 6), choose to move along the edge that leads to the right. (Moving along this edge increases Z, whereas backtracking to move back down the x2 axis decreases Z.) 2. Stop at the first new constraint boundary encountered when moving in that direction: 3x1 2x2 12. (Moving farther in the direction selected in step 1 leaves the feasible region.) 3. Solve for the intersection of the new set of constraint boundaries: (2, 6). (The equations for these constraint boundaries, 3x1 2x2 18 and 2x2 12, immediately yield this solution.) Optimality Test: Conclude that (2, 6) is an optimal solution, so stop. (None of the adjacent CPF solutions are better.) This sequence of CPF solutions examined is shown in Fig. 4.2, where each circled number identifies which iteration obtained that solution. Now let us look at the six key solution concepts of the simplex method that provide the rationale behind the above steps. (Keep in mind that these concepts also apply for solving problems with more than two decision variables where a graph like Fig. 4.2 is not available to help quickly find an optimal solution.) The Key Solution Concepts The first solution concept is based directly on the relationship between optimal solutions and CPF solutions given at the end of Sec. 3.2. FIGURE 4.2 This graph shows the sequence of CPF solutions (, , ) examined by the simplex method for the Wyndor Glass Co. problem. The optimal solution (2, 6) is found after just three solutions are examined.

x2 (0, 6)

Z 30 (2, 6) Z 36 1 2

Feasible region

(4, 3) Z 27

Z 12

0 (0, 0)

Z0

(4, 0)

x1

4.1 THE ESSENCE OF THE SIMPLEX METHOD

113

Solution concept 1: The simplex method focuses solely on CPF solutions. For any problem with at least one optimal solution, finding one requires only finding a best CPF solution.1 Since the number of feasible solutions generally is infinite, reducing the number of solutions that need to be examined to a small finite number ( just three in Fig. 4.2) is a tremendous simplification. The next solution concept defines the flow of the simplex method. Solution concept 2: The simplex method is an iterative algorithm (a systematic solution procedure that keeps repeating a fixed series of steps, called an iteration, until a desired result has been obtained) with the following structure. →

→ Initialization: Optimality test: → If no If yes ↓ Iteration:

Set up to start iterations, including finding an initial CPF solution. Is the current CPF solution optimal? Stop. Perform an iteration to find a better CPF solution.

When the example was solved, note how this flow diagram was followed through two iterations until an optimal solution was found. We next focus on how to get started. Solution concept 3: Whenever possible, the initialization of the simplex method chooses the origin (all decision variables equal to zero) to be the initial CPF solution. When there are too many decision variables to find an initial CPF solution graphically, this choice eliminates the need to use algebraic procedures to find and solve for an initial CPF solution. Choosing the origin commonly is possible when all the decision variables have nonnegativity constraints, because the intersection of these constraint boundaries yields the origin as a corner-point solution. This solution then is a CPF solution unless it is infeasible because it violates one or more of the functional constraints. If it is infeasible, special procedures described in Sec. 4.6 are needed to find the initial CPF solution. The next solution concept concerns the choice of a better CPF solution at each iteration. Solution concept 4: Given a CPF solution, it is much quicker computationally to gather information about its adjacent CPF solutions than about other CPF solutions. Therefore, each time the simplex method performs an iteration to move from the current CPF solution to a better one, it always chooses a CPF solution that is adjacent to the current one. No other CPF solutions are considered. Consequently, the entire path followed to eventually reach an optimal solution is along the edges of the feasible region. 1

The only restriction is that the problem must possess CPF solutions. This is ensured if the feasible region is bounded.

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

The next focus is on which adjacent CPF solution to choose at each iteration. Solution concept 5: After the current CPF solution is identified, the simplex method examines each of the edges of the feasible region that emanate from this CPF solution. Each of these edges leads to an adjacent CPF solution at the other end, but the simplex method does not even take the time to solve for the adjacent CPF solution. Instead, it simply identifies the rate of improvement in Z that would be obtained by moving along the edge. Among the edges with a positive rate of improvement in Z, it then chooses to move along the one with the largest rate of improvement in Z. The iteration is completed by first solving for the adjacent CPF solution at the other end of this one edge and then relabeling this adjacent CPF solution as the current CPF solution for the optimality test and (if needed) the next iteration. At the first iteration of the example, moving from (0, 0) along the edge on the x1 axis would give a rate of improvement in Z of 3 (Z increases by 3 per unit increase in x1), whereas moving along the edge on the x2 axis would give a rate of improvement in Z of 5 (Z increases by 5 per unit increase in x2), so the decision is made to move along the latter edge. At the second iteration, the only edge emanating from (0, 6) that would yield a positive rate of improvement in Z is the edge leading to (2, 6), so the decision is made to move next along this edge. The final solution concept clarifies how the optimality test is performed efficiently. Solution concept 6: Solution concept 5 describes how the simplex method examines each of the edges of the feasible region that emanate from the current CPF solution. This examination of an edge leads to quickly identifying the rate of improvement in Z that would be obtained by moving along the edge toward the adjacent CPF solution at the other end. A positive rate of improvement in Z implies that the adjacent CPF solution is better than the current CPF solution, whereas a negative rate of improvement in Z implies that the adjacent CPF solution is worse. Therefore, the optimality test consists simply of checking whether any of the edges give a positive rate of improvement in Z. If none do, then the current CPF solution is optimal. In the example, moving along either edge from (2, 6) decreases Z. Since we want to maximize Z, this fact immediately gives the conclusion that (2, 6) is optimal.

4.2

SETTING UP THE SIMPLEX METHOD The preceding section stressed the geometric concepts that underlie the simplex method. However, this algorithm normally is run on a computer, which can follow only algebraic instructions. Therefore, it is necessary to translate the conceptually geometric procedure just described into a usable algebraic procedure. In this section, we introduce the algebraic language of the simplex method and relate it to the concepts of the preceding section. The algebraic procedure is based on solving systems of equations. Therefore, the first step in setting up the simplex method is to convert the functional inequality constraints to equivalent equality constraints. (The nonnegativity constraints are left as inequalities because they are treated separately.) This conversion is accomplished by introducing slack

4.2 SETTING UP THE SIMPLEX METHOD

115

variables. To illustrate, consider the first functional constraint in the Wyndor Glass Co. example of Sec. 3.1 x1 4. The slack variable for this constraint is defined to be x3 4 x1, which is the amount of slack in the left-hand side of the inequality. Thus, x1 x3 4. Given this equation, x1 4 if and only if 4 x1 x3 0. Therefore, the original constraint x1 4 is entirely equivalent to the pair of constraints x1 x3 4

and

x3 0.

Upon the introduction of slack variables for the other functional constraints, the original linear programming model for the example (shown below on the left) can now be replaced by the equivalent model (called the augmented form of the model) shown below on the right: Augmented Form of the Model1

Original Form of the Model Maximize

Z 3x1 5x2,

subject to

Z 3x1 5x2,

Maximize subject to

x1 2x2 4

(1)

3x1 2x2 12

(2)

2x2

3x1 2x2 18

(3)

3x1 2x2

and

x3

x1

4 x4

12 x5 18

and x1 0,

x2 0.

xj 0,

for j 1, 2, 3, 4, 5.

Although both forms of the model represent exactly the same problem, the new form is much more convenient for algebraic manipulation and for identification of CPF solutions. We call this the augmented form of the problem because the original form has been augmented by some supplementary variables needed to apply the simplex method. If a slack variable equals 0 in the current solution, then this solution lies on the constraint boundary for the corresponding functional constraint. A value greater than 0 means that the solution lies on the feasible side of this constraint boundary, whereas a value less than 0 means that the solution lies on the infeasible side of this constraint boundary. A demonstration of these properties is provided by the demonstration example in your OR Tutor entitled Interpretation of the Slack Variables. The terminology used in the preceding section (corner-point solutions, etc.) applies to the original form of the problem. We now introduce the corresponding terminology for the augmented form. An augmented solution is a solution for the original variables (the decision variables) that has been augmented by the corresponding values of the slack variables. 1

The slack variables are not shown in the objective function because the coefficients there are 0.

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

For example, augmenting the solution (3, 2) in the example yields the augmented solution (3, 2, 1, 8, 5) because the corresponding values of the slack variables are x3 1, x4 8, and x5 5. A basic solution is an augmented corner-point solution. To illustrate, consider the corner-point infeasible solution (4, 6) in Fig. 4.1. Augmenting it with the resulting values of the slack variables x3 0, x4 0, and x5 6 yields the corresponding basic solution (4, 6, 0, 0, 6). The fact that corner-point solutions (and so basic solutions) can be either feasible or infeasible implies the following definition: A basic feasible (BF) solution is an augmented CPF solution. Thus, the CPF solution (0, 6) in the example is equivalent to the BF solution (0, 6, 4, 0, 6) for the problem in augmented form. The only difference between basic solutions and corner-point solutions (or between BF solutions and CPF solutions) is whether the values of the slack variables are included. For any basic solution, the corresponding corner-point solution is obtained simply by deleting the slack variables. Therefore, the geometric and algebraic relationships between these two solutions are very close, as described in Sec. 5.1. Because the terms basic solution and basic feasible solution are very important parts of the standard vocabulary of linear programming, we now need to clarify their algebraic properties. For the augmented form of the example, notice that the system of functional constraints has 5 variables and 3 equations, so Number of variables number of equations 5 3 2. This fact gives us 2 degrees of freedom in solving the system, since any two variables can be chosen to be set equal to any arbitrary value in order to solve the three equations in terms of the remaining three variables.1 The simplex method uses zero for this arbitrary value. Thus, two of the variables (called the nonbasic variables) are set equal to zero, and then the simultaneous solution of the three equations for the other three variables (called the basic variables) is a basic solution. These properties are described in the following general definitions. A basic solution has the following properties: 1. Each variable is designated as either a nonbasic variable or a basic variable. 2. The number of basic variables equals the number of functional constraints (now equations). Therefore, the number of nonbasic variables equals the total number of variables minus the number of functional constraints. 3. The nonbasic variables are set equal to zero. 4. The values of the basic variables are obtained as the simultaneous solution of the system of equations (functional constraints in augmented form). (The set of basic variables is often referred to as the basis.) 5. If the basic variables satisfy the nonnegativity constraints, the basic solution is a BF solution. 1

This method of determining the number of degrees of freedom for a system of equations is valid as long as the system does not include any redundant equations. This condition always holds for the system of equations formed from the functional constraints in the augmented form of a linear programming model.

4.2 SETTING UP THE SIMPLEX METHOD

117

To illustrate these definitions, consider again the BF solution (0, 6, 4, 0, 6). This solution was obtained before by augmenting the CPF solution (0, 6). However, another way to obtain this same solution is to choose x1 and x4 to be the two nonbasic variables, and so the two variables are set equal to zero. The three equations then yield, respectively, x3 4, x2 6, and x5 6 as the solution for the three basic variables, as shown below (with the basic variables in bold type): (1) (2) (3)

x3

x1 2x2 3x1 2x2

4 x4 12 x5 18

x1 0 and x4 0 so x3 4 x2 6 x5 6

Because all three of these basic variables are nonnegative, this basic solution (0, 6, 4, 0, 6) is indeed a BF solution. Just as certain pairs of CPF solutions are adjacent, the corresponding pairs of BF solutions also are said to be adjacent. Here is an easy way to tell when two BF solutions are adjacent. Two BF solutions are adjacent if all but one of their nonbasic variables are the same. This implies that all but one of their basic variables also are the same, although perhaps with different numerical values.

Consequently, moving from the current BF solution to an adjacent one involves switching one variable from nonbasic to basic and vice versa for one other variable (and then adjusting the values of the basic variables to continue satisfying the system of equations). To illustrate adjacent BF solutions, consider one pair of adjacent CPF solutions in Fig. 4.1: (0, 0) and (0, 6). Their augmented solutions, (0, 0, 4, 12, 18) and (0, 6, 4, 0, 6), automatically are adjacent BF solutions. However, you do not need to look at Fig. 4.1 to draw this conclusion. Another signpost is that their nonbasic variables, (x1, x2) and (x1, x4), are the same with just the one exception—x2 has been replaced by x4. Consequently, moving from (0, 0, 4, 12, 18) to (0, 6, 4, 0, 6) involves switching x2 from nonbasic to basic and vice versa for x4. When we deal with the problem in augmented form, it is convenient to consider and manipulate the objective function equation at the same time as the new constraint equations. Therefore, before we start the simplex method, the problem needs to be rewritten once again in an equivalent way: Maximize

Z,

subject to (0) (1) (2) (3)

Z 3x1 5x2 0 x1 x3 4 2x2 x4 12 3x1 2x2 x5 18

and xj 0,

for j 1, 2, . . . , 5.

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

It is just as if Eq. (0) actually were one of the original constraints; but because it already is in equality form, no slack variable is needed. While adding one more equation, we also have added one more unknown (Z) to the system of equations. Therefore, when using Eqs. (1) to (3) to obtain a basic solution as described above, we use Eq. (0) to solve for Z at the same time. Somewhat fortuitously, the model for the Wyndor Glass Co. problem fits our standard form, and all its functional constraints have nonnegative right-hand sides bi. If this had not been the case, then additional adjustments would have been needed at this point before the simplex method was applied. These details are deferred to Sec. 4.6, and we now focus on the simplex method itself.

4.3

THE ALGEBRA OF THE SIMPLEX METHOD We continue to use the prototype example of Sec. 3.1, as rewritten at the end of Sec. 4.2, for illustrative purposes. To start connecting the geometric and algebraic concepts of the simplex method, we begin by outlining side by side in Table 4.2 how the simplex method solves this example from both a geometric and an algebraic viewpoint. The geometric viewpoint (first presented in Sec. 4.1) is based on the original form of the model (no slack variables), so again refer to Fig. 4.1 for a visualization when you examine the second column of the table. Refer to the augmented form of the model presented at the end of Sec. 4.2 when you examine the third column of the table. We now fill in the details for each step of the third column of Table 4.2. Initialization The choice of x1 and x2 to be the nonbasic variables (the variables set equal to zero) for the initial BF solution is based on solution concept 3 in Sec. 4.1. This choice eliminates the work required to solve for the basic variables (x3, x4, x5) from the following system of equations (where the basic variables are shown in bold type): (1) (2) (3)

x1

x3

2x2 3x1 2x2

4 x4 12 x5 18

x1 0 and x2 0 so x3 4 x4 12 x5 18

Thus, the initial BF solution is (0, 0, 4, 12, 18). Notice that this solution can be read immediately because each equation has just one basic variable, which has a coefficient of 1, and this basic variable does not appear in any other equation. You will soon see that when the set of basic variables changes, the simplex method uses an algebraic procedure (Gaussian elimination) to convert the equations to this same convenient form for reading every subsequent BF solution as well. This form is called proper form from Gaussian elimination. Optimality Test The objective function is Z 3x1 5x2,

4.3 THE ALGEBRA OF THE SIMPLEX METHOD

119

TABLE 4.2 Geometric and algebraic interpretations of how the simplex method solves the Wyndor Glass Co. problem Method Sequence

Geometric Interpretation

Initialization Choose (0, 0) to be the initial CPF solution. Optimality test Iteration 1 Step 1

Step 2 Step 3

Optimality test Iteration 2 Step 1

Step 2 Step 3

Optimality test

Not optimal, because moving along either edge from (0, 0) increases Z. Move up the edge lying on the x2 axis. Stop when the first new constraint boundary (2x2 12) is reached. Find the intersection of the new pair of constraint boundaries: (0, 6) is the new CPF solution. Not optimal, because moving along the edge from (0, 6) to the right increases Z.

Algebraic Interpretation Choose x1 and x2 to be the nonbasic variables ( 0) for the initial BF solution: (0, 0, 4, 12, 18). Not optimal, because increasing either nonbasic variable (x1 or x2) increases Z. Increase x2 while adjusting other variable values to satisfy the system of equations. Stop when the first basic variable (x3, x4, or x5) drops to zero (x4). With x2 now a basic variable and x4 now a nonbasic variable, solve the system of equations: (0, 6, 4, 0, 6) is the new BF solution. Not optimal, because increasing one nonbasic variable (x1) increases Z.

Increase x1 while adjusting other variable values to satisfy the system of equations. Stop when the first new constraint Stop when the first basic variable (x2, boundary (3x1 2x2 18) is reached. x3, or x5) drops to zero (x5). Find the intersection of the new pair With x1 now a basic variable and x5 of constraint boundaries: (2, 6) is the now a nonbasic variable, solve the new CPF solution. system of equations: (2, 6, 2, 0, 0) is the new BF solution. (2, 6) is optimal, because moving (2, 6, 2, 0, 0) is optimal, because along either edge from (2, 6) decreases Z. increasing either nonbasic variable (x4 or x5) decreases Z. Move along this edge to the right.

so Z 0 for the initial BF solution. Because none of the basic variables (x3, x4, x5) have a nonzero coefficient in this objective function, the coefficient of each nonbasic variable (x1, x2) gives the rate of improvement in Z if that variable were to be increased from zero (while the values of the basic variables are adjusted to continue satisfying the system of equations).1 These rates of improvement (3 and 5) are positive. Therefore, based on solution concept 6 in Sec. 4.1, we conclude that (0, 0, 4, 12, 18) is not optimal. For each BF solution examined after subsequent iterations, at least one basic variable has a nonzero coefficient in the objective function. Therefore, the optimality test then will use the new Eq. (0) to rewrite the objective function in terms of just the nonbasic variables, as you will see later. 1 Note that this interpretation of the coefficients of the xj variables is based on these variables being on the righthand side, Z 3x1 5x2. When these variables are brought to the left-hand side for Eq. (0), Z 3x1 5x2 0, the nonzero coefficients change their signs.

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Determining the Direction of Movement (Step 1 of an Iteration) Increasing one nonbasic variable from zero (while adjusting the values of the basic variables to continue satisfying the system of equations) corresponds to moving along one edge emanating from the current CPF solution. Based on solution concepts 4 and 5 in Sec. 4.1, the choice of which nonbasic variable to increase is made as follows: Z 3x1 5x2 Increase x1? Rate of improvement in Z 3. Increase x2? Rate of improvement in Z 5. 5 3, so choose x2 to increase. As indicated next, we call x2 the entering basic variable for iteration 1. At any iteration of the simplex method, the purpose of step 1 is to choose one nonbasic variable to increase from zero (while the values of the basic variables are adjusted to continue satisfying the system of equations). Increasing this nonbasic variable from zero will convert it to a basic variable for the next BF solution. Therefore, this variable is called the entering basic variable for the current iteration (because it is entering the basis).

Determining Where to Stop (Step 2 of an Iteration) Step 2 addresses the question of how far to increase the entering basic variable x2 before stopping. Increasing x2 increases Z, so we want to go as far as possible without leaving the feasible region. The requirement to satisfy the functional constraints in augmented form (shown below) means that increasing x2 (while keeping the nonbasic variable x1 0) changes the values of some of the basic variables as shown on the right. (1) (2) (3)

x1 2x2 3x1 2x2

x3

4 x4 12 x5 18

x1 0, so x3 4 x4 12 2x2 x5 18 2x2.

The other requirement for feasibility is that all the variables be nonnegative. The nonbasic variables (including the entering basic variable) are nonnegative, but we need to check how far x2 can be increased without violating the nonnegativity constraints for the basic variables. x3 4 0

⇒ no upper bound on x2.

12 x4 12 2x2 0 ⇒ x2 6 minimum. 2 18 x5 18 2x2 0 ⇒ x2 9. 2 Thus, x2 can be increased just to 6, at which point x4 has dropped to 0. Increasing x2 beyond 6 would cause x4 to become negative, which would violate feasibility. These calculations are referred to as the minimum ratio test. The objective of this test is to determine which basic variable drops to zero first as the entering basic variable is increased. We can immediately rule out the basic variable in any equation where the coefficient of the entering basic variable is zero or negative, since such a basic variable would not decrease as the entering basic variable is increased. [This is what happened

4.3 THE ALGEBRA OF THE SIMPLEX METHOD

121

with x3 in Eq. (1) of the example.] However, for each equation where the coefficient of the entering basic variable is strictly positive ( 0), this test calculates the ratio of the right-hand side to the coefficient of the entering basic variable. The basic variable in the equation with the minimum ratio is the one that drops to zero first as the entering basic variable is increased. At any iteration of the simplex method, step 2 uses the minimum ratio test to determine which basic variable drops to zero first as the entering basic variable is increased. Decreasing this basic variable to zero will convert it to a nonbasic variable for the next BF solution. Therefore, this variable is called the leaving basic variable for the current iteration (because it is leaving the basis).

Thus, x4 is the leaving basic variable for iteration 1 of the example. Solving for the New BF Solution (Step 3 of an Iteration) Increasing x2 0 to x2 6 moves us from the initial BF solution on the left to the new BF solution on the right. Nonbasic variables: Basic variables:

Initial BF solution x1 0, x2 0 x3 4, x4 12, x5 18

New BF solution x1 0, x4 0 x3 ?, x2 6, x5 ?

The purpose of step 3 is to convert the system of equations to a more convenient form (proper form from Gaussian elimination) for conducting the optimality test and (if needed) the next iteration with this new BF solution. In the process, this form also will identify the values of x3 and x5 for the new solution. Here again is the complete original system of equations, where the new basic variables are shown in bold type (with Z playing the role of the basic variable in the objective function equation): (0) (1) (2) (3)

Z 3x1 5x2 x3 x1 2x2 x4 3x1 2x2 x5

0. 4. 12. 18.

Thus, x2 has replaced x4 as the basic variable in Eq. (2). To solve this system of equations for Z, x2, x3, and x5, we need to perform some elementary algebraic operations to reproduce the current pattern of coefficients of x4 (0, 0, 1, 0) as the new coefficients of x2. We can use either of two types of elementary algebraic operations: 1. Multiply (or divide) an equation by a nonzero constant. 2. Add (or subtract) a multiple of one equation to (or from) another equation. To prepare for performing these operations, note that the coefficients of x2 in the above system of equations are 5, 0, 2, and 3, respectively, whereas we want these coefficients to become 0, 0, 1, and 0, respectively. To turn the coefficient of 2 in Eq. (2) into 1, we use the first type of elementary algebraic operation by dividing Eq. (2) by 2 to obtain (2)

1 x2 x4 6. 2

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To turn the coefficients of 5 and 3 into zeros, we need to use the second type of elementary algebraic operation. In particular, we add 5 times this new Eq. (2) to Eq. (0), and subtract 2 times this new Eq. (2) from Eq. (3). The resulting complete new system of equations is (0)

Z 3x1

(1)

x1

(2) (3)

5 x4 2 x3 x2

3x1

30

4 1 x4 6 2 x4 x5 6.

Since x1 0 and x4 0, the equations in this form immediately yield the new BF solution, (x1, x2, x3, x4, x5) (0, 6, 4, 0, 6), which yields Z 30. This procedure for obtaining the simultaneous solution of a system of linear equations is called the Gauss-Jordan method of elimination, or Gaussian elimination for short.1 The key concept for this method is the use of elementary algebraic operations to reduce the original system of equations to proper form from Gaussian elimination, where each basic variable has been eliminated from all but one equation (its equation) and has a coefficient of 1 in that equation. Optimality Test for the New BF Solution The current Eq. (0) gives the value of the objective function in terms of just the current nonbasic variables 5 Z 30 3x1 x4. 2 Increasing either of these nonbasic variables from zero (while adjusting the values of the basic variables to continue satisfying the system of equations) would result in moving toward one of the two adjacent BF solutions. Because x1 has a positive coefficient, increasing x1 would lead to an adjacent BF solution that is better than the current BF solution, so the current solution is not optimal. Iteration 2 and the Resulting Optimal Solution Since Z 30 3x1 52 x4, Z can be increased by increasing x1, but not x4. Therefore, step 1 chooses x1 to be the entering basic variable. For step 2, the current system of equations yields the following conclusions about how far x1 can be increased (with x4 0): 4 x3 4 x1 0 ⇒ x1 4. 1 x2 6 0 ⇒ no upper bound on x1. 6 x5 6 3x1 0 ⇒ x1 2 3

minimum.

Therefore, the minimum ratio test indicates that x5 is the leaving basic variable. 1

Actually, there are some technical differences between the Gauss-Jordan method of elimination and Gaussian elimination, but we shall not make this distinction.

4.4 THE SIMPLEX METHOD IN TABULAR FORM

123

For step 3, with x1 replacing x5 as a basic variable, we perform elementary algebraic operations on the current system of equations to reproduce the current pattern of coefficients of x5 (0, 0, 0, 1) as the new coefficients of x1. This yields the following new system of equations: (0)

3 x4 x5 36 2 1 1 x3 x4 x5 2 3 3

Z

(1) (2) (3)

x2 x1

1 x4 6 2 1 1 x4 x5 2. 3 3

Therefore, the next BF solution is (x1, x2, x3, x4, x5) (2, 6, 2, 0, 0), yielding Z 36. To apply the optimality test to this new BF solution, we use the current Eq. (0) to express Z in terms of just the current nonbasic variables, 3 Z 36 x4 x5. 2 Increasing either x4 or x5 would decrease Z, so neither adjacent BF solution is as good as the current one. Therefore, based on solution concept 6 in Sec. 4.1, the current BF solution must be optimal. In terms of the original form of the problem (no slack variables), the optimal solution is x1 2, x2 6, which yields Z 3x1 5x2 36. To see another example of applying the simplex method, we recommend that you now view the demonstration entitled Simplex Method—Algebraic Form in your OR Tutor. This vivid demonstration simultaneously displays both the algebra and the geometry of the simplex method as it dynamically evolves step by step. Like the many other demonstration examples accompanying other sections of the book (including the next section), this computer demonstration highlights concepts that are difficult to convey on the printed page. To further help you learn the simplex method efficiently, your OR Courseware includes a procedure entitled Solve Interactively by the Simplex Method. This routine performs nearly all the calculations while you make the decisions step by step, thereby enabling you to focus on concepts rather than get bogged down in a lot of number crunching. Therefore, you probably will want to use this routine for your homework on this section. The software will help you get started by letting you know whenever you make a mistake on the first iteration of a problem. The next section includes a summary of the simplex method for a more convenient tabular form.

4.4

THE SIMPLEX METHOD IN TABULAR FORM The algebraic form of the simplex method presented in Sec. 4.3 may be the best one for learning the underlying logic of the algorithm. However, it is not the most convenient form for performing the required calculations. When you need to solve a problem by hand (or

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interactively with your OR Courseware), we recommend the tabular form described in this section.1 The tabular form of the simplex method records only the essential information, namely, (1) the coefficients of the variables, (2) the constants on the right-hand sides of the equations, and (3) the basic variable appearing in each equation. This saves writing the symbols for the variables in each of the equations, but what is even more important is the fact that it permits highlighting the numbers involved in arithmetic calculations and recording the computations compactly. Table 4.3 compares the initial system of equations for the Wyndor Glass Co. problem in algebraic form (on the left) and in tabular form (on the right), where the table on the right is called a simplex tableau. The basic variable for each equation is shown in bold type on the left and in the first column of the simplex tableau on the right. [Although only the xj variables are basic or nonbasic, Z plays the role of the basic variable for Eq. (0).] All variables not listed in this basic variable column (x1, x2) automatically are nonbasic variables. After we set x1 0, x2 0, the right side column gives the resulting solution for the basic variables, so that the initial BF solution is (x1, x2, x3, x4, x5) (0, 0, 4, 12, 18) which yields Z 0. The tabular form of the simplex method uses a simplex tableau to compactly display the system of equations yielding the current BF solution. For this solution, each variable in the leftmost column equals the corresponding number in the rightmost column (and variables not listed equal zero). When the optimality test or an iteration is performed, the only relevant numbers are those to the right of the Z column. The term row refers to just a row of numbers to the right of the Z column (including the right side number), where row i corresponds to Eq. (i).

We summarize the tabular form of the simplex method below and, at the same time, briefly describe its application to the Wyndor Glass Co. problem. Keep in mind that the logic is identical to that for the algebraic form presented in the preceding section. Only the form for displaying both the current system of equations and the subsequent iteration has changed (plus we shall no longer bother to bring variables to the right-hand side of an equation before drawing our conclusions in the optimality test or in steps 1 and 2 of an iteration). 1

A form more convenient for automatic execution on a computer is presented in Sec. 5.2.

TABLE 4.3 Initial system of equations for the Wyndor Glass Co. problem (a) Algebraic Form

(b) Tabular Form Coefficient of:

(0) (1) (2) (3)

Z 3x1 5x2 x3 x4 x5 0 Z 3x1 5x2 x3 x4 x5 4 Z 3x1 2x2 x3 x4 x5 12 Z 3x1 2x2 x3 x4 x5 18

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

4.4 THE SIMPLEX METHOD IN TABULAR FORM

125

Summary of the Simplex Method (and Iteration 1 for the Example) Initialization. Introduce slack variables. Select the decision variables to be the initial nonbasic variables (set equal to zero) and the slack variables to be the initial basic variables. (See Sec. 4.6 for the necessary adjustments if the model is not in our standard form—maximization, only functional constraints, and all nonnegativity constraints— or if any bi values are negative.) For the Example: This selection yields the initial simplex tableau shown in Table 4.3b, so the initial BF solution is (0, 0, 4, 12, 18). Optimality Test. The current BF solution is optimal if and only if every coefficient in row 0 is nonnegative ( 0). If it is, stop; otherwise, go to an iteration to obtain the next BF solution, which involves changing one nonbasic variable to a basic variable (step 1) and vice versa (step 2) and then solving for the new solution (step 3). For the Example: Just as Z 3x1 5x2 indicates that increasing either x1 or x2 will increase Z, so the current BF solution is not optimal, the same conclusion is drawn from the equation Z 3x1 5x2 0. These coefficients of 3 and 5 are shown in row 0 of Table 4.3b. Iteration. Step 1: Determine the entering basic variable by selecting the variable (automatically a nonbasic variable) with the negative coefficient having the largest absolute value (i.e., the “most negative” coefficient) in Eq. (0). Put a box around the column below this coefficient, and call this the pivot column. For the Example: The most negative coefficient is 5 for x2 (5 3), so x2 is to be changed to a basic variable. (This change is indicated in Table 4.4 by the box around the x2 column below 5.) Step 2: Determine the leaving basic variable by applying the minimum ratio test. Minimum Ratio Test 1. 2. 3. 4.

Pick out each coefficient in the pivot column that is strictly positive ( 0). Divide each of these coefficients into the right side entry for the same row. Identify the row that has the smallest of these ratios. The basic variable for that row is the leaving basic variable, so replace that variable by the entering basic variable in the basic variable column of the next simplex tableau.

TABLE 4.4 Applying the minimum ratio test to determine the first leaving basic variable for the Wyndor Glass Co. problem Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3

(0) (1)

1 0

3 1

5 0

0 1

0 0

0 0

0 4

x4

(2)

0

0

2

0

1

0

12 12 6 minimum 2

x5

(3)

0

3

2

0

0

1

18 18 9 2

Ratio

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Put a box around this row and call it the pivot row. Also call the number that is in both boxes the pivot number. For the Example: The calculations for the minimum ratio test are shown to the right of Table 4.4. Thus, row 2 is the pivot row (see the box around this row in the first simplex tableau of Table 4.5), and x4 is the leaving basic variable. In the next simplex tableau (see the bottom of Table 4.5), x2 replaces x4 as the basic variable for row 2. Step 3: Solve for the new BF solution by using elementary row operations (multiply or divide a row by a nonzero constant; add or subtract a multiple of one row to another row) to construct a new simplex tableau in proper form from Gaussian elimination below the current one, and then return to the optimality test. The specific elementary row operations that need to be performed are listed below. 1. Divide the pivot row by the pivot number. Use this new pivot row in steps 2 and 3. 2. For each other row (including row 0) that has a negative coefficient in the pivot column, add to this row the product of the absolute value of this coefficient and the new pivot row. 3. For each other row that has a positive coefficient in the pivot column, subtract from this row the product of this coefficient and the new pivot row. For the Example: Since x2 is replacing x4 as a basic variable, we need to reproduce the first tableau’s pattern of coefficients in the column of x4 (0, 0, 1, 0) in the second tableau’s column of x2. To start, divide the pivot row (row 2) by the pivot number (2), which gives the new row 2 shown in Table 4.5. Next, we add to row 0 the product, 5 times the new row 2. Then we subtract from row 3 the product, 2 times the new row 2 (or equivalently, subtract from row 3 the old row 2). These calculations yield the new tableau shown in Table 4.6 for iteration 1. Thus, the new BF solution is (0, 6, 4, 0, 6), with Z 30. We next return to the optimality test to check if the new BF solution is optimal. Since the new row 0 still has a negative coefficient (3 for x1), the solution is not optimal, and so at least one more iteration is needed. TABLE 4.5 Simplex tableaux for the Wyndor Glass Co. problem after the first pivot row is divided by the first pivot number Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

0

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

1

Z x3 x2 x5

(0) (1) (2) (3)

1 0 0 0

0

1

0

1 2

0

6

Iteration

4.4 THE SIMPLEX METHOD IN TABULAR FORM

127

TABLE 4.6 First two simplex tableaux for the Wyndor Glass Co. problem Coefficient of: Iteration

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

Z

(0)

1

3

0

0

0

30

x3

(1)

0

1

0

1

0

4

x2

(2)

0

0

1

0

0

6

x5

(3)

0

3

0

0

1

6

0

1

5 2 0 1 2 1

Iteration 2 for the Example and the Resulting Optimal Solution The second iteration starts anew from the second tableau of Table 4.6 to find the next BF solution. Following the instructions for steps 1 and 2, we find x1 as the entering basic variable and x5 as the leaving basic variable, as shown in Table 4.7. For step 3, we start by dividing the pivot row (row 3) in Table 4.7 by the pivot number (3). Next, we add to row 0 the product, 3 times the new row 3. Then we subtract the new row 3 from row 1. We now have the set of tableaux shown in Table 4.8. Therefore, the new BF solution is (2, 6, 2, 0, 0), with Z 36. Going to the optimality test, we find that this solution is optimal because none of the coefficients in row 0 is negative, so the algorithm is finished. Consequently, the optimal solution for the Wyndor Glass Co. problem (before slack variables are introduced) is x1 2, x2 6. Now compare Table 4.8 with the work done in Sec. 4.3 to verify that these two forms of the simplex method really are equivalent. Then note how the algebraic form is superior for learning the logic behind the simplex method, but the tabular form organizes the TABLE 4.7 Steps 1 and 2 of iteration 2 for the Wyndor Glass Co. problem Coefficient of: Iteration

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z

(0)

1

3

0

0

5 2

0

30

x3

(1)

0

1

0

1

0

0

4

x2

(2)

0

0

1

0

1 2

0

6

x5

(3)

0

3

0

0

1

1

6

1

Ratio

4 4 1

6 2 minimum 3

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TABLE 4.8 Complete set of simplex tableaux for the Wyndor Glass Co. problem Coefficient of: Iteration

0

1

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

Z

(0)

1

3

0

0

0

30

x3

(1)

0

1

0

1

0

4

x2

(2)

0

0

1

0

0

6

x5

(3)

0

3

0

0

5 2 0 1 2 1

1

6

Z

(0)

1

0

0

0

3 2

1

36

x3

(1)

0

0

0

1

1 3

1 3

2

x2

(2)

0

0

1

0

1 2

0

6

x1

(3)

0

1

0

0

1 3

1 3

2

2

work being done in a considerably more convenient and compact form. We generally use the tabular form from now on. An additional example of applying the simplex method in tabular form is available to you in the OR Tutor. See the demonstration entitled Simplex Method—Tabular Form.

4.5

TIE BREAKING IN THE SIMPLEX METHOD You may have noticed in the preceding two sections that we never said what to do if the various choice rules of the simplex method do not lead to a clear-cut decision, because of either ties or other similar ambiguities. We discuss these details now. Tie for the Entering Basic Variable Step 1 of each iteration chooses the nonbasic variable having the negative coefficient with the largest absolute value in the current Eq. (0) as the entering basic variable. Now suppose that two or more nonbasic variables are tied for having the largest negative coefficient (in absolute terms). For example, this would occur in the first iteration for the Wyndor Glass Co. problem if its objective function were changed to Z 3x1 3x2, so that the initial Eq. (0) became Z 3x1 3x2 0. How should this tie be broken? The answer is that the selection between these contenders may be made arbitrarily. The optimal solution will be reached eventually, regardless of the tied variable chosen, and there is no convenient method for predicting in advance which choice will lead there

4.5 TIE BREAKING IN THE SIMPLEX METHOD

129

sooner. In this example, the simplex method happens to reach the optimal solution (2, 6) in three iterations with x1 as the initial entering basic variable, versus two iterations if x2 is chosen. Tie for the Leaving Basic Variable—Degeneracy Now suppose that two or more basic variables tie for being the leaving basic variable in step 2 of an iteration. Does it matter which one is chosen? Theoretically it does, and in a very critical way, because of the following sequence of events that could occur. First, all the tied basic variables reach zero simultaneously as the entering basic variable is increased. Therefore, the one or ones not chosen to be the leaving basic variable also will have a value of zero in the new BF solution. (Note that basic variables with a value of zero are called degenerate, and the same term is applied to the corresponding BF solution.) Second, if one of these degenerate basic variables retains its value of zero until it is chosen at a subsequent iteration to be a leaving basic variable, the corresponding entering basic variable also must remain zero (since it cannot be increased without making the leaving basic variable negative), so the value of Z must remain unchanged. Third, if Z may remain the same rather than increase at each iteration, the simplex method may then go around in a loop, repeating the same sequence of solutions periodically rather than eventually increasing Z toward an optimal solution. In fact, examples have been artificially constructed so that they do become entrapped in just such a perpetual loop. Fortunately, although a perpetual loop is theoretically possible, it has rarely been known to occur in practical problems. If a loop were to occur, one could always get out of it by changing the choice of the leaving basic variable. Furthermore, special rules1 have been constructed for breaking ties so that such loops are always avoided. However, these rules frequently are ignored in actual application, and they will not be repeated here. For your purposes, just break this kind of tie arbitrarily and proceed without worrying about the degenerate basic variables that result. No Leaving Basic Variable—Unbounded Z In step 2 of an iteration, there is one other possible outcome that we have not yet discussed, namely, that no variable qualifies to be the leaving basic variable.2 This outcome would occur if the entering basic variable could be increased indefinitely without giving negative values to any of the current basic variables. In tabular form, this means that every coefficient in the pivot column (excluding row 0) is either negative or zero. As illustrated in Table 4.9, this situation arises in the example displayed in Fig. 3.6 on p. 36. In this example, the last two functional constraints of the Wyndor Glass Co. problem have been overlooked and so are not included in the model. Note in Fig. 3.6 how x2 can be increased indefinitely (thereby increasing Z indefinitely) without ever leaving the feasible region. Then note in Table 4.9 that x2 is the entering basic variable but the 1

See R. Bland, “New Finite Pivoting Rules for the Simplex Method,” Mathematics of Operations Research, 2: 103–107, 1977. 2 Note that the analogous case (no entering basic variable) cannot occur in step 1 of an iteration, because the optimality test would stop the algorithm first by indicating that an optimal solution had been reached.

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TABLE 4.9 Initial simplex tableau for the Wyndor Glass Co. problem without the last two functional constraints Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

Right Side

Ratio

Z x3

(0) (1)

1 0

3 1

5 0

0 1

0 4

None

With x1 0 and x2 increasing, x3 4 1x1 0x2 4 0.

only coefficient in the pivot column is zero. Because the minimum ratio test uses only coefficients that are greater than zero, there is no ratio to provide a leaving basic variable. The interpretation of a tableau like the one shown in Table 4.9 is that the constraints do not prevent the value of the objective function Z increasing indefinitely, so the simplex method would stop with the message that Z is unbounded. Because even linear programming has not discovered a way of making infinite profits, the real message for practical problems is that a mistake has been made! The model probably has been misformulated, either by omitting relevant constraints or by stating them incorrectly. Alternatively, a computational mistake may have occurred. Multiple Optimal Solutions We mentioned in Sec. 3.2 (under the definition of optimal solution) that a problem can have more than one optimal solution. This fact was illustrated in Fig. 3.5 by changing the objective function in the Wyndor Glass Co. problem to Z 3x1 2x2, so that every point on the line segment between (2, 6) and (4, 3) is optimal. Thus, all optimal solutions are a weighted average of these two optimal CPF solutions (x1, x2) w1(2, 6) w2(4, 3), where the weights w1 and w2 are numbers that satisfy the relationships w1 w2 1

and

w1 0,

w2 0.

For example, w1 13 and w2 23 give

1 2 2 8 (x1, x2) (2, 6) (4, 3) , 3 3 3 3

6 6 10 , 3 3 3

4

as one optimal solution. In general, any weighted average of two or more solutions (vectors) where the weights are nonnegative and sum to 1 is called a convex combination of these solutions. Thus, every optimal solution in the example is a convex combination of (2, 6) and (4, 3). This example is typical of problems with multiple optimal solutions. As indicated at the end of Sec. 3.2, any linear programming problem with multiple optimal solutions (and a bounded feasible region) has at least two CPF solutions that are optimal. Every optimal solution is a convex combination of these optimal CPF solutions. Consequently, in augmented form, every optimal solution is a convex combination of the optimal BF solutions.

(Problems 4.5-5 and 4.5-6 guide you through the reasoning behind this conclusion.)

4.5 TIE BREAKING IN THE SIMPLEX METHOD

131

The simplex method automatically stops after one optimal BF solution is found. However, for many applications of linear programming, there are intangible factors not incorporated into the model that can be used to make meaningful choices between alternative optimal solutions. In such cases, these other optimal solutions should be identified as well. As indicated above, this requires finding all the other optimal BF solutions, and then every optimal solution is a convex combination of the optimal BF solutions. After the simplex method finds one optimal BF solution, you can detect if there are any others and, if so, find them as follows: Whenever a problem has more than one optimal BF solution, at least one of the nonbasic variables has a coefficient of zero in the final row 0, so increasing any such variable will not change the value of Z. Therefore, these other optimal BF solutions can be identified (if desired) by performing additional iterations of the simplex method, each time choosing a nonbasic variable with a zero coefficient as the entering basic variable.1

To illustrate, consider again the case just mentioned, where the objective function in the Wyndor Glass Co. problem is changed to Z 3x1 2x2. The simplex method obtains the first three tableaux shown in Table 4.10 and stops with an optimal BF solution. How1

If such an iteration has no leaving basic variable, this indicates that the feasible region is unbounded and the entering basic variable can be increased indefinitely without changing the value of Z.

TABLE 4.10 Complete set of simplex tableaux to obtain all optimal BF solutions for the Wyndor Glass Co. problem with c2 2 Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

2 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

No

0

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

2 0 2 2

3 1 0 3

0 0 1 0

0 0 0 1

12 4 12 6

No

1

Z x1 x4 x5 Z x1 x4

(0) (1) (2)

1 0 0

0 1 0

0 0 0

0 0 1

Yes

(3)

0

0

1

1 0 1 1 2

18 4 6

x2

0 1 3 3 2

Z

(0)

1

0

0

0

(1)

0

1

0

0

1 1 3

18

x1

0 1 3

x3

(2)

0

0

0

1

1 3

1 3

2

x2

(3)

0

0

1

0

1 2

0

6

Iteration

2

Extra

0

Solution Optimal?

3

2

Yes

132

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

ever, because a nonbasic variable (x3) then has a zero coefficient in row 0, we perform one more iteration in Table 4.10 to identify the other optimal BF solution. Thus, the two optimal BF solutions are (4, 3, 0, 6, 0) and (2, 6, 2, 0, 0), each yielding Z 18. Notice that the last tableau also has a nonbasic variable (x4) with a zero coefficient in row 0. This situation is inevitable because the extra iteration does not change row 0, so this leaving basic variable necessarily retains its zero coefficient. Making x4 an entering basic variable now would only lead back to the third tableau. (Check this.) Therefore, these two are the only BF solutions that are optimal, and all other optimal solutions are a convex combination of these two. (x1, x2, x3, x4, x5) w1(2, 6, 2, 0, 0) w2(4, 3, 0, 6, 0), w1 w2 1, w1 0, w2 0.

4.6

ADAPTING TO OTHER MODEL FORMS Thus far we have presented the details of the simplex method under the assumptions that the problem is in our standard form (maximize Z subject to functional constraints in form and nonnegativity constraints on all variables) and that bi 0 for all i 1, 2, . . . , m. In this section we point out how to make the adjustments required for other legitimate forms of the linear programming model. You will see that all these adjustments can be made during the initialization, so the rest of the simplex method can then be applied just as you have learned it already. The only serious problem introduced by the other forms for functional constraints (the or forms, or having a negative right-hand side) lies in identifying an initial BF solution. Before, this initial solution was found very conveniently by letting the slack variables be the initial basic variables, so that each one just equals the nonnegative right-hand side of its equation. Now, something else must be done. The standard approach that is used for all these cases is the artificial-variable technique. This technique constructs a more convenient artificial problem by introducing a dummy variable (called an artificial variable) into each constraint that needs one. This new variable is introduced just for the purpose of being the initial basic variable for that equation. The usual nonnegativity constraints are placed on these variables, and the objective function also is modified to impose an exorbitant penalty on their having values larger than zero. The iterations of the simplex method then automatically force the artificial variables to disappear (become zero), one at a time, until they are all gone, after which the real problem is solved. To illustrate the artificial-variable technique, first we consider the case where the only nonstandard form in the problem is the presence of one or more equality constraints. Equality Constraints Any equality constraint ai1x1 ai2x2 ain xn bi actually is equivalent to a pair of inequality constraints: ai1x1 ai2x2 ain xn bi ai1x1 ai2x2 ain xn bi.

4.6 ADAPTING TO OTHER MODEL FORMS

133

However, rather than making this substitution and thereby increasing the number of constraints, it is more convenient to use the artificial-variable technique. We shall illustrate this technique with the following example. Example. Suppose that the Wyndor Glass Co. problem in Sec. 3.1 is modified to require that Plant 3 be used at full capacity. The only resulting change in the linear programming model is that the third constraint, 3x1 2x2 18, instead becomes an equality constraint 3x1 2x2 18, so that the complete model becomes the one shown in the upper right-hand corner of Fig. 4.3. This figure also shows in darker ink the feasible region which now consists of just the line segment connecting (2, 6) and (4, 3). After the slack variables still needed for the inequality constraints are introduced, the system of equations for the augmented form of the problem becomes (0) (1) (2) (3)

Z 3x1 5x2 0. x1 x3 4. 2x2 x4 12. 3x1 2x2 18.

Unfortunately, these equations do not have an obvious initial BF solution because there is no longer a slack variable to use as the initial basic variable for Eq. (3). It is necessary to find an initial BF solution to start the simplex method. This difficulty can be circumvented in the following way.

FIGURE 4.3 When the third functional constraint becomes an equality constraint, the feasible region for the Wyndor Glass Co. problem becomes the line segment between (2, 6) and (4, 3).

x2 10 Maximize subject to 8

6

and

Z 3x1 5x2, 4 x1 2x2 12 3x1 2x2 18 x1 0, x2 0

(2, 6)

4 (4, 3) 2

0

2

4

6

8

x1

134

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

Obtaining an Initial BF Solution. The procedure is to construct an artificial problem that has the same optimal solution as the real problem by making two modifications of the real problem. 1. Apply the artificial-variable technique by introducing a nonnegative artificial variable (call it x5)1 into Eq. (3), just as if it were a slack variable (3)

3x1 2x2 x5 18.

2. Assign an overwhelming penalty to having x5 0 by changing the objective function Z 3x1 5x2 to Z 3x1 5x2 Mx5, where M symbolically represents a huge positive number. (This method of forcing x5 to be x5 0 in the optimal solution is called the Big M method.) Now find the optimal solution for the real problem by applying the simplex method to the artificial problem, starting with the following initial BF solution: Initial BF Solution Nonbasic variables: Basic variables:

x1 0, x3 4,

x2 0 x4 12,

x5 18.

Because x5 plays the role of the slack variable for the third constraint in the artificial problem, this constraint is equivalent to 3x1 2x2 18 (just as for the original Wyndor Glass Co. problem in Sec. 3.1). We show below the resulting artificial problem (before augmenting) next to the real problem. The Real Problem

The Artificial Problem Define x5 18 3x1 2x2.

Maximize Z 3x1 5x2,

Maximize

subject to

subject to

Z 3x1 5x2 M x5,

x1 2x2 4

(so

3x1 2x2 x5 4

3x1 2x2 12

(so

3x1 2x2 x5 12

3x1 2x2 18

(so

3x1 2x2 x5 18

(so

3x1 2x2 x5 18)

and x1 0,

x2 0.

and x1 0,

x2 0,

x5 0.

Therefore, just as in Sec. 3.1, the feasible region for (x1, x2) for the artificial problem is the one shown in Fig. 4.4. The only portion of this feasible region that coincides with the feasible region for the real problem is where x5 0 (so 3x1 2x2 18). Figure 4.4 also shows the order in which the simplex method examines the CPF solutions (or BF solutions after augmenting), where each circled number identifies which iteration obtained that solution. Note that the simplex method moves counterclockwise here 1

We shall always label the artificial variables by putting a bar over them.

4.6 ADAPTING TO OTHER MODEL FORMS

135

x2 Define x5 18 3x1 2x2. Maximize Z 3x1 5x2 Mx5, subject to x1 4 2x2 12 3x1 2x2 18 x1 0, x2 0, x5 0 and

Z 30 6M (2, 6) Z 36 3

(0, 6)

FIGURE 4.4 This graph shows the feasible region and the sequence of CPF solutions (, , , ) examined by the simplex method for the artificial problem that corresponds to the real problem of Fig. 4.3.

Feasible region

(4, 3)

1

0

(0, 0)

2

Z 0 18M

(4, 0)

Z 27

Z 12 6M x1

whereas it moved clockwise for the original Wyndor Glass Co. problem (see Fig. 4.2). The reason for this difference is the extra term Mxx5 in the objective function for the artificial problem. Before applying the simplex method and demonstrating that it follows the path shown in Fig. 4.4, the following preparatory step is needed. Converting Equation (0) to Proper Form. The system of equations after the artificial problem is augmented is (0) (1) (2) (3)

Z 3x1 5x2 Mx5 0 x3 4 x1 2x2 x4 12 3x1 2x2 x5 18

where the initial basic variables (x3, x4, x5) are shown in bold type. However, this system is not yet in proper form from Gaussian elimination because a basic variable x5 has a nonzero coefficient in Eq. (0). Recall that all basic variables must be algebraically eliminated from Eq. (0) before the simplex method can either apply the optimality test or find the entering basic variable. This elimination is necessary so that the negative of the coefficient of each nonbasic variable will give the rate at which Z would increase if that nonbasic variable were to be increased from 0 while adjusting the values of the basic variables accordingly. To algebraically eliminate x5 from Eq. (0), we need to subtract from Eq. (0) the product, M times Eq. (3).

New (0)

Z 3x1 5x2 Mx5 0 M(3x 1 2x2 Mx x5 18) Z (3M 3)x1 (2M 5)x2 18M.

136

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

Application of the Simplex Method. This new Eq. (0) gives Z in terms of just the nonbasic variables (x1, x2), Z 18M (3M 3)x1 (2M 5)x2. Since 3M 3 2M 5 (remember that M represents a huge number), increasing x1 increases Z at a faster rate than increasing x2 does, so x1 is chosen as the entering basic variable. This leads to the move from (0, 0) to (4, 0) at iteration 1, shown in Fig. 4.4, thereby increasing Z by 4(3M 3). The quantities involving M never appear in the system of equations except for Eq. (0), so they need to be taken into account only in the optimality test and when an entering basic variable is determined. One way of dealing with these quantities is to assign some particular (huge) numerical value to M and use the resulting coefficients in Eq. (0) in the usual way. However, this approach may result in significant rounding errors that invalidate the optimality test. Therefore, it is better to do what we have just shown, namely, to express each coefficient in Eq. (0) as a linear function aM b of the symbolic quantity M by separately recording and updating the current numerical value of (1) the multiplicative factor a and (2) the additive term b. Because M is assumed to be so large that b always is negligible compared with M when a 0, the decisions in the optimality test and the choice of the entering basic variable are made by using just the multiplicative factors in the usual way, except for breaking ties with the additive factors. Using this approach on the example yields the simplex tableaux shown in Table 4.11. Note that the artificial variable x5 is a basic variable (xx5 0) in the first two tableaux and a nonbasic variable (xx5 0) in the last two. Therefore, the first two BF solutions for this artificial problem are infeasible for the real problem whereas the last two also are BF solutions for the real problem. This example involved only one equality constraint. If a linear programming model has more than one, each is handled in just the same way. (If the right-hand side is negative, multiply through both sides by 1 first.)

Negative Right-Hand Sides The technique mentioned in the preceding sentence for dealing with an equality constraint with a negative right-hand side (namely, multiply through both sides by 1) also works for any inequality constraint with a negative right-hand side. Multiplying through both sides of an inequality by 1 also reverses the direction of the inequality; i.e., changes to or vice versa. For example, doing this to the constraint x1 x2 1

(that is, x1 x2 1)

gives the equivalent constraint x1 x2 1

(that is, x2 1 x1)

but now the right-hand side is positive. Having nonnegative right-hand sides for all the functional constraints enables the simplex method to begin, because (after augmenting) these right-hand sides become the respective values of the initial basic variables, which must satisfy nonnegativity constraints.

4.6 ADAPTING TO OTHER MODEL FORMS

137

TABLE 4.11 Complete set of simplex tableaux for the problem shown in Fig. 4.4 Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x 5

Right Side

0

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3M 3 1 0 3

2M 5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

18M 4 12 18

1

Z x1 x4 x5

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

2M 5 0 2 2

3M 3 1 0 3

0 0 1 0

0 0 0 1

6M 12 4 12 6

Z

(0)

1

0

0

x1 x4

(1) (2)

0 0

1 0

0 0

x2

(3)

0

0

1

Z

(0)

1

0

x1

(1)

0

x3

(2)

x2

(3)

Iteration

2

9 2 1 3 3 2

0 1

0

0

3 2

M1

36

1

0

0

1 3

1 3

2

0

0

0

1

1 3

1 3

2

0

0

1

0

1 2

0

6

Extra

0

0

5 M 2 0 1 1 2

27 4 6 3

We next focus on how to augment constraints, such as x1 x2 1, with the help of the artificial-variable technique. Functional Constraints in Form To illustrate how the artificial-variable technique deals with functional constraints in form, we will use the model for designing Mary’s radiation therapy, as presented in Sec. 3.4. For your convenience, this model is repeated below, where we have placed a box around the constraint of special interest here. Radiation Therapy Example Minimize

Z 0.4x1 0.5x2,

subject to 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6 and x1 0,

x2 0.

138

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

x2 27 15 Dots corner-point solutions Dark line segment feasible region Optimal solution (7.5, 4.5)

0.6x1 0.4x2 6 10

(6, 6) 5 (7.5, 4.5)

(8, 3)

FIGURE 4.5 Graphical display of the radiation therapy example and its corner-point solutions.

0.5x1 0.5x2 6

0.3x1 0.1x2 2.7

0

5

10

x1

The graphical solution for this example (originally presented in Fig. 3.12) is repeated here in a slightly different form in Fig. 4.5. The three lines in the figure, along with the two axes, constitute the five constraint boundaries of the problem. The dots lying at the intersection of a pair of constraint boundaries are the corner-point solutions. The only two corner-point feasible solutions are (6, 6) and (7.5, 4.5), and the feasible region is the line segment connecting these two points. The optimal solution is (x1, x2) (7.5, 4.5), with Z 5.25. We soon will show how the simplex method solves this problem by directly solving the corresponding artificial problem. However, first we must describe how to deal with the third constraint.

4.6 ADAPTING TO OTHER MODEL FORMS

139

Our approach involves introducing both a surplus variable x5 (defined as x5 0.6x1 0.4x2 6) and an artificial variable x6, as shown next.

0.6x1 0.4x2 6 0.6x1 0.4x2 x5 6 0.6x1 0.4x2 x5 x6 6

(x5 0) (x5 0, x6 0).

Here x5 is called a surplus variable because it subtracts the surplus of the left-hand side over the right-hand side to convert the inequality constraint to an equivalent equality constraint. Once this conversion is accomplished, the artificial variable is introduced just as for any equality constraint. After a slack variable x3 is introduced into the first constraint, an artificial variable x4 is introduced into the second constraint, and the Big M method is applied, so the complete artificial problem (in augmented form) is Z 0.4x1 0.5x2 Mxx4 Mxx6, 0.3x1 0.1x2 x3 2.7 0.5x1 0.5x2 x4 6 0.6x1 0.4x2 x5 x6 6 x2 0, x3 0, x5 0, x1 0, x4 0,

Minimize subject to

and

x6 0.

Note that the coefficients of the artificial variables in the objective function are M, instead of M, because we now are minimizing Z. Thus, even though x4 0 and/or x6 0 is possible for a feasible solution for the artificial problem, the huge unit penalty of M prevents this from occurring in an optimal solution. As usual, introducing artificial variables enlarges the feasible region. Compare below the original constraints for the real problem with the corresponding constraints on (x1, x2) for the artificial problem. Constraints on (x1, x2) for the Real Problem 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6 x1 0, x2 0

Constraints on (x1, x2) for the Artificial Problem 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 ( holds when x4 0) No such constraint (except when x6 0) x1 0, x2 0

Introducing the artificial variable x4 to play the role of a slack variable in the second constraint allows values of (x1, x2) below the 0.5x1 0.5x2 6 line in Fig. 4.5. Introducing x5 and x6 into the third constraint of the real problem (and moving these variables to the right-hand side) yields the equation 0.6x1 0.4x2 6 x5 x6. Because both x5 and x6 are constrained only to be nonnegative, their difference x5 x6 can be any positive or negative number. Therefore, 0.6x1 0.4x2 can have any value, which has the effect of eliminating the third constraint from the artificial problem and allowing points on either side of the 0.6x1 0.4x2 6 line in Fig. 4.5. (We keep the third constraint in the system of equations only because it will become relevant again later, after the Big M method forces x6 to be zero.) Consequently, the feasible region for the ar-

140

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

tificial problem is the entire polyhedron in Fig. 4.5 whose vertices are (0, 0), (9, 0), (7.5, 4.5), and (0, 12). Since the origin now is feasible for the artificial problem, the simplex method starts with (0, 0) as the initial CPF solution, i.e., with (x1, x2, x3, x4, x5, x6) (0, 0, 2.7, 6, 0, 6) as the initial BF solution. (Making the origin feasible as a convenient starting point for the simplex method is the whole point of creating the artificial problem.) We soon will trace the entire path followed by the simplex method from the origin to the optimal solution for both the artificial and real problems. But, first, how does the simplex method handle minimization? Minimization One straightforward way of minimizing Z with the simplex method is to exchange the roles of the positive and negative coefficients in row 0 for both the optimality test and step 1 of an iteration. However, rather than changing our instructions for the simplex method for this case, we present the following simple way of converting any minimization problem to an equivalent maximization problem: n

Minimizing

Z cj xj j1

is equivalent to n

maximizing

Z (cj)xj; j1

i.e., the two formulations yield the same optimal solution(s). The two formulations are equivalent because the smaller Z is, the larger Z is, so the solution that gives the smallest value of Z in the entire feasible region must also give the largest value of Z in this region. Therefore, in the radiation therapy example, we make the following change in the formulation:

Minimize Maximize

Z 0.4x1 0.5x2 Z 0.4x1 0.5x2.

After artificial variables x4 and x6 are introduced and then the Big M method is applied, the corresponding conversion is

Minimize Maximize

Z 0.4x1 0.5x2 Mxx4 Mxx6 Z 0.4x1 0.5x2 Mxx4 Mxx6.

Solving the Radiation Therapy Example We now are nearly ready to apply the simplex method to the radiation therapy example. By using the maximization form just obtained, the entire system of equations is now (0) (1)

Z 0.4x1 0.5x2 Mx 4 0.3x1 0.1x2 x3

Mx 6 0 2.7

4.6 ADAPTING TO OTHER MODEL FORMS

(2) (3)

0.5x1 0.5x2 0.6x1 0.4x2

x4

141

x5

6 x6 6.

The basic variables (x3, x4, x6) for the initial BF solution (for this artificial problem) are shown in bold type. Note that this system of equations is not yet in proper form from Gaussian elimination, as required by the simplex method, since the basic variables x4 and x6 still need to be algebraically eliminated from Eq. (0). Because x4 and x6 both have a coefficient of M, Eq. (0) needs to have subtracted from it both M times Eq. (2) and M times Eq. (3). The calculations for all the coefficients (and the right-hand sides) are summarized below, where the vectors are the relevant rows of the simplex tableau corresponding to the above system of equations. Row 0: M[0.4, M[0.5, M[0.6, New row 0 [1.1M 0.4,

0.5, 0.5, 0.4, 0.9M 0.5,

0, 0, 0, 0,

M, 1, 0, 0,

0, 0, 1, M,

M, 0, 1, 0,

0] 6] 6] 12M]

The resulting initial simplex tableau, ready to begin the simplex method, is shown at the top of Table 4.12. Applying the simplex method in just the usual way then yields the sequence of simplex tableaux shown in the rest of Table 4.12. For the optimality test and the selection of the entering basic variable at each iteration, the quantities involving M are treated just as discussed in connection with Table 4.11. Specifically, whenever M is present, only its multiplicative factor is used, unless there is a tie, in which case the tie is broken by using the corresponding additive terms. Just such a tie occurs in the last selection of an entering basic variable (see the next-to-last tableau), where the coefficients of x3 and x5 in row 0 both have the same multiplicative factor of 53. Comparing the additive terms, 161 73 leads to choosing x5 as the entering basic variable. Note in Table 4.12 the progression of values of the artificial variables x4 and x6 and of Z. We start with large values, x4 6 and x6 6, with Z 12M (Z 12M). The first iteration greatly reduces these values. The Big M method succeeds in driving x6 to zero (as a new nonbasic variable) at the second iteration and then in doing the same to x4 at the next iteration. With both x4 0 and x6 0, the basic solution given in the last tableau is guaranteed to be feasible for the real problem. Since it passes the optimality test, it also is optimal. Now see what the Big M method has done graphically in Fig. 4.6. The feasible region for the artificial problem initially has four CPF solutions—(0, 0), (9, 0), (0, 12), and (7.5, 4.5)—and then replaces the first three with two new CPF solutions—(8, 3), (6, 6)— after x6 decreases to x6 0 so that 0.6x1 0.4x2 6 becomes an additional constraint. (Note that the three replaced CPF solutions—(0, 0), (9, 0), and (0, 12)—actually were corner-point infeasible solutions for the real problem shown in Fig. 4.5.) Starting with the origin as the convenient initial CPF solution for the artificial problem, we move around the boundary to three other CPF solutions—(9, 0), (8, 3), and (7.5, 4.5). The last of these is the first one that also is feasible for the real problem. Fortuitously, this first feasible solution also is optimal, so no additional iterations are needed.

142

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

TABLE 4.12 The Big M method for the radiation therapy example Coefficient of: Iteration

0

Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Right Side

Z x3 x4 x6

(0) (1) (2) (3)

1 0 0 0

1.1M 0.4 0.3 0.5 0.6

0.9M 0.5 0.1 0.5 0.4

0.0 1.0 0.0 0.0

0.0 0.0 1.0 0.0

M 0 0 1

0 0 0 1

12M1 2.7 6.0 6.0

Z

(0)

1

0.0

16 11 M 30 30

11 4 M 3 3

0.0

M

0

2.1M 3.6

x1

(1)

0

1.0

1 3

10 3

0.0

0

0

9.0

x4

(2)

0

0.0

0

0

1.5

(3)

0

0.0

5 3 2

1.0

x6

1 3 0.2

0.0

1

1

0.6

Z

(0)

1

0.0

0.0

5 7 M 3 3

0.0

5 11 M 3 6

8 11 M 3 6

0.5M 4.7

x1

(1)

0

1.0

0.0

20 3

0.0

5 3

5 3

8.0

x4

(2)

0

0.0

0.0

1.0

(3)

0

0.0

1.0

0.0

5 3 5

5 3 5

0.5

x2

5 3 10.0

Z x1 x5 x2

(0) (1) (2) (3)

1 0 0 0

0.0 1.0 0.0 0.0

0.0 0.0 0.0 1.0

0.5 5.0 1.0 5.0

M 1.1 1.0 1 0.6 3.0

0 0 1 0

M 0 1 0

5.25 7.51 0.31 4.51

1

2

3

3.0

For other problems with artificial variables, it may be necessary to perform additional iterations to reach an optimal solution after the first feasible solution is obtained for the real problem. (This was the case for the example solved in Table 4.11.) Thus, the Big M method can be thought of as having two phases. In the first phase, all the artificial variables are driven to zero (because of the penalty of M per unit for being greater than zero) in order to reach an initial BF solution for the real problem. In the second phase, all the artificial variables are kept at zero (because of this same penalty) while the simplex method generates a sequence of BF solutions for the real problem that leads to an optimal solution. The two-phase method described next is a streamlined procedure for performing these two phases directly, without even introducing M explicitly. The Two-Phase Method For the radiation therapy example just solved in Table 4.12, recall its real objective function Real problem:

Minimize

Z 0.4x1 0.5x2.

4.6 ADAPTING TO OTHER MODEL FORMS

x2

Constraints for the artificial problem:

Z 6 1.2M

(0, 12)

143

0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 ( holds when x4 0) (0.6x1 0.4x2 6 when x6 0) x1 0, x2 0 (x4 0, x6 0)

Z 5.4 This dark line segment is the feasible region for the real problem (x4 0, x6 0).

(6, 6)

(7.5, 4.5) optimal 3

Z 5.25 (8, 3)

FIGURE 4.6 This graph shows the feasible region and the sequence of CPF solutions (, , , ) examined by the simplex method (with the Big M method) for the artificial problem that corresponds to the real problem of Fig. 4.5.

Z 4.7 0.5M

2

Feasible region for the artificial problem 0

Z 3.6 2.1M

1

(0, 0) (9, 0)

Z 0 12M

x1

However, the Big M method uses the following objective function (or its equivalent in maximization form) throughout the entire procedure: Big M method:

Minimize

Z 0.4x1 0.5x2 Mx4 Mx6.

Since the first two coefficients are negligible compared to M, the two-phase method is able to drop M by using the following two objective functions with completely different definitions of Z in turn. Two-phase method: Phase 1: Phase 2:

Minimize Minimize

Z x4 x6 Z 0.4x1 0.5x2

(until x4 0, x6 0). (with x4 0, x6 0).

The phase 1 objective function is obtained by dividing the Big M method objective function by M and then dropping the negligible terms. Since phase 1 concludes by obtaining

144

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a BF solution for the real problem (one where x4 0 and x6 0), this solution is then used as the initial BF solution for applying the simplex method to the real problem (with its real objective function) in phase 2. Before solving the example in this way, we summarize the general method. Summary of the Two-Phase Method. Initialization: Revise the constraints of the original problem by introducing artificial variables as needed to obtain an obvious initial BF solution for the artificial problem. Phase 1: The objective for this phase is to find a BF solution for the real problem. To do this, Minimize Z artificial variables, subject to revised constraints. The optimal solution obtained for this problem (with Z 0) will be a BF solution for the real problem. Phase 2: The objective for this phase is to find an optimal solution for the real problem. Since the artificial variables are not part of the real problem, these variables can now be dropped (they are all zero now anyway).1 Starting from the BF solution obtained at the end of phase 1, use the simplex method to solve the real problem. For the example, the problems to be solved by the simplex method in the respective phases are summarized below. Phase 1 Problem (Radiation Therapy Example): Minimize

Z x4 x6,

subject to 0.3x1 0.1x2 x3 2.7 0.5x1 0.5x2 x4 6 0.6x1 0.4x2 x5 x6 6 and x1 0,

x2 0,

x3 0,

x4 0,

x5 0,

x6 0.

Phase 2 Problem (Radiation Therapy Example): Minimize

Z 0.4x1 0.5x2,

subject to 0.3x1 0.1x2 x3 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 x5 6 and x1 0,

x2 0,

x3 0,

x5 0.

We are skipping over three other possibilities here: (1) artificial variables 0 (discussed in the next subsection), (2) artificial variables that are degenerate basic variables, and (3) retaining the artificial variables as nonbasic variables in phase 2 (and not allowing them to become basic) as an aid to subsequent postoptimality analysis. Your OR Courseware allows you to explore these possibilities. 1

4.6 ADAPTING TO OTHER MODEL FORMS

145

The only differences between these two problems are in the objective function and in the inclusion (phase 1) or exclusion (phase 2) of the artificial variables x4 and x6. Without the artificial variables, the phase 2 problem does not have an obvious initial BF solution. The sole purpose of solving the phase 1 problem is to obtain a BF solution with x4 0 and x6 0 so that this solution (without the artificial variables) can be used as the initial BF solution for phase 2. Table 4.13 shows the result of applying the simplex method to this phase 1 problem. [Row 0 in the initial tableau is obtained by converting Minimize Z x4 x6 to Maximize (Z) x4 x6 and then using elementary row operations to eliminate the basic variables x4 and x6 from Z x4 x6 0.] In the next-to-last tableau, there is a tie for the entering basic variable between x3 and x5, which is broken arbitrarily in favor of x3. The solution obtained at the end of phase 1, then, is (x1, x2, x3, x4, x5, x6) (6, 6, 0.3, 0, 0, 0) or, after x4 and x6 are dropped, (x1, x2, x3, x5) (6, 6, 0.3, 0). As claimed in the summary, this solution from phase 1 is indeed a BF solution for the real problem (the phase 2 problem) because it is the solution (after you set x5 0) to the system of equations consisting of the three functional constraints for the phase 2 problem. In fact, after deleting the x4 and x6 columns as well as row 0 for each iteration, Table TABLE 4.13 Phase 1 of the two-phase method for the radiation therapy example Coefficient of: Iteration

0

Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Right Side

Z x3 x4 x6

(0) (1) (2) (3)

1 0 0 0

1.1 0.3 0.5 0.6

0.9 0.1 0.5 0.4

00 01 00 00

0 0 1 0

1 0 0 1

0 0 0 1

12 2.7 6.0 6.0

Z

(0)

1

0.0

16 30

11 3

0

1

0

2.1

x1

(1)

0

1.0

1 3

10 3

0

0

0

9.0

x4

(2)

0

0.0

0

0

1.5

(3)

0

0.0

5 3 2

1

x6

1 3 0.2

0

1

1

0.6

Z

(0)

1

0.0

0.0

5 3

0

5 3

8 3

0.5

x1

(1)

0

1.0

0.0

20 3

0

5 3

5 3

8.0

x4

(2)

0

0.0

0.0

1

(3)

0

0.0

1.0

5 3 5

5 3 5

0.5

x2

5 3 10

Z x1

(0) (1)

1 0

0.0 1.0

0.0 0.0

00 00

0 5

1 5

0.0 6.0

x3

(2)

0

0.0

0.0

01

1

1

0.3

x2

(3)

0

0.0

1.0

00

1 4 3 5 6

5

5

6.0

1

2

3

0

3.0

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4.13 shows one way of using Gaussian elimination to solve this system of equations by reducing the system to the form displayed in the final tableau. Table 4.14 shows the preparations for beginning phase 2 after phase 1 is completed. Starting from the final tableau in Table 4.13, we drop the artificial variables (x4 and x6), substitute the phase 2 objective function (Z 0.4x1 0.5x2 in maximization form) into row 0, and then restore the proper form from Gaussian elimination (by algebraically eliminating the basic variables x1 and x2 from row 0). Thus, row 0 in the last tableau is obtained by performing the following elementary row operations in the next-to-last tableau: from row 0 subtract both the product, 0.4 times row 1, and the product, 0.5 times row 3. Except for the deletion of the two columns, note that rows 1 to 3 never change. The only adjustments occur in row 0 in order to replace the phase 1 objective function by the phase 2 objective function. The last tableau in Table 4.14 is the initial tableau for applying the simplex method to the phase 2 problem, as shown at the top of Table 4.15. Just one iteration then leads to the optimal solution shown in the second tableau: (x1, x2, x3, x5) (7.5, 4.5, 0, 0.3). This solution is the desired optimal solution for the real problem of interest rather than the artificial problem constructed for phase 1. Now we see what the two-phase method has done graphically in Fig. 4.7. Starting at the origin, phase 1 examines a total of four CPF solutions for the artificial problem. The first three actually were corner-point infeasible solutions for the real problem shown in Fig. 4.5. The fourth CPF solution, at (6, 6), is the first one that also is feasible for the real TABLE 4.14 Preparing to begin phase 2 for the radiation therapy example Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Right Side

Z x1

(0) (1)

1 0

00. 10.

0.0 0.0

0 0

0.0 5.0

1 5

0.0 6.0

x3

(2)

0

00.

0.0

1

1.0

1

0.3

x2

(3)

0

00.

1.0

0

1 4 3 5 6

5.0

5

6.0

Z x1

(0) (1)

1 0

00. 10.

0.0 0.0

0 0

0.0 5.0

0.0 6.0

x3 x2

(2) (3)

0 0

00. 00.

0.0 1.0

1 0

1.0 5.0

0.3 6.0

Substitute phase 2 objective function

Z x1 x3 x2

(0) (1) (2) (3)

1 0 0 0

0.4 10. 00. 00.

0.5 0.0 0.0 1.0

0 0 1 0

0.0 5.0 1.0 5.0

0.0 6.0 0.3 6.0

Restore proper form from Gaussian elimination

Z x1 x3 x2

(0) (1) (2) (3)

1 0 0 0

00. 10. 00. 00.

0.0 0.0 0.0 1.0

0 0 1 0

0.5 5.0 1.0 5.0

5.4 6.0 0.3 6.0

Final Phase 1 tableau

Drop x4 and x6

4.6 ADAPTING TO OTHER MODEL FORMS

147

TABLE 4.15 Phase 2 of the two-phase method for the radiation therapy example Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x5

Right Side

0

Z x1 x3 x2

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

0 0 0 1

0.0 0.0 1.0 0.0

0.5 5.0 1.0 5.0

5.40 6.00 0.30 6.00

1

Z x1 x5 x2

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

0 0 0 1

0.5 5.0 1.0 5.0

0.0 0.0 1.0 0.0

5.25 7.50 0.30 4.50

Iteration

FIGURE 4.7 This graph shows the sequence of CPF solutions for phase 1 (, , , ) and then for phase 2 ( 0 , 1 ) when the two-phase method is applied to the radiation therapy example.

x2 (0, 12)

(6, 6) This dark line segment is the feasible region for the real problem (phase 2).

0 3

1 (7.5, 4.5) optimal Feasible region for the artificial problem (phase 1)

0

2

(8, 3)

1

(0, 0) (9, 0)

x1

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

problem, so it becomes the initial CPF solution for phase 2. One iteration in phase 2 leads to the optimal CPF solution at (7.5, 4.5). If the tie for the entering basic variable in the next-to-last tableau of Table 4.13 had been broken in the other way, then phase 1 would have gone directly from (8, 3) to (7.5, 4.5). After (7.5, 4.5) was used to set up the initial simplex tableau for phase 2, the optimality test would have revealed that this solution was optimal, so no iterations would be done. It is interesting to compare the Big M and two-phase methods. Begin with their objective functions. Big M Method: Minimize

Z 0.4x1 0.5x2 Mx4 Mxx6.

Two-Phase Method: Phase 1: Phase 2:

Minimize Minimize

Z x4 x6. Z 0.4x1 0.5x2.

Because the Mx4 and Mx6 terms dominate the 0.4x1 and 0.5x2 terms in the objective function for the Big M method, this objective function is essentially equivalent to the phase 1 objective function as long as x4 and/or x6 is greater than zero. Then, when both x4 0 and x6 0, the objective function for the Big M method becomes completely equivalent to the phase 2 objective function. Because of these virtual equivalencies in objective functions, the Big M and twophase methods generally have the same sequence of BF solutions. The one possible exception occurs when there is a tie for the entering basic variable in phase 1 of the twophase method, as happened in the third tableau of Table 4.13. Notice that the first three tableaux of Tables 4.12 and 4.13 are almost identical, with the only difference being that the multiplicative factors of M in Table 4.12 become the sole quantities in the corresponding spots in Table 4.13. Consequently, the additive terms that broke the tie for the entering basic variable in the third tableau of Table 4.12 were not present to break this same tie in Table 4.13. The result for this example was an extra iteration for the two-phase method. Generally, however, the advantage of having the additive factors is minimal. The two-phase method streamlines the Big M method by using only the multiplicative factors in phase 1 and by dropping the artificial variables in phase 2. (The Big M method could combine the multiplicative and additive factors by assigning an actual huge number to M, but this might create numerical instability problems.) For these reasons, the two-phase method is commonly used in computer codes. No Feasible Solutions So far in this section we have been concerned primarily with the fundamental problem of identifying an initial BF solution when an obvious one is not available. You have seen how the artificial-variable technique can be used to construct an artificial problem and obtain an initial BF solution for this artificial problem instead. Use of either the Big M method or the two-phase method then enables the simplex method to begin its pilgrim-

4.6 ADAPTING TO OTHER MODEL FORMS

149

age toward the BF solutions, and ultimately toward the optimal solution, for the real problem. However, you should be wary of a certain pitfall with this approach. There may be no obvious choice for the initial BF solution for the very good reason that there are no feasible solutions at all! Nevertheless, by constructing an artificial feasible solution, there is nothing to prevent the simplex method from proceeding as usual and ultimately reporting a supposedly optimal solution. Fortunately, the artificial-variable technique provides the following signpost to indicate when this has happened: If the original problem has no feasible solutions, then either the Big M method or phase 1 of the two-phase method yields a final solution that has at least one artificial variable greater than zero. Otherwise, they all equal zero.

To illustrate, let us change the first constraint in the radiation therapy example (see Fig. 4.5) as follows: 0.3x1 0.1x2 2.7

0.3x1 0.1x2 1.8,

so that the problem no longer has any feasible solutions. Applying the Big M method just as before (see Table 4.12) yields the tableaux shown in Table 4.16. (Phase 1 of the twophase method yields the same tableaux except that each expression involving M is replaced by just the multiplicative factor.) Hence, the Big M method normally would be indicating that the optimal solution is (3, 9, 0, 0, 0, 0.6). However, since an artificial variable x6 0.6 0, the real message here is that the problem has no feasible solutions.

TABLE 4.16 The Big M method for the revision of the radiation therapy example that has no feasible solutions Coefficient of: Iteration

0

Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Z x3 x4 x6

(0) (1) (2) (3)

1 0 0 0

1.1M 0.4 0.3 0.5 0.6

0.9M 0.5 0.1 0.5 0.4

0 1 0 0

0.0 0.0 1.0 0.0

M 0 0 1

0 0 0 1

Z

(0)

1

0.0

16 11 M 30 30

11 4 M 3 3

0.0

M

0

5.4M 2.4

x1

(1)

0

1.0

1 3

10 3

0.0

0

0

6.0

x4

(2)

0

0.0

0

0

3.0

(3)

0

0.0

5 3 2

1.0

x6

1 3 0.2

0.0

1

1

2.4

Z x1 x2 x6

(0) (1) (2) (3)

1 0 0 0

0.0 1.0 0.0 0.0

0.0 0.0 1.0 0.0

M 0.5 5 5 1

1.6M 1.1 1.0 3.0 0.6

M 0 0 1

0 0 0 1

0.6M 5.7 3.0 9.0 0.6

1

2

Right Side 12M 1.8 6.0 6.0

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4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

Variables Allowed to Be Negative In most practical problems, negative values for the decision variables would have no physical meaning, so it is necessary to include nonnegativity constraints in the formulations of their linear programming models. However, this is not always the case. To illustrate, suppose that the Wyndor Glass Co. problem is changed so that product 1 already is in production, and the first decision variable x1 represents the increase in its production rate. Therefore, a negative value of x1 would indicate that product 1 is to be cut back by that amount. Such reductions might be desirable to allow a larger production rate for the new, more profitable product 2, so negative values should be allowed for x1 in the model. Since the procedure for determining the leaving basic variable requires that all the variables have nonnegativity constraints, any problem containing variables allowed to be negative must be converted to an equivalent problem involving only nonnegative variables before the simplex method is applied. Fortunately, this conversion can be done. The modification required for each variable depends upon whether it has a (negative) lower bound on the values allowed. Each of these two cases is now discussed. Variables with a Bound on the Negative Values Allowed. Consider any decision variable xj that is allowed to have negative values which satisfy a constraint of the form xj Lj, where Lj is some negative constant. This constraint can be converted to a nonnegativity constraint by making the change of variables x j xj Lj,

so

x j 0.

Thus, x j Lj would be substituted for xj throughout the model, so that the redefined decision variable x j cannot be negative. (This same technique can be used when Lj is positive to convert a functional constraint xj Lj to a nonnegativity constraint x j 0.) To illustrate, suppose that the current production rate for product 1 in the Wyndor Glass Co. problem is 10. With the definition of x1 just given, the complete model at this point is the same as that given in Sec. 3.1 except that the nonnegativity constraint x1 0 is replaced by x1 10. To obtain the equivalent model needed for the simplex method, this decision variable would be redefined as the total production rate of product 1 x j x1 10, which yields the changes in the objective function and constraints as shown: Z 3x1 5x2 3x1 2x2 4 3x1 2x2 12 3x1 2x2 18 x1 10, x2 0

Z 3(x 1 10) 5x2 3(x 1 10) 2x2 4 3(x 1 10) 2x2 12 3(x 1 10) 2x2 18 x 1 10 10, x2 0

Z 30 3x 1 5x2 2x 1 2x2 14 3x 1 2x2 12 3x 1 2x2 48 x 1 0, x2 0

4.6 ADAPTING TO OTHER MODEL FORMS

151

Variables with No Bound on the Negative Values Allowed. In the case where xj does not have a lower-bound constraint in the model formulated, another approach is required: xj is replaced throughout the model by the difference of two new nonnegative variables xj x j xj ,

where x j 0, xj 0.

Since x j and xj can have any nonnegative values, this difference xj xj can have any value (positive or negative), so it is a legitimate substitute for xj in the model. But after such substitutions, the simplex method can proceed with just nonnegative variables. The new variables x j and xj have a simple interpretation. As explained in the next paragraph, each BF solution for the new form of the model necessarily has the property that either x j 0 or xj 0 (or both). Therefore, at the optimal solution obtained by the simplex method (a BF solution),

0 x 0

x j x j

xj

j

if xj 0, otherwise; if xj 0, otherwise;

so that x j represents the positive part of the decision variable xj and xj its negative part (as suggested by the superscripts). For example, if xj 10, the above expressions give x j 10 and xj 0. This same value of xj xj xj 10 also would occur with larger values of xj and x j such that x x 10. Plotting these values of x and x on a two-dimensional graph gives a line j j j j with an endpoint at x 10, x 0 to avoid violating the nonnegativity constraints. This j j endpoint is the only corner-point solution on the line. Therefore, only this endpoint can be part of an overall CPF solution or BF solution involving all the variables of the model. This illustrates why each BF solution necessarily has either x j 0 or xj 0 (or both). To illustrate the use of the xj and xj , let us return to the example on the preceding page where x1 is redefined as the increase over the current production rate of 10 for product 1 in the Wyndor Glass Co. problem. However, now suppose that the x1 10 constraint was not included in the original model because it clearly would not change the optimal solution. (In some problems, certain variables do not need explicit lower-bound constraints because the functional constraints already prevent lower values.) Therefore, before the simplex method is applied, x1 would be replaced by the difference

x1 x1 x1,

where x1 0, x1 0,

as shown: Maximize subject to

Z 3x1 5x2, Z 3x1 5x2 4 2x2 12 3x1 2x2 18 x2 0 (only)

Maximize subject to

Z 3x1 3x1 5x2, Z 3x1 3x1 5x2 4 2x2 12 3x1 3x1 2x2 18 x1 0, x1 0, x2 0

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

From a computational viewpoint, this approach has the disadvantage that the new equivalent model to be used has more variables than the original model. In fact, if all the original variables lack lower-bound constraints, the new model will have twice as many variables. Fortunately, the approach can be modified slightly so that the number of variables is increased by only one, regardless of how many original variables need to be replaced. This modification is done by replacing each such variable xj by xj x j x,

where x j 0, x 0,

instead, where x is the same variable for all relevant j. The interpretation of x in this case is that x is the current value of the largest (in absolute terms) negative original variable, so that x j is the amount by which xj exceeds this value. Thus, the simplex method now can make some of the x j variables larger than zero even when x 0.

4.7

POSTOPTIMALITY ANALYSIS We stressed in Secs. 2.3, 2.4, and 2.5 that postoptimality analysis—the analysis done after an optimal solution is obtained for the initial version of the model—constitutes a very major and very important part of most operations research studies. The fact that postoptimality analysis is very important is particularly true for typical linear programming applications. In this section, we focus on the role of the simplex method in performing this analysis. Table 4.17 summarizes the typical steps in postoptimality analysis for linear programming studies. The rightmost column identifies some algorithmic techniques that involve the simplex method. These techniques are introduced briefly here with the technical details deferred to later chapters. Reoptimization As discussed in Sec. 3.7, linear programming models that arise in practice commonly are very large, with hundreds or thousands of functional constraints and decision variables. In such cases, many variations of the basic model may be of interest for considering different scenarios. Therefore, after having found an optimal solution for one version of a linear programming model, we frequently must solve again (often many times) for the soTABLE 4.17 Postoptimality analysis for linear programming Task

Purpose

Technique

Model debugging Model validation Final managerial decisions on resource allocations (the bi values) Evaluate estimates of model parameters Evaluate trade-offs between model parameters

Find errors and weaknesses in model Demonstrate validity of final model Make appropriate division of organizational resources between activities under study and other important activities Determine crucial estimates that may affect optimal solution for further study Determine best trade-off

Reoptimization See Sec. 2.4 Shadow prices

Sensitivity analysis Parametric linear programming

4.7 POSTOPTIMALITY ANALYSIS

153

lution of a slightly different version of the model. We nearly always have to solve again several times during the model debugging stage (described in Secs. 2.3 and 2.4), and we usually have to do so a large number of times during the later stages of postoptimality analysis as well. One approach is simply to reapply the simplex method from scratch for each new version of the model, even though each run may require hundreds or even thousands of iterations for large problems. However, a much more efficient approach is to reoptimize. Reoptimization involves deducing how changes in the model get carried along to the final simplex tableau (as described in Secs. 5.3 and 6.6). This revised tableau and the optimal solution for the prior model are then used as the initial tableau and the initial basic solution for solving the new model. If this solution is feasible for the new model, then the simplex method is applied in the usual way, starting from this initial BF solution. If the solution is not feasible, a related algorithm called the dual simplex method (described in Sec. 7.1) probably can be applied to find the new optimal solution,1 starting from this initial basic solution. The big advantage of this reoptimization technique over re-solving from scratch is that an optimal solution for the revised model probably is going to be much closer to the prior optimal solution than to an initial BF solution constructed in the usual way for the simplex method. Therefore, assuming that the model revisions were modest, only a few iterations should be required to reoptimize instead of the hundreds or thousands that may be required when you start from scratch. In fact, the optimal solutions for the prior and revised models are frequently the same, in which case the reoptimization technique requires only one application of the optimality test and no iterations. Shadow Prices Recall that linear programming problems often can be interpreted as allocating resources to activities. In particular, when the functional constraints are in form, we interpreted the bi (the right-hand sides) as the amounts of the respective resources being made available for the activities under consideration. In many cases, there may be some latitude in the amounts that will be made available. If so, the bi values used in the initial (validated) model actually may represent management’s tentative initial decision on how much of the organization’s resources will be provided to the activities considered in the model instead of to other important activities under the purview of management. From this broader perspective, some of the bi values can be increased in a revised model, but only if a sufficiently strong case can be made to management that this revision would be beneficial. Consequently, information on the economic contribution of the resources to the measure of performance (Z ) for the current study often would be extremely useful. The simplex method provides this information in the form of shadow prices for the respective resources. The shadow price for resource i (denoted by y*i ) measures the marginal value of this resource, i.e., the rate at which Z could be increased by (slightly) increasing the amount of 1

The one requirement for using the dual simplex method here is that the optimality test is still passed when applied to row 0 of the revised final tableau. If not, then still another algorithm called the primal-dual method can be used instead.

154

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

this resource (bi) being made available.1,2 The simplex method identifies this shadow price by yi* coefficient of the ith slack variable in row 0 of the final simplex tableau.

To illustrate, for the Wyndor Glass Co. problem, Resource i production capacity of Plant i (i 1, 2, 3) being made available to the two new products under consideration, bi hours of production time per week being made available in Plant i for these new products. Providing a substantial amount of production time for the new products would require adjusting production times for the current products, so choosing the bi value is a difficult managerial decision. The tentative initial decision has been b1 4,

b2 12,

b3 18,

as reflected in the basic model considered in Sec. 3.1 and in this chapter. However, management now wishes to evaluate the effect of changing any of the bi values. The shadow prices for these three resources provide just the information that management needs. The final tableau in Table 4.8 (see p. 128) yields y*1 0 shadow price for resource 1, 3 y*2 shadow price for resource 2, 2 y*3 1 shadow price for resource 3. With just two decision variables, these numbers can be verified by checking graphically that individually increasing any bi by 1 indeed would increase the optimal value of Z by y*i . For example, Fig. 4.8 demonstrates this increase for resource 2 by reapplying the graphical method presented in Sec. 3.1. The optimal solution, (2, 6) with Z 36, changes to (53, 123) with Z 3712 when b2 is increased by 1 (from 12 to 13), so that 1 3 y*2 Z 37 36 . 2 2 Since Z is expressed in thousands of dollars of profit per week, y*2 32 indicates that adding 1 more hour of production time per week in Plant 2 for these two new products would increase their total profit by $1,500 per week. Should this actually be done? It depends on the marginal profitability of other products currently using this production time. If there is a current product that contributes less than $1,500 of weekly profit per hour of weekly production time in Plant 2, then some shift of production time to the new products would be worthwhile. We shall continue this story in Sec. 6.7, where the Wyndor OR team uses shadow prices as part of its sensitivity analysis of the model. 1

The increase in bi must be sufficiently small that the current set of basic variables remains optimal since this rate (marginal value) changes if the set of basic variables changes. 2 In the case of a functional constraint in or form, its shadow price is again defined as the rate at which Z could be increased by (slightly) increasing the value of bi, although the interpretation of bi now would normally be something other than the amount of a resource being made available.

4.7 POSTOPTIMALITY ANALYSIS

155

x2 3x1 2x2 18

Z 3x1 5x2

8 5 , 13 3 2

6 (2, 6) FIGURE 4.8 This graph shows that the shadow price is y2* 32 for resource 2 for the Wyndor Glass Co. problem. The two dots are the optimal solutions for b2 12 or b2 13, and plugging these solutions into the objective function reveals that increasing b2 by 1 increases Z by y2* 32.

5

13

Z 3 3 5 2 37 12 Z 3( 2 ) 5( 6 ) 36

2x2 13 2x2 12

Z

3 2

y*2

x1 4 4

2

0

2

4

6

x1

Figure 4.8 demonstrates that y*2 32 is the rate at which Z could be increased by increasing b2 slightly. However, it also demonstrates the common phenomenon that this interpretation holds only for a small increase in b2. Once b2 is increased beyond 18, the optimal solution stays at (0, 9) with no further increase in Z. (At that point, the set of basic variables in the optimal solution has changed, so a new final simplex tableau will be obtained with new shadow prices, including y*2 0.) Now note in Fig. 4.8 why y*1 0. Because the constraint on resource 1, x1 4, is not binding on the optimal solution (2, 6), there is a surplus of this resource. Therefore, increasing b1 beyond 4 cannot yield a new optimal solution with a larger value of Z. By contrast, the constraints on resources 2 and 3, 2x2 12 and 3x1 2x2 18, are binding constraints (constraints that hold with equality at the optimal solution). Because the limited supply of these resources (b2 12, b3 18) binds Z from being increased further, they have positive shadow prices. Economists refer to such resources as scarce goods, whereas resources available in surplus (such as resource 1) are free goods (resources with a zero shadow price). The kind of information provided by shadow prices clearly is valuable to management when it considers reallocations of resources within the organization. It also is very helpful when an increase in bi can be achieved only by going outside the organization to purchase more of the resource in the marketplace. For example, suppose that Z represents profit and that the unit profits of the activities (the cj values) include the costs (at regular prices) of all the resources consumed. Then a positive shadow price of y*i for resource i means that the total profit Z can be increased by y*i by purchasing 1 more unit of this resource at its regular price. Alternatively, if a premium price must be paid for the resource

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in the marketplace, then y*i represents the maximum premium (excess over the regular price) that would be worth paying.1 The theoretical foundation for shadow prices is provided by the duality theory described in Chap. 6. Sensitivity Analysis When discussing the certainty assumption for linear programming at the end of Sec. 3.3, we pointed out that the values used for the model parameters (the ai j, bi, and cj identified in Table 3.3) generally are just estimates of quantities whose true values will not become known until the linear programming study is implemented at some time in the future. A main purpose of sensitivity analysis is to identify the sensitive parameters (i.e., those that cannot be changed without changing the optimal solution). The sensitive parameters are the parameters that need to be estimated with special care to minimize the risk of obtaining an erroneous optimal solution. They also will need to be monitored particularly closely as the study is implemented. If it is discovered that the true value of a sensitive parameter differs from its estimated value in the model, this immediately signals a need to change the solution. How are the sensitive parameters identified? In the case of the bi, you have just seen that this information is given by the shadow prices provided by the simplex method. In particular, if y*i 0, then the optimal solution changes if bi is changed, so bi is a sensitive parameter. However, y*i 0 implies that the optimal solution is not sensitive to at least small changes in bi. Consequently, if the value used for bi is an estimate of the amount of the resource that will be available (rather than a managerial decision), then the bi values that need to be monitored more closely are those with positive shadow prices—especially those with large shadow prices. When there are just two variables, the sensitivity of the various parameters can be analyzed graphically. For example, in Fig. 4.9, c1 3 can be changed to any other value from 0 to 7.5 without the optimal solution changing from (2, 6). (The reason is that any value of c1 within this range keeps the slope of Z c1x1 5x2 between the slopes of the lines 2x2 12 and 3x1 2x2 18.) Similarly, if c2 5 is the only parameter changed, it can have any value greater than 2 without affecting the optimal solution. Hence, neither c1 nor c2 is a sensitive parameter. The easiest way to analyze the sensitivity of each of the aij parameters graphically is to check whether the corresponding constraint is binding at the optimal solution. Because x1 4 is not a binding constraint, any sufficiently small change in its coefficients (a11 1, a12 0) is not going to change the optimal solution, so these are not sensitive parameters. On the other hand, both 2x2 12 and 3x1 2x2 18 are binding constraints, so changing any one of their coefficients (a21 0, a22 2, a31 3, a32 2) is going to change the optimal solution, and therefore these are sensitive parameters. Typically, greater attention is given to performing sensitivity analysis on the bi and cj parameters than on the aij parameters. On real problems with hundreds or thousands of constraints and variables, the effect of changing one aij value is usually negligible, but 1

If the unit profits do not include the costs of the resources consumed, then y*i represents the maximum total unit price that would be worth paying to increase bi.

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157

x2 10

FIGURE 4.9 This graph demonstrates the sensitivity analysis of c1 and c2 for the Wyndor Glass Co. problem. Starting with the original objective function line [where c1 3, c2 5, and the optimal solution is (2, 6)], the other two lines show the extremes of how much the slope of the objective function line can change and still retain (2, 6) as an optimal solution. Thus, with c2 5, the allowable range for c1 is 0 c1 7.5. With c1 3, the allowable range for c2 is c2 2.

8 Z 36 3x1 5x2

Z 45 7.5x1 5x2 (or Z 18 3x1 2x2) (2, 6) optimal

Z 30 0x1 5x2

4

Feasible region

2

0

2

4

6

x1

changing one bi or cj value can have real impact. Furthermore, in many cases, the ai j values are determined by the technology being used (the aij values are sometimes called technological coefficients), so there may be relatively little (or no) uncertainty about their final values. This is fortunate, because there are far more aij parameters than bi and cj parameters for large problems. For problems with more than two (or possibly three) decision variables, you cannot analyze the sensitivity of the parameters graphically as was just done for the Wyndor Glass Co. problem. However, you can extract the same kind of information from the simplex method. Getting this information requires using the fundamental insight described in Sec. 5.3 to deduce the changes that get carried along to the final simplex tableau as a result of changing the value of a parameter in the original model. The rest of the procedure is described and illustrated in Secs. 6.6 and 6.7. Using Excel to Generate Sensitivity Analysis Information Sensitivity analysis normally is incorporated into software packages based on the simplex method. For example, the Excel Solver will generate sensitivity analysis information upon request. As was shown in Fig. 3.19 (see page 72), when the Solver gives the message that it has found a solution, it also gives on the right a list of three reports that can be provided. By selecting the second one (labeled “Sensitivity”) after solving the Wyndor Glass Co. problem, you will obtain the sensitivity report shown in Fig. 4.10. The upper table in this report provides sensitivity analysis information about the decision variables and their coefficients in the objective function. The lower table does the same for the functional constraints and their right-hand sides.

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FIGURE 4.10 The sensitivity report provided by the Excel Solver for the Wyndor Glass Co. problem.

Look first at the upper table in this figure. The “Final Value” column indicates the optimal solution. The next column gives the reduced costs. (We will not discuss these reduced costs now because the information they provide can also be gleaned from the rest of the upper table.) The next three columns provide the information needed to identify the allowable range to stay optimal for each coefficient cj in the objective function. For any cj, its allowable range to stay optimal is the range of values for this coefficient over which the current optimal solution remains optimal, assuming no change in the other coefficients.

The “Objective Coefficient” column gives the current value of each coefficient, and then the next two columns give the allowable increase and the allowable decrease from this value to remain within the allowable range. Therefore, 3 3 c1 3 4.5,

so

0 c1 7.5

is the allowable range for c1 over which the current optimal solution will stay optimal (assuming c2 5), just as was found graphically in Fig. 4.9. Similarly, since Excel uses 1E 30 (1030) to represent infinity, 5 3 c2 5 ,

so

2 c2

is the allowable range to stay optimal for c2. The fact that both the allowable increase and the allowable decrease are greater than zero for the coefficient of both decision variables provides another useful piece of information, as described below. When the upper table in the sensitivity report generated by the Excel Solver indicates that both the allowable increase and the allowable decrease are greater than zero for every objective coefficient, this is a signpost that the optimal solution in the “Final Value” column is the only optimal solution. Conversely, having any allowable increase or allowable decrease equal to zero is a signpost that there are multiple optimal solutions. Changing the corresponding coefficient a tiny amount beyond the zero allowed and re-solving provides another optimal CPF solution for the original model.

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159

Now consider the lower table in Fig. 4.10 that focuses on sensitivity analysis for the three functional constraints. The “Final Value” column gives the value of each constraint’s left-hand side for the optimal solution. The next two columns give the shadow price and the current value of the right-hand side (bi) for each constraint. When just one bi value is then changed, the last two columns give the allowable increase or allowable decrease in order to remain within its allowable range to stay feasible. For any bi, its allowable range to stay feasible is the range of values for this right-hand side over which the current optimal BF solution (with adjusted values1 for the basic variables) remains feasible, assuming no change in the other right-hand sides.

Thus, using the lower table in Fig. 4.10, combining the last two columns with the current values of the right-hand sides gives the following allowable ranges to stay feasible: 2 b1 6 b2 18 12 b3 24. This sensitivity report generated by the Excel Solver is typical of the sensitivity analysis information provided by linear programming software packages. You will see in Appendix 4.1 that LINDO provides essentially the same report. MPL/CPLEX does also when it is requested with the Solution File dialogue box. Once again, this information obtained algebraically also can be derived from graphical analysis for this two-variable problem. (See Prob. 4.7-1.) For example, when b2 is increased from 12 in Fig. 4.8, the originally optimal CPF solution at the intersection of two constraint boundaries 2x2 b2 and 3x1 2x2 18 will remain feasible (including x1 0) only for b2 18. The latter part of Chap. 6 will delve into this type of analysis more deeply. Parametric Linear Programming Sensitivity analysis involves changing one parameter at a time in the original model to check its effect on the optimal solution. By contrast, parametric linear programming (or parametric programming for short) involves the systematic study of how the optimal solution changes as many of the parameters change simultaneously over some range. This study can provide a very useful extension of sensitivity analysis, e.g., to check the effect of “correlated” parameters that change together due to exogenous factors such as the state of the economy. However, a more important application is the investigation of trade-offs in parameter values. For example, if the cj values represent the unit profits of the respective activities, it may be possible to increase some of the cj values at the expense of decreasing others by an appropriate shifting of personnel and equipment among activities. Similarly, if the bi values represent the amounts of the respective resources being made available, it may be possible to increase some of the bi values by agreeing to accept decreases in some of the others. The analysis of such possibilities is discussed and illustrated at the end of Sec. 6.7. 1

Since the values of the basic variables are obtained as the simultaneous solution of a system of equations (the functional constraints in augmented form), at least some of these values change if one of the right-hand sides changes. However, the adjusted values of the current set of basic variables still will satisfy the nonnegativity constraints, and so still will be feasible, as long as the new value of this right-hand side remains within its allowable range to stay feasible. If the adjusted basic solution is still feasible, it also will still be optimal. We shall elaborate further in Sec. 6.7.

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In some applications, the main purpose of the study is to determine the most appropriate trade-off between two basic factors, such as costs and benefits. The usual approach is to express one of these factors in the objective function (e.g., minimize total cost) and incorporate the other into the constraints (e.g., benefits minimum acceptable level), as was done for the Nori & Leets Co. air pollution problem in Sec. 3.4. Parametric linear programming then enables systematic investigation of what happens when the initial tentative decision on the trade-off (e.g., the minimum acceptable level for the benefits) is changed by improving one factor at the expense of the other. The algorithmic technique for parametric linear programming is a natural extension of that for sensitivity analysis, so it, too, is based on the simplex method. The procedure is described in Sec. 7.2.

4.8

COMPUTER IMPLEMENTATION If the electronic computer had never been invented, undoubtedly you would have never heard of linear programming and the simplex method. Even though it is possible to apply the simplex method by hand to solve tiny linear programming problems, the calculations involved are just too tedious to do this on a routine basis. However, the simplex method is ideally suited for execution on a computer. It is the computer revolution that has made possible the widespread application of linear programming in recent decades. Implementation of the Simplex Method Computer codes for the simplex method now are widely available for essentially all modern computer systems. These codes commonly are part of a sophisticated software package for mathematical programming that includes many of the procedures described in subsequent chapters (including those used for postoptimality analysis). These production computer codes do not closely follow either the algebraic form or the tabular form of the simplex method presented in Secs. 4.3 and 4.4. These forms can be streamlined considerably for computer implementation. Therefore, the codes use instead a matrix form (usually called the revised simplex method) that is especially well suited for the computer. This form accomplishes exactly the same things as the algebraic or tabular form, but it does this while computing and storing only the numbers that are actually needed for the current iteration; and then it carries along the essential data in a more compact form. The revised simplex method is described in Sec. 5.2. The simplex method is used routinely to solve surprisingly large linear programming problems. For example, powerful desktop computers (especially workstations) commonly are used to solve problems with many thousand functional constraints and a larger number of decision variables. We now are beginning to hear reports of successfully solved problems ranging into the hundreds of thousands of functional constraints and millions of decision variables.1 For certain special types of linear programming problems (such as the 1

Do not try this at home. Attacking such a massive problem requires an especially sophisticated linear programming system that uses the latest techniques for exploiting sparcity in the coefficient matrix as well as other special techniques (e.g., crashing techniques for quickly finding an advanced initial BF solution). When problems are re-solved periodically after minor updating of the data, much time often is saved by using (or modifying) the last optimal solution to provide the initial BF solution for the new run.

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161

transportation, assignment, and minimum cost flow problems to be described later in the book), even larger problems now can be solved by specialized versions of the simplex method. Several factors affect how long it will take to solve a linear programming problem by the general simplex method. The most important one is the number of ordinary functional constraints. In fact, computation time tends to be roughly proportional to the cube of this number, so that doubling this number may multiply the computation time by a factor of approximately 8. By contrast, the number of variables is a relatively minor factor.1 Thus, doubling the number of variables probably will not even double the computation time. A third factor of some importance is the density of the table of constraint coefficients (i.e., the proportion of the coefficients that are not zero), because this affects the computation time per iteration. (For large problems encountered in practice, it is common for the density to be under 5 percent, or even under 1 percent, and this much “sparcity” tends to greatly accelerate the simplex method.) One common rule of thumb for the number of iterations is that it tends to be roughly twice the number of functional constraints. With large linear programming problems, it is inevitable that some mistakes and faulty decisions will be made initially in formulating the model and inputting it into the computer. Therefore, as discussed in Sec. 2.4, a thorough process of testing and refining the model (model validation) is needed. The usual end product is not a single static model that is solved once by the simplex method. Instead, the OR team and management typically consider a long series of variations on a basic model (sometimes even thousands of variations) to examine different scenarios as part of postoptimality analysis. This entire process is greatly accelerated when it can be carried out interactively on a desktop computer. And, with the help of both mathematical programming modeling languages and improving computer technology, this now is becoming common practice. Until the mid-1980s, linear programming problems were solved almost exclusively on mainframe computers. Since then, there has been an explosion in the capability of doing linear programming on desktop computers, including personal computers as well as workstations. Workstations, including some with parallel processing capabilities, now are commonly used instead of mainframe computers to solve massive linear programming models. The fastest personal computers are not lagging far behind, although solving huge models usually requires additional memory. Linear Programming Software Featured in This Book A considerable number of excellent software packages for linear programming and its extensions now are available to fill a variety of needs. One that is widely regarded to be a particularly powerful package for solving massive problems is CPLEX, a product of ILOG, Inc., located in Silicon Valley. For more than a decade, CPLEX has helped to lead the way in solving larger and larger linear programming problems. An extensive research and development effort has enabled a series of upgrades with dramatic increases in efficiency. CPLEX 6.5 released in March 1999 provided another order-of-magnitude improvement. This software package has successfully solved real linear programming problems arising in industry with as many as 2 million functional constraints and a comparable number of 1

This statement assumes that the revised simplex method described in Sec. 5.2 is being used.

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decision variables! CPLEX 6.5 often uses the simplex method and its variants (such as the dual simplex method presented in Sec. 7.1) to solve these massive problems. In addition to the simplex method, CPLEX 6.5 also features some other powerful weapons for attacking linear programming problems. One is a lightning-fast algorithm that uses the interior-point approach introduced in the next section. This algorithm can solve some huge general linear programming problems that the simplex method cannot (and vice versa). Another feature is the network simplex method (described in Sec. 9.7) that can solve even larger special types of linear programming problems. CPLEX 6.5 also extends beyond linear programming by including state-of-the-art algorithms for integer programming (Chap. 12) and quadratic programming (Sec. 13.7). Because it often is used to solve really large problems, CPLEX normally is used in conjunction with a mathematical programming modeling language. As described in Sec. 3.7, modeling languages are designed for efficiently formulating large linear programming models (and related models) in a compact way, after which a solver is called upon to solve the model. Several of the prominent modeling languages support CPLEX as a solver. ILOG also has recently introduced its own modeling language, called OPL Studio, that can be used with CPLEX. (A trial version of OPL Studio is available at ILOG’s website, www.ilog.com.) As we mentioned in Sec. 3.7, the student version of CPLEX is included in your OR Courseware as the solver for the MPL modeling language. This version features the simplex method for solving linear programming problems. LINDO (short for Linear, INteractive, and Discrete Optimizer) is another prominent software package for linear programming and its extensions. A product of LINDO Systems based in Chicago, LINDO has an even longer history than CPLEX. Although not as powerful as CPLEX, the largest version of LINDO has solved problems with tens of thousands of functional constraints and hundreds of thousands of decision variables. Its longtime popularity is partially due to its ease of use. For relatively small (textbook-sized) problems, the model can be entered and solved in an intuitive straightforward manner, so LINDO provides a convenient tool for students to use. However, LINDO lacks some of the capabilities of modeling languages for dealing with large linear programming problems. For such problems, it may be more efficient to use the LINGO modeling language to formulate the model and then to call the solver it shares with LINDO to solve the model. You can download the student version of LINDO from the website, www.lindo.com. Appendix 4.1 provides an introduction to how to use LINDO. The CD-ROM also includes a LINDO tutorial, as well as LINDO formulations for all the examples in this book to which it can be applied. Spreadsheet-based solvers are becoming increasingly popular for linear programming and its extensions. Leading the way are the solvers produced by Frontline Systems for Microsoft Excel, Lotus 1-2-3, and Corel Quattro Pro. In addition to the basic solver shipped with these packages, two more powerful upgrades—Premium Solver and Premium Solver Plus—also are available. Because of the widespread use of spreadsheet packages such as Microsoft Excel today, these solvers are introducing large numbers of people to the potential of linear programming for the first time. For textbook-sized linear programming problems (and considerably larger problems as well), spreadsheets provide a convenient way to formulate and solve the model, as described in Sec. 3.6. The more powerful spreadsheet solvers can solve fairly large models with many thousand decision variables. How-

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163

ever, when the spreadsheet grows to an unwieldy size, a good modeling language and its solver may provide a more efficient approach to formulating and solving the model. Spreadsheets provide an excellent communication tool, especially when dealing with typical managers who are very comfortable with this format but not with the algebraic formulations of OR models. Therefore, optimization software packages and modeling languages now can commonly import and export data and results in a spreadsheet format. For example, the MPL modeling language now includes an enhancement (called the OptiMax 2000 Component Library) that enables the modeler to create the feel of a spreadsheet model for the user of the model while still using MPL to formulate the model very efficiently. (The student version of OptiMax 2000 is included in your OR Courseware.) Premium Solver is one of the Excel add-ins included on the CD-ROM. You can install this add-in to obtain a much better performance than with the standard Excel Solver. Consequently, all the software, tutorials, and examples packed on the CD-ROM are providing you with several attractive software options for linear programming. Available Software Options for Linear Programming. 1. Demonstration examples (in OR Tutor) and interactive routines for efficiently learning the simplex method. 2. Excel and its Premium Solver for formulating and solving linear programming models in a spreadsheet format. 3. MPL/CPLEX for efficiently formulating and solving large linear programming models. 4. LINGO and its solver (shared with LINDO) for an alternative way of efficiently formulating and solving large linear programming models. 5. LINDO for formulating and solving linear programming models in a straightforward way. Your instructor may specify which software to use. Whatever the choice, you will be gaining experience with the kind of state-of-the-art software that is used by OR professionals.

4.9

THE INTERIOR-POINT APPROACH TO SOLVING LINEAR PROGRAMMING PROBLEMS The most dramatic new development in operations research during the 1980s was the discovery of the interior-point approach to solving linear programming problems. This discovery was made in 1984 by a young mathematician at AT&T Bell Laboratories, Narendra Karmarkar, when he successfully developed a new algorithm for linear programming with this kind of approach. Although this particular algorithm experienced only mixed success in competing with the simplex method, the key solution concept described below appeared to have great potential for solving huge linear programming problems beyond the reach of the simplex method. Many top researchers subsequently worked on modifying Karmarkar’s algorithm to fully tap this potential. Much progress has been made (and continues to be made), and a number of powerful algorithms using the interior-point approach have been developed. Today, the more powerful software packages that are designed for solving really large linear programming problems (such as CPLEX) include at least one algorithm using the interior-point approach along with the simplex method. As research continues on these algorithms, their computer implementations continue to improve. This has spurred renewed research on the simplex method, and its computer implementations continue to improve as well (recall the dramatic advance by CPLEX 6.5

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cited in the preceding section). The competition between the two approaches for supremacy in solving huge problems is continuing. Now let us look at the key idea behind Karmarkar’s algorithm and its subsequent variants that use the interior-point approach. The Key Solution Concept Although radically different from the simplex method, Karmarkar’s algorithm does share a few of the same characteristics. It is an iterative algorithm. It gets started by identifying a feasible trial solution. At each iteration, it moves from the current trial solution to a better trial solution in the feasible region. It then continues this process until it reaches a trial solution that is (essentially) optimal. The big difference lies in the nature of these trial solutions. For the simplex method, the trial solutions are CPF solutions (or BF solutions after augmenting), so all movement is along edges on the boundary of the feasible region. For Karmarkar’s algorithm, the trial solutions are interior points, i.e., points inside the boundary of the feasible region. For this reason, Karmarkar’s algorithm and its variants are referred to as interior-point algorithms. To illustrate, Fig. 4.11 shows the path followed by the interior-point algorithm in your OR Courseware when it is applied to the Wyndor Glass Co. problem, starting from the

FIGURE 4.11 The curve from (1, 2) to (2, 6) shows a typical path followed by an interior-point algorithm, right through the interior of the feasible region for the Wyndor Glass Co. problem.

x2 (2, 6) optimal

6 (1.56, 5.5)

(1.38, 5)

4

2

0

(1.27, 4)

(1, 2)

2

4

x1

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TABLE 4.18 Output of interior-point algorithm in OR Courseware for Wyndor Glass Co. problem Iteration

x1

x2

Z

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1.27298 1.37744 1.56291 1.80268 1.92134 1.96639 1.98385 1.99197 1.99599 1.99799 1.999 1.9995 1.99975 1.99987 1.99994

2 4 5 5.5 5.71816 5.82908 5.90595 5.95199 5.97594 5.98796 5.99398 5.99699 5.9985 5.99925 5.99962 5.99981

13 23.8189 29.1323 32.1887 33.9989 34.9094 35.429 35.7115 35.8556 35.9278 35.9639 35.9819 35.991 35.9955 35.9977 35.9989

initial trial solution (1, 2). Note how all the trial solutions (dots) shown on this path are inside the boundary of the feasible region as the path approaches the optimal solution (2, 6). (All the subsequent trial solutions not shown also are inside the boundary of the feasible region.) Contrast this path with the path followed by the simplex method around the boundary of the feasible region from (0, 0) to (0, 6) to (2, 6). Table 4.18 shows the actual output from your OR Courseware for this problem.1 (Try it yourself.) Note how the successive trial solutions keep getting closer and closer to the optimal solution, but never literally get there. However, the deviation becomes so infinitesimally small that the final trial solution can be taken to be the optimal solution for all practical purposes. Section 7.4 presents the details of the specific interior-point algorithm that is implemented in your OR Courseware. Comparison with the Simplex Method One meaningful way of comparing interior-point algorithms with the simplex method is to examine their theoretical properties regarding computational complexity. Karmarkar has proved that the original version of his algorithm is a polynomial time algorithm; i.e., the time required to solve any linear programming problem can be bounded above by a polynomial function of the size of the problem. Pathological counterexamples have been constructed to demonstrate that the simplex method does not possess this property, so it is an exponential time algorithm (i.e., the required time can be bounded above only by an exponential function of the problem size). This difference in worst-case performance 1 The routine is called Solve Automatically by the Interior-Point Algorithm. The option menu provides two choices for a certain parameter of the algorithm (defined in Sec. 7.4). The choice used here is the default value of 0.5.

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is noteworthy. However, it tells us nothing about their comparison in average performance on real problems, which is the more crucial issue. The two basic factors that determine the performance of an algorithm on a real problem are the average computer time per iteration and the number of iterations. Our next comparisons concern these factors. Interior-point algorithms are far more complicated than the simplex method. Considerably more extensive computations are required for each iteration to find the next trial solution. Therefore, the computer time per iteration for an interior-point algorithm is many times longer than that for the simplex method. For fairly small problems, the numbers of iterations needed by an interior-point algorithm and by the simplex method tend to be somewhat comparable. For example, on a problem with 10 functional constraints, roughly 20 iterations would be typical for either kind of algorithm. Consequently, on problems of similar size, the total computer time for an interior-point algorithm will tend to be many times longer than that for the simplex method. On the other hand, a key advantage of interior-point algorithms is that large problems do not require many more iterations than small problems. For example, a problem with 10,000 functional constraints probably will require well under 100 iterations. Even considering the very substantial computer time per iteration needed for a problem of this size, such a small number of iterations makes the problem quite tractable. By contrast, the simplex method might need 20,000 iterations and so might not finish within a reasonable amount of computer time. Therefore, interior-point algorithms often are faster than the simplex method for such huge problems. The reason for this very large difference in the number of iterations on huge problems is the difference in the paths followed. At each iteration, the simplex method moves from the current CPF solution to an adjacent CPF solution along an edge on the boundary of the feasible region. Huge problems have an astronomical number of CPF solutions. The path from the initial CPF solution to an optimal solution may be a very circuitous one around the boundary, taking only a small step each time to the next adjacent CPF solution, so a huge number of steps may be required to reach an optimal solution. By contrast, an interior-point algorithm bypasses all this by shooting through the interior of the feasible region toward an optimal solution. Adding more functional constraints adds more constraint boundaries to the feasible region, but has little effect on the number of trial solutions needed on this path through the interior. This makes it possible for interior-point algorithms to solve problems with a huge number of functional constraints. A final key comparison concerns the ability to perform the various kinds of postoptimality analysis described in Sec. 4.7. The simplex method and its extensions are very well suited to and are widely used for this kind of analysis. Unfortunately, the interiorpoint approach currently has limited capability in this area.1 Given the great importance of postoptimality analysis, this is a crucial drawback of interior-point algorithms. However, we point out next how the simplex method can be combined with the interior-point approach to overcome this drawback. 1

However, research aimed at increasing this capability continues to make progress. For example, see H. J. Greenberg, “Matrix Sensitivity Analysis from an Interior Solution of a Linear Program,” INFORMS Journal on Computing, 11: 316–327, 1999, and its references.

4.9 THE INTERIOR POINT APPROACH TO SOLVING LINEAR PROGRAMMING PROBLEMS

167

The Complementary Roles of the Simplex Method and the Interior-Point Approach Ongoing research is continuing to provide substantial improvements in computer implementations of both the simplex method (including its variants) and interior-point algorithms. Therefore, any predictions about their future roles are risky. However, we do summarize below our current assessment of their complementary roles. The simplex method (and its variants) continues to be the standard algorithm for the routine use of linear programming. It continues to be the most efficient algorithm for problems with less than a few hundred functional constraints. It also is the most efficient for some (but not all) problems with up to several thousand functional constraints and nearly an unlimited number of decision variables, so most users are continuing to use the simplex method for such problems. However, as the number of functional constraints increases even further, it becomes increasingly likely that an interior-point approach will be the most efficient, so it often is now used instead. As the size grows into the tens of thousands of functional constraints, the interior-point approach may be the only one capable of solving the problem. However, this certainly is not always the case. As mentioned in the preceding section, the latest state-of-the-art software (CPLEX 6.5) is successfully using the simplex method and its variants to solve some truly massive problems with hundreds of thousands, or even millions of functional constraints and decision variables. These generalizations about how the interior-point approach and the simplex method should compare for various problem sizes will not hold across the board. The specific software packages and computer equipment being used have a major impact. The comparison also is affected considerably by the specific type of linear programming problem being solved. As time goes on, we should learn much more about how to identify specific types which are better suited for one kind of algorithm. One of the by-products of the emergence of the interior-point approach has been a major renewal of efforts to improve the efficiency of computer implementations of the simplex method. As we indicated, impressive progress has been made in recent years, and more lies ahead. At the same time, ongoing research and development of the interior-point approach will further increase its power, and perhaps at a faster rate than for the simplex method. Improving computer technology, such as massive parallel processing (a huge number of computer units operating in parallel on different parts of the same problem), also will substantially increase the size of problem that either kind of algorithm can solve. However, it now appears that the interior-point approach has much greater potential to take advantage of parallel processing than the simplex method does. As discussed earlier, a key disadvantage of the interior-point approach is its limited capability for performing postoptimality analysis. To overcome this drawback, researchers have been developing procedures for switching over to the simplex method after an interior-point algorithm has finished. Recall that the trial solutions obtained by an interior-point algorithm keep getting closer and closer to an optimal solution (the best CPF solution), but never quite get there. Therefore, a switching procedure requires identifying a CPF solution (or BF solution after augmenting) that is very close to the final trial solution. For example, by looking at Fig. 4.11, it is easy to see that the final trial solution in Table 4.18 is very near the CPF solution (2, 6). Unfortunately, on problems with thou-

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sands of decision variables (so no graph is available), identifying a nearby CPF (or BF) solution is a very challenging and time-consuming task. However, good progress has been made in developing procedures to do this. Once this nearby BF solution has been found, the optimality test for the simplex method is applied to check whether this actually is the optimal BF solution. If it is not optimal, some iterations of the simplex method are conducted to move from this BF solution to an optimal solution. Generally, only a very few iterations (perhaps one) are needed because the interior-point algorithm has brought us so close to an optimal solution. Therefore, these iterations should be done quite quickly, even on problems that are too huge to be solved from scratch. After an optimal solution is actually reached, the simplex method and its variants are applied to help perform postoptimality analysis. Because of the difficulties involved in applying a switching procedure (including the extra computer time), some practitioners prefer to just use the simplex method from the outset. This makes good sense when you only occasionally encounter problems that are large enough for an interior-point algorithm to be modestly faster (before switching) than the simplex method. This modest speed-up would not justify both the extra computer time for a switching procedure and the high cost of acquiring (and learning to use) a software package based on the interior-point approach. However, for organizations which frequently must deal with extremely large linear programming problems, acquiring a state-of-the-art software package of this kind (including a switching procedure) probably is worthwhile. For sufficiently huge problems, the only available way of solving them may be with such a package. Applications of huge linear programming models sometimes lead to savings of millions of dollars. Just one such application can pay many times over for a state-of-the-art software package based on the interior-point approach plus switching over to the simplex method at the end.

4.10

CONCLUSIONS The simplex method is an efficient and reliable algorithm for solving linear programming problems. It also provides the basis for performing the various parts of postoptimality analysis very efficiently. Although it has a useful geometric interpretation, the simplex method is an algebraic procedure. At each iteration, it moves from the current BF solution to a better, adjacent BF solution by choosing both an entering basic variable and a leaving basic variable and then using Gaussian elimination to solve a system of linear equations. When the current solution has no adjacent BF solution that is better, the current solution is optimal and the algorithm stops. We presented the full algebraic form of the simplex method to convey its logic, and then we streamlined the method to a more convenient tabular form. To set up for starting the simplex method, it is sometimes necessary to use artificial variables to obtain an initial BF solution for an artificial problem. If so, either the Big M method or the two-phase method is used to ensure that the simplex method obtains an optimal solution for the real problem. Computer implementations of the simplex method and its variants have become so powerful that they now are frequently used to solve linear programming problems with

APPENDIX 4.1

AN INTRODUCTION TO USING LINDO

169

many thousand functional constraints and decision variables, and occasionally vastly larger problems. Interior-point algorithms also provide a powerful tool for solving very large problems.

APPENDIX 4.1

AN INTRODUCTION TO USING LINDO The LINDO software is designed to be easy to learn and to use, especially for small problems of the size you will encounter in this book. In addition to linear programming, it also can be used to solve both integer programming problems (Chap. 12) and quadratic programming problems (Sec. 13.7). Our focus in this appendix is on its use for linear programming. LINDO allows you to enter a model in a straightforward algebraic way. For example, here is a nice way of entering the LINDO model for the Wyndor Glass Co. example introduced in Sec. 3.1. ! Wyndor Glass Co. Problem. LINDO model ! X1 batches of product 1 per week ! X2 batches of product 2 per week ! Profit, in 1000 of dollars MAX Profit) 3 X1 5 X2 Subject to ! Production time Plant1) X1 4 Plant2) 2 X2 12 Plant3) 3 X1 2 X2 18 END In addition to the basic model, this formulation includes several clarifying comments, where each comment is indicated by starting with an exclamation point. Thus, the first three lines give the title and the definitions of the decision variables. The decision variables can be either lowercase or uppercase, but uppercase usually is used so the variables won’t be dwarfed by the following “subscripts.” Another option is to use a suggestive word (or abbreviation of a word), such as the name of the product being produced, to represent the decision variable throughout the model, provided the word does not exceed eight letters. The fifth line of the LINDO formulation indicates that the objective of the model is to maximize the objective function, 3x1 5x2. The word Profit followed by a parenthesis clarifies that this quantity being maximized is profit. The comment on the fourth line further clarifies that the objective function is expressed in units of thousands of dollars. The number 1000 in this comment does not have the usual comma in front of the last three digits because LINDO does not accept commas. (It also does not accept parentheses in algebraic expressions.) The comment on the seventh line points out that the following constraints are on the production times being used. The next three lines start by giving a name (followed by a parenthesis) for each of the functional constraints. These constraints are written in the usual way except for the inequality signs. Because many keyboards do not include and signs, LINDO interprets either or as and either or as . (On systems that include and signs, LINDO will not recognize them.) The end of the constraints is signified by the word END. No nonnegativity constraints are stated because LINDO automatically assumes that all the variables have these constraints. If, say, x1 had not had a nonnegativity constraint, this would have to be indicated by typing FREE X1 on the next line below END.

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To solve this model in the Windows version of LINDO, either select the Solve command from the Solve menu or press the Solve button on the toolbar. On a platform other than Windows, simply type GO followed by a return at the colon prompt. Figure A4.1 shows the resulting solution report delivered by LINDO. Both the top line and bottom line in this figure indicate that an optimal solution was found at iteration 2 of the simplex method. Next comes the value of the objective function for this solution. Below this, we have the values of x1 and x2 for the optimal solution. The column to the right of these values gives the reduced costs. We have not discussed reduced costs in this chapter because the information they provide can also be gleaned from the allowable range to stay optimal for the coefficients in the objective function, and these allowable ranges also are readily available (as you will see in the next figure). When the variable is a basic variable in the optimal solution (as for both variables in the Wyndor problem), its reduced cost automatically is 0. When the variable is a nonbasic variable, its reduced cost provides some interesting information. This variable is 0 because its coefficient in the objective function is too small (when maximizing the objective function) or too large (when minimizing) to justify undertaking the activity represented by the variable. The reduced cost indicates how much this coefficient can be increased (when maximizing) or decreased (when minimizing) before the optimal solution would change and this variable would become a basic variable. However, recall that this same information already is available from the allowable range to stay optimal for the coefficient of this variable in the objective function. The reduced cost (for a nonbasic variable) is just the allowable increase (when maximizing) from the current value of this coefficient to remain within its allowable range to stay optimal or the allowable decrease (when minimizing). Below the variable values and reduced costs in Fig. A4.1, we next have information about the three functional constraints. The Slack or Surplus column gives the difference between the two sides of each constraint. The Dual Prices column gives, by another name, the shadow prices discussed in Sec. 4.7 for these constraints.1 (This alternate name comes from the fact found in Sec. 6.1 that these shadow prices are just the optimal values of the dual variables introduced in Chap. 6.) When LINDO provides you with this solution report, it also asks you whether you want to do range (sensitivity) analysis. Answering yes (by pressing the Y key) provides you with the additional range report shown in Fig. A4.2. This report is identical to the last three columns of the 1

However, beware that LINDO uses a different sign convention from the common one adopted here (see the second footnote for the definition of shadow price in Sec. 4.7), so that for minimization problems, its shadow prices (dual prices) are the negative of ours.

FIGURE A4.1 The solution report provided by LINDO for the Wyndor Glass Co. problem.

LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE Profit) 36.00000 VARIABLE X1 X2 ROW Plant1) Plant2) Plant3)

VALUE 2.000000 6.000000

REDUCED COST .000000 .000000

SLACK OR SURPLUS 2.000000 .000000 .000000

NO. ITERATIONS= 2

DUAL PRICES .000000 1.500000 1.000000

SELECTED REFERENCES

171

RANGES IN WHICH THE BASIS IS UNCHANGED: VARIABLE X1 X2 ROW FIGURE A4.2 The range report provided by LINDO for the Wyndor Glass Co. problem.

Plant1 Plant2 Plant3

OBJ COEFFICIENT RANGES CURRENT ALLOWABLE COEF INCREASE 3.000000 4.500000 5.000000 INFINITY RIGHTHAND SIDE RANGES CURRENT ALLOWABLE RHS INCREASE 4.000000 INFINITY 12.000000 6.000000 18.000000 6.000000

ALLOWABLE DECREASE 3.000000 3.000000 ALLOWABLE DECREASE 2.000000 6.000000 6.000000

tables in the sensitivity report generated by the Excel Solver, as shown earlier in Fig. 4.10. Thus, as already discussed in Sec. 4.7, the first two rows of this range report indicate that the allowable range to stay optimal for each coefficient in the objective function (assuming no other change in the model) is 0 c1 7.5 2 c2 Similarly, the last three rows indicate that the allowable range to stay feasible for each right-hand side (assuming no other change in the model) is 2 b1 6 b2 18 12 b3 24 To print your results with the Windows version of LINDO, you simply need to use the Print command to send the contents of the active window to the printer. If you are running LINDO on a platform other than Windows, you can use the DIVERT command (followed by the file name) to send screen output to a file, which can then print from either the operating system or a word processing package. These are the basics for getting started with LINDO. The LINDO tutorial on the CD-ROM also provides some additional details. The LINGO/LINDO files on the CD-ROM for various chapters show the LINDO formulations for numerous examples. In addition, LINDO includes a Help menu to provide guidance. These resources should enable you to apply LINDO to any linear programming problem you will encounter in this book. (We will discuss applications to other problem types in Chaps. 12 and 13.) For more advanced applications, the LINDO User’s Manual (Selected Reference 4 for this chapter) might be needed.

SELECTED REFERENCES 1. Bazaraa, M. S., J. J. Jarvis, and H. D. Sherali: Linear Programming and Network Flows, 2d ed., Wiley, New York, 1990. 2. Calvert, J. E., and W. L. Voxman: Linear Programming, Harcourt Brace Jovanovich, Orlando, FL, 1989.

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4 SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

3. Dantzig, G.B., and M.N. Thapa: Linear Programming 1: Introduction, Springer, New York, 1997. 4. LINDO User’s Manual, LINDO Systems, Inc., Chicago, IL, e-mail: [email protected], 1999. 5. Vanderbei, R. J.: Linear Programming: Foundations and Extensions, Kluwer Academic Publishers, Boston, MA, 1996.

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE Demonstration Examples in OR Tutor: Interpretation of the Slack Variables Simplex Method—Algebraic Form Simplex Method—Tabular Form

Interactive Routines: Enter or Revise a General Linear Programming Model Set Up for the Simplex Method—Interactive Only Solve Interactively by the Simplex Method

An Automatic Routine: Solve Automatically by the Interior-Point Algorithm

An Excel Add-In: Premium Solver

Files (Chapter 3) for Solving the Wyndor and Radiation Therapy Examples: Excel File LINGO/LINDO File MPL/CPLEX File

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The corresponding demonstration example listed above may be helpful. I: We suggest that you use the corresponding interactive routine listed above (the printout records your work). C: Use the computer with any of the software options available to you (or as instructed by your instructor) to solve the problem automatically. (See Sec. 4.8 for a listing of the options featured in this book and on the CD-ROM.) An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.

4.1-1. Consider the following problem. Z x1 2x2,

Maximize subject to 2 x2 2 x1 x2 3

x1

and x1 0,

x2 0.

(a) Plot the feasible region and circle all the CPF solutions.

CHAPTER 4 PROBLEMS

173

(b) For each CPF solution, identify the pair of constraint boundary equations that it satisfies. (c) For each CPF solution, use this pair of constraint boundary equations to solve algebraically for the values of x1 and x2 at the corner point. (d) For each CPF solution, identify its adjacent CPF solutions. (e) For each pair of adjacent CPF solutions, identify the constraint boundary they share by giving its equation.

The objective is to maximize the total profit from the two activities. The unit profit for activity 1 is $1,000 and the unit profit for activity 2 is $2,000. (a) Calculate the total profit for each CPF solution. Use this information to find an optimal solution. (b) Use the solution concepts of the simplex method given in Sec. 4.1 to identify the sequence of CPF solutions that would be examined by the simplex method to reach an optimal solution.

4.1-2. Consider the following problem.

4.1-4.* Consider the linear programming model (given in the back of the book) that was formulated for Prob. 3.2-3. (a) Use graphical analysis to identify all the corner-point solutions for this model. Label each as either feasible or infeasible. (b) Calculate the value of the objective function for each of the CPF solutions. Use this information to identify an optimal solution. (c) Use the solution concepts of the simplex method given in Sec. 4.1 to identify which sequence of CPF solutions might be examined by the simplex method to reach an optimal solution. (Hint: There are two alternative sequences to be identified for this particular model.)

Z 3x1 2x2,

Maximize subject to 2x1 x2 6 x1 2x2 6 and x1 0,

x2 0.

(a) Use the graphical method to solve this problem. Circle all the corner points on the graph. (b) For each CPF solution, identify the pair of constraint boundary equations it satisfies. (c) For each CPF solution, identify its adjacent CPF solutions. (d) Calculate Z for each CPF solution. Use this information to identify an optimal solution. (e) Describe graphically what the simplex method does step by step to solve the problem. 4.1-3. A certain linear programming model involving two activities has the feasible region shown below.

subject to x1 3x2 8 x1 x2 4 and x1 0,

Maximize

(0, 6 23 )

x2 0.

Z 3x1 2x2,

subject to

6

x1 3x2 4 x1 3x2 15 2x1 x2 10

(5, 5)

and

(6, 4)

4

x1 0,

Feasible region

Maximize (8, 0) 2

x2 0.

4.1-7. Describe graphically what the simplex method does step by step to solve the following problem.

2

0

Z x1 2x2,

Maximize

4.1-6. Repeat Prob. 4.1-4 for the following problem.

8

Level of Activity 2

4.1-5. Repeat Prob. 4.1-4 for the following problem.

4 6 Level of Activity 1

8

Z 2x1 3x2,

subject to 3x1 x2 1 4x1 2x2 20

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4x1 x2 10 x1 2x2 5 and x1 0,

x2 0.

4.1-8. Describe graphically what the simplex method does step by step to solve the following problem. Minimize

Z 5x1 7x2,

subject to 2x1 3x2 42 3x1 4x2 60 x1 x2 18 and x1 0,

x2 0.

(b) For each CPF solution, identify the corresponding BF solution by calculating the values of the slack variables. For each BF solution, use the values of the variables to identify the nonbasic variables and the basic variables. (c) For each BF solution, demonstrate (by plugging in the solution) that, after the nonbasic variables are set equal to zero, this BF solution also is the simultaneous solution of the system of equations obtained in part (a). 4.2-2. Reconsider the model in Prob. 4.1-5. Follow the instructions of Prob. 4.2-1 for parts (a), (b), and (c). (d) Repeat part (b) for the corner-point infeasible solutions and the corresponding basic infeasible solutions. (e) Repeat part (c) for the basic infeasible solutions. 4.2-3. Follow the instructions of Prob. 4.2-1 for the model in Prob. 4.1-6. 4.3-1. Work through the simplex method (in algebraic form) step by step to solve the model in Prob. 4.1-4.

D,I

4.1-9. Label each of the following statements about linear programming problems as true or false, and then justify your answer. (a) For minimization problems, if the objective function evaluated at a CPF solution is no larger than its value at every adjacent CPF solution, then that solution is optimal. (b) Only CPF solutions can be optimal, so the number of optimal solutions cannot exceed the number of CPF solutions. (c) If multiple optimal solutions exist, then an optimal CPF solution may have an adjacent CPF solution that also is optimal (the same value of Z). 4.1-10. The following statements give inaccurate paraphrases of the six solution concepts presented in Sec. 4.1. In each case, explain what is wrong with the statement. (a) The best CPF solution always is an optimal solution. (b) An iteration of the simplex method checks whether the current CPF solution is optimal and, if not, moves to a new CPF solution. (c) Although any CPF solution can be chosen to be the initial CPF solution, the simplex method always chooses the origin. (d) When the simplex method is ready to choose a new CPF solution to move to from the current CPF solution, it only considers adjacent CPF solutions because one of them is likely to be an optimal solution. (e) To choose the new CPF solution to move to from the current CPF solution, the simplex method identifies all the adjacent CPF solutions and determines which one gives the largest rate of improvement in the value of the objective function. 4.2-1. Reconsider the model in Prob. 4.1-4. (a) Introduce slack variables in order to write the functional constraints in augmented form.

4.3-2. Reconsider the model in Prob. 4.1-5. (a) Work through the simplex method (in algebraic form) by hand to solve this model. D,I (b) Repeat part (a) with the corresponding interactive routine in your OR Tutor. C (c) Verify the optimal solution you obtained by using a software package based on the simplex method. 4.3-3. Follow the instructions of Prob. 4.3-2 for the model in Prob. 4.1-6. 4.3-4.* Work through the simplex method (in algebraic form) step by step to solve the following problem.

D,I

Maximize

Z 4x1 3x2 6x3,

subject to 3x1 x2 3x3 30 2x1 2x2 3x3 40 and x1 0,

x2 0,

x3 0.

4.3-5. Work through the simplex method (in algebraic form) step by step to solve the following problem.

D,I

Maximize

Z x1 2x2 4x3,

subject to 3x1 x2 5x3 10 x1 4x2 x3 8 2x1 4x2 2x3 7

CHAPTER 4 PROBLEMS

and x1 0,

x2 0,

x3 0.

4.3-6. Work through the simplex method (in algebraic form) step by step to solve the following problem.

D,I

Maximize

Z x1 2x2 2x3,

subject to 5x1 2x2 3x3 15 x1 4x2 2x3 12 2x1 4x2 x3 8 and x1 0,

x2 0,

x3 0.

4.3-7. Consider the following problem. Maximize

Z 5x1 3x2 4x3,

4.4-2. Repeat Prob. 4.3-2, using the tabular form of the simplex method.

D,I,C

and x2 0,

x3 0.

You are given the information that the nonzero variables in the optimal solution are x2 and x3. (a) Describe how you can use this information to adapt the simplex method to solve this problem in the minimum possible number of iterations (when you start from the usual initial BF solution). Do not actually perform any iterations. (b) Use the procedure developed in part (a) to solve this problem by hand. (Do not use your OR Courseware.) 4.3-8. Consider the following problem. Z 2x1 4x2 3x3,

subject to x1 3x2 2x3 30 x1 x2 x3 24 3x1 5x2 3x3 60 and x1 0,

4.3-9. Label each of the following statements as true or false, and then justify your answer by referring to specific statements (with page citations) in the chapter. (a) The simplex method’s rule for choosing the entering basic variable is used because it always leads to the best adjacent BF solution (largest Z). (b) The simplex method’s minimum ratio rule for choosing the leaving basic variable is used because making another choice with a larger ratio would yield a basic solution that is not feasible. (c) When the simplex method solves for the next BF solution, elementary algebraic operations are used to eliminate each nonbasic variable from all but one equation (its equation) and to give it a coefficient of 1 in that one equation. D,I

2x1 x2 x3 20 3x1 x2 2x3 30

Maximize

(b) Use the procedure developed in part (a) to solve this problem by hand. (Do not use your OR Courseware.)

4.4-1. Repeat Prob. 4.3-1, using the tabular form of the simplex method.

subject to

x1 0,

175

x2 0,

x3 0.

You are given the information that x1 0, x2 0, and x3 0 in the optimal solution. (a) Describe how you can use this information to adapt the simplex method to solve this problem in the minimum possible number of iterations (when you start from the usual initial BF solution). Do not actually perform any iterations.

4.4-3. Repeat Prob. 4.3-3, using the tabular form of the simplex method.

D,I,C

4.4-4. Consider the following problem. Maximize

Z 2x1 x2,

subject to x1 x2 40 4x1 x2 100 and x1 0,

x2 0.

(a) Solve this problem graphically in a freehand manner. Also identify all the CPF solutions. (b) Now repeat part (a) when using a ruler to draw the graph carefully. D (c) Use hand calculations to solve this problem by the simplex method in algebraic form. D,I (d) Now use your OR Courseware to solve this problem interactively by the simplex method in algebraic form. D (e) Use hand calculations to solve this problem by the simplex method in tabular form. D,I (f) Now use your OR Courseware to solve this problem interactively by the simplex method in tabular form. C (g) Use a software package based on the simplex method to solve the problem.

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4.4-5. Repeat Prob. 4.4-4 for the following problem. subject to

x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step in algebraic form to solve this problem. D,I (b) Work through the simplex method step by step in tabular form to solve the problem. C (c) Use a computer package based on the simplex method to solve the problem. D,I

x1 2x2 30 x1 x2 20 and x1 0,

and

Z 2x1 3x2,

Maximize

x2 0.

4.4-6. Consider the following problem. Maximize

Z 2x1 4x2 3x3,

4.4-9. Work through the simplex method step by step (in tabular form) to solve the following problem.

D,I

subject to

Maximize

3x1 4x2 2x3 60 2x1 x2 2x3 40 x1 3x2 2x3 80

subject to 3x1 x2 x3 6 x1 x2 2x3 1 x1 x2 x3 2

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step in algebraic form. D,I (b) Work through the simplex method step by step in tabular form. C (c) Use a software package based on the simplex method to solve the problem. D,I

4.4-7. Consider the following problem. Maximize

Z 3x1 5x2 6x3,

and x1 0,

x2 0,

x3 0.

4.4-10. Work through the simplex method step by step to solve the following problem.

D,I

Maximize

Z x1 x2 2x3,

subject to x1 2x2 x3 20 2x1 4x2 2x3 60 2x1 3x2 x3 50

subject to 2x1 x2 x3 x1 2x2 x3 x1 x2 2x3 x1 x2 x3

Z 2x1 x2 x3,

4 4 4 3

and x1 0,

x2 0,

x3 0.

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step in algebraic form. D,I (b) Work through the simplex method in tabular form. C (c) Use a computer package based on the simplex method to solve the problem. D,I

4.4-8. Consider the following problem. Maximize

Z 2x1 x2 x3,

subject to x1 x2 3x3 4 2x1 x2 3x3 10 x1 x2 x3 7

4.5-1. Consider the following statements about linear programming and the simplex method. Label each statement as true or false, and then justify your answer. (a) In a particular iteration of the simplex method, if there is a tie for which variable should be the leaving basic variable, then the next BF solution must have at least one basic variable equal to zero. (b) If there is no leaving basic variable at some iteration, then the problem has no feasible solutions. (c) If at least one of the basic variables has a coefficient of zero in row 0 of the final tableau, then the problem has multiple optimal solutions. (d) If the problem has multiple optimal solutions, then the problem must have a bounded feasible region.

CHAPTER 4 PROBLEMS

4.5-2. Suppose that the following constraints have been provided for a linear programming model with decision variables x1 and x2. x1 3x2 30 3x1 x2 30 and x1 0,

x2 0.

(a) Demonstrate graphically that the feasible region is unbounded. (b) If the objective is to maximize Z x1 x2, does the model have an optimal solution? If so, find it. If not, explain why not. (c) Repeat part (b) when the objective is to maximize Z x1 x2. (d) For objective functions where this model has no optimal solution, does this mean that there are no good solutions according to the model? Explain. What probably went wrong when formulating the model? D,I (e) Select an objective function for which this model has no optimal solution. Then work through the simplex method step by step to demonstrate that Z is unbounded. C (f) For the objective function selected in part (e), use a software package based on the simplex method to determine that Z is unbounded. 4.5-3. Follow the instructions of Prob. 4.5-2 when the constraints are the following: 2x1 x2 20 x1 2x2 20 and

(a) Show that any convex combination of any set of feasible solutions must be a feasible solution (so that any convex combination of CPF solutions must be feasible). (b) Use the result quoted in part (a) to show that any convex combination of BF solutions must be a feasible solution. 4.5-6. Using the facts given in Prob. 4.5-5, show that the following statements must be true for any linear programming problem that has a bounded feasible region and multiple optimal solutions: (a) Every convex combination of the optimal BF solutions must be optimal. (b) No other feasible solution can be optimal. 4.5-7. Consider a two-variable linear programming problem whose CPF solutions are (0, 0), (6, 0), (6, 3), (3, 3), and (0, 2). (See Prob. 3.2-2 for a graph of the feasible region.) (a) Use the graph of the feasible region to identify all the constraints for the model. (b) For each pair of adjacent CPF solutions, give an example of an objective function such that all the points on the line segment between these two corner points are multiple optimal solutions. (c) Now suppose that the objective function is Z x1 2x2. Use the graphical method to find all the optimal solutions. D,I (d) For the objective function in part (c), work through the simplex method step by step to find all the optimal BF solutions. Then write an algebraic expression that identifies all the optimal solutions. 4.5-8. Consider the following problem.

D,I

x1 0,

x2 0.

Maximize

Z 5x1 x2 3x3 4x4,

subject to x1 2x2 4x3 3x4 20 4x1 6x2 5x3 4x4 40 2x1 3x2 3x3 8x4 50

subject to x1 x2 3 x3 x4 2 and xj 0,

for j 1, 2, 3, 4.

Work through the simplex method step by step to find all the optimal BF solutions.

and x1 0,

Z x1 x2 x3 x4,

Maximize

4.5-4. Consider the following problem.

D,I

177

x2 0,

x3 0,

x4 0.

Work through the simplex method step by step to demonstrate that Z is unbounded. 4.5-5. A basic property of any linear programming problem with a bounded feasible region is that every feasible solution can be expressed as a convex combination of the CPF solutions (perhaps in more than one way). Similarly, for the augmented form of the problem, every feasible solution can be expressed as a convex combination of the BF solutions.

4.6-1.* Consider the following problem. Z 2x1 3x2,

Maximize subject to x1 2x2 4 x1 x2 3 and x1 0,

x2 0.

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(a) Solve this problem graphically. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (c) Continue from part (b) to work through the simplex method step by step to solve the problem. 4.6-2. Consider the following problem. Maximize

Z 4x1 2x2 3x3 5x4,

subject to 2x1 3x2 4x3 2x4 300 8x1 x2 x3 5x4 300 and xj 0,

for j 1, 2, 3, 4.

(a) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (b) Work through the simplex method step by step to solve the problem. (c) Using the two-phase method, construct the complete first simplex tableau for phase 1 and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (d) Work through phase 1 step by step. (e) Construct the complete first simplex tableau for phase 2. I (f) Work through phase 2 step by step to solve the problem. (g) Compare the sequence of BF solutions obtained in part (b) with that in parts (d) and ( f ). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (h) Use a software package based on the simplex method to solve the problem. 4.6-3. Consider the following problem. Minimize

Z 3x1 2x2,

4.6-4.* Consider the following problem. Z 2x1 3x2 x3,

Minimize subject to

x1 4x2 2x3 8 3x1 2x2 2x3 6 and x1 0,

x2 0,

x3 0.

(a) Reformulate this problem to fit our standard form for a linear programming model presented in Sec. 3.2. I (b) Using the Big M method, work through the simplex method step by step to solve the problem. I (c) Using the two-phase method, work through the simplex method step by step to solve the problem. (d) Compare the sequence of BF solutions obtained in parts (b) and (c). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (e) Use a software package based on the simplex method to solve the problem. 4.6-5. For the Big M method, explain why the simplex method never would choose an artificial variable to be an entering basic variable once all the artificial variables are nonbasic. 4.6-6. Consider the following problem. Z 90x1 70x2,

Maximize subject to 2x1 x2 2 x1 x2 2 and

subject to 2x1 x2 10 3x1 2x2 6 x1 x2 6 and x1 0,

I

initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. (c) Work through the simplex method step by step to solve the problem.

x2 0.

(a) Solve this problem graphically. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding

x1 0,

x2 0.

(a) Demonstrate graphically that this problem has no feasible solutions. C (b) Use a computer package based on the simplex method to determine that the problem has no feasible solutions. I (c) Using the Big M method, work through the simplex method step by step to demonstrate that the problem has no feasible solutions. I (d) Repeat part (c) when using phase 1 of the two-phase method.

CHAPTER 4 PROBLEMS

4.6-7. Follow the instructions of Prob. 4.6-6 for the following problem. Minimize

Z 5,000x1 7,000x2,

subject to 2x1 x2 1 x1 2x2 1 and x1 0,

(a) Using the two-phase method, work through phase 1 step by step. C (b) Use a software package based on the simplex method to formulate and solve the phase 1 problem. I (c) Work through phase 2 step by step to solve the original problem. C (d) Use a computer code based on the simplex method to solve the original problem. I

4.6-10.* Consider the following problem. x2 0.

Minimize

4.6-8. Consider the following problem. Maximize

179

subject to

Z 2x1 5x2 3x3,

2x1 x2 3x3 60 3x1 3x2 5x3 120

subject to x1 2x2 x3 20 2x1 4x2 x3 50

and x1 0,

and x1 0,

x3 0.

4.6-9. Consider the following problem. Z 2x1 x2 3x3,

subject to

x3 0.

(a) Using the Big M method, work through the simplex method step by step to solve the problem. I (b) Using the two-phase method, work through the simplex method step by step to solve the problem. (c) Compare the sequence of BF solutions obtained in parts (a) and (b). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (d) Use a software package based on the simplex method to solve the problem. 4.6-11. Follow the instructions of Prob. 4.6-10 for the following problem. Minimize

Z 3x1 2x2 7x3,

subject to x1 x2 x3 10 2x1 x2 x3 10 and x1 0,

x2 0,

x3 0.

4.6-12. Follow the instructions of Prob. 4.6-10 for the following problem. Minimize

Z 3x1 2x2 x3,

subject to

5x1 2x2 7x3 420 3x1 2x2 5x3 280

x1 x2 x3 7 3x1 x2 x3 10

and x1 0,

x2 0,

I

x2 0,

(a) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (b) Work through the simplex method step by step to solve the problem. I (c) Using the two-phase method, construct the complete first simplex tableau for phase 1 and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (d) Work through phase 1 step by step. (e) Construct the complete first simplex tableau for phase 2. I (f) Work through phase 2 step by step to solve the problem. (g) Compare the sequence of BF solutions obtained in part (b) with that in parts (d) and ( f ). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (h) Use a software package based on the simplex method to solve the problem.

Minimize

Z 3x1 2x2 4x3,

and x2 0,

x3 0.

x1 0,

x2 0,

x3 0.

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4.6-13. Label each of the following statements as true or false, and then justify your answer. (a) When a linear programming model has an equality constraint, an artificial variable is introduced into this constraint in order to start the simplex method with an obvious initial basic solution that is feasible for the original model. (b) When an artificial problem is created by introducing artificial variables and using the Big M method, if all artificial variables in an optimal solution for the artificial problem are equal to zero, then the real problem has no feasible solutions. (c) The two-phase method is commonly used in practice because it usually requires fewer iterations to reach an optimal solution than the Big M method does. 4.6-14. Consider the following problem. Maximize Z x1 4x2 2x3, subject to 4x1 x2 x3 5 x1 x2 2x3 10 and x3 0 x2 0, (no nonnegativity constraint for x1). (a) Reformulate this problem so all variables have nonnegativity constraints. D,I (b) Work through the simplex method step by step to solve the problem. C (c) Use a software package based on the simplex method to solve the problem. 4.6-15.* Consider the following problem. Maximize Z x1 4x2, subject to 3x1 x2 6 x1 2x2 4 x1 2x2 3 (no lower bound constraint for x1). (a) Solve this problem graphically. (b) Reformulate this problem so that it has only two functional constraints and all variables have nonnegativity constraints. D,I (c) Work through the simplex method step by step to solve the problem. 4.6-16. Consider the following problem. Maximize Z x1 2x2 x3, subject to 3x2 x3 120 x1 x2 4x3 80 3x1 x2 2x3 100 (no nonnegativity constraints).

(a) Reformulate this problem so that all variables have nonnegativity constraints. D,I (b) Work through the simplex method step by step to solve the problem. C (c) Use a computer package based on the simplex method to solve the problem. 4.6-17. This chapter has described the simplex method as applied to linear programming problems where the objective function is to be maximized. Section 4.6 then described how to convert a minimization problem to an equivalent maximization problem for applying the simplex method. Another option with minimization problems is to make a few modifications in the instructions for the simplex method given in the chapter in order to apply the algorithm directly. (a) Describe what these modifications would need to be. (b) Using the Big M method, apply the modified algorithm developed in part (a) to solve the following problem directly by hand. (Do not use your OR Courseware.) Minimize

Z 3x1 8x2 5x3,

subject to 3x1 3x2 4x3 70 3x1 5x2 2x3 70 and x1 0,

x2 0,

x3 0.

4.6-18. Consider the following problem. Maximize

Z 2x1 x2 4x3 3x4,

subject to x1 x2 3x3 2x4 4 x1 x2 x3 x4 1 2x1 x2 x3 x4 2 x1 2x2 x3 2x4 2 and x2 0,

x3 0,

x4 0

(no nonnegativity constraint for x1). (a) Reformulate this problem to fit our standard form for a linear programming model presented in Sec. 3.2. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. (c) Using the two-phase method, construct row 0 of the first simplex tableau for phase 1. C (d) Use a computer package based on the simplex method to solve the problem.

CHAPTER 4 PROBLEMS

I

4.6-19. Consider the following problem. Maximize

and x2 0,

x3 0.

Work through the simplex method step by step to demonstrate that this problem does not possess any feasible solutions. 4.7-1. Refer to Fig. 4.10 and the resulting allowable range to stay feasible for the respective right-hand sides of the Wyndor Glass Co. problem given in Sec. 3.1. Use graphical analysis to demonstrate that each given allowable range is correct. 4.7-2. Reconsider the model in Prob. 4.1-5. Interpret the right-hand side of the respective functional constraints as the amount available of the respective resources. (a) Use graphical analysis as in Fig. 4.8 to determine the shadow prices for the respective resources. (b) Use graphical analysis to perform sensitivity analysis on this model. In particular, check each parameter of the model to determine whether it is a sensitive parameter (a parameter whose value cannot be changed without changing the optimal solution) by examining the graph that identifies the optimal solution. (c) Use graphical analysis as in Fig. 4.9 to determine the allowable range for each cj value (coefficient of xj in the objective function) over which the current optimal solution will remain optimal. (d) Changing just one bi value (the right-hand side of functional constraint i) will shift the corresponding constraint boundary. If the current optimal CPF solution lies on this constraint boundary, this CPF solution also will shift. Use graphical analysis to determine the allowable range for each bi value over which this CPF solution will remain feasible. C (e) Verify your answers in parts (a), (c), and (d) by using a computer package based on the simplex method to solve the problem and then to generate sensitivity analysis information. 4.7-3. Repeat Prob. 4.7-2 for the model in Prob. 4.1-6. 4.7-4. You are given the following linear programming problem. Z 4x1 2x2,

subject to 2x1 3x2 16 x1 3x2 17 x1 3x2 5

x1 0,

x2 0.

(a) Solve this problem graphically. (b) Use graphical analysis to find the shadow prices for the resources. (c) Determine how many additional units of resource 1 would be needed to increase the optimal value of Z by 15.

x1 x2 2x3 20 15x1 6x2 5x3 50 x1 3x2 5x3 30

Maximize

and

Z 4x1 5x2 3x3,

subject to

x1 0,

181

(resource 1) (resource 2) (resource 3)

4.7-5. Consider the following problem. Maximize

Z x1 7x2 3x3,

subject to 2x1 x2 x3 4 4x1 3x2 x3 2 3x1 2x2 x3 3

(resource 1) (resource 2) (resource 3)

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the three resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient, and the allowable range to stay feasible for each right-hand side. D,I

4.7-6.* Consider the following problem. Maximize

Z 2x1 2x2 3x3,

subject to x1 x2 x3 4 2x1 x2 x3 2 x1 x2 3x3 12

(resource 1) (resource 2) (resource 3)

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the three resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient and the allowable range to stay feasible for each right-hand side. D,I

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4.7-7. Consider the following problem. Maximize

and x1 0,

Z 2x1 4x2 x3,

subject to

x2 0,

x3 0,

x4 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the two resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient, and the allowable range to stay feasible for each right-hand side. D,I

2x1 3x2 x3 30 2x1 x2 x3 10 4x1 2x2 2x3 40

(resource 1) (resource 2) (resource 3)

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the three resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient, and the allowable range to stay feasible for each right-hand side. D,I

4.7-8. Consider the following problem. Maximize

4.9.1. Use the interior-point algorithm in your OR Courseware to solve the model in Prob. 4.1-4. Choose 0.5 from the Option menu, use (x1, x2) (0.1, 0.4) as the initial trial solution, and run 15 iterations. Draw a graph of the feasible region, and then plot the trajectory of the trial solutions through this feasible region. 4.9-2. Repeat Prob. 4.9-1 for the model in Prob. 4.1-5. 4.9-3. Repeat Prob. 4.9-1 for the model in Prob. 4.1-6.

Z 5x1 4x2 x3 3x4,

subject to 3x1 2x2 3x3 x4 24 3x1 3x2 x3 3x4 36

CASE 4.1

(resource 1) (resource 2)

FABRICS AND FALL FASHIONS From the tenth floor of her office building, Katherine Rally watches the swarms of New Yorkers fight their way through the streets infested with yellow cabs and the sidewalks littered with hot dog stands. On this sweltering July day, she pays particular attention to the fashions worn by the various women and wonders what they will choose to wear in the fall. Her thoughts are not simply random musings; they are critical to her work since she owns and manages TrendLines, an elite women’s clothing company. Today is an especially important day because she must meet with Ted Lawson, the production manager, to decide upon next month’s production plan for the fall line. Specifically, she must determine the quantity of each clothing item she should produce given the plant’s production capacity, limited resources, and demand forecasts. Accurate planning for next month’s production is critical to fall sales since the items produced next month will appear in stores during September, and women generally buy the majority of the fall fashions when they first appear in September. She turns back to her sprawling glass desk and looks at the numerous papers covering it. Her eyes roam across the clothing patterns designed almost six months ago,

CASE 4.1

FABRICS AND FALL FASHIONS

183

the lists of materials requirements for each pattern, and the lists of demand forecasts for each pattern determined by customer surveys at fashion shows. She remembers the hectic and sometimes nightmarish days of designing the fall line and presenting it at fashion shows in New York, Milan, and Paris. Ultimately, she paid her team of six designers a total of $860,000 for their work on her fall line. With the cost of hiring runway models, hair stylists, and makeup artists, sewing and fitting clothes, building the set, choreographing and rehearsing the show, and renting the conference hall, each of the three fashion shows cost her an additional $2,700,000. She studies the clothing patterns and material requirements. Her fall line consists of both professional and casual fashions. She determined the prices for each clothing item by taking into account the quality and cost of material, the cost of labor and machining, the demand for the item, and the prestige of the TrendLines brand name. The fall professional fashions include:

Clothing Item

Materials Requirements

Price

Labor and Machine Cost

Tailored wool slacks

3 yards of wool 2 yards of acetate for lining 1.5 yards of cashmere 1.5 yards of silk 0.5 yard of silk 2 yards of rayon 1.5 yards of acetate for lining 2.5 yards of wool 1.5 yards of acetate for lining

$300

$160

$450 $180 $120 $270

$150 $100 $ 60 $120

$320

$140

Cashmere sweater Silk blouse Silk camisole Tailored skirt Wool blazer

The fall casual fashions include:

Clothing Item

Materials Requirements

Price

Labor and Machine Cost

Velvet pants

3 yards of velvet 2 yards of acetate for lining 1.5 yards of cotton 0.5 yard of cotton 1.5 yards of velvet 1.5 yards of rayon

$350

$175

$130 $ 75 $200 $120

$ 60 $ 40 $160 $ 90

Cotton sweater Cotton miniskirt Velvet shirt Button-down blouse

She knows that for the next month, she has ordered 45,000 yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere, 18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of velvet, and 30,000 yards of cotton for production. The prices of the materials are listed on the next page.

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Material

Price per yard

Wool Acetate Cashmere Silk Rayon Velvet Cotton

$ 9.00 $ 1.50 $60.00 $13.00 $ 2.25 $12.00 $ 2.50

Any material that is not used in production can be sent back to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler. She knows that the production of both the silk blouse and cotton sweater leaves leftover scraps of material. Specifically, for the production of one silk blouse or one cotton sweater, 2 yards of silk and cotton, respectively, are needed. From these 2 yards, 1.5 yards are used for the silk blouse or the cotton sweater and 0.5 yard is left as scrap material. She does not want to waste the material, so she plans to use the rectangular scrap of silk or cotton to produce a silk camisole or cotton miniskirt, respectively. Therefore, whenever a silk blouse is produced, a silk camisole is also produced. Likewise, whenever a cotton sweater is produced, a cotton miniskirt is also produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton miniskirt without producing a cotton sweater. The demand forecasts indicate that some items have limited demand. Specifically, because the velvet pants and velvet shirts are fashion fads, TrendLines has forecasted that it can sell only 5,500 pairs of velvet pants and 6,000 velvet shirts. TrendLines does not want to produce more than the forecasted demand because once the pants and shirts go out of style, the company cannot sell them. TrendLines can produce less than the forecasted demand, however, since the company is not required to meet the demand. The cashmere sweater also has limited demand because it is quite expensive, and TrendLines knows it can sell at most 4,000 cashmere sweaters. The silk blouses and camisoles have limited demand because many women think silk is too hard to care for, and TrendLines projects that it can sell at most 12,000 silk blouses and 15,000 silk camisoles. The demand forecasts also indicate that the wool slacks, tailored skirts, and wool blazers have a great demand because they are basic items needed in every professional wardrobe. Specifically, the demand for wool slacks is 7,000 pairs of slacks, and the demand for wool blazers is 5,000 blazers. Katherine wants to meet at least 60 percent of the demand for these two items in order to maintain her loyal customer base and not lose business in the future. Although the demand for tailored skirts could not be estimated, Katherine feels she should make at least 2,800 of them. (a) Ted is trying to convince Katherine not to produce any velvet shirts since the demand for this fashion fad is quite low. He argues that this fashion fad alone accounts for $500,000 of the fixed design and other costs. The net contribution (price of clothing item materials cost labor cost) from selling the fashion fad should cover these fixed costs. Each velvet shirt generates a net contribution of $22. He argues that given the net contribution,

CASE 4.2

NEW FRONTIERS

185

even satisfying the maximum demand will not yield a profit. What do you think of Ted’s argument? (b) Formulate and solve a linear programming problem to maximize profit given the production, resource, and demand constraints.

Before she makes her final decision, Katherine plans to explore the following questions independently except where otherwise indicated. (c) The textile wholesaler informs Katherine that the velvet cannot be sent back because the demand forecasts show that the demand for velvet will decrease in the future. Katherine can therefore get no refund for the velvet. How does this fact change the production plan? (d) What is an intuitive economic explanation for the difference between the solutions found in parts (b) and (c)? (e) The sewing staff encounters difficulties sewing the arms and lining into the wool blazers since the blazer pattern has an awkward shape and the heavy wool material is difficult to cut and sew. The increased labor time to sew a wool blazer increases the labor and machine cost for each blazer by $80. Given this new cost, how many of each clothing item should TrendLines produce to maximize profit? (f) The textile wholesaler informs Katherine that since another textile customer canceled his order, she can obtain an extra 10,000 yards of acetate. How many of each clothing item should TrendLines now produce to maximize profit? (g) TrendLines assumes that it can sell every item that was not sold during September and October in a big sale in November at 60 percent of the original price. Therefore, it can sell all items in unlimited quantity during the November sale. (The previously mentioned upper limits on demand concern only the sales during September and October.) What should the new production plan be to maximize profit?

CASE 4.2

NEW FRONTIERS Rob Richman, president of AmeriBank, takes off his glasses, rubs his eyes in exhaustion, and squints at the clock in his study. It reads 3 A.M. For the last several hours, Rob has been poring over AmeriBank’s financial statements from the last three quarters of operation. AmeriBank, a medium-sized bank with branches throughout the United States, is headed for dire economic straits. The bank, which provides transaction, savings, and investment and loan services, has been experiencing a steady decline in its net income over the past year, and trends show that the decline will continue. The bank is simply losing customers to nonbank and foreign bank competitors. AmeriBank is not alone in its struggle to stay out of the red. From his daily industry readings, Rob knows that many American banks have been suffering significant losses because of increasing competition from nonbank and foreign bank competitors offering services typically in the domain of American banks. Because the nonbank and foreign bank competitors specialize in particular services, they are able to better capture the market for those services by offering less expensive, more efficient, more convenient services. For example, large corporations now turn to foreign banks and commercial paper offerings for loans, and affluent Americans now turn to money-market funds for investment. Banks face the daunting challenge of distinguishing themselves from nonbank and foreign bank competitors.

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Rob has concluded that one strategy for distinguishing AmeriBank from its competitors is to improve services that nonbank and foreign bank competitors do not readily provide: transaction services. He has decided that a more convenient transaction method must logically succeed the automatic teller machine, and he believes that electronic banking over the Internet allows this convenient transaction method. Over the Internet, customers are able to perform transactions on their desktop computers either at home or at work. The explosion of the Internet means that many potential customers understand and use the World Wide Web. He therefore feels that if AmeriBank offers Web banking (as the practice of Internet banking is commonly called), the bank will attract many new customers. Before Rob undertakes the project to make Web banking possible, however, he needs to understand the market for Web banking and the services AmeriBank should provide over the Internet. For example, should the bank only allow customers to access account balances and historical transaction information over the Internet, or should the bank develop a strategy to allow customers to make deposits and withdrawals over the Internet? Should the bank try to recapture a portion of the investment market by continuously running stock prices and allowing customers to make stock transactions over the Internet for a minimal fee? Because AmeriBank is not in the business of performing surveys, Rob has decided to outsource the survey project to a professional survey company. He has opened the project up for bidding by several survey companies and will award the project to the company which is willing to perform the survey for the least cost. Sophisticated Surveys is one of three survey companies competing for the project. Rob provided each survey company with a list of survey requirements to ensure that AmeriBank receives the needed information for planning the Web banking project. Because different age groups require different services, AmeriBank is interested in surveying four different age groups. The first group encompasses customers who are 18 to 25 years old. The bank assumes that this age group has limited yearly income and performs minimal transactions. The second group encompasses customers who are 26 to 40 years old. This age group has significant sources of income, performs many transactions, requires numerous loans for new houses and cars, and invests in various securities. The third group encompasses customers who are 41 to 50 years old. These customers typically have the same level of income and perform the same number of transactions as the second age group, but the bank assumes that these customers are less likely to use Web banking since they have not become as comfortable with the explosion of computers or the Internet. Finally, the fourth group encompasses customers who are 51 years of age and over. These customers commonly crave security and require continuous information on retirement funds. The banks believes that it is highly unlikely that customers in this age group will use Web banking, but the bank desires to learn the needs of this age group for the future. AmeriBank wants to interview 2,000 customers with at least 20 percent from the first age group, at least 27.5 percent from the second age group, at least 15 percent from the third age group, and at least 15 percent from the fourth age group. Rob understands that the Internet is a recent phenomenon and that some customers may not have heard of the World Wide Web. He therefore wants to ensure that the sur-

CASE 4.2

NEW FRONTIERS

187

vey includes a mix of customers who know the Internet well and those that have less exposure to the Internet. To ensure that AmeriBank obtains the correct mix, he wants to interview at least 15 percent of customers from the Silicon Valley where Internet use is high, at least 35 percent of customers from big cities where Internet use is medium, and at least 20 percent of customers from small towns where Internet use is low. Sophisticated Surveys has performed an initial analysis of these survey requirements to determine the cost of surveying different populations. The costs per person surveyed are listed in the following table:

Age Group Region Silicon Valley Big cities Small towns

18 to 25

26 to 40

41 to 50

51 and over

$4.75 $5.25 $6.50

$6.50 $5.75 $7.50

$6.50 $6.25 $7.50

$5.00 $6.25 $7.25

Sophisticated Surveys explores the following options cumulatively. (a) Formulate a linear programming model to minimize costs while meeting all survey constraints imposed by AmeriBank. (b) If the profit margin for Sophisticated Surveys is 15 percent of cost, what bid will they submit? (c) After submitting its bid, Sophisticated Surveys is informed that it has the lowest cost but that AmeriBank does not like the solution. Specifically, Rob feels that the selected survey population is not representative enough of the banking customer population. Rob wants at least 50 people of each age group surveyed in each region. What is the new bid made by Sophisticated Surveys? (d) Rob feels that Sophisticated Survey oversampled the 18- to 25-year-old population and the Silicon Valley population. He imposes a new constraint that no more than 600 individuals can be surveyed from the 18- to 25-year-old population and no more than 650 individuals can be surveyed from the Silicon Valley population. What is the new bid? (e) When Sophisticated Surveys calculated the cost of reaching and surveying particular individuals, the company thought that reaching individuals in young populations would be easiest. In a recently completed survey, however, Sophisticated Surveys learned that this assumption was wrong. The new costs for surveying the 18- to 25-year-old population are listed below. Region survey cost per person Silicon Valley Big cities Small towns

$6.50 $6.75 $7.00

Given the new costs, what is the new bid?

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(f) To ensure the desired sampling of individuals, Rob imposes even stricter requirements. He fixes the exact percentage of people that should be surveyed from each population. The requirements are listed below: Population percentage of people surveyed 18 26 41 51

to 25 to 40 to 50 and over

25% 35% 20% 20%

Silicon Valley Big cities Small towns

20% 50% 30%

By how much would these new requirements increase the cost of surveying for Sophisticated Surveys? Given the 15 percent profit margin, what would Sophisticated Surveys bid?

CASE 4.3

ASSIGNING STUDENTS TO SCHOOLS The Springfield school board has made the decision to close one of its middle schools (sixth, seventh, and eighth grades) at the end of this school year and reassign all of next year’s middle school students to the three remaining middle schools. The school district provides bussing for all middle school students who must travel more than approximately a mile, so the school board wants a plan for reassigning the students that will minimize the total bussing cost. The annual cost per student of bussing from each of the six residential areas of the city to each of the schools is shown in the following table (along with other basic data for next year), where 0 indicates that bussing is not needed and a dash indicates an infeasible assignment.

No. of Area Students 1 2 3 4 5 6

450 600 550 350 500 450

Percentage Percentage Percentage Bussing Cost per Student in 6th in 7th in 8th Grade Grade Grade School 1 School 2 School 3 32 37 30 28 39 34

38 28 32 40 34 28

30 35 38 32 27 38 School capacity:

$300 — $600 $200 0 $500

0 $400 $300 $500 — $300

$700 $500 $200 — $400 0

900

1,100

1,000

The school board also has imposed the restriction that each grade must constitute between 30 and 36 percent of each school’s population. The above table shows the percentage of each area’s middle school population for next year that falls into each of

CASE 4.3

ASSIGNING STUDENTS TO SCHOOLS

189

the three grades. The school attendance zone boundaries can be drawn so as to split any given area among more than one school, but assume that the percentages shown in the table will continue to hold for any partial assignment of an area to a school. You have been hired as an operations research consultant to assist the school board in determining how many students in each area should be assigned to each school. (a) Formulate a linear programming model for this problem. (b) Solve the model. (c) What is your resulting recommendation to the school board?

After seeing your recommendation, the school board expresses concern about all the splitting of residential areas among multiple schools. They indicate that they “would like to keep each neighborhood together.” (d) Adjust your recommendation as well as you can to enable each area to be assigned to just one school. (Adding this restriction may force you to fudge on some other constraints.) How much does this increase the total bussing cost? (This line of analysis will be pursued more rigorously in Case 12.4.)

The school board is considering eliminating some bussing to reduce costs. Option 1 is to eliminate bussing only for students traveling 1 to 1.5 miles, where the cost per student is given in the table as $200. Option 2 is to also eliminate bussing for students traveling 1.5 to 2 miles, where the estimated cost per student is $300. (e) Revise the model from part (a) to fit Option 1, and solve. Compare these results with those from part (c), including the reduction in total bussing cost. (f) Repeat part (e) for Option 2.

The school board now needs to choose among the three alternative bussing plans (the current one or Option 1 or Option 2). One important factor is bussing costs. However, the school board also wants to place equal weight on a second factor: the inconvenience and safety problems caused by forcing students to travel by foot or bicycle a substantial distance (more than a mile, and especially more than 1.5 miles). Therefore, they want to choose a plan that provides the best trade-off between these two factors. (g) Use your results from parts (c), (e), and ( f ) to summarize the key information related to these two factors that the school board needs to make this decision. (h) Which decision do you think should be made? Why?

Note: This case will be continued in later chapters (Cases 6.3 and 12.4), so we suggest that you save your analysis, including your basic model.

5 The Theory of the Simplex Method Chapter 4 introduced the basic mechanics of the simplex method. Now we shall delve a little more deeply into this algorithm by examining some of its underlying theory. The first section further develops the general geometric and algebraic properties that form the foundation of the simplex method. We then describe the matrix form of the simplex method (called the revised simplex method), which streamlines the procedure considerably for computer implementation. Next we present a fundamental insight about a property of the simplex method that enables us to deduce how changes that are made in the original model get carried along to the final simplex tableau. This insight will provide the key to the important topics of Chap. 6 (duality theory and sensitivity analysis).

5.1

FOUNDATIONS OF THE SIMPLEX METHOD Section 4.1 introduced corner-point feasible (CPF) solutions and the key role they play in the simplex method. These geometric concepts were related to the algebra of the simplex method in Secs. 4.2 and 4.3. However, all this was done in the context of the Wyndor Glass Co. problem, which has only two decision variables and so has a straightforward geometric interpretation. How do these concepts generalize to higher dimensions when we deal with larger problems? We address this question in this section. We begin by introducing some basic terminology for any linear programming problem with n decision variables. While we are doing this, you may find it helpful to refer to Fig. 5.1 (which repeats Fig. 4.1) to interpret these definitions in two dimensions (n 2). Terminology It may seem intuitively clear that optimal solutions for any linear programming problem must lie on the boundary of the feasible region, and in fact this is a general property. Because boundary is a geometric concept, our initial definitions clarify how the boundary of the feasible region is identified algebraically. The constraint boundary equation for any constraint is obtained by replacing its , , or sign by an sign.

Consequently, the form of a constraint boundary equation is ai1x1 ai2 x2 ain xn bi for functional constraints and xj 0 for nonnegativity constraints. Each such 190

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

Maximize Z 3x1 5x2, subject to 4 x1 2x2 12 2x1 3x2 18 and x1 0, x2 0

x1 0 (0, 9) 3x1 2x2 18

(0, 6)

(2, 6)

191

(4, 6)

2x2 12

x1 4

Feasible region FIGURE 5.1 Constraint boundaries, constraint boundary equations, and corner-point solutions for the Wyndor Glass Co. problem.

(4, 3)

(0, 0) (4, 0)

(6, 0)

x2 0

equation defines a “flat” geometric shape (called a hyperplane) in n-dimensional space, analogous to the line in two-dimensional space and the plane in three-dimensional space. This hyperplane forms the constraint boundary for the corresponding constraint. When the constraint has either a or a sign, this constraint boundary separates the points that satisfy the constraint (all the points on one side up to and including the constraint boundary) from the points that violate the constraint (all those on the other side of the constraint boundary). When the constraint has an sign, only the points on the constraint boundary satisfy the constraint. For example, the Wyndor Glass Co. problem has five constraints (three functional constraints and two nonnegativity constraints), so it has the five constraint boundary equations shown in Fig. 5.1. Because n 2, the hyperplanes defined by these constraint boundary equations are simply lines. Therefore, the constraint boundaries for the five constraints are the five lines shown in Fig. 5.1. The boundary of the feasible region contains just those feasible solutions that satisfy one or more of the constraint boundary equations.

Geometrically, any point on the boundary of the feasible region lies on one or more of the hyperplanes defined by the respective constraint boundary equations. Thus, in Fig. 5.1, the boundary consists of the five darker line segments. Next, we give a general definition of CPF solution in n-dimensional space. A corner-point feasible (CPF) solution is a feasible solution that does not lie on any line segment1 connecting two other feasible solutions. 1

An algebraic expression for a line segment is given in Appendix 2.

192

5

THE THEORY OF THE SIMPLEX METHOD

As this definition implies, a feasible solution that does lie on a line segment connecting two other feasible solutions is not a CPF solution. To illustrate when n 2, consider Fig. 5.1. The point (2, 3) is not a CPF solution, because it lies on various such line segments, e.g., the line segment connecting (0, 3) and (4, 3). Similarly, (0, 3) is not a CPF solution, because it lies on the line segment connecting (0, 0) and (0, 6). However, (0, 0) is a CPF solution, because it is impossible to find two other feasible solutions that lie on completely opposite sides of (0, 0). (Try it.) When the number of decision variables n is greater than 2 or 3, this definition for CPF solution is not a very convenient one for identifying such solutions. Therefore, it will prove most helpful to interpret these solutions algebraically. For the Wyndor Glass Co. example, each CPF solution in Fig. 5.1 lies at the intersection of two (n 2) constraint lines; i.e., it is the simultaneous solution of a system of two constraint boundary equations. This situation is summarized in Table 5.1, where defining equations refer to the constraint boundary equations that yield (define) the indicated CPF solution. For any linear programming problem with n decision variables, each CPF solution lies at the intersection of n constraint boundaries; i.e., it is the simultaneous solution of a system of n constraint boundary equations.

However, this is not to say that every set of n constraint boundary equations chosen from the n m constraints (n nonnegativity and m functional constraints) yields a CPF solution. In particular, the simultaneous solution of such a system of equations might violate one or more of the other m constraints not chosen, in which case it is a corner-point infeasible solution. The example has three such solutions, as summarized in Table 5.2. (Check to see why they are infeasible.) Furthermore, a system of n constraint boundary equations might have no solution at all. This occurs twice in the example, with the pairs of equations (1) x1 0 and x1 4 and (2) x2 0 and 2x2 12. Such systems are of no interest to us. The final possibility (which never occurs in the example) is that a system of n constraint boundary equations has multiple solutions because of redundant equations. You need not be concerned with this case either, because the simplex method circumvents its difficulties. TABLE 5.1 Defining equations for each CPF solution for the Wyndor Glass Co. problem CPF Solution

Defining Equations

(0, 0)

x1 0 x2 0

(0, 6)

x1 0 2x2 12

(2, 6)

2x2 12 3x1 2x2 18

(4, 3)

3x1 2x2 18 x1 4

(4, 0)

x1 4 x2 0

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

193

TABLE 5.2 Defining equations for each corner-point infeasible solution for the Wyndor Glass Co. problem Corner-Point Infeasible Solution

Defining Equations

(0, 9)

x1 0 3x1 2x2 18

(4, 6)

2x2 12 x1 4

(6, 0)

3x1 2x2 18 x2 0

To summarize for the example, with five constraints and two variables, there are 10 pairs of constraint boundary equations. Five of these pairs became defining equations for CPF solutions (Table 5.1), three became defining equations for corner-point infeasible solutions (Table 5.2), and each of the final two pairs had no solution. Adjacent CPF Solutions Section 4.1 introduced adjacent CPF solutions and their role in solving linear programming problems. We now elaborate. Recall from Chap. 4 that (when we ignore slack, surplus, and artificial variables) each iteration of the simplex method moves from the current CPF solution to an adjacent one. What is the path followed in this process? What really is meant by adjacent CPF solution? First we address these questions from a geometric viewpoint, and then we turn to algebraic interpretations. These questions are easy to answer when n 2. In this case, the boundary of the feasible region consists of several connected line segments forming a polygon, as shown in Fig. 5.1 by the five darker line segments. These line segments are the edges of the feasible region. Emanating from each CPF solution are two such edges leading to an adjacent CPF solution at the other end. (Note in Fig. 5.1 how each CPF solution has two adjacent ones.) The path followed in an iteration is to move along one of these edges from one end to the other. In Fig. 5.1, the first iteration involves moving along the edge from (0, 0) to (0, 6), and then the next iteration moves along the edge from (0, 6) to (2, 6). As Table 5.1 illustrates, each of these moves to an adjacent CPF solution involves just one change in the set of defining equations (constraint boundaries on which the solution lies). When n 3, the answers are slightly more complicated. To help you visualize what is going on, Fig. 5.2 shows a three-dimensional drawing of a typical feasible region when n 3, where the dots are the CPF solutions. This feasible region is a polyhedron rather than the polygon we had with n 2 (Fig. 5.1), because the constraint boundaries now are planes rather than lines. The faces of the polyhedron form the boundary of the feasible region, where each face is the portion of a constraint boundary that satisfies the other constraints as well. Note that each CPF solution lies at the intersection of three constraint boundaries (sometimes including some of the x1 0, x2 0, and x3 0 constraint boundaries for the nonnegativity

194

5

THE THEORY OF THE SIMPLEX METHOD

Constraints

x3

x1 4 x2 4 x1 x2 6 x1 2x3 4 x1 0, x2 0, x3 0

(4, 0, 4)

(4, 2, 4)

(0, 0, 2)

(4, 0, 0)

(2, 4, 3)

x1

(0, 0, 0) (0, 4, 2) (4, 2, 0) FIGURE 5.2 Feasible region and CPF solutions for a three-variable linear programming problem.

x2

(0, 4, 0)

(2, 4, 0)

constraints), and the solution also satisfies the other constraints. Such intersections that do not satisfy one or more of the other constraints yield corner-point infeasible solutions instead. The darker line segment in Fig. 5.2 depicts the path of the simplex method on a typical iteration. The point (2, 4, 3) is the current CPF solution to begin the iteration, and the point (4, 2, 4) will be the new CPF solution at the end of the iteration. The point (2, 4, 3) lies at the intersection of the x2 4, x1 x2 6, and x1 2x3 4 constraint boundaries, so these three equations are the defining equations for this CPF solution. If the x2 4 defining equation were removed, the intersection of the other two constraint boundaries (planes) would form a line. One segment of this line, shown as the dark line segment from (2, 4, 3) to (4, 2, 4) in Fig. 5.2, lies on the boundary of the feasible region, whereas the rest of the line is infeasible. This line segment is an edge of the feasible region, and its endpoints (2, 4, 3) and (4, 2, 4) are adjacent CPF solutions. For n 3, all the edges of the feasible region are formed in this way as the feasible segment of the line lying at the intersection of two constraint boundaries, and the two endpoints of an edge are adjacent CPF solutions. In Fig. 5.2 there are 15 edges of the feasible region, and so there are 15 pairs of adjacent CPF solutions. For the current CPF solution (2, 4, 3), there are three ways to remove one of its three defining equations to obtain an intersection of the other two constraint boundaries, so there are three edges emanating from (2, 4, 3). These edges lead to (4, 2, 4), (0, 4, 2), and (2, 4, 0), so these are the CPF solutions that are adjacent to (2, 4, 3). For the next iteration, the simplex method chooses one of these three edges, say, the darker line segment in Fig. 5.2, and then moves along this edge away from (2, 4, 3) until it reaches the first new constraint boundary, x1 4, at its other endpoint. [We cannot continue farther along this line to the next constraint boundary, x2 0, because this leads

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

195

to a corner-point infeasible solution—(6, 0, 5).] The intersection of this first new constraint boundary with the two constraint boundaries forming the edge yields the new CPF solution (4, 2, 4). When n 3, these same concepts generalize to higher dimensions, except the constraint boundaries now are hyperplanes instead of planes. Let us summarize. Consider any linear programming problem with n decision variables and a bounded feasible region. A CPF solution lies at the intersection of n constraint boundaries (and satisfies the other constraints as well). An edge of the feasible region is a feasible line segment that lies at the intersection of n 1 constraint boundaries, where each endpoint lies on one additional constraint boundary (so that these endpoints are CPF solutions). Two CPF solutions are adjacent if the line segment connecting them is an edge of the feasible region. Emanating from each CPF solution are n such edges, each one leading to one of the n adjacent CPF solutions. Each iteration of the simplex method moves from the current CPF solution to an adjacent one by moving along one of these n edges.

When you shift from a geometric viewpoint to an algebraic one, intersection of constraint boundaries changes to simultaneous solution of constraint boundary equations. The n constraint boundary equations yielding (defining) a CPF solution are its defining equations, where deleting one of these equations yields a line whose feasible segment is an edge of the feasible region. We next analyze some key properties of CPF solutions and then describe the implications of all these concepts for interpreting the simplex method. However, while the above summary is fresh in your mind, let us give you a preview of its implications. When the simplex method chooses an entering basic variable, the geometric interpretation is that it is choosing one of the edges emanating from the current CPF solution to move along. Increasing this variable from zero (and simultaneously changing the values of the other basic variables accordingly) corresponds to moving along this edge. Having one of the basic variables (the leaving basic variable) decrease so far that it reaches zero corresponds to reaching the first new constraint boundary at the other end of this edge of the feasible region. Properties of CPF Solutions We now focus on three key properties of CPF solutions that hold for any linear programming problem that has feasible solutions and a bounded feasible region. Property 1: (a) If there is exactly one optimal solution, then it must be a CPF solution. (b) If there are multiple optimal solutions (and a bounded feasible region), then at least two must be adjacent CPF solutions. Property 1 is a rather intuitive one from a geometric viewpoint. First consider Case (a), which is illustrated by the Wyndor Glass Co. problem (see Fig. 5.1) where the one optimal solution (2, 6) is indeed a CPF solution. Note that there is nothing special about this example that led to this result. For any problem having just one optimal solution, it always is possible to keep raising the objective function line (hyperplane) until it just touches one point (the optimal solution) at a corner of the feasible region. We now give an algebraic proof for this case. Proof of Case (a) of Property 1: We set up a proof by contradiction by assuming that there is exactly one optimal solution and that it is not a CPF solution.

196

5 THE THEORY OF THE SIMPLEX METHOD

We then show below that this assumption leads to a contradiction and so cannot be true. (The solution assumed to be optimal will be denoted by x*, and its objective function value by Z*.) Recall the definition of CPF solution (a feasible solution that does not lie on any line segment connecting two other feasible solutions). Since we have assumed that the optimal solution x* is not a CPF solution, this implies that there must be two other feasible solutions such that the line segment connecting them contains the optimal solution. Let the vectors x and x denote these two other feasible solutions, and let Z 1 and Z 2 denote their respective objective function values. Like each other point on the line segment connecting x and x , x* x (1 )x for some value of such that 0 1. Thus, Z* Z2 (1 )Z1. Since the weights and 1 add to 1, the only possibilities for how Z*, Z1, and Z2 compare are (1) Z* Z1 Z2, (2) Z1 Z* Z2, and (3) Z1 Z* Z2. The first possibility implies that x and x also are optimal, which contradicts the assumption that there is exactly one optimal solution. Both the latter possibilities contradict the assumption that x* (not a CPF solution) is optimal. The resulting conclusion is that it is impossible to have a single optimal solution that is not a CPF solution. Now consider Case (b), which was demonstrated in Sec. 3.2 under the definition of optimal solution by changing the objective function in the example to Z 3x1 2x2 (see Fig. 3.5 on page 35). What then happens when you are solving graphically is that the objective function line keeps getting raised until it contains the line segment connecting the two CPF solutions (2, 6) and (4, 3). The same thing would happen in higher dimensions except that an objective function hyperplane would keep getting raised until it contained the line segment(s) connecting two (or more) adjacent CPF solutions. As a consequence, all optimal solutions can be obtained as weighted averages of optimal CPF solutions. (This situation is described further in Probs. 4.5-5 and 4.5-6.) The real significance of Property 1 is that it greatly simplifies the search for an optimal solution because now only CPF solutions need to be considered. The magnitude of this simplification is emphasized in Property 2. Property 2: There are only a finite number of CPF solutions. This property certainly holds in Figs. 5.1 and 5.2, where there are just 5 and 10 CPF solutions, respectively. To see why the number is finite in general, recall that each CPF solution is the simultaneous solution of a system of n out of the m n constraint boundary equations. The number of different combinations of m n equations taken n at a time is mn

(m n)!

, n m!n! which is a finite number. This number, in turn, in an upper bound on the number of CPF solutions. In Fig. 5.1, m 3 and n 2, so there are 10 different systems of two equa-

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

197

tions, but only half of them yield CPF solutions. In Fig. 5.2, m 4 and n 3, which gives 35 different systems of three equations, but only 10 yield CPF solutions. Property 2 suggests that, in principle, an optimal solution can be obtained by exhaustive enumeration; i.e., find and compare all the finite number of CPF solutions. Unfortunately, there are finite numbers, and then there are finite numbers that (for all practical purposes) might as well be infinite. For example, a rather small linear programming problem with only m 50 and n 50 would have 100!/(50!)2 1029 systems of equations to be solved! By contrast, the simplex method would need to examine only approximately 100 CPF solutions for a problem of this size. This tremendous savings can be obtained because of the optimality test given in Sec. 4.1 and restated here as Property 3. Property 3: If a CPF solution has no adjacent CPF solutions that are better (as measured by Z), then there are no better CPF solutions anywhere. Therefore, such a CPF solution is guaranteed to be an optimal solution (by Property 1), assuming only that the problem possesses at least one optimal solution (guaranteed if the problem possesses feasible solutions and a bounded feasible region). To illustrate Property 3, consider Fig. 5.1 for the Wyndor Glass Co. example. For the CPF solution (2, 6), its adjacent CPF solutions are (0, 6) and (4, 3), and neither has a better value of Z than (2, 6) does. This outcome implies that none of the other CPF solutions—(0, 0) and (4, 0)—can be better than (2, 6), so (2, 6) must be optimal. By contrast, Fig. 5.3 shows a feasible region that can never occur for a linear programming problem but that does violate Property 3. The problem shown is identical to the Wyndor Glass Co. example (including the same objective function) except for the en-

FIGURE 5.3 Modification of the Wyndor Glass Co. problem that violates both linear programming and Property 3 for CPF solutions in linear programming.

x2

6

(0, 6)

(2, 6)

( 83 , 5) (4, 5) Z 36 3x1 5x2

4

2

(4, 0) (0, 0)

2

4

x1

198

5

THE THEORY OF THE SIMPLEX METHOD

largement of the feasible region to the right of ( 83 , 5). Consequently, the adjacent CPF solutions for (2, 6) now are (0, 6) and ( 83 , 5), and again neither is better than (2, 6). However, another CPF solution (4, 5) now is better than (2, 6), thereby violating Property 3. The reason is that the boundary of the feasible region goes down from (2, 6) to ( 83 , 5) and then “bends outward” to (4, 5), beyond the objective function line passing through (2, 6). The key point is that the kind of situation illustrated in Fig. 5.3 can never occur in linear programming. The feasible region in Fig. 5.3 implies that the 2x2 12 and 3x1 2x2 18 constraints apply for 0 x1 83 . However, under the condition that 83 x1 4, the 3x1 2x2 18 constraint is dropped and replaced by x2 5. Such “conditional constraints” just are not allowed in linear programming. The basic reason that Property 3 holds for any linear programming problem is that the feasible region always has the property of being a convex set, as defined in Appendix 2 and illustrated in several figures there. For two-variable linear programming problems, this convex property means that the angle inside the feasible region at every CPF solution is less than 180°. This property is illustrated in Fig. 5.1, where the angles at (0, 0), (0, 6), and (4, 0) are 90° and those at (2, 6) and (4, 3) are between 90° and 180°. By contrast, the feasible region in Fig. 5.3 is not a convex set, because the angle at ( 83 , 5) is more than 180°. This is the kind of “bending outward” at an angle greater than 180° that can never occur in linear programming. In higher dimensions, the same intuitive notion of “never bending outward” continues to apply. To clarify the significance of a convex feasible region, consider the objective function hyperplane that passes through a CPF solution that has no adjacent CPF solutions that are better. [In the original Wyndor Glass Co. example, this hyperplane is the objective function line passing through (2, 6).] All these adjacent solutions [(0, 6) and (4, 3) in the example] must lie either on the hyperplane or on the unfavorable side (as measured by Z) of the hyperplane. The feasible region being convex means that its boundary cannot “bend outward” beyond an adjacent CPF solution to give another CPF solution that lies on the favorable side of the hyperplane. So Property 3 holds. Extensions to the Augmented Form of the Problem For any linear programming problem in our standard form (including functional constraints in form), the appearance of the functional constraints after slack variables are introduced is as follows: (1) a11x1 a12x2 a1n xn xn1 (2) a21x1 a22x2 a2n xn xn2

b1 b2

............................................................

(m) am1x1 am2x2 amn xn

xnm bm,

where xn1, xn2, . . . , xnm are the slack variables. For other linear programming problems, Sec. 4.6 described how essentially this same appearance (proper form from Gaussian elimination) can be obtained by introducing artificial variables, etc. Thus, the original solutions (x1, x2, . . . , xn) now are augmented by the corresponding values of the slack or artificial variables (xn1, xn2, . . . , xnm) and perhaps some surplus variables as well. This augmentation led in Sec. 4.2 to defining basic solutions as augmented corner-point solutions and basic feasible solutions (BF solutions) as augmented CPF so-

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

199

lutions. Consequently, the preceding three properties of CPF solutions also hold for BF solutions. Now let us clarify the algebraic relationships between basic solutions and corner-point solutions. Recall that each corner-point solution is the simultaneous solution of a system of n constraint boundary equations, which we called its defining equations. The key question is: How do we tell whether a particular constraint boundary equation is one of the defining equations when the problem is in augmented form? The answer, fortunately, is a simple one. Each constraint has an indicating variable that completely indicates (by whether its value is zero) whether that constraint’s boundary equation is satisfied by the current solution. A summary appears in Table 5.3. For the type of constraint in each row of the table, note that the corresponding constraint boundary equation (fourth column) is satisfied if and only if this constraint’s indicating variable (fifth column) equals zero. In the last row (functional constraint in form), the indicating variable xni xsi actually is the difference between the artificial variable xni and the surplus variable xsi . Thus, whenever a constraint boundary equation is one of the defining equations for a corner-point solution, its indicating variable has a value of zero in the augmented form of the problem. Each such indicating variable is called a nonbasic variable for the corresponding basic solution. The resulting conclusions and terminology (already introduced in Sec. 4.2) are summarized next. Each basic solution has m basic variables, and the rest of the variables are nonbasic variables set equal to zero. (The number of nonbasic variables equals n plus the number of surplus variables.) The values of the basic variables are given by the simultaneous solution of the system of m equations for the problem in augmented form (after the nonbasic variables are set to zero). This basic solution is the augmented corner-point solution whose n defining equations are those indicated by the nonbasic variables. In particular, whenever an indicating variable in the fifth column of Table 5.3 is a nonbasic variable, the constraint boundary equation in the fourth column is a defining equation for the corner-point solution. (For functional constraints in form, at least one of the two supplementary variables xni and xsi always is a nonbasic variable, but the constraint boundary equation becomes a defining equation only if both of these variables are nonbasic variables.)

TABLE 5.3 Indicating variables for constraint boundary equations* Type of Constraint

Form of Constraint

Nonnegativity

xj 0 n

Functional ()

aijxj bi

j1 n

Functional ()

aijxj bi j1

Functional ()

aijxj bi j1

n

Constraint in Augmented Form xj 0 n

Constraint Boundary Equation

Indicating Variable

xj 0

xj

n

aijxj xni bi

aijxj bi

j1

xni

j1

n

n

aijxj xni bi j1

aijxj bi j1

n

aijxj xni xs bi j1 i

*Indicating variable 0 ⇒ constraint boundary equation satisfied; indicating variable 0 ⇒ constraint boundary equation violated.

n

aijxj bi j1

xni xni xsi

200

5

THE THEORY OF THE SIMPLEX METHOD

Now consider the basic feasible solutions. Note that the only requirements for a solution to be feasible in the augmented form of the problem are that it satisfy the system of equations and that all the variables be nonnegative. A BF solution is a basic solution where all m basic variables are nonnegative ( 0). A BF solution is said to be degenerate if any of these m variables equals zero.

Thus, it is possible for a variable to be zero and still not be a nonbasic variable for the current BF solution. (This case corresponds to a CPF solution that satisfies another constraint boundary equation in addition to its n defining equations.) Therefore, it is necessary to keep track of which is the current set of nonbasic variables (or the current set of basic variables) rather than to rely upon their zero values. We noted earlier that not every system of n constraint boundary equations yields a corner-point solution, because either the system has no solution or it has multiple solutions. For analogous reasons, not every set of n nonbasic variables yields a basic solution. However, these cases are avoided by the simplex method. To illustrate these definitions, consider the Wyndor Glass Co. example once more. Its constraint boundary equations and indicating variables are shown in Table 5.4. Augmenting each of the CPF solutions (see Table 5.1) yields the BF solutions listed in Table 5.5. This table places adjacent BF solutions next to each other, except for the pair consisting of the first and last solutions listed. Notice that in each case the nonbasic variables necessarily are the indicating variables for the defining equations. Thus, adjacent BF solutions differ by having just one different nonbasic variable. Also notice that each BF solution is the simultaneous solution of the system of equations for the problem in augmented form (see Table 5.4) when the nonbasic variables are set equal to zero. Similarly, the three corner-point infeasible solutions (see Table 5.2) yield the three basic infeasible solutions shown in Table 5.6. The other two sets of nonbasic variables, (1) x1 and x3 and (2) x2 and x4, do not yield a basic solution, because setting either pair of variables equal to zero leads to having no solution for the system of Eqs. (1) to (3) given in Table 5.4. This conclusion parallels the observation we made early in this section that the corresponding sets of constraint boundary equations do not yield a solution.

TABLE 5.4 Indicating variables for the constraint boundary equations of the Wyndor Glass Co. problem* Constraint

Constraint in Augmented Form

Constraint Boundary Equation

Indicating Variable

x1 0 x2 0 x1 4 2x2 12 3x1 x2 18

x1 0 x2 0 (1) 2x1 2x2 x3x3x3 24 (2) 3x1 2x2 x3x4x3 12 (3) 3x1 2x2 x3x3x5 18

x1 0 x2 0 x1 4 2x2 12 3x1 2x2 18

x1 x2 x3 x4 x5

*Indicating variable 0 ⇒ constraint boundary equation satisfied; indicating variable 0 ⇒ constraint boundary equation violated.

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

201

TABLE 5.5 BF solutions for the Wyndor Glass Co. problem CPF Solution

Defining Equations

(0, 0)

x1 0 x2 0

(0, 6)

BF Solution

Nonbasic Variables

(0, 0, 4, 12, 18)

x1 x2

x1 0 2x2 12

(0, 6, 4, 0, 6)

x1 x4

(2, 6)

2x2 12 3x1 2x2 18

(2, 6, 2, 0, 0)

x4 x5

(4, 3)

3x1 2x2 18 x1 4

(4, 3, 0, 6, 0)

x5 x3

(4, 0)

x1 4 x2 0

(4, 0, 0, 12, 6)

x3 x2

The simplex method starts at a BF solution and then iteratively moves to a better adjacent BF solution until an optimal solution is reached. At each iteration, how is the adjacent BF solution reached? For the original form of the problem, recall that an adjacent CPF solution is reached from the current one by (1) deleting one constraint boundary (defining equation) from the set of n constraint boundaries defining the current solution, (2) moving away from the current solution in the feasible direction along the intersection of the remaining n 1 constraint boundaries (an edge of the feasible region), and (3) stopping when the first new constraint boundary (defining equation) is reached. Equivalently, in our new terminology, the simplex method reaches an adjacent BF solution from the current one by (1) deleting one variable (the entering basic variable) from the set of n nonbasic variables defining the current solution, (2) moving away from the current solution by increasing this one variable from zero (and adjusting the other basic variables to still satisfy the system of equations) while keeping the remaining n 1 nonbasic variables at zero, and (3) stopping when the first of the basic variables (the leaving basic variable) reaches a value of zero (its constraint boundary). With either interpretation, the choice among the n alternatives in step 1 is made by selecting the one that would give the best rate of improvement in Z (per unit increase in the entering basic variable) during step 2. TABLE 5.6 Basic infeasible solutions for the Wyndor Glass Co. problem Corner-Point Infeasible Solution

Defining Equations

Basic Infeasible Solution

Nonbasic Variables

(0, 9)

x1 0 3x1 2x2 18

(0, 9, 4, 6, 0)

x1 x5

(4, 6)

2x2 12 x1 4

(4, 6, 0, 0, 6)

x4 x3

(6, 0)

3x1 2x2 18 x2 0

(6, 0, 2, 12, 0)

x5 x2

202

5

THE THEORY OF THE SIMPLEX METHOD

TABLE 5.7 Sequence of solutions obtained by the simplex method for the Wyndor Glass Co. problem Iteration

CPF Solution

Defining Equations

0

(0, 0)

x1 0 x2 0

1

(0, 6)

2

(2, 6)

BF Solution

Nonbasic Variables

Functional Constraints in Augmented Form

(0, 0, 4, 12, 18)

x1 0 x2 0

x1 2x2 x3 4 2x2 x4 12 3x1 2x2 x5 18

x1 0 2x2 12

(0, 6, 4, 0, 6)

x1 0 x4 0

x1 2x2 x3 4 2x2 x4 12 3x1 2x2 x5 18

2x2 12 3x1 2x2 18

(2, 6, 2, 0, 0)

x4 0 x5 0

x1 2x2 x3 4 2x2 x4 12 3x1 2x2 x5 18

Table 5.7 illustrates the close correspondence between these geometric and algebraic interpretations of the simplex method. Using the results already presented in Secs. 4.3 and 4.4, the fourth column summarizes the sequence of BF solutions found for the Wyndor Glass Co. problem, and the second column shows the corresponding CPF solutions. In the third column, note how each iteration results in deleting one constraint boundary (defining equation) and substituting a new one to obtain the new CPF solution. Similarly, note in the fifth column how each iteration results in deleting one nonbasic variable and substituting a new one to obtain the new BF solution. Furthermore, the nonbasic variables being deleted and added are the indicating variables for the defining equations being deleted and added in the third column. The last column displays the initial system of equations [excluding Eq. (0)] for the augmented form of the problem, with the current basic variables shown in bold type. In each case, note how setting the nonbasic variables equal to zero and then solving this system of equations for the basic variables must yield the same solution for (x1, x2) as the corresponding pair of defining equations in the third column.

5.2

THE REVISED SIMPLEX METHOD The simplex method as described in Chap. 4 (hereafter called the original simplex method ) is a straightforward algebraic procedure. However, this way of executing the algorithm (in either algebraic or tabular form) is not the most efficient computational procedure for computers because it computes and stores many numbers that are not needed at the current iteration and that may not even become relevant for decision making at subsequent iterations. The only pieces of information relevant at each iteration are the coefficients of the nonbasic variables in Eq. (0), the coefficients of the entering basic variable in the other equations, and the right-hand sides of the equations. It would be very useful to have a procedure that could obtain this information efficiently without computing and storing all the other coefficients. As mentioned in Sec. 4.8, these considerations motivated the development of the revised simplex method. This method was designed to accomplish exactly the same things as the original simplex method, but in a way that is more efficient for execution on a computer. Thus, it is a streamlined version of the original procedure. It computes and stores

5.2 THE REVISED SIMPLEX METHOD

203

only the information that is currently needed, and it carries along the essential data in a more compact form. The revised simplex method explicitly uses matrix manipulations, so it is necessary to describe the problem in matrix notation. (See Appendix 4 for a review of matrices.) To help you distinguish between matrices, vectors, and scalars, we consistently use BOLDFACE CAPITAL letters to represent matrices, boldface lowercase letters to represent vectors, and italicized letters in ordinary print to represent scalars. We also use a boldface zero (0) to denote a null vector (a vector whose elements all are zero) in either column or row form (which one should be clear from the context), whereas a zero in ordinary print (0) continues to represent the number zero. Using matrices, our standard form for the general linear programming model given in Sec. 3.2 becomes Z cx,

Maximize subject to Ax b

x 0,

and

where c is the row vector c [c1, c2, . . . , cn], x, b, and 0 are the column vectors such that x1 x x 2 , xn

b1 b b 2 , bm

0 0 0 , 0

and A is the matrix a11 a12 … a1n a a22 … a2n A 21 . ……………………… am1 am2 … amn To obtain the augmented form of the problem, introduce the column vector of slack variables xn1 x xs n2 xnm so that the constraints become [A, I]

x b x

s

and

x 0, x

s

204

5

THE THEORY OF THE SIMPLEX METHOD

where I is the m m identity matrix, and the null vector 0 now has n m elements. (We comment at the end of the section about how to deal with problems that are not in our standard form.) Solving for a Basic Feasible Solution Recall that the general approach of the simplex method is to obtain a sequence of improving BF solutions until an optimal solution is reached. One of the key features of the revised simplex method involves the way in which it solves for each new BF solution after identifying its basic and nonbasic variables. Given these variables, the resulting basic solution is the solution of the m equations [A, I]

x b, x

s

in which the n nonbasic variables from the n m elements of

x x

s

are set equal to zero. Eliminating these n variables by equating them to zero leaves a set of m equations in m unknowns (the basic variables). This set of equations can be denoted by BxB b, where the vector of basic variables xB1 x xB B2 xBm is obtained by eliminating the nonbasic variables from

x , x

s

and the basis matrix B11 B12 … B1m B B22 … B2m 21 B ………………………… Bm1 Bm2 … Bmm is obtained by eliminating the columns corresponding to coefficients of nonbasic variables from [A, I]. (In addition, the elements of xB and, therefore, the columns of B may be placed in a different order when the simplex method is executed.) The simplex method introduces only basic variables such that B is nonsingular, so that B1 always will exist. Therefore, to solve BxB b, both sides are premultiplied by B1: B1BxB B1b.

5.2 THE REVISED SIMPLEX METHOD

205

Since B1B I, the desired solution for the basic variables is xB B1b. Let cB be the vector whose elements are the objective function coefficients (including zeros for slack variables) for the corresponding elements of xB. The value of the objective function for this basic solution is then Z cBxB cBB1b. Example. To illustrate this method of solving for a BF solution, consider again the Wyndor Glass Co. problem presented in Sec. 3.1 and solved by the original simplex method in Table 4.8. In this case, c [3, 5],

1 [A, I] 0 3

0 2 2

1 0 0

0 1 0

0 0 , 1

4 b 12 , 18

x x 1 , x2

x3 xs x4 . x5

Referring to Table 4.8, we see that the sequence of BF solutions obtained by the simplex method (original or revised) is the following: Iteration 0 x3 xB x4 , x 5

1 B 0 0

cB [0, 0, 0],

0 1 0

0 0 B1, 1

so

1 x3 x4 0 0 x5

0 1 0

4 0 4 0 12 12 , 1 18 18

4 Z [0, 0, 0] 12 0. 18

so

Iteration 1 x3 xB x2 , x5

1 B 0 0

1 x3 x2 0 0 x5

0

0 2 2

0 0 , 1

1 B1 0 0

so 4 0 4 0 12 6 , 1 18 6

1 2

1

cB [0, 5, 0],

so

4 Z [0, 5, 0] 6 30. 6

0 1 2

1

0 0 , 1

206

5

THE THEORY OF THE SIMPLEX METHOD

Iteration 2 x3 xB x2 , x1

1 B 0 0

1 x3 x2 0 0 x1

1 3 1 2 1 3

0 2 2

1 0 , 3

1 1 B 0 0

1 3 1 2 1 3

13 0 , 1 3

so

cB [0, 5, 3],

1 3 4 2 0 12 6 , 1 2 3 18 2 Z [0, 5, 3] 6 36. 2

so

Matrix Form of the Current Set of Equations The last preliminary before we summarize the revised simplex method is to show the matrix form of the set of equations appearing in the simplex tableau for any iteration of the original simplex method. For the original set of equations, the matrix form is

1 0

Z c 0 0 . x A I b xs

This set of equations also is exhibited in the first simplex tableau of Table 5.8. The algebraic operations performed by the simplex method (multiply an equation by a constant and add a multiple of one equation to another equation) are expressed in maTABLE 5.8 Initial and later simplex tableaux in matrix form Coefficient of: Basic Variable

Eq.

Z

Original Variables

Slack Variables

Right Side

0

Z xB

(0) (1, 2, . . . , m)

1 0

c A

0 I

0 b

Any

Z xB

(0) (1, 2, . . . m)

1 0

cBB1A c B1 A

cBB1 B1

Iteration

cBB1b B1b

5.2 THE REVISED SIMPLEX METHOD

207

trix form by premultiplying both sides of the original set of equations by the appropriate matrix. This matrix would have the same elements as the identity matrix, except that each multiple for an algebraic operation would go into the spot needed to have the matrix multiplication perform this operation. Even after a series of algebraic operations over several iterations, we still can deduce what this matrix must be (symbolically) for the entire series by using what we already know about the right-hand sides of the new set of equations. In particular, after any iteration, xB B1b and Z cBB1b, so the right-hand sides of the new set of equations have become cBB1b cBB1 0 . B1 b B1b

Z 1 xB 0

Because we perform the same series of algebraic operations on both sides of the original set of operations, we use this same matrix that premultiplies the original right-hand side to premultiply the original left-hand side. Consequently, since 1 cBB1 1 0 B1 0

cBB1A c cBB1 , B1A B1

c 0 1 A I 0

the desired matrix form of the set of equations after any iteration is

1 0

cBB1A c B1A

Z cBB1 cBB1b . x 1 B B1b x s

The second simplex tableau of Table 5.8 also exhibits this same set of equations. Example. To illustrate this matrix form for the current set of equations, we will show how it yields the final set of equations resulting from iteration 2 for the Wyndor Glass Co. problem. Using the B1 and cB given for iteration 2 at the end of the preceding subsection, we have 1 B A 0 0

1 3 1 2 1 3

13 1 0 0 1 3 3

1 3 1 2 1 3

13 0 [0, 32 , 1], 1 3

1

1

cBB

1 [0, 5, 3] 0 0

0 cBB1A c [0, 5, 3] 0 1

0 0 2 0 2 1

0 1 [3, 5] [0, 0]. 0

0 1 , 0

208

5

THE THEORY OF THE SIMPLEX METHOD

Also, by using the values of xB B1b and Z cBB1b calculated at the end of the preceding subsection, these results give the following set of equations: 1 0 0 0

0 0 0 1

0 0 1 0

0 1 0 0

3 2 1 3 1 2 1 3

Z 1 x1 1 3 x2 0 x3 1 x4 3 x5

36 2 , 6 2

as shown in the final simplex tableau in Table 4.8.

The Overall Procedure There are two key implications from the matrix form of the current set of equations shown at the bottom of Table 5.8. The first is that only B1 needs to be derived to be able to calculate all the numbers in the simplex tableau from the original parameters (A, b, cB) of the problem. (This implication is the essence of the fundamental insight described in the next section.) The second is that any one of these numbers can be obtained individually, usually by performing only a vector multiplication (one row times one column) instead of a complete matrix multiplication. Therefore, the required numbers to perform an iteration of the simplex method can be obtained as needed without expending the computational effort to obtain all the numbers. These two key implications are incorporated into the following summary of the overall procedure. Summary of the Revised Simplex Method. 1. Initialization: Same as for the original simplex method. 2. Iteration: Step 1 Determine the entering basic variable: Same as for the original simplex method. Step 2 Determine the leaving basic variable: Same as for the original simplex method, except calculate only the numbers required to do this [the coefficients of the entering basic variable in every equation but Eq. (0), and then, for each strictly positive coefficient, the right-hand side of that equation].1 Step 3 Determine the new BF solution: Derive B1 and set xB B1b. 3. Optimality test: Same as for the original simplex method, except calculate only the numbers required to do this test, i.e., the coefficients of the nonbasic variables in Eq. (0). In step 3 of an iteration, B1 could be derived each time by using a standard computer routine for inverting a matrix. However, since B (and therefore B1) changes so little from one iteration to the next, it is much more efficient to derive the new B1 (denote it by B1 new) from the B1 at the preceding iteration (denote it by B1 old). (For the initial BF solution, 1

Because the value of xB is the entire vector of right-hand sides except for Eq. (0), the relevant right-hand sides need not be calculated here if xB was calculated in step 3 of the preceding iteration.

5.2 THE REVISED SIMPLEX METHOD

209

B I B1.) One method for doing this derivation is based directly upon the interpretation of the elements of B1 [the coefficients of the slack variables in the current Eqs. (1), (2), . . . , (m)] presented in the next section, as well as upon the procedure used by the original simplex method to obtain the new set of equations from the preceding set. To describe this method formally, let xk entering basic variable, a ik coefficient of xk in current Eq. (i), for i 1, 2, . . . , m (calculated in step 2 of an iteration), r number of equation containing the leaving basic variable. Recall that the new set of equations [excluding Eq. (0)] can be obtained from the preceding set by subtracting a ik /a rk times Eq. (r) from Eq. (i), for all i 1, 2, . . . , m except i r, and then dividing Eq. (r) by a rk. Therefore, the element in row i and column j of B1 new is (B1 new)ij

a ik 1 (B1 old)rj old)ij a r (B k 1 (B1 ) a rk old rj

if i r, if i r.

These formulas are expressed in matrix notation as 1 B1 new EB old,

where matrix E is an identity matrix except that its rth column is replaced by the vector 1 2 , m

where

a i k a i rk 1 a rk

if i r, if i r.

Thus, E [U1, U2, . . . , Ur1, , Ur1, . . . , Um], where the m elements of each of the Ui column vectors are 0 except for a 1 in the ith position. Example. We shall illustrate the revised simplex method by applying it to the Wyndor Glass Co. problem. The initial basic variables are the slack variables x3 xB x4 . x5 Iteration 1 Because the initial B1 I, no calculations are needed to obtain the numbers required to identify the entering basic variable x2 (c2 5 3 c1) and the leaving basic variable x4 (a12 0, b2/a22 1 2 2 1 2 8 b3/a32, so r 2). Thus, the new set of basic variables is x3 xB x2 . x5

210

5

THE THEORY OF THE SIMPLEX METHOD

To obtain the new B1, a12 0 a22 1 1 , 2 a22 a32 1 a22 so

B

1

1 0 0

0 1 0 0 1 0

0 1 2

1

1 0 0 0 1 0

0 1 0

0 1 2

1

0 0 , 1

so that 1 xB 0 0

4 0 4 0 12 6 . 1 18 6

0 1 2

1

To test whether this solution is optimal, we calculate the coefficients of the nonbasic variables (x1 and x4) in Eq. (0). Performing only the relevant parts of the matrix multiplications, we obtain 1 cBB1A c [0, 5, 0] 0 0 1

cBB

— [0, 5, 0] — —

0 1 0 0 1 3

0 1 2

1 0 1 2

1

— — [3, —] [3, —], —

— — [—, 52 , —], —

so the coefficients of x1 and x4 are 3 and 52 , respectively. Since x1 has a negative coefficient, this solution is not optimal. Iteration 2 Using these coefficients of the nonbasic variables in Eq. (0), since only x1 has a negative coefficient, we begin the next iteration by identifying x1 as the entering basic variable. To determine the leaving basic variable, we must calculate the other coefficients of x1: 1 B1A 0 0

0 1 2

1

0 1 0 0 1 3

1 — — 0 — 3

— — . —

5.2 THE REVISED SIMPLEX METHOD

211

By using the right side column for the current BF solution (the value of xB) just given for iteration 1, the ratios 4/1 6/3 indicate that x5 is the leaving basic variable, so the new set of basic variables is x3 xB x2 x1

a 11 1 a 31 3 a 21 0 . a 31 1 1 3 a 31

with

Therefore, the new B1 is 1 B1 0 0

0 1 3 1 1 0 0 1 0 3 0

0 1 2

1

1 0 0 0 1 0

1 3 1 2 1 3

1 3 0 , 1 3

so that 1 xB 0 0

1 3 1 2 1 3

2 1 3 4 0 12 6 . 1 2 3 18

Applying the optimality test, we find that the coefficients of the nonbasic variables (x4 and x5) in Eq. (0) are — cBB1 [0, 5, 3] — —

1 3 1 2 1 3

1 3 0 [—, 32 , 1]. 1 3

Because both coefficients ( and 1) are nonnegative, the current solution (x1 2, x2 6, x3 2, x4 0, x5 0) is optimal and the procedure terminates. 3 2

General Observations The preceding discussion was limited to the case of linear programming problems fitting our standard form given in Sec. 3.2. However, the modifications for other forms are relatively straightforward. The initialization would be conducted just as it would for the original simplex method (see Sec. 4.6). When this step involves introducing artificial variables to obtain an initial BF solution (and thereby to obtain an identity matrix as the initial basis matrix), these variables are included among the m elements of xs. Let us summarize the advantages of the revised simplex method over the original simplex method. One advantage is that the number of arithmetic computations may be reduced. This is especially true when the A matrix contains a large number of zero elements (which is usually the case for the large problems arising in practice). The amount of information that must be stored at each iteration is less, sometimes considerably so. The revised simplex method also permits the control of the rounding errors inevitably generated

212

5 THE THEORY OF THE SIMPLEX METHOD

by computers. This control can be exercised by periodically obtaining the current B1 by directly inverting B. Furthermore, some of the postoptimality analysis problems discussed in Sec. 4.7 can be handled more conveniently with the revised simplex method. For all these reasons, the revised simplex method is usually preferable to the original simplex method for computer execution.

5.3

A FUNDAMENTAL INSIGHT We shall now focus on a property of the simplex method (in any form) that has been revealed by the revised simplex method in the preceding section.1 This fundamental insight provides the key to both duality theory and sensitivity analysis (Chap. 6), two very important parts of linear programming. The insight involves the coefficients of the slack variables and the information they give. It is a direct result of the initialization, where the ith slack variable xni is given a coefficient of 1 in Eq. (i) and a coefficient of 0 in every other equation [including Eq. (0)] for i 1, 2, . . . , m, as shown by the null vector 0 and the identity matrix I in the slack variables column for iteration 0 in Table 5.8. (For most of this section, we are assuming that the problem is in our standard form, with bi 0 for all i 1, 2, . . . , m, so that no additional adjustments are needed in the initialization.) The other key factor is that subsequent iterations change the initial equations only by 1. Multiplying (or dividing) an entire equation by a nonzero constant 2. Adding (or subtracting) a multiple of one entire equation to another entire equation As already described in the preceding section, a sequence of these kinds of elementary algebraic operations is equivalent to premultiplying the initial simplex tableau by some matrix. (See Appendix 4 for a review of matrices.) The consequence can be summarized as follows. Verbal description of fundamental insight: After any iteration, the coefficients of the slack variables in each equation immediately reveal how that equation has been obtained from the initial equations. As one example of the importance of this insight, recall from Table 5.8 that the matrix formula for the optimal solution obtained by the simplex method is xB B1b, where xB is the vector of basic variables, B1 is the matrix of coefficients of slack variables for rows 1 to m of the final tableau, and b is the vector of original right-hand sides (resource availabilities). (We soon will denote this particular B1 by S*.) Postoptimality analysis normally includes an investigation of possible changes in b. By using this formula, you can see exactly how the optimal BF solution changes (or whether it becomes infeasible because of negative variables) as a function of b. You do not have to reapply the simplex method over and over for each new b, because the coefficients of the slack 1

However, since some instructors do not cover the preceding section, we have written this section in a way that can be understood without first reading Sec. 5.2. It is helpful to take a brief look at the matrix notation introduced at the beginning of Sec. 5.2, including the resulting key equation, xB B1b.

5.3 A FUNDAMENTAL INSIGHT

213

variables tell all! In a similar fashion, this fundamental insight provides a tremendous computational saving for the rest of sensitivity analysis as well. To spell out the how and the why of this insight, let us look again at the Wyndor Glass Co. example. (The OR Tutor also includes another demonstration example.) Example. Table 5.9 shows the relevant portion of the simplex tableau for demonstrating this fundamental insight. Light lines have been drawn around the coefficients of the slack variables in all the tableaux in this table because these are the crucial coefficients for applying the insight. To avoid clutter, we then identify the pivot row and pivot column by a single box around the pivot number only. Iteration 1 To demonstrate the fundamental insight, our focus is on the algebraic operations performed by the simplex method while using Gaussian elimination to obtain the new BF solution. If we do all the algebraic operations with the old row 2 (the pivot row) rather than the new one, then the algebraic operations spelled out in Chap. 4 for iteration 1 are New New New New

row row row row

0 old row 0 ( 52 )(old 1 old row 1 (0)(old 2 ( 12 )(old 3 old row 3 (1)(old

row row row row

2), 2), 2), 2).

TABLE 5.9 Simplex tableaux without leftmost columns for the Wyndor Glass Co. problem Coefficient of: Iteration

x1

x2

x3

0

3 1 0 3

5 0 2 2

0 1 0 0

3

0

0

1

0

1

0

1

0

3

0

0

0

0

0

3 2

0

0

1

1 3

0

1

0

1

0

0

1

2

x4 0 0 1 0 5 2 0 1 2 1

x5

Right Side

0 0 0 1

0 4 12 18

0

30

0

4

0

6

1

6

1

36

1

3

2

1 2

0

6

1 3

1 3

2

214

5

THE THEORY OF THE SIMPLEX METHOD

Ignoring row 0 for the moment, we see that these algebraic operations amount to premultiplying rows 1 to 3 of the initial tableau by the matrix 1 0 0

0 1 2

1

0 0 . 1

Rows 1 to 3 of the initial tableau are 1 Old rows 1–3 0 3

0 2 2

1 0 0

0 1 0

4 12 , 18

0 0 1

where the third, fourth, and fifth columns (the coefficients of the slack variables) form an identity matrix. Therefore, 1 New rows 1–3 0 0 1 0 3

0 1 0

1 0 0

0 1 2

1 0 1 2

1

0 1 0 0 1 3 0 0 1

0 2 2

1 0 0

0 1 0

4 12 18

0 0 1

4 6 . 6

Note how the first matrix is reproduced exactly in the box below it as the coefficients of the slack variables in rows 1 to 3 of the new tableau, because the coefficients of the slack variables in rows 1 to 3 of the initial tableau form an identity matrix. Thus, just as stated in the verbal description of the fundamental insight, the coefficients of the slack variables in the new tableau do indeed provide a record of the algebraic operations performed. This insight is not much to get excited about after just one iteration, since you can readily see from the initial tableau what the algebraic operations had to be, but it becomes invaluable after all the iterations are completed. For row 0, the algebraic operation performed amounts to the following matrix calculations, where now our focus is on the vector [0, 52, 0] that premultiplies rows 1 to 3 of the initial tableau. New row 0 [3,

5 0,

0,

0 0] [0,

[3,

0,

0,

5 2

, 5 2

,

1 0] 0 3

0 2 2

0,

30].

1 0 0

0 1 0

0 0 1

4 12 18

Note how this vector is reproduced exactly in the box below it as the coefficients of the slack variables in row 0 of the new tableau, just as was claimed in the statement of the fundamental insight. (Once again, the reason is the identity matrix for the coefficients of the slack variables in rows 1 to 3 of the initial tableau, along with the zeros for these coefficients in row 0 of the initial tableau.)

5.3 A FUNDAMENTAL INSIGHT

215

Iteration 2 The algebraic operations performed on the second tableau of Table 5.9 for iteration 2 are New New New New

row row row row

0 old row 0 (1)(old row 3), 1 old row 1 (1 3 )(old row 3), 2 old row 2 (0)(old row 3), 3 (1 3 )(old row 3).

Ignoring row 0 for the moment, we see that these operations amount to premultiplying rows 1 to 3 of this tableau by the matrix 1 0 0

0 1 3 1 0 . 1 0 3

Writing this second tableau as the matrix product shown for iteration 1 (namely, the corresponding matrix times rows 1 to 3 of the initial tableau) then yields 1 Final rows 1–3 0 0

0 0 1

0 1 3 1 0 0 1 1 0 3 0 1 0 0 0 1 0

1 0 0

1

0 1 0 0 1 3

0 2 2

1 0 0

0 1 0

0 0 1

4 12 18

1 3 1 2 1 3

1 3 1 0 0 1 3 3

0 2 2

1 0 0

0 1 0

0 0 1

4 12 18

1 3 1 2 1 3

13 0

0 1 2

1 3

2 6 . 2

The first two matrices shown on the first line of these calculations summarize the algebraic operations of the second and first iterations, respectively. Their product, shown as the first matrix on the second line, then combines the algebraic operations of the two iterations. Note how this matrix is reproduced exactly in the box below it as the coefficients of the slack variables in rows 1 to 3 of the new (final) tableau shown on the third line. What this portion of the tableau reveals is how the entire final tableau (except row 0) has been obtained from the initial tableau, namely, Final row 1 (1)(initial row 1) ( 13 )(initial row 2) ( 13 )(initial row 3), Final row 2 (0)(initial row 1) (1 2 )(initial row 2) (0)(initial row 3), Final row 3 (0)(initial row 1) (1 3 )(initial row 2) (1 3 )(initial row 3). To see why these multipliers of the initial rows are correct, you would have to trace through all the algebraic operations of both iterations. For example, why does final row 1 include ( 13 )(initial row 2), even though a multiple of row 2 has never been added directly to row 1? The reason is that initial row 2 was subtracted from initial row 3 in iteration 1, and then ( 13 )(old row 3) was subtracted from old row 1 in iteration 2.

216

5

THE THEORY OF THE SIMPLEX METHOD

However, there is no need for you to trace through. Even when the simplex method has gone through hundreds or thousands of iterations, the coefficients of the slack variables in the final tableau will reveal how this tableau has been obtained from the initial tableau. Furthermore, the same algebraic operations would give these same coefficients even if the values of some of the parameters in the original model (initial tableau) were changed, so these coefficients also reveal how the rest of the final tableau changes with changes in the initial tableau. To complete this story for row 0, the fundamental insight reveals that the entire final row 0 can be calculated from the initial tableau by using just the coefficients of the slack variables in the final row 0—[0, 32, 1]. This calculation is shown below, where the first vector is row 0 of the initial tableau and the matrix is rows 1 to 3 of the initial tableau. Final row 0 [3,

5 0,

0,

0 0] [0, [0,

0,

0,

3 2

, 3 2

,

1 1] 0 3

0 2 2

1,

36].

1 0 0

0 1 0

0 0 1

4 12 18

Note again how the vector premultiplying rows 1 to 3 of the initial tableau is reproduced exactly as the coefficients of the slack variables in the final row 0. These quantities must be identical because of the coefficients of the slack variables in the initial tableau (an identity matrix below a null vector). This conclusion is the row 0 part of the fundamental insight. Mathematical Summary Because its primary applications involve the final tableau, we shall now give a general mathematical expression for the fundamental insight just in terms of this tableau, using matrix notation. If you have not read Sec. 5.2, you now need to know that the parameters of the model are given by the matrix A aij and the vectors b bi and c cj, as displayed at the beginning of that section. The only other notation needed is summarized and illustrated in Table 5.10. Notice how vector t (representing row 0) and matrix T (representing the other rows) together correspond to the rows of the initial tableau in Table 5.9, whereas vector t* and matrix T* together correspond to the rows of the final tableau in Table 5.9. This table also shows these vectors and matrices partitioned into three parts: the coefficients of the original variables, the coefficients of the slack variables (our focus), and the right-hand side. Once again, the notation distinguishes between parts of the initial tableau and the final tableau by using an asterisk only in the latter case. For the coefficients of the slack variables (the middle part) in the initial tableau of Table 5.10, notice the null vector 0 in row 0 and the identity matrix I below, which provide the keys for the fundamental insight. The vector and matrix in the same location of the final tableau, y* and S*, then play a prominent role in the equations for the fundamental insight. A and b in the initial tableau turn into A* and b* in the final tableau. For row 0 of the final tableau, the coefficients of the decision variables are z* c (so the vector z* is what has been added to the vector of initial coefficients, c), and the right-hand side Z* denotes the optimal value of Z.

5.3 A FUNDAMENTAL INSIGHT

217

TABLE 5.10 General notation for initial and final simplex tableaux in matrix form, illustrated by the Wyndor Glass Co. problem Initial Tableau Row 0:

t [3, 5

Other rows:

T

Combined:

T A

1 0 3

t

0, 0, 0

0 2 2

1 0 0

0 1 0

0 0 1

c

0 I

0 . b

0] [c 0 0]. 4 12 [A I b]. 18

Final Tableau Row 0:

t* [0, 0

1 13 2 1 0 0 6 [A* S* b*]. 2 1 1 0 3 2 3

Other rows:

0 T* 0 1

0 1 0

Combined:

T*

z* c A*

t*

36] [z* c y* Z*].

0, 3 2 , 1 1 3

y* S*

Z* b* .

It is helpful at this point to look back at Table 5.8 in Sec. 5.2 and compare it with Table 5.10. (If you haven’t previously studied Sec. 5.2, you will need to read the definition of the basis matrix B and the vectors xB and cB given early in that section before looking at Table 5.8.) The notation for the components of the initial simplex tableau is the same in the two tables. The lower part of Table 5.8 shows any later simplex tableau in matrix form, whereas the lower part of Table 5.10 gives the final tableau in matrix form. Note that the matrix B1 in Table 5.8 is in the same location as S* in Table 5.10. Thus, S* B1 when B is the basis matrix for the optimal solution found by the simplex method. Referring to Table 5.10 again, suppose now that you are given the initial tableau, t and T, and just y* and S* from the final tableau. How can this information alone be used to calculate the rest of the final tableau? The answer is provided by Table 5.8. This table includes some information that is not directly relevant to our current discussion, namely, how y* and S* themselves can be calculated (y* cBB1 and S* B1) by knowing the set of basic variables and so the basis matrix B for the optimal solution found by the simplex method. However, the lower part of this table also shows how the rest of the final tableau can be obtained from the coefficients of the slack variables, which is summarized as follows. Fundamental Insight (1) t* t y*T [y*A c y* y*b]. (2) T* S*T [S*A S* S*b].

218

5

THE THEORY OF THE SIMPLEX METHOD

Thus, by knowing the parameters of the model in the initial tableau (c, A, and b) and only the coefficients of the slack variables in the final tableau (y* and S*), these equations enable calculating all the other numbers in the final tableau. We already used these two equations when dealing with iteration 2 for the Wyndor Glass Co. problem in the preceding subsection. In particular, the right-hand side of the expression for final row 0 for iteration 2 is just t y*T, and the second line of the expression for final rows 1 to 3 is just S*T. Now let us summarize the mathematical logic behind the two equations for the fundamental insight. To derive Eq. (2), recall that the entire sequence of algebraic operations performed by the simplex method (excluding those involving row 0) is equivalent to premultiplying T by some matrix, call it M. Therefore, T* MT, but now we need to identify M. By writing out the component parts of T and T*, this equation becomes [A* S* b*] M [A I b] ↑ [MA M Mb]. ↑ Because the middle (or any other) component of these equal matrices must be the same, it follows that M S*, so Eq. (2) is a valid equation. Equation (1) is derived in a similar fashion by noting that the entire sequence of algebraic operations involving row 0 amounts to adding some linear combination of the rows in T to t, which is equivalent to adding to t some vector times T. Denoting this vector by v, we thereby have t* t vT, but v still needs to be identified. Writing out the component parts of t and t* yields [z* c y* Z*] [c 0 0] v [A I b] [c vA v vb]. ↑ ↑ Equating the middle component of these equal vectors gives v y*, which validates Eq. (1). Adapting to Other Model Forms Thus far, the fundamental insight has been described under the assumption that the original model is in our standard form, described in Sec. 3.2. However, the above mathematical logic now reveals just what adjustments are needed for other forms of the original model. The key is the identity matrix I in the initial tableau, which turns into S* in the final tableau. If some artificial variables must be introduced into the initial tableau to serve as initial basic variables, then it is the set of columns (appropriately ordered) for all the initial basic variables (both slack and artificial) that forms I in this tableau. (The columns for any surplus variables are extraneous.) The same columns in the final tableau provide S* for the T* S*T equation and y* for the t* t y*T equation. If M’s were introduced into the

5.3 A FUNDAMENTAL INSIGHT

219

preliminary row 0 as coefficients for artificial variables, then the t for the t* t y*T equation is the row 0 for the initial tableau after these nonzero coefficients for basic variables are algebraically eliminated. (Alternatively, the preliminary row 0 can be used for t, but then these M’s must be subtracted from the final row 0 to give y*.) (See Prob. 5.3-11.) Applications The fundamental insight has a variety of important applications in linear programming. One of these applications involves the revised simplex method. As described in the preceding section (see Table 5.8), this method used B1 and the initial tableau to calculate all the relevant numbers in the current tableau for every iteration. It goes even further than the fundamental insight by using B1 to calculate y* itself as y* cBB1. Another application involves the interpretation of the shadow prices ( y1*, y2*, . . . , y*m) described in Sec. 4.7. The fundamental insight reveals that Z* (the value of Z for the optimal solution) is m

Z* y*b yi*bi, i1

so, e.g., 3 Z* 0b1 b2 b3 2 for the Wyndor Glass Co. problem. This equation immediately yields the interpretation for the yi* values given in Sec. 4.7. Another group of extremely important applications involves various postoptimality tasks (reoptimization technique, sensitivity analysis, parametric linear programming— described in Sec. 4.7) that investigate the effect of making one or more changes in the original model. In particular, suppose that the simplex method already has been applied to obtain an optimal solution (as well as y* and S*) for the original model, and then these changes are made. If exactly the same sequence of algebraic operations were to be applied to the revised initial tableau, what would be the resulting changes in the final tableau? Because y* and S* don’t change, the fundamental insight reveals the answer immediately. For example, consider the change from b2 12 to b2 13 as illustrated in Fig. 4.8 for the Wyndor Glass Co. problem. It is not necessary to solve for the new optimal solution (x1, x2) ( 53 , 123 ) because the values of the basic variables in the final tableau (b*) are immediately revealed by the fundamental insight: x3 x2 b* S*b x1

73 13 4 0 13 1 23 . 5 1 3 3 18 There is an even easier way to make this calculation. Since the only change is in the second component of b (b2 1), which gets premultiplied by only the second column of S*, the change in b* can be calculated as simply 1 3 b* 12 b2 1 3

1 0 0

13 1 2 , 1 3

1 3

1 2 1 3

220

5

THE THEORY OF THE SIMPLEX METHOD

so the original values of the basic variables in the final tableau (x3 2, x2 6, x1 2) now become x3 2 1 3 17 3 1 13 x2 6 2 2 . 1 5 x1 2 3 3 (If any of these new values were negative, and thus infeasible, then the reoptimization technique described in Sec. 4.7 would be applied, starting from this revised final tableau.) Applying incremental analysis to the preceding equation for Z* also immediately yields 3 3 Z* b2 . 2 2 The fundamental insight can be applied to investigating other kinds of changes in the original model in a very similar fashion; it is the crux of the sensitivity analysis procedure described in the latter part of Chap. 6. You also will see in the next chapter that the fundamental insight plays a key role in the very useful duality theory for linear programming.

5.4

CONCLUSIONS Although the simplex method is an algebraic procedure, it is based on some fairly simple geometric concepts. These concepts enable one to use the algorithm to examine only a relatively small number of BF solutions before reaching and identifying an optimal solution. Chapter 4 describes how elementary algebraic operations are used to execute the algebraic form of the simplex method, and then how the tableau form of the simplex method uses the equivalent elementary row operations in the same way. Studying the simplex method in these forms is a good way of getting started in learning its basic concepts. However, these forms of the simplex method do not provide the most efficient form for execution on a computer. Matrix operations are a faster way of combining and executing elementary algebraic operations or row operations. Therefore, by using the matrix form of the simplex method, the revised simplex method provides an effective way of adapting the simplex method for computer implementation. The final simplex tableau includes complete information on how it can be algebraically reconstructed directly from the initial simplex tableau. This fundamental insight has some very important applications, especially for postoptimality analysis.

SELECTED REFERENCES 1. Bazaraa, M. S., J. J. Jarvis, and H. D. Sherali: Linear Programming and Network Flows, 2d ed., Wiley, New York, 1990. 2. Dantzig, G. B., and M. N. Thapa: Linear Programming 1: Introduction, Springer, New York, 1997. 3. Schriver, A: Theory of Linear and Integer Programming, Wiley, New York, 1986. 4. Vanderbei, R. J.: Linear Programming: Foundations and Extensions, Kluwer Academic Publishers, Boston, MA, 1996.

CHAPTER 5 PROBLEMS

221

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE A Demonstration Example in OR Tutor: Fundamental Insight

Interactive Routines: Enter or Revise a General Linear Programming Model Set Up for the Simplex Method—Interactive Only Solve Interactively by the Simplex Method

Files (Chapter 3) for Solving the Wyndor Example: Excel File LINGO/LINDO File MPL/CPLEX File

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The demonstration example listed above may be helpful. I: You can check some of your work by using the interactive routines listed above for the original simplex method. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 5.1-1.* Consider the following problem. Z 3x1 2x2,

Maximize subject to 2x1 x2 6 x1 2x2 6

5.1-2. Repeat Prob. 5.1-1 for the model in Prob. 3.1-5.

and x1 0,

(d) Do the following for each set of two defining equations from part (b): Identify the indicating variable for each defining equation. Display the set of equations from part (c) after deleting these two indicating (nonbasic) variables. Then use the latter set of equations to solve for the two remaining variables (the basic variables). Compare the resulting basic solution to the corresponding basic solution obtained in part (c). (e) Without executing the simplex method, use its geometric interpretation (and the objective function) to identify the path (sequence of CPF solutions) it would follow to reach the optimal solution. For each of these CPF solutions in turn, identify the following decisions being made for the next iteration: (i) which defining equation is being deleted and which is being added; (ii) which indicating variable is being deleted (the entering basic variable) and which is being added (the leaving basic variable).

x2 0.

(a) Solve this problem graphically. Identify the CPF solutions by circling them on the graph. (b) Identify all the sets of two defining equations for this problem. For each set, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or corner-point infeasible solution. (c) Introduce slack variables in order to write the functional constraints in augmented form. Use these slack variables to identify the basic solution that corresponds to each corner-point solution found in part (b).

5.1-3. Consider the following problem. Maximize

Z 2x1 3x2,

subject to 3x1 x2 1 4x1 2x2 20 4x1 x2 10 x1 2x2 5 and x1 0,

x2 0.

222

5

THE THEORY OF THE SIMPLEX METHOD

(a) Solve this problem graphically. Identify the CPF solutions by circling them on the graph. (b) Develop a table giving each of the CPF solutions and the corresponding defining equations, BF solution, and nonbasic variables. Calculate Z for each of these solutions, and use just this information to identify the optimal solution. (c) Develop the corresponding table for the corner-point infeasible solutions, etc. Also identify the sets of defining equations and nonbasic variables that do not yield a solution.

x1 x2 15 2x1 x2 90 2x1 x2 30

3x1 x2 x3 60 x1 x2 2x3 10 x1 x2 x3 20

and x1 0,

and x2 0,

x3 0.

After slack variables are introduced and then one complete iteration of the simplex method is performed, the following simplex tableau is obtained.

Coefficient of: Iteration

1

Z x1 2x2,

subject to

Z 2x1 x2 x3,

subject to

x1 0,

5.1-7. Consider the following problem. Minimize

5.1-4. Consider the following problem. Maximize

(a) Identify the 10 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or cornerpoint infeasible solution. (b) For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables. (Compare with Table 6.9.)

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

x6

Right Side

Z x4 x1 x6

(0) (1) (2) (3)

1 0 0 0

0 0 1 0

1 4 1 2

3 5 2 3

0 1 0 0

2 3 1 1

0 0 0 1

20 30 10 10

(a) Identify the CPF solution obtained at iteration 1. (b) Identify the constraint boundary equations that define this CPF solution. 5.1-5. Consider the three-variable linear programming problem shown in Fig. 5.2. (a) Construct a table like Table 5.1, giving the set of defining equations for each CPF solution. (b) What are the defining equations for the corner-point infeasible solution (6, 0, 5)? (c) Identify one of the systems of three constraint boundary equations that yields neither a CPF solution nor a corner-point infeasible solution. Explain why this occurs for this system. 5.1-6. Consider the linear programming problem given in Table 6.1 as the dual problem for the Wyndor Glass Co. example.

x2 0.

(a) Solve this problem graphically. (b) Develop a table giving each of the CPF solutions and the corresponding defining equations, BF solution, and nonbasic variables. 5.1-8. Reconsider the model in Problem 4.6-3. (a) Identify the 10 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or a corner-point infeasible solution. (b) For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables. 5.1-9. Reconsider the model in Prob. 3.1-4. (a) Identify the 15 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or a corner-point infeasible solution. (b) For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables. 5.1-10. Each of the following statements is true under most circumstances, but not always. In each case, indicate when the statement will not be true and why. (a) The best CPF solution is an optimal solution. (b) An optimal solution is a CPF solution. (c) A CPF solution is the only optimal solution if none of its adjacent CPF solutions are better (as measured by the value of the objective function). 5.1-11. Consider the original form (before augmenting) of a linear programming problem with n decision variables (each with a nonnegativity constraint) and m functional constraints. Label each of the following statements as true or false, and then justify your

CHAPTER 5 PROBLEMS

answer with specific references (including page citations) to material in the chapter. (a) If a feasible solution is optimal, it must be a CPF solution. (b) The number of CPF solutions is at least (m n)! . m!n! (c) If a CPF solution has adjacent CPF solutions that are better (as measured by Z), then one of these adjacent CPF solutions must be an optimal solution. 5.1-12. Label each of the following statements about linear programming problems as true or false, and then justify your answer. (a) If a feasible solution is optimal but not a CPF solution, then infinitely many optimal solutions exist. (b) If the value of the objective function is equal at two different feasible points x* and x**, then all points on the line segment connecting x* and x** are feasible and Z has the same value at all those points. (c) If the problem has n variables (before augmenting), then the simultaneous solution of any set of n constraint boundary equations is a CPF solution. 5.1-13. Consider the augmented form of linear programming problems that have feasible solutions and a bounded feasible region. Label each of the following statements as true or false, and then justify your answer by referring to specific statements (with page citations) in the chapter. (a) There must be at least one optimal solution. (b) An optimal solution must be a BF solution. (c) The number of BF solutions is finite. 5.1-14.* Reconsider the model in Prob. 4.6-10. Now you are given the information that the basic variables in the optimal solution are x2 and x3. Use this information to identify a system of three constraint boundary equations whose simultaneous solution must be this optimal solution. Then solve this system of equations to obtain this solution. 5.1-15. Reconsider Prob. 4.3-7. Now use the given information and the theory of the simplex method to identify a system of three constraint boundary equations (in x1, x2, x3) whose simultaneous solution must be the optimal solution, without applying the simplex method. Solve this system of equations to find the optimal solution. 5.1-16. Reconsider Prob. 4.3-8. Using the given information and the theory of the simplex method, analyze the constraints of the problem in order to identify a system of three constraint boundary equations whose simultaneous solution must be the optimal solution (not augmented). Then solve this system of equations to obtain this solution.

223

5.1-17. Consider the following problem. Maximize

Z 2x1 2x2 3x3,

subject to 2x1 x2 2x3 4 x1 x2 x3 3 and x1 0,

x2 0,

x3 0.

Let x4 and x5 be the slack variables for the respective functional constraints. Starting with these two variables as the basic variables for the initial BF solution, you now are given the information that the simplex method proceeds as follows to obtain the optimal solution in two iterations: (1) In iteration 1, the entering basic variable is x3 and the leaving basic variable is x4; (2) in iteration 2, the entering basic variable is x2 and the leaving basic variable is x5. (a) Develop a three-dimensional drawing of the feasible region for this problem, and show the path followed by the simplex method. (b) Give a geometric interpretation of why the simplex method followed this path. (c) For each of the two edges of the feasible region traversed by the simplex method, give the equation of each of the two constraint boundaries on which it lies, and then give the equation of the additional constraint boundary at each endpoint. (d) Identify the set of defining equations for each of the three CPF solutions (including the initial one) obtained by the simplex method. Use the defining equations to solve for these solutions. (e) For each CPF solution obtained in part (d ), give the corresponding BF solution and its set of nonbasic variables. Explain how these nonbasic variables identify the defining equations obtained in part (d ). 5.1-18. Consider the following problem. Maximize

Z 3x1 4x2 2x3,

subject to x1 x2 x3 20 x1 2x2 x3 30 and x1 0,

x2 0,

x3 0.

Let x4 and x5 be the slack variables for the respective functional constraints. Starting with these two variables as the basic variables for the initial BF solution, you now are given the information that the simplex method proceeds as follows to obtain the optimal solution in two iterations: (1) In iteration 1, the entering basic vari-

224

5

THE THEORY OF THE SIMPLEX METHOD

able is x2 and the leaving basic variable is x5; (2) in iteration 2, the entering basic variable is x1 and the leaving basic variable is x4. Follow the instructions of Prob. 5.1-17 for this situation. 5.1-19. By inspecting Fig. 5.2, explain why Property 1b for CPF solutions holds for this problem if it has the following objective function. (a) Maximize Z x3. (b) Maximize Z x1 2x3. 5.1-20. Consider the three-variable linear programming problem shown in Fig. 5.2. (a) Explain in geometric terms why the set of solutions satisfying any individual constraint is a convex set, as defined in Appendix 2. (b) Use the conclusion in part (a) to explain why the entire feasible region (the set of solutions that simultaneously satisfies every constraint) is a convex set. 5.1-21. Suppose that the three-variable linear programming problem given in Fig. 5.2 has the objective function Maximize

Z 3x1 4x2 3x3.

method as it goes through one iteration in moving from (2, 4, 3) to (4, 2, 4). (You are given the information that it is moving along this line segment.) (a) What is the entering basic variable? (b) What is the leaving basic variable? (c) What is the new BF solution? 5.1-24. Consider a two-variable mathematical programming problem that has the feasible region shown on the graph, where the six dots correspond to CPF solutions. The problem has a linear objective function, and the two dashed lines are objective function lines passing through the optimal solution (4, 5) and the secondbest CPF solution (2, 5). Note that the nonoptimal solution (2, 5) is better than both of its adjacent CPF solutions, which violates Property 3 in Sec. 5.1 for CPF solutions in linear programming. Demonstrate that this problem cannot be a linear programming problem by constructing the feasible region that would result if the six line segments on the boundary were constraint boundaries for linear programming constraints. x2

Without using the algebra of the simplex method, apply just its geometric reasoning (including choosing the edge giving the maximum rate of increase of Z) to determine and explain the path it would follow in Fig. 5.2 from the origin to the optimal solution. 5.1-22. Consider the three-variable linear programming problem shown in Fig. 5.2. (a) Construct a table like Table 5.4, giving the indicating variable for each constraint boundary equation and original constraint. (b) For the CPF solution (2, 4, 3) and its three adjacent CPF solutions (4, 2, 4), (0, 4, 2), and (2, 4, 0), construct a table like Table 5.5, showing the corresponding defining equations, BF solution, and nonbasic variables. (c) Use the sets of defining equations from part (b) to demonstrate that (4, 2, 4), (0, 4, 2), and (2, 4, 0) are indeed adjacent to (2, 4, 3), but that none of these three CPF solutions are adjacent to each other. Then use the sets of nonbasic variables from part (b) to demonstrate the same thing. 5.1-23. The formula for the line passing through (2, 4, 3) and (4, 2, 4) in Fig. 5.2 can be written as (2, 4, 3) [(4, 2, 4) (2, 4, 3)] (2, 4, 3) (2, 2, 1), where 0 1 for just the line segment between these points. After augmenting with the slack variables x4, x5, x6, x7 for the respective functional constraints, this formula becomes (2, 4, 3, 2, 0, 0, 0) (2, 2, 1, 2, 2, 0, 0). Use this formula directly to answer each of the following questions, and thereby relate the algebra and geometry of the simplex

(2, 5)

5

(4, 5)

4 3 2 1

0

1

2

3

4

x1

5.2-1. Consider the following problem. Maximize

Z 8x1 4x2 6x3 3x4 9x5,

subject to x1 2x2 3x3 3x4 x5 180 4x1 3x2 2x3 x4 x5 270 x1 3x2 2x3 x4 3x5 180

(resource 1) (resource 2) (resource 3)

CHAPTER 5 PROBLEMS

and

225

and

xj 0,

j 1, . . . , 5.

You are given the facts that the basic variables in the optimal solution are x3, x1, and x5 and that 1 3 1 0 1 11 3 2 4 1 1 6 9 3 . 27 0 1 3 2 3 10

x1 0,

Coefficient of: Basic Variable

Eq.

Z

Z

(0)

x2 x6 x3

(1) (2) (3)

5.2-2.* Work through the revised simplex method step by step to solve the following problem.

I

Z 5x1 8x2 7x3 4x4 6x5,

subject to 2x1 3x2 3x3 2x4 2x5 20 3x1 5x2 4x3 2x4 4x5 30 and xj 0,

j 1, 2, 3, 4, 5.

5.2-3. Work through the revised simplex method step by step to solve the model given in Prob. 4.3-4.

I

5.2-4. Reconsider Prob. 5.1-1. For the sequence of CPF solutions identified in part (e), construct the basis matrix B for each of the corresponding BF solutions. For each one, invert B manually, use this B1 to calculate the current solution, and then perform the next iteration (or demonstrate that the current solution is optimal). 5.2-5. Work through the revised simplex method step by step to solve the model given in Prob. 4.1-5.

I

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

(a) Use the given information to identify the optimal solution. (b) Use the given information to identify the shadow prices for the three resources.

Maximize

x2 0,

x1

x2

x3

x4

x5

x6

1

1

1

0

0 0 0

1 0 1

3 1 2

0 1 0

Right Side

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau. D

5.3-2. Consider the following problem. Z 4x1 3x2 x3 2x4,

Maximize subject to

4x1 2x2 x3 x4 5 3x1 x2 2x3 x4 4 and x1 0,

x2 0,

x3 0,

x4 0.

Let x5 and x6 denote the slack variables for the respective constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

5.2-6. Work through the revised simplex method step by step to solve the model given in Prob. 4.7-6.

I

5.2-7. Work through the revised simplex method step by step to solve each of the following models: (a) Model given in Prob. 3.1-5. (b) Model given in Prob. 4.7-8.

I

D

5.3-1.* Consider the following problem. Maximize

Z x1 x2 2x3,

subject to 2x1 2x2 3x3 5 x1 x2 x3 3 x1 x2 x3 2

Coefficient of: Basic Variable

Eq.

Z

Z

(0)

x2 x4

(1) (2)

x1

x2

x3

x4

x5

x6

1

1

1

0 0

1 1

1 2

Right Side

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau.

226

D

5

THE THEORY OF THE SIMPLEX METHOD

5.3-3. Consider the following problem. Coefficient of:

Z 6x1 x2 2x3,

Maximize

Basic Variable

Eq.

Z

Z

(0)

x4

(1)

x1

x2

x3

x4

x5

x6

1

0

3 2

1 2

0

1

1

2

1 2 1 2

1 2 1 2

subject to 1 2x1 2x2 x3 2 2 3 4x1 2x2 x3 3 2 1 2x1 2x2 x3 1 2 and x1 0,

x2 0,

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

Coefficient of: Eq.

Z

Z

(0)

x5 x3 x1

(1) (2) (3)

x1

x2

x3

x4

x5

x6

1

2

0

2

0 0 0

1 2 1

1 0 0

2 4 1

Right Side

(2)

0

0

x2

(3)

0

0

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau. D

Basic Variable

x3

Right Side

5.3-5. Consider the following problem. Z 20x1 6x2 8x3,

Maximize subject to

8x1 2x2 3x3 200 4x1 3x2 3x3 100 2x1 3x2 x3 50 2x1 3x2 x3 20 and x1 0,

Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. D

5.3-4. Consider the following problem. Maximize

Z x1 x2 2x3,

subject to x1 x2 3x3 15 2x1 x2 x3 2 x1 x2 x3 4 and x1 0,

x2 0,

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After the simplex method is applied, a portion of the final simplex tableau is as follows:

x2 0,

x3 0.

Let x4, x5, x6, and x7 denote the slack variables for the first through fourth constraints, respectively. Suppose that after some number of iterations of the simplex method, a portion of the current simplex tableau is as follows: Coefficient of: Basic Variable

Eq.

Z

Z

(0)

1

x1

(1)

0

x2

(2)

0

x6

(3)

0

x7

(4)

0

x1

x2

x3

x4

x5

x6

x7

9 4

1 2

0

0

0

0

0

0

1

0

0

1

3 1 16 8 1 1 4 2 3 1 8 4 0 0

Right Side

CHAPTER 5 PROBLEMS

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the current simplex tableau. Show your calculations. (b) Indicate which of these missing numbers would be generated by the revised simplex method in order to perform the next iteration. (c) Identify the defining equations of the CPF solution corresponding to the BF solution in the current simplex tableau. 5.3-6. You are using the simplex method to solve the following linear programming problem. D

Z 6x1 5x2 x3 4x4,

Maximize

227

Now suppose that your boss has inserted her best estimate of the values of c1, c2, c3, and b without informing you and then has run the simplex method. You are given the resulting final simplex tableau below (where x4 and x5 are the slack variables for the respective functional constraints), but you are unable to read the value of Z*. Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Z

(0)

1

7 10

0

0

3 5

4 5

x2

(1)

0

1

0

x3

(2)

0

0

1

3 5 1 5

1 5 2 5

subject to 3x1 2x2 3x3 x4 120 3x1 3x2 x3 3x4 180 and x1 0,

x2 0,

x3 0,

x4 0.

You have obtained the following final simplex tableau where x5 and x6 are the slack variables for the respective constraints. Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Z

(0)

1

0

1 4

0

1 2

3 4

5 4

x1

(1)

0

1

x3

(2)

0

0

5 6 1 2

1 12 1 4

1 4 1 4

11 12 1 4

0 1

x5

x6

Right Side Z*

b*1 b*2

Use the fundamental insight presented in Sec. 5.3 to identify Z*, b*1, and b*2. Show your calculations. D

5.3-7. Consider the following problem. Maximize Z c1x1 c2x2 c3x3,

subject to x1 2x2 x3 b 2x1 x2 3x3 2b and x1 0,

x2 0,

x3 0.

Note that values have not been assigned to the coefficients in the objective function (c1, c2, c3), and that the only specification for the right-hand side of the functional constraints is that the second one (2b) be twice as large as the first (b).

1 5 3 5

Right Side Z*

1 3

(a) Use the fundamental insight presented in Sec. 5.3 to identify the value of (c1, c2, c3) that was used. (b) Use the fundamental insight presented in Sec. 5.3 to identify the value of b that was used. (c) Calculate the value of Z* in two ways, where one way uses your results from part (a) and the other way uses your result from part (b). Show your two methods for finding Z*. 5.3-8. For iteration 2 of the example in Sec. 5.3, the following expression was shown: Final row 0 [3,

5 0, [0,

0,

0 0]

3 2

1 1] 0 3

,

0 2 2

1 0 0

0 1 0

0 0 1

4 12 . 18

Derive this expression by combining the algebraic operations (in matrix form) for iterations 1 and 2 that affect row 0. 5.3-9. Most of the description of the fundamental insight presented in Sec. 5.3 assumes that the problem is in our standard form. Now consider each of the following other forms, where the additional adjustments in the initialization step are those presented in Sec. 4.6, including the use of artificial variables and the Big M method where appropriate. Describe the resulting adjustments in the fundamental insight. (a) Equality constraints (b) Functional constraints in form (c) Negative right-hand sides (d) Variables allowed to be negative (with no lower bound) 5.3-10. Reconsider the model in Prob. 4.6-6. Use artificial variables and the Big M method to construct the complete first sim-

228

5

THE THEORY OF THE SIMPLEX METHOD

plex tableau for the simplex method, and then identify the columns that will contains S* for applying the fundamental insight in the final tableau. Explain why these are the appropriate columns. 5.3-11. Consider the following problem. Z 2x1 3x2 2x3,

Minimize subject to

x1 4x2 2x3 8 3x1 2x2 2x3 6 and

(c) When you apply the t* t vT equation, another option is to use t [2, 3, 2, 0, M, 0, M, 0], which is the preliminary row 0 before the algebraic elimination of the nonzero coefficients of the initial basic variables x5 and x7. Repeat part (b) for this equation with this new t. After you derive the new v, show that this equation yields the same final row 0 for this problem as the equation derived in part (b). (d) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau. 5.3-12. Consider the following problem.

x1 0,

x2 0,

x3 0.

Let x4 and x6 be the surplus variables for the first and second constraints, respectively. Let x5 and x7 be the corresponding artificial variables. After you make the adjustments described in Sec. 4.6 for this model form when using the Big M method, the initial simplex tableau ready to apply the simplex method is as follows:

Z

x1

x2

x3

x4 x 5 x6 x 7

Z

(0) 1 4M 2 6M 3 2M 2 M

x5 x7

(1) 0 (2) 0

1 3

4 2

2 0

subject to x1 3x2 2x3 20 x1 5x2 2x3 10 and x1 0,

Coefficient of: Basic Variable Eq.

0

M

Right Side

0 14M

1 1 0 0 0 0 1 1

8 6

After you apply the simplex method, a portion of the final simplex tableau is as follows:

Z

Z

(0) 1

x2 x1

(1) 0 (2) 0

x1 x2 x3 x4

x 5 M 0.5

0.3 0.2

x6

x 7

Eq.

Z

Z

(0)

x1 x5

(1) (2)

M 0.5

(a) Based on the above tableaux, use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Examine the mathematical logic presented in Sec. 5.3 to validate the fundamental insight (see the T* MT and t* t vT equations and the subsequent derivations of M and v). This logic assumes that the original model fits our standard form, whereas the current problem does not fit this form. Show how, with minor adjustments, this same logic applies to the current problem when t is row 0 and T is rows 1 and 2 in the initial simplex tableau given above. Derive M and v for this problem.

x3 0.

Coefficient of: Basic Variable

Right Side

0.1 0.4

x2 0,

Let x4 be the artificial variable for the first constraint. Let x5 and x6 be the surplus variable and artificial variable, respectively, for the second constraint. You are now given the information that a portion of the final simplex tableau is as follows:

Coefficient of: Basic Variable Eq.

Z 2x1 4x2 3x3,

Maximize

x1

x2

x3

x 4

x5

x 6

1

M2

0

M

0 0

1 1

0 1

0 1

Right Side

(a) Extend the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal solution in the final simplex tableau. 5.3-13. Consider the following problem. Maximize

Z 3x1 7x2 2x3,

subject to 2x1 2x2 x3 10 3x1 x2 x3 20 and x1 0,

x2 0,

x3 0.

CHAPTER 5 PROBLEMS

You are given the fact that the basic variables in the optimal solution are x1 and x3. (a) Introduce slack variables, and then use the given information to find the optimal solution directly by Gaussian elimination. (b) Extend the work in part (a) to find the shadow prices. (c) Use the given information to identify the defining equations of the optimal CPF solution, and then solve these equations to obtain the optimal solution.

229

(d) Construct the basis matrix B for the optimal BF solution, invert B manually, and then use this B1 to solve for the optimal solution and the shadow prices y*. Then apply the optimality test for the revised simplex method to verify that this solution is optimal. (e) Given B1 and y* from part (d ), use the fundamental insight presented in Sec. 5.3 to construct the complete final simplex tableau.

6 Duality Theory and Sensitivity Analysis

One of the most important discoveries in the early development of linear programming was the concept of duality and its many important ramifications. This discovery revealed that every linear programming problem has associated with it another linear programming problem called the dual. The relationships between the dual problem and the original problem (called the primal) prove to be extremely useful in a variety of ways. For example, you soon will see that the shadow prices described in Sec. 4.7 actually are provided by the optimal solution for the dual problem. We shall describe many other valuable applications of duality theory in this chapter as well. One of the key uses of duality theory lies in the interpretation and implementation of sensitivity analysis. As we already mentioned in Secs. 2.3, 3.3, and 4.7, sensitivity analysis is a very important part of almost every linear programming study. Because most of the parameter values used in the original model are just estimates of future conditions, the effect on the optimal solution if other conditions prevail instead needs to be investigated. Furthermore, certain parameter values (such as resource amounts) may represent managerial decisions, in which case the choice of the parameter values may be the main issue to be studied, which can be done through sensitivity analysis. For greater clarity, the first three sections discuss duality theory under the assumption that the primal linear programming problem is in our standard form (but with no restriction that the bi values need to be positive). Other forms are then discussed in Sec. 6.4. We begin the chapter by introducing the essence of duality theory and its applications. We then describe the economic interpretation of the dual problem (Sec. 6.2) and delve deeper into the relationships between the primal and dual problems (Sec. 6.3). Section 6.5 focuses on the role of duality theory in sensitivity analysis. The basic procedure for sensitivity analysis (which is based on the fundamental insight of Sec. 5.3) is summarized in Sec. 6.6 and illustrated in Sec. 6.7. 230

6.1 THE ESSENCE OF DUALITY THEORY

6.1

231

THE ESSENCE OF DUALITY THEORY Given our standard form for the primal problem at the left (perhaps after conversion from another form), its dual problem has the form shown to the right. Primal Problem

Dual Problem m

n

Maximize

Z c j x j,

Minimize

W bi yi, i1

j1

subject to

subject to m

n

aij x j bi, j1

for i 1, 2, . . . , m

aij yi cj, i1

for j 1, 2, . . . , n

and

and xj 0,

yi 0,

for j 1, 2, . . . , n.

for i 1, 2, . . . , m.

Thus, the dual problem uses exactly the same parameters as the primal problem, but in different locations. To highlight the comparison, now look at these same two problems in matrix notation (as introduced at the beginning of Sec. 5.2), where c and y [y1, y2, . . . , ym] are row vectors but b and x are column vectors. Primal Problem Maximize

Z cx,

subject to

Dual Problem Minimize

W yb,

subject to yA c

Ax b and

and x 0.

y 0.

To illustrate, the primal and dual problems for the Wyndor Glass Co. example of Sec. 3.1 are shown in Table 6.1 in both algebraic and matrix form. The primal-dual table for linear programming (Table 6.2) also helps to highlight the correspondence between the two problems. It shows all the linear programming parameters (the aij, bi, and cj) and how they are used to construct the two problems. All the headings for the primal problem are horizontal, whereas the headings for the dual problem are read by turning the book sideways. For the primal problem, each column (except the Right Side column) gives the coefficients of a single variable in the respective constraints and then in the objective function, whereas each row (except the bottom one) gives the parameters for a single contraint. For the dual problem, each row (except the Right Side row) gives the coefficients of a single variable in the respective constraints and then in the objective function, whereas each column (except the rightmost one) gives the parameters for a single constraint. In addition, the Right Side column gives the right-hand sides for the primal problem and the objective function coefficients for the dual problem, whereas the bottom row gives the objective function coefficients for the primal problem and the righthand sides for the dual problem.

232

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.1 Primal and dual problems for the Wyndor Glass Co. example Primal Problem in Algebraic Form

Dual Problem in Algebraic Form

Z 3x1 5x2,

Maximize subject to

W 4y1 12y2 18y3,

Minimize subject to

3x1 2x2 4

y12y2 3y3 3

3x1 2x2 12

2y2 2y3 5

3x1 2x2 18 and

and

x2 0.

x1 0,

y1 0,

Primal Problem in Matrix Form Z [3, 5]

Maximize

x1

x , 2

0 x1 2 x2 2

Minimize

4 12 18

0

x 0. 2

4 W [y1,y2,y3] 12 18

subject to [y1, y2, y3]

and x1

y3 0.

Dual Problem in Matrix Form

subject to

1 0 3

y2 0,

1 0 3

0 2 [3, 5] 2

and [y1, y2, y3] [0, 0, 0].

Consequently, (1) the parameters for a constraint in either problem are the coefficients of a variable in the other problem and (2) the coefficients for the objective function of either problem are the right sides for the other problem. Thus, there is a direct correspondence between these entities in the two problems, as summarized in Table 6.3. These correspondences are a key to some of the applications of duality theory, including sensitivity analysis. Origin of the Dual Problem Duality theory is based directly on the fundamental insight (particularly with regard to row 0) presented in Sec. 5.3. To see why, we continue to use the notation introduced in Table 5.10 for row 0 of the final tableau, except for replacing Z* by W* and dropping the asterisks from z* and y* when referring to any tableau. Thus, at any given iteration of the simplex method for the primal problem, the current numbers in row 0 are denoted as shown in the (partial) tableau given in Table 6.4. For the coefficients of x1, x2, . . . , xn, recall that z (z1, z2, . . . , zn) denotes the vector that the simplex method added to the vector of initial coefficients, c, in the process of reaching the current tableau. (Do not confuse z with the value of the objective function Z.) Similarly, since the initial coefficients of xn1, xn2, . . . , xnm in row 0 all are 0, y (y1, y2, . . . , ym) denotes the vector that the simplex method has added to these coefficients. Also recall [see Eq. (1) in the

6.1 THE ESSENCE OF DUALITY THEORY

233

TABLE 6.2 Primal-dual table for linear programming, illustrated by the Wyndor Glass Co. example (a) General Case Primal Problem Coefficient of:

…

xn

a11 a21

ym

a12 a22

VI c1

VI c2

… …

Right Side b1 b2

bm

Coefficients for Objective Function (Minimize)

x2

… a1n … a2n ……………………………… … am1 am2 amn

y1 y2

Right Side

Dual Problem

Coefficient of:

x1

VI cn

Coefficients for Objective Function (Maximize) (b) Wyndor Glass Co. Example

y1 y2 y3

x1

x2

1 0 3

0 2 2

VI 3

VI 5

4 12 18

“Mathematical Summary” subsection of Sec. 5.3] that the fundamental insight led to the following relationships between these quantities and the parameters of the original model: m

W yb bi yi , i1

m

z yA,

so

zj aij yi ,

for j 1, 2, . . . , n.

i1

To illustrate these relationships with the Wyndor example, the first equation gives W 4y1 12y2 18y3, which is just the objective function for the dual problem shown TABLE 6.3 Correspondence between entities in primal and dual problems One Problem

Other Problem

Constraint i ←→ Variable i Objective function ←→ Right sides

234

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.4 Notation for entries in row 0 of a simplex tableau Coefficient of: Iteration

Basic Variable

Eq.

Z

x1

x2

…

xn

xn1

xn2

…

xnm

Right Side

Any

Z

(0)

1

z1 c1

z2 c2

…

zn cn

y1

y2

…

ym

W

in the upper right-hand box of Table 6.1. The second set of equations give z1 y1 3y3 and z2 2y2 2y3, which are the left-hand sides of the functional constraints for this dual problem. Thus, by subtracting the right-hand sides of these constraints (c1 3 and c2 5), (z1 c1) and (z2 c2) can be interpreted as being the surplus variables for these functional constraints. The remaining key is to express what the simplex method tries to accomplish (according to the optimality test) in terms of these symbols. Specifically, it seeks a set of basic variables, and the corresponding BF solution, such that all coefficients in row 0 are nonnegative. It then stops with this optimal solution. Using the notation in Table 6.4, this goal is expressed symbolically as follows: Condition for Optimality: for j 1, 2, . . . , n, zj cj 0 yi 0 for i 1, 2, . . . , m. After we substitute the preceding expression for zj, the condition for optimality says that the simplex method can be interpreted as seeking values for y1, y2, . . . , ym such that m

W biyi, i1

subject to m

aijyi cj, i1

for j 1, 2, . . . , n

and yi 0,

for i 1, 2, . . . , m.

But, except for lacking an objective for W, this problem is precisely the dual problem! To complete the formulation, let us now explore what the missing objective should be. Since W is just the current value of Z, and since the objective for the primal problem is to maximize Z, a natural first reaction is that W should be maximized also. However, this is not correct for the following rather subtle reason: The only feasible solutions for this new problem are those that satisfy the condition for optimality for the primal problem. Therefore, it is only the optimal solution for the primal problem that corresponds to a feasible solution for this new problem. As a consequence, the optimal value of Z in the primal problem is the minimum feasible value of W in the new problem, so W should be minimized. (The full justification for this conclusion is provided by the relationships we develop in Sec. 6.3.) Adding this objective of minimizing W gives the complete dual problem.

6.1 THE ESSENCE OF DUALITY THEORY

235

Consequently, the dual problem may be viewed as a restatement in linear programming terms of the goal of the simplex method, namely, to reach a solution for the primal problem that satisfies the optimality test. Before this goal has been reached, the corresponding y in row 0 (coefficients of slack variables) of the current tableau must be infeasible for the dual problem. However, after the goal is reached, the corresponding y must be an optimal solution (labeled y*) for the dual problem, because it is a feasible solution that attains the minimum feasible value of W. This optimal solution (y1*, y2*, . . . , *) provides for the primal problem the shadow prices that were described in Sec. 4.7. ym Furthermore, this optimal W is just the optimal value of Z, so the optimal objective function values are equal for the two problems. This fact also implies that cx yb for any x and y that are feasible for the primal and dual problems, respectively. To illustrate, the left-hand side of Table 6.5 shows row 0 for the respective iterations when the simplex method is applied to the Wyndor Glass Co. example. In each case, row 0 is partitioned into three parts: the coefficients of the decision variables (x1, x2), the coefficients of the slack variables (x3, x4, x5), and the right-hand side (value of Z). Since the coefficients of the slack variables give the corresponding values of the dual variables (y1, y2, y3), each row 0 identifies a corresponding solution for the dual problem, as shown in the y1, y2, and y3 columns of Table 6.5. To interpret the next two columns, recall that (z1 c1) and (z2 c2) are the surplus variables for the functional constraints in the dual problem, so the full dual problem after augmenting with these surplus variables is W 4y1 12y2 18y3,

Minimize subject to y1

3y3 (z1 c1) 3 2y2 2y3 (z2 c2) 5

and y1 0,

y2 0,

y3 0.

Therefore, by using the numbers in the y1, y2, and y3 columns, the values of these surplus variables can be calculated as z1 c1 y1 3y3 3, z2 c2 2y2 2y3 5.

TABLE 6.5 Row 0 and corresponding dual solution for each iteration for the Wyndor Glass Co. example Primal Problem Iteration

Dual Problem

Row 0

0

[3,

5

0,

1

[3,

0

0,

2

[0,

0

0,

0, 5 , 2 3 , 2

y1

y2

y3

z1 c1

z2 c2

0 5 2 3 2

0

3

5

0

0

3

0

30

1

0

0

36

0

0]

0

0

30]

0

1

36]

0

W

236

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Thus, a negative value for either surplus variable indicates that the corresponding constraint is violated. Also included in the rightmost column of the table is the calculated value of the dual objective function W 4y1 12y2 18y3. As displayed in Table 6.4, all these quantities to the right of row 0 in Table 6.5 already are identified by row 0 without requiring any new calculations. In particular, note in Table 6.5 how each number obtained for the dual problem already appears in row 0 in the spot indicated by Table 6.4. For the initial row 0, Table 6.5 shows that the corresponding dual solution (y1, y2, y3) (0, 0, 0) is infeasible because both surplus variables are negative. The first iteration succeeds in eliminating one of these negative values, but not the other. After two iterations, the optimality test is satisfied for the primal problem because all the dual variables and surplus variables are nonnegative. This dual solution (y1*, y2*, y3*) (0, 32, 1) is optimal (as could be verified by applying the simplex method directly to the dual problem), so the optimal value of Z and W is Z* 36 W*. Summary of Primal-Dual Relationships Now let us summarize the newly discovered key relationships between the primal and dual problems. Weak duality property: If x is a feasible solution for the primal problem and y is a feasible solution for the dual problem, then cx yb. For example, for the Wyndor Glass Co. problem, one feasible solution is x1 3, x2 3, which yields Z cx 24, and one feasible solution for the dual problem is y1 1, y2 1, y3 2, which yields a larger objective function value W yb 52. These are just sample feasible solutions for the two problems. For any such pair of feasible solutions, this inequality must hold because the maximum feasible value of Z cx (36) equals the minimum feasible value of the dual objective function W yb, which is our next property. Strong duality property: If x* is an optimal solution for the primal problem and y* is an optimal solution for the dual problem, then cx* y*b. Thus, these two properties imply that cx yb for feasible solutions if one or both of them are not optimal for their respective problems, whereas equality holds when both are optimal. The weak duality property describes the relationship between any pair of solutions for the primal and dual problems where both solutions are feasible for their respective problems. At each iteration, the simplex method finds a specific pair of solutions for the two problems, where the primal solution is feasible but the dual solution is not feasible (except at the final iteration). Our next property describes this situation and the relationship between this pair of solutions. Complementary solutions property: At each iteration, the simplex method simultaneously identifies a CPF solution x for the primal problem and a complementary solution y for the dual problem (found in row 0, the coefficients of the slack variables), where cx yb.

6.1 THE ESSENCE OF DUALITY THEORY

237

If x is not optimal for the primal problem, then y is not feasible for the dual problem. To illustrate, after one iteration for the Wyndor Glass Co. problem, x1 0, x2 6, and y1 0, y2 52, y3 0, with cx 30 yb. This x is feasible for the primal problem, but this y is not feasible for the dual problem (since it violates the constraint, y1 3y3 3). The complementary solutions property also holds at the final iteration of the simplex method, where an optimal solution is found for the primal problem. However, more can be said about the complementary solution y in this case, as presented in the next property. Complementary optimal solutions property: At the final iteration, the simplex method simultaneously identifies an optimal solution x* for the primal problem and a complementary optimal solution y* for the dual problem (found in row 0, the coefficients of the slack variables), where cx* y*b. The y*i are the shadow prices for the primal problem. For the example, the final iteration yields x1* 2, x2* 6, and y1* 0, y2* 32, y3* 1, with cx* 36 y*b. We shall take a closer look at some of these properties in Sec. 6.3. There you will see that the complementary solutions property can be extended considerably further. In particular, after slack and surplus variables are introduced to augment the respective problems, every basic solution in the primal problem has a complementary basic solution in the dual problem. We already have noted that the simplex method identifies the values of the surplus variables for the dual problem as zj cj in Table 6.4. This result then leads to an additional complementary slackness property that relates the basic variables in one problem to the nonbasic variables in the other (Tables 6.7 and 6.8), but more about that later. In Sec. 6.4, after describing how to construct the dual problem when the primal problem is not in our standard form, we discuss another very useful property, which is summarized as follows: Symmetry property: For any primal problem and its dual problem, all relationships between them must be symmetric because the dual of this dual problem is this primal problem. Therefore, all the preceding properties hold regardless of which of the two problems is labeled as the primal problem. (The direction of the inequality for the weak duality property does require that the primal problem be expressed or reexpressed in maximization form and the dual problem in minimization form.) Consequently, the simplex method can be applied to either problem, and it simultaneously will identify complementary solutions (ultimately a complementary optimal solution) for the other problem. So far, we have focused on the relationships between feasible or optimal solutions in the primal problem and corresponding solutions in the dual problem. However, it is possible that the primal (or dual) problem either has no feasible solutions or has feasible solutions but no optimal solution (because the objective function is unbounded). Our final property summarizes the primal-dual relationships under all these possibilities.

238

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Duality theorem: The following are the only possible relationships between the primal and dual problems. 1. If one problem has feasible solutions and a bounded objective function (and so has an optimal solution), then so does the other problem, so both the weak and strong duality properties are applicable. 2. If one problem has feasible solutions and an unbounded objective function (and so no optimal solution), then the other problem has no feasible solutions. 3. If one problem has no feasible solutions, then the other problem has either no feasible solutions or an unbounded objective function. Applications As we have just implied, one important application of duality theory is that the dual problem can be solved directly by the simplex method in order to identify an optimal solution for the primal problem. We discussed in Sec. 4.8 that the number of functional constraints affects the computational effort of the simplex method far more than the number of variables does. If m n, so that the dual problem has fewer functional constraints (n) than the primal problem (m), then applying the simplex method directly to the dual problem instead of the primal problem probably will achieve a substantial reduction in computational effort. The weak and strong duality properties describe key relationships between the primal and dual problems. One useful application is for evaluating a proposed solution for the primal problem. For example, suppose that x is a feasible solution that has been proposed for implementation and that a feasible solution y has been found by inspection for the dual problem such that cx yb. In this case, x must be optimal without the simplex method even being applied! Even if cx yb, then yb still provides an upper bound on the optimal value of Z, so if yb cx is small, intangible factors favoring x may lead to its selection without further ado. One of the key applications of the complementary solutions property is its use in the dual simplex method presented in Sec. 7.1. This algorithm operates on the primal problem exactly as if the simplex method were being applied simultaneously to the dual problem, which can be done because of this property. Because the roles of row 0 and the right side in the simplex tableau have been reversed, the dual simplex method requires that row 0 begin and remain nonnegative while the right side begins with some negative values (subsequent iterations strive to reach a nonnegative right side). Consequently, this algorithm occasionally is used because it is more convenient to set up the initial tableau in this form than in the form required by the simplex method. Furthermore, it frequently is used for reoptimization (discussed in Sec. 4.7), because changes in the original model lead to the revised final tableau fitting this form. This situation is common for certain types of sensitivity analysis, as you will see later in the chapter. In general terms, duality theory plays a central role in sensitivity analysis. This role is the topic of Sec. 6.5. Another important application is its use in the economic interpretation of the dual problem and the resulting insights for analyzing the primal problem. You already have seen one example when we discussed shadow prices in Sec. 4.7. The next section describes how this interpretation extends to the entire dual problem and then to the simplex method.

6.2 ECONOMIC INTERPRETATION OF DUALITY

6.2

239

ECONOMIC INTERPRETATION OF DUALITY The economic interpretation of duality is based directly upon the typical interpretation for the primal problem (linear programming problem in our standard form) presented in Sec. 3.2. To refresh your memory, we have summarized this interpretation of the primal problem in Table 6.6. Interpretation of the Dual Problem To see how this interpretation of the primal problem leads to an economic interpretation for the dual problem,1 note in Table 6.4 that W is the value of Z (total profit) at the current iteration. Because W b1y1 b2 y2 … bm ym , each bi yi can thereby be interpreted as the current contribution to profit by having bi units of resource i available for the primal problem. Thus, The dual variable yi is interpreted as the contribution to profit per unit of resource i (i 1, 2, . . . , m), when the current set of basic variables is used to obtain the primal solution.

In other words, the yi values (or y*i values in the optimal solution) are just the shadow prices discussed in Sec. 4.7. For example, when iteration 2 of the simplex method finds the optimal solution for the Wyndor problem, it also finds the optimal values of the dual variables (as shown in the bottom row of Table 6.5) to be y1* 0, y2* 32, and y3* 1. These are precisely the shadow prices found in Sec. 4.7 for this problem through graphical analysis. Recall that the resources for the Wyndor problem are the production capacities of the three plants being made available to the two new products under consideration, so that bi is the number of hours of production time per week being made available in Plant i for these new products, where i 1, 2, 3. As discussed in Sec. 4.7, the shadow prices indicate that individually increasing any bi by 1 would increase the optimal value of the objective function (total weekly profit in units of thousands of dollars) by y*i . Thus, y*i can be interpreted as the contribution to profit per unit of resource i when using the optimal solution.

1

Actually, several slightly different interpretations have been proposed. The one presented here seems to us to be the most useful because it also directly interprets what the simplex method does in the primal problem.

TABLE 6.6 Economic interpretation of the primal problem Quantity xj cj Z bi aij

Interpretation Level of activity j (j 1, 2, . . . , n) Unit profit from activity j Total profit from all activities Amount of resource i available (i 1, 2, . . . , m) Amount of resource i consumed by each unit of activity j

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DUALITY THEORY AND SENSITIVITY ANALYSIS

This interpretation of the dual variables leads to our interpretation of the overall dual problem. Specifically, since each unit of activity j in the primal problem consumes aij units of resource i,

m i1 ai j yi is interpreted as the current contribution to profit of the mix of resources that would be consumed if 1 unit of activity j were used ( j 1, 2, . . . , n).

For the Wyndor problem, 1 unit of activity j corresponds to producing 1 batch of product j per week, where j 1, 2. The mix of resources consumed by producing 1 batch of product 1 is 1 hour of production time in Plant 1 and 3 hours in Plant 3. The corresponding mix per batch of product 2 is 2 hours each in Plants 2 and 3. Thus, y1 3y3 and 2y2 2y3 are interpreted as the current contributions to profit (in thousands of dollars per week) of these respective mixes of resources per batch produced per week of the respective products. For each activity j, this same mix of resources (and more) probably can be used in other ways as well, but no alternative use should be considered if it is less profitable than 1 unit of activity j. Since cj is interpreted as the unit profit from activity j, each functional constraint in the dual problem is interpreted as follows:

m i1 aij yi cj says that the actual contribution to profit of the above mix of resources must be at least as much as if they were used by 1 unit of activity j; otherwise, we would not be making the best possible use of these resources.

For the Wyndor problem, the unit profits (in thousands of dollars per week) are c1 3 and c2 5, so the dual functional constraints with this interpretation are y1 y3 3 and 2y2 2y3 5. Similarly, the interpretation of the nonnegativity constraints is the following: yi 0 says that the contribution to profit of resource i (i 1, 2, . . . , m) must be nonnegative: otherwise, it would be better not to use this resource at all.

The objective m

Minimize

W bi yi i1

can be viewed as minimizing the total implicit value of the resources consumed by the activities. For the Wyndor problem, the total implicit value (in thousands of dollars per week) of the resources consumed by the two products is W 4y1 12y2 18y3. This interpretation can be sharpened somewhat by differentiating between basic and nonbasic variables in the primal problem for any given BF solution (x1, x2, . . . , xnm). Recall that the basic variables (the only variables whose values can be nonzero) always have a coefficient of zero in row 0. Therefore, referring again to Table 6.4 and the accompanying equation for zj, we see that m

aij yi cj, i1 yi 0,

if xj 0

( j 1, 2, . . . , n),

if xni 0

(i 1, 2, . . . , m).

(This is one version of the complementary slackness property discussed in the next section.) The economic interpretation of the first statement is that whenever an activity j op-

6.2 ECONOMIC INTERPRETATION OF DUALITY

241

erates at a strictly positive level (xj 0), the marginal value of the resources it consumes must equal (as opposed to exceeding) the unit profit from this activity. The second statement implies that the marginal value of resource i is zero (yi 0) whenever the supply of this resource is not exhausted by the activities (xni 0). In economic terminology, such a resource is a “free good”; the price of goods that are oversupplied must drop to zero by the law of supply and demand. This fact is what justifies interpreting the objective for the dual problem as minimizing the total implicit value of the resources consumed, rather than the resources allocated. To illustrate these two statements, consider the optimal BF solution (2, 6, 2, 0, 0) for the Wyndor problem. The basic variables are x1, x2, and x3, so their coefficients in row 0 are zero, as shown in the bottom row of Table 6.5. This bottom row also gives the corresponding dual solution: y1* 0, y2* 32, y3* 1, with surplus variables (z1* c1) 0 and (z2* c2) 0. Since x1 0 and x2 0, both these surplus variables and direct calculations indicate that y1* 3y3* c1 3 and 2y2* 2y3* c2 5. Therefore, the value of the resources consumed per batch of the respective products produced does indeed equal the respective unit profits. The slack variable for the constraint on the amount of Plant 1 capacity used is x3 0, so the marginal value of adding any Plant 1 capacity would be zero (y1* 0). Interpretation of the Simplex Method The interpretation of the dual problem also provides an economic interpretation of what the simplex method does in the primal problem. The goal of the simplex method is to find how to use the available resources in the most profitable feasible way. To attain this goal, we must reach a BF solution that satisfies all the requirements on profitable use of the resources (the constraints of the dual problem). These requirements comprise the condition for optimality for the algorithm. For any given BF solution, the requirements (dual constraints) associated with the basic variables are automatically satisfied (with equality). However, those associated with nonbasic variables may or may not be satisfied. In particular, if an original variable xj is nonbasic so that activity j is not used, then the current contribution to profit of the resources that would be required to undertake each unit of activity j m

aij yi

i1

may be smaller than, larger than, or equal to the unit profit cj obtainable from the activity. If it is smaller, so that zj cj 0 in row 0 of the simplex tableau, then these resources can be used more profitably by initiating this activity. If it is larger (zj cj 0), then these resources already are being assigned elsewhere in a more profitable way, so they should not be diverted to activity j. If zj cj 0, there would be no change in profitability by initiating activity j. Similarly, if a slack variable xni is nonbasic so that the total allocation bi of resource i is being used, then yi is the current contribution to profit of this resource on a marginal basis. Hence, if yi 0, profit can be increased by cutting back on the use of this resource (i.e., increasing xni). If yi 0, it is worthwhile to continue fully using this resource, whereas this decision does not affect profitability if yi 0.

242

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Therefore, what the simplex method does is to examine all the nonbasic variables in the current BF solution to see which ones can provide a more profitable use of the resources by being increased. If none can, so that no feasible shifts or reductions in the current proposed use of the resources can increase profit, then the current solution must be optimal. If one or more can, the simplex method selects the variable that, if increased by 1, would improve the profitability of the use of the resources the most. It then actually increases this variable (the entering basic variable) as much as it can until the marginal values of the resources change. This increase results in a new BF solution with a new row 0 (dual solution), and the whole process is repeated. The economic interpretation of the dual problem considerably expands our ability to analyze the primal problem. However, you already have seen in Sec. 6.1 that this interpretation is just one ramification of the relationships between the two problems. In the next section, we delve into these relationships more deeply.

6.3

PRIMAL-DUAL RELATIONSHIPS Because the dual problem is a linear programming problem, it also has corner-point solutions. Furthermore, by using the augmented form of the problem, we can express these corner-point solutions as basic solutions. Because the functional constraints have the form, this augmented form is obtained by subtracting the surplus (rather than adding the slack) from the left-hand side of each constraint j ( j 1, 2, . . . , n).1 This surplus is m

zj cj aijyi cj ,

for j 1, 2, . . . , n.

i1

Thus, zjcj plays the role of the surplus variable for constraint j (or its slack variable if the constraint is multiplied through by 1). Therefore, augmenting each corner-point solution (y1, y2, . . . , ym) yields a basic solution (y1, y2, . . . , ym , z1 c1, z2 c2, . . . , zn cn) by using this expression for zj cj. Since the augmented form of the dual problem has n functional constraints and n m variables, each basic solution has n basic variables and m nonbasic variables. (Note how m and n reverse their previous roles here because, as Table 6.3 indicates, dual constraints correspond to primal variables and dual variables correspond to primal constraints.) Complementary Basic Solutions One of the important relationships between the primal and dual problems is a direct correspondence between their basic solutions. The key to this correspondence is row 0 of the simplex tableau for the primal basic solution, such as shown in Table 6.4 or 6.5. Such a row 0 can be obtained for any primal basic solution, feasible or not, by using the formulas given in the bottom part of Table 5.8. Note again in Tables 6.4 and 6.5 how a complete solution for the dual problem (including the surplus variables) can be read directly from row 0. Thus, because of its coefficient in 1

You might wonder why we do not also introduce artificial variables into these constraints as discussed in Sec. 4.6. The reason is that these variables have no purpose other than to change the feasible region temporarily as a convenience in starting the simplex method. We are not interested now in applying the simplex method to the dual problem, and we do not want to change its feasible region.

6.3 PRIMAL-DUAL RELATIONSHIPS

243

TABLE 6.7 Association between variables in primal and dual problems Primal Variable

Associated Dual Variable

Any problem

(Decision variable) xj (Slack variable) xni

zj cj (surplus variable) j 1, 2, . . . , n yi (decision variable) i 1, 2, . . . , m

Wyndor problem

Decision variables: Decision variables: Slack variables: Decision variables: Decision variables:

z1 c1 (surplus variables) z2 c2 y1 (decision variables) y2 y3

x1 x2 x3 x4 x5

row 0, each variable in the primal problem has an associated variable in the dual problem, as summarized in Table 6.7, first for any problem and then for the Wyndor problem. A key insight here is that the dual solution read from row 0 must also be a basic solution! The reason is that the m basic variables for the primal problem are required to have a coefficient of zero in row 0, which thereby requires the m associated dual variables to be zero, i.e., nonbasic variables for the dual problem. The values of the remaining n (basic) variables then will be the simultaneous solution to the system of equations given at the beginning of this section. In matrix form, this system of equations is z c yA c, and the fundamental insight of Sec. 5.3 actually identifies its solution for z c and y as being the corresponding entries in row 0. Because of the symmetry property quoted in Sec. 6.1 (and the direct association between variables shown in Table 6.7), the correspondence between basic solutions in the primal and dual problems is a symmetric one. Furthermore, a pair of complementary basic solutions has the same objective function value, shown as W in Table 6.4. Let us now summarize our conclusions about the correspondence between primal and dual basic solutions, where the first property extends the complementary solutions property of Sec. 6.1 to the augmented forms of the two problems and then to any basic solution (feasible or not) in the primal problem. Complementary basic solutions property: Each basic solution in the primal problem has a complementary basic solution in the dual problem, where their respective objective function values (Z and W) are equal. Given row 0 of the simplex tableau for the primal basic solution, the complementary dual basic solution (y, z c) is found as shown in Table 6.4. The next property shows how to identify the basic and nonbasic variables in this complementary basic solution. Complementary slackness property: Given the association between variables in Table 6.7, the variables in the primal basic solution and the complementary dual basic solution satisfy the complementary slackness relationship shown in Table 6.8. Furthermore, this relationship is a symmetric one, so that these two basic solutions are complementary to each other. The reason for using the name complementary slackness for this latter property is that it says (in part) that for each pair of associated variables, if one of them has slack in its

244

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DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.8 Complementary slackness relationship for complementary basic solutions Primal Variable

Associated Dual Variable

Basic Nonbasic

Nonbasic Basic

(m variables) (n variables)

nonnegativity constraint (a basic variable 0), then the other one must have no slack (a nonbasic variable 0). We mentioned in Sec. 6.2 that this property has a useful economic interpretation for linear programming problems. Example. To illustrate these two properties, again consider the Wyndor Glass Co. problem of Sec. 3.1. All eight of its basic solutions (five feasible and three infeasible) are shown in Table 6.9. Thus, its dual problem (see Table 6.1) also must have eight basic solutions, each complementary to one of these primal solutions, as shown in Table 6.9. The three BF solutions obtained by the simplex method for the primal problem are the first, fifth, and sixth primal solutions shown in Table 6.9. You already saw in Table 6.5 how the complementary basic solutions for the dual problem can be read directly from row 0, starting with the coefficients of the slack variables and then the original variables. The other dual basic solutions also could be identified in this way by constructing row 0 for each of the other primal basic solutions, using the formulas given in the bottom part of Table 5.8. Alternatively, for each primal basic solution, the complementary slackness property can be used to identify the basic and nonbasic variables for the complementary dual basic solution, so that the system of equations given at the beginning of the section can be TABLE 6.9 Complementary basic solutions for the Wyndor Glass Co. example Primal Problem

Dual Problem ZW

No.

Basic Solution

Feasible?

Feasible?

1 2 3

(0, 0, 4, 12, 18) (4, 0, 0, 12, 6) (6, 0, 2, 12, 0)

Yes Yes No

0 12 18

No No No

4

(4, 3, 0, 6, 0)

Yes

27

No

5

(0, 6, 4, 0, 6)

Yes

30

No

6

(2, 6, 2, 0, 0)

Yes

36

Yes

7

(4, 6, 0, 0, 6)

No

42

Yes

8

(0, 9, 4, 6, 0)

No

45

Yes

Basic Solution (0, 0, 0, 3, 5) (3, 0, 0, 0, 5) (0, 0, 1, 0, 3) 9 5 , 0, , 0, 0 2 2 5 0, , 0, 3, 0 2 3 0, , 1, 0, 0 2 5 3, , 0, 0, 0 2 5 9 0, 0, , , 0 2 2

6.3 PRIMAL-DUAL RELATIONSHIPS

245

solved directly to obtain this complementary solution. For example, consider the next-tolast primal basic solution in Table 6.9, (4, 6, 0, 0, 6). Note that x1, x2, and x5 are basic variables, since these variables are not equal to 0. Table 6.7 indicates that the associated dual variables are (z1 c1), (z2 c2), and y3. Table 6.8 specifies that these associated dual variables are nonbasic variables in the complementary basic solution, so z1 c1 0,

z2 c2 0,

y3 0.

Consequently, the augmented form of the functional constraints in the dual problem, y1

3y3 (z1 c1) 3 2y2 2y3 (z2 c2) 5,

reduce to y1

003 2y2 0 0 5,

so that y1 3 and y2 52. Combining these values with the values of 0 for the nonbasic variables gives the basic solution (3, 52, 0, 0, 0), shown in the rightmost column and nextto-last row of Table 6.9. Note that this dual solution is feasible for the dual problem because all five variables satisfy the nonnegativity constraints. Finally, notice that Table 6.9 demonstrates that (0, 32, 1, 0, 0) is the optimal solution for the dual problem, because it is the basic feasible solution with minimal W (36).

Relationships between Complementary Basic Solutions We now turn our attention to the relationships between complementary basic solutions, beginning with their feasibility relationships. The middle columns in Table 6.9 provide some valuable clues. For the pairs of complementary solutions, notice how the yes or no answers on feasibility also satisfy a complementary relationship in most cases. In particular, with one exception, whenever one solution is feasible, the other is not. (It also is possible for neither solution to be feasible, as happened with the third pair.) The one exception is the sixth pair, where the primal solution is known to be optimal. The explanation is suggested by the Z W column. Because the sixth dual solution also is optimal (by the complementary optimal solutions property), with W 36, the first five dual solutions cannot be feasible because W 36 (remember that the dual problem objective is to minimize W). By the same token, the last two primal solutions cannot be feasible because Z 36. This explanation is further supported by the strong duality property that optimal primal and dual solutions have Z W. Next, let us state the extension of the complementary optimal solutions property of Sec. 6.1 for the augmented forms of the two problems. Complementary optimal basic solutions property: Each optimal basic solution in the primal problem has a complementary optimal basic solution in the dual problem, where their respective objective function values (Z and W) are equal. Given row 0 of the simplex tableau for the optimal primal solution, the complementary optimal dual solution (y*, z* c) is found as shown in Table 6.4.

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TABLE 6.10 Classification of basic solutions Satisfies Condition for Optimality?

Feasible?

Yes

No

Yes

Optimal

Suboptimal

No

Superoptimal

Neither feasible nor superoptimal

To review the reasoning behind this property, note that the dual solution (y*, z* c) must be feasible for the dual problem because the condition for optimality for the primal problem requires that all these dual variables (including surplus variables) be nonnegative. Since this solution is feasible, it must be optimal for the dual problem by the weak duality property (since W Z, so y*b cx* where x* is optimal for the primal problem). Basic solutions can be classified according to whether they satisfy each of two conditions. One is the condition for feasibility, namely, whether all the variables (including slack variables) in the augmented solution are nonnegative. The other is the condition for optimality, namely, whether all the coefficients in row 0 (i.e., all the variables in the complementary basic solution) are nonnegative. Our names for the different types of basic solutions are summarized in Table 6.10. For example, in Table 6.9, primal basic solutions 1, 2, 4, and 5 are suboptimal, 6 is optimal, 7 and 8 are superoptimal, and 3 is neither feasible nor superoptimal. Given these definitions, the general relationships between complementary basic solutions are summarized in Table 6.11. The resulting range of possible (common) values for the objective functions (Z W) for the first three pairs given in Table 6.11 (the last pair can have any value) is shown in Fig. 6.1. Thus, while the simplex method is dealing directly with suboptimal basic solutions and working toward optimality in the primal problem, it is simultaneously dealing indirectly with complementary superoptimal solutions and working toward feasibility in the dual problem. Conversely, it sometimes is more convenient (or necessary) to work directly with superoptimal basic solutions and to move toward feasibility in the primal problem, which is the purpose of the dual simplex method described in Sec. 7.1. The third and fourth columns of Table 6.11 introduce two other common terms that are used to describe a pair of complementary basic solutions. The two solutions are said to be primal feasible if the primal basic solution is feasible, whereas they are called dual feasible if the complementary dual basic solution is feasible for the dual problem. Using TABLE 6.11 Relationships between complementary basic solutions Both Basic Solutions Primal Basic Solution

Complementary Dual Basic Solution

Suboptimal Optimal Superoptimal Neither feasible nor superoptimal

Superoptimal Optimal Suboptimal Neither feasible nor superoptimal

Primal Feasible?

Dual Feasible?

Yes Yes No No

No Yes Yes No

6.4 ADAPTING TO OTHER PRIMAL FORMS

Primal problem

Dual problem

n

cj xj Z

j1

Superoptimal

247

m

W

bi yi

i 1

Suboptimal

(optimal) Z*

W* (optimal)

Suboptimal

Superoptimal

FIGURE 6.1 Range of possible values of Z W for certain types of complementary basic solutions.

this terminology, the simplex method deals with primal feasible solutions and strives toward achieving dual feasibility as well. When this is achieved, the two complementary basic solutions are optimal for their respective problems. These relationships prove very useful, particularly in sensitivity analysis, as you will see later in the chapter.

6.4

ADAPTING TO OTHER PRIMAL FORMS Thus far it has been assumed that the model for the primal problem is in our standard form. However, we indicated at the beginning of the chapter that any linear programming problem, whether in our standard form or not, possesses a dual problem. Therefore, this section focuses on how the dual problem changes for other primal forms. Each nonstandard form was discussed in Sec. 4.6, and we pointed out how it is possible to convert each one to an equivalent standard form if so desired. These conversions are summarized in Table 6.12. Hence, you always have the option of converting any model to our standard form and then constructing its dual problem in the usual way. To illustrate, we do this for our standard dual problem (it must have a dual also) in Table 6.13. Note that what we end up with is just our standard primal problem! Since any pair of primal and dual problems can be converted to these forms, this fact implies that the dual of the dual problem always is the primal problem. Therefore, for any primal problem and its dual problem, all relationships between them must be symmetric. This is just the symmetry property already stated in Sec. 6.1 (without proof), but now Table 6.13 demonstrates why it holds.

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.12 Conversions to standard form for linear programming models Nonstandard Form

Equivalent Standard Form

Minimize

Maximize

Z

n

(Z)

n

aij xj bi

aij xj bi

j1

j1

n

n

aij xj bi j1

aij xj bi j1

xj unconstrained in sign

x j

x j ,

n

and

aij xj bi j1

x j

0,

x j 0

One consequence of the symmetry property is that all the statements made earlier in the chapter about the relationships of the dual problem to the primal problem also hold in reverse. Another consequence is that it is immaterial which problem is called the primal and which is called the dual. In practice, you might see a linear programming problem fitting our standard form being referred to as the dual problem. The convention is that the model formulated to fit the actual problem is called the primal problem, regardless of its form. Our illustration of how to construct the dual problem for a nonstandard primal problem did not involve either equality constraints or variables unconstrained in sign. Actually, for these two forms, a shortcut is available. It is possible to show (see Probs. 6.4-7 and 6.4-2a) that an equality constraint in the primal problem should be treated just like a constraint in TABLE 6.13 Constructing the dual of the dual problem Dual Problem Minimize

Converted to Standard Form

W yb,

Maximize

(W) yb,

subject to

subject to yA c

→

yA c and

and

y 0.

Converted to Standard Form

Its Dual Problem

Maximize

→

y 0.

Z cx,

Minimize

Ax b

(Z) cx,

subject to

subject to

→

248

Ax b and

and x 0.

x 0.

6.4 ADAPTING TO OTHER PRIMAL FORMS

249

constructing the dual problem except that the nonnegativity constraint for the corresponding dual variable should be deleted (i.e., this variable is unconstrained in sign). By the symmetry property, deleting a nonnegativity constraint in the primal problem affects the dual problem only by changing the corresponding inequality constraint to an equality constraint. Another shortcut involves functional constraints in form for a maximization problem. The straightforward (but longer) approach would begin by converting each such constraint to form n

n

aij xj bi → j1 aij xj bi. j1 Constructing the dual problem in the usual way then gives aij as the coefficient of yi in functional constraint j (which has form) and a coefficient of bi in the objective function (which is to be minimized), where yi also has a nonnegativity constraint yi 0. Now suppose we define a new variable yi yi. The changes caused by expressing the dual problem in terms of yi instead of yi are that (1) the coefficients of the variable become ai j for functional constraint j and bi for the objective function and (2) the constraint on the variable becomes yi 0 (a nonpositivity constraint). The shortcut is to use yi instead of yi as a dual variable so that the parameters in the original constraint (aij and bi) immediately become the coefficients of this variable in the dual problem. Here is a useful mnemonic device for remembering what the forms of dual constraints should be. With a maximization problem, it might seem sensible for a functional constraint to be in form, slightly odd to be in form, and somewhat bizarre to be in form. Similarly, for a minimization problem, it might seem sensible to be in form, slightly odd to be in form, and somewhat bizarre to be in form. For the constraint on an individual variable in either kind of problem, it might seem sensible to have a nonnegativity constraint, somewhat odd to have no constraint (so the variable is unconstrained in sign), and quite bizarre for the variable to be restricted to be less than or equal to zero. Now recall the correspondence between entities in the primal and dual problems indicated in Table 6.3; namely, functional constraint i in one problem corresponds to variable i in the other problem, and vice versa. The sensible-odd-bizarre method, or SOB method for short, says that the form of a functional constraint or the constraint on a variable in the dual problem should be sensible, odd, or bizarre, depending on whether the form for the corresponding entity in the primal problem is sensible, odd, or bizarre. Here is a summary. The SOB Method for Determining the Form of Constraints in the Dual.1 1. Formulate the primal problem in either maximization form or minimization form, and then the dual problem automatically will be in the other form. 2. Label the different forms of functional constraints and of constraints on individual variables in the primal problem as being sensible, odd, or bizarre according to Table 6.14. 1

This particular mnemonic device (and a related one) for remembering what the forms of dual constraints should be has been suggested by Arthur T. Benjamin, a mathematics professor at Harvey Mudd College. An interesting and wonderfully bizarre fact about Professor Benjamin himself is that he is one of the world’s great human calculators who can perform such feats as quickly multiplying six-digit numbers in his head.

250

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.14 Corresponding primal-dual forms Label

Primal Problem (or Dual Problem)

Dual Problem (or Primal Problem)

Maximize

Minimize

Z (or W)

W (or Z)

Sensible Odd Bizarre

Constraint i: form form form

Variable yi (or xi): yi 0 Unconstrained yi 0

Sensible Odd Bizarre

Variable xj (or yj): Constraint j: xj 0 ←→ form Unconstrained ←→ form xj 0 ←→ form

←→ ←→ ←→

The labeling of the functional constraints depends on whether the problem is a maximization problem (use the second column) or a minimization problem (use the third column). 3. For each constraint on an individual variable in the dual problem, use the form that has the same label as for the functional constraint in the primal problem that corresponds to this dual variable (as indicated by Table 6.3). 4. For each functional constraint in the dual problem, use the form that has the same label as for the constraint on the corresponding individual variable in the primal problem (as indicated by Table 6.3). The arrows between the second and third columns of Table 6.14 spell out the correspondence between the forms of constraints in the primal and dual. Note that the correspondence always is between a functional constraint in one problem and a constraint on an individual variable in the other problem. Since the primal problem can be either a maximization or minimization problem, where the dual then will be of the opposite type, the second column of the table gives the form for whichever is the maximization problem and the third column gives the form for the other problem (a minimization problem). To illustrate, consider the radiation therapy example presented in Sec. 3.4. (Its model is shown on p. 46.) To show the conversion in both directions in Table 6.14, we begin with the maximization form of this model as the primal problem, before using the (original) minimization form. The primal problem in maximization form is shown on the left side of Table 6.15. By using the second column of Table 6.14 to represent this problem, the arrows in this table indicate the form of the dual problem in the third column. These same arrows are used in Table 6.15 to show the resulting dual problem. (Because of these arrows, we have placed the functional constraints last in the dual problem rather than in their usual top position.) Beside each constraint in both problems, we have inserted (in parentheses) an S, O, or B to label the form as sensible, odd, or bizarre. As prescribed by the SOB method, the label for each dual constraint always is the same as for the corresponding primal constraint.

6.4 ADAPTING TO OTHER PRIMAL FORMS

251

TABLE 6.15 One primal-dual form for the radiation therapy example Primal Problem Maximize

Dual Problem

Z 0.4x1 0.5x2,

(S) (O) (B)

Minimize

W 2.7y1 6y2 6y3,

subject to

subject to 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6

←→ ←→ ←→

y1 0 y2 unconstrained in sign y3 0

(S) (O) (B)

and

and x1 0 x2 0

(S) (S)

←→ ←→

0.3y1 0.5y2 0.6y3 0.4 0.1y1 0.5y2 0.4y3 0.5

(S) (S)

However, there was no need (other than for illustrative purposes) to convert the primal problem to maximization form. Using the original minimization form, the equivalent primal problem is shown on the left side of Table 6.16. Now we use the third column of Table 6.14 to represent this primal problem, where the arrows indicate the form of the dual problem in the second column. These same arrows in Table 6.16 show the resulting dual problem on the right side. Again, the labels on the constraints show the application of the SOB method. Just as the primal problems in Tables 6.15 and 6.16 are equivalent, the two dual problems also are completely equivalent. The key to recognizing this equivalency lies in the fact that the variables in each version of the dual problem are the negative of those in the other version (y1 y1, y2 y2, y3 y3). Therefore, for each version, if the variables in the other version are used instead, and if both the objective function and the constraints are multiplied through by 1, then the other version is obtained. (Problem 6.4-5 asks you to verify this.) If the simplex method is to be applied to either a primal or a dual problem that has any variables constrained to be nonpositive (for example, y3 0 in the dual problem of Table 6.15), this variable may be replaced by its nonnegative counterpart (for example, y3 y3). TABLE 6.16 The other primal-dual form for the radiation therapy example Primal Problem Minimize

Z 0.4x1 0.5x2,

0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6

W 2.7y1 6y2 6y3,

←→ ←→ ←→

y1 0 y2 unconstrained in sign y3 0

(B) (O) (S)

and

and (S) (S)

Maximize subject to

subject to (B) (O) (S)

Dual Problem

x1 0 x2 0

←→ ←→

0.3y1 0.5y2 0.6y3 0.4 0.1y1 0.5y2 0.4y3 0.6

(S) (S)

252

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

When artificial variables are used to help the simplex method solve a primal problem, the duality interpretation of row 0 of the simplex tableau is the following: Since artificial variables play the role of slack variables, their coefficients in row 0 now provide the values of the corresponding dual variables in the complementary basic solution for the dual problem. Since artificial variables are used to replace the real problem with a more convenient artificial problem, this dual problem actually is the dual of the artificial problem. However, after all the artificial variables become nonbasic, we are back to the real primal and dual problems. With the two-phase method, the artificial variables would need to be retained in phase 2 in order to read off the complete dual solution from row 0. With the Big M method, since M has been added initially to the coefficient of each artificial variable in row 0, the current value of each corresponding dual variable is the current coefficient of this artificial variable minus M. For example, look at row 0 in the final simplex tableau for the radiation therapy example, given at the bottom of Table 4.12 on p. 142. After M is subtracted from the coefficients of the artificial variables x4 and x6, the optimal solution for the corresponding dual problem given in Table 6.15 is read from the coefficients of x3, x4, and x6 as (y1, y2, y3) (0.5, 1.1, 0). As usual, the surplus variables for the two functional constraints are read from the coefficients of x1 and x2 as z1 c1 0 and z2 c2 0.

6.5

THE ROLE OF DUALITY THEORY IN SENSITIVITY ANALYSIS As described further in the next two sections, sensitivity analysis basically involves investigating the effect on the optimal solution of making changes in the values of the model parameters aij , bi, and cj. However, changing parameter values in the primal problem also changes the corresponding values in the dual problem. Therefore, you have your choice of which problem to use to investigate each change. Because of the primal-dual relationships presented in Secs. 6.1 and 6.3 (especially the complementary basic solutions property), it is easy to move back and forth between the two problems as desired. In some cases, it is more convenient to analyze the dual problem directly in order to determine the complementary effect on the primal problem. We begin by considering two such cases. Changes in the Coefficients of a Nonbasic Variable Suppose that the changes made in the original model occur in the coefficients of a variable that was nonbasic in the original optimal solution. What is the effect of these changes on this solution? Is it still feasible? Is it still optimal? Because the variable involved is nonbasic (value of zero), changing its coefficients cannot affect the feasibility of the solution. Therefore, the open question in this case is whether it is still optimal. As Tables 6.10 and 6.11 indicate, an equivalent question is whether the complementary basic solution for the dual problem is still feasible after these changes are made. Since these changes affect the dual problem by changing only one constraint, this question can be answered simply by checking whether this complementary basic solution still satisfies this revised constraint. We shall illustrate this case in the corresponding subsection of Sec. 6.7 after developing a relevant example.

6.5 THE ROLE OF DUALITY THEORY IN SENSITIVITY ANALYSIS

253

Introduction of a New Variable As indicated in Table 6.6, the decision variables in the model typically represent the levels of the various activities under consideration. In some situations, these activities were selected from a larger group of possible activities, where the remaining activities were not included in the original model because they seemed less attractive. Or perhaps these other activities did not come to light until after the original model was formulated and solved. Either way, the key question is whether any of these previously unconsidered activities are sufficiently worthwhile to warrant initiation. In other words, would adding any of these activities to the model change the original optimal solution? Adding another activity amounts to introducing a new variable, with the appropriate coefficients in the functional constraints and objective function, into the model. The only resulting change in the dual problem is to add a new constraint (see Table 6.3). After these changes are made, would the original optimal solution, along with the new variable equal to zero (nonbasic), still be optimal for the primal problem? As for the preceding case, an equivalent question is whether the complementary basic solution for the dual problem is still feasible. And, as before, this question can be answered simply by checking whether this complementary basic solution satisfies one constraint, which in this case is the new constraint for the dual problem. To illustrate, suppose for the Wyndor Glass Co. problem of Sec. 3.1 that a possible third new product now is being considered for inclusion in the product line. Letting xnew represent the production rate for this product, we show the resulting revised model as follows: Maximize

Z 3x1 5x2 4xnew,

subject to x1 2x2 2xnew 4 3x1 2x2 3xnew 12 3x1 2x2 xnew 18 and x1 0,

x2 0,

xnew 0.

After we introduced slack variables, the original optimal solution for this problem without xnew (given by Table 4.8) was (x1, x2, x3, x4, x5) (2, 6, 2, 0, 0). Is this solution, along with xnew 0, still optimal? To answer this question, we need to check the complementary basic solution for the dual problem. As indicated by the complementary optimal basic solutions property in Sec. 6.3, this solution is given in row 0 of the final simplex tableau for the primal problem, using the locations shown in Table 6.4 and illustrated in Table 6.5. Therefore, as given in both the bottom row of Table 6.5 and the sixth row of Table 6.9, the solution is

3 (y1, y2, y3, z1 c1, z2 c2) 0, , 1, 0, 0 . 2 (Alternatively, this complementary basic solution can be derived in the way that was illustrated in Sec. 6.3 for the complementary basic solution in the next-to-last row of Table 6.9.)

254

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Since this solution was optimal for the original dual problem, it certainly satisfies the original dual constraints shown in Table 6.1. But does it satisfy this new dual constraint? 2y1 3y2 y3 4 Plugging in this solution, we see that

3 2(0) 3 (1) 4 2 is satisfied, so this dual solution is still feasible (and thus still optimal). Consequently, the original primal solution (2, 6, 2, 0, 0), along with xnew 0, is still optimal, so this third possible new product should not be added to the product line. This approach also makes it very easy to conduct sensitivity analysis on the coefficients of the new variable added to the primal problem. By simply checking the new dual constraint, you can immediately see how far any of these parameter values can be changed before they affect the feasibility of the dual solution and so the optimality of the primal solution. Other Applications Already we have discussed two other key applications of duality theory to sensitivity analysis, namely, shadow prices and the dual simplex method. As described in Secs. 4.7 and 6.2, the optimal dual solution (y1*, y2*, . . . , ym*) provides the shadow prices for the respective resources that indicate how Z would change if (small) changes were made in the bi (the resource amounts). The resulting analysis will be illustrated in some detail in Sec. 6.7. In more general terms, the economic interpretation of the dual problem and of the simplex method presented in Sec. 6.2 provides some useful insights for sensitivity analysis. When we investigate the effect of changing the bi or the aij values (for basic variables), the original optimal solution may become a superoptimal basic solution (as defined in Table 6.10) instead. If we then want to reoptimize to identify the new optimal solution, the dual simplex method (discussed at the end of Secs. 6.1 and 6.3) should be applied, starting from this basic solution. We mentioned in Sec. 6.1 that sometimes it is more efficient to solve the dual problem directly by the simplex method in order to identify an optimal solution for the primal problem. When the solution has been found in this way, sensitivity analysis for the primal problem then is conducted by applying the procedure described in the next two sections directly to the dual problem and then inferring the complementary effects on the primal problem (e.g., see Table 6.11). This approach to sensitivity analysis is relatively straightforward because of the close primal-dual relationships described in Secs. 6.1 and 6.3. (See Prob. 6.6-3.)

6.6

THE ESSENCE OF SENSITIVITY ANALYSIS The work of the operations research team usually is not even nearly done when the simplex method has been successfully applied to identify an optimal solution for the model. As we pointed out at the end of Sec. 3.3, one assumption of linear programming is that all the parameters of the model (aij, bi, and cj) are known constants. Actually, the parameter values used in the model normally are just estimates based on a prediction of future conditions. The data obtained to develop these estimates often are rather crude or non-

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

255

existent, so that the parameters in the original formulation may represent little more than quick rules of thumb provided by harassed line personnel. The data may even represent deliberate overestimates or underestimates to protect the interests of the estimators. Thus, the successful manager and operations research staff will maintain a healthy skepticism about the original numbers coming out of the computer and will view them in many cases as only a starting point for further analysis of the problem. An “optimal” solution is optimal only with respect to the specific model being used to represent the real problem, and such a solution becomes a reliable guide for action only after it has been verified as performing well for other reasonable representations of the problem. Furthermore, the model parameters (particularly bi) sometimes are set as a result of managerial policy decisions (e.g., the amount of certain resources to be made available to the activities), and these decisions should be reviewed after their potential consequences are recognized. For these reasons it is important to perform sensitivity analysis to investigate the effect on the optimal solution provided by the simplex method if the parameters take on other possible values. Usually there will be some parameters that can be assigned any reasonable value without the optimality of this solution being affected. However, there may also be parameters with likely alternative values that would yield a new optimal solution. This situation is particularly serious if the original solution would then have a substantially inferior value of the objective function, or perhaps even be infeasible! Therefore, one main purpose of sensitivity analysis is to identify the sensitive parameters (i.e., the parameters whose values cannot be changed without changing the optimal solution). For certain parameters that are not categorized as sensitive, it is also very helpful to determine the range of values of the parameter over which the optimal solution will remain unchanged. (We call this range of values the allowable range to stay optimal.) In some cases, changing a parameter value can affect the feasibility of the optimal BF solution. For such parameters, it is useful to determine the range of values over which the optimal BF solution (with adjusted values for the basic variables) will remain feasible. (We call this range of values the allowable range to stay feasible.) In the next section, we will describe the specific procedures for obtaining this kind of information. Such information is invaluable in two ways. First, it identifies the more important parameters, so that special care can be taken to estimate them closely and to select a solution that performs well for most of their likely values. Second, it identifies the parameters that will need to be monitored particularly closely as the study is implemented. If it is discovered that the true value of a parameter lies outside its allowable range, this immediately signals a need to change the solution. For small problems, it would be straightforward to check the effect of a variety of changes in parameter values simply by reapplying the simplex method each time to see if the optimal solution changes. This is particularly convenient when using a spreadsheet formulation. Once the Solver has been set up to obtain an optimal solution, all you have to do is make any desired change on the spreadsheet and then click on the Solve button again. However, for larger problems of the size typically encountered in practice, sensitivity analysis would require an exorbitant computational effort if it were necessary to reapply the simplex method from the beginning to investigate each new change in a parameter value. Fortunately, the fundamental insight discussed in Sec. 5.3 virtually eliminates computational effort. The basic idea is that the fundamental insight immediately reveals

256

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

just how any changes in the original model would change the numbers in the final simplex tableau (assuming that the same sequence of algebraic operations originally performed by the simplex method were to be duplicated ). Therefore, after making a few simple calculations to revise this tableau, we can check easily whether the original optimal BF solution is now nonoptimal (or infeasible). If so, this solution would be used as the initial basic solution to restart the simplex method (or dual simplex method) to find the new optimal solution, if desired. If the changes in the model are not major, only a very few iterations should be required to reach the new optimal solution from this “advanced” initial basic solution. To describe this procedure more specifically, consider the following situation. The simplex method already has been used to obtain an optimal solution for a linear programming model with specified values for the bi , cj , and aij parameters. To initiate sensitivity analysis, at least one of the parameters is changed. After the changes are made, let bi , cj , and aij denote the values of the various parameters. Thus, in matrix notation, bb ,

c c,

AA ,

for the revised model. The first step is to revise the final simplex tableau to reflect these changes. Continuing to use the notation presented in Table 5.10, as well as the accompanying formulas for the fundamental insight [(1) t* t y*T and (2) T* S*T], we see that the revised final tableau is calculated from y* and S* (which have not changed) and the new initial tableau, as shown in Table 6.17. Example (Variation 1 of the Wyndor Model). To illustrate, suppose that the first revision in the model for the Wyndor Glass Co. problem of Sec. 3.1 is the one shown in Table 6.18. Thus, the changes from the original model are c1 3 4, a31 3 2, and b2 12 24. Figure 6.2 shows the graphical effect of these changes. For the original model, the simplex method already has identified the optimal CPF solution as (2, 6), lying at the intersection of the two constraint boundaries, shown as dashed lines 2x2 12 and 3x1 2x2 18. Now the revision of the model has shifted both of these constraint boundaries as shown by the dark lines 2x2 24 and 2x1 2x2 18. Consequently, the previous TABLE 6.17 Revised final simplex tableau resulting from changes in original model Coefficient of: Eq.

Z

Original Variables

Slack Variables

Right Side

(0)

1

c

0

0

(1, 2, . . . , m)

0

A

I

b

(0)

1

z* c y*A c

y*

Z* y*b

(1, 2, . . . , m)

0

A* S*A

S*

b* S*b

New initial tableau

Revised final tableau

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

257

TABLE 6.18 The original model and the first revised model (variation 1) for conducting sensitivity analysis on the Wyndor Glass Co. model Original Model Maximize

Revised Model

Z [3, 5]

x , x1

Maximize

Z [4, 5]

0 x1 2 x2 2

1 0 2

4 12 18

0 x1 2 x2 2

4 24 18

and

and

x 0.

x 0.

FIGURE 6.2 Shift of the final corner-point solution from (2, 6) to (3, 12) for Variation 1 of the Wyndor Glass Co. model where c1 3 4, a31 3 2, and b2 12 24.

x1

subject to

subject to

1 0 3

x , 2

2

x2 x1 0

(3, 12)

2x2 24

10 x1 4

(0, 9) optimal 8

6

2x2 12

(2, 6)

4 2x1 2x2 18 2

3x1 2x2 18 x2 0

0

2

4

6

8

x1

258

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

CPF solution (2, 6) now shifts to the new intersection (3, 12), which is a corner-point infeasible solution for the revised model. The procedure described in the preceding paragraphs finds this shift algebraically (in augmented form). Furthermore, it does so in a manner that is very efficient even for huge problems where graphical analysis is impossible. To carry out this procedure, we begin by displaying the parameters of the revised model in matrix form: c [4, 5],

1 A 0 2

0 2 , 2

4 b 24 . 18

The resulting new initial simplex tableau is shown at the top of Table 6.19. Below this tableau is the original final tableau (as first given in Table 4.8). We have drawn dark boxes around the portions of this final tableau that the changes in the model definitely do not change, namely, the coefficients of the slack variables in both row 0 (y*) and the rest of the rows (S*). Thus, y* [0, , 1], 3 2

1 S* 0 0

1 3 1 2 1 3

13 0 . 1 3

TABLE 6.19 Obtaining the revised final simplex tableau for Variation 1 of the Wyndor Glass Co. model Coefficient of:

New initial tableau

Final tableau for original model

Basic Variable

Eq.

Z

x1

x2

x3

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

4 1 0 2

5 0 2 2

0 1 0 0

Z

(0)

1

0

0

0

x3

(1)

0

0

0

1

x2

(2)

0

0

1

0

x1

(3)

0

1

0

0

Z

(0)

1

2

0

0

x3

(1)

0

1 3

0

1

x2

(2)

0

0

1

0

x1

(3)

0

2 3

0

0

Revised final tableau

x4 0 0 1 0

3 2 1 3 1 2 1 3 3 2 1 3 1 2 1 3

x5

Right Side

0 0 0 1

0 4 24 18

1

36

1 3

2

0

6

1 3

2

1

54

1 3

6

0

12

1 3

2

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

259

These coefficients of the slack variables necessarily are unchanged with the same algebraic operations originally performed by the simplex method because the coefficients of these same variables in the initial tableau are unchanged. However, because other portions of the initial tableau have changed, there will be changes in the rest of the final tableau as well. Using the formulas in Table 6.17, we calculate the revised numbers in the rest of the final tableau as follows: 1 z* c [0, 32, 1] 0 2

0 2 [4, 5] [2, 0], 2

1 1 13 1 3 1 0 0 A* 0 2 1 1 0 3 3 2

13 0 2 0 2 2 3

4 Z* [0, 32, 1] 24 54, 18

0 1 , 0

1 6 1 13 4 3 1 0 24 12 . b* 0 2 1 1 2 0 3 3 18 The resulting revised final tableau is shown at the bottom of Table 6.19. Actually, we can substantially streamline these calculations for obtaining the revised final tableau. Because none of the coefficients of x2 changed in the original model (tableau), none of them can change in the final tableau, so we can delete their calculation. Several other original parameters (a11, a21, b1, b3) also were not changed, so another shortcut is to calculate only the incremental changes in the final tableau in terms of the incremental changes in the initial tableau, ignoring those terms in the vector or matrix multiplication that involve zero change in the initial tableau. In particular, the only incremental changes in the initial tableau are c1 1, a31 1, and b2 12, so these are the only terms that need be considered. This streamlined approach is shown below, where a zero or dash appears in each spot where no calculation is needed.

0 (z* c) y* A c [0, 32, 1] 0 1

— — [1, —] [2, —]. —

0 Z* y* b [0, , 1] 12 18. 0 3 2

1 A* S* A 0 0

1 3 1 2 1 3

13 0 0 0 1 3 1

13 — — 0 1 — 3

— — . —

1 1 4 13 0 3 1 0 12 6 . b* S* b 0 2 1 1 0 3 4 3 0 Adding these increments to the original quantities in the final tableau (middle of Table 6.19) then yields the revised final tableau (bottom of Table 6.19).

260

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

This incremental analysis also provides a useful general insight, namely, that changes in the final tableau must be proportional to each change in the initial tableau. We illustrate in the next section how this property enables us to use linear interpolation or extrapolation to determine the range of values for a given parameter over which the final basic solution remains both feasible and optimal. After obtaining the revised final simplex tableau, we next convert the tableau to proper form from Gaussian elimination (as needed). In particular, the basic variable for row i must have a coefficient of 1 in that row and a coefficient of 0 in every other row (including row 0) for the tableau to be in the proper form for identifying and evaluating the current basic solution. Therefore, if the changes have violated this requirement (which can occur only if the original constraint coefficients of a basic variable have been changed), further changes must be made to restore this form. This restoration is done by using Gaussian elimination, i.e., by successively applying step 3 of an iteration for the simplex method (see Chap. 4) as if each violating basic variable were an entering basic variable. Note that these algebraic operations may also cause further changes in the right side column, so that the current basic solution can be read from this column only when the proper form from Gaussian elimination has been fully restored. For the example, the revised final simplex tableau shown in the top half of Table 6.20 is not in proper form from Gaussian elimination because of the column for the basic variable x1. Specifically, the coefficient of x1 in its row (row 3) is 23 instead of 1, and it has nonzero coefficients (2 and 13) in rows 0 and 1. To restore proper form, row 3 is multiplied by 32; then 2 times this new row 3 is added to row 0 and 13 times new row 3 is subtracted from row 1. This yields the proper form from Gaussian elimination shown in

TABLE 6.20 Converting the revised final simplex tableau to proper form from Gaussian elimination for Variation 1 of the Wyndor Glass Co. model Coefficient of:

Revised final tableau

Converted to proper form

Basic Variable

Eq.

Z

x1

x2

Z

(0)

1

2

0

0

x3

(1)

0

1 3

0

1

x2

(2)

0

0

1

0

x1

(3)

0

2 3

0

0

Z

(0)

1

0

0

0

x3

(1)

0

0

0

1

x2

(2)

0

0

1

0

x1

(3)

0

1

0

0

x3

x4 3 2 1 3 1 2 1 3 1 2 1 2 1 2 1 2

x5

Right Side

1

54

1 3

6

0

12

1 3

2

2

48

1 2

7

0

12

1 2

3

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

261

the bottom half of Table 6.20, which now can be used to identify the new values for the current (previously optimal) basic solution: (x1, x2, x3, x4, x5) (3, 12, 7, 0, 0). Because x1 is negative, this basic solution no longer is feasible. However, it is superoptimal (as defined in Table 6.10), and so dual feasible, because all the coefficients in row 0 still are nonnegative. Therefore, the dual simplex method can be used to reoptimize (if desired), by starting from this basic solution. (The sensitivity analysis routine in the OR Courseware includes this option.) Referring to Fig. 6.2 (and ignoring slack variables), the dual simplex method uses just one iteration to move from the corner-point solution (3, 12) to the optimal CPF solution (0, 9). (It is often useful in sensitivity analysis to identify the solutions that are optimal for some set of likely values of the model parameters and then to determine which of these solutions most consistently performs well for the various likely parameter values.) If the basic solution (3, 12, 7, 0, 0) had been neither primal feasible nor dual feasible (i.e., if the tableau had negative entries in both the right side column and row 0), artificial variables could have been introduced to convert the tableau to the proper form for an initial simplex tableau.1 The General Procedure. When one is testing to see how sensitive the original optimal solution is to the various parameters of the model, the common approach is to check each parameter (or at least cj and bi) individually. In addition to finding allowable ranges as described in the next section, this check might include changing the value of the parameter from its initial estimate to other possibilities in the range of likely values (including the endpoints of this range). Then some combinations of simultaneous changes of parameter values (such as changing an entire functional constraint) may be investigated. Each time one (or more) of the parameters is changed, the procedure described and illustrated here would be applied. Let us now summarize this procedure. Summary of Procedure for Sensitivity Analysis 1. Revision of model: Make the desired change or changes in the model to be investigated next. 2. Revision of final tableau: Use the fundamental insight (as summarized by the formulas on the bottom of Table 6.17) to determine the resulting changes in the final simplex tableau. (See Table 6.19 for an illustration.) 3. Conversion to proper form from Gaussian elimination: Convert this tableau to the proper form for identifying and evaluating the current basic solution by applying (as necessary) Gaussian elimination. (See Table 6.20 for an illustration.) 4. Feasibility test: Test this solution for feasibility by checking whether all its basic variable values in the right-side column of the tableau still are nonnegative. 5. Optimality test: Test this solution for optimality (if feasible) by checking whether all its nonbasic variable coefficients in row 0 of the tableau still are nonnegative. 6. Reoptimization: If this solution fails either test, the new optimal solution can be obtained (if desired) by using the current tableau as the initial simplex tableau (and making any necessary conversions) for the simplex method or dual simplex method. 1

There also exists a primal-dual algorithm that can be directly applied to such a simplex tableau without any conversion.

262

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

The interactive routine entitled sensitivity analysis in the OR Courseware will enable you to efficiently practice applying this procedure. In addition, a demonstration in OR Tutor (also entitled sensitivity analysis) provides you with another example. In the next section, we shall discuss and illustrate the application of this procedure to each of the major categories of revisions in the original model. This discussion will involve, in part, expanding upon the example introduced in this section for investigating changes in the Wyndor Glass Co. model. In fact, we shall begin by individually checking each of the preceding changes. At the same time, we shall integrate some of the applications of duality theory to sensitivity analysis discussed in Sec. 6.5.

6.7

APPLYING SENSITIVITY ANALYSIS Sensitivity analysis often begins with the investigation of changes in the values of bi, the amount of resource i (i 1, 2, . . . , m) being made available for the activities under consideration. The reason is that there generally is more flexibility in setting and adjusting these values than there is for the other parameters of the model. As already discussed in Secs. 4.7 and 6.2, the economic interpretation of the dual variables (the yi) as shadow prices is extremely useful for deciding which changes should be considered. Case 1—Changes in bi Suppose that the only changes in the current model are that one or more of the bi parameters (i 1, 2, . . . , m) has been changed. In this case, the only resulting changes in the final simplex tableau are in the right-side column. Consequently, the tableau still will be in proper form from Gaussian elimination and all the nonbasic variable coefficients in row 0 still will be nonnegative. Therefore, both the conversion to proper form from Gaussian elimination and the optimality test steps of the general procedure can be skipped. After revising the right-side column of the tableau, the only question will be whether all the basic variable values in this column still are nonnegative (the feasibility test). As shown in Table 6.17, when the vector of the bi values is changed from b to b, the formulas for calculating the new right-side column in the final tableau are Right side of final row 0: Right side of final rows 1, 2, . . . , m:

Z* y*b , b* S*b .

(See the bottom of Table 6.17 for the location of the unchanged vector y* and matrix S* in the final tableau.) Example (Variation 2 of the Wyndor Model). Sensitivity analysis is begun for the original Wyndor Glass Co. problem of Sec. 3.1 by examining the optimal values of the yi dual variables ( y1* 0, y2* 32, y3* 1). These shadow prices give the marginal value of each resource i for the activities (two new products) under consideration, where marginal value is expressed in the units of Z (thousands of dollars of profit per week). As discussed in Sec. 4.7 (see Fig. 4.8), the total profit from these activities can be increased $1,500 per week ( y2* times $1,000 per week) for each additional unit of resource 2 (hour of production time per week in Plant 2) that is made available. This increase in profit holds for relatively small changes that do not affect the feasibility of the current basic solution (and so do not affect the yi* values).

6.7 APPLYING SENSITIVITY ANALYSIS

263

Consequently, the OR team has investigated the marginal profitability from the other current uses of this resource to determine if any are less than $1,500 per week. This investigation reveals that one old product is far less profitable. The production rate for this product already has been reduced to the minimum amount that would justify its marketing expenses. However, it can be discontinued altogether, which would provide an additional 12 units of resource 2 for the new products. Thus, the next step is to determine the profit that could be obtained from the new products if this shift were made. This shift changes b2 from 12 to 24 in the linear programming model. Figure 6.3 shows the graphical effect of this change, including the shift in the final corner-point solution from (2, 6) to (2, 12). (Note that this figure differs from Fig. 6.2, which depicts Variation 1 of the Wyndor model, because the constraint 3x1 2x2 18 has not been changed here.) Thus, for Variation 2 of the Wyndor model, the only revision in the original model is the following change in the vector of the bi values: 4 b 12 → b 18

4 24 . 18

so only b2 has a new value.

FIGURE 6.3 Feasible region for Variation 2 of the Wyndor Glass Co. model where b2 12 → 24.

x2 14

x1 0

(2, 12) 2x2 24

10 x1 4

(0, 9) optimal 8 (2, 6)

2x2 12

6

Z 45 3x1 5x2

4 Feasible region

3x1 2x2 18

2

x2 0 0

2

4

6

8

x1

264

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Analysis of Variation 2. When the fundamental insight (Table 6.17) is applied, the effect of this change in b2 on the original final simplex tableau (middle of Table 6.19) is that the entries in the right-side column change to the following values: 4 Z* y*b [0, , 1] 24 54, 18 3 2

1 b* S*b 0 0

1 3 1 2 1 3

6 13 4 0 24 12 , 1 2 3 18

x3 6 so x2 12 . x1 2

Equivalently, because the only change in the original model is b2 24 12 12, incremental analysis can be used to calculate these same values more quickly. Incremental analysis involves calculating just the increments in the tableau values caused by the change (or changes) in the original model, and then adding these increments to the original values. In this case, the increments in Z* and b* are b1 Z* y* b y* b2 y* b3

0 12 , 0 b1 b* S* b S* b2 S* b3

0 12 . 0

Therefore, using the second component of y* and the second column of S*, the only calculations needed are 3 Z* (12) 18, 2 1 b1* (12) 4, 3 1 b2* (12) 6, 2 1 b3* (12) 4, 3

so Z* 36 18 54, so b1* 2 4 6, so b2* 6 6 12, so b3* 2 4 2,

where the original values of these quantities are obtained from the right-side column in the original final tableau (middle of Table 6.19). The resulting revised final tableau corresponds completely to this original final tableau except for replacing the right-side column with these new values. Therefore, the current (previously optimal) basic solution has become (x1, x2, x3, x4, x5) (2, 12, 6, 0, 0), which fails the feasibility test because of the negative value. The dual simplex method now can be applied, starting with this revised simplex tableau, to find the new optimal so-

6.7 APPLYING SENSITIVITY ANALYSIS

265

TABLE 6.21 Data for Variation 2 of the Wyndor Glass Co. model Final Simplex Tableau after Reoptimization Coefficient of: Basic Variable

Eq.

Z

Z

(0)

1

Model Parameters c1 3, a11 1, a21 0, a31 3,

c2 5 a12 0, a22 2, a32 2,

(n 2) b1 4 b2 24 b3 18

x3

(1)

0

x2

(2)

0

x4

(3)

0

x1 9 2 1 3 2 3

x2

x3

x4

0

0

0

0

1

0

1

0

0

0

0

1

x5 5 2 0 1 2 1

Right Side 45 4 9 6

lution. This method leads in just one iteration to the new final simplex tableau shown in Table 6.21. (Alternatively, the simplex method could be applied from the beginning, which also would lead to this final tableau in just one iteration in this case.) This tableau indicates that the new optimal solution is (x1, x2, x3, x4, x5) (0, 9, 4, 6, 0), with Z 45, thereby providing an increase in profit from the new products of 9 units ($9,000 per week) over the previous Z 36. The fact that x4 6 indicates that 6 of the 12 additional units of resource 2 are unused by this solution. Based on the results with b2 24, the relatively unprofitable old product will be discontinued and the unused 6 units of resource 2 will be saved for some future use. Since y3* still is positive, a similar study is made of the possibility of changing the allocation of resource 3, but the resulting decision is to retain the current allocation. Therefore, the current linear programming model at this point (Variation 2) has the parameter values and optimal solution shown in Table 6.21. This model will be used as the starting point for investigating other types of changes in the model later in this section. However, before turning to these other cases, let us take a broader look at the current case. The Allowable Range to Stay Feasible. Although b2 12 proved to be too large an increase in b2 to retain feasibility (and so optimality) with the basic solution where x1, x2, and x3 are the basic variables (middle of Table 6.19), the above incremental analysis shows immediately just how large an increase is feasible. In particular, note that 1 b1* 2 b2, 3 1 b2* 6 b2, 2 1 b3* 2 b2, 3

266

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

where these three quantities are the values of x3, x2, and x1, respectively, for this basic solution. The solution remains feasible, and so optimal, as long as all three quantities remain nonnegative. 1 1 2 b2 0 ⇒ b2 2 ⇒ b2 6, 3 3 1 1 6 b2 0 ⇒ b2 6 ⇒ b2 12, 2 2 1 1 2 b2 0 ⇒ 2 b2 3 3

⇒ b2

6.

Therefore, since b2 12 b2, the solution remains feasible only if 6 b2 6,

that is,

6 b2 18.

(Verify this graphically in Fig. 6.3.) As introduced in Sec. 4.7, this range of values for b2 is referred to as its allowable range to stay feasible. For any bi, recall from Sec. 4.7 that its allowable range to stay feasible is the range of values over which the current optimal BF solution1 (with adjusted values for the basic variables) remains feasible. Thus, the shadow price for bi remains valid for evaluating the effect on Z of changing bi only as long as bi remains within this allowable range. (It is assumed that the change in this one bi value is the only change in the model.) The adjusted values for the basic variables are obtained from the formula b* S*b . The calculation of the allowable range to stay feasible then is based on finding the range of values of bi such that b* 0. Many linear programming software packages use this same technique for automatically generating the allowable range to stay feasible for each bi. (A similar technique, discussed under Cases 2a and 3, also is used to generate an allowable range to stay optimal for each cj.) In Chap. 4, we showed the corresponding output for the Excel Solver and LINDO in Figs. 4.10 and 4.13, respectively. Table 6.22 summarizes this same output with respect to the bi for the original Wyndor Glass Co. model. For example, both the allowable increase and allowable decrease for b2 are 6, that is, 6 b2 6. The above analysis shows how these quantities were calculated. 1

When there is more than one optimal BF solution for the current model (before changing bi), we are referring here to the one obtained by the simplex method.

TABLE 6.22 Typical software output for sensitivity analysis of the right-hand sides for the original Wyndor Glass Co. model Constraint Plant 1 Plant 2 Plant 3

Shadow Price

Current RHS

Allowable Increase

Allowable Decrease

0.0 1.5 1.0

4 12 18

6 6

2 6 6

6.7 APPLYING SENSITIVITY ANALYSIS

267

Analyzing Simultaneous Changes in Right-Hand Sides. When multiple bi values are changed simultaneously, the formula b* S*b can again be used to see how the righthand sides change in the final tableau. If all these right-hand sides still are nonnegative, the feasibility test will indicate that the revised solution provided by this tableau still is feasible. Since row 0 has not changed, being feasible implies that this solution also is optimal. Although this approach works fine for checking the effect of a specific set of changes in the bi, it does not give much insight into how far the bi can be simultaneously changed from their original values before the revised solution will no longer be feasible. As part of postoptimality analysis, the management of an organization often is interested in investigating the effect of various changes in policy decisions (e.g., the amounts of resources being made available to the activities under consideration) that determine the right-hand sides. Rather than considering just one specific set of changes, management may want to explore directions of changes where some right-hand sides increase while others decrease. Shadow prices are invaluable for this kind of exploration. However, shadow prices remain valid for evaluating the effect of such changes on Z only within certain ranges of changes. For each bi, the allowable range to stay feasible gives this range if none of the other bi are changing at the same time. What do these allowable ranges become when some of the bi are changing simultaneously? A partial answer to this question is provided by the following 100 percent rule, which combines the allowable changes (increase or decrease) for the individual bi that are given by the last two columns of a table like Table 6.22. The 100 Percent Rule for Simultaneous Changes in Right-Hand Sides: The shadow prices remain valid for predicting the effect of simultaneously changing the right-hand sides of some of the functional constraints as long as the changes are not too large. To check whether the changes are small enough, calculate for each change the percentage of the allowable change (increase or decrease) for that right-hand side to remain within its allowable range to stay feasible. If the sum of the percentage changes does not exceed 100 percent, the shadow prices definitely will still be valid. (If the sum does exceed 100 percent, then we cannot be sure.) Example (Variation 3 of the Wyndor Model). To illustrate this rule, consider Variation 3 of the Wyndor Glass Co. model, which revises the original model by changing the right-hand side vector as follows: 4 b 12 b 18

4 15 . 15

The calculations for the 100 percent rule in this case are b2: 12 15. b3: 18 15.

15 12 Percentage of allowable increase 100 50% 6 18 15 Percentage of allowable decrease 100 50% 6 Sum 100%

Since the sum of 100 percent barely does not exceed 100 percent, the shadow prices definitely are valid for predicting the effect of these changes on Z. In particular, since

268

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

the shadow prices of b2 and b3 are 1.5 and 1, respectively, the resulting change in Z would be Z 1.5(3) 1(3) 1.5, so Z* would increase from 36 to 37.5. Figure 6.4 shows the feasible region for this revised model. (The dashed lines show the original locations of the revised constraint boundary lines.) The optimal solution now is the CPF solution (0, 7.5), which gives Z 3x1 5x2 0 5(7.5) 37.5, just as predicted by the shadow prices. However, note what would happen if either b2 were further increased above 15 or b3 were further decreased below 15, so that the sum of the percentages of allowable changes would exceed 100 percent. This would cause the previously optimal corner-point solution to slide to the left of the x2 axis (x1 0), so this infeasible solution would no longer be optimal. Consequently, the old shadow prices would no longer be valid for predicting the new value of Z*.

FIGURE 6.4 Feasible region for Variation 3 of the Wyndor Glass Co. model where b2 12 15 and b3 18 15.

x2

8 (0, 7.5) optimal

2x2 15

6 x1 4 4 Feasible region 2 3x1 2x2 15

0

2

4

6

8

x1

6.7 APPLYING SENSITIVITY ANALYSIS

269

Case 2a—Changes in the Coefficients of a Nonbasic Variable Consider a particular variable xj (fixed j) that is a nonbasic variable in the optimal solution shown by the final simplex tableau. In Case 2a, the only change in the current model is that one or more of the coefficients of this variable—cj , a1j , a2j , . . . , amj —have been changed. Thus, letting cj and aij denote the new values of these parameters, with Aj (column j of matrix A ) as the vector containing the aij , we have cj → cj ,

Aj → A j

for the revised model. As described at the beginning of Sec. 6.5, duality theory provides a very convenient way of checking these changes. In particular, if the complementary basic solution y* in the dual problem still satisfies the single dual constraint that has changed, then the original optimal solution in the primal problem remains optimal as is. Conversely, if y* violates this dual constraint, then this primal solution is no longer optimal. If the optimal solution has changed and you wish to find the new one, you can do so rather easily. Simply apply the fundamental insight to revise the xj column (the only one that has changed) in the final simplex tableau. Specifically, the formulas in Table 6.17 reduce to the following: Coefficient of xj in final row 0: Coefficient of xj in final rows 1 to m:

z j* cj y*A j cj , Aj* S*A j.

With the current basic solution no longer optimal, the new value of zj* cj now will be the one negative coefficient in row 0, so restart the simplex method with xj as the initial entering basic variable. Note that this procedure is a streamlined version of the general procedure summarized at the end of Sec. 6.6. Steps 3 and 4 (conversion to proper form from Gaussian elimination and the feasibility test) have been deleted as irrelevant, because the only column being changed in the revision of the final tableau (before reoptimization) is for the nonbasic variable xj. Step 5 (optimality test) has been replaced by a quicker test of optimality to be performed right after step 1 (revision of model). It is only if this test reveals that the optimal solution has changed, and you wish to find the new one, that steps 2 and 6 (revision of final tableau and reoptimization) are needed. Example (Variation 4 of the Wyndor Model). Since x1 is nonbasic in the current optimal solution (see Table 6.21) for Variation 2 of the Wyndor Glass Co. model, the next step in its sensitivity analysis is to check whether any reasonable changes in the estimates of the coefficients of x1 could still make it advisable to introduce product 1. The set of changes that goes as far as realistically possible to make product 1 more attractive would be to reset c1 4 and a31 2. Rather than exploring each of these changes independently (as is often done in sensitivity analysis), we will consider them together. Thus, the changes under consideration are c1 3 → c1 4,

1 A1 0 → A1 3

1 0 . 2

270

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

These two changes in Variation 2 give us Variation 4 of the Wyndor model. Variation 4 actually is equivalent to Variation 1 considered in Sec. 6.6 and depicted in Fig. 6.2, since Variation 1 combined these two changes with the change in the original Wyndor model (b2 12 24) that gave Variation 2. However, the key difference from the treatment of Variation 1 in Sec. 6.6 is that the analysis of Variation 4 treats Variation 2 as being the original model, so our starting point is the final simplex tableau given in Table 6.21 where x1 now is a nonbasic variable. The change in a31 revises the feasible region from that shown in Fig. 6.3 to the corresponding region in Fig. 6.5. The change in c1 revises the objective function from Z 3x1 5x2 to Z 4x1 5x2. Figure 6.5 shows that the optimal objective function line Z 45 4x1 5x2 still passes through the current optimal solution (0, 9), so this solution remains optimal after these changes in a31 and c1. To use duality theory to draw this same conclusion, observe that the changes in c1 and a31 lead to a single revised constraint for the dual problem, namely, the constraint that a11y1 a21y2 a31y3 c1. Both this revised constraint and the current y* (coefficients of the slack variables in row 0 of Table 6.21) are shown below. 5 y2* 0, y3* , y1* 0, 2 y1 3y3 3 → y1 2y3 4, 5 0 2 4. 2 Since y* still satisfies the revised constraint, the current primal solution (Table 6.21) is still optimal. Because this solution is still optimal, there is no need to revise the xj column in the final tableau (step 2). Nevertheless, we do so below for illustrative purposes. 1 5 * 1 c1 [0, 0, 2] 0 4 1. z 1 c1 y*A 2

1 1 0 0 1 1 0 A1* S*A 0 1 0 1 . 2 1 1 2 0 2 The fact that z1* c1 0 again confirms the optimality of the current solution. Since z1* c1 is the surplus variable for the revised constraint in the dual problem, this way of testing for optimality is equivalent to the one used above. This completes the analysis of the effect of changing the current model (Variation 2) to Variation 4. Because any larger changes in the original estimates of the coefficients of x1 would be unrealistic, the OR team concludes that these coefficients are insensitive parameters in the current model. Therefore, they will be kept fixed at their best estimates shown in Table 6.21—c1 3 and a31 3—for the remainder of the sensitivity analysis. The Allowable Range to Stay Optimal. We have just described and illustrated how to analyze simultaneous changes in the coefficients of a nonbasic variable xj. It is common practice in sensitivity analysis to also focus on the effect of changing just one param-

6.7 APPLYING SENSITIVITY ANALYSIS

271

x2 12

2x2 24

x1 4

10 (0, 9) optimal 8

6

4

FIGURE 6.5 Feasible region for Variation 4 of the Wyndor model where Variation 2 (Fig. 6.3) has been revised so a31 3 2 and c1 3 4.

Feasible region

Z 45 4x1 5x2

2

0

2x1 2x2 18

2

4

6

8

10 x1

eter, cj. As introduced in Sec. 4.7, this involves streamlining the above approach to find the allowable range to stay optimal for cj. For any cj, recall from Sec. 4.7 that its allowable range to stay optimal is the range of values over which the current optimal solution (as obtained by the simplex method for the current model before cj is changed) remains optimal. (It is assumed that the change in this one cj is the only change in the current model.) When xj is a nonbasic variable for this solution, the solution remains optimal as long as z*j cj 0, where z*j y*Aj is a constant unaffected by any change in the value of cj. Therefore, the allowable range to stay optimal for cj can be calculated as cj y*Aj. For example, consider the current model (Variation 2) for the Wyndor Glass Co. problem summarized on the left side of Table 6.21, where the current optimal solution (with c1 3) is given on the right side. When considering only the decision variables, x1 and x2, this optimal solution is (x1, x2) = (0, 9), as displayed in Fig. 6.3. When just c1 is changed, this solution remains optimal as long as 1 c1 y*A1 [0, 0, 52] 0 712, 3 1 so c1 72 is the allowable range to stay optimal.

272

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

An alternative to performing this vector multiplication is to note in Table 6.21 that z1* c1 92 (the coefficient of x1 in row 0) when c1 3, so z1* 3 92 712. Since z1* y*A1, this immediately yields the same allowable range. Figure 6.3 provides graphical insight into why c1 712 is the allowable range. At c1 712, the objective function becomes Z 7.5x1 5x2 2.5(3x1 2x2), so the optimal objective line will lie on top of the constraint boundary line 3x1 2x2 18 shown in the figure. Thus, at this endpoint of the allowable range, we have multiple optimal solutions consisting of the line segment between (0, 9) and (4, 3). If c1 were to be increased any further (c1 712 ), only (4, 3) would be optimal. Consequently, we need c1 712 for (0, 9) to remain optimal. For any nonbasic decision variable xj , the value of z*j cj sometimes is referred to as the reduced cost for xj , because it is the minimum amount by which the unit cost of activity j would have to be reduced to make it worthwhile to undertake activity j (increase xj from zero). Interpreting cj as the unit profit of activity j (so reducing the unit cost increases cj by the same amount), the value of z*j cj thereby is the maximum allowable increase in cj to keep the current BF solution optimal. The sensitivity analysis information generated by linear programming software packages normally includes both the reduced cost and the allowable range to stay optimal for each coefficient in the objective function (along with the types of information displayed in Table 6.22). This was illustrated in Figs. 4.10, 4.12, and 4.13 for the Excel Solver and LINDO. Table 6.23 displays this information in a typical form for our current model (Variation 2 of the Wyndor Glass Co. model). The last three columns are used to calculate the allowable range to stay optimal for each coefficient, so these allowable ranges are c1 3 4.5 7.5, c2 5 3 2. As was discussed in Sec. 4.7, if any of the allowable increases or decreases had turned out to be zero, this would have been a signpost that the optimal solution given in the table is only one of multiple optimal solutions. In this case, changing the corresponding coefficient a tiny amount beyond the zero allowed and re-solving would provide another optimal CPF solution for the original model. Thus far, we have described how to calculate the type of information in Table 6.23 for only nonbasic variables. For a basic variable like x2, the reduced cost automatically is 0. We will discuss how to obtain the allowable range to stay optimal for cj when xj is a basic variable under Case 3.

TABLE 6.23 Typical software output for sensitivity analysis of the objective function coefficients for Variation 2 of the Wyndor Glass Co. model Variable

Value

Reduced Cost

Current Coefficient

Allowable Increase

Allowable Decrease

x1 x2

0 9

4.5 0.0

3 5

4.5

3

6.7 APPLYING SENSITIVITY ANALYSIS

273

Analyzing Simultaneous Changes in Objective Function Coefficients. Regardless of whether xj is a basic or nonbasic variable, the allowable range to stay optimal for cj is valid only if this objective function coefficient is the only one being changed. However, when simultaneous changes are made in the coefficients of the objective function, a 100 percent rule is available for checking whether the original solution must still be optimal. Much like the 100 percent rule for simultaneous changes in right-hand sides, this 100 percent rule combines the allowable changes (increase or decrease) for the individual cj that are given by the last two columns of a table like Table 6.23, as described below. The 100 Percent Rule for Simultaneous Changes in Objective Function Coefficients: If simultaneous changes are made in the coefficients of the objective function, calculate for each change the percentage of the allowable change (increase or decrease) for that coefficient to remain within its allowable range to stay optimal. If the sum of the percentage changes does not exceed 100 percent, the original optimal solution definitely will still be optimal. (If the sum does exceed 100 percent, then we cannot be sure.) Using Table 6.23 (and referring to Fig. 6.3 for visualization), this 100 percent rule says that (0, 9) will remain optimal for Variation 2 of the Wyndor Glass Co. model even if we simultaneously increase c1 from 3 and decrease c2 from 5 as long as these changes are not too large. For example, if c1 is increased by 1.5 (3313 percent of the allowable change), then c2 can be decreased by as much as 2 (6623 percent of the allowable change). Similarly, if c1 is increased by 3 (6623 percent of the allowable change), then c2 can only be decreased by as much as 1 (3313 percent of the allowable change). These maximum changes revise the objective function to either Z 4.5x1 3x2 or Z 6x1 4x2, which causes the optimal objective function line in Fig. 6.3 to rotate clockwise until it coincides with the constraint boundary equation 3x1 2x2 18. In general, when objective function coefficients change in the same direction, it is possible for the percentages of allowable changes to sum to more than 100 percent without changing the optimal solution. We will give an example at the end of the discussion of Case 3. Case 2b—Introduction of a New Variable After solving for the optimal solution, we may discover that the linear programming formulation did not consider all the attractive alternative activities. Considering a new activity requires introducing a new variable with the appropriate coefficients into the objective function and constraints of the current model—which is Case 2b. The convenient way to deal with this case is to treat it just as if it were Case 2a! This is done by pretending that the new variable xj actually was in the original model with all its coefficients equal to zero (so that they still are zero in the final simplex tableau) and that xj is a nonbasic variable in the current BF solution. Therefore, if we change these zero coefficients to their actual values for the new variable, the procedure (including any reoptimization) does indeed become identical to that for Case 2a. In particular, all you have to do to check whether the current solution still is optimal is to check whether the complementary basic solution y* satisfies the one new

274

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

dual constraint that corresponds to the new variable in the primal problem. We already have described this approach and then illustrated it for the Wyndor Glass Co. problem in Sec. 6.5. Case 3—Changes in the Coefficients of a Basic Variable Now suppose that the variable xj (fixed j) under consideration is a basic variable in the optimal solution shown by the final simplex tableau. Case 3 assumes that the only changes in the current model are made to the coefficients of this variable. Case 3 differs from Case 2a because of the requirement that a simplex tableau be in proper form from Gaussian elimination. This requirement allows the column for a nonbasic variable to be anything, so it does not affect Case 2a. However, for Case 3, the basic variable xj must have a coefficient of 1 in its row of the simplex tableau and a coefficient of 0 in every other row (including row 0). Therefore, after the changes in the xj column of the final simplex tableau have been calculated,1 it probably will be necessary to apply Gaussian elimination to restore this form, as illustrated in Table 6.20. In turn, this step probably will change the value of the current basic solution and may make it either infeasible or nonoptimal (so reoptimization may be needed). Consequently, all the steps of the overall procedure summarized at the end of Sec. 6.6 are required for Case 3. Before Gaussian elimination is applied, the formulas for revising the xj column are the same as for Case 2a, as summarized below. Coefficient of xj in final row 0: Coefficient of xj in final rows 1 to m:

z j* cj y*A j cj. A*j S*A j.

Example (Variation 5 of the Wyndor Model). Because x2 is a basic variable in Table 6.21 for Variation 2 of the Wyndor Glass Co. model, sensitivity analysis of its coefficients fits Case 3. Given the current optimal solution (x1 0, x2 9), product 2 is the only new product that should be introduced, and its production rate should be relatively large. Therefore, the key question now is whether the initial estimates that led to the coefficients of x2 in the current model (Variation 2) could have overestimated the attractiveness of product 2 so much as to invalidate this conclusion. This question can be tested by checking the most pessimistic set of reasonable estimates for these coefficients, which turns out to be c2 3, a22 3, and a32 4. Consequently, the changes to be investigated (Variation 5 of the Wyndor model) are c2 5 → c2 3,

0 A2 2 → A 2 2

0 3 . 4

The graphical effect of these changes is that the feasible region changes from the one shown in Fig. 6.3 to the one in Fig. 6.6. The optimal solution in Fig. 6.3 is (x1, x2) (0, 9), which is the corner-point solution lying at the intersection of the x1 0 and 3x1 2x2 18 constraint boundaries. With the revision of the constraints, the corre1

For the relatively sophisticated reader, we should point out a possible pitfall for Case 3 that would be discovered at this point. Specifically, the changes in the initial tableau can destroy the linear independence of the columns of coefficients of basic variables. This event occurs only if the unit coefficient of the basic variable xj in the final tableau has been changed to zero at this point, in which case more extensive simplex method calculations must be used for Case 3.

6.7 APPLYING SENSITIVITY ANALYSIS

275

sponding corner-point solution in Fig. 6.6 is (0, 92 ). However, this solution no longer is optimal, because the revised objective function of Z 3x1 3x2 now yields a new optimal solution of (x1, x2) (4, 32 ). Analysis of Variation 5. Now let us see how we draw these same conclusions algebraically. Because the only changes in the model are in the coefficients of x2, the only resulting changes in the final simplex tableau (Table 6.21) are in the x2 column. Therefore, the above formulas are used to recompute just this column. 0 z2 c2 y*A 2 c2 [0, 0, 52] 3 3 7. 4 1 A2* S*A 2 0 0

FIGURE 6.6 Feasible region for Variation 5 of the Wyndor model where Variation 2 (Fig. 6.3) has been revised so c2 5 3, a22 2 3, and a32 2 4.

0 0 1

x2 x1 0

2x2 24

12

10 x1 4

(0, 9)

3x2 24

8

6

(0, 92)

3x1 2x2 18

4

3x1 4x2 18 3 (4, 2 )optimal

2 Feasible region

x2 0 0

2

4

6

8

x1

0 0 0 1 3 2 . 2 1 4 1

276

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.24 Sensitivity analysis procedure applied to Variation 5 of the Wyndor Glass Co. model Coefficient of:

Revised final tableau

Converted to proper form

New final tableau after reoptimization (only one iteration of the simplex method needed in this case)

Basic Variable

Eq.

Z

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4

(3)

0

x1 9 2 1 3 2 3 3 4 1 3 4 9 4

x2

x3

x4

7

0

0

0

1

0

2

0

0

1

0

1

Z

(0)

1

x3

(1)

0

0

0

0

0

1

0

x2

(2)

0

1

0

0

x4

(3)

0

0

0

1

Z

(0)

1

0

0

x1

(1)

0

1

0

x2

(2)

0

0

1

x4

(3)

0

0

0

3 4 1 3 4 9 4

0 0 0 1

x5 5 2 0 1 2

Right Side 45 4 9

1

6

3 4 0 1 4 3 4

27 2 4 9 2 21 2

3 4 0 1 4 3 4

33 2 4 3 2 39 2

(Equivalently, incremental analysis with c2 2, a22 1, and a32 2 can be used in the same way to obtain this column.) The resulting revised final tableau is shown at the top of Table 6.24. Note that the new coefficients of the basic variable x2 do not have the required values, so the conversion to proper form from Gaussian elimination must be applied next. This step involves dividing row 2 by 2, subtracting 7 times the new row 2 from row 0, and adding the new row 2 to row 3. The resulting second tableau in Table 6.24 gives the new value of the current basic solution, namely, x3 4, x2 92, x4 221 (x1 0, x5 0). Since all these variables are nonnegative, the solution is still feasible. However, because of the negative coefficient of x1 in row 0, we know that it is no longer optimal. Therefore, the simplex method would be applied to this tableau, with this solution as the initial BF solution, to find the new optimal solution. The initial entering basic variable is x1, with x3 as the leaving basic variable. Just one iteration is needed in this case to reach the new optimal solution x1 4, x2 32, x4 329 (x3 0, x5 0), as shown in the last tableau of Table 6.24. All this analysis suggests that c2, a22, and a32 are relatively sensitive parameters. However, additional data for estimating them more closely can be obtained only by conducting a pilot run. Therefore, the OR team recommends that production of product 2 be ini-

6.7 APPLYING SENSITIVITY ANALYSIS

277

tiated immediately on a small scale (x2 32) and that this experience be used to guide the decision on whether the remaining production capacity should be allocated to product 2 or product 1. The Allowable Range to Stay Optimal. For Case 2a, we described how to find the allowable range to stay optimal for any cj such that xj is a nonbasic variable for the current optimal solution (before cj is changed). When xj is a basic variable instead, the procedure is somewhat more involved because of the need to convert to proper form from Gaussian elimination before testing for optimality. To illustrate the procedure, consider Variation 5 of the Wyndor Glass Co. model (with c2 3, a22 3, a23 4) that is graphed in Fig. 6.6 and solved in Table 6.24. Since x2 is a basic variable for the optimal solution (with c2 3) given at the bottom of this table, the steps needed to find the allowable range to stay optimal for c2 are the following: 1. Since x2 is a basic variable, note that its coefficient in the new final row 0 (see the bottom tableau in Table 6.24) is automatically z2* c2 0 before c2 is changed from its current value of 3. 2. Now increment c2 3 by c2 (so c2 3 c2). This changes the coefficient noted in step 1 to z2* c2 c2, which changes row 0 to

3 3 33 Row 0 0, c2, , 0, . 4 4 2 3. With this coefficient now not zero, we must perform elementary row operations to restore proper form from Gaussian elimination. In particular, add to row 0 the product, c2 times row 2, to obtain the new row 0, as shown below.

0, c , 34, c 0, 34 c 3 1 0, c , c , 0, c 4 4 2

2

2

2

2

2

33 2 3 c2 2

3 3 3 1 33 3 New row 0 0, 0, c2, 0, c2 c2 4 4 4 4 2 2

4. Using this new row 0, solve for the range of values of c2 that keeps the coefficients of the nonbasic variables (x3 and x5) nonnegative. 3 3 c2 0 4 4 3 1 c2 0 4 4

3 3 ⇒ c2 4 4 1 3 ⇒ c2 4 4

⇒ c2 1. ⇒ c2 3.

Thus, the range of values is 3 c2 1. 5. Since c2 3 c2, add 3 to this range of values, which yields 0 c2 4 as the allowable range to stay optimal for c2.

278

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

With just two decision variables, this allowable range can be verified graphically by using Fig. 6.6 with an objective function of Z 3x1 c2 x2. With the current value of c2 3, the optimal solution is (4, 23). When c2 is increased, this solution remains optimal only for c2 4. For c2 4, (0, 29) becomes optimal (with a tie at c2 4), because of the constraint boundary 3x1 4x2 18. When c2 is decreased instead, (4, 23) remains optimal only for c2 0. For c2 0, (4, 0) becomes optimal because of the constraint boundary x1 4. In a similar manner, the allowable range to stay optimal for c1 (with c2 fixed at 3) can be derived either algebraically or graphically to be c1 94. (Problem 6.7-13 asks you to verify this both ways.) Thus, the allowable decrease for c1 from its current value of 3 is only 34. However, it is possible to decrease c1 by a larger amount without changing the optimal solution if c2 also decreases sufficiently. For example, suppose that both c1 and c2 are decreased by 1 from their current value of 3, so that the objective function changes from Z 3x1 3x2 to Z 2x1 2x2. According to the 100 percent rule for simultaneous changes in objective function coefficients, the percentages of allowable changes are 13313 percent and 3313 percent, respectively, which sum to far over 100 percent. However, the slope of the objective function line has not changed at all, so (4, 32) still is optimal. Case 4—Introduction of a New Constraint In this case, a new constraint must be introduced to the model after it has already been solved. This case may occur because the constraint was overlooked initially or because new considerations have arisen since the model was formulated. Another possibility is that the constraint was deleted purposely to decrease computational effort because it appeared to be less restrictive than other constraints already in the model, but now this impression needs to be checked with the optimal solution actually obtained. To see if the current optimal solution would be affected by a new constraint, all you have to do is to check directly whether the optimal solution satisfies the constraint. If it does, then it would still be the best feasible solution (i.e., the optimal solution), even if the constraint were added to the model. The reason is that a new constraint can only eliminate some previously feasible solutions without adding any new ones. If the new constraint does eliminate the current optimal solution, and if you want to find the new solution, then introduce this constraint into the final simplex tableau (as an additional row) just as if this were the initial tableau, where the usual additional variable (slack variable or artificial variable) is designated to be the basic variable for this new row. Because the new row probably will have nonzero coefficients for some of the other basic variables, the conversion to proper form from Gaussian elimination is applied next, and then the reoptimization step is applied in the usual way. Just as for some of the preceding cases, this procedure for Case 4 is a streamlined version of the general procedure summarized at the end of Sec. 6.6. The only question to be addressed for this case is whether the previously optimal solution still is feasible, so step 5 (optimality test) has been deleted. Step 4 (feasibility test) has been replaced by a much quicker test of feasibility (does the previously optimal solution satisfy the new constraint?) to be performed right after step 1 (revision of model). It is only if this test provides a negative answer, and you wish to reoptimize, that steps 2, 3, and 6 are used (revision of final tableau, conversion to proper form from Gaussian elimination, and reoptimization).

6.7 APPLYING SENSITIVITY ANALYSIS

279

Example (Variation 6 of the Wyndor Model). To illustrate this case, we consider Variation 6 of the Wyndor Glass Co. model, which simply introduces the new constraint 2x1 3x2 24 into the Variation 2 model given in Table 6.21. The graphical effect is shown in Fig. 6.7. The previous optimal solution (0, 9) violates the new constraint, so the optimal solution changes to (0, 8). To analyze this example algebraically, note that (0, 9) yields 2x1 3x2 27 24, so this previous optimal solution is no longer feasible. To find the new optimal solution, add the new constraint to the current final simplex tableau as just described, with the slack variable x6 as its initial basic variable. This step yields the first tableau shown in Table 6.25. The conversion to proper form from Gaussian elimination then requires subtracting from the new row the product, 3 times row 2, which identifies the current basic solution x3 4, x2 9, x4 6, x6 3 (x1 0, x5 0), as shown in the second tableau. Applying the dual simplex method (described in Sec. 7.1) to this tableau then leads in just one iteration (more are sometimes needed) to the new optimal solution in the last tableau of Table 6.25.

FIGURE 6.7 Feasible region for Variation 6 of the Wyndor model where Variation 2 (Fig. 6.3) has been revised by adding the new constraint, 2x1 3x2 24.

x2

x1 0

14

2x2 24

12

10 x1 4

(0, 9) 8

(0, 8) optimal

6

2x1 3x2 24

4

2

Feasible region 3x1 2x2 18

0

2

4

6

8

x2 0 10

12

14

x1

280

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.25 Sensitivity analysis procedure applied to Variation 6 of the Wyndor Glass Co. model Coefficient of:

Revised final tableau

Converted to proper form

New final tableau after reoptimization (only one iteration of dual simplex method needed in this case)

Basic Variable

Eq.

Z

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4 x6

(3) New

0 0

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4

(3)

0

x6

New

0

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4

(3)

0

x5

New

0

x1 9 2 1 3 2 3 2 9 2 1 3 2 3 5 2 1 3 1 2 3 4 3 5 3

x2

x3

x4

0

0

0

0

1

0

x5 5 2 0 1 2 1 0

1

0

0

0 3

0 0

1 0

0

0

0

0

1

0

1

0

0

0

0

1

0

0

0

0

0

0

0

0

1

0

0

1

0

0

0

0

0

1

0

0

0

0

1

5 2 0 1 2 1 3 2

x6

Right Side

0

45

0

4

0

9

0 1

6 24

0

45

0

4

0

9

0

6

1

3

5 3 0 1 3 2 3 2 3

40 4 8 8 2

Systematic Sensitivity Analysis—Parametric Programming So far we have described how to test specific changes in the model parameters. Another common approach to sensitivity analysis is to vary one or more parameters continuously over some interval(s) to see when the optimal solution changes. For example, with Variation 2 of the Wyndor Glass Co. model, rather than beginning by testing the specific change from b2 12 to b2 24, we might instead set b2 12 and then vary continuously from 0 to 12 (the maximum value of interest). The geometric interpretation in Fig. 6.3 is that the 2x2 12 constraint line is being shifted upward to 2x2 12 , with being increased from 0 to 12. The result is that the original optimal CPF solution (2, 6) shifts up the 3x1 2x2 18 constraint line toward (2, 12). This corner-point solution remains optimal as long as it is still feasible (x1 0), after which (0, 9) becomes the optimal solution. The algebraic calculations of the effect of having b2 are directly analogous to those for the Case 1 example where b2 12. In particular, we use the expressions for Z* and b* given for Case 1,

6.7 APPLYING SENSITIVITY ANALYSIS

281

Z* y*b b* S*b where b now is 4 12 b 18 and where y* and S* are given in the boxes in the middle tableau in Table 6.19. These equations indicate that the optimal solution is 3 Z* 36 2 1 x3 2 3 1 x2 6 2 1 x1 2 3

(x4 0, x5 0)

for small enough that this solution still is feasible, i.e., for 6. For 6, the dual simplex method (described in Sec. 7.1) yields the tableau shown in Table 6.21 except for the value of x4. Thus, Z 45, x3 4, x2 9 (along with x1 0, x5 0), and the expression for b* yields x4 b3* 0(4) 1(12 ) 1(18) 6 . This information can then be used (along with other data not incorporated into the model on the effect of increasing b2) to decide whether to retain the original optimal solution and, if not, how much to increase b2. In a similar way, we can investigate the effect on the optimal solution of varying several parameters simultaneously. When we vary just the bi parameters, we express the new value bi in terms of the original value bi as follows: bi bi i,

for i 1, 2, . . . , m,

where the i values are input constants specifying the desired rate of increase (positive or negative) of the corresponding right-hand side as is increased. For example, suppose that it is possible to shift some of the production of a current Wyndor Glass Co. product from Plant 2 to Plant 3, thereby increasing b2 by decreasing b3. Also suppose that b3 decreases twice as fast as b2 increases. Then b2 12 b3 18 2, where the (nonnegative) value of measures the amount of production shifted. (Thus, 1 0, 2 1, and 3 2 in this case.) In Fig. 6.3, the geometric interpretation is that as is increased from 0, the 2x2 12 constraint line is being pushed up to 2x2 12 (ignore the 2x2 24 line) and simultaneously the 3x1 2x2 18 constraint line is being

282

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

pushed down to 3x1 2x2 18 2. The original optimal CPF solution (2, 6) lies at the intersection of the 2x2 12 and 3x1 2x2 18 lines, so shifting these lines causes this corner-point solution to shift. However, with the objective function of Z 3x1 5x2, this corner-point solution will remain optimal as long as it is still feasible (x1 0). An algebraic investigation of simultaneously changing b2 and b3 in this way again involves using the formulas for Case 1 (treating as representing an unknown number) to calculate the resulting changes in the final tableau (middle of Table 6.19), namely, 4 [0, , 1] 12 36 12, Z* y*b 18 2 3 2

1 b* S*b 0 0

1 3 1 2 1 3

2 13 4 0 12 6 12 . 1 2 3 18 2

Therefore, the optimal solution becomes 1 Z* 36 2 x3 2

(x4 0,

1 x2 6 2 x1 2

x5 0)

for small enough that this solution still is feasible, i.e., for 2. (Check this conclusion in Fig. 6.3.) However, the fact that Z decreases as increases from 0 indicates that the best choice for is 0, so none of the possible shifting of production should be done. The approach to varying several cj parameters simultaneously is similar. In this case, we express the new value cj in terms of the original value of cj as cj cj j,

for j 1, 2, . . . , n,

where the j are input constants specifying the desired rate of increase (positive or negative) of cj as is increased. To illustrate this case, reconsider the sensitivity analysis of c1 and c2 for the Wyndor Glass Co. problem that was performed earlier in this section. Starting with Variation 2 of the Wyndor model presented in Table 6.21 and Fig. 6.3, we separately considered the effect of changing c1 from 3 to 4 (its most optimistic estimate) and c2 from 5 to 3 (its most pessimistic estimate). Now we can simultaneously consider both changes, as well as various intermediate cases with smaller changes, by setting c1 3

and

c2 5 2,

where the value of measures the fraction of the maximum possible change that is made. The result is to replace the original objective function Z 3x1 5x2 by a function of Z() (3 )x1 (5 2)x2,

6.7 APPLYING SENSITIVITY ANALYSIS

283

so the optimization now can be performed for any desired (fixed) value of between 0 and 1. By checking the effect as increases from 0 to 1, we can determine just when and how the optimal solution changes as the error in the original estimates of these parameters increases. Considering these changes simultaneously is especially appropriate if there are factors that cause the parameters to change together. Are the two products competitive in some sense, so that a larger-than-expected unit profit for one implies a smaller-thanexpected unit profit for the other? Are they both affected by some exogenous factor, such as the advertising emphasis of a competitor? Is it possible to simultaneously change both unit profits through appropriate shifting of personnel and equipment? In the feasible region shown in Fig. 6.3, the geometric interpretation of changing the objective function from Z 3x1 5x2 to Z() (3 )x1 (5 2)x2 is that we are changing the slope of the original objective function line (Z 45 3x1 5x2) that passes through the optimal solution (0, 9). If is increased enough, this slope will change sufficiently that the optimal solution will switch from (0, 9) to another CPF solution (4, 3). (Check graphically whether this occurs for 1.) The algebraic procedure for dealing simultaneously with these two changes ( c1 and c2 2) is shown in Table 6.26. Although the changes now are expressed in terms of rather than specific numerical amounts, is treated just as an unknown number. The table displays just the relevant rows of the tableaux involved (row 0 and the row for the basic variable x2). The first tableau shown is just the final tableau for the current version of the model (before c1 and c2 are changed) as given in Table 6.21. Refer to the formulas in Table 6.17. The only changes in the revised final tableau shown next are that c1 and c2 are subtracted from the row 0 coefficients of x1 and x2, respectively. To convert this tableau to proper form from Gaussian elimination, we subtract 2 times row 2 from row 0, which yields the last tableau shown. The expressions in terms of for the coeffiTABLE 6.26 Dealing with c1 and c2 2 for Variation 2 of the Wyndor model as given in Table 6.21 Coefficient of: Basic Variable

Eq.

Z

Z

(0)

1

x2

(2)

0

Z()

(0)

1

x2

(2)

0

Z()

(0)

1

x2

(2)

0

Final tableau

Revised final tableau when c1 and c2 2

Converted to proper form

x1 9 2 3 2 9 2 3 2 9 4 2 3 2

x2

x3

x4

0

0

0

1

0

0

2

0

0

1

0

0

0

0

0

1

0

0

x5 5 2 1 2 5 2 1 2 5 2 1 2

Right Side 45 9

45 9 45 18 9

284

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

cients of nonbasic variables x1 and x5 in row 0 of this tableau show that the current BF solution remains optimal for 98. Because 1 is the maximum realistic value of , this indicates that c1 and c2 together are insensitive parameters with respect to the Variation 2 model in Table 6.21. There is no need to try to estimate these parameters more closely unless other parameters change (as occurred for Variation 5 of the Wyndor model). As we discussed in Sec. 4.7, this way of continuously varying several parameters simultaneously is referred to as parametric linear programming. Section 7.2 presents the complete parametric linear programming procedure (including identifying new optimal solutions for larger values of ) when just the cj parameters are being varied and then when just the bi parameters are being varied. Some linear programming software packages also include routines for varying just the coefficients of a single variable or just the parameters of a single constraint. In addition to the other applications discussed in Sec. 4.7, these procedures provide a convenient way of conducting sensitivity analysis systematically.

6.8

CONCLUSIONS Every linear programming problem has associated with it a dual linear programming problem. There are a number of very useful relationships between the original (primal) problem and its dual problem that enhance our ability to analyze the primal problem. For example, the economic interpretation of the dual problem gives shadow prices that measure the marginal value of the resources in the primal problem and provides an interpretation of the simplex method. Because the simplex method can be applied directly to either problem in order to solve both of them simultaneously, considerable computational effort sometimes can be saved by dealing directly with the dual problem. Duality theory, including the dual simplex method for working with superoptimal basic solutions, also plays a major role in sensitivity analysis. The values used for the parameters of a linear programming model generally are just estimates. Therefore, sensitivity analysis needs to be performed to investigate what happens if these estimates are wrong. The fundamental insight of Sec. 5.3 provides the key to performing this investigation efficiently. The general objectives are to identify the sensitive parameters that affect the optimal solution, to try to estimate these sensitive parameters more closely, and then to select a solution that remains good over the range of likely values of the sensitive parameters. This analysis is a very important part of most linear programming studies.

SELECTED REFERENCES 1. Bazaraa, M. S., J. J. Jarvis, and H. D. Sherali: Linear Programming and Network Flows, 2d ed., Wiley, New York, 1990. 2. Dantzig, G. B., and M. N. Thapa: Linear Programming 1: Introduction, Springer, New York, 1997. 3. Hillier, F. S., M. S. Hillier, and G. J. Lieberman: Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Irwin/McGraw-Hill, Burr Ridge, IL, 2000, chap. 4. 4. Vanderbei, R. J.: Linear Programming: Foundations and Extensions, Kluwer Academic Publishers, Boston, MA, 1996.

CHAPTER 6 PROBLEMS

285

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE A Demonstration Example in OR Tutor: Sensitivity Analysis

Interactive Routines: Enter or Revise a General Linear Programming Model Solve Interactively by the Simplex Method Sensitivity Analysis

An Excel Add-In: Premium Solver

Files (Chapter 3) for Solving the Wyndor Example: Excel File LINGO/LINDO File MPL/CPLEX File

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The demonstration example listed above may be helpful. I: We suggest that you use the corresponding interactive routine listed above (the printout records your work). C: Use the computer with any of the software options available to you (or as instructed by your instructor) to solve the problem automatically. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 6.1-1. Construct the primal-dual table and the dual problem for each of the following linear programming models fitting our standard form. (a) Model in Prob. 4.1-6 (b) Model in Prob. 4.7-8 6.1-2.* Construct the dual problem for each of the following linear programming models fitting our standard form. (a) Model in Prob. 3.1-5 (b) Model in Prob. 4.7-6

6.1-3. Consider the linear programming model in Prob. 4.5-4. (a) Construct the primal-dual table and the dual problem for this model. (b) What does the fact that Z is unbounded for this model imply about its dual problem? 6.1-4. For each of the following linear programming models, give your recommendation on which is the more efficient way (probably) to obtain an optimal solution: by applying the simplex method directly to this primal problem or by applying the simplex method directly to the dual problem instead. Explain. (a) Maximize Z 10x1 4x2 7x3, subject to 3x1 x1 5x1 x1 2x1

x2 2x2 x2 x2 x2

2x3 3x3 2x3 x3 x3

25 25 40 90 20

and x1 0,

x2 0,

x3 0.

286

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Z 2x1 5x2 3x3 4x4 x5,

(b) Maximize

Maximize

subject to x1 3x2 2x3 3x4 x5 6 4x1 6x2 5x3 7x4 x5 15 and xj 0,

for j 1, 2, 3, 4, 5.

6.1-5. Consider the following problem. Maximize

Z x1 2x2 x3,

subject to x1 x2 2x3 12 x1 x2 x3 1 and x1 0,

x2 0,

x3 0.

(a) Construct the dual problem. (b) Use duality theory to show that the optimal solution for the primal problem has Z 0. 6.1-6. Consider the following problem. Maximize

6.1-8. Consider the following problem.

Z 2x1 6x2 9x3,

subject to x1x1 x3 3 x1x2 2x3 5

(resource 1) (resource 2)

Z x1 2x2,

subject to x1 x2 2 4x1 x2 4 and x1 0,

x2 0.

(a) Demonstrate graphically that this problem has no feasible solutions. (b) Construct the dual problem. (c) Demonstrate graphically that the dual problem has an unbounded objective function. 6.1-9. Construct and graph a primal problem with two decision variables and two functional constraints that has feasible solutions and an unbounded objective function. Then construct the dual problem and demonstrate graphically that it has no feasible solutions. 6.1-10. Construct a pair of primal and dual problems, each with two decision variables and two functional constraints, such that both problems have no feasible solutions. Demonstrate this property graphically. 6.1-11. Construct a pair of primal and dual problems, each with two decision variables and two functional constraints, such that the primal problem has no feasible solutions and the dual problem has an unbounded objective function.

and x1 0,

x2 0,

x3 0.

(a) Construct the dual problem for this primal problem. (b) Solve the dual problem graphically. Use this solution to identify the shadow prices for the resources in the primal problem. C (c) Confirm your results from part (b) by solving the primal problem automatically by the simplex method and then identifying the shadow prices. 6.1-7. Follow the instructions of Prob. 6.1-6 for the following problem. Maximize

Z x1 3x2 2x3,

subject to 2x1 2x2 2x3 6 2x1 x2 2x3 4

(resource 1) (resource 2)

and x1 0,

x2 0,

x3 0.

6.1-12. Use the weak duality property to prove that if both the primal and the dual problem have feasible solutions, then both must have an optimal solution. 6.1-13. Consider the primal and dual problems in our standard form presented in matrix notation at the beginning of Sec. 6.1. Use only this definition of the dual problem for a primal problem in this form to prove each of the following results. (a) The weak duality property presented in Sec. 6.1. (b) If the primal problem has an unbounded feasible region that permits increasing Z indefinitely, then the dual problem has no feasible solutions. 6.1-14. Consider the primal and dual problems in our standard form presented in matrix notation at the beginning of Sec. 6.1. Let y* denote the optimal solution for this dual problem. Suppose that b is then replaced by b . Let x denote the optimal solution for the new primal problem. Prove that cx y*b .

CHAPTER 6 PROBLEMS

287

6.1-15. For any linear programming problem in our standard form and its dual problem, label each of the following statements as true or false and then justify your answer. (a) The sum of the number of functional constraints and the number of variables (before augmenting) is the same for both the primal and the dual problems. (b) At each iteration, the simplex method simultaneously identifies a CPF solution for the primal problem and a CPF solution for the dual problem such that their objective function values are the same. (c) If the primal problem has an unbounded objective function, then the optimal value of the objective function for the dual problem must be zero.

6.3-3. Consider the primal and dual problems for the Wyndor Glass Co. example given in Table 6.1. Using Tables 5.5, 5.6, 6.8, and 6.9, construct a new table showing the eight sets of nonbasic variables for the primal problem in column 1, the corresponding sets of associated variables for the dual problem in column 2, and the set of nonbasic variables for each complementary basic solution in the dual problem in column 3. Explain why this table demonstrates the complementary slackness property for this example.

6.2-1. Consider the simplex tableaux for the Wyndor Glass Co. problem given in Table 4.8. For each tableau, give the economic interpretation of the following items: (a) Each of the coefficients of the slack variables (x3, x4, x5) in row 0 (b) Each of the coefficients of the decision variables (x1, x2) in row 0 (c) The resulting choice for the entering basic variable (or the decision to stop after the final tableau)

6.3-5. Consider the following problem.

6.3-1.* Consider the following problem. Maximize

Z 6x1 8x2,

subject to 5x1 2x2 20 x1 2x2 10 and x1 0,

x2 0.

(a) Construct the dual problem for this primal problem. (b) Solve both the primal problem and the dual problem graphically. Identify the CPF solutions and corner-point infeasible solutions for both problems. Calculate the objective function values for all these solutions. (c) Use the information obtained in part (b) to construct a table listing the complementary basic solutions for these problems. (Use the same column headings as for Table 6.9.) I (d) Work through the simplex method step by step to solve the primal problem. After each iteration (including iteration 0), identify the BF solution for this problem and the complementary basic solution for the dual problem. Also identify the corresponding corner-point solutions. 6.3-2. Consider the model with two functional constraints and two variables given in Prob. 4.1-5. Follow the instructions of Prob. 6.3-1 for this model.

6.3-4. Suppose that a primal problem has a degenerate BF solution (one or more basic variables equal to zero) as its optimal solution. What does this degeneracy imply about the dual problem? Why? Is the converse also true?

Z 2x1 4x2,

Maximize subject to x1 x2 1 and x1 0,

x2 0.

(a) Construct the dual problem, and then find its optimal solution by inspection. (b) Use the complementary slackness property and the optimal solution for the dual problem to find the optimal solution for the primal problem. (c) Suppose that c1, the coefficient of x1 in the primal objective function, actually can have any value in the model. For what values of c1 does the dual problem have no feasible solutions? For these values, what does duality theory then imply about the primal problem? 6.3-6. Consider the following problem. Maximize

Z 2x1 7x2 4x3,

subject to x1 2x2 x3 10 3x1 3x2 2x3 10 and x1 0,

x2 0,

x3 0.

(a) Construct the dual problem for this primal problem. (b) Use the dual problem to demonstrate that the optimal value of Z for the primal problem cannot exceed 25. (c) It has been conjectured that x2 and x3 should be the basic variables for the optimal solution of the primal problem. Directly derive this basic solution (and Z) by using Gaussian elimination. Simultaneously derive and identify the complementary ba-

288

I

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

sic solution for the dual problem by using Eq. (0) for the primal problem. Then draw your conclusions about whether these two basic solutions are optimal for their respective problems. (d) Solve the dual problem graphically. Use this solution to identify the basic variables and the nonbasic variables for the optimal solution of the primal problem. Directly derive this solution, using Gaussian elimination.

6.3-7.* Reconsider the model of Prob. 6.1-4b. (a) Construct its dual problem. (b) Solve this dual problem graphically. (c) Use the result from part (b) to identify the nonbasic variables and basic variables for the optimal BF solution for the primal problem. (d) Use the results from part (c) to obtain the optimal solution for the primal problem directly by using Gaussian elimination to solve for its basic variables, starting from the initial system of equations [excluding Eq. (0)] constructed for the simplex method and setting the nonbasic variables to zero. (e) Use the results from part (c) to identify the defining equations (see Sec. 5.1) for the optimal CPF solution for the primal problem, and then use these equations to find this solution. 6.3-8. Consider the model given in Prob. 5.3-13. (a) Construct the dual problem. (b) Use the given information about the basic variables in the optimal primal solution to identify the nonbasic variables and basic variables for the optimal dual solution. (c) Use the results from part (b) to identify the defining equations (see Sec. 5.1) for the optimal CPF solution for the dual problem, and then use these equations to find this solution. (d) Solve the dual problem graphically to verify your results from part (c). 6.3-9. Consider the model given in Prob. 3.1-4. (a) Construct the dual problem for this model. (b) Use the fact that (x1, x2) (13, 5) is optimal for the primal problem to identify the nonbasic variables and basic variables for the optimal BF solution for the dual problem. (c) Identify this optimal solution for the dual problem by directly deriving Eq. (0) corresponding to the optimal primal solution identified in part (b). Derive this equation by using Gaussian elimination. (d) Use the results from part (b) to identify the defining equations (see Sec. 5.1) for the optimal CPF solution for the dual problem. Verify your optimal dual solution from part (c) by checking to see that it satisfies this system of equations. 6.3-10. Suppose that you also want information about the dual problem when you apply the revised simplex method (see Sec. 5.2) to the primal problem in our standard form.

(a) How would you identify the optimal solution for the dual problem? (b) After obtaining the BF solution at each iteration, how would you identify the complementary basic solution in the dual problem? 6.4-1. Consider the following problem. Maximize

Z x1 x2,

subject to x1 2x2 10 2x1 x2 2 and x2 0

(x1 unconstrained in sign).

(a) Use the SOB method to construct the dual problem. (b) Use Table 6.12 to convert the primal problem to our standard form given at the beginning of Sec. 6.1, and construct the corresponding dual problem. Then show that this dual problem is equivalent to the one obtained in part (a). 6.4-2. Consider the primal and dual problems in our standard form presented in matrix notation at the beginning of Sec. 6.1. Use only this definition of the dual problem for a primal problem in this form to prove each of the following results. (a) If the functional constraints for the primal problem Ax b are changed to Ax b, the only resulting change in the dual problem is to delete the nonnegativity constraints, y 0. (Hint: The constraints Ax b are equivalent to the set of constraints Ax b and Ax b.) (b) If the functional constraints for the primal problem Ax b are changed to Ax b, the only resulting change in the dual problem is that the nonnegativity constraints y 0 are replaced by nonpositivity constraints y 0, where the current dual variables are interpreted as the negative of the original dual variables. (Hint: The constraints Ax b are equivalent to Ax b.) (c) If the nonnegativity constraints for the primal problem x 0 are deleted, the only resulting change in the dual problem is to replace the functional constraints yA c by yA c. (Hint: A variable unconstrained in sign can be replaced by the difference of two nonnegative variables.) 6.4-3.* Construct the dual problem for the linear programming problem given in Prob. 4.6-4. 6.4-4. Consider the following problem. Minimize

Z x1 2x2,

subject to 2x1 x2 1 2x1 2x2 1

CHAPTER 6 PROBLEMS

and x1 0,

289

and x2 0.

(a) Construct the dual problem. (b) Use graphical analysis of the dual problem to determine whether the primal problem has feasible solutions and, if so, whether its objective function is bounded.

x1 0,

x2 0.

(a) Demonstrate graphically that this problem has an unbounded objective function. (b) Construct the dual problem. (c) Demonstrate graphically that the dual problem has no feasible solutions.

6.4-5. Consider the two versions of the dual problem for the radiation therapy example that are given in Tables 6.15 and 6.16. Review in Sec. 6.4 the general discussion of why these two versions are completely equivalent. Then fill in the details to verify this equivalency by proceeding step by step to convert the version in Table 6.15 to equivalent forms until the version in Table 6.16 is obtained.

6.5-1. Consider the model of Prob. 6.7-1. Use duality theory directly to determine whether the current basic solution remains optimal after each of the following independent changes. (a) The change in part (e) of Prob. 6.7-1 (b) The change in part (g) of Prob. 6.7-1

6.4-6. For each of the following linear programming models, use the SOB method to construct its dual problem. (a) Model in Prob. 4.6-3 (b) Model in Prob. 4.6-8 (c) Model in Prob. 4.6-18

6.5-2. Consider the model of Prob. 6.7-3. Use duality theory directly to determine whether the current basic solution remains optimal after each of the following independent changes. (a) The change in part (c) of Prob. 6.7-3 (b) The change in part ( f ) of Prob. 6.7-3

6.4-7. Consider the model with equality constraints given in Prob. 4.6-2. (a) Construct its dual problem. (b) Demonstrate that the answer in part (a) is correct (i.e., equality constraints yield dual variables without nonnegativity constraints) by first converting the primal problem to our standard form (see Table 6.12), then constructing its dual problem, and next converting this dual problem to the form obtained in part (a).

6.5-3. Consider the model of Prob. 6.7-4. Use duality theory directly to determine whether the current basic solution remains optimal after each of the following independent changes. (a) The change in part (b) of Prob. 6.7-4 (b) The change in part (d ) of Prob. 6.7-4

6.4-8.* Consider the model without nonnegativity constraints given in Prob. 4.6-16. (a) Construct its dual problem. (b) Demonstrate that the answer in part (a) is correct (i.e., variables without nonnegativity constraints yield equality constraints in the dual problem) by first converting the primal problem to our standard form (see Table 6.12), then constructing its dual problem, and finally converting this dual problem to the form obtained in part (a). 6.4-9. Consider the dual problem for the Wyndor Glass Co. example given in Table 6.1. Demonstrate that its dual problem is the primal problem given in Table 6.1 by going through the conversion steps given in Table 6.13. 6.4-10. Consider the following problem. Minimize

6.5-4. Reconsider part (d) of Prob. 6.7-6. Use duality theory directly to determine whether the original optimal solution is still optimal. 6.6-1.* Consider the following problem. Maximize subject to 6x1 3x2 5x3 25 3x1 4x2 5x3 20 and x1 0,

x1 2x2 2 x1 x2 4

x2 0,

x3 0.

The corresponding final set of equations yielding the optimal solution is (0) (1)

Z x1 3x2,

subject to

Z 3x1 x2 4x3,

(2)

Z

2x2

1 3 x4 x5 17 5 5

1 x1 x2 3

1 1 5 x4 x5 3 3 3

1 2 x2 x3 x4 x5 3. 5 5

(a) Identify the optimal solution from this set of equations. (b) Construct the dual problem.

290

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

(c) Identify the optimal solution for the dual problem from the final set of equations. Verify this solution by solving the dual problem graphically. (d) Suppose that the original problem is changed to Maximize

subject to 4y1 2y1 y1 y1

subject to

x2 0,

3y2 4 y2 3 2y2 1 y2 2

y1 0,

x3 0.

Use duality theory to determine whether the previous optimal solution is still optimal. (e) Use the fundamental insight presented in Sec. 5.3 to identify the new coefficients of x2 in the final set of equations after it has been adjusted for the changes in the original problem given in part (d ). (f) Now suppose that the only change in the original problem is that a new variable xnew has been introduced into the model as follows: Maximize

and

and x1 0,

W 5y1 4y2,

Minimize

Z 3x1 3x2 4x3,

6x1 2x2 5x3 25 3x1 3x2 5x3 20

6.6-3. Consider the following problem.

D,I

y2 0.

Because this primal problem has more functional constraints than variables, suppose that the simplex method has been applied directly to its dual problem. If we let x5 and x6 denote the slack variables for this dual problem, the resulting final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

x6

Right Side

Z x2 x4

(0) (1) (2)

1 0 0

3 1 2

0 1 0

2 1 3

0 0 1

1 1 1

1 1 2

9 1 3

Z 3x1 x2 4x3 2xnew,

subject to 6x1 3x2 5x3 3xnew 25 3x1 4x2 5x3 2xnew 20 and x1 0,

x2 0,

x3 0,

xnew 0.

Use duality theory to determine whether the previous optimal solution, along with xnew 0, is still optimal. (g) Use the fundamental insight presented in Sec. 5.3 to identify the coefficients of xnew as a nonbasic variable in the final set of equations resulting from the introduction of xnew into the original model as shown in part ( f ). 6.6-2. Reconsider the model of Prob. 6.6-1. You are now to conduct sensitivity analysis by independently investigating each of the following six changes in the original model. For each change, use the sensitivity analysis procedure to revise the given final set of equations (in tableau form) and convert it to proper form from Gaussian elimination. Then test this solution for feasibility and for optimality. (Do not reoptimize.) (a) Change the right-hand side of constraint 1 to b1 15. (b) Change the right-hand side of constraint 2 to b2 5. (c) Change the coefficient of x2 in the objective function to c2 4. (d) Change the coefficient of x3 in the objective function to c3 3. (e) Change the coefficient of x2 in constraint 2 to a22 1. (f) Change the coefficient of x1 in constraint 1 to a11 10. D,I

For each of the following independent changes in the original primal model, you now are to conduct sensitivity analysis by directly investigating the effect on the dual problem and then inferring the complementary effect on the primal problem. For each change, apply the procedure for sensitivity analysis summarized at the end of Sec. 6.6 to the dual problem (do not reoptimize), and then give your conclusions as to whether the current basic solution for the primal problem still is feasible and whether it still is optimal. Then check your conclusions by a direct graphical analysis of the primal problem. (a) Change the objective function to W 3y1 5y2. (b) Change the right-hand sides of the functional constraints to 3, 5, 2, and 3, respectively. (c) Change the first constraint to 2y1 4y2 7. (d) Change the second constraint to 5y1 2y2 10. 6.7-1.* Consider the following problem.

D,I

Maximize

Z 5x1 5x2 13x3,

subject to x1 x2 3x3 20 12x1 4x2 10x3 90 and xj 0

( j 1, 2, 3).

CHAPTER 6 PROBLEMS

If we let x4 and x5 be the slack variables for the respective constraints, the simplex method yields the following final set of equations: (0) (1) (2)

Z

2x3 5x4 100. x1 x2 3x3 x4 20. 16x1 2x3 4x4 x5 = 10.

Now you are to conduct sensitivity analysis by independently investigating each of the following nine changes in the original model. For each change, use the sensitivity analysis procedure to revise this set of equations (in tableau form) and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. (Do not reoptimize.) (a) Change the right-hand side of constraint 1 to b1 30. (b) Change the right-hand side of constraint 2 to b2 70. (c) Change the right-hand sides to

b1 10 . 100 b2

(d) Change the coefficient of x3 in the objective function to c3 8.

291

6.7-2.* Reconsider the model of Prob. 6.7-1. Suppose that we now want to apply parametric linear programming analysis to this problem. Specifically, the right-hand sides of the functional constraints are changed to 20 2

(for constraint 1)

and 90

(for constraint 2),

where can be assigned any positive or negative values. Express the basic solution (and Z) corresponding to the original optimal solution as a function of . Determine the lower and upper bounds on before this solution would become infeasible. 6.7-3. Consider the following problem.

D,I

Z 2x1 x2 x3,

Maximize subject to

3x1 x2 x3 60 x1 x2 2x3 10 x1 x2 x3 20 and x1 0,

x2 0,

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After we apply the simplex method, the final simplex tableau is

(e) Change the coefficients of x1 to

c1 2 a11 0 . a21 5 (f) Change the coefficients of x2 to

c2 6 a12 2 . a22 5 (g) Introduce a new variable x6 with coefficients

c6 10 a16 3 . a26 5 (h) Introduce a new constraint 2x1 3x2 5x3 50. (Denote its slack variable by x6.) (i) Change constraint 2 to 10x1 5x2 10x3 100.

Coefficient of: Basic Variable

Eq.

Z

x1

x2

Z

(0)

1

0

0

x4

(1)

0

0

0

x1

(2)

0

1

0

x2

(3)

0

0

1

x3 3 2 1 1 2 3 2

x4 0 1 0 0

x5

x6

3 2 1 1 2 1 2

1 2 2 1 2 1 2

Right Side 25 10 15 5

Now you are to conduct sensitivity analysis by independently investigating each of the following six changes in the original model. For each change, use the sensitivity analysis procedure to revise this final tableau and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. If either test fails, reoptimize to find a new optimal solution.

292

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

(a) Change the right-hand sides

b1 60 b2 10 b3 20

from

to

b1 70 b2 20 . b3 10

form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. If either test fails, reoptimize to find a new optimal solution. (a) Change the right-hand sides to

(b) Change the coefficients of x1

c1 2 a11 3 a 1 21 a31 1

from

b 30. b1

c1 1 a11 2 a 2 . 21 a31 0

to

(b) Change the coefficients of x3 to

c3 2 a13 3 . a23 2

(c) Change the coefficients of x3

c3 1 a13 1 a 2 23 a33 1

from

to

c3 2 a13 3 a 1 . 23 a33 2

(d) Change the objective function to Z 3x1 2x2 3x3. (e) Introduce a new constraint 3x1 2x2 x3 30. (Denote its slack variable by x7.) (f) Introduce a new variable x8 with coefficients

c8 1 a18 2 a 1 . 28 a38 2

Maximize

(c) Change the coefficients of x1 to

c1 4 a11 3 . a21 2 (d) Introduce a new variable x6 with coefficients

c6 3 a16 1 . a26 2 (e) Change the objective function to Z x1 5x2 2x3. (f) Introduce a new constraint 3x1 2x2 3x3 25. (g) Change constraint 2 to x1 2x2 2x3 35.

6.7-4. Consider the following problem.

D,I

Z 2x1 7x2 3x3,

subject to

6.7-5. Reconsider the model of Prob. 6.7-4. Suppose that we now want to apply parametric linear programming analysis to this problem. Specifically, the right-hand sides of the functional constraints are changed to 30 3

x1 3x2 4x3 30 x1 4x2 x3 10

10

x1 0,

x2 0,

(for constraint 1)

and

and x3 0.

By letting x4 and x5 be the slack variables for the respective constraints, the simplex method yields the following final set of equations: (0) (1) (2)

20

2

x2 x3 2x5 20, x2 5x3 x4 x5 20, x1 4x2 x3 x5 10.

Z

Now you are to conduct sensitivity analysis by independently investigating each of the following seven changes in the original model. For each change, use the sensitivity analysis procedure to revise this set of equations (in tableau form) and convert it to proper

(for constraint 2),

where can be assigned any positive or negative values. Express the basic solution (and Z) corresponding to the original optimal solution as a function of . Determine the lower and upper bounds on before this solution would become infeasible. D,I

6.7-6. Consider the following problem. Maximize

Z 2x1 x2 x3,

subject to 3x1 2x2 2x3 15 x1 x2 x3 3 x1 x2 x3 4

CHAPTER 6 PROBLEMS

and x1 0,

x2 0,

x3 0.

If we let x4, x5, and x6 be the slack variables for the respective constraints, the simplex method yields the following final set of equations: (0) (1) (2) (3)

2x3 x4 x5 18, x2 5x3 x4 3x5 24, 2x3 x5 x6 7, x1 4x3 x4 2x5 21.

Z

Now you are to conduct sensitivity analysis by independently investigating each of the following eight changes in the original model. For each change, use the sensitivity analysis procedure to revise this set of equations (in tableau form) and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. If either test fails, reoptimize to find a new optimal solution. (a) Change the right-hand sides to

b1 10 b2 4 . b3 2

of type A, but the vendor providing these subassemblies would only be able to increase its supply rate from the current 2,000 per day to a maximum of 3,000 per day. Each toy requires only one subassembly of type B, but the vendor providing these subassemblies would be unable to increase its supply rate above the current level of 1,000 per day. Because no other vendors currently are available to provide these subassemblies, management is considering initiating a new production process internally that would simultaneously produce an equal number of subassemblies of the two types to supplement the supply from the two vendors. It is estimated that the company’s cost for producing one subassembly of each type would be $2.50 more than the cost of purchasing these subassemblies from the two vendors. Management wants to determine both the production rate of the toy and the production rate of each pair of subassemblies (one A and one B) that would maximize the total profit. The following table summarizes the data for the problem. Resource Usage per Unit of Each Activity

Resource

(b) Change the coefficient of x3 in the objective function to c3 2. (c) Change the coefficient of x1 in the objective function to c1 3. (d) Change the coefficients of x3 to

c3 4 a13 3 a 2 . 23 a33 1 (e) Change the coefficients of x1 and x2 to

c1 1 a11 1 a 2 21 a31 3

293

and

c2 2 a12 2 a 3 , 22 a32 2

respectively. (f) Change the objective function to Z 5x1 x2 3x3. (g) Change constraint 1 to 2x1 x2 4x3 12. (h) Introduce a new constraint 2x1 x2 3x3 60. 6.7-7. One of the products of the G. A. Tanner Company is a special kind of toy that provides an estimated unit profit of $3. Because of a large demand for this toy, management would like to increase its production rate from the current level of 1,000 per day. However, a limited supply of two subassemblies (A and B) from vendors makes this difficult. Each toy requires two subassemblies

Produce Produce Amount of Resource Toys Subassemblies Available

Subassembly A Subassembly B

$2 $1

.00$1 .00$1

Unit profit

$3

$2.50

3,000 1,000

(a) Formulate a linear programming model for this problem and use the graphical method to obtain its optimal solution. C (b) Use a software package based on the simplex method to solve for an optimal solution. C (c) Since the stated unit profits for the two activities are only estimates, management wants to know how much each of these estimates can be off before the optimal solution would change. Begin exploring this question for the first activity (producing toys) by using the same software package to resolve for an optimal solution and total profit as the unit profit for this activity increases in 50-cent increments from $2.00 to $4.00. What conclusion can be drawn about how much the estimate of this unit profit can differ in each direction from its original value of $3.00 before the optimal solution would change? C (d) Repeat part (c) for the second activity (producing subassemblies) by re-solving as the unit profit for this activity increases in 50-cent increments from $3.50 to $1.50 (with the unit profit for the first activity fixed at $3). C (e) Use the same software package to generate the usual output (as in Table 6.23) for sensitivity analysis of the unit profits.

294

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Use this output to obtain the allowable range to stay optimal for each unit profit. (f) Use graphical analysis to verify the allowable ranges obtained in part (e). (g) For each of the 16 combinations of unit profits considered in parts (c) and (d ) where both unit profits differ from their original estimates, use the 100 percent rule for simultaneous changes in objective function coefficients to determine if the original optimal solution must still be optimal. (h) For each of the combinations of unit profits considered in part (g) where it was found that the original optimal solution is not guaranteed to still be optimal, use graphical analysis to determine whether this solution is still optimal. 6.7-8. Reconsider Prob. 6.7-7. After further negotiations with each vendor, management of the G. A. Tanner Co. has learned that either of them would be willing to consider increasing their supply of their respective subassemblies over the previously stated maxima (3,000 subassemblies of type A per day and 1,000 of type B per day) if the company would pay a small premium over the regular price for the extra subassemblies. The size of the premium for each type of subassembly remains to be negotiated. The demand for the toy being produced is sufficiently high that 2,500 per day could be sold if the supply of subassemblies could be increased enough to support this production rate. Assume that the original estimates of unit profits given in Prob. 6.7-7 are accurate. (a) Formulate a linear programming model for this problem with the original maximum supply levels and the additional constraint that no more than 2,500 toys should be produced per day. Then use the graphical method to obtain its optimal solution. C (b) Use a software package based on the simplex method to solve for an optimal solution. C (c) Without considering the premium, use the same software package to determine the shadow price for the subassembly A constraint by solving the model again after increasing the maximum supply by 1. Use this shadow price to determine the maximum premium that the company should be willing to pay for each subassembly of this type. C (d) Repeat part (c) for the subassembly B constraint. C (e) Estimate how much the maximum supply of subassemblies of type A could be increased before the shadow price (and the corresponding premium) found in part (c) would no longer be valid by using the same software package to resolve for an optimal solution and the total profit (excluding the premium) as the maximum supply increases in increments of 100 from 3,000 to 4,000. C (f) Repeat part (e) for subassemblies of type B by re-solving as the maximum supply increases in increments of 100 from 1,000 to 2,000.

(g) Use the same software package to generate the usual output (as in Table 6.23) for sensitivity analysis of the supplies being made available of the subassemblies. Use this output to obtain the allowable range to stay feasible for each subassembly supply. (h) Use graphical analysis to verify the allowable ranges obtained in part (g). (i) For each of the four combinations where the maximum supply of subassembly A is either 3,500 or 4,000 and the maximum supply of subassembly B is either 1,500 or 2,000, use the 100 percent rule for simultaneous changes in right-hand sides to determine whether the original shadow prices definitely will still be valid. (j) For each of the combinations considered in part (i) where it was found that the original shadow prices are not guaranteed to still be valid, use graphical analysis to determine whether these shadow prices actually are still valid for predicting the effect of changing the right-hand sides. C

6.7-9 Consider the Distribution Unlimited Co. problem presented in Sec. 3.4 and summarized in Fig. 3.13. Although Fig. 3.13 gives estimated unit costs for shipping through the various shipping lanes, there actually is some uncertainty about what these unit costs will turn out to be. Therefore, before adopting the optimal solution given at the end of Sec. 3.4, management wants additional information about the effect of inaccuracies in estimating these unit costs. Use a computer package based on the simplex method to generate sensitivity analysis information preparatory to addressing the following questions. (a) Which of the unit shipping costs given in Fig. 3.13 has the smallest margin for error without invalidating the optimal solution given in Sec. 3.4? Where should the greatest effort be placed in estimating the unit shipping costs? (b) What is the allowable range to stay optimal for each of the unit shipping costs? (c) How should these allowable ranges be interpreted to management? (d) If the estimates change for more than one of the unit shipping costs, how can you use the generated sensitivity analysis information to determine whether the optimal solution might change?

C

6.7-10. Consider the Union Airways problem presented in Sec. 3.4, including the data given in Table 3.19. Management is about to begin negotiations on a new contract with the union that represents the company’s customer service agents. This might result in some small changes in the daily costs per agent given in Table 3.19 for the various shifts. Several possible changes listed below are being considered separately. In each case, management would like to know whether the change might

C

CHAPTER 6 PROBLEMS

result in the original optimal solution (given in Sec. 3.4) no longer being optimal. Answer this question in parts (a) to (e) by using a software package based on the simplex method to generate sensitivity analysis information. If the optimal solution might change, use the software package to re-solve for the optimal solution. (a) The daily cost per agent for Shift 2 changes from $160 to $165. (b) The daily cost per agent for Shift 4 changes from $180 to $170. (c) The changes in parts (a) and (b) both occur. (d) The daily cost per agent increases by $4 for shifts 2, 4, and 5, but decreases by $4 for shifts 1 and 3. (e) The daily cost per agent increases by 2 percent for each shift. 6.7-11. Consider the following problem.

6.7-13. Consider Variation 5 of the Wyndor Glass Co. model (see Fig. 6.6 and Table 6.24), where the changes in the parameter values given in Table 6.21 are c2 3, a22 3, and a32 4. Verify both algebraically and graphically that the allowable range to stay optimal for c1 is c1 94. 6.7-14. Consider the following problem. Z 3x1 x2 2x3,

Maximize subject to

x1 x2 2x3 20 2x1 x2 x3 10 and

Z c1x1 c2x2,

Maximize

295

x1 0,

subject to

x2 0,

x3 0.

Let x4 and x5 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is

2x1 x2 b1 x1 x2 b2 and x1 0,

x2 0.

Coefficient of:

Let x3 and x4 denote the slack variables for the respective functional constraints. When c1 3, c2 2, b1 30, and b2 10, the simplex method yields the following final simplex tableau.

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Right Side

Z x2 x1

(0) (1) (2)

1 0 0

0 0 1

0 1 0

1 1 1

1 2 1

40 10 20

(a) Use graphical analysis to determine the allowable range to stay optimal for c1 and c2. (b) Use algebraic analysis to derive and verify your answers in part (a). (c) Use graphical analysis to determine the allowable range to stay feasible for b1 and b2. (d) Use algebraic analysis to derive and verify your answers in part (c) C (e) Use a software package based on the simplex method to find these allowable ranges. 6.7-12. Consider Variation 5 of the Wyndor Glass Co. model (see Fig. 6.6 and Table 6.24), where the changes in the parameter values given in Table 6.21 are c2 3, a22 3, and a32 4. Use the formula b* S*b to find the allowable range to stay feasible for each bi. Then interpret each allowable range graphically.

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x2

(0) (1) (2)

1 0 0

8 3 5

0 0 1

0 1 0

3 1 1

4 1 2

100 30 40

(a) Perform sensitivity analysis to determine which of the 11 parameters of the model are sensitive parameters in the sense that any change in just that parameter’s value will change the optimal solution. (b) Use algebraic analysis to find the allowable range to stay optimal for each cj. (c) Use algebraic analysis to find the allowable range to stay feasible for each bi. C (d) Use a software package based on the simplex method to find these allowable ranges. 6.7-15. For the problem given in Table 6.21, find the allowable range to stay optimal for c2. Show your work algebraically, using the tableau given in Table 6.21. Then justify your answer from a geometric viewpoint, referring to Fig. 6.3. 6.7-16.* For the original Wyndor Glass Co. problem, use the last tableau in Table 4.8 to do the following. (a) Find the allowable range to stay feasible for each bi. (b) Find the allowable range to stay optimal for c1 and c2. C (c) Use a software package based on the simplex method to find these allowable ranges.

296

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

6.7-17. For Variation 6 of the Wyndor Glass Co. model presented in Sec. 6.7, use the last tableau in Table 6.25 to do the following. (a) Find the allowable range to stay feasible for each bi. (b) Find the allowable range to stay optimal for c1 and c2. C (c) Use a software package based on the simplex method to find these allowable ranges.

gallons of cream left in its inventory. The linear programming formulation for this problem is shown below in algebraic form. Let

C gallons of chocolate ice cream produced, V gallons of vanilla ice cream produced, B gallons of banana ice cream produced.

Maximize 6.7-18. Ken and Larry, Inc., supplies its ice cream parlors with three flavors of ice cream: chocolate, vanilla, and banana. Because of extremely hot weather and a high demand for its products, the company has run short of its supply of ingredients: milk, sugar, and cream. Hence, they will not be able to fill all the orders received from their retail outlets, the ice cream parlors. Owing to these circumstances, the company has decided to choose the amount of each flavor to produce that will maximize total profit, given the constraints on supply of the basic ingredients. The chocolate, vanilla, and banana flavors generate, respectively, $1.00, $0.90, and $0.95 of profit per gallon sold. The company has only 200 gallons of milk, 150 pounds of sugar, and 60

profit 1.00 C 0.90 V 0.95 B,

subject to Milk: Sugar: Cream:

0.45 C 0.50 V 0.40 B 200 gallons 0.50 C 0.40 V 0.40 B 150 pounds 0.10 C 0.15 V 0.20 B 60 gallons

and C 0,

V 0,

B 0.

This problem was solved using the Excel Solver. The spreadsheet (already solved) and the sensitivity report are shown below. [Note: The numbers in the sensitivity report for the milk constraint are missing on purpose, since you will be asked to fill in these numbers in part ( f ).]

CHAPTER 6 PROBLEMS

For each of the following parts, answer the question as specifically and completely as is possible without solving the problem again on the Excel Solver. Note: Each part is independent (i.e., any change made to the model in one part does not apply to any other parts). (a) What is the optimal solution and total profit? (b) Suppose the profit per gallon of banana changes to $1.00. Will the optimal solution change, and what can be said about the effect on total profit? (c) Suppose the profit per gallon of banana changes to 92 cents. Will the optimal solution change, and what can be said about the effect on total profit? (d) Suppose the company discovers that 3 gallons of cream have gone sour and so must be thrown out. Will the optimal solution change, and what can be said about the effect on total profit? (e) Suppose the company has the opportunity to buy an additional 15 pounds of sugar at a total cost of $15. Should they? Explain. (f) Fill in all the sensitivity report information for the milk constraint, given just the optimal solution for the problem. Explain how you were able to deduce each number. 6.7-19. David, LaDeana, and Lydia are the sole partners and workers in a company which produces fine clocks. David and LaDeana each are available to work a maximum of 40 hours per week at the company, while Lydia is available to work a maximum of 20 hours per week. The company makes two different types of clocks: a grandfather clock and a wall clock. To make a clock, David (a mechanical engineer) assembles the inside mechanical parts of the clock while LaDeana (a woodworker) produces the hand-carved wood casings. Lydia is responsible for taking orders and shipping the clocks. The amount of time required for each of these tasks is shown below.

Time Required Task Assemble clock mechanism Carve wood casing Shipping

Grandfather Clock

Wall Clock

6 hours 8 hours 3 hours

4 hours 4 hours 3 hours

Each grandfather clock built and shipped yields a profit of $300, while each wall clock yields a profit of $200. The three partners now want to determine how many clocks of each type should be produced per week to maximize the total profit.

297

(a) Formulate a linear programming model for this problem. (b) Use the graphical method to solve the model. C (c) Use a software package based on the simplex method to solve the model. C (d) Use this same software package to generate sensitivity analysis information. (e) Use this sensitivity analysis information to determine whether the optimal solution must remain optimal if the estimate of the unit profit for grandfather clocks is changed from $300 to $375 (with no other changes in the model). (f) Repeat part (e) if, in addition to this change in the unit profit for grandfather clocks, the estimated unit profit for wall clocks also changes from $200 to $175. (g) Use graphical analysis to verify your answers in parts (e) and ( f ). (h) To increase the total profit, the three partners have agreed that one of them will slightly increase the maximum number of hours available to work per week. The choice of which one will be based on which one would increase the total profit the most. Use the sensitivity analysis information to make this choice. (Assume no change in the original estimates of the unit profits.) (i) Explain why one of the shadow prices is equal to zero. (j) Can the shadow prices given in the sensitivity analysis information be validly used to determine the effect if Lydia were to change her maximum number of hours available to work per week from 20 to 25? If so, what would be the increase in the total profit? (k) Repeat part ( j) if, in addition to the change for Lydia, David also were to change his maximum number of hours available to work per week from 40 to 35. (l) Use graphical analysis to verify your answer in part (k). 6.7-20. Consider the Union Airways problem presented in Sec. 3.4, including the data given in Table 3.19. Management now is considering increasing the level of service provided to customers by increasing one or more of the numbers in the rightmost column of Table 3.19 for the minimum number of agents needed in the various time periods. To guide them in making this decision, they would like to know what impact this change would have on total cost. Use a software package based on the simplex method to generate sensitivity analysis information in preparation for addressing the following questions. (a) Which of the numbers in the rightmost column of Table 3.19 can be increased without increasing total cost? In each case, indicate how much it can be increased (if it is the only one being changed) without increasing total cost. (b) For each of the other numbers, how much would the total cost increase per increase of 1 in the number? For each answer, in-

C

298

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

dicate how much the number can be increased (if it is the only one being changed) before the answer is no longer valid. (c) Do your answers in part (b) definitely remain valid if all the numbers considered in part (b) are simultaneously increased by 1? (d) Do your answers in part (b) definitely remain valid if all 10 numbers are simultaneously increased by 1? (e) How far can all 10 numbers be simultaneously increased by the same amount before your answers in part (b) may no longer be valid? 6.7-21. Consider the following problem. Z 2x1 5x2,

Maximize subject to x1 2x2 10 x1 3x2 12 and x1 0,

x2 0.

Let x3 and x4 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Right Side

Z x1 x2

(0) (1) (2)

1 0 0

0 1 0

0 0 1

1 3 1

1 2 1

22 6 2

While doing postoptimality analysis, you learn that all four bi and cj values used in the original model just given are accurate only to within 50 percent. In other words, their ranges of likely values are 5 b1 15, 6 b2 18, 1 c1 3, and 2.5 c2 7.5. Your job now is to perform sensitivity analysis to determine for each parameter individually (assuming the other three parameters equal their values in the original model) whether this uncertainty might affect either the feasibility or the optimality of the above basic solution (perhaps with new values for the basic variables). Specifically, determine the allowable range to stay feasible for each bi and the allowable range to stay optimal for each cj. Then, for each parameter and its range of likely values, indicate which part of this range lies within the allowable range and which parts correspond to values for which the current basic solution will no longer be both feasible and optimal. (a) Perform this sensitivity analysis graphically on the original model.

(b) Now perform this sensitivity analysis as described and illustrated in Sec. 6.7 for b1 and c1. (c) Repeat part (b) for b2. (d) Repeat part (b) for c2. 6.7-22. Reconsider Prob. 6.7-21. Now use a software package based on the simplex method to generate sensitivity analysis information preparatory to doing parts (a) and (c) below. C (a) Suppose that the estimates for c1 and c2 are correct but the estimates for both b1 and b2 are incorrect. Consider the following four cases where the true values of b1 and b2 differ from their estimates by the same percentage: (1) both b1 and b2 are smaller than their estimates, (2) both b1 and b2 are larger than their estimates, (3) b1 is smaller and b2 is larger than their estimates, and (4) b1 is larger and b2 is smaller than their estimates. For each of these cases, use the 100 percent rule for simultaneous changes in right-hand sides to determine how large the percentage error can be while guaranteeing that the original shadow prices still will be valid. (b) For each of the four cases considered in part (a), start with the final simplex tableau given in Prob. 6.7-21 and use algebraic analysis based on the fundamental insight presented in Sec. 5.3 to determine how large the percentage error can be without invalidating the original shadow prices. C (c) Suppose that the estimates for b1 and b2 are correct but the estimates for both c1 and c2 are incorrect. Consider the following four cases where the true values of c1 and c2 differ from their estimates by the same percentage: (1) both c1 and c2 are smaller than their estimates, (2) both c1 and c2 are larger than their estimates, (3) c1 is smaller and c2 is larger than their estimates, and (4) c1 is larger and c2 is smaller than their estimates. For each of these cases, use the 100 percent rule for simultaneous changes in objective function coefficients to determine how large the percentage error can be while guaranteeing that the original optimal solution must still be optimal. (d) For each of the four cases considered in part (c), start with the final simplex tableau given in Prob. 6.7-21 and use algebraic analysis based on the fundamental insight presented in Sec. 5.3 to determine how large the percentage error can be without invalidating the original optimal solution. 6.7-23. Consider the following problem. Maximize

Z 3x1 4x2 8x3,

subject to 2x1 3x2 5x3 9 x1 2x2 3x3 5 and x1 0,

x2 0,

x3 0.

CHAPTER 6 PROBLEMS

Let x4 and x5 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

1 0 0

0 1 0

1 1 1

0 0 1

1 3 1

1 5 2

14 2 1

While doing postoptimality analysis, you learn that some of the parameter values used in the original model just given are just rough estimates, where the range of likely values in each case is within 50 percent of the value used here. For each of these following parameters, perform sensitivity analysis to determine whether this uncertainty might affect either the feasibility or the optimality of the above basic solution. Specifically, for each parameter, determine the allowable range of values for which the current basic solution (perhaps with new values for the basic variables) will remain both feasible and optimal. Then, for each parameter and its range of likely values, indicate which part of this range lies within the allowable range and which parts correspond to values for which the current basic solution will no longer be both feasible and optimal. (a) Parameter b2 (b) Parameter c2 (c) Parameter a22 (d) Parameter c3 (e) Parameter a12 (f) Parameter b1 6.7-24. Consider Variation 5 of the Wyndor Glass Co. model presented in Sec. 6.7, where c2 3, a22 3, a32 4, and where the other parameters are given in Table 6.21. Starting from the resulting final tableau given at the bottom of Table 6.24, construct a table like Table 6.26 to perform parametric linear programming analysis, where c1 3

and

c2 3 2.

How far can be increased above 0 before the current basic solution is no longer optimal? 6.7-25. Reconsider the model of Prob. 6.7-6. Suppose that you now have the option of making trade-offs in the profitability of the first two activities, whereby the objective function coefficient of x1 can be increased by any amount by simultaneously decreasing the objective function coefficient of x2 by the same amount. Thus, the alternative choices of the objective function are Z() (2 )x1 (1 )x2 x3, where any nonnegative value of can be chosen.

299

Construct a table like Table 6.26 to perform parametric linear programming analysis on this problem. Determine the upper bound on before the original optimal solution would become nonoptimal. Then determine the best choice of over this range. 6.7-26. Consider the following parametric linear programming problem. Z() (10 4)x1 (4 )x2 (7 )x3,

Maximize subject to

3x1 x2 2x3 7 2x1 x2 3x3 5

(resource 1), (resource 2),

and x1 0,

x2 0,

x3 0,

where can be assigned any positive or negative values. Let x4 and x5 be the slack variables for the respective constraints. After we apply the simplex method with 0, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x2

(0) (1) (2)

1 0 0

0 1 0

0 0 1

3 1 5

2 1 2

2 1 3

24 2 1

(a) Determine the range of values of over which the above BF solution will remain optimal. Then find the best choice of within this range. (b) Given that is within the range of values found in part (a), find the allowable range to stay feasible for b1 (the available amount of resource 1). Then do the same for b2 (the available amount of resource 2). (c) Given that is within the range of values found in part (a), identify the shadow prices (as a function of ) for the two resources. Use this information to determine how the optimal value of the objective function would change (as a function of ) if the available amount of resource 1 were decreased by 1 and the available amount of resource 2 simultaneously were increased by 1. (d) Construct the dual of this parametric linear programming problem. Set 0 and solve this dual problem graphically to find the corresponding shadow prices for the two resources of the primal problem. Then find these shadow prices as a function of [within the range of values found in part (a)] by algebraically solving for this same optimal CPF solution for the dual problem as a function of .

300

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

6.7-27. Consider the following parametric linear programming problem. Z() 2x1 4x2 5x3,

Maximize

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x 5

x 6

Right Side

Z x2 x4

(0) (1) (2)

1 0 0

1 3 1

0 1 0

1 2 2

0 0 1

M 0 1

M2 1 1

20 10 5

subject to x1 3x2 2x3 5 x1 2x2 3x3 6 2 and x1 0,

x2 0,

x3 0,

where can be assigned any positive or negative values. Let x4 and x5 be the slack variables for the respective functional constraints. After we apply the simplex method with 0, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

0 1 2

0 1 0

1 5 1

0 0 1

1 3 1

1 2 1

11 3 1

(a) Use the fundamental insight (Sec. 5.3) to revise this tableau to reflect the inclusion of the parameter in the original model. Show the complete tableau needed to apply the feasibility test and the optimality test for any value of . Express the corresponding basic solution (and Z) as a function of . (b) Determine the range of nonnegative values of over which this basic solution is feasible. (c) Determine the range of nonnegative values of over which this basic solution is both feasible and optimal. Determine the best choice of over this range. 6.7-29. Consider the following problem. Z 10x1 4x2,

Maximize subject to 3x1 x2 30 2x1 x2 25 and (a) Express the BF solution (and Z) given in this tableau as a function of . Determine the lower and upper bounds on before this optimal solution would become infeasible. Then determine the best choice of between these bounds. (b) Given that is between the bounds found in part (a), determine the allowable range to stay optimal for c1 (the coefficient of x1 in the objective function). 6.7-28. Consider the following parametric linear programming problem, where the parameter must be nonnegative: Maximize

Z() (5 2)x1 (2 )x2 (3 )x3,

subject to 4x1 x2 2x3 5 5 3x1 x2 2x3 10 10 and x1 0,

x2 0,

x3 0.

Let x4 be the surplus variable for the first functional constraint, and let x5 and x6 be the artificial variables for the respective functional constraints. After we apply the simplex method with the Big M method and with 0, the final simplex tableau is

x1 0,

x2 0.

Let x3 and x4 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Right Side

Z x2 x1

(0) (1) (2)

1 0 0

0 0 1

0 1 0

2 2 1

2 3 1

110 15 5

Now suppose that both of the following changes are made simultaneously in the original model: 1. The first constraint is changed to 4x1 x2 40. 2. Parametric programming is introduced to change the objective function to the alternative choices of Z() (10 2)x1 (4 )x2, where any nonnegative value of can be chosen.

CHAPTER 6 PROBLEMS

301

nology, so 0 1. Given , the coefficients of x1 in the model become

(a) Construct the resulting revised final tableau (as a function of ), and then convert this tableau to proper form from Gaussian elimination. Use this tableau to identify the new optimal solution that applies for either 0 or sufficiently small values of . (b) What is the upper bound on before this optimal solution would become nonoptimal? (c) Over the range of from zero to this upper bound, which choice of gives the largest value of the objective function?

c1 9 9 a11 2 . a21 5 Construct the resulting revised final tableau (as a function of ), and convert this tableau to proper form from Gaussian elimination. Use this tableau to identify the current basic solution as a function of . Over the allowable values of 0 1, give the range of values of for which this solution is both feasible and optimal. What is the best choice of within this range?

6.7-30. Consider the following problem. Z 9x1 8x2 5x3,

Maximize subject to

6.7-31. Consider the following problem.

2x1 3x2 x3 4 5x1 4x2 3x3 11

subject to

and x1 0,

x2 0,

2x1 2x2 x3 5 3x1 x2 x3 10

x3 0.

Let x4 and x5 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

1 0 0

0 1 0

2 5 7

0 0 1

2 3 5

1 1 2

19 1 2

D,I

Z 3x1 5x2 2x3,

Maximize

(a) Suppose that a new technology has become available for conducting the first activity considered in this problem. If the new technology were adopted to replace the existing one, the coefficients of x1 in the model would change from

c1 9 a11 2 a21 5

to

c1 18 a11 3 . a21 6

Use the sensitivity analysis procedure to investigate the potential effect and desirability of adopting the new technology. Specifically, assuming it were adopted, construct the resulting revised final tableau, convert this tableau to proper form from Gaussian elimination, and then reoptimize (if necessary) to find the new optimal solution. (b) Now suppose that you have the option of mixing the old and new technologies for conducting the first activity. Let denote the fraction of the technology used that is from the new tech-

and x1 0,

x2 0,

x3 0.

Let x4 and x5 be the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

1 0 0

0 1 0

20 3 8

0 0 1

9 1 3

7 1 2

115 15 35

Parametric linear programming analysis now is to be applied simultaneously to the objective function and right-hand sides, where the model in terms of the new parameter is the following: Maximize

Z() (3 2)x1 (5 )x2 (2 )x3,

subject to 2x1 2x2 x3 5 6 3x1 x2 x3 10 8 and x1 0,

x2 0,

x3 0.

Construct the resulting revised final tableau (as a function of ), and convert this tableau to proper form from Gaussian elimination.

302

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

Use this tableau to identify the current basic solution as a function of . For 0, give the range of values of for which this solution is both feasible and optimal. What is the best choice of within this range?

(d) If the unit profit is below this breakeven point, how much can the old product’s production rate be decreased (assuming its previous rate was larger than this decrease) before the final BF solution would become infeasible?

6.7-32. Consider the Wyndor Glass Co. problem described in Sec. 3.1. Suppose that, in addition to considering the introduction of two new products, management now is considering changing the production rate of a certain old product that is still profitable. Refer to Table 3.1. The number of production hours per week used per unit production rate of this old product is 1, 4, and 3 for Plants 1, 2, and 3, respectively. Therefore, if we let denote the change (positive or negative) in the production rate of this old product, the right-hand sides of the three functional constraints in Sec. 3.1 become 4 , 12 4, and 18 3, respectively. Thus, choosing a negative value of would free additional capacity for producing more of the two new products, whereas a positive value would have the opposite effect. (a) Use a parametric linear programming formulation to determine the effect of different choices of on the optimal solution for the product mix of the two new products given in the final tableau of Table 4.8. In particular, use the fundamental insight of Sec. 5.3 to obtain expressions for Z and the basic variables x3, x2, and x1 in terms of , assuming that is sufficiently close to zero that this “final” basic solution still is feasible and thus optimal for the given value of . (b) Now consider the broader question of the choice of along with the product mix for the two new products. What is the breakeven unit profit for the old product (in comparison with the two new products) below which its production rate should be decreased ( 0) in favor of the new products and above which its production rate should be increased ( 0)? (c) If the unit profit is above this breakeven point, how much can the old product’s production rate be increased before the final BF solution would become infeasible?

6.7-33. Consider the following problem. Maximize

Z 2x1 x2 3x3,

subject to x1 x2 x3 3 x1 2x2 x3 1 x1 2x2 x3 2 and x1 0,

x2 0,

x3 0.

Suppose that the Big M method (see Sec. 4.6) is used to obtain the initial (artificial) BF solution. Let x4 be the artificial slack variable for the first constraint, x5 the surplus variable for the second constraint, x6 the artificial variable for the second constraint, and x7 the slack variable for the third constraint. The corresponding final set of equations yielding the optimal solution is (0) (1) (2) (3)

Z 5x2 (M 2)x4 Mx6 x7 8, x1 x2 x4 x7 1, 2x2 x3 x7 2, 3x2 x4 x5 x6 2.

Suppose that the original objective function is changed to Z 2x1 3x2 4x3 and that the original third constraint is changed to 2x2 x3 1. Use the sensitivity analysis procedure to revise the final set of equations (in tableau form) and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. (Do not reoptimize.)

CASE 6.1 CONTROLLING AIR POLLUTION Refer to Sec. 3.4 (subsection entitled “Controlling Air Pollution”) for the Nori & Leets Co. problem. After the OR team obtained an optimal solution, we mentioned that the team then conducted sensitivity analysis. We now continue this story by having you retrace the steps taken by the OR team, after we provide some additional background. The values of the various parameters in the original formulation of the model are given in Tables 3.12, 3.13, and 3.14. Since the company does not have much prior experience with the pollution abatement methods under consideration, the cost estimates given in Table 3.14 are fairly rough, and each one could easily be off by as much as 10 percent in either direction. There also is some uncertainty about the parameter val-

CASE 6.1 CONTROLLING AIR POLLUTION

303

ues given in Table 3.13, but less so than for Table 3.14. By contrast, the values in Table 3.12 are policy standards, and so are prescribed constants. However, there still is considerable debate about where to set these policy standards on the required reductions in the emission rates of the various pollutants. The numbers in Table 3.12 actually are preliminary values tentatively agreed upon before learning what the total cost would be to meet these standards. Both the city and company officials agree that the final decision on these policy standards should be based on the tradeoff between costs and benefits. With this in mind, the city has concluded that each 10 percent increase in the policy standards over the current values (all the numbers in Table 3.12) would be worth $3.5 million to the city. Therefore, the city has agreed to reduce the company’s tax payments to the city by $3.5 million for each 10 percent reduction in the policy standards (up to 50 percent) that is accepted by the company. Finally, there has been some debate about the relative values of the policy standards for the three pollutants. As indicated in Table 3.12, the required reduction for particulates now is less than half of that for either sulfur oxides or hydrocarbons. Some have argued for decreasing this disparity. Others contend that an even greater disparity is justified because sulfur oxides and hydrocarbons cause considerably more damage than particulates. Agreement has been reached that this issue will be reexamined after information is obtained about which trade-offs in policy standards (increasing one while decreasing another) are available without increasing the total cost. (a) Use any available linear programming software to solve the model for this problem as formulated in Sec. 3.4. In addition to the optimal solution, obtain the additional output provided for performing postoptimality analysis (e.g., the Sensitivity Report when using Excel). This output provides the basis for the following steps. (b) Ignoring the constraints with no uncertainty about their parameter values (namely, xj 1 for j 1, 2, . . . , 6), identify the parameters of the model that should be classified as sensitive parameters. (Hint: See the subsection “Sensitivity Analysis” in Sec. 4.7.) Make a resulting recommendation about which parameters should be estimated more closely, if possible. (c) Analyze the effect of an inaccuracy in estimating each cost parameter given in Table 3.14. If the true value is 10 percent less than the estimated value, would this alter the optimal solution? Would it change if the true value were 10 percent more than the estimated value? Make a resulting recommendation about where to focus further work in estimating the cost parameters more closely. (d) Consider the case where your model has been converted to maximization form before applying the simplex method. Use Table 6.14 to construct the corresponding dual problem, and use the output from applying the simplex method to the primal problem to identify an optimal solution for this dual problem. If the primal problem had been left in minimization form, how would this affect the form of the dual problem and the sign of the optimal dual variables? (e) For each pollutant, use your results from part (d) to specify the rate at which the total cost of an optimal solution would change with any small change in the required reduction in the annual emission rate of the pollutant. Also specify how much this required reduction can be changed (up or down) without affecting the rate of change in the total cost. (f) For each unit change in the policy standard for particulates given in Table 3.12, determine the change in the opposite direction for sulfur oxides that would keep the total cost of an optimal solution unchanged. Repeat this for hydrocarbons instead of sulfur oxides. Then do

304

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

it for a simultaneous and equal change for both sulfur oxides and hydrocarbons in the opposite direction from particulates. (g) Letting denote the percentage increase in all the policy standards given in Table 3.12, formulate the problem of analyzing the effect of simultaneous proportional increases in these standards as a parametric linear programming problem. Then use your results from part (e) to determine the rate at which the total cost of an optimal solution would increase with a small increase in from zero. (h) Use the simplex method to find an optimal solution for the parametric linear programming problem formulated in part (g) for each 10, 20, 30, 40, 50. Considering the tax incentive offered by the city, use these results to determine which value of (including the option of 0) should be chosen to minimize the company’s total cost of both pollution abatement and taxes. (i) For the value of chosen in part (h), repeat parts (e) and ( f ) so that the decision makers can make a final decision on the relative values of the policy standards for the three pollutants.

CASE 6.2

FARM MANAGEMENT The Ploughman family owns and operates a 640-acre farm that has been in the family for several generations. The Ploughmans always have had to work hard to make a decent living from the farm and have had to endure some occasional difficult years. Stories about earlier generations overcoming hardships due to droughts, floods, etc., are an important part of the family history. However, the Ploughmans enjoy their selfreliant lifestyle and gain considerable satisfaction from continuing the family tradition of successfully living off the land during an era when many family farms are being abandoned or taken over by large agricultural corporations. John Ploughman is the current manager of the farm while his wife Eunice runs the house and manages the farm’s finances. John’s father, Grandpa Ploughman, lives with them and still puts in many hours working on the farm. John and Eunice’s older children, Frank, Phyllis, and Carl, also are given heavy chores before and after school. The entire famiy can produce a total of 4,000 person-hours worth of labor during the winter and spring months and 4,500 person-hours during the summer and fall. If any of these person-hours are not needed, Frank, Phyllis, and Carl will use them to work on a neighboring farm for $5 per hour during the winter and spring months and $5.50 per hour during the summer and fall. The farm supports two types of livestock: dairy cows and laying hens, as well as three crops: soybeans, corn, and wheat. (All three are cash crops, but the corn also is a feed crop for the cows and the wheat also is used for chicken feed.) The crops are harvested during the late summer and fall. During the winter months, John, Eunice, and Grandpa make a decision about the mix of livestock and crops for the coming year. Currently, the family has just completed a particularly successful harvest which has provided an investment fund of $20,000 that can be used to purchase more livestock. (Other money is available for ongoing expenses, including the next planting of crops.) The family currently has 30 cows valued at $35,000 and 2,000 hens valued at $5,000. They wish to keep all this livestock and perhaps purchase more. Each new cow would cost $1,500, and each new hen would cost $3.

CASE 6.2

FARM MANAGEMENT

305

Over a year’s time, the value of a herd of cows will decrease by about 10 percent and the value of a flock of hens will decrease by about 25 percent due to aging. Each cow will require 2 acres of land for grazing and 10 person-hours of work per month, while producing a net annual cash income of $850 for the family. The corresponding figures for each hen are: no significant acreage, 0.05 person-hour per month, and an annual net cash income of $4.25. The chicken house can accommodate a maximum of 5,000 hens, and the size of the barn limits the herd to a maximum of 42 cows. For each acre planted in each of the three crops, the following table gives the number of person-hours of work that will be required during the first and second halves of the year, as well as a rough estimate of the crop’s net value (in either income or savings in purchasing feed for the livestock).

Data per acre planted

Winter and spring, person-hours Summer and fall, person-hours Net value

Soybeans

Corn

Wheat

1.0 1.4 $70

0.9 1.2 $60

0.6 0.7 $40

To provide much of the feed for the livestock, John wants to plant at least 1 acre of corn for each cow in the coming year’s herd and at least 0.05 acre of wheat for each hen in the coming year’s flock. John, Eunice, and Grandpa now are discussing how much acreage should be planted in each of the crops and how many cows and hens to have for the coming year. Their objective is to maximize the family’s monetary worth at the end of the coming year (the sum of the net income from the livestock for the coming year plus the net value of the crops for the coming year plus what remains from the investment fund plus the value of the livestock at the end of the coming year plus any income from working on a neighboring farm, minus living expenses of $40,000 for the year). (a) Identify verbally the components of a linear programming model for this problem. (b) Formulate this model. (Either an algebraic or a spreadsheet formulation is acceptable.) (c) Obtain an optimal solution and generate the additional output provided for performing postoptimality analysis (e.g., the Sensitivity Report when using Excel). What does the model predict regarding the family’s monetary worth at the end of the coming year? (d) Find the allowable range to stay optimal for the net value per acre planted for each of the three crops.

The above estimates of the net value per acre planted in each of the three crops assumes good weather conditions. Adverse weather conditions would harm the crops and greatly reduce the resulting value. The scenarios particularly feared by the family are a drought, a flood, an early frost, both a drought and an early frost, and both a flood and an early frost. The estimated net values for the year under these scenarios are shown on the next page.

306

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Net Value per Acre Planted Scenario Drought Flood Early frost Drought and early frost Flood and early frost

Soybeans

Corn

Wheat

$10 $15 $50 $15 $10

$15 $20 $40 $20 $10

0 $10 $30 $10 $ 5

(e) Find an optimal solution under each scenario after making the necessary adjustments to the linear programming model formulated in part (b). In each case, what is the prediction regarding the family’s monetary worth at the end of the year? (f) For the optimal solution obtained under each of the six scenarios [including the good weather scenario considered in parts (a) to (d )], calculate what the family’s monetary worth would be at the end of the year if each of the other five scenarios occur instead. In your judgment, which solution provides the best balance between yielding a large monetary worth under good weather conditions and avoiding an overly small monetary worth under adverse weather conditions.

Grandpa has researched what the weather conditions were in past years as far back as weather records have been kept, and obtained the following data. Scenario Good weather Drought Flood Early frost Drought and early frost Flood and early frost

Frequency 40% 20% 10% 15% 10% 5%

With these data, the family has decided to use the following approach to making its planting and livestock decisions. Rather than the optimistic approach of assuming that good weather conditions will prevail [as done in parts (a) to (d)], the average net value under all weather conditions will be used for each crop (weighting the net values under the various scenarios by the frequencies in the above table). (g) Modify the linear programming model formulated in part (b) to fit this new approach. (h) Repeat part (c) for this modified model. (i) Use a shadow price obtained in part (h) to analyze whether it would be worthwhile for the family to obtain a bank loan with a 10 percent interest rate to purchase more livestock now beyond what can be obtained with the $20,000 from the investment fund. (j) For each of the three crops, use the postoptimality analysis information obtained in part (h) to identify how much latitude for error is available in estimating the net value per acre planted for that crop without changing the optimal solution. Which two net values need to be estimated most carefully? If both estimates are incorrect simultaneously, how close do the estimates need to be to guarantee that the optimal solution will not change?

CASE 6.3

ASSIGNING STUDENTS TO SCHOOLS (REVISITED)

307

This problem illustrates a kind of situation that is frequently faced by various kinds of organizations. To describe the situation in general terms, an organization faces an uncertain future where any one of a number of scenarios may unfold. Which one will occur depends on conditions that are outside the control of the organization. The organization needs to choose the levels of various activities, but the unit contribution of each activity to the overall measure of performance is greatly affected by which scenario unfolds. Under these circumstances, what is the best mix of activities? (k) Think about specific situations outside of farm management that fit this description. Describe one.

CASE 6.3

ASSIGNING STUDENTS TO SCHOOLS (REVISITED) Reconsider Case 4.3. The Springfield School Board still has the policy of providing bussing for all middle school students who must travel more than approximately 1 mile. Another current policy is to allow splitting residential areas among multiple schools if this will reduce the total bussing cost. (This latter policy will be reversed in Case 12.4.) However, before adopting a bussing plan based on parts (a) and (b) of Case 4.3, the school board now wants to conduct some postoptimality analysis. (a) If you have not already done so for parts (a) and (b) of Case 4.3, formulate and solve a linear programming model for this problem. (Either an algebraic or a spreadsheet formulation is acceptable.) (b) Generate a sensitivity analysis report with the same software package as used in part (a).

One concern of the school board is the ongoing road construction in area 6. These construction projects have been delaying traffic considerably and are likely to affect the cost of bussing students from area 6, perhaps increasing them as much as 10 percent. (c) Use the report from part (b) to check how much the bussing cost from area 6 to school 1 can increase (assuming no change in the costs for the other schools) before the current optimal solution would no longer be optimal. If the allowable increase is less than 10 percent, re-solve to find the new optimal solution with a 10 percent increase. (d) Repeat part (c) for school 2 (assuming no change in the costs for the other schools). (e) Now assume that the bussing cost from area 6 would increase by the same percentage for all the schools. Use the report from part (b) to determine how large this percentage can be before the current optimal solution might no longer be optimal. If the allowable increase is less than 10 percent, re-solve to find the new optimal solution with a 10 percent increase.

The school board has the option of adding portable classrooms to increase the capacity of one or more of the middle schools for a few years. However, this is a costly move that the board would consider only if it would significantly decrease bussing costs. Each portable classroom holds 20 students and has a leasing cost of $2,500 per year. To analyze this option, the school board decides to assume that the road construction in area 6 will wind down without significantly increasing the bussing costs from that area.

308

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

(f) For each school, use the corresponding shadow price from the report obtained in part (b) to determine whether it would be worthwhile to add any portable classrooms. (g) For each school where it is worthwhile to add any portable classrooms, use the report from part (b) to determine how many could be added before the shadow price would no longer be valid (assuming this is the only school receiving portable classrooms). (h) If it would be worthwhile to add portable classrooms to more than one school, use the report from part (b) to determine the combinations of the number to add for which the shadow prices definitely would still be valid. Then use the shadow prices to determine which of these combinations is best in terms of minimizing the total cost of bussing students and leasing portable classrooms. Re-solve to find the corresponding optimal solution for assigning students to schools. (i) If part (h) was applicable, modify the best combination of portable classrooms found there by adding one more to the school with the most favorable shadow price. Find the corresponding optimal solution for assigning students to schools and generate the corresponding sensitivity analysis report. Use this information to assess whether the plan developed in part (h) is the best one available for minimizing the total cost of bussing students and leasing portable classrooms. If not, find the best plan.

7 Other Algorithms for Linear Programming The key to the extremely widespread use of linear programming is the availability of an exceptionally efficient algorithm—the simplex method—that will routinely solve the largesize problems that typically arise in practice. However, the simplex method is only part of the arsenal of algorithms regularly used by linear programming practitioners. We now turn to these other algorithms. This chapter focuses first on three particularly important algorithms that are, in fact, variants of the simplex method. In particular, the next three sections present the dual simplex method (a modification particularly useful for sensitivity analysis), parametric linear programming (an extension for systematic sensitivity analysis), and the upper bound technique (a streamlined version of the simplex method for dealing with variables having upper bounds). Section 4.9 introduced another algorithmic approach to linear programming—a type of algorithm that moves through the interior of the feasible region. We describe this interior-point approach further in Sec. 7.4. We next introduce linear goal programming where, rather than having a single objective (maximize or minimize Z) as for linear programming, the problem instead has several goals toward which we must strive simultaneously. Certain formulation techniques enable converting a linear goal programming problem back into a linear programming problem so that solution procedures based on the simplex method can still be used. Section 7.5 describes these techniques and procedures.

7.1

THE DUAL SIMPLEX METHOD The dual simplex method is based on the duality theory presented in the first part of Chap. 6. To describe the basic idea behind this method, it is helpful to use some terminology introduced in Tables 6.10 and 6.11 of Sec. 6.3 for describing any pair of complementary basic solutions in the primal and dual problems. In particular, recall that both solutions are said to be primal feasible if the primal basic solution is feasible, whereas they are called dual feasible if the complementary dual basic solution is feasible for the dual problem. Also recall (as indicated on the right side of Table 6.11) that each complementary basic solution is optimal for its problem only if it is both primal feasible and dual feasible. The dual simplex method can be thought of as the mirror image of the simplex method. The simplex method deals directly with basic solutions in the primal problem that are primal feasible but not dual feasible. It then moves toward an optimal solution by striving 309

310

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

to achieve dual feasibility as well (the optimality test for the simplex method). By contrast, the dual simplex method deals with basic solutions in the primal problem that are dual feasible but not primal feasible. It then moves toward an optimal solution by striving to achieve primal feasibility as well. Furthermore, the dual simplex method deals with a problem as if the simplex method were being applied simultaneously to its dual problem. If we make their initial basic solutions complementary, the two methods move in complete sequence, obtaining complementary basic solutions with each iteration. The dual simplex method is very useful in certain special types of situations. Ordinarily it is easier to find an initial basic solution that is feasible than one that is dual feasible. However, it is occasionally necessary to introduce many artificial variables to construct an initial BF solution artificially. In such cases it may be easier to begin with a dual feasible basic solution and use the dual simplex method. Furthermore, fewer iterations may be required when it is not necessary to drive many artificial variables to zero. As we mentioned several times in Chap. 6 as well as in Sec. 4.7, another important primary application of the dual simplex method is its use in conjunction with sensitivity analysis. Suppose that an optimal solution has been obtained by the simplex method but that it becomes necessary (or of interest for sensitivity analysis) to make minor changes in the model. If the formerly optimal basic solution is no longer primal feasible (but still satisfies the optimality test), you can immediately apply the dual simplex method by starting with this dual feasible basic solution. Applying the dual simplex method in this way usually leads to the new optimal solution much more quickly than would solving the new problem from the beginning with the simplex method. The dual simplex method also can be useful in solving huge linear programming problems from scratch because it is such an efficient algorithm. Recent computational experience with the latest versions of CPLEX indicates that the dual simplex method often is more efficient than the simplex method for solving particularly massive problems encountered in practice. The rules for the dual simplex method are very similar to those for the simplex method. In fact, once the methods are started, the only difference between them is in the criteria used for selecting the entering and leaving basic variables and for stopping the algorithm. To start the dual simplex method (for a maximization problem), we must have all the coefficients in Eq. (0) nonnegative (so that the basic solution is dual feasible). The basic solutions will be infeasible (except for the last one) only because some of the variables are negative. The method continues to decrease the value of the objective function, always retaining nonnegative coefficients in Eq. (0), until all the variables are nonnegative. Such a basic solution is feasible (it satisfies all the equations) and is, therefore, optimal by the simplex method criterion of nonnegative coefficients in Eq. (0). The details of the dual simplex method are summarized next. Summary of the Dual Simplex Method. 1. Initialization: After converting any functional constraints in form to form (by multiplying through both sides by 1), introduce slack variables as needed to construct a set of equations describing the problem. Find a basic solution such that the coefficients in Eq. (0) are zero for basic variables and nonnegative for nonbasic variables (so the solution is optimal if it is feasible). Go to the feasibility test.

7.1 THE DUAL SIMPLEX METHOD

311

2. Feasibility test: Check to see whether all the basic variables are nonnegative. If they are, then this solution is feasible, and therefore optimal, so stop. Otherwise, go to an iteration. 3. Iteration: Step 1 Determine the leaving basic variable: Select the negative basic variable that has the largest absolute value. Step 2 Determine the entering basic variable: Select the nonbasic variable whose coefficient in Eq. (0) reaches zero first as an increasing multiple of the equation containing the leaving basic variable is added to Eq. (0). This selection is made by checking the nonbasic variables with negative coefficients in that equation (the one containing the leaving basic variable) and selecting the one with the smallest absolute value of the ratio of the Eq. (0) coefficient to the coefficient in that equation. Step 3 Determine the new basic solution: Starting from the current set of equations, solve for the basic variables in terms of the nonbasic variables by Gaussian elimination. When we set the nonbasic variables equal to zero, each basic variable (and Z) equals the new right-hand side of the one equation in which it appears (with a coefficient of 1). Return to the feasibility test. To fully understand the dual simplex method, you must realize that the method proceeds just as if the simplex method were being applied to the complementary basic solutions in the dual problem. (In fact, this interpretation was the motivation for constructing the method as it is.) Step 1 of an iteration, determining the leaving basic variable, is equivalent to determining the entering basic variable in the dual problem. The negative variable with the largest absolute value corresponds to the negative coefficient with the largest absolute value in Eq. (0) of the dual problem (see Table 6.3). Step 2, determining the entering basic variable, is equivalent to determining the leaving basic variable in the dual problem. The coefficient in Eq. (0) that reaches zero first corresponds to the variable in the dual problem that reaches zero first. The two criteria for stopping the algorithm are also complementary. We shall now illustrate the dual simplex method by applying it to the dual problem for the Wyndor Glass Co. (see Table 6.1). Normally this method is applied directly to the problem of concern (a primal problem). However, we have chosen this problem because you have already seen the simplex method applied to its dual problem (namely, the primal problem1) in Table 4.8 so you can compare the two. To facilitate the comparison, we shall continue to denote the decision variables in the problem being solved by yi rather than xj. In maximization form, the problem to be solved is Maximize

Z 4y1 12y2 18y3,

subject to y1

3y3 3 2y2 2y3 5

and y1 0, 1

y2 0,

y3 0.

Recall that the symmetry property in Sec. 6.1 points out that the dual of a dual problem is the original primal problem.

312

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

TABLE 7.1 Dual simplex method applied to the Wyndor Glass Co. dual problem Coefficient of: Iteration

0

1

2

Basic Variable

Eq.

Z

y1

y2

y3

Z y4 y5

(0) (1) (2)

1 0 0

4 1 0

12 0 2

18 3 2

0 1 0

0 0 1

0 3 5

Z y4

(0) (1)

1 0

4 1

0 0

6 3

0 1

y2

(2)

0

0

1

1

0

6 0 1 2

30 3 5 2

2 1 3 1 3

0

0

36

1

0

1

1

0

2 1 3 1 3

6

0

1 2

3 2

Z

(0)

1

y3

(1)

0

y2

(2)

0

y4

y5

Right Side

Since negative right-hand sides are now allowed, we do not need to introduce artificial variables to be the initial basic variables. Instead, we simply convert the functional constraints to form and introduce slack variables to play this role. The resulting initial set of equations is that shown for iteration 0 in Table 7.1. Notice that all the coefficients in Eq. (0) are nonnegative, so the solution is optimal if it is feasible. The initial basic solution is y1 0, y2 0, y3 0, y4 3, y5 5, with Z 0, which is not feasible because of the negative values. The leaving basic variable is y5 (5 3), and the entering basic variable is y2 (12/2 18/2), which leads to the second set of equations, labeled as iteration 1 in Table 7.1. The corresponding basic solution is y1 0, y2 5 2 , y3 0, y4 3, y5 0, with Z 30, which is not feasible. The next leaving basic variable is y4, and the entering basic variable is y3 (6/3 4/1), which leads to the final set of equations in Table 7.1. The corresponding basic solution is y1 0, y2 3 2 , y3 1, y4 0, y5 0, with Z 36, which is feasible and therefore optimal. Notice that the optimal solution for the dual of this problem1 is x*1 2, x*2 6, x*3 2, x*4 0, x*5 0, as was obtained in Table 4.8 by the simplex method. We suggest that you now trace through Tables 7.1 and 4.8 simultaneously and compare the complementary steps for the two mirror-image methods.

7.2

PARAMETRIC LINEAR PROGRAMMING At the end of Sec. 6.7 we described parametric linear programming and its use for conducting sensitivity analysis systematically by gradually changing various model parameters simultaneously. We shall now present the algorithmic procedure, first for the case where the cj parameters are being changed and then where the bi parameters are varied. 1 The complementary optimal basic solutions property presented in Sec. 6.3 indicates how to read the optimal solution for the dual problem from row 0 of the final simplex tableau for the primal problem. This same conclusion holds regardless of whether the simplex method or the dual simplex method is used to obtain the final tableau.

7.2 PARAMETRIC LINEAR PROGRAMMING

313

Systematic Changes in the cj Parameters For the case where the cj parameters are being changed, the objective function of the ordinary linear programming model n

Z cj xj j1

is replaced by n

Z( ) (cj j )xj, j1

where the j are given input constants representing the relative rates at which the coefficients are to be changed. Therefore, gradually increasing from zero changes the coefficients at these relative rates. The values assigned to the j may represent interesting simultaneous changes of the cj for systematic sensitivity analysis of the effect of increasing the magnitude of these changes. They may also be based on how the coefficients (e.g., unit profits) would change together with respect to some factor measured by . This factor might be uncontrollable, e.g., the state of the economy. However, it may also be under the control of the decision maker, e.g., the amount of personnel and equipment to shift from some of the activities to others. For any given value of , the optimal solution of the corresponding linear programming problem can be obtained by the simplex method. This solution may have been obtained already for the original problem where 0. However, the objective is to find the optimal solution of the modified linear programming problem [maximize Z( ) subject to the original constraints] as a function of . Therefore, in the solution procedure you need to be able to determine when and how the optimal solution changes (if it does) as increases from zero to any specified positive number. Figure 7.1 illustrates how Z*( ), the objective function value for the optimal solution (given ), changes as increases. In fact, Z*( ) always has this piecewise linear and con-

FIGURE 7.1 The objective function value for an optimal solution as a function of for parametric linear programming with systematic changes in the cj parameters.

Z* ( )

0

1

2

314

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

vex1 form (see Prob. 7.2-7). The corresponding optimal solution changes (as increases) just at the values of where the slope of the Z*( ) function changes. Thus, Fig. 7.1 depicts a problem where three different solutions are optimal for different values of , the first for 0 1, the second for 1 2, and the third for 2. Because the value of each xj remains the same within each of these intervals for , the value of Z*( ) varies with only because the coefficients of the xj are changing as a linear function of

. The solution procedure is based directly upon the sensitivity analysis procedure for investigating changes in the cj parameters (Cases 2a and 3, Sec. 6.7). As described in the last subsection of Sec. 6.7, the only basic difference with parametric linear programming is that the changes now are expressed in terms of rather than as specific numbers. To illustrate, suppose that 1 2 and 2 1 for the original Wyndor Glass Co. problem presented in Sec. 3.1, so that Z( ) (3 2 )x1 (5 )x2. Beginning with the final simplex tableau for 0 (Table 4.8), we see that its Eq. (0) (0)

3 Z x4 x5 36 2

would first have these changes from the original ( 0) coefficients added into it on the left-hand side: (0)

3 Z 2 x1 x2 x4 x5 36. 2

Because both x1 and x2 are basic variables [appearing in Eqs. (3) and (2), respectively], they both need to be eliminated algebraically from Eq. (0): 3 Z 2 x1 x2 x4 x5 36 2 2 times Eq. (3) times Eq. (2) (0)

3 7 2 Z x4 1 x5 36 2 . 2 6 3

The optimality test says that the current BF solution will remain optimal as long as these coefficients of the nonbasic variables remain nonnegative: 3 7 0, 2 6

9 for 0 , 7

2 1 0, 3

for all 0.

Therefore, after is increased past 9 7 , x4 would need to be the entering basic variable for another iteration of the simplex method to find the new optimal solution. Then would be increased further until another coefficient goes negative, and so on until has been increased as far as desired. This entire procedure is now summarized, and the example is completed in Table 7.2. 1

See Appendix 2 for a definition and discussion of convex functions.

7.2 PARAMETRIC LINEAR PROGRAMMING

315

TABLE 7.2 The cj parametric linear programming procedure applied to the Wyndor Glass Co. example Coefficient of: Range of

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Optimal Solution

Z( )

(0)

1

0

0

0

9 7

6

3 2

3

36 2

x4 0

1 3 1 2 1 3

1 3

2

x3 2

0

6

x2 6

1 3

2

x1 2

5

2

27 5

x3 0

x5 0 9 0 7

x3

(1)

0

0

0

1

x2

(2)

0

0

1

0

x1

(3)

0

1

0

0

Z( )

(0)

1

0

0

9 7

2

0

x5 0 9 5 7

3

x4

(1)

0

0

0

x2

(2)

0

0

1

x1

(3)

0

1

0

3 2 1

Z( )

(0)

1

0

5

x4 x5 x1

(1) (2) (3)

0 0 0

0 0 1

2 2 0

5

1

1

6

x4 6

3

x2 3

4

x1 4 x2 0 x3 0 x4 12 x5 6 x1 4

0

1 2 0

3 2

0

0

12 8

0 3 1

1 0 0

0 1 0

12 6 4

0

Summary of the Parametric Linear Programming Procedure for Systematic Changes in the cj Parameters. 1. Solve the problem with 0 by the simplex method. 2. Use the sensitivity analysis procedure (Cases 2a and 3, Sec. 6.7) to introduce the cj j changes into Eq. (0). 3. Increase until one of the nonbasic variables has its coefficient in Eq. (0) go negative (or until has been increased as far as desired). 4. Use this variable as the entering basic variable for an iteration of the simplex method to find the new optimal solution. Return to step 3. Systematic Changes in the bi Parameters For the case where the bi parameters change systematically, the one modification made in the original linear programming model is that bi is replaced by bi i , for i 1, 2, . . . , m, where the i are given input constants. Thus, the problem becomes n

Maximize

Z( ) cj xj, j1

316

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

subject to n

aij xj bi i

for i 1, 2, . . . , m

j1

and xj 0

for j 1, 2, . . . , n.

The goal is to identify the optimal solution as a function of . With this formulation, the corresponding objective function value Z*( ) always has the piecewise linear and concave1 form shown in Fig. 7.2. (See Prob. 7.2-8.) The set of basic variables in the optimal solution still changes (as increases) only where the slope of Z*( ) changes. However, in contrast to the preceding case, the values of these variables now change as a (linear) function of between the slope changes. The reason is that increasing changes the right-hand sides in the initial set of equations, which then causes changes in the right-hand sides in the final set of equations, i.e., in the values of the final set of basic variables. Figure 7.2 depicts a problem with three sets of basic variables that are optimal for different values of , the first for 0 1, the second for 1 2, and the third for 2. Within each of these intervals of , the value of Z*( ) varies with

despite the fixed coefficients cj because the xj values are changing. The following solution procedure summary is very similar to that just presented for systematic changes in the cj parameters. The reason is that changing the bi values is equivalent to changing the coefficients in the objective function of the dual model. Therefore, the procedure for the primal problem is exactly complementary to applying simultaneously the procedure for systematic changes in the cj parameters to the dual problem. Consequently, the dual simplex method (see Sec. 7.1) now would be used to obtain each new optimal solution, and the applicable sensitivity analysis case (see Sec. 6.7) now is Case 1, but these differences are the only major differences.

1

See Appendix 2 for a definition and discussion of concave functions.

FIGURE 7.2 The objective function value for an optimal solution as a function of for parametric linear programming with systematic changes in the bi parameters.

Z* ( )

0

1

2

7.3 THE UPPER BOUND TECHNIQUE

317

Summary of the Parametric Linear Programming Procedure for Systematic Changes in the bi Parameters. 1. Solve the problem with 0 by the simplex method. 2. Use the sensitivity analysis procedure (Case 1, Sec. 6.7) to introduce the bi i

changes to the right side column. 3. Increase until one of the basic variables has its value in the right side column go negative (or until has been increased as far as desired). 4. Use this variable as the leaving basic variable for an iteration of the dual simplex method to find the new optimal solution. Return to step 3. To illustrate this procedure in a way that demonstrates its duality relationship with the procedure for systematic changes in the cj parameters, we now apply it to the dual problem for the Wyndor Glass Co. (see Table 6.1). In particular, suppose that 1 2 and 2 1 so that the functional constraints become y1 3y3 3 2

2y2 2y3 5

or or

y1

3y3 3 2

2y2 2y3 5 .

Thus, the dual of this problem is just the example considered in Table 7.2. This problem with 0 has already been solved in Table 7.1, so we begin with the final simplex tableau given there. Using the sensitivity analysis procedure for Case 1, Sec. 6.7, we find that the entries in the right side column of the tableau change to the values given below. [2, 6] Z* y*b

3 2

5 36 2 ,

2 1 1 0 3 3 2

3 . b* S*b 3 1 5

1 7 2 6 3 2

Therefore, the two basic variables in this tableau 3 2

y3 3

and

9 7

y2 6

remain nonnegative for 0 9 7 . Increasing past 9 7 requires making y2 a leaving basic variable for another iteration of the dual simplex method, and so on, as summarized in Table 7.3. We suggest that you now trace through Tables 7.2 and 7.3 simultaneously to note the duality relationship between the two procedures.

7.3

THE UPPER BOUND TECHNIQUE It is fairly common in linear programming problems for some of or all the individual xj variables to have upper bound constraints xj uj, where uj is a positive constant representing the maximum feasible value of xj. We pointed out in Sec. 4.8 that the most important determinant of computation time for the simplex

318

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

TABLE 7.3 The bi parametric linear programming procedure applied to the dual of the Wyndor Glass Co. example Coefficient of: Range of 9 0 7

9 5 7

5

Basic Variable

Eq.

Z

y1

Z( )

(0)

1

y3

(1)

0

y2

(2)

0

Z( )

(0)

1

y2

y3

2 1 3 1 3

0

0

0

1

1

0

0

6

0

y4 2 1 3 1 3 4

y3

(1)

0

0

1

1

0

y1

(2)

0

1

3

0

1

Right Side

Optimal Solution

1 2

36 2

3 2

3 9 7

6

y1 y4 y5 0 3 2

y3 3 9 7

y2 6

y5 6 0

3

27 5

y2 y4 y5 0

1 2 3 2

5

2 9 7

2

5

y3 2 9 7

y1 2 y2 y3 y4 0

Z( )

(0)

1

0

12

6

4

0

12 8

y5 y1

(1) (2)

0 0

0 1

2 0

2 3

0 1

1 0

5

3 2

y5 5

y1 3 2

method is the number of functional constraints, whereas the number of nonnegativity constraints is relatively unimportant. Therefore, having a large number of upper bound constraints among the functional constraints greatly increases the computational effort required. The upper bound technique avoids this increased effort by removing the upper bound constraints from the functional constraints and treating them separately, essentially like nonnegativity constraints. Removing the upper bound constraints in this way causes no problems as long as none of the variables gets increased over its upper bound. The only time the simplex method increases some of the variables is when the entering basic variable is increased to obtain a new BF solution. Therefore, the upper bound technique simply applies the simplex method in the usual way to the remainder of the problem (i.e., without the upper bound constraints) but with the one additional restriction that each new BF solution must satisfy the upper bound constraints in addition to the usual lower bound (nonnegativity) constraints. To implement this idea, note that a decision variable xj with an upper bound constraint xj uj can always be replaced by xj uj yj, where yj would then be the decision variable. In other words, you have a choice between letting the decision variable be the amount above zero (xj) or the amount below uj (yj uj xj). (We shall refer to xj and yj as complementary decision variables.) Because 0 xj uj

7.3 THE UPPER BOUND TECHNIQUE

319

it also follows that 0 yj uj. Thus, at any point during the simplex method, you can either 1. Use xj, where 0 xj uj, or 2. Replace xj by uj yj , where 0 yj uj. The upper bound technique uses the following rule to make this choice: Rule: Begin with choice 1. Whenever xj 0, use choice 1, so xj is nonbasic. Whenever xj uj , use choice 2, so yj 0 is nonbasic. Switch choices only when the other extreme value of xj is reached. Therefore, whenever a basic variable reaches its upper bound, you should switch choices and use its complementary decision variable as the new nonbasic variable (the leaving basic variable) for identifying the new BF solution. Thus, the one substantive modification being made in the simplex method is in the rule for selecting the leaving basic variable. Recall that the simplex method selects as the leaving basic variable the one that would be the first to become infeasible by going negative as the entering basic variable is increased. The modification now made is to select instead the variable that would be the first to become infeasible in any way, either by going negative or by going over the upper bound, as the entering basic variable is increased. (Notice that one possibility is that the entering basic variable may become infeasible first by going over its upper bound, so that its complementary decision variable becomes the leaving basic variable.) If the leaving basic variable reaches zero, then proceed as usual with the simplex method. However, if it reaches its upper bound instead, then switch choices and make its complementary decision variable the leaving basic variable. To illustrate, consider this problem: Maximize

Z 2x1 x2 2x3,

subject to 4x1 x2 12 2x1 x3 4 and 0 x1 4,

0 x2 15,

0 x3 6.

Thus, all three variables have upper bound constraints (u1 4, u2 15, u3 6). The two equality constraints are already in proper form from Gaussian elimination for identifying the initial BF solution (x1 0, x2 12, x3 4), and none of the variables in this solution exceeds its upper bound, so x2 and x3 can be used as the initial basic variables without artificial variables being introduced. However, these variables then need to be eliminated algebraically from the objective function to obtain the initial Eq. (0), as follows:

(0)

Z 2( (2x1 x2 2x3 0 Z 2( (4x1 x2 2x3 12) Z 2( (2x1 x2 x3 4) Z 2( (2x1 x2 2x3 20.

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OTHER ALGORITHMS FOR LINEAR PROGRAMMING

TABLE 7.4 Equations and calculations for the initial leaving basic variable in the example for the upper bound technique Initial Set of Equations

Maximum Feasible Value of x1

(0) Z 2x1 x2 x3 20

x1 4 (since u1 4) 12 x1 3 4 64 x1 1 minimum (because u3 6) 2

(1) Z 4x1 x2 x3 12 (2) Z 2x1 x2 x3 4

To start the first iteration, this initial Eq. (0) indicates that the initial entering basic variable is x1. Since the upper bound constraints are not to be included, the entire initial set of equations and the corresponding calculations for selecting the leaving basic variables are those shown in Table 7.4. The second column shows how much the entering basic variable x1 can be increased from zero before some basic variable (including x1) becomes infeasible. The maximum value given next to Eq. (0) is just the upper bound constraint for x1. For Eq. (1), since the coefficient of x1 is positive, increasing x1 to 3 decreases the basic variable in this equation (x2) from 12 to its lower bound of zero. For Eq. (2), since the coefficient of x1 is negative, increasing x1 to 1 increases the basic variable in this equation (x3) from 4 to its upper bound of 6. Because Eq. (2) has the smallest maximum feasible value of x1 in Table 7.4, the basic variable in this equation (x3) provides the leaving basic variable. However, because x3 reached its upper bound, replace x3 by 6 y3, so that y3 0 becomes the new nonbasic variable for the next BF solution and x1 becomes the new basic variable in Eq. (2). This replacement leads to the following changes in this equation: (2)

2x1 x3 4 → 2x1 6 y3 4 → 2x1 y3 2 1 → x1 y3 1 2

Therefore, after we eliminate x1 algebraically from the other equations, the second complete set of equations becomes (0) (1) (2)

Zx2x2 y3 22 Zx2x2 2y3 8 1 Zx1x2 y3 1. 2

The resulting BF solution is x1 1, x2 8, y3 0. By the optimality test, it also is an optimal solution, so x1 1, x2 8, x3 6 y3 6 is the desired solution for the original problem.

7.4

AN INTERIOR-POINT ALGORITHM In Sec. 4.9 we discussed a dramatic development in linear programming that occurred in 1984, the invention by Narendra Karmarkar of AT&T Bell Laboratories of a powerful algorithm for solving huge linear programming problems with an approach very different

7.4 AN INTERIOR-POINT ALGORITHM

321

from the simplex method. We now introduce the nature of Karmarkar’s approach by describing a relatively elementary variant (the “affine” or “affine-scaling” variant) of his algorithm.1 (Your OR Courseware also includes this variant under the title, Solve Automatically by the Interior-Point Algorithm.) Throughout this section we shall focus on Karmarkar’s main ideas on an intuitive level while avoiding mathematical details. In particular, we shall bypass certain details that are needed for the full implementation of the algorithm (e.g., how to find an initial feasible trial solution) but are not central to a basic conceptual understanding. The ideas to be described can be summarized as follows: Concept 1: Shoot through the interior of the feasible region toward an optimal solution. Concept 2: Move in a direction that improves the objective function value at the fastest possible rate. Concept 3: Transform the feasible region to place the current trial solution near its center, thereby enabling a large improvement when concept 2 is implemented. To illustrate these ideas throughout the section, we shall use the following example: Maximize

Z x1 2x2,

subject to x1 x2 8 and x1 0,

x2 0.

This problem is depicted graphically in Fig. 7.3, where the optimal solution is seen to be (x1, x2) (0, 8) with Z 16. The Relevance of the Gradient for Concepts 1 and 2 The algorithm begins with an initial trial solution that (like all subsequent trial solutions) lies in the interior of the feasible region, i.e., inside the boundary of the feasible region. Thus, for the example, the solution must not lie on any of the three lines (x1 0, x2 0, x1 x2 8) that form the boundary of this region in Fig. 7.3. (A trial solution that lies on the boundary cannot be used because this would lead to the undefined mathematical operation of division by zero at one point in the algorithm.) We have arbitrarily chosen (x1, x2) (2, 2) to be the initial trial solution. To begin implementing concepts 1 and 2, note in Fig. 7.3 that the direction of movement from (2, 2) that increases Z at the fastest possible rate is perpendicular to (and toward) the objective function line Z 16 x1 2x2. We have shown this direction by the arrow from (2, 2) to (3, 4). Using vector addition, we have (3, 4) (2, 2) (1, 2), 1

The basic approach for this variant actually was proposed in 1967 by a Russian mathematician I. I. Dikin and then rediscovered soon after the appearance of Karmarkar’s work by a number of researchers, including E. R. Barnes, T. M. Cavalier, and A. L. Soyster. Also see R. J. Vanderbei, M. S. Meketon, and B. A. Freedman, “A Modification of Karmarkar’s Linear Programming Algorithm,” Algorithmica, 1(4) (Special Issue on New Approaches to Linear Programming): 395–407, 1986.

322

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

x2 8

(0, 8) optimal

Z 16 x1 2x2

6

4

(3, 4)

(2, 2)

2

FIGURE 7.3 Example for the interior-point algorithm.

2

0

4

6

8

x1

where the vector (1, 2) is the gradient of the objective function. (We will discuss gradients further in Sec. 13.5 in the broader context of nonlinear programming, where algorithms similar to Karmarkar’s have long been used.) The components of (1, 2) are just the coefficients in the objective function. Thus, with one subsequent modification, the gradient (1, 2) defines the ideal direction to which to move, where the question of the distance to move will be considered later. The algorithm actually operates on linear programming problems after they have been rewritten in augmented form. Letting x3 be the slack variable for the functional constraint of the example, we see that this form is Maximize

Z x1 2x2,

subject to x1 x2 x3 8 and x1 0,

x2 0,

x3 0.

In matrix notation (slightly different from Chap. 5 because the slack variable now is incorporated into the notation), the augmented form can be written in general as Maximize subject to Ax b

Z cTx,

7.4 AN INTERIOR-POINT ALGORITHM

323

and x 0, where 1 c 2 , 0

x1 x x2 , x3

A [1,

1,

1],

b [8],

0 0 0 0

for the example. Note that cT [1, 2, 0] now is the gradient of the objective function. The augmented form of the example is depicted graphically in Fig. 7.4. The feasible region now consists of the triangle with vertices (8, 0, 0), (0, 8, 0), and (0, 0, 8). Points in the interior of this feasible region are those where x1 0, x2 0, and x3 0. Each of these three xj 0 conditions has the effect of forcing (x1, x2) away from one of the three lines forming the boundary of the feasible region in Fig. 7.3. Using the Projected Gradient to Implement Concepts 1 and 2 In augmented form, the initial trial solution for the example is (x1, x2, x3) (2, 2, 4). Adding the gradient (1, 2, 0) leads to (3, 4, 4) (2, 2, 4) (1, 2, 0). However, now there is a complication. The algorithm cannot move from (2, 2, 4) toward (3, 4, 4), because (3, 4, 4) is infeasible! When x1 3 and x2 4, then x3 8 x1 x2 1 instead of 4. The point (3, 4, 4) lies on the near side as you look down on the feasible triangle in Fig. 7.4. Therefore, to remain feasible, the algorithm (indirectly) projects the point (3, 4, 4) down onto the feasible triangle by dropping a line that is perpendicular to this triangle. A vector from (0, 0, 0) to (1, 1, 1) is perpendicular to this triangle, so the perpendicular line through (3, 4, 4) is given by the equation (x1, x2, x3) (3, 4, 4) (1, 1, 1), where is a scalar. Since the triangle satisfies the equation x1 x2 x3 8, this perpendicular line intersects the triangle at (2, 3, 3). Because (2, 3, 3) (2, 2, 4) (0, 1, 1), the projected gradient of the objective function (the gradient projected onto the feasible region) is (0, 1, 1). It is this projected gradient that defines the direction of movement for the algorithm, as shown by the arrow in Fig. 7.4. A formula is available for computing the projected gradient directly. By defining the projection matrix P as P I AT(AAT)1A, the projected gradient (in column form) is cp Pc.

324

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

x3 8

(2, 2, 4)

(3, 3, 4)

(2, 3, 3) 0

8 x1

FIGURE 7.4 Example in augmented form for the interior-point algorithm.

8 x2

(0, 8, 0) optimal

Thus, for the example,

1 P 0 0

0 1 0

1 0 0 1 [1 1 1

1

1 1] 1 1

1 0 0

0 1 0

1 0 1 0 1 [1 3 1 1

1

1]

1 0 0

0 1 0

1 0 1 0 1 3 1 1

23 1 1 1 3 1 1 3

1 1 1

1

[1

1 3 2 3 1 3

1

1]

1 3 1 3 , 2 3

7.4 AN INTERIOR-POINT ALGORITHM

325

so 23 cp 13 1 3

1 3 2 3 1 3

0 1 3 1 1 3 2 1 . 2 1 3 0

Moving from (2, 2, 4) in the direction of the projected gradient (0, 1, 1) involves increasing from zero in the formula 2 x 2 4cp 4

2 0 2 4 1 , 4 1

where the coefficient 4 is used simply to give an upper bound of 1 for to maintain feasibility (all xj 0). Note that increasing to 1 would cause x3 to decrease to x3 4 4(1)(1) 0, where 1 yields x3 0. Thus, measures the fraction used of the distance that could be moved before the feasible region is left. How large should be made for moving to the next trial solution? Because the increase in Z is proportional to , a value close to the upper bound of 1 is good for giving a relatively large step toward optimality on the current iteration. However, the problem with a value too close to 1 is that the next trial solution then is jammed against a constraint boundary, thereby making it difficult to take large improving steps during subsequent iterations. Therefore, it is very helpful for trial solutions to be near the center of the feasible region (or at least near the center of the portion of the feasible region in the vicinity of an optimal solution), and not too close to any constraint boundary. With this in mind, Karmarkar has stated for his algorithm that a value as large as 0.25 should be “safe.” In practice, much larger values (for example, 0.9) sometimes are used. For the purposes of this example (and the problems at the end of the chapter), we have chosen 0.5. (Your OR Courseware uses 0.5 as the default value, but also has 0.9 available.) A Centering Scheme for Implementing Concept 3 We now have just one more step to complete the description of the algorithm, namely, a special scheme for transforming the feasible region to place the current trial solution near its center. We have just described the benefit of having the trial solution near the center, but another important benefit of this centering scheme is that it keeps turning the direction of the projected gradient to point more nearly toward an optimal solution as the algorithm converges toward this solution. The basic idea of the centering scheme is straightforward—simply change the scale (units) for each of the variables so that the trial solution becomes equidistant from the constraint boundaries in the new coordinate system. (Karmarkar’s original algorithm uses a more sophisticated centering scheme.) For the example, there are three constraint boundaries in Fig. 7.3, each one corresponding to a zero value for one of the three variables of the problem in augmented form, namely, x1 0, x2 0, and x3 0. In Fig. 7.4, see how these three constraint boundaries intersect the Ax b (x1 x2 x3 8) plane to form the boundary of the feasible re-

326

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

gion. The initial trial solution is (x1, x2, x3) (2, 2, 4), so this solution is 2 units away from the x1 0 and x2 0 constraint boundaries and 4 units away from the x3 0 constraint boundary, when the units of the respective variables are used. However, whatever these units are in each case, they are quite arbitrary and can be changed as desired without changing the problem. Therefore, let us rescale the variables as follows: 1 ~x x , 1 2

2 ~x x , 2 2

~x x 3 3 4

in order to make the current trial solution of (x1, x2, x3) (2, 2, 4) become ~ , ~x , ~x ) (1, 1, 1). (x 1

2

3

~ for x , 2x ~ for x , and 4x ~ for x ), the problem In these new coordinates (substituting 2x 1 1 2 2 3 3 becomes ~ 4x ~, Maximize Z 2x 1

2

subject to ~ 2x ~x 4x ~ 8 2x 1 2 3 and ~x 0, 1

~x 0, 2

~x 0, 3

as depicted graphically in Fig. 7.5. Note that the trial solution (1, 1, 1) in Fig. 7.5 is equidistant from the three constraint boundaries ~x 1 0, ~x 2 0, ~x 3 0. For each subsequent iteration as well, the problem is rescaled again to achieve this same property, so that the current trial solution always is (1, 1, 1) in the current coordinates.

~ x3

FIGURE 7.5 Example after rescaling for iteration 1.

2

(1, 1, 1) 0

( 54, 74, 12) 4 ~ x2

(0, 4, 0) optimal

4

~ x1

7.4 AN INTERIOR-POINT ALGORITHM

327

Summary and Illustration of the Algorithm Now let us summarize and illustrate the algorithm by going through the first iteration for the example, then giving a summary of the general procedure, and finally applying this summary to a second iteration. Iteration 1. Given the initial trial solution (x1, x2, x3) (2, 2, 4), let D be the corresponding diagonal matrix such that x Dx~, so that 2 D 0 0

0 2 0

0 0 . 4

The rescaled variables then are the components of 1 0 0 x 1 2 2 x1 1 x ~ x D1x 0 0 x2 2 . 2 2 x x3 1 3 0 0 4 4 In these new coordinates, A and c have become Ã AD [1

1

2 1] 0 0

2 ~c Dc 0 0

0 2 0

0 0 4

0 0 [2 4

0 2 0

2

4],

1 2 2 4 . 0 0

Therefore, the projection matrix is P I ÃT(ÃÃT)1Ã

1 P 0 0

0 1 0

2 0 0 2 [2 1 4

1 P 0 0

0 1 0

4 4 0 0 21 4 4 4 1 8 8

2

2 4] 2 4

1

56 8 8 1 6 1 16 3

[2

2

4]

1 6 13 5 13 , 6 1 1 3 3

so that the projected gradient is 56 cp Pc~ 16 1 3

1 6 5 6

1 3

1 1 3 2 1 3 4 3 . 1 2 3 0

Define v as the absolute value of the negative component of cp having the largest absolute value, so that v 2 2 in this case. Consequently, in the current coordinates, the

328

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

~ , ~x , ~x ) (1, 1, 1) to the next algorithm now moves from the current trial solution (x 1 2 3 trial solution 5 4 1 1 1 7 0.5 ~ x 1 cp 1 3 , v 2 4 1 1 2 1 2 as shown in Fig. 7.5. (The definition of v has been chosen to make the smallest component of ~ x equal to zero when 1 in this equation for the next trial solution.) In the original coordinates, this solution is x1 2 0 0 5 4 5 2 7 7 x2 Dx~ 0 2 0 4 2 . 1 x3 0 0 4 2 2 This completes the iteration, and this new solution will be used to start the next iteration. These steps can be summarized as follows for any iteration. Summary of the Interior-Point Algorithm. 1. Given the current trial solution (x1, x2, . . . , xn), set 0

0 x1 0 0

0 0 x2 D 0 0 x3

0

0

xn 0 0 2. Calculate Ã AD and ~ c Dc. 3. Calculate P I ÃT(ÃÃT)1Ã and cp Pc~. 4. Identify the negative component of cp having the largest absolute value, and set v equal to this absolute value. Then calculate 1 1 ~ x cp, v 1 where is a selected constant between 0 and 1 (for example, 0.5). 5. Calculate x Dx~ as the trial solution for the next iteration (step 1). (If this trial solution is virtually unchanged from the preceding one, then the algorithm has virtually converged to an optimal solution, so stop.) Now let us apply this summary to iteration 2 for the example. Iteration 2. Step 1: Given the current trial solution (x1, x2, x3) (5 2 , 72 , 2), set 52 0 0 D 0 7 2 0 . 0 0 2

7.4 AN INTERIOR-POINT ALGORITHM

329

(Note that the rescaled variables are ~x 1 2 5 x1 2 5 0 0 x1 ~ 2 2 1 x 2 D x 0 7 0 x2 7 x2 , ~ 1 1 x3 2 x3 0 0 2 x3 so that the BF solutions in these new coordinates are 8 1 56 ~ x D1 0 0 , 0 0

0 0 1 ~ x D 8 176 , 0 0

and 0 0 1 ~ x D 0 0 , 8 4 as depicted in Fig. 7.6.) Step 2: Ã AD [ , , 2] 5 7 2 2

and

52 ~c Dc 7 . 0

~ x3

FIGURE 7.6 Example after rescaling for iteration 2.

4

2

(1, 1, 1) 0 1 16 7 3

~ x2

2 16 (0.83, 1.40, 0.5) 5

(0, 167 , 0) optimal

4

~ x1

330

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

Step 3: 1138 P 178 2 9

178 41 90 14 45

29 1445 37 45

and

1 11 2 cp 1630 3 . 4 1 15

Step 4: 4 11 5 1 11 2, so v 4 115 and 1 11 2 2372 38 1 0.83 461 0.5 1 3 3 ~ x 1 41 60 328 1.40 . 1 41 15 1 5 2 1 0.50 Step 5: 1 6356 65 2.08 3227 ~ x Dx 656 4.92 1 1.00 is the trial solution for iteration 3. Since there is little to be learned by repeating these calculations for additional iterations, we shall stop here. However, we do show in Fig. 7.7 the reconfigured feasible region after rescaling based on the trial solution just obtained for iteration 3. As always, the ~ , ~x , ~x ) (1, 1, 1), equidistant from the ~x rescaling has placed the trial solution at (x 1 2 3 1 ~ ~ 0, x 2 0, and x 3 0 constraint boundaries. Note in Figs. 7.5, 7.6, and 7.7 how the sequence of iterations and rescaling have the effect of “sliding” the optimal solution toward (1, 1, 1) while the other BF solutions tend to slide away. Eventually, after enough itera~ , ~x , ~x ) (0, 1, 0) after rescaling, while tions, the optimal solution will lie very near (x 1 2 3 the other two BF solutions will be very far from the origin on the ~x 1 and ~x 3 axes. Step 5 of that iteration then will yield a solution in the original coordinates very near the optimal solution of (x1, x2, x3) (0, 8, 0). Figure 7.8 shows the progress of the algorithm in the original x1 x2 coordinate system before the problem is augmented. The three points—(x1, x2) (2, 2), (2.5, 3.5), and (2.08, 4.92)—are the trial solutions for initiating iterations 1, 2, and 3, respectively. We then have drawn a smooth curve through and beyond these points to show the trajectory of the algorithm in subsequent iterations as it approaches (x1, x2) (0, 8). The functional constraint for this particular example happened to be an inequality constraint. However, equality constraints cause no difficulty for the algorithm, since it deals with the constraints only after any necessary augmenting has been done to convert them to equality form (Ax b) anyway. To illustrate, suppose that the only change in the example is that the constraint x1 x2 8 is changed to x1 x2 8. Thus, the feasible region in Fig. 7.3 changes to just the line segment between (8, 0) and (0, 8). Given an initial feasible trial solution in the interior (x1 0 and x2 0) of this line segment—say, (x1, x2) (4, 4)—the algorithm can proceed just as presented in the five-step summary with just the two variables and A [1, 1]. For each iteration, the projected gradient points along this line segment in the direction of (0, 8). With 12 , iteration 1 leads from (4, 4)

7.4 AN INTERIOR-POINT ALGORITHM

331

~ x3 8

(1, 1, 1) 0 3.85 1.63

FIGURE 7.7 Example after rescaling for iteration 3.

~ x1

(0, 1.63, 0) optimal

~ x2

to (2, 6), iteration 2 leads from (2, 6) to (1, 7), etc. (Problem 7.4-3 asks you to verify these results.) Although either version of the example has only one functional constraint, having more than one leads to just one change in the procedure as already illustrated (other than more extensive calculations). Having a single functional constraint in the example meant that A had only a single row, so the (ÃÃT)1 term in step 3 only involved taking the reciprocal of the number obtained from the vector product ÃÃT. Multiple functional constraints mean that A has multiple rows, so then the (ÃÃT)1 term involves finding the inverse of the matrix obtained from the matrix product ÃÃT. To conclude, we need to add a comment to place the algorithm into better perspective. For our extremely small example, the algorithm requires relatively extensive calcul

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INTRODUCTION TO OPERATIONS RESEARCH

McGraw-Hill Series in Industrial Engineering and Management Science CONSULTING EDITORS Kenneth E. Case, Department of Industrial Engineering and Management, Oklahoma State University Philip M. Wolfe, Department of Industrial and Management Systems Engineering, Arizona State University Barnes Statistical Analysis for Engineers and Scientists: A Computer-Based Approach Bedworth, Henderson, and Wolfe Computer-Integrated Design and Manufacturing Blank and Tarquin Engineering Economy Ebeling Reliability and Maintainability Engineering Grant and Leavenworth Statistical Quality Control Harrell, Ghosh, and Bowden Simulation Using PROMODEL Hillier and Lieberman Introduction to Operations Research Gryna Quality Planning and Analysis: From Product Development through Use Kelton, Sadowski, and Sadowski Simulation with ARENA Khalil Management of Technology Kolarik Creating Quality: Concepts, Systems, Strategies, and Tools Creating Quality: Process Design for Results Law and Kelton Simulation Modeling and Analysis Nash and Sofer Linear and Nonlinear Programming Nelson Stochastic Modeling: Analysis and Simulation Niebel and Freivalds Methods, Standards, and Work Design Pegden Introduction to Simulation Using SIMAN Riggs, Bedworth, and Randhawa Engineering Economics Sipper and Bulfin Production: Planning, Control, and Integration Steiner Engineering Economics Principles

INTRODUCTION TO OPERATIONS RESEARCH Seventh Edition

FREDERICK S. HILLIER, Stanford University

GERALD J. LIEBERMAN, Late of Stanford University

Cases developed by Karl Schmedders and Molly Stephens Tutorial software developed by Mark Hillier and Michael O’Sullivan

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INTRODUCTION TO OPERATIONS RESEARCH Published by McGraw-Hill, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY, 10020. Copyright © 2001, 1995, 1990, 1986, 1980, 1974, 1967, by The McGraw-Hill Companies, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGrawHill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 ISBN

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Vice president/Editor-in-chief: Kevin Kane Publisher: Thomas Casson Executive editor: Eric M. Munson Developmental editor: Maja Lorkovic Marketing manager: John Wannemacher Project manager: Christine A. Vaughan Manager, new book production: Melonie Salvati Coordinator, freelance design: Gino Cieslik Supplement coordinator: Cathy Tepper Media technology producer: Judi David Cover design: Gino Cieslik Cover Illustration: Paul Turnbaugh Compositor: York Graphic Services, Inc. Typeface: 10/12 Times Printer: R. R. Donnelley & Sons Company Library of Congress Cataloging-in-Publication Data Hillier, Frederick S. Introduction to operations research/Frederick S. Hillier, Gerald J. Lieberman; cases developed by Karl Schmedders and Molly Stephens; tutorial software developed by Mark Hillier and Michael O’Sullivan.—7th ed. p. cm. ISBN 0-07-232169-5 1. Operations research. I. Lieberman, Gerald J. II. Title. T57.6. H53 2001 658.4034—dc21 00-025683 www.mhhe.com

ABOUT THE AUTHORS

Frederick S. Hillier was born and raised in Aberdeen, Washington, where he was an award winner in statewide high school contests in essay writing, mathematics, debate, and music. As an undergraduate at Stanford University he ranked first in his engineering class of over 300 students. He also won the McKinsey Prize for technical writing, won the Outstanding Sophomore Debater award, played in the Stanford Woodwind Quintet, and won the Hamilton Award for combining excellence in engineering with notable achievements in the humanities and social sciences. Upon his graduation with a B.S. degree in Industrial Engineering, he was awarded three national fellowships (National Science Foundation, Tau Beta Pi, and Danforth) for graduate study at Stanford with specialization in operations research. After receiving his Ph.D. degree, he joined the faculty of Stanford University, and also received visiting appointments at Cornell University, Carnegie-Mellon University, the Technical University of Denmark, the University of Canterbury (New Zealand), and the University of Cambridge (England). After 35 years on the Stanford faculty, he took early retirement from his faculty responsibilities in 1996 in order to focus full time on textbook writing, and so now is Professor Emeritus of Operations Research at Stanford. Dr. Hillier’s research has extended into a variety of areas, including integer programming, queueing theory and its application, statistical quality control, and the application of operations research to the design of production systems and to capital budgeting. He has published widely, and his seminal papers have been selected for republication in books of selected readings at least ten times. He was the first-prize winner of a research contest on “Capital Budgeting of Interrelated Projects” sponsored by The Institute of Management Sciences (TIMS) and the U.S. Office of Naval Research. He and Dr. Lieberman also received the honorable mention award for the 1995 Lanchester Prize (best English-language publication of any kind in the field of operations research), which was awarded by the Institute of Operations Research and the Management Sciences (INFORMS) for the 6th edition of this book. Dr. Hillier has held many leadership positions with the professional societies in his field. For example, he has served as Treasurer of the Operations Research Society of America (ORSA), Vice President for Meetings of TIMS, Co-General Chairman of the 1989 TIMS International Meeting in Osaka, Japan, Chair of the TIMS Publications Committee, Chair of the ORSA Search Committee for Editor of Operations Research, Chair of the ORSA Resources Planning Committee, Chair of the ORSA/TIMS Combined Meetings Committee, and Chair of the John von Neumann Theory Prize Selection Committee for INFORMS. vii

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ABOUT THE AUTHORS

He currently is serving as the Series Editor for the International Series in Operations Research and Management Science being published by Kluwer Academic Publishers. In addition to Introduction to Operations Research and the two companion volumes, Introduction to Mathematical Programming and Introduction to Stochastic Models in Operations Research, his books are The Evaluation of Risky Interrelated Investments (NorthHolland, 1969), Queueing Tables and Graphs (Elsevier North-Holland, 1981, co-authored by O. S. Yu, with D. M. Avis, L. D. Fossett, F. D. Lo, and M. I. Reiman), and Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets (Irwin/McGraw-Hill, co-authored by M. S. Hillier and G. J. Lieberman). The late Gerald J. Lieberman sadly passed away shortly before the completion of this edition. He had been Professor Emeritus of Operations Research and Statistics at Stanford University, where he was the founding chair of the Department of Operations Research. He was both an engineer (having received an undergraduate degree in mechanical engineering from Cooper Union) and an operations research statistician (with an A.M. from Columbia University in mathematical statistics, and a Ph.D. from Stanford University in statistics). Dr. Lieberman was one of Stanford’s most eminent leaders in recent decades. After chairing the Department of Operations Research, he served as Associate Dean of the School of Humanities and Sciences, Vice Provost and Dean of Research, Vice Provost and Dean of Graduate Studies, Chair of the Faculty Senate, member of the University Advisory Board, and Chair of the Centennial Celebration Committee. He also served as Provost or Acting Provost under three different Stanford presidents. Throughout these years of university leadership, he also remained active professionally. His research was in the stochastic areas of operations research, often at the interface of applied probability and statistics. He published extensively in the areas of reliability and quality control, and in the modeling of complex systems, including their optimal design, when resources are limited. Highly respected as a senior statesman of the field of operations research, Dr. Lieberman served in numerous leadership roles, including as the elected President of The Institute of Management Sciences. His professional honors included being elected to the National Academy of Engineering, receiving the Shewhart Medal of the American Society for Quality Control, receiving the Cuthbertson Award for exceptional service to Stanford University, and serving as a fellow at the Center for Advanced Study in the Behavioral Sciences. In addition, the Institute of Operations Research and the Management Sciences (INFORMS) awarded him and Dr. Hillier the honorable mention award for the 1995 Lanchester Prize for the 6th edition of this book. In 1996, INFORMS also awarded him the prestigious Kimball Medal for his exceptional contributions to the field of operations research and management science. In addition to Introduction to Operations Research and the two companion volumes, Introduction to Mathematical Programming and Introduction to Stochastic Models in Operations Research, his books are Handbook of Industrial Statistics (Prentice-Hall, 1955, co-authored by A. H. Bowker), Tables of the Non-Central t-Distribution (Stanford University Press, 1957, co-authored by G. J. Resnikoff), Tables of the Hypergeometric Probability Distribution (Stanford University Press, 1961, co-authored by D. Owen), Engineering Statistics, Second Edition (Prentice-Hall, 1972, co-authored by A. H. Bowker), and Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets (Irwin/McGraw-Hill, 2000, co-authored by F. S. Hillier and M. S. Hillier).

ABOUT THE CASE WRITERS

Karl Schmedders is assistant professor in the Department of Managerial Economics and Decision Sciences at the Kellogg Graduate School of Management (Northwestern University), where he teaches quantitative methods for managerial decision making. His research interests include applications of operations research in economic theory, general equilibrium theory with incomplete markets, asset pricing, and computational economics. Dr. Schmedders received his doctorate in operations research from Stanford University, where he taught both undergraduate and graduate classes in operations research. Among the classes taught was a case studies course in operations research, and he subsequently was invited to speak at a conference sponsored by the Institute of Operations Research and the Management Sciences (INFORMS) about his successful experience with this course. He received several teaching awards at Stanford, including the university’s prestigious Walter J. Gores Teaching Award. Molly Stephens is currently pursuing a J.D. degree with a concentration in technology and law. She graduated from Stanford University with a B.S. in Industrial Engineering and an M.S. in Operations Research. A champion debater in both high school and college, and president of the Stanford Debating Society, Ms. Stephens taught public speaking in Stanford’s School of Engineering and served as a teaching assistant for a case studies course in operations research. As a teaching assistant, she analyzed operations research problems encountered in the real world and the transformation of these problems into classroom case studies. Her research was rewarded when she won an undergraduate research grant from Stanford to continue her work and was invited to speak at an INFORMS conference to present her conclusions regarding successful classroom case studies. Following graduation, Ms. Stephens worked at Andersen Consulting as a systems integrator, experiencing real cases from the inside, before resuming her graduate studies.

ix

DEDICATION

To the memory of our parents and To the memory of one of the true giants of our field, Jerry Lieberman, whose recent passing prevented him from seeing the publication of this edition

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It now is 33 years since the first edition of this book was published in 1967. We have been humbled by having had both the privilege and the responsibility of introducing so many students around the world to our field over such a long span of time. With each new edition, we have worked toward the goal of meeting the changing needs of new generations of students by helping to define the modern approach to teaching the current status of operations research effectively at the introductory level. Over 33 years, much has changed in both the field and the pedagogical needs of the students being introduced to the field. These changes have been reflected in the substantial revisions of successive editions of this book. We believe that this is true for the current 7th edition as well. The enthusiastic response to our first six editions has been most gratifying. It was a particular pleasure to have the 6th edition receive honorable mention for the 1995 INFORMS Lanchester Prize (the prize awarded for the year’s most outstanding Englishlanguage publication of any kind in the field of operations research), including receiving the following citation. “This is the latest edition of the textbook that has introduced approximately one-half million students to the methods and models of Operations Research. While adding material on a variety of new topics, the sixth edition maintains the high standard of clarity and expositional excellence for which the authors have long been known. In honoring this work, the prize committee noted the enormous cumulative impact that the Hillier-Lieberman text has had on the development of our field, not only in the United States but also around the world through its many foreign-language editions.” As we enter a new millennium, the particular challenge for this new edition was to revise a book with deep roots in the 20th century so thoroughly that it would become fully suited for the 21st century. We made a special effort to meet this challenge, especially in regard to the software and pedagogy in the book.

A WEALTH OF SOFTWARE OPTIONS The new CD-ROM that accompanies the book provides an exciting array of software options that reflect current practice. One option is to use the increasingly popular spreadsheet approach with Excel and its Solver. Using spreadsheets as a key medium of instruction clearly is one new wave in xxiii

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the teaching of operations research. The new Sec. 3.6 describes and illustrates how to use Excel and its Solver to formulate and solve linear programming models on a spreadsheet. Similar discussions and examples also are included in several subsequent chapters for other kinds of models. In addition, the CD-ROM provides an Excel file for many of the chapters that displays the spreadsheet formulation and solution for the relevant examples in the chapter. Several of the Excel files also include a number of Excel templates for solving the models in the chapter. Another key resource is a collection of Excel add-ins on the CD-ROM (Premium Solver, TreePlan, SensIt, and RiskSim) that are integrated into the corresponding chapters. In addition, Sec. 22.6 describes how some simulations can be performed efficiently on spreadsheets by using another popular Excel add-in (@RISK) that can be downloaded temporarily from a website. Practitioners of operations research now usually use a modeling language to formulate and manage models of the very large size commonly encountered in practice. A modeling language system also will support one or more sophisticated software packages that can be called to solve a model once it has been formulated appropriately. The new Sec. 3.7 discusses the application of modeling languages and illustrates it with one modeling language (MPL) that is relatively amenable to student use. The student version of MPL is provided on the CD-ROM, along with an extensive MPL tutorial. Accompanying MPL as its primary solver is the student version of the renowned state-of-the-art software package, CPLEX. The student version of CONOPT also is provided as the solver for nonlinear programming. We are extremely pleased to be able to provide such powerful and popular software to students using this book. To further assist students, many of the chapters include an MPL/CPLEX file (or MPL/CPLEX/CONOPT file in the case of the nonlinear programming chapter) on the CD-ROM that shows how MPL and CPLEX would formulate and solve the relevant examples in the chapter. These files also illustrate how MPL and CPLEX can be integrated with spreadsheets. As described in the appendix to Chaps. 3 and 4, a third attractive option is to employ the student version of the popular and student-friendly software package LINDO and its modeling language companion LINGO. Both packages can be downloaded free from the LINDO Systems website. Associated tutorial material is included on the CD-ROM, along with a LINDO/LINGO file for many of the chapters showing how LINDO and LINGO would formulate and solve the relevant examples in the chapter. Once again, integration with spreadsheets also is illustrated. Complementing all these options on the CD-ROM is an updated version of the tutorial software that many instructors have found so useful for their students with the 5th and 6th editions. A program called OR Tutor provides 16 demonstration examples from the 6th edition, but now with an attractive new design based on JavaScript. These demos vividly demonstrate the evolution of an algorithm in ways that cannot be duplicated on the printed page. Most of the interactive routines from the 6th edition also are included on the CD-ROM, but again with an attractive new design. This design features a spreadsheet format based on VisualBasic. Each of the interactive routines enables the student to interactively execute one of the algorithms of operations research, making the needed decision at each step while the computer does the needed arithmetic. By enabling the student to focus on concepts rather than mindless number crunching when doing homework to learn an algorithm, we have found that these interactive routines make the learning process far more efficient and effective as well as more stimulating. In addition to these

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routines, the CD-ROM includes a few of the automatic routines from the 6th edition (again redesigned with VisualBasic) for those cases that are not covered by the software options described above. We were very fortunate to have the services of Michael O’Sullivan, a talented programmer and an advanced Ph.D. student in operations research at Stanford, to do all this updating of the software that had been developed by Mark S. Hillier for the 5th and 6th editions. Microsoft Project is introduced in Chap. 10 as a useful tool for project management. This software package also is included on the CD-ROM.

NEW EMPHASES Today’s students in introductory operations research courses tend to be very interested in learning more about the relevance of the material being covered, including how it is actually being used in practice. Therefore, without diluting any of the features of the 6th edition, the focus of the revision for this edition has been on increasing the motivation and excitement of the students by making the book considerably more “real world” oriented and accessible. The new emphasis on the kinds of software that practitioners use is one thrust in this direction. Other major new features are outlined below. Twenty-five elaborate new cases, embedded in a realistic setting and employing a stimulating storytelling approach, have been added at the end of the problem sections. All but one of these cases were developed jointly by two talented case writers, Karl Schmedders (a faculty member at the Kellogg Graduate School of Management at Northwestern University) and Molly Stephens (recently an operations research consultant with Andersen Consulting). We also have further fleshed out six cases that were in the 6th edition. The cases generally require relatively challenging and comprehensive analyses with substantial use of the computer. Therefore, they are suitable for student projects, working either individually or in teams, and can then lead to class discussion of the analysis. A complementary new feature is that many new problems embedded in a realistic setting have been added to the problem section of many chapters. Some of the current problems also have been fleshed out in a more interesting way. This edition also places much more emphasis on providing perspective in terms of what is actually happening in the practice of operations research. What kinds of applications are occurring? What sizes of problems are being solved? Which models and techniques are being used most widely? What are their shortcomings and what new developments are beginning to address these shortcomings? These kinds of questions are being addressed to convey the relevance of the techniques under discussion. Eight new sections (Secs. 10.7, 12.2, 15.6, 18.5, 19.8, 20.1, 20.10, and 22.2) are fully devoted to discussing the practice of operations research in such ways, along with briefer mentions elsewhere. The new emphases described above benefited greatly from our work in developing our recent new textbook with Mark S. Hillier (Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Irwin/McGraw-Hill, 2000). That book has a very different orientation from this one. It is aimed directly at business students rather than students who may be in engineering and the mathematical sciences, and it provides almost no coverage of the mathematics and algorithms of operations research. Nevertheless, its applied orientation enabled us to adapt some excellent material developed for that book to provide a more well-rounded coverage in this edition.

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OTHER FEATURES In addition to all the new software and new emphases just described, this edition received a considerable number of other enhancements as well. The previous section on project planning and control with PERT/CPM has been replaced by a complete new chapter (Chap. 10) with an applied orientation. Using the activity-on-node (AON) convention, this chapter provides an extensive modern treatment of the topic in a very accessible way. Other new topics not yet mentioned include the SOB mnemonic device for determining the form of constraints in the dual problem (in Sec. 6.4), 100 percent rules for simultaneous changes when conducting sensitivity analysis (in Sec. 6.7), sensitivity analysis with Bayes’ decision rule (in Sec. 15.2), a probability tree diagram for calculating posterior probabilities (in Sec. 15.3), a single-server variation of the nonpreemptive priorities model where the service for different priority classes of customers now have different mean service rates (in Sec. 17.8), a new simpler analysis of a stochastic continuous-review inventory model (Sec. 19.5), the mean absolute deviation as a measure of performance for forecasting methods (in Sec. 20.7), and the elements of a major simulation study (Sec. 22.5). We also have added much supplementary text material on the book’s new website, www.mhhe.com/hillier. Some of these supplements are password protected, but are available to all instructors who adopt this textbook. For the most part, this material appeared in previous editions of this book and then was subsequently deleted (for space reasons), to the disappointment of some instructors. Some also appeared in our Introduction to Mathematical Programming textbook. As delineated in the table of contents, this supplementary material includes a chapter on additional special types of linear programming problems, a review or primer chapter on probability theory, and a chapter on reliability, along with supplements to a few chapters in the book. In addition to providing this supplementary text material, the website will give updates about the book, including an errata, as the need arises. We made two changes in the order of the chapters. The decision analysis chapter has been moved forward to Chap. 15 in front of the stochastic chapters. The game theory chapter has been moved backward to Chap. 14 to place it next to the related decision analysis chapter. We believe that these changes provide a better transition from topics that are mainly deterministic to those that are mainly stochastic. Every chapter has received significant revision and updating, ranging from modest refining to extensive rewriting. Chapters receiving a particularly major revision and reorganization included Chaps. 15 (Decision Analysis), 19 (Inventory Theory), 20 (Forecasting), and 22 (Simulation). Many sections in the linear programming and mathematical programming chapters also received major revisions and updating. The overall thrust of all the revision efforts has been to build upon the strengths of previous editions while thoroughly updating and clarifying the material in a contemporary setting to fully meet the needs of today’s students. We think that the net effect has been to make this edition even more of a “student’s book”—clear, interesting, and well-organized with lots of helpful examples and illustrations, good motivation and perspective, easy-to-find important material, and enjoyable homework, without too much notation, terminology, and dense mathematics. We believe

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and trust that the numerous instructors who have used previous editions will agree that this is the best edition yet. This feeling has been reinforced by the generally enthusiastic reviews of drafts of this edition. The prerequisites for a course using this book can be relatively modest. As with previous editions, the mathematics has been kept at a relatively elementary level. Most of Chaps. 1 to 14 (introduction, linear programming, and mathematical programming) require no mathematics beyond high school algebra. Calculus is used only in Chaps. 13 (Nonlinear Programming) and in one example in Chap. 11 (Dynamic Programming). Matrix notation is used in Chap. 5 (The Theory of the Simplex Method), Chap. 6 (Duality Theory and Sensitivity Analysis), Sec. 7.4 (An Interior-Point Algorithm), and Chap. 13, but the only background needed for this is presented in Appendix 4. For Chaps. 15 to 22 (probabilistic models), a previous introduction to probability theory is assumed, and calculus is used in a few places. In general terms, the mathematical maturity that a student achieves through taking an elementary calculus course is useful throughout Chaps. 15 to 22 and for the more advanced material in the preceding chapters. The content of the book is aimed largely at the upper-division undergraduate level (including well-prepared sophomores) and at first-year (master’s level) graduate students. Because of the book’s great flexibility, there are many ways to package the material into a course. Chapters 1 and 2 give an introduction to the subject of operations research. Chapters 3 to 14 (on linear programming and on mathematical programming) may essentially be covered independently of Chaps. 15 to 22 (on probabilistic models), and vice versa. Furthermore, the individual chapters among Chaps. 3 to 14 are almost independent, except that they all use basic material presented in Chap. 3 and perhaps in Chap. 4. Chapter 6 and Sec. 7.2 also draw upon Chap. 5. Sections 7.1 and 7.2 use parts of Chap. 6. Section 9.6 assumes an acquaintance with the problem formulations in Secs. 8.1 and 8.3, while prior exposure to Secs. 7.3 and 8.2 is helpful (but not essential) in Sec. 9.7. Within Chaps. 15 to 22, there is considerable flexibility of coverage, although some integration of the material is available. An elementary survey course covering linear programming, mathematical programming, and some probabilistic models can be presented in a quarter (40 hours) or semester by selectively drawing from material throughout the book. For example, a good survey of the field can be obtained from Chaps. 1, 2, 3, 4, 15, 17, 19, 20, and 22, along with parts of Chaps. 9, 11, 12, and 13. A more extensive elementary survey course can be completed in two quarters (60 to 80 hours) by excluding just a few chapters, for example, Chaps. 7, 14, and 21. Chapters 1 to 8 (and perhaps part of Chap. 9) form an excellent basis for a (one-quarter) course in linear programming. The material in Chaps. 9 to 14 covers topics for another (one-quarter) course in other deterministic models. Finally, the material in Chaps. 15 to 22 covers the probabilistic (stochastic) models of operations research suitable for presentation in a (one-quarter) course. In fact, these latter three courses (the material in the entire text) can be viewed as a basic one-year sequence in the techniques of operations research, forming the core of a master’s degree program. Each course outlined has been presented at either the undergraduate or the graduate level at Stanford University, and this text has been used in the manner suggested. To assist the instructor who will be covering only a portion of the chapters and who prefers a slimmer book containing only those chapters, all the material (including the supplementary text material on the book’s website) has been placed in McGraw-Hill’s PRIMIS

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system. This system enables an instructor to pick and choose precisely which material to include in a self-designed book, and then to order copies for the students at an economical price. For example, this enables instructors who previously used our Introduction to Mathematical Programming or Introduction to Stochastic Models in Operations Research textbooks to obtain updated versions of the same material from the PRIMIS system. For this reason, we will not be publishing new separate editions of these other books. Again, as in previous editions, we thank our wives, Ann and Helen, for their encouragement and support during the long process of preparing this 7th edition. Our children, David, John, and Mark Hillier, Janet Lieberman Argyres, and Joanne, Michael, and Diana Lieberman, have literally grown up with the book and our periodic hibernations to prepare a new edition. Now, most of them have used the book as a text in their own college courses, given considerable advice, and even (in the case of Mark Hillier) become a software collaborator. It is a joy to see them and (we trust) the book reach maturity together. And now I must add a very sad note. My close friend and co-author, Jerry Lieberman, passed away on May 18, 1999, while this edition was in preparation, so I am writing this preface on behalf of both of us. Jerry was one of the great leaders of our field and he had a profound influence on my life. More than a third of a century ago, we embarked on a mission together to attempt to develop a path-breaking book for teaching operations research at the introductory level. Ever since, we have striven to meet and extend the same high standards for each new edition. Having worked so closely with Jerry for so many years, I believe I understand well how he would want the book to evolve to meet the needs of each new generation of students. As the substantially younger co-author, I am grateful that I am able to carry on our joint mission to continue to update and improve the book, both with this edition and with future editions as well. It is the least I can do to honor Jerry. I welcome your comments, suggestions, and errata to help me improve the book in the future.

ACKNOWLEDGMENTS We are indebted to an excellent group of reviewers who provided sage advice throughout the revision process. This group included Jeffery Cochran, Arizona State University; Yahya Fathi, North Carolina State University; Yasser Hosni and Charles Reilly, University of Central Florida; Cerry Klein, University of Missouri—Columbia; Robert Lipset, Ohio University; Mark Parker, United States Air Force Academy; Christopher Rump, State University of New York at Buffalo; and Ahmad Seifoddini, California Polytechnic State University—San Luis Obispo. We also received helpful advice from Judith Liebman, Siegfried Schaible, David Sloan, and Arthur F. Veinott, Jr., as well as many instructors who sent us letters or e-mail messages. In addition, we also thank many dozens of Stanford students and many students at other universities who gave us helpful written suggestions. This edition was very much of a team effort. Our case writers, Karl Schmedders and Molly Stephens (both graduates of our department), made a vital contribution. One of our department’s current Ph.D. students, Roberto Szechtman, did an excellent job in preparing the solutions manual. Another Ph.D. student, Michael O’Sullivan, was very skillful in updating the software that Mark Hillier had developed for the 5th and 6th editions. Mark

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(who was born the same year as the first edition and now is a tenured faculty member in the Management Science Department at the University of Washington) helped to oversee this updating and also provided both the spreadsheets and the Excel files (including many Excel templates) for this edition. Linus Schrage of the University of Chicago and LINDO Systems (and who took an introductory operations research course from me 37 years ago) supervised the development of LINGO/LINDO files for the various chapters as well as providing tutorial material for the CD-ROM. Another long-time friend, Bjarni Kristjansson (who heads Maximal Software), did the same thing for the MPL/CPLEX files and MPL tutorial material, as well as arranging to provide student versions of MPL, CPLEX, CONOPT, and OptiMax 2000 for the CD-ROM. One of our department’s Ph.D. graduates, Irv Lustig, was the ILOG project manager for providing CPLEX. Linus, Bjarni, and Irv all were helpful in checking material going into this edition regarding their software. Ann Hillier devoted numerous long days and nights to sitting with a Macintosh, doing word processing and constructing many figures and tables, in addition to endless cutting and pasting, photocopying, and FedExing of material. Helen Lieberman also carried a heavy burden in supporting Jerry. They all were vital members of the team. The inside back cover lists the various companies and individuals who have provided software for the CD-ROM. We greatly appreciate their key contributions. It was a real pleasure working with McGraw-Hill’s thoroughly professional editorial and production staff, including Eric Munson (executive editor), Maja Lorkovic (developmental editor), and Christine Vaughan (project manager). Frederick S. Hillier Stanford University ([email protected])

January 2000

TABLE OF CONTENTS

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CHAPTER 1 Introduction

1

1.1 The Origins of Operations Research 1 1.2 The Nature of Operations Research 2 1.3 The Impact of Operations Research 3 1.4 Algorithms and OR Courseware 5 Problems 6 CHAPTER 2 Overview of the Operations Research Modeling Approach 2.1 Defining the Problem and Gathering Data 2.2 Formulating a Mathematical Model 10 2.3 Deriving Solutions from the Model 14 2.4 Testing the Model 16 2.5 Preparing to Apply the Model 18 2.6 Implementation 20 2.7 Conclusions 21 Selected References 22 Problems 22 CHAPTER 3 Introduction to Linear Programming

7

7

24

3.1 Prototype Example 25 3.2 The Linear Programming Model 31 3.3 Assumptions of Linear Programming 36 3.4 Additional Examples 44 3.5 Some Case Studies 61 3.6 Displaying and Solving Linear Programming Models on a Spreadsheet 3.7 Formulating Very Large Linear Programming Models 73 3.8 Conclusions 79 Appendix 3.1 The LINGO Modeling Language 79

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Selected References 89 Learning Aids for This Chapter in Your OR Courseware 90 Problems 90 Case 3.1 Auto Assembly 103 Case 3.2 Cutting Cafeteria Costs 104 Case 3.3 Staffing a Call Center 106 CHAPTER 4 Solving Linear Programming Problems: The Simplex Method

109

4.1 The Essence of the Simplex Method 109 4.2 Setting Up the Simplex Method 114 4.3 The Algebra of the Simplex Method 118 4.4 The Simplex Method in Tabular Form 123 4.5 Tie Breaking in the Simplex Method 128 4.6 Adapting to Other Model Forms 132 4.7 Postoptimality Analysis 152 4.8 Computer Implementation 160 4.9 The Interior-Point Approach to Solving Linear Programming Problems 4.10 Conclusions 168 Appendix 4.1 An Introduction to Using LINDO 169 Selected References 171 Learning Aids for This Chapter in Your OR Courseware 172 Problems 172 Case 4.1 Fabrics and Fall Fashions 182 Case 4.2 New Frontiers 185 Case 4.3 Assigning Students to Schools 188 CHAPTER 5 The Theory of the Simplex Method

190

5.1 Foundations of the Simplex Method 190 5.2 The Revised Simplex Method 202 5.3 A Fundamental Insight 212 5.4 Conclusions 220 Selected References 220 Learning Aids for This Chapter in Your OR Courseware 221 Problems 221 CHAPTER 6 Duality Theory and Sensitivity Analysis 6.1 6.2 6.3 6.4 6.5 6.6

230

The Essence of Duality Theory 231 Economic Interpretation of Duality 239 Primal-Dual Relationships 242 Adapting to Other Primal Forms 247 The Role of Duality Theory in Sensitivity Analysis The Essence of Sensitivity Analysis 254

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6.7 Applying Sensitivity Analysis 262 6.8 Conclusions 284 Selected References 284 Learning Aids for This Chapter in Your OR Courseware 285 Problems 285 Case 6.1 Controlling Air Pollution 302 Case 6.2 Farm Management 304 Case 6.3 Assigning Students to Schools (Revisited) 307 CHAPTER 7 Other Algorithms for Linear Programming

309

7.1 The Dual Simplex Method 309 7.2 Parametric Linear Programming 312 7.3 The Upper Bound Technique 317 7.4 An Interior-Point Algorithm 320 7.5 Linear Goal Programming and Its Solution Procedures 332 7.6 Conclusions 339 Selected References 340 Learning Aids for This Chapter in Your OR Courseware 340 Problems 341 Case 7.1 A Cure for Cuba 347 CHAPTER 8 The Transportation and Assignment Problems

350

8.1 The Transportation Problem 351 8.2 A Streamlined Simplex Method for the Transportation Problem 8.3 The Assignment Problem 381 8.4 Conclusions 391 Selected References 391 Learning Aids for This Chapter in Your OR Courseware 392 Problems 392 Case 8.1 Shipping Wood to Market 401 Case 8.2 Project Pickings 402 CHAPTER 9 Network Optimization Models

405

9.1 Prototype Example 406 9.2 The Terminology of Networks 407 9.3 The Shortest-Path Problem 411 9.4 The Minimum Spanning Tree Problem 415 9.5 The Maximum Flow Problem 420 9.6 The Minimum Cost Flow Problem 429 9.7 The Network Simplex Method 438 9.8 Conclusions 448 Selected References 449

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Learning Aids for This Chapter in Your OR Courseware 449 Problems 450 Case 9.1 Aiding Allies 458 Case 9.2 Money in Motion 464 CHAPTER 10 Project Management with PERT/CPM

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10.1 A Prototype Example—The Reliable Construction Co. Project 10.2 Using a Network to Visually Display a Project 470 10.3 Scheduling a Project with PERT/CPM 475 10.4 Dealing with Uncertain Activity Durations 485 10.5 Considering Time-Cost Trade-Offs 492 10.6 Scheduling and Controlling Project Costs 502 10.7 An Evaluation of PERT/CPM 508 10.8 Conclusions 512 Selected References 513 Learning Aids for This Chapter in Your OR Courseware 514 Problems 514 Case 10.1 Steps to Success 524 Case 10.2 “School’s out forever . . .” 527 CHAPTER 11 Dynamic Programming

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11.1 A Prototype Example for Dynamic Programming 533 11.2 Characteristics of Dynamic Programming Problems 538 11.3 Deterministic Dynamic Programming 541 11.4 Probabilistic Dynamic Programming 562 11.5 Conclusions 568 Selected References 568 Learning Aids for This Chapter in Your OR Courseware 568 Problems 569 CHAPTER 12 Integer Programming

576

12.1 Prototype Example 577 12.2 Some BIP Applications 580 12.3 Innovative Uses of Binary Variables in Model Formulation 585 12.4 Some Formulation Examples 591 12.5 Some Perspectives on Solving Integer Programming Problems 600 12.6 The Branch-and-Bound Technique and Its Application to Binary Integer Programming 604 12.7 A Branch-and-Bound Algorithm for Mixed Integer Programming 616 12.8 Other Developments in Solving BIP Problems 622 12.9 Conclusions 630 Selected References 631

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Learning Aids for This Chapter in Your OR Courseware Problems 632 Case 12.1 Capacity Concerns 642 Case 12.2 Assigning Art 645 Case 12.3 Stocking Sets 649 Case 12.4 Assigning Students to Schools (Revisited Again) CHAPTER 13 Nonlinear Programming

631

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13.1 Sample Applications 655 13.2 Graphical Illustration of Nonlinear Programming Problems 659 13.3 Types of Nonlinear Programming Problems 664 13.4 One-Variable Unconstrained Optimization 670 13.5 Multivariable Unconstrained Optimization 673 13.6 The Karush-Kuhn-Tucker (KKT) Conditions for Constrained Optimization 13.7 Quadratic Programming 683 13.8 Separable Programming 690 13.9 Convex Programming 697 13.10 Nonconvex Programming 702 13.11 Conclusions 706 Selected References 706 Learning Aids for This Chapter in Your OR Courseware 707 Problems 708 Case 13.1 Savvy Stock Selection 720 CHAPTER 14 Game Theory 726 14.1 The Formulation of Two-Person, Zero-Sum Games 726 14.2 Solving Simple Games—A Prototype Example 728 14.3 Games with Mixed Strategies 733 14.4 Graphical Solution Procedure 735 14.5 Solving by Linear Programming 738 14.6 Extensions 741 14.7 Conclusions 742 Selected References 743 Learning Aids for This Chapter in Your OR Courseware 743 Problems 743 CHAPTER 15 Decision Analysis 15.1 15.2 15.3 15.4 15.5

749

A Prototype Example 750 Decision Making without Experimentation 751 Decision Making with Experimentation 758 Decision Trees 764 Utility Theory 770

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15.6 The Practical Application of Decision Analysis 778 15.7 Conclusions 781 Selected References 781 Learning Aids for This Chapter in Your OR Courseware 782 Problems 782 Case 15.1 Brainy Business 795 Case 15.2 Smart Steering Support 798 CHAPTER 16 Markov Chains

802

16.1 Stochastic Processes 802 16.2 Markov Chains 803 16.3 Chapman-Kolmogorov Equations 808 16.4 Classification of States of a Markov Chain 810 16.5 Long-Run Properties of Markov Chains 812 16.6 First Passage Times 818 16.7 Absorbing States 820 16.8 Continuous Time Markov Chains 822 Selected References 827 Learning Aids for This Chapter in Your OR Courseware 828 Problems 828 CHAPTER 17 Queueing Theory 834 17.1 Prototype Example 835 17.2 Basic Structure of Queueing Models 835 17.3 Examples of Real Queueing Systems 840 17.4 The Role of the Exponential Distribution 841 17.5 The Birth-and-Death Process 848 17.6 Queueing Models Based on the Birth-and-Death Process 852 17.7 Queueing Models Involving Nonexponential Distributions 871 17.8 Priority-Discipline Queueing Models 879 17.9 Queueing Networks 885 17.10 Conclusions 889 Selected References 890 Learning Aids for This Chapter in Your OR Courseware 890 Problems 891 Case 17.1 Reducing In-Process Inventory 905 CHAPTER 18 The Application of Queueing Theory 18.1 Examples 907 18.2 Decision Making 909 18.3 Formulation of Waiting-Cost Functions

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18.4 Decision Models 917 18.5 Some Award-Winning Applications of Queueing Theory 18.6 Conclusions 926 Selected References 926 Learning Aids for This Chapter in Your OR Courseware 926 Problems 927 Case 18.1 Queueing Quandary 932

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CHAPTER 19 Inventory Theory 935 19.1 Examples 936 19.2 Components of Inventory Models 938 19.3 Deterministic Continuous-Review Models 941 19.4 A Deterministic Periodic-Review Model 951 19.5 A Stochastic Continuous-Review Model 956 19.6 A Stochastic Single-Period Model for Perishable Products 19.7 Stochastic Periodic-Review Models 975 19.8 Larger Inventory Systems in Practice 983 19.9 Conclusions 987 Selected References 987 Learning Aids for This Chapter in Your OR Courseware 987 Problems 988 Case 19.1 Brushing Up on Inventory Control 1000 Case 19.2 TNT: Tackling Newsboy’s Teachings 1002 Case 19.3 Jettisoning Surplus Stock 1004

961

CHAPTER 20 Forecasting 1009 20.1 Some Applications of Forecasting 1010 20.2 Judgmental Forecasting Methods 1013 20.3 Time Series 1014 20.4 Forecasting Methods for a Constant-Level Model 1016 20.5 Incorporating Seasonal Effects into Forecasting Methods 1018 20.6 An Exponential Smoothing Method for a Linear Trend Model 1021 20.7 Forecasting Errors 1025 20.8 Box-Jenkins Method 1026 20.9 Causal Forecasting with Linear Regression 1028 20.10 Forecasting in Practice 1036 20.11 Conclusions 1038 Selected References 1038 Learning Aids for This Chapter in Your OR Courseware 1038 Problems 1039 Case 20.1 Finagling the Forecasts 1048

xx

TABLE OF CONTENTS

CHAPTER 21 Markov Decision Processes

1053

21.1 A Prototype Example 1053 21.2 A Model for Markov Decision Processes 1056 21.3 Linear Programming and Optimal Policies 1059 21.4 Policy Improvement Algorithm for Finding Optimal Policies 21.5 Discounted Cost Criterion 1069 21.6 Conclusions Selected References 1077 Learning Aids for This Chapter in Your OR Courseware 1078 Problems 1078

1064

CHAPTER 22 Simulation 1084 22.1 The Essence of Simulation 1084 22.2 Some Common Types of Applications of Simulation 1097 22.3 Generation of Random Numbers 1101 22.4 Generation of Random Observations from a Probability Distribution 22.5 Outline of a Major Simulation Study 1110 22.6 Performing Simulations on Spreadsheets 1115 22.7 Variance-Reducing Techniques 1126 22.8 Regenerative Method of Statistical Analysis 1131 22.9 Conclusions 1138 Selected References 1140 Learning Aids for This Chapter in Your OR Courseware 1140 Problems 1141 Case 22.1 Planning Planers 1151 Case 22.2 Pricing under Pressure 1153 APPENDIXES 1. Documentation for the OR Courseware 1156 2. Convexity 1159 3. Classical Optimization Methods 1165 4. Matrices and Matrix Operations 1169 5. Tables 1174 PARTIAL ANSWERS TO SELECTED PROBLEMS INDEXES Author Index 1195 Subject Index 1199

1176

1105

1 Introduction

1.1

THE ORIGINS OF OPERATIONS RESEARCH Since the advent of the industrial revolution, the world has seen a remarkable growth in the size and complexity of organizations. The artisans’ small shops of an earlier era have evolved into the billion-dollar corporations of today. An integral part of this revolutionary change has been a tremendous increase in the division of labor and segmentation of management responsibilities in these organizations. The results have been spectacular. However, along with its blessings, this increasing specialization has created new problems, problems that are still occurring in many organizations. One problem is a tendency for the many components of an organization to grow into relatively autonomous empires with their own goals and value systems, thereby losing sight of how their activities and objectives mesh with those of the overall organization. What is best for one component frequently is detrimental to another, so the components may end up working at cross purposes. A related problem is that as the complexity and specialization in an organization increase, it becomes more and more difficult to allocate the available resources to the various activities in a way that is most effective for the organization as a whole. These kinds of problems and the need to find a better way to solve them provided the environment for the emergence of operations research (commonly referred to as OR). The roots of OR can be traced back many decades, when early attempts were made to use a scientific approach in the management of organizations. However, the beginning of the activity called operations research has generally been attributed to the military services early in World War II. Because of the war effort, there was an urgent need to allocate scarce resources to the various military operations and to the activities within each operation in an effective manner. Therefore, the British and then the U.S. military management called upon a large number of scientists to apply a scientific approach to dealing with this and other strategic and tactical problems. In effect, they were asked to do research on (military) operations. These teams of scientists were the first OR teams. By developing effective methods of using the new tool of radar, these teams were instrumental in winning the Air Battle of Britain. Through their research on how to better manage convoy and antisubmarine operations, they also played a major role in winning the Battle of the North Atlantic. Similar efforts assisted the Island Campaign in the Pacific. When the war ended, the success of OR in the war effort spurred interest in applying OR outside the military as well. As the industrial boom following the war was run1

2

1 INTRODUCTION

ning its course, the problems caused by the increasing complexity and specialization in organizations were again coming to the forefront. It was becoming apparent to a growing number of people, including business consultants who had served on or with the OR teams during the war, that these were basically the same problems that had been faced by the military but in a different context. By the early 1950s, these individuals had introduced the use of OR to a variety of organizations in business, industry, and government. The rapid spread of OR soon followed. At least two other factors that played a key role in the rapid growth of OR during this period can be identified. One was the substantial progress that was made early in improving the techniques of OR. After the war, many of the scientists who had participated on OR teams or who had heard about this work were motivated to pursue research relevant to the field; important advancements in the state of the art resulted. A prime example is the simplex method for solving linear programming problems, developed by George Dantzig in 1947. Many of the standard tools of OR, such as linear programming, dynamic programming, queueing theory, and inventory theory, were relatively well developed before the end of the 1950s. A second factor that gave great impetus to the growth of the field was the onslaught of the computer revolution. A large amount of computation is usually required to deal most effectively with the complex problems typically considered by OR. Doing this by hand would often be out of the question. Therefore, the development of electronic digital computers, with their ability to perform arithmetic calculations thousands or even millions of times faster than a human being can, was a tremendous boon to OR. A further boost came in the 1980s with the development of increasingly powerful personal computers accompanied by good software packages for doing OR. This brought the use of OR within the easy reach of much larger numbers of people. Today, literally millions of individuals have ready access to OR software. Consequently, a whole range of computers from mainframes to laptops now are being routinely used to solve OR problems.

1.2

THE NATURE OF OPERATIONS RESEARCH As its name implies, operations research involves “research on operations.” Thus, operations research is applied to problems that concern how to conduct and coordinate the operations (i.e., the activities) within an organization. The nature of the organization is essentially immaterial, and, in fact, OR has been applied extensively in such diverse areas as manufacturing, transportation, construction, telecommunications, financial planning, health care, the military, and public services, to name just a few. Therefore, the breadth of application is unusually wide. The research part of the name means that operations research uses an approach that resembles the way research is conducted in established scientific fields. To a considerable extent, the scientific method is used to investigate the problem of concern. (In fact, the term management science sometimes is used as a synonym for operations research.) In particular, the process begins by carefully observing and formulating the problem, including gathering all relevant data. The next step is to construct a scientific (typically mathematical) model that attempts to abstract the essence of the real problem. It is then hypothesized that this model is a sufficiently precise representation of the essential features of the situation that the conclusions (solutions) obtained from the model are also

1.3 THE IMPACT OF OPERATIONS RESEARCH

3

valid for the real problem. Next, suitable experiments are conducted to test this hypothesis, modify it as needed, and eventually verify some form of the hypothesis. (This step is frequently referred to as model validation.) Thus, in a certain sense, operations research involves creative scientific research into the fundamental properties of operations. However, there is more to it than this. Specifically, OR is also concerned with the practical management of the organization. Therefore, to be successful, OR must also provide positive, understandable conclusions to the decision maker(s) when they are needed. Still another characteristic of OR is its broad viewpoint. As implied in the preceding section, OR adopts an organizational point of view. Thus, it attempts to resolve the conflicts of interest among the components of the organization in a way that is best for the organization as a whole. This does not imply that the study of each problem must give explicit consideration to all aspects of the organization; rather, the objectives being sought must be consistent with those of the overall organization. An additional characteristic is that OR frequently attempts to find a best solution (referred to as an optimal solution) for the problem under consideration. (We say a best instead of the best solution because there may be multiple solutions tied as best.) Rather than simply improving the status quo, the goal is to identify a best possible course of action. Although it must be interpreted carefully in terms of the practical needs of management, this “search for optimality” is an important theme in OR. All these characteristics lead quite naturally to still another one. It is evident that no single individual should be expected to be an expert on all the many aspects of OR work or the problems typically considered; this would require a group of individuals having diverse backgrounds and skills. Therefore, when a full-fledged OR study of a new problem is undertaken, it is usually necessary to use a team approach. Such an OR team typically needs to include individuals who collectively are highly trained in mathematics, statistics and probability theory, economics, business administration, computer science, engineering and the physical sciences, the behavioral sciences, and the special techniques of OR. The team also needs to have the necessary experience and variety of skills to give appropriate consideration to the many ramifications of the problem throughout the organization.

1.3

THE IMPACT OF OPERATIONS RESEARCH Operations research has had an impressive impact on improving the efficiency of numerous organizations around the world. In the process, OR has made a significant contribution to increasing the productivity of the economies of various countries. There now are a few dozen member countries in the International Federation of Operational Research Societies (IFORS), with each country having a national OR society. Both Europe and Asia have federations of OR societies to coordinate holding international conferences and publishing international journals in those continents. It appears that the impact of OR will continue to grow. For example, according to the U.S. Bureau of Labor Statistics, OR currently is one of the fastest-growing career areas for U.S. college graduates. To give you a better notion of the wide applicability of OR, we list some actual awardwinning applications in Table 1.1. Note the diversity of organizations and applications in the first two columns. The curious reader can find a complete article describing each application in the January–February issue of Interfaces for the year cited in the third col-

4

1 INTRODUCTION

TABLE 1.1 Some applications of operations research Organization

Nature of Application

The Netherlands Rijkswaterstaat

Develop national water management policy, including mix of new facilities, operating procedures, and pricing. Optimize production operations in chemical plants to meet production targets with minimum cost. Schedule shift work at reservation offices and airports to meet customer needs with minimum cost. Optimize refinery operations and the supply, distribution, and marketing of products. Optimally schedule and deploy police patrol officers with a computerized system. Optimally blend available ingredients into gasoline products to meet quality and sales requirements. Integrate a national network of spare parts inventories to improve service support.

Monsanto Corp.

United Airlines

Citgo Petroleum Corp. San Francisco Police Department Texaco, Inc.

IBM

Yellow Freight System, Inc. New Haven Health Department AT&T

Delta Airlines

Digital Equipment Corp. China

South African defense force Proctor and Gamble

Taco Bell

Hewlett-Packard

Optimize the design of a national trucking network and the routing of shipments. Design an effective needle exchange program to combat the spread of HIV/AIDS. Develop a PC-based system to guide business customers in designing their call centers. Maximize the profit from assigning airplane types to over 2500 domestic flights. Restructure the global supply chain of suppliers, plants, distribution centers, potential sites, and market areas. Optimally select and schedule massive projects for meeting the country’s future energy needs. Optimally redesign the size and shape of the defense force and its weapons systems. Redesign the North American production and distribution system to reduce costs and improve speed to market. Optimally schedule employees to provide desired customer service at a minimum cost. Redesign the sizes and locations of buffers in a printer production line to meet production goals.

Year of Publication*

Related Chapters†

1985

2–8, 13, 22

$15 million

1985

2, 12

$2 million

1986

2–9, 12, 17, 18, 20

$6 million

1987

2–9, 20

$70 million

1989

2–4, 12, 20

$11 million

1989

2, 13

$30 million

1990

2, 19, 22

1992 1993

2, 9, 13, 20, 22 2

$20 million $250 million less inventory $17.3 million

1993

17, 18, 22

33% less HIV/AIDS $750 million

1994

12

$100 million

1995

12

$800 million

1995

12

$425 million

1997

12

$1.1 billion

1997

8

$200 million

1998

12, 20, 22

$13 million

1998

17, 18

$280 million more revenue

*Pertains to a January–February issue of Interfaces in which a complete article can be found describing the application. † Refers to chapters in this book that describe the kinds of OR techniques used in the application.

Annual Savings

1.4 ALGORITHMS AND OR COURSEWARE

5

umn of the table. The fourth column lists the chapters in this book that describe the kinds of OR techniques that were used in the application. (Note that many of the applications combine a variety of techniques.) The last column indicates that these applications typically resulted in annual savings in the millions (or even tens of millions) of dollars. Furthermore, additional benefits not recorded in the table (e.g., improved service to customers and better managerial control) sometimes were considered to be even more important than these financial benefits. (You will have an opportunity to investigate these less tangible benefits further in Probs. 1.3-1 and 1.3-2.) Although most routine OR studies provide considerably more modest benefits than these award-winning applications, the figures in the rightmost column of Table 1.1 do accurately reflect the dramatic impact that large, well-designed OR studies occasionally can have. We will briefly describe some of these applications in the next chapter, and then we present two in greater detail as case studies in Sec. 3.5.

1.4

ALGORITHMS AND OR COURSEWARE An important part of this book is the presentation of the major algorithms (systematic solution procedures) of OR for solving certain types of problems. Some of these algorithms are amazingly efficient and are routinely used on problems involving hundreds or thousands of variables. You will be introduced to how these algorithms work and what makes them so efficient. You then will use these algorithms to solve a variety of problems on a computer. The CD-ROM called OR Courseware that accompanies the book will be a key tool for doing all this. One special feature in your OR Courseware is a program called OR Tutor. This program is intended to be your personal tutor to help you learn the algorithms. It consists of many demonstration examples that display and explain the algorithms in action. These “demos” supplement the examples in the book. In addition, your OR Courseware includes many interactive routines for executing the algorithms interactively in a convenient spreadsheet format. The computer does all the routine calculations while you focus on learning and executing the logic of the algorithm. You should find these interactive routines a very efficient and enlightening way of doing many of your homework problems. In practice, the algorithms normally are executed by commercial software packages. We feel that it is important to acquaint students with the nature of these packages that they will be using after graduation. Therefore, your OR Courseware includes a wealth of material to introduce you to three particularly popular software packages described below. Together, these packages will enable you to solve nearly all the OR models encountered in this book very efficiently. We have added our own automatic routines to the OR Courseware only in a few cases where these packages are not applicable. A very popular approach now is to use today’s premier spreadsheet package, Microsoft Excel, to formulate small OR models in a spreadsheet format. The Excel Solver then is used to solve the models. Your OR Courseware includes a separate Excel file for nearly every chapter in this book. Each time a chapter presents an example that can be solved using Excel, the complete spreadsheet formulation and solution is given in that chapter’s Excel file. For many of the models in the book, an Excel template also is pro-

6

1 INTRODUCTION

vided that already includes all the equations necessary to solve the model. Some Excel add-ins also are included on the CD-ROM. After many years, LINDO (and its companion modeling language LINGO) continues to be a dominant OR software package. Student versions of LINDO and LINGO now can be downloaded free from the Web. As for Excel, each time an example can be solved with this package, all the details are given in a LINGO/LINDO file for that chapter in your OR Courseware. CPLEX is an elite state-of-the-art software package that is widely used for solving large and challenging OR problems. When dealing with such problems, it is common to also use a modeling system to efficiently formulate the mathematical model and enter it into the computer. MPL is a user-friendly modeling system that uses CPLEX as its main solver. A student version of MPL and CPLEX is available free by downloading it from the Web. For your convenience, we also have included this student version in your OR Courseware. Once again, all the examples that can be solved with this package are detailed in MPL/CPLEX files for the corresponding chapters in your OR Courseware. We will further describe these three software packages and how to use them later (especially near the end of Chaps. 3 and 4). Appendix 1 also provides documentation for the OR Courseware, including OR Tutor. To alert you to relevant material in OR Courseware, the end of each chapter from Chap. 3 onward has a list entitled Learning Aids for This Chapter in Your OR Courseware. As explained at the beginning of the problem section for each of these chapters, symbols also are placed to the left of each problem number or part where any of this material (including demonstration examples and interactive routines) can be helpful.

PROBLEMS 1.3-1. Select one of the applications of operations research listed in Table 1.1. Read the article describing the application in the January–February issue of Interfaces for the year indicated in the third column. Write a two-page summary of the application and the benefits (including nonfinancial benefits) it provided.

1.3-2. Select three of the applications of operations research listed in Table 1.1. Read the articles describing the applications in the January–February issue of Interfaces for the years indicated in the third column. For each one, write a one-page summary of the application and the benefits (including nonfinancial benefits) it provided.

2 Overview of the Operations Research Modeling Approach The bulk of this book is devoted to the mathematical methods of operations research (OR). This is quite appropriate because these quantitative techniques form the main part of what is known about OR. However, it does not imply that practical OR studies are primarily mathematical exercises. As a matter of fact, the mathematical analysis often represents only a relatively small part of the total effort required. The purpose of this chapter is to place things into better perspective by describing all the major phases of a typical OR study. One way of summarizing the usual (overlapping) phases of an OR study is the following: 1. Define the problem of interest and gather relevant data. 2. Formulate a mathematical model to represent the problem. 3. Develop a computer-based procedure for deriving solutions to the problem from the model. 4. Test the model and refine it as needed. 5. Prepare for the ongoing application of the model as prescribed by management. 6. Implement. Each of these phases will be discussed in turn in the following sections. Most of the award-winning OR studies introduced in Table 1.1 provide excellent examples of how to execute these phases well. We will intersperse snippets from these examples throughout the chapter, with references to invite your further reading.

2.1

DEFINING THE PROBLEM AND GATHERING DATA In contrast to textbook examples, most practical problems encountered by OR teams are initially described to them in a vague, imprecise way. Therefore, the first order of business is to study the relevant system and develop a well-defined statement of the problem to be considered. This includes determining such things as the appropriate objectives, constraints on what can be done, interrelationships between the area to be studied and other areas of the organization, possible alternative courses of action, time limits for making a decision, and so on. This process of problem definition is a crucial one because it greatly affects how relevant the conclusions of the study will be. It is difficult to extract a “right” answer from the “wrong” problem! 7

8

2

OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

The first thing to recognize is that an OR team is normally working in an advisory capacity. The team members are not just given a problem and told to solve it however they see fit. Instead, they are advising management (often one key decision maker). The team performs a detailed technical analysis of the problem and then presents recommendations to management. Frequently, the report to management will identify a number of alternatives that are particularly attractive under different assumptions or over a different range of values of some policy parameter that can be evaluated only by management (e.g., the tradeoff between cost and benefits). Management evaluates the study and its recommendations, takes into account a variety of intangible factors, and makes the final decision based on its best judgment. Consequently, it is vital for the OR team to get on the same wavelength as management, including identifying the “right” problem from management’s viewpoint, and to build the support of management for the course that the study is taking. Ascertaining the appropriate objectives is a very important aspect of problem definition. To do this, it is necessary first to identify the member (or members) of management who actually will be making the decisions concerning the system under study and then to probe into this individual’s thinking regarding the pertinent objectives. (Involving the decision maker from the outset also is essential to build her or his support for the implementation of the study.) By its nature, OR is concerned with the welfare of the entire organization rather than that of only certain of its components. An OR study seeks solutions that are optimal for the overall organization rather than suboptimal solutions that are best for only one component. Therefore, the objectives that are formulated ideally should be those of the entire organization. However, this is not always convenient. Many problems primarily concern only a portion of the organization, so the analysis would become unwieldy if the stated objectives were too general and if explicit consideration were given to all side effects on the rest of the organization. Instead, the objectives used in the study should be as specific as they can be while still encompassing the main goals of the decision maker and maintaining a reasonable degree of consistency with the higher-level objectives of the organization. For profit-making organizations, one possible approach to circumventing the problem of suboptimization is to use long-run profit maximization (considering the time value of money) as the sole objective. The adjective long-run indicates that this objective provides the flexibility to consider activities that do not translate into profits immediately (e.g., research and development projects) but need to do so eventually in order to be worthwhile. This approach has considerable merit. This objective is specific enough to be used conveniently, and yet it seems to be broad enough to encompass the basic goal of profitmaking organizations. In fact, some people believe that all other legitimate objectives can be translated into this one. However, in actual practice, many profit-making organizations do not use this approach. A number of studies of U.S. corporations have found that management tends to adopt the goal of satisfactory profits, combined with other objectives, instead of focusing on long-run profit maximization. Typically, some of these other objectives might be to maintain stable profits, increase (or maintain) one’s share of the market, provide for product diversification, maintain stable prices, improve worker morale, maintain family control of the business, and increase company prestige. Fulfilling these objectives might achieve long-run profit maximization, but the relationship may be sufficiently obscure that it may not be convenient to incorporate them all into this one objective.

2.1 DEFINING THE PROBLEM AND GATHERING DATA

9

Furthermore, there are additional considerations involving social responsibilities that are distinct from the profit motive. The five parties generally affected by a business firm located in a single country are (1) the owners (stockholders, etc.), who desire profits (dividends, stock appreciation, and so on); (2) the employees, who desire steady employment at reasonable wages; (3) the customers, who desire a reliable product at a reasonable price; (4) the suppliers, who desire integrity and a reasonable selling price for their goods; and (5) the government and hence the nation, which desire payment of fair taxes and consideration of the national interest. All five parties make essential contributions to the firm, and the firm should not be viewed as the exclusive servant of any one party for the exploitation of others. By the same token, international corporations acquire additional obligations to follow socially responsible practices. Therefore, while granting that management’s prime responsibility is to make profits (which ultimately benefits all five parties), we note that its broader social responsibilities also must be recognized. OR teams typically spend a surprisingly large amount of time gathering relevant data about the problem. Much data usually are needed both to gain an accurate understanding of the problem and to provide the needed input for the mathematical model being formulated in the next phase of study. Frequently, much of the needed data will not be available when the study begins, either because the information never has been kept or because what was kept is outdated or in the wrong form. Therefore, it often is necessary to install a new computer-based management information system to collect the necessary data on an ongoing basis and in the needed form. The OR team normally needs to enlist the assistance of various other key individuals in the organization to track down all the vital data. Even with this effort, much of the data may be quite “soft,” i.e., rough estimates based only on educated guesses. Typically, an OR team will spend considerable time trying to improve the precision of the data and then will make do with the best that can be obtained. Examples. An OR study done for the San Francisco Police Department1 resulted in the development of a computerized system for optimally scheduling and deploying police patrol officers. The new system provided annual savings of $11 million, an annual $3 million increase in traffic citation revenues, and a 20 percent improvement in response times. In assessing the appropriate objectives for this study, three fundamental objectives were identified: 1. Maintain a high level of citizen safety. 2. Maintain a high level of officer morale. 3. Minimize the cost of operations. To satisfy the first objective, the police department and city government jointly established a desired level of protection. The mathematical model then imposed the requirement that this level of protection be achieved. Similarly, the model imposed the requirement of balancing the workload equitably among officers in order to work toward the second objective. Finally, the third objective was incorporated by adopting the long-term goal of minimizing the number of officers needed to meet the first two objectives. 1

P. E. Taylor and S. J. Huxley, “A Break from Tradition for the San Francisco Police: Patrol Officer Scheduling Using an Optimization-Based Decision Support System,” Interfaces, 19(1): 4–24, Jan.–Feb. 1989. See especially pp. 4–11.

10

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OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

The Health Department of New Haven, Connecticut used an OR team1 to design an effective needle exchange program to combat the spread of the virus that causes AIDS (HIV), and succeeded in reducing the HIV infection rate among program clients by 33 percent. The key part of this study was an innovative data collection program to obtain the needed input for mathematical models of HIV transmission. This program involved complete tracking of each needle (and syringe), including the identity, location, and date for each person receiving the needle and each person returning the needle during an exchange, as well as testing whether the returned needle was HIVpositive or HIV-negative. An OR study done for the Citgo Petroleum Corporation2 optimized both refinery operations and the supply, distribution, and marketing of its products, thereby achieving a profit improvement of approximately $70 million per year. Data collection also played a key role in this study. The OR team held data requirement meetings with top Citgo management to ensure the eventual and continual quality of data. A state-of-the-art management database system was developed and installed on a mainframe computer. In cases where needed data did not exist, LOTUS 1-2-3 screens were created to help operations personnel input the data, and then the data from the personal computers (PCs) were uploaded to the mainframe computer. Before data was inputted to the mathematical model, a preloader program was used to check for data errors and inconsistencies. Initially, the preloader generated a paper log of error messages 1 inch thick! Eventually, the number of error and warning messages (indicating bad or questionable numbers) was reduced to less than 10 for each new run. We will describe the overall Citgo study in much more detail in Sec. 3.5.

2.2

FORMULATING A MATHEMATICAL MODEL After the decision maker’s problem is defined, the next phase is to reformulate this problem in a form that is convenient for analysis. The conventional OR approach for doing this is to construct a mathematical model that represents the essence of the problem. Before discussing how to formulate such a model, we first explore the nature of models in general and of mathematical models in particular. Models, or idealized representations, are an integral part of everyday life. Common examples include model airplanes, portraits, globes, and so on. Similarly, models play an important role in science and business, as illustrated by models of the atom, models of genetic structure, mathematical equations describing physical laws of motion or chemical reactions, graphs, organizational charts, and industrial accounting systems. Such models are invaluable for abstracting the essence of the subject of inquiry, showing interrelationships, and facilitating analysis. 1

E. H. Kaplan and E. O’Keefe, “Let the Needles Do the Talking! Evaluating the New Haven Needle Exchange,” Interfaces, 23(1): 7–26, Jan.–Feb. 1993. See especially pp. 12–14. 2 D. Klingman, N. Phillips, D. Steiger, R. Wirth, and W. Young, “The Challenges and Success Factors in Implementing an Integrated Products Planning System for Citgo,” Interfaces, 16(3): 1–19, May–June 1986. See especially pp. 11–14. Also see D. Klingman, N. Phillips, D. Steiger, and W. Young, “The Successful Deployment of Management Science throughout Citgo Petroleum Corporation,” Interfaces, 17(1): 4–25, Jan.–Feb. 1987. See especially pp. 13–15. This application will be described further in Sec. 3.5.

2.2 FORMULATING A MATHEMATICAL MODEL

11

Mathematical models are also idealized representations, but they are expressed in terms of mathematical symbols and expressions. Such laws of physics as F ma and E mc2 are familiar examples. Similarly, the mathematical model of a business problem is the system of equations and related mathematical expressions that describe the essence of the problem. Thus, if there are n related quantifiable decisions to be made, they are represented as decision variables (say, x1, x2, . . . , xn) whose respective values are to be determined. The appropriate measure of performance (e.g., profit) is then expressed as a mathematical function of these decision variables (for example, P 3x1 2x2 + 5xn). This function is called the objective function. Any restrictions on the values that can be assigned to these decision variables are also expressed mathematically, typically by means of inequalities or equations (for example, x1 3x1x2 2x2 10). Such mathematical expressions for the restrictions often are called constraints. The constants (namely, the coefficients and right-hand sides) in the constraints and the objective function are called the parameters of the model. The mathematical model might then say that the problem is to choose the values of the decision variables so as to maximize the objective function, subject to the specified constraints. Such a model, and minor variations of it, typifies the models used in OR. Determining the appropriate values to assign to the parameters of the model (one value per parameter) is both a critical and a challenging part of the model-building process. In contrast to textbook problems where the numbers are given to you, determining parameter values for real problems requires gathering relevant data. As discussed in the preceding section, gathering accurate data frequently is difficult. Therefore, the value assigned to a parameter often is, of necessity, only a rough estimate. Because of the uncertainty about the true value of the parameter, it is important to analyze how the solution derived from the model would change (if at all) if the value assigned to the parameter were changed to other plausible values. This process is referred to as sensitivity analysis, as discussed further in the next section (and much of Chap. 6). Although we refer to “the” mathematical model of a business problem, real problems normally don’t have just a single “right” model. Section 2.4 will describe how the process of testing a model typically leads to a succession of models that provide better and better representations of the problem. It is even possible that two or more completely different types of models may be developed to help analyze the same problem. You will see numerous examples of mathematical models throughout the remainder of this book. One particularly important type that is studied in the next several chapters is the linear programming model, where the mathematical functions appearing in both the objective function and the constraints are all linear functions. In the next chapter, specific linear programming models are constructed to fit such diverse problems as determining (1) the mix of products that maximizes profit, (2) the design of radiation therapy that effectively attacks a tumor while minimizing the damage to nearby healthy tissue, (3) the allocation of acreage to crops that maximizes total net return, and (4) the combination of pollution abatement methods that achieves air quality standards at minimum cost. Mathematical models have many advantages over a verbal description of the problem. One advantage is that a mathematical model describes a problem much more concisely. This tends to make the overall structure of the problem more comprehensible, and it helps to reveal important cause-and-effect relationships. In this way, it indicates more clearly what additional data are relevant to the analysis. It also facilitates dealing with the problem in its

12

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OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

entirety and considering all its interrelationships simultaneously. Finally, a mathematical model forms a bridge to the use of high-powered mathematical techniques and computers to analyze the problem. Indeed, packaged software for both personal computers and mainframe computers has become widely available for solving many mathematical models. However, there are pitfalls to be avoided when you use mathematical models. Such a model is necessarily an abstract idealization of the problem, so approximations and simplifying assumptions generally are required if the model is to be tractable (capable of being solved). Therefore, care must be taken to ensure that the model remains a valid representation of the problem. The proper criterion for judging the validity of a model is whether the model predicts the relative effects of the alternative courses of action with sufficient accuracy to permit a sound decision. Consequently, it is not necessary to include unimportant details or factors that have approximately the same effect for all the alternative courses of action considered. It is not even necessary that the absolute magnitude of the measure of performance be approximately correct for the various alternatives, provided that their relative values (i.e., the differences between their values) are sufficiently precise. Thus, all that is required is that there be a high correlation between the prediction by the model and what would actually happen in the real world. To ascertain whether this requirement is satisfied, it is important to do considerable testing and consequent modifying of the model, which will be the subject of Sec. 2.4. Although this testing phase is placed later in the chapter, much of this model validation work actually is conducted during the modelbuilding phase of the study to help guide the construction of the mathematical model. In developing the model, a good approach is to begin with a very simple version and then move in evolutionary fashion toward more elaborate models that more nearly reflect the complexity of the real problem. This process of model enrichment continues only as long as the model remains tractable. The basic trade-off under constant consideration is between the precision and the tractability of the model. (See Selected Reference 6 for a detailed description of this process.) A crucial step in formulating an OR model is the construction of the objective function. This requires developing a quantitative measure of performance relative to each of the decision maker’s ultimate objectives that were identified while the problem was being defined. If there are multiple objectives, their respective measures commonly are then transformed and combined into a composite measure, called the overall measure of performance. This overall measure might be something tangible (e.g., profit) corresponding to a higher goal of the organization, or it might be abstract (e.g., utility). In the latter case, the task of developing this measure tends to be a complex one requiring a careful comparison of the objectives and their relative importance. After the overall measure of performance is developed, the objective function is then obtained by expressing this measure as a mathematical function of the decision variables. Alternatively, there also are methods for explicitly considering multiple objectives simultaneously, and one of these (goal programming) is discussed in Chap. 7. Examples. An OR study done for Monsanto Corp.1 was concerned with optimizing production operations in Monsanto’s chemical plants to minimize the cost of meeting the target for the amount of a certain chemical product (maleic anhydride) to be produced in a given 1

R. F. Boykin, “Optimizing Chemical Production at Monsanto,” Interfaces, 15(1): 88–95, Jan.–Feb. 1985. See especially pp. 92–93.

2.2 FORMULATING A MATHEMATICAL MODEL

13

month. The decisions to be made are the dial setting for each of the catalytic reactors used to produce this product, where the setting determines both the amount produced and the cost of operating the reactor. The form of the resulting mathematical model is as follows: Choose the values of the decision variables Rij (i 1, 2, . . . , r; j 1, 2, . . . , s) so as to r

s

cij Rij , i1 j1

Minimize subject to r

s

pijRij T

i1 j1 s

Rij 1, j1

for i 1, 2, . . . , r Rij 0 or 1,

0

if reactor i is operated at setting j otherwise cij cost for reactor i at setting j pij production of reactor i at setting j T production target r number of reactors s number of settings (including off position)

where Rij

1

The objective function for this model is cijRij. The constraints are given in the three lines below the objective function. The parameters are cij, pij, and T. For Monsanto’s application, this model has over 1,000 decision variables Rij (that is, rs 1,000). Its use led to annual savings of approximately $2 million. The Netherlands government agency responsible for water control and public works, the Rijkswaterstaat, commissioned a major OR study1 to guide the development of a new national water management policy. The new policy saved hundreds of millions of dollars in investment expenditures and reduced agricultural damage by about $15 million per year, while decreasing thermal and algae pollution. Rather than formulating one mathematical model, this OR study developed a comprehensive, integrated system of 50 models! Furthermore, for some of the models, both simple and complex versions were developed. The simple version was used to gain basic insights, including trade-off analyses. The complex version then was used in the final rounds of the analysis or whenever greater accuracy or more detailed outputs were desired. The overall OR study directly involved over 125 person-years of effort (more than one-third in data gathering), created several dozen computer programs, and structured an enormous amount of data. 1

B. F. Goeller and the PAWN team: “Planning the Netherlands’ Water Resources,” Interfaces, 15(1): 3–33, Jan.–Feb. 1985. See especially pp. 7–18.

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OVERVIEW OF THE OPERATIONS RESEARCH MODELING APPROACH

DERIVING SOLUTIONS FROM THE MODEL After a mathematical model is formulated for the problem under consideration, the next phase in an OR study is to develop a procedure (usually a computer-based procedure) for deriving solutions to the problem from this model. You might think that this must be the major part of the study, but actually it is not in most cases. Sometimes, in fact, it is a relatively simple step, in which one of the standard algorithms (systematic solution procedures) of OR is applied on a computer by using one of a number of readily available software packages. For experienced OR practitioners, finding a solution is the fun part, whereas the real work comes in the preceding and following steps, including the postoptimality analysis discussed later in this section. Since much of this book is devoted to the subject of how to obtain solutions for various important types of mathematical models, little needs to be said about it here. However, we do need to discuss the nature of such solutions. A common theme in OR is the search for an optimal, or best, solution. Indeed, many procedures have been developed, and are presented in this book, for finding such solutions for certain kinds of problems. However, it needs to be recognized that these solutions are optimal only with respect to the model being used. Since the model necessarily is an idealized rather than an exact representation of the real problem, there cannot be any utopian guarantee that the optimal solution for the model will prove to be the best possible solution that could have been implemented for the real problem. There just are too many imponderables and uncertainties associated with real problems. However, if the model is well formulated and tested, the resulting solution should tend to be a good approximation to an ideal course of action for the real problem. Therefore, rather than be deluded into demanding the impossible, you should make the test of the practical success of an OR study hinge on whether it provides a better guide for action than can be obtained by other means. Eminent management scientist and Nobel Laureate in economics Herbert Simon points out that satisficing is much more prevalent than optimizing in actual practice. In coining the term satisficing as a combination of the words satisfactory and optimizing, Simon is describing the tendency of managers to seek a solution that is “good enough” for the problem at hand. Rather than trying to develop an overall measure of performance to optimally reconcile conflicts between various desirable objectives (including well-established criteria for judging the performance of different segments of the organization), a more pragmatic approach may be used. Goals may be set to establish minimum satisfactory levels of performance in various areas, based perhaps on past levels of performance or on what the competition is achieving. If a solution is found that enables all these goals to be met, it is likely to be adopted without further ado. Such is the nature of satisficing. The distinction between optimizing and satisficing reflects the difference between theory and the realities frequently faced in trying to implement that theory in practice. In the words of one of England’s OR leaders, Samuel Eilon, “Optimizing is the science of the ultimate; satisficing is the art of the feasible.”1 OR teams attempt to bring as much of the “science of the ultimate” as possible to the decision-making process. However, the successful team does so in full recognition of the 1

S. Eilon, “Goals and Constraints in Decision-making,” Operational Research Quarterly, 23: 3–15, 1972—address given at the 1971 annual conference of the Canadian Operational Research Society.

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overriding need of the decision maker to obtain a satisfactory guide for action in a reasonable period of time. Therefore, the goal of an OR study should be to conduct the study in an optimal manner, regardless of whether this involves finding an optimal solution for the model. Thus, in addition to pursuing the science of the ultimate, the team should also consider the cost of the study and the disadvantages of delaying its completion, and then attempt to maximize the net benefits resulting from the study. In recognition of this concept, OR teams occasionally use only heuristic procedures (i.e., intuitively designed procedures that do not guarantee an optimal solution) to find a good suboptimal solution. This is most often the case when the time or cost required to find an optimal solution for an adequate model of the problem would be very large. In recent years, great progress has been made in developing efficient and effective heuristic procedures (including so-called metaheuristics), so their use is continuing to grow. The discussion thus far has implied that an OR study seeks to find only one solution, which may or may not be required to be optimal. In fact, this usually is not the case. An optimal solution for the original model may be far from ideal for the real problem, so additional analysis is needed. Therefore, postoptimality analysis (analysis done after finding an optimal solution) is a very important part of most OR studies. This analysis also is sometimes referred to as what-if analysis because it involves addressing some questions about what would happen to the optimal solution if different assumptions are made about future conditions. These questions often are raised by the managers who will be making the ultimate decisions rather than by the OR team. The advent of powerful spreadsheet software now has frequently given spreadsheets a central role in conducting postoptimality analysis. One of the great strengths of a spreadsheet is the ease with which it can be used interactively by anyone, including managers, to see what happens to the optimal solution when changes are made to the model. This process of experimenting with changes in the model also can be very helpful in providing understanding of the behavior of the model and increasing confidence in its validity. In part, postoptimality analysis involves conducting sensitivity analysis to determine which parameters of the model are most critical (the “sensitive parameters”) in determining the solution. A common definition of sensitive parameter (used throughout this book) is the following. For a mathematical model with specified values for all its parameters, the model’s sensitive parameters are the parameters whose value cannot be changed without changing the optimal solution.

Identifying the sensitive parameters is important, because this identifies the parameters whose value must be assigned with special care to avoid distorting the output of the model. The value assigned to a parameter commonly is just an estimate of some quantity (e.g., unit profit) whose exact value will become known only after the solution has been implemented. Therefore, after the sensitive parameters are identified, special attention is given to estimating each one more closely, or at least its range of likely values. One then seeks a solution that remains a particularly good one for all the various combinations of likely values of the sensitive parameters. If the solution is implemented on an ongoing basis, any later change in the value of a sensitive parameter immediately signals a need to change the solution.

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In some cases, certain parameters of the model represent policy decisions (e.g., resource allocations). If so, there frequently is some flexibility in the values assigned to these parameters. Perhaps some can be increased by decreasing others. Postoptimality analysis includes the investigation of such trade-offs. In conjunction with the study phase discussed in the next section (testing the model), postoptimality analysis also involves obtaining a sequence of solutions that comprises a series of improving approximations to the ideal course of action. Thus, the apparent weaknesses in the initial solution are used to suggest improvements in the model, its input data, and perhaps the solution procedure. A new solution is then obtained, and the cycle is repeated. This process continues until the improvements in the succeeding solutions become too small to warrant continuation. Even then, a number of alternative solutions (perhaps solutions that are optimal for one of several plausible versions of the model and its input data) may be presented to management for the final selection. As suggested in Sec. 2.1, this presentation of alternative solutions would normally be done whenever the final choice among these alternatives should be based on considerations that are best left to the judgment of management. Example. Consider again the Rijkswaterstaat OR study of national water management policy for the Netherlands, introduced at the end of the preceding section. This study did not conclude by recommending just a single solution. Instead, a number of attractive alternatives were identified, analyzed, and compared. The final choice was left to the Dutch political process, culminating with approval by Parliament. Sensitivity analysis played a major role in this study. For example, certain parameters of the models represented environmental standards. Sensitivity analysis included assessing the impact on water management problems if the values of these parameters were changed from the current environmental standards to other reasonable values. Sensitivity analysis also was used to assess the impact of changing the assumptions of the models, e.g., the assumption on the effect of future international treaties on the amount of pollution entering the Netherlands. A variety of scenarios (e.g., an extremely dry year and an extremely wet year) also were analyzed, with appropriate probabilities assigned.

2.4

TESTING THE MODEL Developing a large mathematical model is analogous in some ways to developing a large computer program. When the first version of the computer program is completed, it inevitably contains many bugs. The program must be thoroughly tested to try to find and correct as many bugs as possible. Eventually, after a long succession of improved programs, the programmer (or programming team) concludes that the current program now is generally giving reasonably valid results. Although some minor bugs undoubtedly remain hidden in the program (and may never be detected), the major bugs have been sufficiently eliminated that the program now can be reliably used. Similarly, the first version of a large mathematical model inevitably contains many flaws. Some relevant factors or interrelationships undoubtedly have not been incorporated into the model, and some parameters undoubtedly have not been estimated correctly. This is inevitable, given the difficulty of communicating and understanding all the aspects and

2.4 TESTING THE MODEL

17

subtleties of a complex operational problem as well as the difficulty of collecting reliable data. Therefore, before you use the model, it must be thoroughly tested to try to identify and correct as many flaws as possible. Eventually, after a long succession of improved models, the OR team concludes that the current model now is giving reasonably valid results. Although some minor flaws undoubtedly remain hidden in the model (and may never be detected), the major flaws have been sufficiently eliminated that the model now can be reliably used. This process of testing and improving a model to increase its validity is commonly referred to as model validation. It is difficult to describe how model validation is done, because the process depends greatly on the nature of the problem being considered and the model being used. However, we make a few general comments, and then we give some examples. (See Selected Reference 2 for a detailed discussion.) Since the OR team may spend months developing all the detailed pieces of the model, it is easy to “lose the forest for the trees.” Therefore, after the details (“the trees”) of the initial version of the model are completed, a good way to begin model validation is to take a fresh look at the overall model (“the forest”) to check for obvious errors or oversights. The group doing this review preferably should include at least one individual who did not participate in the formulation of the model. Reexamining the definition of the problem and comparing it with the model may help to reveal mistakes. It is also useful to make sure that all the mathematical expressions are dimensionally consistent in the units used. Additional insight into the validity of the model can sometimes be obtained by varying the values of the parameters and/or the decision variables and checking to see whether the output from the model behaves in a plausible manner. This is often especially revealing when the parameters or variables are assigned extreme values near their maxima or minima. A more systematic approach to testing the model is to use a retrospective test. When it is applicable, this test involves using historical data to reconstruct the past and then determining how well the model and the resulting solution would have performed if they had been used. Comparing the effectiveness of this hypothetical performance with what actually happened then indicates whether using this model tends to yield a significant improvement over current practice. It may also indicate areas where the model has shortcomings and requires modifications. Furthermore, by using alternative solutions from the model and estimating their hypothetical historical performances, considerable evidence can be gathered regarding how well the model predicts the relative effects of alternative courses of actions. On the other hand, a disadvantage of retrospective testing is that it uses the same data that guided the formulation of the model. The crucial question is whether the past is truly representative of the future. If it is not, then the model might perform quite differently in the future than it would have in the past. To circumvent this disadvantage of retrospective testing, it is sometimes useful to continue the status quo temporarily. This provides new data that were not available when the model was constructed. These data are then used in the same ways as those described here to evaluate the model. Documenting the process used for model validation is important. This helps to increase confidence in the model for subsequent users. Furthermore, if concerns arise in the

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future about the model, this documentation will be helpful in diagnosing where problems may lie. Examples. Consider once again the Rijkswaterstaat OR study of national water management policy for the Netherlands, discussed at the end of Secs. 2.2 and 2.3. The process of model validation in this case had three main parts. First, the OR team checked the general behavior of the models by checking whether the results from each model moved in reasonable ways when changes were made in the values of the model parameters. Second, retrospective testing was done. Third, a careful technical review of the models, methodology, and results was conducted by individuals unaffiliated with the project, including Dutch experts. This process led to a number of important new insights and improvements in the models. Many new insights also were gleaned during the model validation phase of the OR study for the Citgo Petroleum Corp., discussed at the end of Sec. 2.1. In this case, the model of refinery operations was tested by collecting the actual inputs and outputs of the refinery for a series of months, using these inputs to fix the model inputs, and then comparing the model outputs with the actual refinery outputs. The process of properly calibrating and recalibrating the model was a lengthy one, but ultimately led to routine use of the model to provide critical decision information. As already mentioned in Sec. 2.1, the validation and correction of input data for the models also played an important role in this study. Our next example concerns an OR study done for IBM1 to integrate its national network of spare-parts inventories to improve service support for IBM’s customers. This study resulted in a new inventory system that improved customer service while reducing the value of IBM’s inventories by over $250 million and saving an additional $20 million per year through improved operational efficiency. A particularly interesting aspect of the model validation phase of this study was the way that future users of the inventory system were incorporated into the testing process. Because these future users (IBM managers in functional areas responsible for implementation of the inventory system) were skeptical about the system being developed, representatives were appointed to a user team to serve as advisers to the OR team. After a preliminary version of the new system had been developed (based on a multiechelon inventory model), a preimplementation test of the system was conducted. Extensive feedback from the user team led to major improvements in the proposed system.

2.5

PREPARING TO APPLY THE MODEL What happens after the testing phase has been completed and an acceptable model has been developed? If the model is to be used repeatedly, the next step is to install a welldocumented system for applying the model as prescribed by management. This system will include the model, solution procedure (including postoptimality analysis), and oper1

M. Cohen, P. V. Kamesam, P. Kleindorfer, H. Lee, and A. Tekerian, “Optimizer: IBM’s Multi-Echelon Inventory System for Managing Service Logistics,” Interfaces, 20(1): 65–82, Jan.–Feb. 1990. See especially pp. 73–76. This application will be described further in Sec. 19.8.

2.5 PREPARING TO APPLY THE MODEL

19

ating procedures for implementation. Then, even as personnel changes, the system can be called on at regular intervals to provide a specific numerical solution. This system usually is computer-based. In fact, a considerable number of computer programs often need to be used and integrated. Databases and management information systems may provide up-to-date input for the model each time it is used, in which case interface programs are needed. After a solution procedure (another program) is applied to the model, additional computer programs may trigger the implementation of the results automatically. In other cases, an interactive computer-based system called a decision support system is installed to help managers use data and models to support (rather than replace) their decision making as needed. Another program may generate managerial reports (in the language of management) that interpret the output of the model and its implications for application. In major OR studies, several months (or longer) may be required to develop, test, and install this computer system. Part of this effort involves developing and implementing a process for maintaining the system throughout its future use. As conditions change over time, this process should modify the computer system (including the model) accordingly. Examples. The IBM OR study introduced at the end of Sec. 2.4 provides a good example of a particularly large computer system for applying a model. The system developed, called Optimizer, provides optimal control of service levels and spare-parts inventories throughout IBM’s U.S. parts distribution network, which includes two central automated warehouses, dozens of field distribution centers and parts stations, and many thousands of outside locations. The parts inventory maintained in this network is valued in the billions of dollars. Optimizer consists of four major modules. A forecasting system module contains a few programs for estimating the failure rates of individual types of parts. A data delivery system module consists of approximately 100 programs that process over 15 gigabytes of data to provide the input for the model. A decision system module then solves the model on a weekly basis to optimize control of the inventories. The fourth module includes six programs that integrate Optimizer into IBM’s Parts Inventory Management System (PIMS). PIMS is a sophisticated information and control system that contains millions of lines of code. Our next example also involves a large computer system for applying a model to control operations over a national network. This system, called SYSNET, was developed as the result of an OR study done for Yellow Freight System, Inc.1 Yellow Freight annually handles over 15 million shipments by motor carrier over a network of 630 terminals throughout the United States. SYSNET is used to optimize both the routing of shipments and the design of the network. Because SYSNET requires extensive information about freight flows and forecasts, transportation and handling costs, and so on, a major part of the OR study involved integrating SYSNET into the corporate management information system. This integration enabled periodic updating of all the input for the model. The implementation of SYSNET resulted in annual savings of approximately $17.3 million as well as improved service to customers. 1

J. W. Braklow, W. W. Graham, S. M. Hassler, K. E. Peck, and W. B. Powell, “Interactive Optimization Improves Service and Performance for Yellow Freight System,” Interfaces, 22(1): 147–172, Jan.–Feb. 1992. See especially p. 163.

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Our next example illustrates a decision support system. A system of this type was developed for Texaco1 to help plan and schedule its blending operations at its various refineries. Called OMEGA (Optimization Method for the Estimation of Gasoline Attributes), it is an interactive system based on a nonlinear optimization model that is implemented on both personal computers and larger computers. Input data can be entered either manually or by interfacing with refinery databases. The user has considerable flexibility in choosing an objective function and constraints to fit the current situation as well as in asking a series of what-if questions (i.e., questions about what would happen if the assumed conditions change). OMEGA is maintained centrally by Texaco’s information technology department, which enables constant updating to reflect new government regulations, other business changes, and changes in refinery operations. The implementation of OMEGA is credited with annual savings of more than $30 million as well as improved planning, quality control, and marketing information.

2.6

IMPLEMENTATION After a system is developed for applying the model, the last phase of an OR study is to implement this system as prescribed by management. This phase is a critical one because it is here, and only here, that the benefits of the study are reaped. Therefore, it is important for the OR team to participate in launching this phase, both to make sure that model solutions are accurately translated to an operating procedure and to rectify any flaws in the solutions that are then uncovered. The success of the implementation phase depends a great deal upon the support of both top management and operating management. The OR team is much more likely to gain this support if it has kept management well informed and encouraged management’s active guidance throughout the course of the study. Good communications help to ensure that the study accomplishes what management wanted and so deserves implementation. They also give management a greater sense of ownership of the study, which encourages their support for implementation. The implementation phase involves several steps. First, the OR team gives operating management a careful explanation of the new system to be adopted and how it relates to operating realities. Next, these two parties share the responsibility for developing the procedures required to put this system into operation. Operating management then sees that a detailed indoctrination is given to the personnel involved, and the new course of action is initiated. If successful, the new system may be used for years to come. With this in mind, the OR team monitors the initial experience with the course of action taken and seeks to identify any modifications that should be made in the future. Throughout the entire period during which the new system is being used, it is important to continue to obtain feedback on how well the system is working and whether the assumptions of the model continue to be satisfied. When significant deviations from the original assumptions occur, the model should be revisited to determine if any modifications should be made in the system. The postoptimality analysis done earlier (as described in Sec. 2.3) can be helpful in guiding this review process. 1

C. W. DeWitt, L. S. Lasdon, A. D. Waren, D. A. Brenner, and S. A. Melhem, “OMEGA: An Improved Gasoline Blending System for Texaco,” Interfaces, 19(1): 85–101, Jan.–Feb. 1989. See especially pp. 93–95.

2.7 CONCLUSIONS

21

Upon culmination of a study, it is appropriate for the OR team to document its methodology clearly and accurately enough so that the work is reproducible. Replicability should be part of the professional ethical code of the operations researcher. This condition is especially crucial when controversial public policy issues are being studied. Examples. This last point about documenting an OR study is illustrated by the Rijkswaterstaat study of national water management policy for the Netherlands discussed at the end of Secs. 2.2, 2.3, and 2.4. Management wanted unusually thorough and extensive documentation, both to support the new policy and to use in training new analysts or in performing new studies. Requiring several years to complete, this documentation aggregated 4000 single-spaced pages and 21 volumes! Our next example concerns the IBM OR study discussed at the end of Secs. 2.4 and 2.5. Careful planning was required to implement the complex Optimizer system for controlling IBM’s national network of spare-parts inventories. Three factors proved to be especially important in achieving a successful implementation. As discussed in Sec. 2.4, the first was the inclusion of a user team (consisting of operational managers) as advisers to the OR team throughout the study. By the time of the implementation phase, these operational managers had a strong sense of ownership and so had become ardent supporters for installing Optimizer in their functional areas. A second success factor was a very extensive user acceptance test whereby users could identify problem areas that needed rectifying prior to full implementation. The third key was that the new system was phased in gradually, with careful testing at each phase, so the major bugs could be eliminated before the system went live nationally. Our final example concerns Yellow Freight’s SYSNET system for routing shipments over a national network, as described at the end of the preceding section. In this case, there were four key elements to the implementation process. The first was selling the concept to upper management. This was successfully done through validating the accuracy of the cost model and then holding interactive sessions for upper management that demonstrated the effectiveness of the system. The second element was the development of an implementation strategy for gradually phasing in the new system while identifying and eliminating its flaws. The third involved working closely with operational managers to install the system properly, provide the needed support tools, train the personnel who will use the system, and convince them of the usefulness of the system. The final key element was the provision of management incentives and enforcement for the effective implementation of the system.

2.7

CONCLUSIONS Although the remainder of this book focuses primarily on constructing and solving mathematical models, in this chapter we have tried to emphasize that this constitutes only a portion of the overall process involved in conducting a typical OR study. The other phases described here also are very important to the success of the study. Try to keep in perspective the role of the model and the solution procedure in the overall process as you move through the subsequent chapters. Then, after gaining a deeper understanding of mathematical models, we suggest that you plan to return to review this chapter again in order to further sharpen this perspective.

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OR is closely intertwined with the use of computers. In the early years, these generally were mainframe computers, but now personal computers and workstations are being widely used to solve OR models. In concluding this discussion of the major phases of an OR study, it should be emphasized that there are many exceptions to the “rules” prescribed in this chapter. By its very nature, OR requires considerable ingenuity and innovation, so it is impossible to write down any standard procedure that should always be followed by OR teams. Rather, the preceding description may be viewed as a model that roughly represents how successful OR studies are conducted.

SELECTED REFERENCES 1. Fortuin, L., P. van Beek, and L. van Wassenhove (eds.): OR at wORk: Practical Experiences of Operational Research, Taylor & Francis, Bristol, PA, 1996. 2. Gass, S. I.: “Decision-Aiding Models: Validation, Assessment, and Related Issues for Policy Analysis,” Operations Research, 31: 603–631, 1983. 3. Gass, S. I.: “Model World: Danger, Beware the User as Modeler,” Interfaces, 20(3): 60–64, May–June 1990. 4. Hall, R. W.: “What’s So Scientific about MS/OR?” Interfaces, 15(2): 40–45, March–April 1985. 5. Miser, H. J.: “The Easy Chair: Observation and Experimentation,” Interfaces, 19(5): 23–30, Sept.–Oct. 1989. 6. Morris, W. T.: “On the Art of Modeling,” Management Science, 13: B707–717, 1967. 7. Murthy, D. N. P., N. W. Page, and E. Y. Rodin: Mathematical Modeling: A Tool for Problem Solving in Engineering, Physical, Biological and Social Sciences, Pergamon Press, Oxford, England, 1990. 8. Simon, H. A.: “Prediction and Prescription in Systems Modeling,” Operations Research, 38: 7–14, 1990. 9. Tilanus, C. B., O. B. DeGans, and J. K. Lenstra (eds.): Quantitative Methods in Management: Case Studies of Failures and Successes, Wiley, New York, 1986. 10. Williams, H. P.: Model Building in Mathematical Programming, 3d ed., Wiley, New York, 1990.

PROBLEMS 2.1-1. Read the article footnoted in Sec. 2.1 that describes an OR study done for the San Francisco Police Department. (a) Summarize the background that led to undertaking this study. (b) Define part of the problem being addressed by identifying the six directives for the scheduling system to be developed. (c) Describe how the needed data were gathered. (d) List the various tangible and intangible benefits that resulted from the study. 2.1-2. Read the article footnoted in Sec. 2.1 that describes an OR study done for the Health Department of New Haven, Connecticut. (a) Summarize the background that led to undertaking this study.

(b) Outline the system developed to track and test each needle and syringe in order to gather the needed data. (c) Summarize the initial results from this tracking and testing system. (d) Describe the impact and potential impact of this study on public policy. 2.2-1. Read the article footnoted in Sec. 2.2 that describes an OR study done for the Rijkswaterstaat of the Netherlands. (Focus especially on pp. 3–20 and 30–32.) (a) Summarize the background that led to undertaking this study. (b) Summarize the purpose of each of the five mathematical models described on pp. 10–18.

CHAPTER 2 PROBLEMS

(c) Summarize the “impact measures” (measures of performance) for comparing policies that are described on pp. 6–7 of this article. (d) List the various tangible and intangible benefits that resulted from the study. 2.2-2. Read Selected Reference 4. (a) Identify the author’s example of a model in the natural sciences and of a model in OR. (b) Describe the author’s viewpoint about how basic precepts of using models to do research in the natural sciences can also be used to guide research on operations (OR). 2.3-1. Refer to Selected Reference 4. (a) Describe the author’s viewpoint about whether the sole goal in using a model should be to find its optimal solution. (b) Summarize the author’s viewpoint about the complementary roles of modeling, evaluating information from the model, and then applying the decision maker’s judgment when deciding on a course of action. 2.4-1. Refer to pp. 18–20 of the article footnoted in Sec. 2.2 that describes an OR study done for the Rijkswaterstaat of the Netherlands. Describe an important lesson that was gained from model validation in this study. 2.4-2. Read Selected Reference 5. Summarize the author’s viewpoint about the roles of observation and experimentation in the model validation process.

23

(c) OMEGA is constantly being updated and extended to reflect changes in the operating environment. Briefly describe the various kinds of changes involved. (d) Summarize how OMEGA is used. (e) List the various tangible and intangible benefits that resulted from the study. 2.5-2. Refer to the article footnoted in Sec. 2.5 that describes an OR study done for Yellow Freight System, Inc. (a) Referring to pp. 147–149 of this article, summarize the background that led to undertaking this study. (b) Referring to p. 150, briefly describe the computer system SYSNET that was developed as a result of this study. Also summarize the applications of SYSNET. (c) Referring to pp. 162–163, describe why the interactive aspects of SYSNET proved important. (d) Referring to p. 163, summarize the outputs from SYSNET. (e) Referring to pp. 168–172, summarize the various benefits that have resulted from using SYSNET. 2.6-1. Refer to pp. 163–167 of the article footnoted in Sec. 2.5 that describes an OR study done for Yellow Freight System, Inc., and the resulting computer system SYSNET. (a) Briefly describe how the OR team gained the support of upper management for implementing SYSNET. (b) Briefly describe the implementation strategy that was developed. (c) Briefly describe the field implementation. (d) Briefly describe how management incentives and enforcement were used in implementing SYSNET.

2.4-3. Read pp. 603–617 of Selected Reference 2. (a) What does the author say about whether a model can be completely validated? (b) Summarize the distinctions made between model validity, data validity, logical/mathematical validity, predictive validity, operational validity, and dynamic validity. (c) Describe the role of sensitivity analysis in testing the operational validity of a model. (d) What does the author say about whether there is a validation methodology that is appropriate for all models? (e) Cite the page in the article that lists basic validation steps.

2.6-2. Read the article footnoted in Sec. 2.4 that describes an OR study done for IBM and the resulting computer system Optimizer. (a) Summarize the background that led to undertaking this study. (b) List the complicating factors that the OR team members faced when they started developing a model and a solution algorithm. (c) Briefly describe the preimplementation test of Optimizer. (d) Briefly describe the field implementation test. (e) Briefly describe national implementation. (f) List the various tangible and intangible benefits that resulted from the study.

2.5-1. Read the article footnoted in Sec. 2.5 that describes an OR study done for Texaco. (a) Summarize the background that led to undertaking this study. (b) Briefly describe the user interface with the decision support system OMEGA that was developed as a result of this study.

2.7-1. Read Selected Reference 3. The author describes 13 detailed phases of any OR study that develops and applies a computer-based model, whereas this chapter describes six broader phases. For each of these broader phases, list the detailed phases that fall partially or primarily within the broader phase.

3 Introduction to Linear Programming The development of linear programming has been ranked among the most important scientific advances of the mid-20th century, and we must agree with this assessment. Its impact since just 1950 has been extraordinary. Today it is a standard tool that has saved many thousands or millions of dollars for most companies or businesses of even moderate size in the various industrialized countries of the world; and its use in other sectors of society has been spreading rapidly. A major proportion of all scientific computation on computers is devoted to the use of linear programming. Dozens of textbooks have been written about linear programming, and published articles describing important applications now number in the hundreds. What is the nature of this remarkable tool, and what kinds of problems does it address? You will gain insight into this topic as you work through subsequent examples. However, a verbal summary may help provide perspective. Briefly, the most common type of application involves the general problem of allocating limited resources among competing activities in a best possible (i.e., optimal) way. More precisely, this problem involves selecting the level of certain activities that compete for scarce resources that are necessary to perform those activities. The choice of activity levels then dictates how much of each resource will be consumed by each activity. The variety of situations to which this description applies is diverse, indeed, ranging from the allocation of production facilities to products to the allocation of national resources to domestic needs, from portfolio selection to the selection of shipping patterns, from agricultural planning to the design of radiation therapy, and so on. However, the one common ingredient in each of these situations is the necessity for allocating resources to activities by choosing the levels of those activities. Linear programming uses a mathematical model to describe the problem of concern. The adjective linear means that all the mathematical functions in this model are required to be linear functions. The word programming does not refer here to computer programming; rather, it is essentially a synonym for planning. Thus, linear programming involves the planning of activities to obtain an optimal result, i.e., a result that reaches the specified goal best (according to the mathematical model) among all feasible alternatives. Although allocating resources to activities is the most common type of application, linear programming has numerous other important applications as well. In fact, any problem whose mathematical model fits the very general format for the linear programming model is a linear programming problem. Furthermore, a remarkably efficient solution pro24

3.1 PROTOTYPE EXAMPLE

25

cedure, called the simplex method, is available for solving linear programming problems of even enormous size. These are some of the reasons for the tremendous impact of linear programming in recent decades. Because of its great importance, we devote this and the next six chapters specifically to linear programming. After this chapter introduces the general features of linear programming, Chaps. 4 and 5 focus on the simplex method. Chapter 6 discusses the further analysis of linear programming problems after the simplex method has been initially applied. Chapter 7 presents several widely used extensions of the simplex method and introduces an interior-point algorithm that sometimes can be used to solve even larger linear programming problems than the simplex method can handle. Chapters 8 and 9 consider some special types of linear programming problems whose importance warrants individual study. You also can look forward to seeing applications of linear programming to other areas of operations research (OR) in several later chapters. We begin this chapter by developing a miniature prototype example of a linear programming problem. This example is small enough to be solved graphically in a straightforward way. The following two sections present the general linear programming model and its basic assumptions. Sections 3.4 and 3.5 give some additional examples of linear programming applications, including three case studies. Section 3.6 describes how linear programming models of modest size can be conveniently displayed and solved on a spreadsheet. However, some linear programming problems encountered in practice require truly massive models. Section 3.7 illustrates how a massive model can arise and how it can still be formulated successfully with the help of a special modeling language such as MPL (described in this section) or LINGO (described in the appendix to this chapter).

3.1

PROTOTYPE EXAMPLE The WYNDOR GLASS CO. produces high-quality glass products, including windows and glass doors. It has three plants. Aluminum frames and hardware are made in Plant 1, wood frames are made in Plant 2, and Plant 3 produces the glass and assembles the products. Because of declining earnings, top management has decided to revamp the company’s product line. Unprofitable products are being discontinued, releasing production capacity to launch two new products having large sales potential: Product 1: An 8-foot glass door with aluminum framing Product 2: A 4 6 foot double-hung wood-framed window Product 1 requires some of the production capacity in Plants 1 and 3, but none in Plant 2. Product 2 needs only Plants 2 and 3. The marketing division has concluded that the company could sell as much of either product as could be produced by these plants. However, because both products would be competing for the same production capacity in Plant 3, it is not clear which mix of the two products would be most profitable. Therefore, an OR team has been formed to study this question. The OR team began by having discussions with upper management to identify management’s objectives for the study. These discussions led to developing the following definition of the problem: Determine what the production rates should be for the two products in order to maximize their total profit, subject to the restrictions imposed by the limited production capacities

26

3

INTRODUCTION TO LINEAR PROGRAMMING

available in the three plants. (Each product will be produced in batches of 20, so the production rate is defined as the number of batches produced per week.) Any combination of production rates that satisfies these restrictions is permitted, including producing none of one product and as much as possible of the other.

The OR team also identified the data that needed to be gathered: 1. Number of hours of production time available per week in each plant for these new products. (Most of the time in these plants already is committed to current products, so the available capacity for the new products is quite limited.) 2. Number of hours of production time used in each plant for each batch produced of each new product. 3. Profit per batch produced of each new product. (Profit per batch produced was chosen as an appropriate measure after the team concluded that the incremental profit from each additional batch produced would be roughly constant regardless of the total number of batches produced. Because no substantial costs will be incurred to initiate the production and marketing of these new products, the total profit from each one is approximately this profit per batch produced times the number of batches produced.) Obtaining reasonable estimates of these quantities required enlisting the help of key personnel in various units of the company. Staff in the manufacturing division provided the data in the first category above. Developing estimates for the second category of data required some analysis by the manufacturing engineers involved in designing the production processes for the new products. By analyzing cost data from these same engineers and the marketing division, along with a pricing decision from the marketing division, the accounting department developed estimates for the third category. Table 3.1 summarizes the data gathered. The OR team immediately recognized that this was a linear programming problem of the classic product mix type, and the team next undertook the formulation of the corresponding mathematical model. Formulation as a Linear Programming Problem To formulate the mathematical (linear programming) model for this problem, let x1 number of batches of product 1 produced per week x2 number of batches of product 2 produced per week Z total profit per week (in thousands of dollars) from producing these two products Thus, x1 and x2 are the decision variables for the model. Using the bottom row of Table 3.1, we obtain Z 3x1 5x2. The objective is to choose the values of x1 and x2 so as to maximize Z 3x1 5x2, subject to the restrictions imposed on their values by the limited production capacities available in the three plants. Table 3.1 indicates that each batch of product 1 produced per week uses 1 hour of production time per week in Plant 1, whereas only 4 hours per week are available. This restriction is expressed mathematically by the inequality x1 4. Similarly, Plant 2 imposes the restriction that 2x2 12. The number of hours of production

3.1 PROTOTYPE EXAMPLE

27

TABLE 3.1 Data for the Wyndor Glass Co. problem Production Time per Batch, Hours Product Plant

1

2

Production Time Available per Week, Hours

1 2 3

1 0 3

0 2 2

4 12 18

Profit per batch

$3,000

$5,000

time used per week in Plant 3 by choosing x1 and x2 as the new products’ production rates would be 3x1 2x2. Therefore, the mathematical statement of the Plant 3 restriction is 3x1 2x2 18. Finally, since production rates cannot be negative, it is necessary to restrict the decision variables to be nonnegative: x1 0 and x2 0. To summarize, in the mathematical language of linear programming, the problem is to choose values of x1 and x2 so as to Maximize

Z 3x1 5x2,

subject to the restrictions 3x1 2x2 4 3x1 2x2 12 3x1 2x2 18 and x1 0,

x2 0.

(Notice how the layout of the coefficients of x1 and x2 in this linear programming model essentially duplicates the information summarized in Table 3.1.) Graphical Solution This very small problem has only two decision variables and therefore only two dimensions, so a graphical procedure can be used to solve it. This procedure involves constructing a two-dimensional graph with x1 and x2 as the axes. The first step is to identify the values of (x1, x2) that are permitted by the restrictions. This is done by drawing each line that borders the range of permissible values for one restriction. To begin, note that the nonnegativity restrictions x1 0 and x2 0 require (x1, x2) to lie on the positive side of the axes (including actually on either axis), i.e., in the first quadrant. Next, observe that the restriction x1 4 means that (x1, x2) cannot lie to the right of the line x1 4. These results are shown in Fig. 3.1, where the shaded area contains the only values of (x1, x2) that are still allowed. In a similar fashion, the restriction 2x2 12 (or, equivalently, x2 6) implies that the line 2x2 12 should be added to the boundary of the permissible region. The final restriction, 3x1 2x2 18, requires plotting the points (x1, x2) such that 3x1 2x2 18

28

3

INTRODUCTION TO LINEAR PROGRAMMING

x2

5 4 3 2 FIGURE 3.1 Shaded area shows values of (x1, x2) allowed by x1 0, x2 0, x1 4.

1 0

1

2

3

4

5

6

7

x1

(another line) to complete the boundary. (Note that the points such that 3x1 2x2 18 are those that lie either underneath or on the line 3x1 2x2 18, so this is the limiting line above which points do not satisfy the inequality.) The resulting region of permissible values of (x1, x2), called the feasible region, is shown in Fig. 3.2. (The demo called Graphical Method in your OR Tutor provides a more detailed example of constructing a feasible region.)

FIGURE 3.2 Shaded area shows the set of permissible values of (x1, x2), called the feasible region.

x2 10 3x1 2x2 18 8 x1 4 2x2 12

6

4 Feasible region 2

0

2

4

6

8

x1

3.1 PROTOTYPE EXAMPLE

29

The final step is to pick out the point in this feasible region that maximizes the value of Z 3x1 5x2. To discover how to perform this step efficiently, begin by trial and error. Try, for example, Z 10 3x1 5x2 to see if there are in the permissible region any values of (x1, x2) that yield a value of Z as large as 10. By drawing the line 3x1 5x2 10 (see Fig. 3.3), you can see that there are many points on this line that lie within the region. Having gained perspective by trying this arbitrarily chosen value of Z 10, you should next try a larger arbitrary value of Z, say, Z 20 3x1 5x2. Again, Fig. 3.3 reveals that a segment of the line 3x1 5x2 20 lies within the region, so that the maximum permissible value of Z must be at least 20. Now notice in Fig. 3.3 that the two lines just constructed are parallel. This is no coincidence, since any line constructed in this way has the form Z 3x1 5x2 for the chosen value of Z, which implies that 5x2 3x1 Z or, equivalently, 3 1 x2 x1 Z 5 5 This last equation, called the slope-intercept form of the objective function, demonstrates that the slope of the line is 35 (since each unit increase in x1 changes x2 by 35), whereas the intercept of the line with the x2 axis is 15 Z (since x2 15 Z when x1 0). The fact that the slope is fixed at 35 means that all lines constructed in this way are parallel. Again, comparing the 10 3x1 5x2 and 20 3x1 5x2 lines in Fig. 3.3, we note that the line giving a larger value of Z (Z 20) is farther up and away from the origin than the other line (Z 10). This fact also is implied by the slope-intercept form of the objective function, which indicates that the intercept with the x1 axis ( 15 Z) increases when the value chosen for Z is increased.

FIGURE 3.3 The value of (x1, x2) that maximizes 3x1 5x2 is (2, 6).

x2

8 Z 36 3x1 5x2 6

Z 20 3x1 5x2

(2, 6)

4

Z 10 3x1 5x2 2

0

2

4

6

8

10

x1

30

3 INTRODUCTION TO LINEAR PROGRAMMING

These observations imply that our trial-and-error procedure for constructing lines in Fig. 3.3 involves nothing more than drawing a family of parallel lines containing at least one point in the feasible region and selecting the line that corresponds to the largest value of Z. Figure 3.3 shows that this line passes through the point (2, 6), indicating that the optimal solution is x1 2 and x2 6. The equation of this line is 3x1 5x2 3(2) 5(6) 36 Z, indicating that the optimal value of Z is Z 36. The point (2, 6) lies at the intersection of the two lines 2x2 12 and 3x1 2x2 18, shown in Fig. 3.2, so that this point can be calculated algebraically as the simultaneous solution of these two equations. Having seen the trial-and-error procedure for finding the optimal point (2, 6), you now can streamline this approach for other problems. Rather than draw several parallel lines, it is sufficient to form a single line with a ruler to establish the slope. Then move the ruler with fixed slope through the feasible region in the direction of improving Z. (When the objective is to minimize Z, move the ruler in the direction that decreases Z.) Stop moving the ruler at the last instant that it still passes through a point in this region. This point is the desired optimal solution. This procedure often is referred to as the graphical method for linear programming. It can be used to solve any linear programming problem with two decision variables. With considerable difficulty, it is possible to extend the method to three decision variables but not more than three. (The next chapter will focus on the simplex method for solving larger problems.) Conclusions The OR team used this approach to find that the optimal solution is x1 2, x2 6, with Z 36. This solution indicates that the Wyndor Glass Co. should produce products 1 and 2 at the rate of 2 batches per week and 6 batches per week, respectively, with a resulting total profit of $36,000 per week. No other mix of the two products would be so profitable—according to the model. However, we emphasized in Chap. 2 that well-conducted OR studies do not simply find one solution for the initial model formulated and then stop. All six phases described in Chap. 2 are important, including thorough testing of the model (see Sec. 2.4) and postoptimality analysis (see Sec. 2.3). In full recognition of these practical realities, the OR team now is ready to evaluate the validity of the model more critically (to be continued in Sec. 3.3) and to perform sensitivity analysis on the effect of the estimates in Table 3.1 being different because of inaccurate estimation, changes of circumstances, etc. (to be continued in Sec. 6.7). Continuing the Learning Process with Your OR Courseware This is the first of many points in the book where you may find it helpful to use your OR Courseware in the CD-ROM that accompanies this book. A key part of this courseware is a program called OR Tutor. This program includes a complete demonstration example of the graphical method introduced in this section. Like the many other demonstration examples accompanying other sections of the book, this computer demonstration highlights concepts that are difficult to convey on the printed page. You may refer to Appendix 1 for documentation of the software. When you formulate a linear programming model with more than two decision variables (so the graphical method cannot be used), the simplex method described in Chap. 4

3.2 THE LINEAR PROGRAMMING MODEL

31

enables you to still find an optimal solution immediately. Doing so also is helpful for model validation, since finding a nonsensical optimal solution signals that you have made a mistake in formulating the model. We mentioned in Sec. 1.4 that your OR Courseware introduces you to three particularly popular commercial software packages—the Excel Solver, LINGO/LINDO, and MPL/CPLEX—for solving a variety of OR models. All three packages include the simplex method for solving linear programming models. Section 3.6 describes how to use Excel to formulate and solve linear programming models in a spreadsheet format. Descriptions of the other packages are provided in Sec. 3.7 (MPL and LINGO), Appendix 3.1 (LINGO), Sec. 4.8 (CPLEX and LINDO), and Appendix 4.1 (LINDO). In addition, your OR Courseware includes a file for each of the three packages showing how it can be used to solve each of the examples in this chapter.

3.2

THE LINEAR PROGRAMMING MODEL The Wyndor Glass Co. problem is intended to illustrate a typical linear programming problem (miniature version). However, linear programming is too versatile to be completely characterized by a single example. In this section we discuss the general characteristics of linear programming problems, including the various legitimate forms of the mathematical model for linear programming. Let us begin with some basic terminology and notation. The first column of Table 3.2 summarizes the components of the Wyndor Glass Co. problem. The second column then introduces more general terms for these same components that will fit many linear programming problems. The key terms are resources and activities, where m denotes the number of different kinds of resources that can be used and n denotes the number of activities being considered. Some typical resources are money and particular kinds of machines, equipment, vehicles, and personnel. Examples of activities include investing in particular projects, advertising in particular media, and shipping goods from a particular source to a particular destination. In any application of linear programming, all the activities may be of one general kind (such as any one of these three examples), and then the individual activities would be particular alternatives within this general category. As described in the introduction to this chapter, the most common type of application of linear programming involves allocating resources to activities. The amount available of each resource is limited, so a careful allocation of resources to activities must be made. Determining this allocation involves choosing the levels of the activities that achieve the best possible value of the overall measure of performance. TABLE 3.2 Common terminology for linear programming Prototype Example

General Problem

Production capacities of plants 3 plants

Resources m resources

Production of products 2 products Production rate of product j, xj

Activities n activities Level of activity j, xj

Profit Z

Overall measure of performance Z

32

3

INTRODUCTION TO LINEAR PROGRAMMING

Certain symbols are commonly used to denote the various components of a linear programming model. These symbols are listed below, along with their interpretation for the general problem of allocating resources to activities. Z value of overall measure of performance. xj level of activity j (for j 1, 2, . . . , n). cj increase in Z that would result from each unit increase in level of activity j. bi amount of resource i that is available for allocation to activities (for i 1, 2, . . . , m). aij amount of resource i consumed by each unit of activity j. The model poses the problem in terms of making decisions about the levels of the activities, so x1, x2, . . . , xn are called the decision variables. As summarized in Table 3.3, the values of cj, bi, and aij (for i 1, 2, . . . , m and j 1, 2, . . . , n) are the input constants for the model. The cj, bi, and aij are also referred to as the parameters of the model. Notice the correspondence between Table 3.3 and Table 3.1. A Standard Form of the Model Proceeding as for the Wyndor Glass Co. problem, we can now formulate the mathematical model for this general problem of allocating resources to activities. In particular, this model is to select the values for x1, x2, . . . , xn so as to Maximize

Z c1x1 c2x2 cnxn,

subject to the restrictions a11x1 a12x2 a1nxn b1 a21x1 a22x2 a2nxn b2 am1x1 am2x2 amnxn bm, TABLE 3.3 Data needed for a linear programming model involving the allocation of resources to activities Resource Usage per Unit of Activity Activity Resource

1

2

...

n

1 2 . . . m

a11 a21

a12 a22

... ...

a1n a2n

...

...

...

...

am1

am2

...

amn

c1

c2

...

cn

Contribution to Z per unit of activity

Amount of Resource Available b1 b2 . . . bm

3.2 THE LINEAR PROGRAMMING MODEL

33

and x1 0,

x2 0,

...,

xn 0.

We call this our standard form1 for the linear programming problem. Any situation whose mathematical formulation fits this model is a linear programming problem. Notice that the model for the Wyndor Glass Co. problem fits our standard form, with m 3 and n 2. Common terminology for the linear programming model can now be summarized. The function being maximized, c1x1 c2x2 cn xn, is called the objective function. The restrictions normally are referred to as constraints. The first m constraints (those with a function of all the variables ai1x1 ai2x2 ain xn on the left-hand side) are sometimes called functional constraints (or structural constraints). Similarly, the xj 0 restrictions are called nonnegativity constraints (or nonnegativity conditions). Other Forms We now hasten to add that the preceding model does not actually fit the natural form of some linear programming problems. The other legitimate forms are the following: 1. Minimizing rather than maximizing the objective function: Minimize

Z c1x1 c2 x2 cn xn.

2. Some functional constraints with a greater-than-or-equal-to inequality: ai1x1 ai2x2 ain xn bi

for some values of i.

3. Some functional constraints in equation form: ai1x1 ai2x2 ain xn bi

for some values of i.

4. Deleting the nonnegativity constraints for some decision variables: xj unrestricted in sign

for some values of j.

Any problem that mixes some of or all these forms with the remaining parts of the preceding model is still a linear programming problem. Our interpretation of the words allocating limited resources among competing activities may no longer apply very well, if at all; but regardless of the interpretation or context, all that is required is that the mathematical statement of the problem fit the allowable forms. Terminology for Solutions of the Model You may be used to having the term solution mean the final answer to a problem, but the convention in linear programming (and its extensions) is quite different. Here, any specification of values for the decision variables (x1, x2, . . . , xn) is called a solution, regardless of whether it is a desirable or even an allowable choice. Different types of solutions are then identified by using an appropriate adjective. 1

This is called our standard form rather than the standard form because some textbooks adopt other forms.

34

3

INTRODUCTION TO LINEAR PROGRAMMING

A feasible solution is a solution for which all the constraints are satisfied. An infeasible solution is a solution for which at least one constraint is violated. In the example, the points (2, 3) and (4, 1) in Fig. 3.2 are feasible solutions, while the points (1, 3) and (4, 4) are infeasible solutions. The feasible region is the collection of all feasible solutions. The feasible region in the example is the entire shaded area in Fig. 3.2. It is possible for a problem to have no feasible solutions. This would have happened in the example if the new products had been required to return a net profit of at least $50,000 per week to justify discontinuing part of the current product line. The corresponding constraint, 3x1 5x2 50, would eliminate the entire feasible region, so no mix of new products would be superior to the status quo. This case is illustrated in Fig. 3.4. Given that there are feasible solutions, the goal of linear programming is to find a best feasible solution, as measured by the value of the objective function in the model. An optimal solution is a feasible solution that has the most favorable value of the objective function. The most favorable value is the largest value if the objective function is to be maximized, whereas it is the smallest value if the objective function is to be minimized. Most problems will have just one optimal solution. However, it is possible to have more than one. This would occur in the example if the profit per batch produced of product 2 were changed to $2,000. This changes the objective function to Z 3x1 2x2, so that all the points

FIGURE 3.4 The Wyndor Glass Co. problem would have no feasible solutions if the constraint 3x1 5x2 50 were added to the problem.

x2 Maximize Z 3x1 5x2, 4 x1 subject to 2x2 12 3x1 2x2 18 3x1 5x2 50 x1 0, and x2 0

10 3x1 5x2 50 8

6 2x2 12 4

3x1 2x2 18 x1 0

2

x1 4 x2 0

0

2

4

6

8

10

x1

3.2 THE LINEAR PROGRAMMING MODEL

35

FIGURE 3.5 The Wyndor Glass Co. problem would have multiple optimal solutions if the objective function were changed to Z 3x1 2x2.

on the line segment connecting (2, 6) and (4, 3) would be optimal. This case is illustrated in Fig. 3.5. As in this case, any problem having multiple optimal solutions will have an infinite number of them, each with the same optimal value of the objective function. Another possibility is that a problem has no optimal solutions. This occurs only if (1) it has no feasible solutions or (2) the constraints do not prevent improving the value of the objective function (Z) indefinitely in the favorable direction (positive or negative). The latter case is referred to as having an unbounded Z. To illustrate, this case would result if the last two functional constraints were mistakenly deleted in the example, as illustrated in Fig. 3.6. We next introduce a special type of feasible solution that plays the key role when the simplex method searches for an optimal solution. A corner-point feasible (CPF) solution is a solution that lies at a corner of the feasible region. Figure 3.7 highlights the five CPF solutions for the example. Sections 4.1 and 5.1 will delve into the various useful properties of CPF solutions for problems of any size, including the following relationship with optimal solutions. Relationship between optimal solutions and CPF solutions: Consider any linear programming problem with feasible solutions and a bounded feasible region. The problem must possess CPF solutions and at least one optimal solution. Furthermore, the best CPF solution must be an optimal solution. Thus, if a problem has exactly one optimal solution, it must be a CPF solution. If the problem has multiple optimal solutions, at least two must be CPF solutions.

36

3

INTRODUCTION TO LINEAR PROGRAMMING

(4, ), Z

x2 10

(4, 10), Z 62

8

(4, 8), Z 52

6

FIGURE 3.6 The Wyndor Glass Co. problem would have no optimal solutions if the only functional constraint were x1 4, because x2 then could be increased indefinitely in the feasible region without ever reaching the maximum value of Z 3x1 5x2.

(4, 6), Z 42 Feasible region

4

(4, 4), Z 32

2

(4, 2), Z 22

0

Maximize Z 3x1 5x2, subject to x1 4 x1 0, and x2 0

2

4

6

8

10

x1

The example has exactly one optimal solution, (x1, x2) (2, 6), which is a CPF solution. (Think about how the graphical method leads to the one optimal solution being a CPF solution.) When the example is modified to yield multiple optimal solutions, as shown in Fig. 3.5, two of these optimal solutions—(2, 6) and (4, 3)—are CPF solutions.

3.3

ASSUMPTIONS OF LINEAR PROGRAMMING All the assumptions of linear programming actually are implicit in the model formulation given in Sec. 3.2. However, it is good to highlight these assumptions so you can more easily evaluate how well linear programming applies to any given problem. Furthermore, we still need to see why the OR team for the Wyndor Glass Co. concluded that a linear programming formulation provided a satisfactory representation of the problem. Proportionality Proportionality is an assumption about both the objective function and the functional constraints, as summarized below. Proportionality assumption: The contribution of each activity to the value of the objective function Z is proportional to the level of the activity xj, as represented by the cj xj term in the objective function. Similarly, the contribution of each activity to the left-hand side of each functional constraint is proportional to the level of the activity xj, as represented by the aij xj term in the constraint.

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

FIGURE 3.7 The five dots are the five CPF solutions for the Wyndor Glass Co. problem.

37

x2 (0, 6)

(2, 6)

Feasible region

(4, 3)

(0, 0)

(4, 0)

x1

Consequently, this assumption rules out any exponent other than 1 for any variable in any term of any function (whether the objective function or the function on the left-hand side of a functional constraint) in a linear programming model.1 To illustrate this assumption, consider the first term (3x1) in the objective function (Z 3x1 5x2) for the Wyndor Glass Co. problem. This term represents the profit generated per week (in thousands of dollars) by producing product 1 at the rate of x1 batches per week. The proportionality satisfied column of Table 3.4 shows the case that was assumed in Sec. 3.1, namely, that this profit is indeed proportional to x1 so that 3x1 is the appropriate term for the objective function. By contrast, the next three columns show different hypothetical cases where the proportionality assumption would be violated. Refer first to the Case 1 column in Table 3.4. This case would arise if there were start-up costs associated with initiating the production of product 1. For example, there 1 When the function includes any cross-product terms, proportionality should be interpreted to mean that changes in the function value are proportional to changes in each variable (xj) individually, given any fixed values for all the other variables. Therefore, a cross-product term satisfies proportionality as long as each variable in the term has an exponent of 1. (However, any cross-product term violates the additivity assumption, discussed next.)

TABLE 3.4 Examples of satisfying or violating proportionality Profit from Product 1 ($000 per Week) Proportionality Violated x1

Proportionality Satisfied

Case 1

Case 2

Case 3

0 1 2 3 4

0 3 6 9 12

0 2 5 8 11

0 3 7 12 18

0 3 5 6 6

38

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INTRODUCTION TO LINEAR PROGRAMMING

might be costs involved with setting up the production facilities. There might also be costs associated with arranging the distribution of the new product. Because these are one-time costs, they would need to be amortized on a per-week basis to be commensurable with Z (profit in thousands of dollars per week). Suppose that this amortization were done and that the total start-up cost amounted to reducing Z by 1, but that the profit without considering the start-up cost would be 3x1. This would mean that the contribution from product 1 to Z should be 3x1 1 for x1 0, whereas the contribution would be 3x1 0 when x1 0 (no start-up cost). This profit function,1 which is given by the solid curve in Fig. 3.8, certainly is not proportional to x1. At first glance, it might appear that Case 2 in Table 3.4 is quite similar to Case 1. However, Case 2 actually arises in a very different way. There no longer is a start-up cost, and the profit from the first unit of product 1 per week is indeed 3, as originally assumed. However, there now is an increasing marginal return; i.e., the slope of the profit function for product 1 (see the solid curve in Fig. 3.9) keeps increasing as x1 is increased. This violation of proportionality might occur because of economies of scale that can sometimes be achieved at higher levels of production, e.g., through the use of more efficient highvolume machinery, longer production runs, quantity discounts for large purchases of raw materials, and the learning-curve effect whereby workers become more efficient as they gain experience with a particular mode of production. As the incremental cost goes down, the incremental profit will go up (assuming constant marginal revenue). If the contribution from product 1 to Z were 3x1 1 for all x1 0, including x1 0, then the fixed constant, 1, could be deleted from the objective function without changing the optimal solution and proportionality would be restored. However, this “fix” does not work here because the 1 constant does not apply when x1 0.

1

FIGURE 3.8 The solid curve violates the proportionality assumption because of the start-up cost that is incurred when x1 is increased from 0. The values at the dots are given by the Case 1 column of Table 3.4.

Contribution of x1 to Z 12

9 Satisfies proportionality assumption 6

Violates proportionality assumption

3

0 Start-up cost 3

1

2

3

4

x1

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

39

Contribution of x1 to Z 18

15 12 9 FIGURE 3.9 The solid curve violates the proportionality assumption because its slope (the marginal return from product 1) keeps increasing as x1 is increased. The values at the dots are given by the Case 2 column of Table 3.4.

Violates proportionality assumption Satisfies proportionality assumption

6 3

0

1

2

3

4

x1

Referring again to Table 3.4, the reverse of Case 2 is Case 3, where there is a decreasing marginal return. In this case, the slope of the profit function for product 1 (given by the solid curve in Fig. 3.10) keeps decreasing as x1 is increased. This violation of proportionality might occur because the marketing costs need to go up more than proportionally to attain increases in the level of sales. For example, it might be possible to sell product 1 at the rate of 1 per week (x1 1) with no advertising, whereas attaining sales to sustain a production rate of x1 2 might require a moderate amount of advertising, x1 3 might necessitate an extensive advertising campaign, and x1 4 might require also lowering the price. All three cases are hypothetical examples of ways in which the proportionality assumption could be violated. What is the actual situation? The actual profit from produc-

FIGURE 3.10 The solid curve violates the proportionality assumption because its slope (the marginal return from product 1) keeps decreasing as x1 is increased. The values at the dots are given by the Case 3 column in Table 3.4.

Contribution of x1 to Z 12 9

Satisfies proportionality assumption

6 Violates proportionality assumption

3

0

1

2

3

4

x1

40

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INTRODUCTION TO LINEAR PROGRAMMING

ing product 1 (or any other product) is derived from the sales revenue minus various direct and indirect costs. Inevitably, some of these cost components are not strictly proportional to the production rate, perhaps for one of the reasons illustrated above. However, the real question is whether, after all the components of profit have been accumulated, proportionality is a reasonable approximation for practical modeling purposes. For the Wyndor Glass Co. problem, the OR team checked both the objective function and the functional constraints. The conclusion was that proportionality could indeed be assumed without serious distortion. For other problems, what happens when the proportionality assumption does not hold even as a reasonable approximation? In most cases, this means you must use nonlinear programming instead (presented in Chap. 13). However, we do point out in Sec. 13.8 that a certain important kind of nonproportionality can still be handled by linear programming by reformulating the problem appropriately. Furthermore, if the assumption is violated only because of start-up costs, there is an extension of linear programming (mixed integer programming) that can be used, as discussed in Sec. 12.3 (the fixed-charge problem). Additivity Although the proportionality assumption rules out exponents other than 1, it does not prohibit cross-product terms (terms involving the product of two or more variables). The additivity assumption does rule out this latter possibility, as summarized below. Additivity assumption: Every function in a linear programming model (whether the objective function or the function on the left-hand side of a functional constraint) is the sum of the individual contributions of the respective activities. To make this definition more concrete and clarify why we need to worry about this assumption, let us look at some examples. Table 3.5 shows some possible cases for the objective function for the Wyndor Glass Co. problem. In each case, the individual contributions from the products are just as assumed in Sec. 3.1, namely, 3x1 for product 1 and 5x2 for product 2. The difference lies in the last row, which gives the function value for Z when the two products are produced jointly. The additivity satisfied column shows the case where this function value is obtained simply by adding the first two rows (3 5 8), so that Z 3x1 5x2 as previously assumed. By contrast, the next two columns show hypothetical cases where the additivity assumption would be violated (but not the proportionality assumption). TABLE 3.5 Examples of satisfying or violating additivity for the objective function Value of Z Additivity Violated (x1, x2)

Additivity Satisfied

Case 1

Case 2

(1, 0) (0, 1)

3 5

3 5

3 5

(1, 1)

8

9

7

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

41

Referring to the Case 1 column of Table 3.5, this case corresponds to an objective function of Z 3x1 5x2 x1x2, so that Z 3 5 1 9 for (x1, x2) (1, 1), thereby violating the additivity assumption that Z 3 5. (The proportionality assumption still is satisfied since after the value of one variable is fixed, the increment in Z from the other variable is proportional to the value of that variable.) This case would arise if the two products were complementary in some way that increases profit. For example, suppose that a major advertising campaign would be required to market either new product produced by itself, but that the same single campaign can effectively promote both products if the decision is made to produce both. Because a major cost is saved for the second product, their joint profit is somewhat more than the sum of their individual profits when each is produced by itself. Case 2 in Table 3.5 also violates the additivity assumption because of the extra term in the corresponding objective function, Z 3x1 5x2 x1x2, so that Z 3 5 1 7 for (x1, x2) (1, 1). As the reverse of the first case, Case 2 would arise if the two products were competitive in some way that decreased their joint profit. For example, suppose that both products need to use the same machinery and equipment. If either product were produced by itself, this machinery and equipment would be dedicated to this one use. However, producing both products would require switching the production processes back and forth, with substantial time and cost involved in temporarily shutting down the production of one product and setting up for the other. Because of this major extra cost, their joint profit is somewhat less than the sum of their individual profits when each is produced by itself. The same kinds of interaction between activities can affect the additivity of the constraint functions. For example, consider the third functional constraint of the Wyndor Glass Co. problem: 3x1 2x2 18. (This is the only constraint involving both products.) This constraint concerns the production capacity of Plant 3, where 18 hours of production time per week is available for the two new products, and the function on the left-hand side (3x1 2x2) represents the number of hours of production time per week that would be used by these products. The additivity satisfied column of Table 3.6 shows this case as is, whereas the next two columns display cases where the function has an extra crossproduct term that violates additivity. For all three columns, the individual contributions from the products toward using the capacity of Plant 3 are just as assumed previously, namely, 3x1 for product 1 and 2x2 for product 2, or 3(2) 6 for x1 2 and 2(3) 6 for

TABLE 3.6 Examples of satisfying or violating additivity for a functional constraint Amount of Resource Used Additivity Violated (x1, x2)

Additivity Satisfied

Case 3

Case 4

(2, 0) (0, 3)

6 6

6 6

6 6

(2, 3)

12

15

10.8

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INTRODUCTION TO LINEAR PROGRAMMING

x2 3. As was true for Table 3.5, the difference lies in the last row, which now gives the total function value for production time used when the two products are produced jointly. For Case 3 (see Table 3.6), the production time used by the two products is given by the function 3x1 2x2 0.5x1x2, so the total function value is 6 6 3 15 when (x1, x2) (2, 3), which violates the additivity assumption that the value is just 6 6 12. This case can arise in exactly the same way as described for Case 2 in Table 3.5; namely, extra time is wasted switching the production processes back and forth between the two products. The extra cross-product term (0.5x1x2) would give the production time wasted in this way. (Note that wasting time switching between products leads to a positive crossproduct term here, where the total function is measuring production time used, whereas it led to a negative cross-product term for Case 2 because the total function there measures profit.) For Case 4 in Table 3.6, the function for production time used is 3x1 2x2 0.1x 21x2, so the function value for (x1, x2) (2, 3) is 6 6 1.2 10.8. This case could arise in the following way. As in Case 3, suppose that the two products require the same type of machinery and equipment. But suppose now that the time required to switch from one product to the other would be relatively small. Because each product goes through a sequence of production operations, individual production facilities normally dedicated to that product would incur occasional idle periods. During these otherwise idle periods, these facilities can be used by the other product. Consequently, the total production time used (including idle periods) when the two products are produced jointly would be less than the sum of the production times used by the individual products when each is produced by itself. After analyzing the possible kinds of interaction between the two products illustrated by these four cases, the OR team concluded that none played a major role in the actual Wyndor Glass Co. problem. Therefore, the additivity assumption was adopted as a reasonable approximation. For other problems, if additivity is not a reasonable assumption, so that some of or all the mathematical functions of the model need to be nonlinear (because of the crossproduct terms), you definitely enter the realm of nonlinear programming (Chap. 13). Divisibility Our next assumption concerns the values allowed for the decision variables. Divisibility assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values. Since each decision variable represents the level of some activity, it is being assumed that the activities can be run at fractional levels. For the Wyndor Glass Co. problem, the decision variables represent production rates (the number of batches of a product produced per week). Since these production rates can have any fractional values within the feasible region, the divisibility assumption does hold. In certain situations, the divisibility assumption does not hold because some of or all the decision variables must be restricted to integer values. Mathematical models with this restriction are called integer programming models, and they are discussed in Chap. 12.

3.3 ASSUMPTIONS OF LINEAR PROGRAMMING

43

Certainty Our last assumption concerns the parameters of the model, namely, the coefficients in the objective function cj, the coefficients in the functional constraints aij, and the right-hand sides of the functional constraints bi. Certainty assumption: The value assigned to each parameter of a linear programming model is assumed to be a known constant. In real applications, the certainty assumption is seldom satisfied precisely. Linear programming models usually are formulated to select some future course of action. Therefore, the parameter values used would be based on a prediction of future conditions, which inevitably introduces some degree of uncertainty. For this reason it is usually important to conduct sensitivity analysis after a solution is found that is optimal under the assumed parameter values. As discussed in Sec. 2.3, one purpose is to identify the sensitive parameters (those whose value cannot be changed without changing the optimal solution), since any later change in the value of a sensitive parameter immediately signals a need to change the solution being used. Sensitivity analysis plays an important role in the analysis of the Wyndor Glass Co. problem, as you will see in Sec. 6.7. However, it is necessary to acquire some more background before we finish that story. Occasionally, the degree of uncertainty in the parameters is too great to be amenable to sensitivity analysis. In this case, it is necessary to treat the parameters explicitly as random variables. Formulations of this kind have been developed, as discussed in Secs. 23.6 and 23.7 on the book’s web site, www.mhhe.com/hillier. The Assumptions in Perspective We emphasized in Sec. 2.2 that a mathematical model is intended to be only an idealized representation of the real problem. Approximations and simplifying assumptions generally are required in order for the model to be tractable. Adding too much detail and precision can make the model too unwieldy for useful analysis of the problem. All that is really needed is that there be a reasonably high correlation between the prediction of the model and what would actually happen in the real problem. This advice certainly is applicable to linear programming. It is very common in real applications of linear programming that almost none of the four assumptions hold completely. Except perhaps for the divisibility assumption, minor disparities are to be expected. This is especially true for the certainty assumption, so sensitivity analysis normally is a must to compensate for the violation of this assumption. However, it is important for the OR team to examine the four assumptions for the problem under study and to analyze just how large the disparities are. If any of the assumptions are violated in a major way, then a number of useful alternative models are available, as presented in later chapters of the book. A disadvantage of these other models is that the algorithms available for solving them are not nearly as powerful as those for linear programming, but this gap has been closing in some cases. For some applications, the powerful linear programming approach is used for the initial analysis, and then a more complicated model is used to refine this analysis.

44

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As you work through the examples in the next section, you will find it good practice to analyze how well each of the four assumptions of linear programming applies.

3.4

ADDITIONAL EXAMPLES The Wyndor Glass Co. problem is a prototype example of linear programming in several respects: It involves allocating limited resources among competing activities, its model fits our standard form, and its context is the traditional one of improved business planning. However, the applicability of linear programming is much wider. In this section we begin broadening our horizons. As you study the following examples, note that it is their underlying mathematical model rather than their context that characterizes them as linear programming problems. Then give some thought to how the same mathematical model could arise in many other contexts by merely changing the names of the activities and so forth. These examples are scaled-down versions of actual applications (including two that are included in the case studies presented in the next section). Design of Radiation Therapy

FIGURE 3.11 Cross section of Mary’s tumor (viewed from above), nearby critical tissues, and the radiation beams being used. Beam 2 1 3

2

3

Beam 1 1. Bladder and tumor 2. Rectum, coccyx, etc. 3. Femur, part of pelvis, etc.

MARY has just been diagnosed as having a cancer at a fairly advanced stage. Specifically, she has a large malignant tumor in the bladder area (a “whole bladder lesion”). Mary is to receive the most advanced medical care available to give her every possible chance for survival. This care will include extensive radiation therapy. Radiation therapy involves using an external beam treatment machine to pass ionizing radiation through the patient’s body, damaging both cancerous and healthy tissues. Normally, several beams are precisely administered from different angles in a twodimensional plane. Due to attenuation, each beam delivers more radiation to the tissue near the entry point than to the tissue near the exit point. Scatter also causes some delivery of radiation to tissue outside the direct path of the beam. Because tumor cells are typically microscopically interspersed among healthy cells, the radiation dosage throughout the tumor region must be large enough to kill the malignant cells, which are slightly more radiosensitive, yet small enough to spare the healthy cells. At the same time, the aggregate dose to critical tissues must not exceed established tolerance levels, in order to prevent complications that can be more serious than the disease itself. For the same reason, the total dose to the entire healthy anatomy must be minimized. Because of the need to carefully balance all these factors, the design of radiation therapy is a very delicate process. The goal of the design is to select the combination of beams to be used, and the intensity of each one, to generate the best possible dose distribution. (The dose strength at any point in the body is measured in units called kilorads.) Once the treatment design has been developed, it is administered in many installments, spread over several weeks. In Mary’s case, the size and location of her tumor make the design of her treatment an even more delicate process than usual. Figure 3.11 shows a diagram of a cross section of the tumor viewed from above, as well as nearby critical tissues to avoid. These tissues include critical organs (e.g., the rectum) as well as bony structures (e.g., the femurs and pelvis) that will attenuate the radiation. Also shown are the entry point and direction for the only two beams that can be used with any modicum of safety in this case. (Actually,

3.4 ADDITIONAL EXAMPLES

45

we are simplifying the example at this point, because normally dozens of possible beams must be considered.) For any proposed beam of given intensity, the analysis of what the resulting radiation absorption by various parts of the body would be requires a complicated process. In brief, based on careful anatomical analysis, the energy distribution within the twodimensional cross section of the tissue can be plotted on an isodose map, where the contour lines represent the dose strength as a percentage of the dose strength at the entry point. A fine grid then is placed over the isodose map. By summing the radiation absorbed in the squares containing each type of tissue, the average dose that is absorbed by the tumor, healthy anatomy, and critical tissues can be calculated. With more than one beam (administered sequentially), the radiation absorption is additive. After thorough analysis of this type, the medical team has carefully estimated the data needed to design Mary’s treatment, as summarized in Table 3.7. The first column lists the areas of the body that must be considered, and then the next two columns give the fraction of the radiation dose at the entry point for each beam that is absorbed by the respective areas on average. For example, if the dose level at the entry point for beam 1 is 1 kilorad, then an average of 0.4 kilorad will be absorbed by the entire healthy anatomy in the two-dimensional plane, an average of 0.3 kilorad will be absorbed by nearby critical tissues, an average of 0.5 kilorad will be absorbed by the various parts of the tumor, and 0.6 kilorad will be absorbed by the center of the tumor. The last column gives the restrictions on the total dosage from both beams that is absorbed on average by the respective areas of the body. In particular, the average dosage absorption for the healthy anatomy must be as small as possible, the critical tissues must not exceed 2.7 kilorads, the average over the entire tumor must equal 6 kilorads, and the center of the tumor must be at least 6 kilorads. Formulation as a Linear Programming Problem. The two decision variables x1 and x2 represent the dose (in kilorads) at the entry point for beam 1 and beam 2, respectively. Because the total dosage reaching the healthy anatomy is to be minimized, let Z denote this quantity. The data from Table 3.7 can then be used directly to formulate the following linear programming model.1 1

Actually, Table 3.7 simplifies the real situation, so the real model would be somewhat more complicated than this one and would have dozens of variables and constraints. For details about the general situation, see D. Sonderman and P. G. Abrahamson, “Radiotherapy Treatment Design Using Mathematical Programming Models,” Operations Research, 33:705–725, 1985, and its ref. 1.

TABLE 3.7 Data for the design of Mary’s radiation therapy Fraction of Entry Dose Absorbed by Area (Average) Area Healthy anatomy Critical tissues Tumor region Center of tumor

Beam 1

Beam 2

Restriction on Total Average Dosage, Kilorads

0.4 0.3 0.5 0.6

0.5 0.1 0.5 0.4

Minimize 2.7 6 6

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INTRODUCTION TO LINEAR PROGRAMMING

Minimize

Z 0.4x1 0.5x2,

subject to 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6 and x1 0,

x2 0.

Notice the differences between this model and the one in Sec. 3.1 for the Wyndor Glass Co. problem. The latter model involved maximizing Z, and all the functional constraints were in form. This new model does not fit this same standard form, but it does incorporate three other legitimate forms described in Sec. 3.2, namely, minimizing Z, functional constraints in form, and functional constraints in form. However, both models have only two variables, so this new problem also can be solved by the graphical method illustrated in Sec. 3.1. Figure 3.12 shows the graphical solution. The feasible region consists of just the dark line segment between (6, 6) and (7.5, 4.5), because the points on this segment are the only ones that simultaneously satisfy all the constraints. (Note that the equality constraint limits the feasible region to the line containing this line segment, and then the other two functional constraints determine the two endpoints of the line segment.) The dashed line is the objective function line that passes through the optimal solution (x1, x2) (7.5, 4.5) with Z 5.25. This solution is optimal rather than the point (6, 6) because decreasing Z (for positive values of Z) pushes the objective function line toward the origin (where Z 0). And Z 5.25 for (7.5, 4.5) is less than Z 5.4 for (6, 6). Thus, the optimal design is to use a total dose at the entry point of 7.5 kilorads for beam 1 and 4.5 kilorads for beam 2. Regional Planning The SOUTHERN CONFEDERATION OF KIBBUTZIM is a group of three kibbutzim (communal farming communities) in Israel. Overall planning for this group is done in its Coordinating Technical Office. This office currently is planning agricultural production for the coming year. The agricultural output of each kibbutz is limited by both the amount of available irrigable land and the quantity of water allocated for irrigation by the Water Commissioner (a national government official). These data are given in Table 3.8.

TABLE 3.8 Resource data for the Southern Confederation of Kibbutzim Kibbutz

Usable Land (Acres)

Water Allocation (Acre Feet)

1 2 3

400 600 300

600 800 375

3.4 ADDITIONAL EXAMPLES

47

x2 15

0.6x1 0.4x2 6

10

(6, 6)

5 (7.5, 4.5) Z 5.25 0.4x1 0.5x2

0.3x1 0.1x2 2.7 FIGURE 3.12 Graphical solution for the design of Mary’s radiation therapy.

0.5x1 0.5x2 6 0

5

10

x1

The crops suited for this region include sugar beets, cotton, and sorghum, and these are the three being considered for the upcoming season. These crops differ primarily in their expected net return per acre and their consumption of water. In addition, the Ministry of Agriculture has set a maximum quota for the total acreage that can be devoted to each of these crops by the Southern Confederation of Kibbutzim, as shown in Table 3.9. TABLE 3.9 Crop data for the Southern Confederation of Kibbutzim Crop Sugar beets Cotton Sorghum

Maximum Quota (Acres)

Water Consumption (Acre Feet/Acre)

Net Return ($/Acre)

600 500 325

3 2 1

1,000 750 250

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INTRODUCTION TO LINEAR PROGRAMMING

Because of the limited water available for irrigation, the Southern Confederation of Kibbutzim will not be able to use all its irrigable land for planting crops in the upcoming season. To ensure equity between the three kibbutzim, it has been agreed that every kibbutz will plant the same proportion of its available irrigable land. For example, if kibbutz 1 plants 200 of its available 400 acres, then kibbutz 2 must plant 300 of its 600 acres, while kibbutz 3 plants 150 acres of its 300 acres. However, any combination of the crops may be grown at any of the kibbutzim. The job facing the Coordinating Technical Office is to plan how many acres to devote to each crop at the respective kibbutzim while satisfying the given restrictions. The objective is to maximize the total net return to the Southern Confederation of Kibbutzim as a whole. Formulation as a Linear Programming Problem. The quantities to be decided upon are the number of acres to devote to each of the three crops at each of the three kibbutzim. The decision variables xj ( j 1, 2, . . . , 9) represent these nine quantities, as shown in Table 3.10. Since the measure of effectiveness Z is the total net return, the resulting linear programming model for this problem is Maximize

Z 1,000(x1 x2 x3) 750(x4 x5 x6) 250(x7 x8 x9),

subject to the following constraints: 1. Usable land for each kibbutz: x1 x4 x7 400 x2 x5 x8 600 x3 x6 x9 300 2. Water allocation for each kibbutz: 3x1 2x4 x7 600 3x2 2x5 x8 800 3x3 2x6 x9 375 3. Total acreage for each crop: x1 x2 x3 600 x4 x5 x6 500 x7 x8 x9 325 TABLE 3.10 Decision variables for the Southern Confederation of Kibbutzim problem Allocation (Acres) Kibbutz Crop

1

2

3

Sugar beets Cotton Sorghum

x1 x4 x7

x2 x5 x8

x3 x6 x9

3.4 ADDITIONAL EXAMPLES

49

4. Equal proportion of land planted: x1 x4 x7 x2 x5 x8 400 600 x2 x5 x8 x3 x6 x9 600 300 x3 x6 x9 x1 x4 x7 300 400 5. Nonnegativity: xj 0,

for j 1, 2, . . . , 9.

This completes the model, except that the equality constraints are not yet in an appropriate form for a linear programming model because some of the variables are on the righthand side. Hence, their final form1 is 3(x1 x4 x7) 2(x2 x5 x8) 0 (x2 x5 x8) 2(x3 x6 x9) 0 4(x3 x6 x9) 3(x1 x4 x7) 0 The Coordinating Technical Office formulated this model and then applied the simplex method (developed in the next chapter) to find an optimal solution

1 (x1, x2, x3, x4, x5, x6, x7, x8, x9) 133, 100, 25, 100, 250, 150, 0, 0, 0 , 3 as shown in Table 3.11. The resulting optimal value of the objective function is Z 633,33313, that is, a total net return of $633,333.33.

1

Actually, any one of these equations is redundant and can be deleted if desired. Also, because of these equations, any two of the usable land constraints also could be deleted because they automatically would be satisfied when both the remaining usable land constraint and these equations are satisfied. However, no harm is done (except a little more computational effort) by including unnecessary constraints, so you don’t need to worry about identifying and deleting them in models you formulate.

TABLE 3.11 Optimal solution for the Southern Confederation of Kibbutzim problem Best Allocation (Acres) Kibbutz Crop Sugar beets Cotton Sorghum

1

2

3

13313 100 0

100 250 0

25 150 0

50

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INTRODUCTION TO LINEAR PROGRAMMING

Controlling Air Pollution The NORI & LEETS CO., one of the major producers of steel in its part of the world, is located in the city of Steeltown and is the only large employer there. Steeltown has grown and prospered along with the company, which now employs nearly 50,000 residents. Therefore, the attitude of the townspeople always has been, “What’s good for Nori & Leets is good for the town.” However, this attitude is now changing; uncontrolled air pollution from the company’s furnaces is ruining the appearance of the city and endangering the health of its residents. A recent stockholders’ revolt resulted in the election of a new enlightened board of directors for the company. These directors are determined to follow socially responsible policies, and they have been discussing with Steeltown city officials and citizens’ groups what to do about the air pollution problem. Together they have worked out stringent air quality standards for the Steeltown airshed. The three main types of pollutants in this airshed are particulate matter, sulfur oxides, and hydrocarbons. The new standards require that the company reduce its annual emission of these pollutants by the amounts shown in Table 3.12. The board of directors has instructed management to have the engineering staff determine how to achieve these reductions in the most economical way. The steelworks has two primary sources of pollution, namely, the blast furnaces for making pig iron and the open-hearth furnaces for changing iron into steel. In both cases the engineers have decided that the most effective types of abatement methods are (1) increasing the height of the smokestacks,1 (2) using filter devices (including gas traps) in the smokestacks, and (3) including cleaner, high-grade materials among the fuels for the furnaces. Each of these methods has a technological limit on how heavily it can be used (e.g., a maximum feasible increase in the height of the smokestacks), but there also is considerable flexibility for using the method at a fraction of its technological limit. Table 3.13 shows how much emission (in millions of pounds per year) can be eliminated from each type of furnace by fully using any abatement method to its technological limit. For purposes of analysis, it is assumed that each method also can be used less fully to achieve any fraction of the emission-rate reductions shown in this table. Furthermore, the fractions can be different for blast furnaces and for open-hearth furnaces. For either type of furnace, the emission reduction achieved by each method is not substantially affected by whether the other methods also are used. 1

Subsequent to this study, this particular abatement method has become a controversial one. Because its effect is to reduce ground-level pollution by spreading emissions over a greater distance, environmental groups contend that this creates more acid rain by keeping sulfur oxides in the air longer. Consequently, the U.S. Environmental Protection Agency adopted new rules in 1985 to remove incentives for using tall smokestacks.

TABLE 3.12 Clean air standards for the Nori & Leets Co. Pollutant Particulates Sulfur oxides Hydrocarbons

Required Reduction in Annual Emission Rate (Million Pounds) 60 150 125

3.4 ADDITIONAL EXAMPLES

51

TABLE 3.13 Reduction in emission rate (in millions of pounds per year) from the maximum feasible use of an abatement method for Nori & Leets Co. Taller Smokestacks

Pollutant

Filters

Better Fuels

Blast Open-Hearth Blast Open-Hearth Blast Open-Hearth Furnaces Furnaces Furnaces Furnaces Furnaces Furnaces

Particulates Sulfur oxides Hydrocarbons

12 35 37

9 42 53

25 18 28

20 31 24

17 56 29

13 49 20

After these data were developed, it became clear that no single method by itself could achieve all the required reductions. On the other hand, combining all three methods at full capacity on both types of furnaces (which would be prohibitively expensive if the company’s products are to remain competitively priced) is much more than adequate. Therefore, the engineers concluded that they would have to use some combination of the methods, perhaps with fractional capacities, based upon the relative costs. Furthermore, because of the differences between the blast and the open-hearth furnaces, the two types probably should not use the same combination. An analysis was conducted to estimate the total annual cost that would be incurred by each abatement method. A method’s annual cost includes increased operating and maintenance expenses as well as reduced revenue due to any loss in the efficiency of the production process caused by using the method. The other major cost is the start-up cost (the initial capital outlay) required to install the method. To make this one-time cost commensurable with the ongoing annual costs, the time value of money was used to calculate the annual expenditure (over the expected life of the method) that would be equivalent in value to this start-up cost. This analysis led to the total annual cost estimates (in millions of dollars) given in Table 3.14 for using the methods at their full abatement capacities. It also was determined that the cost of a method being used at a lower level is roughly proportional to the fraction of the abatement capacity given in Table 3.13 that is achieved. Thus, for any given fraction achieved, the total annual cost would be roughly that fraction of the corresponding quantity in Table 3.14. The stage now was set to develop the general framework of the company’s plan for pollution abatement. This plan specifies which types of abatement methods will be used and at what fractions of their abatement capacities for (1) the blast furnaces and (2) the open-hearth furnaces. Because of the combinatorial nature of the problem of finding a TABLE 3.14 Total annual cost from the maximum feasible use of an abatement method for Nori & Leets Co. ($ millions) Abatement Method Taller smokestacks Filters Better fuels

Blast Furnaces

Open-Hearth Furnaces

8 7 11

10 6 9

52

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INTRODUCTION TO LINEAR PROGRAMMING

plan that satisfies the requirements with the smallest possible cost, an OR team was formed to solve the problem. The team adopted a linear programming approach, formulating the model summarized next. Formulation as a Linear Programming Problem. This problem has six decision variables xj, j 1, 2, . . . , 6, each representing the use of one of the three abatement methods for one of the two types of furnaces, expressed as a fraction of the abatement capacity (so xj cannot exceed 1). The ordering of these variables is shown in Table 3.15. Because the objective is to minimize total cost while satisfying the emission reduction requirements, the data in Tables 3.12, 3.13, and 3.14 yield the following model: Minimize

Z 8x1 10x2 7x3 6x4 11x5 9x6,

subject to the following constraints: 1. Emission reduction: 12x1 9x2 25x3 20x4 17x5 13x6 60 35x1 42x2 18x3 31x4 56x5 49x6 150 37x1 53x2 28x3 24x4 29x5 20x6 125 2. Technological limit: xj 1,

for j 1, 2, . . . , 6

xj 0,

for j 1, 2, . . . , 6.

3. Nonnegativity: The OR team used this model1 to find a minimum-cost plan (x1, x2, x3, x4, x5, x6) (1, 0.623, 0.343, 1, 0.048, 1), with Z 32.16 (total annual cost of $32.16 million). Sensitivity analysis then was conducted to explore the effect of making possible adjustments in the air standards given in Table 3.12, as well as to check on the effect of any inaccuracies in the cost data given in Table 3.14. (This story is continued in Case 6.1 at the end of Chap. 6.) Next came detailed planning and managerial review. Soon after, this program for controlling air pollution was fully implemented by the company, and the citizens of Steeltown breathed deep (cleaner) sighs of relief. 1

An equivalent formulation can express each decision variable in natural units for its abatement method; for example, x1 and x2 could represent the number of feet that the heights of the smokestacks are increased.

TABLE 3.15 Decision variables (fraction of the maximum feasible use of an abatement method) for Nori & Leets Co. Abatement Method Taller smokestacks Filters Better fuels

Blast Furnaces

Open-Hearth Furnaces

x1 x3 x5

x2 x4 x6

3.4 ADDITIONAL EXAMPLES

53

Reclaiming Solid Wastes The SAVE-IT COMPANY operates a reclamation center that collects four types of solid waste materials and treats them so that they can be amalgamated into a salable product. (Treating and amalgamating are separate processes.) Three different grades of this product can be made (see the first column of Table 3.16), depending upon the mix of the materials used. Although there is some flexibility in the mix for each grade, quality standards may specify the minimum or maximum amount allowed for the proportion of a material in the product grade. (This proportion is the weight of the material expressed as a percentage of the total weight for the product grade.) For each of the two higher grades, a fixed percentage is specified for one of the materials. These specifications are given in Table 3.16 along with the cost of amalgamation and the selling price for each grade. The reclamation center collects its solid waste materials from regular sources and so is normally able to maintain a steady rate for treating them. Table 3.17 gives the quantities available for collection and treatment each week, as well as the cost of treatment, for each type of material. The Save-It Co. is solely owned by Green Earth, an organization devoted to dealing with environmental issues, so Save-It’s profits are used to help support Green Earth’s activities. Green Earth has raised contributions and grants, amounting to $30,000 per week, to be used exclusively to cover the entire treatment cost for the solid waste materials. The board of directors of Green Earth has instructed the management of Save-It to divide this money among the materials in such a way that at least half of the amount available of each material is actually collected and treated. These additional restrictions are listed in Table 3.17. Within the restrictions specified in Tables 3.16 and 3.17, management wants to determine the amount of each product grade to produce and the exact mix of materials to be used for each grade. The objective is to maximize the net weekly profit (total sales income minus total amalgamation cost), exclusive of the fixed treatment cost of $30,000 per week that is being covered by gifts and grants. Formulation as a Linear Programming Problem. Before attempting to construct a linear programming model, we must give careful consideration to the proper definition of the decision variables. Although this definition is often obvious, it sometimes becomes TABLE 3.16 Product data for Save-It Co. Grade

Specification 1: 2: 3: 4:

Amalgamation Cost per Pound ($)

Selling Price per Pound ($)

A

Material Material Material Material

Not more than 30% of total Not less than 40% of total Not more than 50% of total Exactly 20% of total

3.00

8.50

B

Material 1: Not more than 50% of total Material 2: Not less than 10% of total Material 4: Exactly 10% of total

2.50

7.00

C

Material 1: Not more than 70% of total

2.00

5.50

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INTRODUCTION TO LINEAR PROGRAMMING

TABLE 3.17 Solid waste materials data for the Save-It Co. Material

Pounds per Week Available

Treatment Cost per Pound ($)

1 2 3 4

3,000 2,000 4,000 1,000

3.00 6.00 4.00 5.00

Additional Restrictions 1. For each material, at least half of the pounds per week available should be collected and treated. 2. $30,000 per week should be used to treat these materials.

the crux of the entire formulation. After clearly identifying what information is really desired and the most convenient form for conveying this information by means of decision variables, we can develop the objective function and the constraints on the values of these decision variables. In this particular problem, the decisions to be made are well defined, but the appropriate means of conveying this information may require some thought. (Try it and see if you first obtain the following inappropriate choice of decision variables.) Because one set of decisions is the amount of each product grade to produce, it would seem natural to define one set of decision variables accordingly. Proceeding tentatively along this line, we define yi number of pounds of product grade i produced per week

(i A, B, C).

The other set of decisions is the mix of materials for each product grade. This mix is identified by the proportion of each material in the product grade, which would suggest defining the other set of decision variables as zij proportion of material j in product grade i

(i A, B, C; j 1, 2, 3, 4).

However, Table 3.17 gives both the treatment cost and the availability of the materials by quantity (pounds) rather than proportion, so it is this quantity information that needs to be recorded in some of the constraints. For material j ( j 1, 2, 3, 4), Number of pounds of material j used per week zAj yA zBj yB zCj yC. For example, since Table 3.17 indicates that 3,000 pounds of material 1 is available per week, one constraint in the model would be zA1 yA zB1 yB zC1 yC 3,000. Unfortunately, this is not a legitimate linear programming constraint. The expression on the left-hand side is not a linear function because it involves products of variables. Therefore, a linear programming model cannot be constructed with these decision variables. Fortunately, there is another way of defining the decision variables that will fit the linear programming format. (Do you see how to do it?) It is accomplished by merely replacing each product of the old decision variables by a single variable! In other words, define (for i A, B, C; j 1, 2, 3, 4) xij zij yi xij number of pounds of material j allocated to product grade i per week,

3.4 ADDITIONAL EXAMPLES

55

and then we let the xij be the decision variables. Combining the xij in different ways yields the following quantities needed in the model (for i A, B, C; j 1, 2, 3, 4). xi1 xi2 xi3 xi4 number of pounds of product grade i produced per week. xAj xBj xCj number of pounds of material j used per week. xij proportion of material j in product grade i. xi1 xi2 xi3 xi4 The fact that this last expression is a nonlinear function does not cause a complication. For example, consider the first specification for product grade A in Table 3.16 (the proportion of material 1 should not exceed 30 percent). This restriction gives the nonlinear constraint xA1 0.3. xA1 xA2 xA3 xA4 However, multiplying through both sides of this inequality by the denominator yields an equivalent constraint xA1 0.3(xA1 xA2 xA3 xA4), so 0.7xA1 0.3xA2 0.3xA3 0.3xA4 0, which is a legitimate linear programming constraint. With this adjustment, the three quantities given above lead directly to all the functional constraints of the model. The objective function is based on management’s objective of maximizing net weekly profit (total sales income minus total amalgamation cost) from the three product grades. Thus, for each product grade, the profit per pound is obtained by subtracting the amalgamation cost given in the third column of Table 3.16 from the selling price in the fourth column. These differences provide the coefficients for the objective function. Therefore, the complete linear programming model is Maximize

Z 5.5(xA1 xA2 xA3 xA4) 4.5(xB1 xB2 xB3 xB4) 3.5(xC1 xC2 xC3 xC4),

subject to the following constraints: 1. Mixture specifications (second column of Table 3.16): xA1 0.3(xA1 xA2 xA3 xA4) xA2 0.4(xA1 xA2 xA3 xA4) xA3 0.5(xA1 xA2 xA3 xA4) xA4 0.2(xA1 xA2 xA3 xA4)

(grade (grade (grade (grade

xB1 0.5(xB1 xB2 xB3 xB4) xB2 0.1(xB1 xB2 xB3 xB4) xB4 0.1(xB1 xB2 xB3 xB4)

(grade B, material 1) (grade B, material 2) (grade B, material 4).

xC1 0.7(xC1 xC2 xC3 xC4)

(grade C, material 1).

A, A, A, A,

material material material material

1) 2) 3) 4).

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INTRODUCTION TO LINEAR PROGRAMMING

2. Availability of materials (second column of Table 3.17): xA1 xB1 xC1 xA2 xB2 xC2 xA3 xB3 xC3 xA4 xB4 xC4

3,000 2,000 4,000 1,000

(material (material (material (material

1) 2) 3) 4).

3. Restrictions on amounts treated (right side of Table 3.17): xA1 xB1 xC1 1,500 xA2 xB2 xC2 1,000 xA3 xB3 xC3 2,000 xA4 xB4 xC4 500

(material (material (material (material

1) 2) 3) 4).

4. Restriction on treatment cost (right side of Table 3.17): 3(xA1 xB1 xC1) 6(xA2 xB2 xC2) 4(xA3 xB3 xC3) 5(xA4 xB4 xC4) 30,000. 5. Nonnegativity constraints: xA1 0,

xA2 0,

...,

xC4 0.

This formulation completes the model, except that the constraints for the mixture specifications need to be rewritten in the proper form for a linear programming model by bringing all variables to the left-hand side and combining terms, as follows: Mixture specifications: 0.7xA1 0.3xA2 0.3xA3 0.3xA4 0 0.4xA1 0.6xA2 0.4xA3 0.4xA4 0 0.5xA1 0.5xA2 0.5xA3 0.5xA4 0 0.2xA1 0.2xA2 0.2xA3 0.8xA4 0

(grade A, material 1) (grade A, material 2) (grade A, material 3)

0.5xB1 0.5xB2 0.5xB3 0.5xB4 0 0.1xB1 0.9xB2 0.1xB3 0.1xB4 0 0.1xB1 0.1xB2 0.1xB3 0.9xB4 0

(grade B, material 1) (grade B, material 2) (grade B, material 4).

0.3xC1 0.7xC2 0.7xC3 0.7xC4 0

(grade C, material 1).

(grade A, material 4).

An optimal solution for this model is shown in Table 3.18, and then these xij values are used to calculate the other quantities of interest given in the table. The resulting optimal value of the objective function is Z 35,108.90 (a total weekly profit of $35,108.90). The Save-It Co. problem is an example of a blending problem. The objective for a blending problem is to find the best blend of ingredients into final products to meet certain specifications. Some of the earliest applications of linear programming were for gasoline blending, where petroleum ingredients were blended to obtain various grades of gasoline. The award-winning OR study at Texaco discussed at the end of

3.4 ADDITIONAL EXAMPLES

57

TABLE 3.18 Optimal solution for the Save-It Co. problem Pounds Used per Week Material Grade

1

A

412.3 (19.2%) 2587.7 (50%) 0

B C Total

2 859.6 (40%) 517.5 (10%) 0

3000

1377

Number of Pounds Produced per Week

3

4

447.4 (20.8%) 1552.6 (30%) 0

429.8 (20%) 517.5 (10%) 0

2000

947

2149 5175 0

Sec. 2.5 dealt with gasoline blending (although Texaco used a nonlinear programming model). Other blending problems involve such final products as steel, fertilizer, and animal feed. Personnel Scheduling UNION AIRWAYS is adding more flights to and from its hub airport, and so it needs to hire additional customer service agents. However, it is not clear just how many more should be hired. Management recognizes the need for cost control while also consistently providing a satisfactory level of service to customers. Therefore, an OR team is studying how to schedule the agents to provide satisfactory service with the smallest personnel cost. Based on the new schedule of flights, an analysis has been made of the minimum number of customer service agents that need to be on duty at different times of the day to provide a satisfactory level of service. The rightmost column of Table 3.19 shows the number of agents needed for the time periods given in the first column. The other entries TABLE 3.19 Data for the Union Airways personnel scheduling problem Time Periods Covered Shift Time Period

1

6:00 A.M. to 8:00 A.M. 8:00 A.M. to 10:00 A.M. 10:00 A.M. to noon Noon to 2:00 P.M. 2:00 P.M. to 4:00 P.M. 4:00 P.M. to 6:00 P.M. 6:00 P.M. to 8:00 P.M. 8:00 P.M. to 10:00 P.M. 10:00 P.M. to midnight Midnight to 6:00 A.M.

✔ ✔ ✔ ✔

Daily cost per agent

$170

2 ✔ ✔ ✔ ✔

$160

3

✔ ✔ ✔ ✔

$175

4

✔ ✔ ✔ ✔ $180

5

Minimum Number of Agents Needed

✔ ✔

48 79 65 87 64 73 82 43 52 15

$195

58

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INTRODUCTION TO LINEAR PROGRAMMING

in this table reflect one of the provisions in the company’s current contract with the union that represents the customer service agents. The provision is that each agent work an 8-hour shift 5 days per week, and the authorized shifts are Shift Shift Shift Shift Shift

1: 2: 3: 4: 5:

6:00 A.M. to 2:00 P.M. 8:00 A.M. to 4:00 P.M. Noon to 8:00 P.M. 4:00 P.M. to midnight 10:00 P.M. to 6:00 A.M.

Checkmarks in the main body of Table 3.19 show the hours covered by the respective shifts. Because some shifts are less desirable than others, the wages specified in the contract differ by shift. For each shift, the daily compensation (including benefits) for each agent is shown in the bottom row. The problem is to determine how many agents should be assigned to the respective shifts each day to minimize the total personnel cost for agents, based on this bottom row, while meeting (or surpassing) the service requirements given in the rightmost column. Formulation as a Linear Programming Problem. Linear programming problems always involve finding the best mix of activity levels. The key to formulating this particular problem is to recognize the nature of the activities. Activities correspond to shifts, where the level of each activity is the number of agents assigned to that shift. Thus, this problem involves finding the best mix of shift sizes. Since the decision variables always are the levels of the activities, the five decision variables here are xj number of agents assigned to shift j,

for j 1, 2, 3, 4, 5.

The main restrictions on the values of these decision variables are that the number of agents working during each time period must satisfy the minimum requirement given in the rightmost column of Table 3.19. For example, for 2:00 P.M. to 4:00 P.M., the total number of agents assigned to the shifts that cover this time period (shifts 2 and 3) must be at least 64, so x2 x3 64 is the functional constraint for this time period. Because the objective is to minimize the total cost of the agents assigned to the five shifts, the coefficients in the objective function are given by the last row of Table 3.19. Therefore, the complete linear programming model is Minimize

Z 170x1 160x2 175x3 180x4 195x5,

subject to x1 x1 x2 x1 x2 x1 x2 x3 x2 x3 x3 x4

48 79 65 87 64 73

(6–8 A.M.) (8–10 A.M.) (10 A.M. to noon) (Noon–2 P.M.) (2–4 P.M.) (4–6 P.M.)

3.4 ADDITIONAL EXAMPLES

59

x3 x4 82 x4 43 x4 x5 52 x5 15

(6–8 P.M.) (8–10 P.M.) (10 P.M.–midnight) (Midnight–6 A.M.)

and xj 0,

for j 1, 2, 3, 4, 5.

With a keen eye, you might have noticed that the third constraint, x1 x2 65, actually is not necessary because the second constraint, x1 x2 79, ensures that x1 x2 will be larger than 65. Thus, x1 x2 65 is a redundant constraint that can be deleted. Similarly, the sixth constraint, x3 x4 73, also is a redundant constraint because the seventh constraint is x3 x4 82. (In fact, three of the nonnegativity constraints—x1 0, x4 0, x5 0—also are redundant constraints because of the first, eighth, and tenth functional constraints: x1 48, x4 43, and x5 15. However, no computational advantage is gained by deleting these three nonnegativity constraints.) The optimal solution for this model is (x1, x2, x3, x4, x5) (48, 31, 39, 43, 15). This yields Z 30,610, that is, a total daily personnel cost of $30,610. This problem is an example where the divisibility assumption of linear programming actually is not satisfied. The number of agents assigned to each shift needs to be an integer. Strictly speaking, the model should have an additional constraint for each decision variable specifying that the variable must have an integer value. Adding these constraints would convert the linear programming model to an integer programming model (the topic of Chap. 12). Without these constraints, the optimal solution given above turned out to have integer values anyway, so no harm was done by not including the constraints. (The form of the functional constraints made this outcome a likely one.) If some of the variables had turned out to be noninteger, the easiest approach would have been to round up to integer values. (Rounding up is feasible for this example because all the functional constraints are in form with nonnegative coefficients.) Rounding up does not ensure obtaining an optimal solution for the integer programming model, but the error introduced by rounding up such large numbers would be negligible for most practical situations. Alternatively, integer programming techniques described in Chap. 12 could be used to solve exactly for an optimal solution with integer values. Section 3.5 includes a case study of how United Airlines used linear programming to develop a personnel scheduling system on a vastly larger scale than this example. Distributing Goods through a Distribution Network The Problem. The DISTRIBUTION UNLIMITED CO. will be producing the same new product at two different factories, and then the product must be shipped to two warehouses, where either factory can supply either warehouse. The distribution network available for shipping this product is shown in Fig. 3.13, where F1 and F2 are the two factories, W1 and W2 are the two warehouses, and DC is a distribution center. The amounts to be shipped from F1 and F2 are shown to their left, and the amounts to be received at W1 and W2 are shown to their right. Each arrow represents a feasible shipping lane. Thus, F1 can ship directly to W1 and has three possible routes (F1 DC W2, F1 F2

60

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INTRODUCTION TO LINEAR PROGRAMMING

DC W2, and F1 W1 W2) for shipping to W2. Factory F2 has just one route to W2 (F2 DC W2) and one to W1 (F2 DC W2 W1). The cost per unit shipped through each shipping lane is shown next to the arrow. Also shown next to F1 F2 and DC W2 are the maximum amounts that can be shipped through these lanes. The other lanes have sufficient shipping capacity to handle everything these factories can send. The decision to be made concerns how much to ship through each shipping lane. The objective is to minimize the total shipping cost. Formulation as a Linear Programming Problem. With seven shipping lanes, we need seven decision variables (xF1-F2, xF1-DC, xF1-W1, xF2-DC, xDC-W2, xW1-W2, xW2-W1) to represent the amounts shipped through the respective lanes. There are several restrictions on the values of these variables. In addition to the usual nonnegativity constraints, there are two upper-bound constraints, xF1-F2 10 and xDC-W2 80, imposed by the limited shipping capacities for the two lanes, F1 F2 and DC W2. All the other restrictions arise from five net flow constraints, one for each of the five locations. These constraints have the following form. Net flow constraint for each location: Amount shipped out amount shipped in required amount. As indicated in Fig. 3.13, these required amounts are 50 for F1, 40 for F2, 30 for W1, and 60 for W2.

50 units produced

$900/unit

F1

W1

30 units needed

$4 00 /u ni t

DC

ni

00 /u t

its

ni

un

/u

$300/unit

$1

00

$200/unit

80

t

$200/unit 10 units max.

m

$3

FIGURE 3.13 The distribution network for Distribution Unlimited Co.

ax .

40 units produced

F2

W2

60 units needed

3.5 SOME CASE STUDIES

61

What is the required amount for DC? All the units produced at the factories are ultimately needed at the warehouses, so any units shipped from the factories to the distribution center should be forwarded to the warehouses. Therefore, the total amount shipped from the distribution center to the warehouses should equal the total amount shipped from the factories to the distribution center. In other words, the difference of these two shipping amounts (the required amount for the net flow constraint) should be zero. Since the objective is to minimize the total shipping cost, the coefficients for the objective function come directly from the unit shipping costs given in Fig. 3.13. Therefore, by using money units of hundreds of dollars in this objective function, the complete linear programming model is Z 2xF1-F2 4xF1-DC 9xF1-W1 3xF2-DC xDC-W2 3xW1-W2 2xW2-W1,

Minimize

subject to the following constraints: 1. Net flow constraints: xF1-F2 xF1-DC xF1-W1 xF1-F2 xF2-DC xF1-DC xF2-DC xDC-W2 xF1-W1

xW1-W2 xDC-W2 xW1-W2

50 (factory 1) 40 (factory 2) 0 (distribution center) xW2-W1 30 (warehouse 1) xW2-W1 60 (warehouse 2)

2. Upper-bound constraints: xF1-F2 10,

xDC-W2 80

3. Nonnegativity constraints: xF1-F2 0,

xF1-DC 0, xF1-W1 0, xF2-DC 0, xW1-W2 0, xW2-W1 0.

xDC-W2 0,

You will see this problem again in Sec. 9.6, where we focus on linear programming problems of this type (called the minimum cost flow problem). In Sec. 9.7, we will solve for its optimal solution: xF1-F2 0, xW1-W2 0,

xF1-DC 40, xW2-W1 20.

xF1-W1 10,

xF2-DC 40,

xDC-W2 80,

The resulting total shipping cost is $49,000. You also will see a case study involving a much larger problem of this same type at the end of the next section.

3.5

SOME CASE STUDIES To give you a better perspective about the great impact linear programming can have, we now present three case studies of real applications. Each of these is a classic application, initiated in the early 1980s, that has come to be regarded as a standard of excellence for future applications of linear programming. The first one will bear some strong similarities to the Wyndor Glass Co. problem, but on a realistic scale. Similarly, the second and

62

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third are realistic versions of the last two examples presented in the preceding section (the Union Airways and Distribution Unlimited examples). Choosing the Product Mix at Ponderosa Industrial1 Until its sale in 1988, PONDEROSA INDUSTRIAL was a plywood manufacturer based in Anhuac, Chihuahua, that supplied 25 percent of the plywood in Mexico. Like any plywood manufacturer, Ponderosa’s many products were differentiated by thickness and by the quality of the wood. The plywood market in Mexico is competitive, so the market establishes the prices of the products. The prices can fluctuate considerably from month to month, and there may be great differences between the products in their price movements from even one month to the next. As a result, each product’s contribution to Ponderosa’s total profit was continually varying, and in different ways for different products. Because of its pronounced effect on profits, a critical issue facing management was the choice of product mix—how much to produce of each product—on a monthly basis. This choice was a very complex one, since it had to take into account the current amounts available of various resources needed to produce the products. The most important resources were logs in four quality categories and production capacities for both the pressing operation and the polishing operation. Beginning in 1980, linear programming was used on a monthly basis to guide the product-mix decision. The linear programming model had an objective of maximizing the total profit from all products. The model’s constraints included the various resource constraints as well as other relevant restrictions such as the minimum amount of a product that must be provided to regular customers and the maximum amount that can be sold. (To aid planning for the procurement of raw materials, the model also considered the impact of the product-mix decision for the upcoming month on production in the following month.) The model had 90 decision variables and 45 functional constraints. This model was used each month to find the product mix for the upcoming month that would be optimal if the estimated values of the various parameters of the model prove to be accurate. However, since some of the parameter values could change quickly (e.g., the unit profits of the products), sensitivity analysis was done to determine the effect if the estimated values turned out to be inaccurate. The results indicated when adjustments in the product mix should be made (if time permitted) as unanticipated market changes occurred that affected the price (and so the unit profit) of certain products. One key decision each month concerned the number of logs in each of the four quality categories to purchase. The amounts available for the upcoming month’s production actually were parameters of the model. Therefore, after the purchase decision was made and then the corresponding optimal product mix was determined, postoptimality analysis was conducted to investigate the effect of adjusting the purchase decision. For example, it is very easy with linear programming to check what the impact on total profit would be if a quick purchase were to be made of additional logs in a certain quality category to enable increasing production for the upcoming month. Ponderosa’s linear programming system was interactive, so management received an immediate response to its “what-if questions” about the impact of encountering parame1

A. Roy, E. E. DeFalomir, and L. Lasdon: “An Optimization-Based Decision Support System for a Product Mix Problem,” Interfaces, 12(2):26–33, April 1982.

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ter values that differ from those in the original model. What if a quick purchase of logs of a certain kind were made? What if product prices were to fluctuate in a certain way? A variety of such scenarios can be investigated. Management effectively used this power to reach better decisions than the “optimal” product mix from the original model. The impact of linear programming at Ponderosa was reported to be “tremendous.” It led to a dramatic shift in the types of plywood products emphasized by the company. The improved product-mix decisions were credited with increasing the overall profitability of the company by 20 percent. Other contributions of linear programming included better utilization of raw material, capital equipment, and personnel. Two factors helped make this application of linear programming so successful. One factor is that a natural language financial planning system was interfaced with the codes for finding an optimal solution for the linear programming model. Using natural language rather than mathematical symbols to display the components of the linear programming model and its output made the process understandable and meaningful for the managers making the product-mix decisions. Reporting to management in the language of managers is necessary for the successful application of linear programming. The other factor was that the linear programming system used was interactive. As mentioned earlier, after an optimal solution was obtained for one version of the model, this feature enabled managers to ask a variety of “what-if” questions and receive immediate responses. Better decisions frequently were reached by exploring other plausible scenarios, and this process also gave managers more confidence that their decision would perform well under most foreseeable circumstances. In any application, this ability to respond quickly to management’s needs and queries through postoptimality analysis (whether interactive or not) is a vital part of a linear programming study. Personnel Scheduling at United Airlines1 Despite unprecedented industry competition in 1983 and 1984, UNITED AIRLINES managed to achieve substantial growth with service to 48 new airports. In 1984, it became the only airline with service to cities in all 50 states. Its 1984 operating profit reached $564 million, with revenues of $6.2 billion, an increase of 6 percent over 1983, while costs grew by less than 2 percent. Cost control is essential to competing successfully in the airline industry. In 1982, upper management of United Airlines initiated an OR study of its personnel scheduling as part of the cost control measures associated with the airline’s 1983–1984 expansion. The goal was to schedule personnel at the airline’s reservations offices and airports so as to minimize the cost of providing the necessary service to customers. At the time, United Airlines employed over 4,000 reservations sales representatives and support personnel at its 11 reservations offices and about 1,000 customer service agents at its 10 largest airports. Some were part-time, working shifts from 2 to 8 hours; most were full-time, working 8- or 10-hour-shifts. Shifts start at several different times. Each reservations office was open (by telephone) 24 hours a day, as was each of the major airports. However, the number of employees needed at each location to provide the re1

T. J. Holloran and J. E. Bryn, “United Airlines Station Manpower Planning System,” Interfaces, 16(1): 39–50, Jan.–Feb. 1986.

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quired level of service varied greatly during the 24-hour day, and might fluctuate considerably from one half-hour to the next. Trying to design the work schedules for all the employees at a given location to meet these service requirements most efficiently is a nightmare of combinatorial considerations. Once an employee begins working, he or she will be there continuously for the entire shift (2 to 10 hours, depending on the employee), except for either a meal break or short rest breaks every 2 hours. Given the minimum number of employees needed on duty for each half-hour interval over a 24-hour day (where these requirements change from day to day over a 7-day week), how many employees of each shift length should begin work at what start time over each 24-hour day of a 7-day week? Fortunately, linear programming thrives on such combinatorial nightmares. Actually, several OR techniques described in this book were used in the computerized planning system developed to attack this problem. Both forecasting (Chap. 20) and queuing theory (Chaps. 17 and 18) were used to determine the minimum number of employees needed on duty for each half-hour interval. Integer programming (Chap. 12) was used to determine the times of day at which shifts would be allowed to start. However, the core of the planning system was linear programming, which did all the actual scheduling to provide the needed service with the smallest possible labor cost. A complete work schedule was developed for the first full week of a month, and then it was reused for the remainder of the month. This process was repeated each month to reflect changing conditions. Although the details about the linear programming model have not been published, it is clear that the basic approach used is the one illustrated by the Union Airways example of personnel scheduling in Sec. 3.4. The objective function being minimized represents the total personnel cost for the location being scheduled. The main functional constraints require that the number of employees on duty during each time period will not fall below minimum acceptable levels. However, the Union Airways example has only five decision variables. By contrast, the United Airlines model for some locations has over 20,000 decision variables! The difference is that a real application must consider myriad important details that can be ignored in a textbook example. For example, the United Airlines model takes into account such things as the meal and break assignment times for each employee scheduled, differences in shift lengths for different employees, and days off over a weekly schedule, among other scheduling details. This application of linear programming was reported to have had “an overwhelming impact not only on United management and members of the manpower planning group, but also for many who had never before heard of management science (OR) or mathematical modeling.” It earned rave reviews from upper management, operating managers, and affected employees alike. For example, one manager described the scheduling system as Magical, . . . just as the [customer] lines begin to build, someone shows up for work, and just as you begin to think you’re overstaffed, people start going home.1

In more tangible terms, this application was credited with saving United Airlines more than $6 million annually in just direct salary and benefit costs. Other benefits included improved customer service and reduced need for support staff. 1

Holloran and Bryn, “United Airlines Station Manpower Planning System,” p. 49.

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After some updating in the early 1990s, the system is providing similar benefits today. One factor that helped make this application of linear programming so successful was “the support of operational managers and their staffs.” This was a lesson learned by experience, because the OR team initially failed to establish a good line of communication with the operating managers, who then resisted the team’s initial recommendations. The team leaders described their mistake as follows: The cardinal rule for earning the trust and respect of operating managers and support staffs—”getting them involved in the development process”—had been violated.1

The team then worked much more closely with the operating managers—with outstanding results. Planning Supply, Distribution, and Marketing at Citgo Petroleum Corporation2 CITGO PETROLEUM CORPORATION specializes in refining and marketing petroleum. In the mid-1980s, it had annual sales of several billion dollars, ranking it among the 150 largest industrial companies in the United States. After several years of financial losses, Citgo was acquired in 1983 by Southland Corporation, the owner of the 7-Eleven convenience store chain (whose sales include 2 billion gallons of quality motor fuels annually). To turn Citgo’s financial losses around, Southland created a task force composed of Southland personnel, Citgo personnel, and outside consultants. An eminent OR consultant was appointed director of the task force to report directly to both the president of Citgo and the chairman of the board of Southland. During 1984 and 1985, this task force applied various OR techniques (as well as information systems technologies) throughout the corporation. It was reported that these OR applications “have changed the way Citgo does business and resulted in approximately $70 million per year profit improvement.”3 The two most important applications were both linear programming systems that provided management with powerful planning support. One, called the refinery LP system, led to great improvements in refinery yield, substantial reductions in the cost of labor, and other important cost savings. This system contributed approximately $50 million to profit improvement in 1985. (See the end of Sec. 2.4 for discussion of the key role that model validation played in the development of this system.) However, we will focus here on the other linear programming system, called the supply, distribution, and marketing modeling system (or just the SDM system), that Citgo is continuing to use. The SDM system is particularly interesting because it is based on a special kind of linear programming model that uses networks, just like the model for the Distribution Unlimited example presented at the end of Sec. 3.4. The model for the SDM system provides a representation of Citgo’s entire marketing and distribution network. At the time the task force conducted its OR study, Citgo owned or leased 36 product storage terminals which were supplied through five distribution center terminals via a dis1

Ibid, p. 47. See the references cited in footnote 2 on p. 10. 3 See p. 4 of the second reference cited in footnote 2 on p. 10. 2

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tribution network of pipelines, tankers, and barges. Citgo also sold product from over 350 exchange terminals that were shared with other petroleum marketers. To supply its customers, product might be acquired by Citgo from its refinery in Lake Charles, LA, or from spot purchases on one of five major spot markets, product exchanges, and trades with other industry refiners. These product acquisition decisions were made daily. However, the time from such a decision until the product reached the intended customers could be as long as 11 weeks. Therefore, the linear programming model used an 11-week planning horizon. The SDM system is used to coordinate the supply, distribution, and marketing of each of Citgo’s major products (originally four grades of motor fuel and No. 2 fuel oil) throughout the United States. Management uses the system to make decisions such as where to sell, what price to charge, where to buy or trade, how much to buy or trade, how much to hold in inventory, and how much to ship by each mode of transportation. Linear programming guides these decisions and when to implement them so as to minimize total cost or maximize total profit. The SDM system also is used in “what-if” sessions, where management asks what-if questions about scenarios that differ from those assumed in the original model. The linear programming model in the SDM system has the same form as the model for the Distribution Unlimited example presented at the end of Sec. 3.4. In fact, both models fit an important special kind of linear programming problem, called the minimum cost flow problem, that will be discussed in Sec. 9.6. The main functional constraints for such models are equality constraints, where each one prescribes what the net flow of goods out of a specific location must be. The Distribution Unlimited model has just seven decision variables and five equality constraints. By contrast, the Citgo model for each major product has about 15,000 decision variables and 3,000 equality constraints! At the end of Sec. 2.1, we described the important role that data collection and data verification played in developing the Citgo models. With such huge models, a massive amount of data must be gathered to determine all the parameter values. A state-of-the-art management database system was developed for this purpose. Before using the data, a preloader program was used to check for data errors and inconsistencies. The importance of doing so was brought forcefully home to the task force when, as mentioned in Sec. 2.1, the initial run of the preloader program generated a paper log of error messages an inch thick! It was clear that the data collection process needed to be thoroughly debugged to help ensure the validity of the models. The SDM linear programming system has greatly improved the efficiency of Citgo’s supply, distribution, and marketing operations, enabling a huge reduction in product inventory with no drop in service levels. During its first year, the value of petroleum products held in inventory was reduced by $116.5 million. This huge reduction in capital tied up in carrying inventory resulted in saving about $14 million annually in interest expenses for borrowed capital, adding $14 million to Citgo’s annual profits. Improvements in coordination, pricing, and purchasing decisions have been estimated to add at least another $2.5 million to annual profits. Many indirect benefits also are attributed to this application of linear programming, including improved data, better pricing strategies, and elimination of unnecessary product terminals, as well as improved communication and coordination between supply, distribution, marketing, and refinery groups.

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Some of the factors that contributed to the success of this OR study are the same as for the two preceding case studies. Like Ponderosa Industrial, one factor was developing output reports in the language of managers to really meet their needs. These output reports are designed to be easy for managers to understand and use, and they address the issues that are important to management. Also like Ponderosa, another factor was enabling management to respond quickly to the dynamics of the industry by using the linear programming system extensively in “what-if” sessions. As in so many applications of linear programming, postoptimality analysis proved more important than the initial optimal solution obtained for one version of the model. Much as in the United Airlines application, another factor was the enthusiastic support of operational managers during the development and implementation of this linear programming system. However, the most important factor was the unlimited support provided to the task force by top management, ranging right up to the chief executive officer and the chairman of the board of Citgo’s parent company, Southland Corporation. As mentioned earlier, the director of the task force (an eminent OR consultant) reported directly to both the president of Citgo and the chairman of the board of Southland. This backing by top management included strong organizational and financial support. The organizational support took a variety of forms. One example was the creation and staffing of the position of senior vice-president of operations coordination to evaluate and coordinate recommendations based on the models which spanned organizational boundaries. When discussing both this linear programming system and other OR applications implemented by the task force, team members described the financial support of top management as follows: The total cost of the systems implemented, $20 million to $30 million, was the greatest obstacle to this project. However, because of the information explosion in the petroleum industry, top management realized that numerous information systems were essential to gather, store, and analyze data. The incremental cost of adding management science (OR) technologies to these computers and systems was small, in fact very small in light of the enormous benefits they provided.1

3.6

DISPLAYING AND SOLVING LINEAR PROGRAMMING MODELS ON A SPREADSHEET Spreadsheet software, such as Excel, is a popular tool for analyzing and solving small linear programming problems. The main features of a linear programming model, including all its parameters, can be easily entered onto a spreadsheet. However, spreadsheet software can do much more than just display data. If we include some additional information, the spreadsheet can be used to quickly analyze potential solutions. For example, a potential solution can be checked to see if it is feasible and what Z value (profit or cost) it achieves. Much of the power of the spreadsheet lies in its ability to immediately see the results of any changes made in the solution. In addition, the Excel Solver can quickly apply the simplex method to find an optimal solution for the model. 1

Ibid, p. 21.

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To illustrate this process, we now return to the Wyndor example introduced in Sec. 3.1. Displaying the Model on a Spreadsheet After expressing profits in units of thousands of dollars, Table 3.1 in Sec. 3.1 gives all the parameters of the model for the Wyndor problem. Figure 3.14 shows the necessary additions to this table for an Excel spreadsheet. In particular, a row is added (row 9, labeled “Solution”) to store the values of the decision variables. Next, a column is added (column E, labeled “Totals”). For each functional constraint, the number in column E is the numerical value of the left-hand side of that constraint. Recall that the left-hand side represents the actual amount of the resource used, given the values of the decision variables in row 9. For example, for the Plant 3 constraint in row 7, the amount of this resource used (in hours of production time per week) is Production time used in Plant 3 3x1 2x2. In the language of Excel, the equivalent equation for the number in cell E7 is E7 C7*C9 D7*D9. Notice that this equation involves the sum of two products. There is a function in Excel, called SUMPRODUCT, that will sum up the product of each of the individual terms in two different ranges of cells. For instance, SUMPRODUCT(C7:D7,C9:D9) takes each of the individual terms in the range C7:D7, multiplies them by the corresponding term in the range C9:D9, and then sums up these individual products, just as shown in the above equation. Although optional with such short equations, this function is especially handy as a shortcut for entering longer linear programming equations. Next, signs are entered in cells F5, F6, and F7 to indicate the form of the functional constraints. (When using a trial-and-error approach, the spreadsheet still will allow you to enter infeasible trial solutions that violate the signs, but these signs serve as a reminder to reject such trial solutions if no changes are made in the numbers in column G.)

FIGURE 3.14 The spreadsheet for the Wyndor problem before using the Excel Solver, so the values of the decision variables and the objective function are just entered as zeros.

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Finally, the value of the objective function is entered in cell E8. Much like the other values in column E, it is the sum of products. The equation for cell E8 is SUMPRODUCT(C8:D8,C9:D9). The lower right-hand side of Fig. 3.14 shows all the formulas that need to be entered in the “Totals” column (column E) for the Wyndor problem. Once the model is entered in this spreadsheet format, it is easy to analyze any potential solution. When values for the decision variables are entered in the spreadsheet, the “Totals” column immediately calculates the total amount of each resource used, as well as the total profit. Hence, by comparing column E with column G, it can be seen immediately whether the potential solution is feasible. If so, cell E8 shows how much profit it would generate. One approach to trying to solve a linear programming model would be trial and error, using the spreadsheet to analyze a variety of solutions. However, you will see next how Excel also can be used to quickly find an optimal solution. Using the Excel Solver to Solve the Model Excel includes a tool called Solver that uses the simplex method to find an optimal solution. (A more powerful version of Solver, called Premium Solver, also is available in your OR Courseware.) Before using Solver, all the following components of the model need to be included on the spreadsheet: 1. Each decision variable 2. The objective function and its value 3. Each functional constraint The spreadsheet layout shown in Fig. 3.14 includes all these components. The parameters for the functional constraints are in rows 5, 6, and 7, and the coefficients for the objective function are in row 8. The values of the decision variables are in cells C9 and D9, and the value of the objective function is in cell E8. Since we don’t know what the values of the decision variables should be, they are just entered as zeros. The Solver will then change these to the optimal values after solving the problem. The Solver can be started by choosing “Solver” in the Tools menu. The Solver dialogue box is shown in Fig. 3.15. The “Target Cell” is the cell containing the value of the objective function, while the “Changing Cells” are the cells containing the values of the decision variables. Before the Solver can apply the simplex method, it needs to know exactly where each component of the model is located on the spreadsheet. You can either type in the cell addresses or click on them. Since the target cell is cell E8 and the changing cells are in the range C9:D9, these addresses are entered into the Solver dialogue box as shown in Fig. 3.15. (Excel then automatically enters the dollar signs shown in the figure to fix these addresses.) Since the goal is to maximize the objective function, “Max” also has been selected. Next, the addresses for the functional constraints need to be added. This is done by clicking on the “Add . . .” button on the Solver dialogue box. This brings up the “Add Constraint” dialogue box shown in Fig. 3.16. The location of the values of the left-hand sides and the right-hand sides of the functional constraints are specified in this dialogue box. The cells E5 through E7 all need to be less than or equal to the corresponding cells in G5 through G7. There also is a menu to choose between , , or , so has been chosen for these constraints. (This choice is needed even though signs were pre-

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FIGURE 3.15 The Solver dialogue box after specifying which cells in Fig. 3.14 contain the values of the objective function and the decision variables, plus indicating that the objective function is to be maximized.

viously entered in column F of the spreadsheet because Solver uses only the functional constraints that are specified with the Add Constraint dialogue box.) If there were more functional constraints to add, you would click on Add to bring up a new Add Constraint dialogue box. However, since there are no more in this example, the next step is to click on OK to go back to the Solver dialogue box. The Solver dialogue box now summarizes the complete model (see Fig. 3.17) in terms of the spreadsheet in Fig. 3.14. However, before asking Solver to solve the model, one more step should be taken. Clicking on the Options . . . button brings up the dialogue box shown in Fig. 3.18. This box allows you to specify a number of options about how the problem will be solved. The most important of these are the Assume Linear Model option and the Assume Non-Negative option. Be sure that both options are checked as shown in the figure. This tells Solver that the problem is a linear programming problem with nonnegativity constraints for all the decision variables, and that the simplex method

FIGURE 3.16 The Add Constraint dialogue box after specifying that cells E5, E6, and E7 in Fig. 3.14 are required to be less than or equal to cells G5, G6, and G7, respectively.

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FIGURE 3.17 The Solver dialogue box after specifying the entire model in terms of the spreadsheet.

FIGURE 3.18 The Solver Options dialogue box after checking the Assume Linear Model and Assume Non-Negative options to indicate that we are dealing with a linear programming model with nonnegativity constraints that needs to be solved by the simplex method.

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FIGURE 3.19 The Solver Results dialogue box that indicates that an optimal solution has been found.

should be used to solve the problem.1 Regarding the other options, accepting the default values shown in the figure usually is fine for small problems. Clicking on the OK button then returns you to the Solver dialogue box. Now you are ready to click on Solve in the Solver dialogue box, which will cause the Solver to execute the simplex method in the background. After a few seconds (for a small problem), Solver will then indicate the results. Typically, it will indicate that it has found an optimal solution, as specified in the Solver Results dialogue box shown in Fig. 3.19. If the model has no feasible solutions or no optimal solution, the dialogue box will indicate that instead by stating that “Solver could not find a feasible solution” or that “the Set Cell values do not converge.” The dialogue box also presents the option of generating various reports. One of these (the Sensitivity Report) will be discussed in detail in Sec. 4.7. After solving the model, the Solver replaces the original value of the decision variables in the spreadsheet with the optimal values, as shown in Fig. 3.20. The spreadsheet also indicates the value of the objective function, as well as the amount of each resource that is being used. 1

In older versions of Excel prior to Excel 97, the Assume Non-Negative option is not available, so nonnegativity constraints have to be added with the Add Constraint dialogue box.

FIGURE 3.20 The spreadsheet obtained after solving the Wyndor problem.

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FORMULATING VERY LARGE LINEAR PROGRAMMING MODELS Linear programming models come in many different sizes. For the examples in Secs. 3.1 and 3.4, the model sizes range from three functional constraints and two decision variables (for the Wyndor and radiation therapy problems) up to 17 functional constraints and 12 decision variables (for the Save-It Company problem). The latter case may seem like a rather large model. After all, it does take a substantial amount of time just to write down a model of this size. However, by contrast, the models for the classic case studies presented in Sec. 3.5 are much, much larger. For example, the models in the Citgo case study typically have about 3,000 functional constraints and 15,000 decision variables. The Citgo model sizes are not at all unusual. Linear programming models in practice commonly have hundreds or thousands of functional constraints. In fact, there have been some recently reported cases of a few hundred thousand constraints. The number of decision variables frequently is even larger than the number of functional constraints, and occasionally will range into the millions. Formulating such monstrously large models can be a daunting task. Even a “mediumsized” model with a thousand functional constraints and a thousand decision variables has over a million parameters (including the million coefficients in these constraints). It simply is not practical to write out the algebraic formulation, or even to fill in the parameters on a spreadsheet, for such a model. So how are these very large models formulated in practice? It requires the use of a modeling language. Modeling Languages A mathematical programming modeling language is software that has been specifically designed for efficiently formulating large linear programming models (and related models). Even with thousands of functional constraints, they typically are of a relatively few types where the constraints of the same type follow the same pattern. Similarly, the decision variables will fall into a small number of categories. Therefore, using large blocks of data in databases, a modeling language will simultaneously formulate all the constraints of the same type by simultaneously dealing with the variables of each type. We will illustrate this process soon. In addition to efficiently formulating large models, a modeling language will expedite a number of model management tasks, including accessing data, transforming data into model parameters, modifying the model whenever desired, and analyzing solutions from the model. It also may produce summary reports in the vernacular of the decision makers, as well as document the model’s contents. Several excellent modeling languages have been developed over the last couple of decades. These include AMPL, MPL, GAMS, and LINGO. The student version of one of these, MPL (short for mathematical programming language), is provided for you on the CD-ROM along with extensive tutorial material. The latest student version also can be downloaded from the website, maximal-usa.com. MPL is a product of Maximal Software, Inc. A new feature is extensive support for Excel in MPL. This includes both importing and exporting Excel ranges from MPL. Full support also is provided for the Excel VBA macro language through OptiMax 2000. (The student version of OptiMax 2000 is on the CD-ROM as well.) This product allows the user to

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fully integrate MPL models into Excel and solve with any of the powerful solvers that MPL supports, including CPLEX (described in Sec. 4.8). LINGO is a product of LINDO Systems, Inc. The latest student version of LINGO is available by downloading it from the website, www.lindo.com. LINDO Systems also provides a completely spreadsheet-oriented optimizer called What’sBest, also available on this website. The CD-ROM includes MPL, LINGO, and What’sBest formulations for essentially every example in this book to which these modeling languages can be applied. Now let us look at a simplified example that illustrates how a very large linear programming model can arise. An Example of a Problem with a Huge Model Management of the WORLDWIDE CORPORATION needs to address a product-mix problem, but one that is vastly more complex than the Wyndor product-mix problem introduced in Sec. 3.1. This corporation has 10 plants in various parts of the world. Each of these plants produces the same 10 products and then sells them within its region. The demand (sales potential) for each of these products from each plant is known for each of the next 10 months. Although the amount of a product sold by a plant in a given month cannot exceed the demand, the amount produced can be larger, where the excess amount would be stored in inventory (at some unit cost per month) for sale in a later month. Each unit of each product takes the same amount of space in inventory, and each plant has some upper limit on the total number of units that can be stored (the inventory capacity). Each plant has the same 10 production processes (we’ll refer to them as machines), each of which can be used to produce any of the 10 products. Both the production cost per unit of a product and the production rate of the product (number of units produced per day devoted to that product) depend on the combination of plant and machine involved (but not the month). The number of working days ( production days available) varies somewhat from month to month. Since some plants and machines can produce a particular product either less expensively or at a faster rate than other plants and machines, it is sometimes worthwhile to ship some units of the product from one plant to another for sale by the latter plant. For each combination of a plant being shipped from (the fromplant) and a plant being shipped to (the toplant), there is a certain cost per unit shipped of any product, where this unit shipping cost is the same for all the products. Management now needs to determine how much of each product should be produced by each machine in each plant during each month, as well as how much each plant should sell of each product in each month and how much each plant should ship of each product in each month to each of the other plants. Considering the worldwide price for each product, the objective is to find the feasible plan that maximizes the total profit (total sales revenue minus the sum of the total production costs, inventory costs, and shipping costs). The Structure of the Resulting Model Because of the inventory costs and the limited inventory capacities, it is necessary to keep track of the amount of each product kept in inventory in each plant during each month. Consequently, the linear programming model has four types of decision variables: pro-

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duction quantities, inventory quantities, sales quantities, and shipping quantities. With 10 plants, 10 machines, 10 products, and 10 months, this gives a total of 21,000 decision variables, as outlined below. Decision Variables. 10,000 production variables: one for each combination of a plant, machine, product, and month 1,000 inventory variables: one for each combination of a plant, product, and month 1,000 sales variables: one for each combination of a plant, product, and month 9,000 shipping variables: one for each combination of a product, month, plant (the fromplant), and another plant (the toplant) Multiplying each of these decision variables by the corresponding unit cost or unit revenue, and then summing over each type, the following objective function can be calculated: Objective Function. Maximize

profit total sales revenue total cost,

where Total cost total production cost total inventory cost total shipping cost. When maximizing this objective function, the 21,000 decision variables need to satisfy nonnegativity constraints as well as four types of functional constraints—production capacity constraints, plant balance constraints (equality constraints that provide appropriate values to the inventory variables), maximum inventory constraints, and maximum sales constraints. As enumerated below, there are a total of 3,100 functional constraints, but all the constraints of each type follow the same pattern. Functional Constraints. 1,000 production capacity constraints (one for each combination of a plant, machine, and month): Production days used production days available, where the left-hand side is the sum of 10 fractions, one for each product, where each fraction is that product’s production quantity (a decision variable) divided by the product’s production rate (a given constant). 1,000 plant balance constraints (one for each combination of a plant, product, and month): Amount produced inventory last month amount shipped in sales current inventory amount shipped out, where the amount produced is the sum of the decision variables representing the production quantities at the machines, the amount shipped in is the sum of the decision variables representing the shipping quantities in from the other plants, and the amount shipped out is the sum of the decision variables representing the shipping quantities out to the other plants.

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100 maximum inventory constraints (one for each combination of a plant and month): Total inventory inventory capacity, where the left-hand side is the sum of the decision variables representing the inventory quantities for the individual products. 1,000 maximum sales constraints (one for each combination of a plant, product, and month): Sales demand. Now let us see how the MPL modeling language, a product of Maximal Software, Inc., can formulate this huge model very compactly.

Formulation of the Model in MPL The modeler begins by assigning a title to the model and listing an index for each of the entities of the problem, as illustrated below. TITLE Production_Planning; INDEX product month plant fromplant toplant machine

: (A1, A2, A3, A4, A5, : (Jan, Feb, Mar, Apr, : (p1, p2, p3, p4, p5, : plant; : plant; : (m1, m2, m3, m4, m5,

A6, A7, A8, A9, A10); May, Jun, Jul, Aug, Sep, Oct); p6, p7, p8, p9, p10);

m6, m7, m8, m9, m10);

Except for the months, the entries on the right-hand side are arbitrary labels for the respective products, plants, and machines, where these same labels are used in the data files. Note that a colon is placed after the name of each entry and a semicolon is placed at the end of each statement (but a statement is allowed to extend over more than one line). A big job with any large model is collecting and organizing the various types of data into data files. In this case, eight data files are needed to hold the product prices, demands, production costs, production rates, production days available, inventory costs, inventory capacities, and shipping costs. Numbering these data files as 1, 2, 3, . . . , 8, the next step is to give a brief suggestive name to each one and to identify (inside square brackets) the index or indexes over which the data in the file run, as shown below. DATA Price [product] Demand [plant, product, month] ProdCost [plant, machine, product] ProdRate [plant, machine, product] ProdDaysAvail [month] InvtCost [product] InvtCapacity [plant] ShipCost [fromplant, toplant]

: DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE : DATAFILE

1; 2; 3; 4; 5; 6; 7; 8;

Next, the modeler gives a short name to each type of decision variable. Following the name, inside square brackets, is the index or indexes over which the subscripts run.

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VARIABLES Produce [plant, machine, product, month] Inventory [plant, product, month] Sales [plant, product, month] Ship [product, month, fromplant, toplant] WHERE (fromplant toplant);

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Prod; Invt; Sale;

In the case of the decision variables with names longer than four letters, the arrows on the right point to four-letter abbreviations to fit the size limitations of many solvers. The last line indicates that the fromplant subscript and toplant subscript are not allowed to have the same value. There is one more step before writing down the model. To make the model easier to read, it is useful first to introduce macros to represent the summations in the objective function. MACROS Total Revenue : SUM (plant, product, month: Price*Sales); TotalProdCost : SUM (plant, machine, product, month: ProdCost*Produce); TotalInvtCost : SUM (plant, product, month: InvtCost*Inventory); TotalShipCost : SUM (product, month, fromplant, toplant: ShipCost*Ship); TotalCost : TotalProdCost TotalInvtCost TotalShipCost;

The first four macros use the MPL keyword SUM to execute the summation involved. Following each SUM keyword (inside the parentheses) is, first, the index or indexes over which the summation runs. Next (after the colon) is the vector product of a data vector (one of the data files) times a variable vector (one of the four types of decision variables). Now this model with 3,100 functional constraints and 21,000 decision variables can be written down in the following compact form. MODEL MAX Profit TotalRevenue TotalCost; SUBJECT TO ProdCapacity [plant, machine, month] PCap; SUM (product: Produce/ProdRate) ProdDaysAvail; PlantBal [plant, product, month] PBal; SUM (machine: Produce) Inventory [month 1] SUM (fromplant: Ship[fromplant, toplant: plant]) Sales Inventory SUM (toplant: Ship[from plant: plant, toplant]); MaxInventory [plant, month] MaxI: SUM (product: Inventory) InvtCapacity; BOUNDS Sales Demand; END

For each of the four types of constraints, the first line gives the name for this type. There is one constraint of this type for each combination of values for the indexes inside

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the square brackets following the name. To the right of the brackets, the arrow points to a four-letter abbreviation of the name that a solver can use. Below the first line, the general form of constraints of this type is shown by using the SUM operator. For each production capacity constraint, each term in the summation consists of a decision variable (the production quantity of that product on that machine in that plant during that month) divided by the corresponding production rate, which gives the number of production days being used. Summing over the products then gives the total number of production days being used on that machine in that plant during that month, so this number must not exceed the number of production days available. The purpose of the plant balance constraint for each plant, product, and month is to give the correct value to the current inventory variable, given the values of all the other decision variables including the inventory level for the preceding month. Each of the SUM operators in these constraints involves simply a sum of decision variables rather than a vector product. This is the case also for the SUM operator in the maximum inventory constraints. By contrast, the left-hand side of the maximum sales constraints is just a single decision variable for each of the 1,000 combinations of a plant, product, and month. (Separating these upper-bound constraints on individual variables from the regular functional constraints is advantageous because of the computational efficiencies that can be obtained by using the upper bound technique described in Sec. 7.3.) No lower-bound constraints are shown here because MPL automatically assumes that all 21,000 decision variables have nonnegativity constraints unless nonzero lower bounds are specified. For each of the 3,100 functional constraints, note that the left-hand side is a linear function of the decision variables and the right-hand side is a constant taken from the appropriate data file. Since the objective function also is a linear function of the decision variables, this model is a legitimate linear programming model. To solve the model, MPL supports various leading solvers (software packages for solving linear programming models and related models) that can be installed into MPL. As discussed in Sec. 4.8, CPLEX is a particularly prominent and powerful solver. The version of MPL in your OR Courseware already has installed the student version of CPLEX, which uses the simplex method to solve linear programming models. Therefore, to solve such a model formulated with MPL, all you have to do is choose Solve CPLEX from the Run menu or press the Run Solve button in the Toolbar. You then can display the solution file in a view window by pressing the View button at the bottom of the Status Window. This brief introduction to MPL illustrates the ease with which modelers can use modeling languages to formulate huge linear programming models in a clear, concise way. To assist you in using MPL, an MPL Tutorial is included on the CD-ROM. This tutorial goes through all the details of formulating smaller versions of the production planning example considered here. You also can see elsewhere on the CD-ROM how all the other linear programming examples in this chapter and subsequent chapters would be formulated with MPL and solved by CPLEX. The LINGO Modeling Language LINGO is another popular modeling language that is featured in this book. The company that produces LINGO, LINDO Systems, also produces a widely used solver called LINDO as well as a spreadsheet solver, What’sBest. All three share a common set of solvers based

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79

on the simplex method and, in more advanced versions, on the kind of algorithmic techniques introduced in Secs. 4.9 and 7.4 as well. (We will discuss LINDO further in Sec. 4.8 and Appendix 4.1.) As mentioned earlier, the student version of LINGO is available to you through downloading from the website, www.lindo.com. Like MPL, LINGO enables a modeler to efficiently formulate a huge linear programming model in a clear, concise way. It also can be used for a wide variety of other models. LINGO uses sets as its fundamental building block. For example, in the Worldwide Corp. production planning problem, the sets of interest include the collections of products, plants, machines, and months. Each member of a set may have one or more attributes associated with it, such as the price of a product, the inventory capacity of a plant, the production rate of a machine, and the number of production days available in a month. These attributes provide data for the model. Some set attributes, such as production quantities and shipping quantities, can be decision variables for the model. As with MPL, the SUM operator is commonly used to write the objective function and each constraint type in a compact form. After completing the formulation, the model can be solved by selecting the Solve command from the LINGO menu or pressing the Solve button on the toolbar. An appendix to this chapter describes LINGO further and illustrates its use on a couple of small examples. A supplement on the CD-ROM shows how LINGO can be used to formulate the model for the Worldwide Corp. production planning example. A LINGO tutorial on the CD-ROM provides the details needed for doing basic modeling with this modeling language. The LINGO formulations and solutions for the various examples in both this chapter and many other chapters also are included on the CD-ROM.

3.8

CONCLUSIONS Linear programming is a powerful technique for dealing with the problem of allocating limited resources among competing activities as well as other problems having a similar mathematical formulation. It has become a standard tool of great importance for numerous business and industrial organizations. Furthermore, almost any social organization is concerned with allocating resources in some context, and there is a growing recognition of the extremely wide applicability of this technique. However, not all problems of allocating limited resources can be formulated to fit a linear programming model, even as a reasonable approximation. When one or more of the assumptions of linear programming is violated seriously, it may then be possible to apply another mathematical programming model instead, e.g., the models of integer programming (Chap. 12) or nonlinear programming (Chap. 13).

APPENDIX 3.1 THE LINGO MODELING LANGUAGE LINGO is a mathematical modeling language designed particularly for formulating and solving a wide variety of optimization problems, including linear programming, integer programming (Chap. 12), and nonlinear programming (Chap. 13) problems. Extensive details and a downloadable student version can be found at www.lindo.com.

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Simple problems are entered into LINGO in a fairly natural fashion. To illustrate, consider the following linear programming problem. Maximize

Z 20x 31y,

subject to 2x 5y 16 4x 3y 6

FIGURE A3.1 Screen shots showing the LINGO formulation and the LINGO solution report for a simple linear programming problem.

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81

and x 0,

y 0.

The screen shot in the top half of Fig. A3.1 shows how this problem would be formulated with LINGO. The first line of this formulation is just a comment describing the model. Note that the comment is preceded by an exclamation point and ended by a semicolon. This is a requirement for all comments in a LINGO formulation. The second line gives the objective function (without bothering to include the Z variable) and indicates that it is to be maximized. Note that each multiplication needs to be indicated by an asterisk. The objective function is ended by a semicolon, as is each of the functional constraints on the next two lines. The nonnegativity constraints are not shown in this formulation because these constraints are automatically assumed by LINGO. (If some variable x did not have a nonnegativity constraint, you would need to add @FREE(x); at the end of the formulation.) Variables can be shown as either lowercase or uppercase, since LINGO is case-insensitive. For example, a variable x1 can be typed in as either x1 or X1. Similarly, words can be either lowercase or uppercase (or a combination). For clarity, we will use uppercase for all reserved words that have a predefined meaning in LINGO. Notice the menu bar at the top of the LINGO window in Fig. A3.1. The ‘File’ and ‘Edit’ menu items behave in a standard Windows fashion. To solve a model once it has been entered, click on the ‘bullseye’ icon. (If you are using a platform other than a Windows-based PC, instead type the GO command at the colon prompt and press the enter key.) Before attempting to solve the model, LINGO will first check whether your model has any syntax errors and, if so, will indicate where they occur. Assuming no such errors, a solver will begin solving the problem, during which time a solver status window will appear on the screen. (For linear programming models, the solver used is LINDO, which will be described in some detail in the appendix to the next chapter.) When the solver finishes, a Solution Report will appear on the screen. The bottom half of Fig. A3.1 shows the solution report for our example. The Value column gives the optimal values of the decision variables. The first entry in the Slack or Surplus column shows the corresponding value of the objective function. The next two entries indicate the difference between the two sides of the respective constraints. The Reduced Cost and Dual Price columns provide some sensitivity analysis information for the problem. After discussing postoptimality analysis (including sensitivity analysis) in Sec. 4.7, we will explain what reduced costs and dual prices are while describing LINDO in Appendix 4.1. These quantities provide only a portion of the useful sensitivity analysis information. To generate a full sensitivity analysis report (such as shown in Appendix 4.1 for LINDO), the Range command in the LINGO menu would need to be chosen next. Just as was illustrated with MPL in Sec. 3.7, LINGO is designed mainly for efficiently formulating very large models by simultaneously dealing with all constraints or variables of the same type. We soon will use the following example to illustrate how LINGO does this.

Example. Consider a production-mix problem where we are concerned with what mix of four products we should produce during the upcoming week. For each product, each unit produced requires a known amount of production time on each of three machines. Each machine has a certain number of hours of production time available per week. Each product provides a certain profit per unit produced. Table A3.1 shows three types of data: machine-related data, product-related data, and data related to combinations of a machine and product. The objective is to determine how much to produce of each product so that total profit is maximized while not exceeding the limited production capacity of each machine.

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TABLE A3.1 Data needed for the product-mix example Production Time per Unit, Hours Product Machine

P01

P02

P03

P04

Production Time Available per Week, Hours

Roll Cut Weld

1.7 1.1 1.6

2.1 2.5 1.3

1.4 1.7 1.6

2.4 2.6 0.8

28 34 21

Profit per unit

26

35

25

37

In standard algebraic form, the structure of the linear programming model for this problem is to choose the nonnegative production levels (number of units produced during the upcoming week) for the four products so as to 4

Maximize

cj x j , j1

subject to 4

aij x j bj j1

for i 1, 2, 3;

where xj production level for product P0j cj unit profit for product P0j aij production time on machine i per unit of product P0j bi production time available per week on machine i. This model is small enough, with just 4 decision variables and 3 functional constraints, that it could be written out completely, term by term, but it would be tedious. In some similar applications, there might instead be hundreds of decision variables and functional constraints, so writing out a term-by-term version of this model each week would not be practical. LINGO provides a much more efficient and compact formulation, comparable to the above summary of the model, as we will see next.

Formulation of the Model in LINGO This model has a repetitive nature. All the decision variables are of the same type and all the functional constraints are of the same type. LINGO uses sets to describe this repetitive nature.1 The simple sets of interest in this case are 1. The set of machines, {Roll, Cut, Weld}. 2. The set of products, {P01, P02, P03, P04}. 1

Order is implied in LINGO sets so, strictly speaking, they are not truly sets in the usual mathematical sense.

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83

The attributes of interest for the members of these sets are 1. Attribute for each machine: Number of hours of production time available per week. 2. Attributes for each product: Profit per unit produced; Number of units produced per week. Thus, the first two types of attributes are input data that will become parameters of the model, whereas the last type (number of units produced per week of the respective products) provides the decision variables for the model. Let us abbreviate these attributes as follows. machine: ProdHoursAvail product: Profit, Produce. One other key type of information is the number of hours of production time that each unit of each product would use on each of the machines. This number can be viewed as an attribute for the members of the set of all combinations of a product and a machine. Since this set is derived from the two simple sets, it is referred to as a derived set. Let us abbreviate the attribute for members of this set as follows. MaPr (machine, product): ProdHoursUsed A LINGO formulation typically has three sections. 1. A SETS section that specifies the sets and their attributes. You can think of it as describing the structure of the data. 2. A DATA section that either provides the data to be used or indicates where it is to be obtained. 3. A section that provides the mathematical model itself. We begin by showing the first two sections for the example below. ! LINGO3h; ! Product mix example; ! Notice: the SETS section says nothing about the number or names of the machines or products. That information is determined completely by supplied data; SETS: ! The simple sets; Machine: ProdHoursAvail; Product: Profit, Produce; ! A derived set; MaPr (Machine, Product): ProdHoursUsed; ENDSETS DATA: ! Get the names of the machines; Machine Roll Cut Weld; ! Hours available on each machine; ProdHoursAvail 28 34 21; ! Get the names of the products; Product P01 P02 P03 P04; ! Profit contribution per unit; Profit 26 35 25 37; ! Hours needed per ProdHoursUsed 1.7 1.1 1.6 ENDDATA

unit 2.1 2.5 1.3

of product; 1.4 2.4 ! Roll; 1.7 2.6 ! Cut; 1.6 0.8; ! Weld;

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Before presenting the mathematical model itself, we need to introduce two key set looping functions that enable applying an operation to all members of a set by using a single statement. One is the @SUM function, which computes the sum of an expression over all members of a set. The general form of @SUM is @SUM( set: expression). For every member of the set, the expression is computed, and then they are all added up. For example, @SUM( Product(j): Profit(j)*Produce(j))

sums the expression following the colon—the unit profit of a product times the production rate of the product—over all members of the set preceding the colon. In particular, since this set is the set of products {Product( j) for j 1, 2, 3, 4}, the sum is over the index j. Therefore, this specific @SUM function provides the objective function, 4

cj xj, j1 given earlier for the model. The second key set looping function is the @FOR function. This function is used to generate constraints over members of a set. The general form is @FOR( set: constraint). For example, @FOR(Machine(i): @SUM( Product(i): ProdHoursUsed(i, j)*Produce (j)) ProdHoursAvail (i, j); );

says to generate the constraint following the colon for each member of the set preceding the colon. (The “less than or equal to” symbol, , is not on the standard keyboard, so LINGO treats the standard keyboard symbols as equivalent to .) This set is the set of machines {Machine (i) for i 1, 2, 3}, so this function loops over the index i. For each i, the constraint following the colon was expressed algebraically earlier as 4

aij x j bj. j1 Therefore, after the third section of the LINGO formulation (the mathematical model itself) is added, we obtain the complete formulation shown below: ! LINGO3h; ! Product mix example; SETS: !The simple sets; Machine: ProdHoursAvail; Product: Profit, Produce; !A derived set; MaPr( Machine, Product): ProdHoursUsed; ENDSETS DATA: !Get the names of the machines; Machine Roll Cut Weld; ! Hours available on each machine; ProdHoursAvail 28 34 21;

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THE LINGO MODELING LANGUAGE

85

! Get the names of the products; Product P01 P02 P03 P04; ! Profit contribution per unit; Profit 26 35 25 37; ! Hours needed per unit of product; ProdHoursUsed 1.7 2.1 1.4 2.4 ! Roll; 1.1 2.5 1.7 2.6 ! Cut; 1.6 1.3 1.6 0.8; ! Weld; ENDDATA ! Maximize total profit contribution; MAX @SUM( Product(i): Profit(i) * Produce(i)); ! For each machine i; @FOR( Machine( i): ! Hours used must be hours available; @SUM( Product( j): ProdHoursUsed( i, j) * Produce( j)) ProdHoursAvail; );

The model is solved by pressing the ‘bullseye’ button on the LINGO command bar. Pressing the ‘x ’ button on the command bar produces a report that looks in part as follows: Variable PRODUCE( P01) PRODUCE( P02) PRODUCE( P03) PRODUCE( P04) Row 1 2 3 4

Value 0.0000000 10.00000 5.000000 0.0000000

Slack or Surplus 475.0000 0.0000000 0.5000000 0.0000000

Reduced Cost 3.577921 0.0000000 0.0000000 1.441558 Dual Price 1.000000 15.25974 0.0000000 2.272727

Thus, we should produce 10 units of product P02 and 5 units of product P03, where Row 1 gives the resulting total profit of 475. Notice that this solution exactly uses the available capacity on the first and third machines (since Rows 2 and 4 give a Slack or Surplus of 0) and leaves the second machine with 0.5 hour of idleness. (We will discuss reduced costs and dual prices in Appendix 4.1 in conjunction with LINDO.) The rows section of this report is slightly ambiguous in that you need to remember that Row 1 in the model concerns the objective function and the subsequent rows involve the constraints on machine capacities. This association can be made more clear in the report by giving names to each constraint in the model. This is done by enclosing the name in [ ], placed just in front of the constraint. See the following modified fragment of the model. [Totprof] MAX @SUM( Product: Profit * Produce); ! For each machine i; @FOR( Machine( i): ! Hours used must be hours available; [Capc] @SUM( Product( j): ProdHoursUsed( i, j) * Produce( j)) ProdHoursAvail; );

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The solution report now contains these row names. Row TOTPROF CAPC( ROLL) CAPC( CUT) CAPC( WELD)

Slack or Surplus 475.0000 0.0000000 0.5000000 0.0000000

Dual Price 1.000000 15.25974 0.0000000 2.272727

An important feature of a LINGO model like this one is that it is completely “scalable” in products and machines. In other words, if you wanted to solve another version of this product-mix problem with a different number of machines and products, you would only have to enter the new data in the DATA section. You would not need to change the SETS section or any of the equations. This conversion could be done by clerical personnel without any understanding of the model equations.

Importing and Exporting Spreadsheet Data with LINGO The above example was completely self-contained in the sense that all the data were directly incorporated into the LINGO formulation. In some other applications, a large body of data will be stored in some source and will need to be entered into the model from that source. One popular place for storing data is spreadsheets. LINGO has a simple function, @OLE(), for retrieving and placing data from and into spreadsheets. To illustrate, let us suppose the data for our product-mix problem were originally entered into a spreadsheet as shown in Fig. A3.2. For the moment we are interested only in the shaded cells

FIGURE A3.2 Screen shot showing data for the product-mix example entered in a spreadsheet.

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THE LINGO MODELING LANGUAGE

87

in columns A-B and E-H. The data in these cells completely describe our little product-mix example. We want to avoid retyping these data into our LINGO model. Suppose that this spreadsheet is stored in the file d:\dirfred7\wbest03i.xls. The only part of the LINGO model that needs to be changed is the DATA section as shown below. DATA: ! Get the names of the machines; Machine @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Hours available on each machine; ProdHoursAvail @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Get the names of the products; Product @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Profit contribution per unit; Profit @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Hours needed per unit of product; ProdHoursUsed @OLE( ‘d:\dirfred7\wbest03i.xls’); ! Send the solution values back; @OLE( ‘d:\dirfred7\wbest03i.xls’) Produce; ENDDATA

The @OLE() function acts as your “plumbing contractor.” It lets the data flow from the spreadsheet to LINGO and back to the spreadsheet. So-called Object Linking and Embedding (OLE) is a feature of the Windows operating system. LINGO exploits this feature to make a link between the LINGO model and a spreadsheet. The first five uses of @OLE() above illustrate that this function can be used on the right of an assignment statement to retrieve data from a spreadsheet. The last use above illustrates that this function can be placed on the left of an assignment statement to place solution results into the spreadsheet instead. Notice from Fig. A3.2 that the optimal solution has been placed back into the spreadsheet in cells E6:H6. One simple but hidden step that had to be done beforehand in the spreadsheet was to define range names for the various collections of cells containing the data. Range names can be defined in Excel by using the mouse and the Insert, Name, Define menu item. For example, the set of cells A9:A11 was given the range name of Machine. Similarly, the set of cells E4:H4 was given the range name Product.

Importing and Exporting from a Database with LINGO Another common repository for data in a large firm is in a database. In a manner similar to @OLE(), LINGO has a connection function, @ODBC(), for transferring data from and to a database. This function is based around the Open DataBase Connectivity (ODBC) standard for communicating with SQL (Structured Query Language) databases. Most popular databases, such as Oracle, Paradox, DB/2, MS Access, and SQL Server, support the ODBC convention. Let us illustrate the ODBC connection for our little product-mix example. Suppose that all the data describing our problem are stored in a database called acces03j. The modification required in the LINGO model is almost trivial. Only the DATA section needs to be changed, as illustrated in the following fragment from the LINGO model. DATA: ! Get the names of the machines and available hours; Machine, ProdHoursAvail @ODBC( ‘acces03j’); ! Get the names of the products and profits; Product, Profit @ODBC( ‘acces03j’);

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! Hours needed per unit of product; ProdHoursUsed @ODBC( ‘acces03j’); ! Send the solution values back; @ODBC( ‘acces03j’) Produce; ENDDATA

Notice that, similar to the spreadsheet-based model, the size of the model in terms of the number of variables and constraints is determined completely by what is found in the database. The LINGO model automatically adjusts to what is found in the database. Now let us show what is in the database considered above. It contains three related tables. We give these tables names to match those in the LINGO model, namely, ‘Machine,’ to hold machinerelated data, ‘Product,’ to hold product-related data, and ‘MaPr,’ to hold data related to combinations of machines and products. Here is what the tables look like on the screen:

Machine Machine

ProdHoursAvail

Roll Cut Weld

28 34 21

Product Product

Profit

P01 P02 P03 P04

26 35 25 37

Produce

MaPr Machine Roll Roll Roll Roll Cut Cut Cut Cut Weld Weld Weld Weld

Product

ProdHoursUsed

P01 P02 P03 P04 P01 P02 P03 P04 P01 P02 P03 P04

1.7 2.1 1.4 2.4 1.1 2.5 1.7 2.6 1.6 1.3 1.6 0.8

SELECTED REFERENCES

89

Notice that the ‘Produce’ column has been left blank in the Product table. Once we solve the model, the ‘Produce’ amounts get inserted into the database and the Product table looks as follows: Product Product

Profit

Produce

P01 P02 P03 P04

26 35 25 37

0 10 5 0

There is one complication in using ODBC in Windows 95. The user must “register” the database with the Windows ODBC administrator. One does this by accessing (with mouse clicks) the My Computer/Control Panel/ODBC32 window. Once there, the user must give a name to the database (which may differ from the actual name of the file in which the data tables reside) and specify the directory in which the database file resides. It is this registered name that should be used in the LINGO model. Because the database has been registered, you did not see a directory specification in the @ODBC( ‘acces03j’) in the LINGO model. The ODBC manager knows the location of the database just from its name.

More about LINGO Only some of the capabilities of LINGO have been illustrated in this appendix. More details can be found in the documentation that accompanies LINGO when it is downloaded. LINGO is available in a variety of sizes. The smallest version is the demo version that can be downloaded from www.lindo.com. It is designed for textbook-sized problems (currently a maximum of 150 functional constraints and 300 decision variables). However, the largest version (called the extended version) is limited only by the storage space available. Tens of thousands of functional constraints and hundreds of thousands of decision variables are not unusual. If you would like to see how LINGO can formulate a huge model like the production planning example introduced in Sec. 3.7, a supplement to this appendix on the book’s website, www.mhhe.com/hillier, shows the LINGO formulation of this example. By reducing the number of products, plants, machines, and months, the supplement also introduces actual data into the formulation and then shows the complete solution. The supplement goes on to discuss and illustrate the debugging and verification of this large model. The supplement also describes further how to retrieve data from external files (including spreadsheets) and how to insert results in existing files. In addition to this supplement, the CD-ROM includes both a LINGO tutorial and LINGO/LINDO files with numerous examples of LINGO formulations.

SELECTED REFERENCES 1. Anderson, D. R., D. J. Sweeney, and T. A. Williams: An Introduction to Management Science, 9th ed., West, St. Paul, MN, 2000, chaps. 2, 4. 2. Gass, S.: An Illustrated Guide to Linear Programming, Dover Publications, New York, 1990. 3. Hillier, F. S., M. S. Hillier, and G. J. Lieberman: Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Irwin/McGraw-Hill, Burr Ridge, IL, 2000, chaps. 2, 3.

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4. LINGO User’s Guide, LINDO Systems, Inc., Chicago, IL, e-mail: [email protected], 1999. 5. MPL Modeling System (Release 4.0) manual, Maximal Software, Inc., Arlington, VA, e-mail: [email protected], 1998. 6. Williams, H. P.: Model Building in Mathematical Programming, 3d ed., Wiley, New York, 1990.

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE A Demonstration Example in OR Tutor: Graphical Method

An Excel Add-In: Premium Solver

“Ch. 3—Intro to LP” Files for Solving the Examples: Excel File LINGO/LINDO File MPL/CPLEX File

Supplement to Appendix 3.1: More about LINGO (appears on the book’s website, www.mhhe.com/hillier).

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The demonstration example listed above may be helpful. C: Use the computer to solve the problem by applying the simplex method. The available software options for doing this include the Excel Solver or Premium Solver (Sec. 3.6), MPL/CPLEX (Sec. 3.7), LINGO (Appendix 3.1), and LINDO (Appendix 4.1), but follow any instructions given by your instructor regarding the option to use. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 3.1-1.* For each of the following constraints, draw a separate graph to show the nonnegative solutions that satisfy this constraint. (a) x1 3x2 6 (b) 4x1 3x2 12 (c) 4x1 x2 8

(d) Now combine these constraints into a single graph to show the feasible region for the entire set of functional constraints plus nonnegativity constraints. 3.1-2. Consider the following objective function for a linear programming model: D

Maximize Z 2x1 3x2 (a) Draw a graph that shows the corresponding objective function lines for Z 6, Z 12, and Z 18. (b) Find the slope-intercept form of the equation for each of these three objective function lines. Compare the slope for these three lines. Also compare the intercept with the x2 axis.

D

3.1-3. Consider the following equation of a line: 20x1 40x2 400 (a) Find the slope-intercept form of this equation.

CHAPTER 3 PROBLEMS

(b) Use this form to identify the slope and the intercept with the x2 axis for this line. (c) Use the information from part (b) to draw a graph of this line. D

3.1-4.* Use the graphical method to solve the problem: Z 2x1 x2,

Maximize subject to x2 2x1 5x2 x1 x2 3x1 x2

10 60 18 44

and x1 0, D

x2 0.

3.1-5. Use the graphical method to solve the problem: Maximize

Z 10x1 20x2,

subject to x1 2x2 15 x1 x2 12 5x1 3x2 45 and x1 0,

x2 0.

3.1-6. The Whitt Window Company is a company with only three employees which makes two different kinds of hand-crafted windows: a wood-framed and an aluminum-framed window. They earn $60 profit for each wood-framed window and $30 profit for each aluminum-framed window. Doug makes the wood frames, and can make 6 per day. Linda makes the aluminum frames, and can make 4 per day. Bob forms and cuts the glass, and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass. The company wishes to determine how many windows of each type to produce per day to maximize total profit. (a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources. (b) Formulate a linear programming model for this problem. D (c) Use the graphical model to solve this model. (d) A new competitor in town has started making wood-framed windows as well. This may force the company to lower the price they charge and so lower the profit made for each woodframed window. How would the optimal solution change (if at

91

all) if the profit per wood-framed window decreases from $60 to $40? From $60 to $20? (e) Doug is considering lowering his working hours, which would decrease the number of wood frames he makes per day. How would the optimal solution change if he makes only 5 wood frames per day? 3.1-7. The Apex Television Company has to decide on the number of 27- and 20-inch sets to be produced at one of its factories. Market research indicates that at most 40 of the 27-inch sets and 10 of the 20-inch sets can be sold per month. The maximum number of work-hours available is 500 per month. A 27-inch set requires 20 work-hours and a 20-inch set requires 10 work-hours. Each 27-inch set sold produces a profit of $120 and each 20-inch set produces a profit of $80. A wholesaler has agreed to purchase all the television sets produced if the numbers do not exceed the maxima indicated by the market research. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. 3.1-8. The WorldLight Company produces two light fixtures (products 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce so as to maximize profit. For each unit of product 1, 1 unit of frame parts and 2 units of electrical components are required. For each unit of product 2, 3 units of frame parts and 2 units of electrical components are required. The company has 200 units of frame parts and 300 units of electrical components. Each unit of product 1 gives a profit of $1, and each unit of product 2, up to 60 units, gives a profit of $2. Any excess over 60 units of product 2 brings no profit, so such an excess has been ruled out. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. What is the resulting total profit? 3.1-9. The Primo Insurance Company is introducing two new product lines: special risk insurance and mortgages. The expected profit is $5 per unit on special risk insurance and $2 per unit on mortgages. Management wishes to establish sales quotas for the new product lines to maximize total expected profit. The work requirements are as follows: Work-Hours per Unit Department

Special Risk

Mortgage

Work-Hours Available

Underwriting Administration Claims

3 0 2

2 1 0

2400 800 1200

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INTRODUCTION TO LINEAR PROGRAMMING

(a) Formulate a linear programming model for this problem. (b) Use the graphical method to solve this model. (c) Verify the exact value of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations. D

3.1-10. Weenies and Buns is a food processing plant which manufactures hot dogs and hot dog buns. They grind their own flour for the hot dog buns at a maximum rate of 200 pounds per week. Each hot dog bun requires 0.1 pound of flour. They currently have a contract with Pigland, Inc., which specifies that a delivery of 800 pounds of pork product is delivered every Monday. Each hot dog requires 14 pound of pork product. All the other ingredients in the hot dogs and hot dog buns are in plentiful supply. Finally, the labor force at Weenies and Buns consists of 5 employees working full time (40 hours per week each). Each hot dog requires 3 minutes of labor, and each hot dog bun requires 2 minutes of labor. Each hot dog yields a profit of $0.20, and each bun yields a profit of $0.10. Weenies and Buns would like to know how many hot dogs and how many hot dog buns they should produce each week so as to achieve the highest possible profit. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. 3.1-11.* The Omega Manufacturing Company has discontinued the production of a certain unprofitable product line. This act created considerable excess production capacity. Management is considering devoting this excess capacity to one or more of three products; call them products 1, 2, and 3. The available capacity on the machines that might limit output is summarized in the following table:

Machine Type

Available Time (Machine Hours per Week)

Milling machine Lathe Grinder

500 350 150

The sales department indicates that the sales potential for products 1 and 2 exceeds the maximum production rate and that the sales potential for product 3 is 20 units per week. The unit profit would be $50, $20, and $25, respectively, on products 1, 2, and 3. The objective is to determine how much of each product Omega should produce to maximize profit. (a) Formulate a linear programming model for this problem. C (b) Use a computer to solve this model by the simplex method. 3.1-12. Consider the following problem, where the value of c1 has not yet been ascertained. D

Z c1x1 x2,

Maximize subject to x1 x2 6 x1 2x2 10 and x1 0,

x2 0.

Use graphical analysis to determine the optimal solution(s) for (x1, x2) for the various possible values of c1( c1 ). 3.1-13. Consider the following problem, where the value of k has not yet been ascertained.

D

Maximize

Z x1 2x2,

subject to x1 x2 2 x2 3 kx1 x2 2k 3,

where k 0

and x1 0,

x2 0.

The solution currently being used is x1 2, x2 3. Use graphical analysis to determine the values of k such that this solution actually is optimal. 3.1-14. Consider the following problem, where the values of c1 and c2 have not yet been ascertained. D

The number of machine hours required for each unit of the respective products is Productivity coefficient (in machine hours per unit) Machine Type

Product 1

Product 2

Product 3

Milling machine Lathe Grinder

9 5 3

3 4 0

5 0 2

Maximize

Z c1x1 c2x2,

subject to 2x1 x2 11 x1 2x2 2 and x1 0,

x2 0.

Use graphical analysis to determine the optimal solution(s) for (x1, x2) for the various possible values of c1 and c2. (Hint: Sepa-

CHAPTER 3 PROBLEMS

rate the cases where c2 0, c2 0, and c2 0. For the latter two cases, focus on the ratio of c1 to c2.) 3.2-1. The following table summarizes the key facts about two products, A and B, and the resources, Q, R, and S, required to produce them. Resource Usage per Unit Produced Resource

Product A Product B

Q R S

2 1 3

1 2 3

Profit per unit

3

2

Amount of Resource Available 2 2 4

All the assumptions of linear programming hold. (a) Formulate a linear programming model for this problem. D (b) Solve this model graphically. (c) Verify the exact value of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations. 3.2-2. The shaded area in the following graph represents the feasible region of a linear programming problem whose objective function is to be maximized. x2

(3, 3) (6, 3)

93

3.2-3.* This is your lucky day. You have just won a $10,000 prize. You are setting aside $4,000 for taxes and partying expenses, but you have decided to invest the other $6,000. Upon hearing this news, two different friends have offered you an opportunity to become a partner in two different entrepreneurial ventures, one planned by each friend. In both cases, this investment would involve expending some of your time next summer as well as putting up cash. Becoming a full partner in the first friend’s venture would require an investment of $5,000 and 400 hours, and your estimated profit (ignoring the value of your time) would be $4,500. The corresponding figures for the second friend’s venture are $4,000 and 500 hours, with an estimated profit to you of $4,500. However, both friends are flexible and would allow you to come in at any fraction of a full partnership you would like. If you choose a fraction of a full partnership, all the above figures given for a full partnership (money investment, time investment, and your profit) would be multiplied by this same fraction. Because you were looking for an interesting summer job anyway (maximum of 600 hours), you have decided to participate in one or both friends’ ventures in whichever combination would maximize your total estimated profit. You now need to solve the problem of finding the best combination. (a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources. (b) Formulate a linear programming model for this problem. D (c) Use the graphical method to solve this model. What is your total estimated profit? 3.2-4. Use the graphical method to find all optimal solutions for the following model: D

(0, 2)

Maximize

Z 500x1 300x2,

subject to

(0, 0) (6, 0)

x1

Label each of the following statements as True or False, and then justify your answer based on the graphical method. In each case, give an example of an objective function that illustrates your answer. (a) If (3, 3) produces a larger value of the objective function than (0, 2) and (6, 3), then (3, 3) must be an optimal solution. (b) If (3, 3) is an optimal solution and multiple optimal solutions exist, then either (0, 2) or (6, 3) must also be an optimal solution. (c) The point (0, 0) cannot be an optimal solution.

15x1 5x2 300 10x1 6x2 240 8x1 12x2 450 and x1 0,

x2 0.

3.2-5. Use the graphical method to demonstrate that the following model has no feasible solutions. D

Maximize

Z 5x1 7x2,

subject to 2x1 x2 1 x1 2x2 1

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INTRODUCTION TO LINEAR PROGRAMMING

and x1 0,

(a) Design of radiation therapy (Mary). (b) Regional planning (Southern Confederation of Kibbutzim). (c) Controlling air pollution (Nori & Leets Co.).

x2 0.

D 3.2-6. Suppose that the following constraints have been provided for a linear programming model.

x1 3x2 30 3x1 x2 30 and x1 0,

x2 0.

(a) Demonstrate that the feasible region is unbounded. (b) If the objective is to maximize Z x1 x2, does the model have an optimal solution? If so, find it. If not, explain why not. (c) Repeat part (b) when the objective is to maximize Z x1 x2. (d) For objective functions where this model has no optimal solution, does this mean that there are no good solutions according to the model? Explain. What probably went wrong when formulating the model?

3.4-2. For each of the four assumptions of linear programming discussed in Sec. 3.3, write a one-paragraph analysis of how well it applies to each of the following examples given in Sec. 3.4. (a) Reclaiming solid wastes (Save-It Co.). (b) Personnel scheduling (Union Airways). (c) Distributing goods through a distribution network (Distribution Unlimited Co.). D

3.4-3. Use the graphical method to solve this problem: Z 15x1 20x2,

Maximize subject to

x1 2x2 10 2x1 3x2 6 x1 x2 6

3.3-1. Reconsider Prob. 3.2-3. Indicate why each of the four assumptions of linear programming (Sec. 3.3) appears to be reasonably satisfied for this problem. Is one assumption more doubtful than the others? If so, what should be done to take this into account?

and

3.3-2. Consider a problem with two decision variables, x1 and x2, which represent the levels of activities 1 and 2, respectively. For each variable, the permissible values are 0, 1, and 2, where the feasible combinations of these values for the two variables are determined from a variety of constraints. The objective is to maximize a certain measure of performance denoted by Z. The values of Z for the possibly feasible values of (x1, x2) are estimated to be those given in the following table:

subject to

x1 0, D

x2 0.

3.4-4. Use the graphical method to solve this problem: Minimize

Z 3x1 2x2,

x1 2x2 12 2x1 3x2 12 2x1 x2 8 and x1 0,

x2 0.

3.4-5. Consider the following problem, where the value of c1 has not yet been ascertained. D

x2 x1

0

1

2

0 1 2

0 3 6

4 8 12

8 13 18

Z c1x1 2x2,

Maximize subject to 4x1 x2 12 x1 x2 2 and

Based on this information, indicate whether this problem completely satisfies each of the four assumptions of linear programming. Justify your answers. 3.4-1.* For each of the four assumptions of linear programming discussed in Sec. 3.3, write a one-paragraph analysis of how well you feel it applies to each of the following examples given in Sec. 3.4:

x1 0,

x2 0.

Use graphical analysis to determine the optimal solution(s) for (x1, x2) for the various possible values of c1. D

3.4-6. Consider the following model: Minimize

Z 40x1 50x2,

CHAPTER 3 PROBLEMS

subject to

Each pig requires at least 8,000 calories per day and at least 700 units of vitamins. A further constraint is that no more than one-third of the diet (by weight) can consist of Feed Type A, since it contains an ingredient which is toxic if consumed in too large a quantity. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. What is the resulting daily cost per pig?

2x1 3x2 30 x1 x2 12 2x1 x2 20 and x1 0,

x2 0.

(a) Use the graphical method to solve this model. (b) How does the optimal solution change if the objective function is changed to Z 40x1 70x2? (c) How does the optimal solution change if the third functional constraint is changed to 2x1 x2 15? 3.4-7. Ralph Edmund loves steaks and potatoes. Therefore, he has decided to go on a steady diet of only these two foods (plus some liquids and vitamin supplements) for all his meals. Ralph realizes that this isn’t the healthiest diet, so he wants to make sure that he eats the right quantities of the two foods to satisfy some key nutritional requirements. He has obtained the following nutritional and cost information: Grams of Ingredient per Serving Steak

Potatoes

Daily Requirement (Grams)

Carbohydrates Protein Fat

5 20 15

15 5 2

50 40 60

Cost per serving

$4

$2

Ingredient

95

Ralph wishes to determine the number of daily servings (may be fractional) of steak and potatoes that will meet these requirements at a minimum cost. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. C (c) Use a computer to solve this model by the simplex method. 3.4-8. Dwight is an elementary school teacher who also raises pigs for supplemental income. He is trying to decide what to feed his pigs. He is considering using a combination of pig feeds available from local suppliers. He would like to feed the pigs at minimum cost while also making sure each pig receives an adequate supply of calories and vitamins. The cost, calorie content, and vitamin content of each feed is given in the table below.

3.4-9. Web Mercantile sells many household products through an on-line catalog. The company needs substantial warehouse space for storing its goods. Plans now are being made for leasing warehouse storage space over the next 5 months. Just how much space will be required in each of these months is known. However, since these space requirements are quite different, it may be most economical to lease only the amount needed each month on a monthby-month basis. On the other hand, the additional cost for leasing space for additional months is much less than for the first month, so it may be less expensive to lease the maximum amount needed for the entire 5 months. Another option is the intermediate approach of changing the total amount of space leased (by adding a new lease and/or having an old lease expire) at least once but not every month. The space requirement and the leasing costs for the various leasing periods are as follows:

Month

Required Space (Sq. Ft.)

Leasing Period (Months)

Cost per Sq. Ft. Leased

1 2 3 4 5

30,000 20,000 40,000 10,000 50,000

1 2 3 4 5

$ 65 $100 $135 $160 $190

The objective is to minimize the total leasing cost for meeting the space requirements. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-10. Larry Edison is the director of the Computer Center for Buckly College. He now needs to schedule the staffing of the center. It is open from 8 A.M. until midnight. Larry has monitored the usage of the center at various times of the day, and determined that the following number of computer consultants are required:

Time of Day Contents Calories (per pound) Vitamins (per pound) Cost (per pound)

Feed Type A

Feed Type B

800 140 units $0.40

1,000 70 units $0.80

8 A.M.–noon Noon–4 P.M. 4 P.M.–8 P.M. 8 P.M.–midnight

Minimum Number of Consultants Required to Be on Duty 4 8 10 6

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Two types of computer consultants can be hired: full-time and part-time. The full-time consultants work for 8 consecutive hours in any of the following shifts: morning (8 A.M.–4 P.M.), afternoon (noon–8 P.M.), and evening (4 P.M.–midnight). Full-time consultants are paid $14 per hour. Part-time consultants can be hired to work any of the four shifts listed in the above table. Part-time consultants are paid $12 per hour. An additional requirement is that during every time period, there must be at least 2 full-time consultants on duty for every parttime consultant on duty. Larry would like to determine how many full-time and how many part-time workers should work each shift to meet the above requirements at the minimum possible cost. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-11.* The Medequip Company produces precision medical diagnostic equipment at two factories. Three medical centers have placed orders for this month’s production output. The table to the right shows what the cost would be for shipping each unit from each factory to each of these customers. Also shown are the number of units that will be produced at each factory and the number of units ordered by each customer. (Go to the next column.)

40 tons produced

$2,000/ton

M1

30 tons max.

Unit Shipping Cost To From

Customer 1 Customer 2 Customer 3

Factory 1 Factory 2

$600 $400

$800 $900

$700 $600

Order size

300 units

200 units

400 units

3.4-12. The Fagersta Steelworks currently is working two mines to obtain its iron ore. This iron ore is shipped to either of two storage facilities. When needed, it then is shipped on to the company’s steel plant. The diagram below depicts this distribution network, where M1 and M2 are the two mines, S1 and S2 are the two storage facilities, and P is the steel plant. The diagram also shows the monthly amounts produced at the mines and needed at the plant, as well as the shipping cost and the maximum amount that can be shipped per month through each shipping lane. (Go to the left column below the diagram.)

S1

$1

70

n /to x. a

m ns

to

00

,7

30

$40

0/to n sm ax.

ton

$1 ,6 to 00/t ns o m n ax . $1,100/ton 50 tons max.

Management now wants to determine the most economical plan for shipping the iron ore from the mines through the distribution network to the steel plant. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

100 tons needed

n 0/to $80 max. s ton 70

50 M2

400 units 500 units

A decision now needs to be made about the shipping plan for how many units to ship from each factory to each customer. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

P

60 tons produced

Output

S2

3.4-13.* Al Ferris has $60,000 that he wishes to invest now in order to use the accumulation for purchasing a retirement annuity in 5 years. After consulting with his financial adviser, he has been offered four types of fixed-income investments, which we will label as investments A, B, C, D.

CHAPTER 3 PROBLEMS

97

Investments A and B are available at the beginning of each of the next 5 years (call them years 1 to 5). Each dollar invested in A at the beginning of a year returns $1.40 (a profit of $0.40) 2 years later (in time for immediate reinvestment). Each dollar invested in B at the beginning of a year returns $1.70 three years later. Investments C and D will each be available at one time in the future. Each dollar invested in C at the beginning of year 2 returns $1.90 at the end of year 5. Each dollar invested in D at the beginning of year 5 returns $1.30 at the end of year 5. Al wishes to know which investment plan maximizes the amount of money that can be accumulated by the beginning of year 6. (a) All the functional constraints for this problem can be expressed as equality constraints. To do this, let At, Bt, Ct, and Dt be the amount invested in investment A, B, C, and D, respectively, at the beginning of year t for each t where the investment is available and will mature by the end of year 5. Also let Rt be the number of available dollars not invested at the beginning of year t (and so available for investment in a later year). Thus, the amount invested at the beginning of year t plus Rt must equal the number of dollars available for investment at that time. Write such an equation in terms of the relevant variables above for the beginning of each of the 5 years to obtain the five functional constraints for this problem. (b) Formulate a complete linear programming model for this problem. C (c) Solve this model by the simplex model. 3.4-14. The Metalco Company desires to blend a new alloy of 40 percent tin, 35 percent zinc, and 25 percent lead from several available alloys having the following properties: Alloy Property

1

2

3

4

5

Percentage of tin Percentage of zinc Percentage of lead

60 10 30

25 15 60

45 45 10

20 50 30

50 40 10

Cost ($/lb)

22

20

25

24

27

The objective is to determine the proportions of these alloys that should be blended to produce the new alloy at a minimum cost. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-15. The Weigelt Corporation has three branch plants with excess production capacity. Fortunately, the corporation has a new product ready to begin production, and all three plants have this capability, so some of the excess capacity can be used in this way. This product can be made in three sizes—large, medium, and

small—that yield a net unit profit of $420, $360, and $300, respectively. Plants 1, 2, and 3 have the excess capacity to produce 750, 900, and 450 units per day of this product, respectively, regardless of the size or combination of sizes involved. The amount of available in-process storage space also imposes a limitation on the production rates of the new product. Plants 1, 2, and 3 have 13,000, 12,000, and 5,000 square feet, respectively, of in-process storage space available for a day’s production of this product. Each unit of the large, medium, and small sizes produced per day requires 20, 15, and 12 square feet, respectively. Sales forecasts indicate that if available, 900, 1,200, and 750 units of the large, medium, and small sizes, respectively, would be sold per day. At each plant, some employees will need to be laid off unless most of the plant’s excess production capacity can be used to produce the new product. To avoid layoffs if possible, management has decided that the plants should use the same percentage of their excess capacity to produce the new product. Management wishes to know how much of each of the sizes should be produced by each of the plants to maximize profit. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-16* A cargo plane has three compartments for storing cargo: front, center, and back. These compartments have capacity limits on both weight and space, as summarized below:

Compartment Front Center Back

Weight Capacity (Tons)

Space Capacity (Cubic Feet)

12 18 10

7,000 9,000 5,000

Furthermore, the weight of the cargo in the respective compartments must be the same proportion of that compartment’s weight capacity to maintain the balance of the airplane. The following four cargoes have been offered for shipment on an upcoming flight as space is available:

Cargo

Weight (Tons)

Volume (Cubic Feet/Ton)

Profit ($/Ton)

1 2 3 4

20 16 25 13

500 700 600 400

320 400 360 290

Any portion of these cargoes can be accepted. The objective is to determine how much (if any) of each cargo should be accepted and

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how to distribute each among the compartments to maximize the total profit for the flight. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method to find one of its multiple optimal solutions. 3.4-17. Comfortable Hands is a company which features a product line of winter gloves for the entire family—men, women, and children. They are trying to decide what mix of these three types of gloves to produce. Comfortable Hands’ manufacturing labor force is unionized. Each full-time employee works a 40-hour week. In addition, by union contract, the number of full-time employees can never drop below 20. Nonunion part-time workers can also be hired with the following union-imposed restrictions: (1) each part-time worker works 20 hours per week, and (2) there must be at least 2 full-time employees for each part-time employee. All three types of gloves are made out of the same 100 percent genuine cowhide leather. Comfortable Hands has a long-term contract with a supplier of the leather, and receives a 5,000 square feet shipment of the material each week. The material requirements and labor requirements, along with the gross profit per glove sold (not considering labor costs) is given in the following table.

Glove Men’s Women’s Children’s

Material Required Labor Required Gross Profit (Square Feet) (Minutes) (per Pair) 2 1.5 1

30 45 40

$8 $10 $6

Maximum Hours of Availability Operators

Wage Rate

Mon.

Tue.

Wed.

Thurs.

Fri.

K. C. D. H. H. B. S. C. K. S. N. K.

$10.00/hour $10.10/hour $ 9.90/hour $ 9.80/hour $10.80/hour $11.30/hour

6 0 4 5 3 0

0 6 8 5 0 0

6 0 4 5 3 0

0 6 0 0 8 6

6 0 4 5 0 2

There are six operators (four undergraduate students and two graduate students). They all have different wage rates because of differences in their experience with computers and in their programming ability. The above table shows their wage rates, along with the maximum number of hours that each can work each day. Each operator is guaranteed a certain minimum number of hours per week that will maintain an adequate knowledge of the operation. This level is set arbitrarily at 8 hours per week for the undergraduate students (K. C., D. H., H. B., and S. C.) and 7 hours per week for the graduate students (K. S. and N. K.). The computer facility is to be open for operation from 8 A.M. to 10 P.M. Monday through Friday with exactly one operator on duty during these hours. On Saturdays and Sundays, the computer is to be operated by other staff. Because of a tight budget, Beryl has to minimize cost. She wishes to determine the number of hours she should assign to each operator on each day. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

Each full-time employee earns $13 per hour, while each parttime employee earns $10 per hour. Management wishes to know what mix of each of the three types of gloves to produce per week, as well as how many full-time and how many part-time workers to employ. They would like to maximize their net profit—their gross profit from sales minus their labor costs. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method.

3.4-19. Slim-Down Manufacturing makes a line of nutritionally complete, weight-reduction beverages. One of their products is a strawberry shake which is designed to be a complete meal. The strawberry shake consists of several ingredients. Some information about each of these ingredients is given below.

3.4-18. Oxbridge University maintains a powerful mainframe computer for research use by its faculty, Ph.D. students, and research associates. During all working hours, an operator must be available to operate and maintain the computer, as well as to perform some programming services. Beryl Ingram, the director of the computer facility, oversees the operation. It is now the beginning of the fall semester, and Beryl is confronted with the problem of assigning different working hours to her operators. Because all the operators are currently enrolled in the university, they are available to work only a limited number of hours each day, as shown in the following table.

Ingredient Strawberry flavoring Cream Vitamin supplement Artificial sweetener Thickening agent

Calories Total Vitamin from Fat Calories Content Thickeners Cost (per (per (mg/ (mg/ (¢/ tbsp) tbsp) tbsp) tbsp) tbsp)

1 75

50 100

20 0

3 8

10 8

0

0

50

1

25

0

120

0

2

15

30

80

2

25

6

CHAPTER 3 PROBLEMS

99

The nutritional requirements are as follows. The beverage must total between 380 and 420 calories (inclusive). No more than 20 percent of the total calories should come from fat. There must be at least 50 milligrams (mg) of vitamin content. For taste reasons, there must be at least 2 tablespoons (tbsp) of strawberry flavoring for each tablespoon of artificial sweetener. Finally, to maintain proper thickness, there must be exactly 15 mg of thickeners in the beverage. Management would like to select the quantity of each ingredient for the beverage which would minimize cost while meeting the above requirements. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.4-20. Joyce and Marvin run a day care for preschoolers. They are trying to decide what to feed the children for lunches. They would like to keep their costs down, but also need to meet the nutritional requirements of the children. They have already decided to go with peanut butter and jelly sandwiches, and some combination of graham crackers, milk, and orange juice. The nutritional content of each food choice and its cost are given in the table below.

Food Item Bread (1 slice) Peanut butter (1 tbsp) Strawberry jelly (1 tbsp) Graham cracker (1 cracker) Milk (1 cup) Juice (1 cup)

Calories Total Vitamin C Protein Cost from Fat Calories (mg) (g) (¢) 10

70

0

3

5

75

100

0

4

4

0

50

3

0

7

20 70 0

60 150 100

0 2 120

1 8 1

8 15 35

The nutritional requirements are as follows. Each child should receive between 400 and 600 calories. No more than 30 percent of the total calories should come from fat. Each child should consume at least 60 milligrams (mg) of vitamin C and 12 grams (g) of protein. Furthermore, for practical reasons, each child needs exactly 2 slices of bread (to make the sandwich), at least twice as much peanut butter as jelly, and at least 1 cup of liquid (milk and/or juice). Joyce and Marvin would like to select the food choices for each child which minimize cost while meeting the above requirements. (a) Formulate a linear programming model for this problem. C (b) Solve this model by the simplex method. 3.5-1. Read the article footnoted in Sec. 3.5 that describes the first case study presented in that section: “Choosing the Product Mix at Ponderosa Industrial.”

(a) Describe the two factors which, according to the article, often hinder the use of optimization models by managers. (b) Section 3.5 indicates without elaboration that using linear programming at Ponderosa “led to a dramatic shift in the types of plywood products emphasized by the company.” Identify this shift. (c) With the success of this application, management then was eager to use optimization for other problems as well. Identify these other problems. (d) Photocopy the two pages of appendixes that give the mathematical formulation of the problem and the structure of the linear programming model. 3.5-2. Read the article footnoted in Sec. 3.5 that describes the second case study presented in that section: “Personnel Scheduling at United Airlines.” (a) Describe how United Airlines prepared shift schedules at airports and reservations offices prior to this OR study. (b) When this study began, the problem definition phase defined five specific project requirements. Identify these project requirements. (c) At the end of the presentation of the corresponding example in Sec. 3.4 (personnel scheduling at Union Airways), we pointed out that the divisibility assumption does not hold for this kind of application. An integer solution is needed, but linear programming may provide an optimal solution that is noninteger. How does United Airlines deal with this problem? (d) Describe the flexibility built into the scheduling system to satisfy the group culture at each office. Why was this flexibility needed? (e) Briefly describe the tangible and intangible benefits that resulted from the study. 3.5-3. Read the 1986 article footnoted in Sec. 2.1 that describes the third case study presented in Sec. 3.5: “Planning Supply, Distribution, and Marketing at Citgo Petroleum Corporation.” (a) What happened during the years preceding this OR study that made it vastly more important to control the amount of capital tied up in inventory? (b) What geographical area is spanned by Citgo’s distribution network of pipelines, tankers, and barges? Where do they market their products? (c) What time periods are included in the model? (d) Which computer did Citgo use to solve the model? What were typical run times? (e) Who are the four types of model users? How does each one use the model? (f) List the major types of reports generated by the SDM system. (g) What were the major implementation challenges for this study? (h) List the direct and indirect benefits that were realized from this study.

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3.6-1.* You are given the following data for a linear programming problem where the objective is to maximize the profit from allocating three resources to two nonnegative activities. Resource Usage per Unit of Each Activity Resource

Activity 1

Activity 2

Amount of Resource Available

1 2 3

2 3 2

1 3 4

10 20 20

Contribution per unit

$20

$30

(c) Make three guesses of your own choosing for the optimal solution. Use the spreadsheet to check each one for feasibility and, if feasible, to find the value of the objective function. Which feasible guess has the best objective function value? (d) Use the Excel Solver to solve the model by the simplex method. 3.6-3. You are given the following data for a linear programming problem where the objective is to minimize the cost of conducting two nonnegative activities so as to achieve three benefits that do not fall below their minimum levels. Benefit Contribution per Unit of Each Activity Benefit

Contribution per unit profit per unit of the activity.

(a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. (c) Display the model on an Excel spreadsheet. (d) Use the spreadsheet to check the following solutions: (x1, x2) (2, 2), (3, 3), (2, 4), (4, 2), (3, 4), (4, 3). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? C (e) Use the Excel Solver to solve the model by the simplex method. 3.6-2. Ed Butler is the production manager for the Bilco Corporation, which produces three types of spare parts for automobiles. The manufacture of each part requires processing on each of two machines, with the following processing times (in hours): Part Machine

A

B

C

1 2

0.02 0.05

0.03 0.02

0.05 0.04

Each machine is available 40 hours per month. Each part manufactured will yield a unit profit as follows:

Activity 1

1 2 3 Unit cost

Activity 2

5 2 7

3 2 9

$60

$50

Minimum Acceptable Level 60 30 126

(a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. (c) Display the model on an Excel spreadsheet. (d) Use the spreadsheet to check the following solutions: (x1, x2) (7, 7), (7, 8), (8, 7), (8, 8), (8, 9), (9, 8). Which of these solutions are feasible? Which of these feasible solutions has the best value of the objective function? C (e) Use the Excel Solver to solve this model by the simplex method. 3.6-4.* Fred Jonasson manages a family-owned farm. To supplement several food products grown on the farm, Fred also raises pigs for market. He now wishes to determine the quantities of the available types of feed (corn, tankage, and alfalfa) that should be given to each pig. Since pigs will eat any mix of these feed types, the objective is to determine which mix will meet certain nutritional requirements at a minimum cost. The number of units of each type of basic nutritional ingredient contained within a kilogram of each feed type is given in the following table, along with the daily nutritional requirements and feed costs:

Part

Profit

A

B

C

$50

$40

$30

Ed wants to determine the mix of spare parts to produce in order to maximize total profit. (a) Formulate a linear programming model for this problem. (b) Display the model on an Excel spreadsheet.

Nutritional Ingredient

Kilogram Kilogram Kilogram Minimum of of of Daily Corn Tankage Alfalfa Requirement

Carbohydrates Protein Vitamins

90 30 10

20 80 20

40 60 60

Cost (¢)

84

72

60

200 180 150

CHAPTER 3 PROBLEMS

(a) Formulate a linear programming model for this problem. (b) Display the model on an Excel spreadsheet. (c) Use the spreadsheet to check if (x1, x2, x3) (1, 2, 2) is a feasible solution and, if so, what the daily cost would be for this diet. How many units of each nutritional ingredient would this diet provide daily? (d) Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the daily cost for your solution? C (e) Use the Excel Solver to solve the model by the simplex method. 3.6-5. Maureen Laird is the chief financial officer for the Alva Electric Co., a major public utility in the midwest. The company has scheduled the construction of new hydroelectric plants 5, 10, and 20 years from now to meet the needs of the growing population in the region served by the company. To cover at least the construction costs, Maureen needs to invest some of the company’s money now to meet these future cash-flow needs. Maureen may purchase only three kinds of financial assets, each of which costs $1 million per unit. Fractional units may be purchased. The assets produce income 5, 10, and 20 years from now, and that income is needed to cover at least minimum cash-flow requirements in those years. (Any excess income above the minimum requirement for each time period will be used to increase dividend payments to shareholders rather than saving it to help meet the minimum cash-flow requirement in the next time period.) The following table shows both the amount of income generated by each unit of each asset and the minimum amount of income needed for each of the future time periods when a new hydroelectric plant will be constructed.

101

(d) Take a few minutes to use a trial-and-error approach with the spreadsheet to develop your best guess for the optimal solution. What is the total amount invested for your solution? C (e) Use the Excel Solver to solve the model by the simplex method. 3.7-1. The Philbrick Company has two plants on opposite sides of the United States. Each of these plants produces the same two products and then sells them to wholesalers within its half of the country. The orders from wholesalers have already been received for the next 2 months (February and March), where the number of units requested are shown below. (The company is not obligated to completely fill these orders but will do so if it can without decreasing its profits.) Plant 1

Plant 2

Product

February

March

February

March

1 2

3,600 4,500

6,300 5,400

4,900 5,100

4,200 6,000

Each plant has 20 production days available in February and 23 production days available in March to produce and ship these products. Inventories are depleted at the end of January, but each plant has enough inventory capacity to hold 1,000 units total of the two products if an excess amount is produced in February for sale in March. In either plant, the cost of holding inventory in this way is $3 per unit of product 1 and $4 per unit of product 2. Each plant has the same two production processes, each of which can be used to produce either of the two products. The production cost per unit produced of each product is shown below for each process in each plant.

Income per Unit of Asset Year 5 10 20

Asset 1

Asset 2

Asset 3

$2 million $1 million $0.5 million $0.5 million $0.5 million $1 million 0 $1.5 million $2 million

Minimum Cash Flow Required $400 million $100 million $300 million

Maureen wishes to determine the mix of investments in these assets that will cover the cash-flow requirements while minimizing the total amount invested. (a) Formulate a linear programming model for this problem. (b) Display the model on a spreadsheet. (c) Use the spreadsheet to check the possibility of purchasing 100 units of Asset 1, 100 units of Asset 2, and 200 units of Asset 3. How much cash flow would this mix of investments generate 5, 10, and 20 years from now? What would be the total amount invested?

Plant 1

Plant 2

Product

Process 1

Process 2

Process 1

Process 2

1 2

$62 $78

$59 $85

$61 $89

$65 $86

The production rate for each product (number of units produced per day devoted to that product) also is given below for each process in each plant. Plant 1

Plant 2

Product

Process 1

Process 2

Process 1

Process 2

1 2

100 120

140 150

130 160

110 130

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The net sales revenue (selling price minus normal shipping costs) the company receives when a plant sells the products to its own customers (the wholesalers in its half of the country) is $83 per unit of product 1 and $112 per unit of product 2. However, it also is possible (and occasionally desirable) for a plant to make a shipment to the other half of the country to help fill the sales of the other plant. When this happens, an extra shipping cost of $9 per unit of product 1 and $7 per unit of product 2 is incurred. Management now needs to determine how much of each product should be produced by each production process in each plant during each month, as well as how much each plant should sell of each product in each month and how much each plant should ship of each product in each month to the other plant’s customers. The objective is to determine which feasible plan would maximize the total profit (total net sales revenue minus the sum of the production costs, inventory costs, and extra shipping costs). (a) Formulate a complete linear programming model in algebraic form that shows the individual constraints and decision variables for this problem. C (b) Formulate this same model on an Excel spreadsheet instead. Then use the Excel Solver to solve the model. C (c) Use MPL to formulate this model in a compact form. Then use the MPL solver CPLEX to solve the model. C (d) Use LINGO to formulate this model in a compact form. Then use the LINGO solver to solve the model. 3.7-2. Reconsider Prob. 3.1-11. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-3. Reconsider Prob. 3.4-11. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-4. Reconsider Prob. 3.4-15. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-5. Reconsider Prob. 3.4-18. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-6. Reconsider Prob. 3.6-4. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-7. Reconsider Prob. 3.6-5. (a) Use MPL/CPLEX to formulate and solve the model for this problem. (b) Use LINGO to formulate and solve this model.

C

3.7-8. A large paper manufacturing company, the Quality Paper Corporation, has 10 paper mills from which it needs to supply 1,000 customers. It uses three alternative types of machines and four types of raw materials to make five different types of paper. Therefore, the company needs to develop a detailed production distribution plan on a monthly basis, with an objective of minimizing the total cost of producing and distributing the paper during the month. Specifically, it is necessary to determine jointly the amount of each type of paper to be made at each paper mill on each type of machine and the amount of each type of paper to be shipped from each paper mill to each customer. The relevant data can be expressed symbolically as follows: Djk number of units of paper type k demanded by customer j, rklm number of units of raw material m needed to produce 1 unit of paper type k on machine type l, Rim number of units of raw material m available at paper mill i, ckl number of capacity units of machine type l that will produce 1 unit of paper type k, Cil number of capacity units of machine type l available at paper mill i, Pikl production cost for each unit of paper type k produced on machine type l at paper mill i, Tijk transportation cost for each unit of paper type k shipped from paper mill i to customer j. (a) Using these symbols, formulate a linear programming model for this problem by hand. (b) How many functional constraints and decision variables does this model have? C (c) Use MPL to formulate this problem. C (d) Use LINGO to formulate this problem.

CASE 3.1

CASE 3.1

AUTO ASSEMBLY

103

AUTO ASSEMBLY Automobile Alliance, a large automobile manufacturing company, organizes the vehicles it manufactures into three families: a family of trucks, a family of small cars, and a family of midsized and luxury cars. One plant outside Detroit, MI, assembles two models from the family of midsized and luxury cars. The first model, the Family Thrillseeker, is a four-door sedan with vinyl seats, plastic interior, standard features, and excellent gas mileage. It is marketed as a smart buy for middle-class families with tight budgets, and each Family Thrillseeker sold generates a modest profit of $3,600 for the company. The second model, the Classy Cruiser, is a two-door luxury sedan with leather seats, wooden interior, custom features, and navigational capabilities. It is marketed as a privilege of affluence for upper-middle-class families, and each Classy Cruiser sold generates a healthy profit of $5,400 for the company. Rachel Rosencrantz, the manager of the assembly plant, is currently deciding the production schedule for the next month. Specifically, she must decide how many Family Thrillseekers and how many Classy Cruisers to assemble in the plant to maximize profit for the company. She knows that the plant possesses a capacity of 48,000 laborhours during the month. She also knows that it takes 6 labor-hours to assemble one Family Thrillseeker and 10.5 labor-hours to assemble one Classy Cruiser. Because the plant is simply an assembly plant, the parts required to assemble the two models are not produced at the plant. They are instead shipped from other plants around the Michigan area to the assembly plant. For example, tires, steering wheels, windows, seats, and doors all arrive from various supplier plants. For the next month, Rachel knows that she will be able to obtain only 20,000 doors (10,000 left-hand doors and 10,000 right-hand doors) from the door supplier. A recent labor strike forced the shutdown of that particular supplier plant for several days, and that plant will not be able to meet its production schedule for the next month. Both the Family Thrillseeker and the Classy Cruiser use the same door part. In addition, a recent company forecast of the monthly demands for different automobile models suggests that the demand for the Classy Cruiser is limited to 3,500 cars. There is no limit on the demand for the Family Thrillseeker within the capacity limits of the assembly plant. (a) Formulate and solve a linear programming problem to determine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled.

Before she makes her final production decisions, Rachel plans to explore the following questions independently except where otherwise indicated. (b) The marketing department knows that it can pursue a targeted $500,000 advertising campaign that will raise the demand for the Classy Cruiser next month by 20 percent. Should the campaign be undertaken? (c) Rachel knows that she can increase next month’s plant capacity by using overtime labor. She can increase the plant’s labor-hour capacity by 25 percent. With the new assembly plant capacity, how many Family Thrillseekers and how many Classy Cruisers should be assembled? (d) Rachel knows that overtime labor does not come without an extra cost. What is the maximum amount she should be willing to pay for all overtime labor beyond the cost of this labor at regular time rates? Express your answer as a lump sum.

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(e) Rachel explores the option of using both the targeted advertising campaign and the overtime labor-hours. The advertising campaign raises the demand for the Classy Cruiser by 20 percent, and the overtime labor increases the plant’s labor-hour capacity by 25 percent. How many Family Thrillseekers and how many Classy Cruisers should be assembled using the advertising campaign and overtime labor-hours if the profit from each Classy Cruiser sold continues to be 50 percent more than for each Family Thrillseeker sold? (f) Knowing that the advertising campaign costs $500,000 and the maximum usage of overtime labor-hours costs $1,600,000 beyond regular time rates, is the solution found in part (e) a wise decision compared to the solution found in part (a)? (g) Automobile Alliance has determined that dealerships are actually heavily discounting the price of the Family Thrillseekers to move them off the lot. Because of a profit-sharing agreement with its dealers, the company is therefore not making a profit of $3,600 on the Family Thrillseeker but is instead making a profit of $2,800. Determine the number of Family Thrillseekers and the number of Classy Cruisers that should be assembled given this new discounted price. (h) The company has discovered quality problems with the Family Thrillseeker by randomly testing Thrillseekers at the end of the assembly line. Inspectors have discovered that in over 60 percent of the cases, two of the four doors on a Thrillseeker do not seal properly. Because the percentage of defective Thrillseekers determined by the random testing is so high, the floor supervisor has decided to perform quality control tests on every Thrillseeker at the end of the line. Because of the added tests, the time it takes to assemble one Family Thrillseeker has increased from 6 to 7.5 hours. Determine the number of units of each model that should be assembled given the new assembly time for the Family Thrillseeker. (i) The board of directors of Automobile Alliance wishes to capture a larger share of the luxury sedan market and therefore would like to meet the full demand for Classy Cruisers. They ask Rachel to determine by how much the profit of her assembly plant would decrease as compared to the profit found in part (a). They then ask her to meet the full demand for Classy Cruisers if the decrease in profit is not more than $2,000,000. (j) Rachel now makes her final decision by combining all the new considerations described in parts ( f ), (g), and (h). What are her final decisions on whether to undertake the advertising campaign, whether to use overtime labor, the number of Family Thrillseekers to assemble, and the number of Classy Cruisers to assemble?

CASE 3.2

CUTTING CAFETERIA COSTS A cafeteria at All-State University has one special dish it serves like clockwork every Thursday at noon. This supposedly tasty dish is a casserole that contains sautéed onions, boiled sliced potatoes, green beans, and cream of mushroom soup. Unfortunately, students fail to see the special quality of this dish, and they loathingly refer to it as the Killer Casserole. The students reluctantly eat the casserole, however, because the cafeteria provides only a limited selection of dishes for Thursday’s lunch (namely, the casserole). Maria Gonzalez, the cafeteria manager, is looking to cut costs for the coming year, and she believes that one sure way to cut costs is to buy less expensive and perhaps lower-quality ingredients. Because the casserole is a weekly staple of the cafeteria menu, she concludes that if she can cut costs on the ingredients purchased for the casserole, she can significantly reduce overall cafeteria operating costs. She therefore de-

CASE 3.2

CUTTING CAFETERIA COSTS

105

cides to invest time in determining how to minimize the costs of the casserole while maintaining nutritional and taste requirements. Maria focuses on reducing the costs of the two main ingredients in the casserole, the potatoes and green beans. These two ingredients are responsible for the greatest costs, nutritional content, and taste of the dish. Maria buys the potatoes and green beans from a wholesaler each week. Potatoes cost $0.40 per pound, and green beans cost $1.00 per pound. All-State University has established nutritional requirements that each main dish of the cafeteria must meet. Specifically, the total amount of the dish prepared for all the students for one meal must contain 180 grams (g) of protein, 80 milligrams (mg) of iron, and 1,050 mg of vitamin C. (There are 453.6 g in 1 lb and 1,000 mg in 1 g.) For simplicity when planning, Maria assumes that only the potatoes and green beans contribute to the nutritional content of the casserole. Because Maria works at a cutting-edge technological university, she has been exposed to the numerous resources on the World Wide Web. She decides to surf the Web to find the nutritional content of potatoes and green beans. Her research yields the following nutritional information about the two ingredients:

Protein Iron Vitamin C

Potatoes

Green Beans

1.5 g per 100 g 0.3 mg per 100 g 12 mg per 100 g

5.67 g per 10 ounces 3.402 mg per 10 ounces 28.35 mg per 10 ounces

(There are 28.35 g in 1 ounce.)

Edson Branner, the cafeteria cook who is surprisingly concerned about taste, informs Maria that an edible casserole must contain at least a six to five ratio in the weight of potatoes to green beans. Given the number of students who eat in the cafeteria, Maria knows that she must purchase enough potatoes and green beans to prepare a minimum of 10 kilograms (kg) of casserole each week. (There are 1,000 g in 1 kg.) Again for simplicity in planning, she assumes that only the potatoes and green beans determine the amount of casserole that can be prepared. Maria does not establish an upper limit on the amount of casserole to prepare, since she knows all leftovers can be served for many days thereafter or can be used creatively in preparing other dishes. (a) Determine the amount of potatoes and green beans Maria should purchase each week for the casserole to minimize the ingredient costs while meeting nutritional, taste, and demand requirements.

Before she makes her final decision, Maria plans to explore the following questions independently except where otherwise indicated. (b) Maria is not very concerned about the taste of the casserole; she is only concerned about meeting nutritional requirements and cutting costs. She therefore forces Edson to change the recipe to allow for only at least a one to two ratio in the weight of potatoes to green beans. Given the new recipe, determine the amount of potatoes and green beans Maria should purchase each week.

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(c) Maria decides to lower the iron requirement to 65 mg since she determines that the other ingredients, such as the onions and cream of mushroom soup, also provide iron. Determine the amount of potatoes and green beans Maria should purchase each week given this new iron requirement. (d) Maria learns that the wholesaler has a surplus of green beans and is therefore selling the green beans for a lower price of $0.50 per lb. Using the same iron requirement from part (c) and the new price of green beans, determine the amount of potatoes and green beans Maria should purchase each week. (e) Maria decides that she wants to purchase lima beans instead of green beans since lima beans are less expensive and provide a greater amount of protein and iron than green beans. Maria again wields her absolute power and forces Edson to change the recipe to include lima beans instead of green beans. Maria knows she can purchase lima beans for $0.60 per lb from the wholesaler. She also knows that lima beans contain 22.68 g of protein per 10 ounces of lima beans, 6.804 mg of iron per 10 ounces of lima beans, and no vitamin C. Using the new cost and nutritional content of lima beans, determine the amount of potatoes and lima beans Maria should purchase each week to minimize the ingredient costs while meeting nutritional, taste, and demand requirements. The nutritional requirements include the reduced iron requirement from part (c). (f) Will Edson be happy with the solution in part (e)? Why or why not? (g) An All-State student task force meets during Body Awareness Week and determines that AllState University’s nutritional requirements for iron are too lax and that those for vitamin C are too stringent. The task force urges the university to adopt a policy that requires each serving of an entrée to contain at least 120 mg of iron and at least 500 mg of vitamin C. Using potatoes and lima beans as the ingredients for the dish and using the new nutritional requirements, determine the amount of potatoes and lima beans Maria should purchase each week.

CASE 3.3

STAFFING A CALL CENTER1 California Children’s Hospital has been receiving numerous customer complaints because of its confusing, decentralized appointment and registration process. When customers want to make appointments or register child patients, they must contact the clinic or department they plan to visit. Several problems exist with this current strategy. Parents do not always know the most appropriate clinic or department they must visit to address their children’s ailments. They therefore spend a significant amount of time on the phone being transferred from clinic to clinic until they reach the most appropriate clinic for their needs. The hospital also does not publish the phone numbers of all clinic and departments, and parents must therefore invest a large amount of time in detective work to track down the correct phone number. Finally, the various clinics and departments do not communicate with each other. For example, when a doctor schedules a referral with a colleague located in another department or clinic, that department or clinic almost never receives word of the referral. The parent must contact the correct department or clinic and provide the needed referral information. 1

This case is based on an actual project completed by a team of master’s students in the Department of Engineering-Economic Systems and Operations Research at Stanford University.

CASE 3.3

STAFFING A CALL CENTER

107

In efforts to reengineer and improve its appointment and registration process, the children’s hospital has decided to centralize the process by establishing one call center devoted exclusively to appointments and registration. The hospital is currently in the middle of the planning stages for the call center. Lenny Davis, the hospital manager, plans to operate the call center from 7 A.M. to 9 P.M. during the weekdays. Several months ago, the hospital hired an ambitious management consulting firm, Creative Chaos Consultants, to forecast the number of calls the call center would receive each hour of the day. Since all appointment and registration-related calls would be received by the call center, the consultants decided that they could forecast the calls at the call center by totaling the number of appointment and registration-related calls received by all clinics and departments. The team members visited all the clinics and departments, where they diligently recorded every call relating to appointments and registration. They then totaled these calls and altered the totals to account for calls missed during data collection. They also altered totals to account for repeat calls that occurred when the same parent called the hospital many times because of the confusion surrounding the decentralized process. Creative Chaos Consultants determined the average number of calls the call center should expect during each hour of a weekday. The following table provides the forecasts. Work Shift 7 9 11 1 3 5 7

A.M.–9 A.M. A.M.–11 A.M. A.M.–1 P.M. P.M.–3 P.M. P.M.–5 P.M. P.M.–7 P.M. P.M.–9 P.M.

Average Number of Calls 40 85 70 95 80 35 10

calls calls calls calls calls calls calls

per per per per per per per

hour hour hour hour hour hour hour

After the consultants submitted these forecasts, Lenny became interested in the percentage of calls from Spanish speakers since the hospital services many Spanish patients. Lenny knows that he has to hire some operators who speak Spanish to handle these calls. The consultants performed further data collection and determined that on average, 20 percent of the calls were from Spanish speakers. Given these call forecasts, Lenny must now decide how to staff the call center during each 2 hour shift of a weekday. During the forecasting project, Creative Chaos Consultants closely observed the operators working at the individual clinics and departments and determined the number of calls operators process per hour. The consultants informed Lenny that an operator is able to process an average of six calls per hour. Lenny also knows that he has both full-time and part-time workers available to staff the call center. A full-time employee works 8 hours per day, but because of paperwork that must also be completed, the employee spends only 4 hours per day on the phone. To balance the schedule, the employee alternates the 2-hour shifts between answering phones and completing paperwork. Full-time employees can start their day either by answering phones or by completing paperwork on the first shift. The full-time em-

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ployees speak either Spanish or English, but none of them are bilingual. Both Spanish-speaking and English-speaking employees are paid $10 per hour for work before 5 P.M. and $12 per hour for work after 5 P.M. The full-time employees can begin work at the beginning of the 7 A.M. to 9 A.M. shift, 9 A.M. to 11 A.M. shift, 11 A.M. to 1 P.M. shift, or 1 P.M. to 3 P.M. shift. The part-time employees work for 4 hours, only answer calls, and only speak English. They can start work at the beginning of the 3 P.M. to 5 P.M. shift or the 5 P.M. to 7 P.M. shift, and like the full-time employees, they are paid $10 per hour for work before 5 P.M. and $12 per hour for work after 5 P.M. For the following analysis consider only the labor cost for the time employees spend answering phones. The cost for paperwork time is charged to other cost centers. (a) How many Spanish-speaking operators and how many English-speaking operators does the hospital need to staff the call center during each 2-hour shift of the day in order to answer all calls? Please provide an integer number since half a human operator makes no sense. (b) Lenny needs to determine how many full-time employees who speak Spanish, full-time employees who speak English, and part-time employees he should hire to begin on each shift. Creative Chaos Consultants advise him that linear programming can be used to do this in such a way as to minimize operating costs while answering all calls. Formulate a linear programming model of this problem. (c) Obtain an optimal solution for the linear programming model formulated in part (b) to guide Lenny’s decision. (d) Because many full-time workers do not want to work late into the evening, Lenny can find only one qualified English-speaking operator willing to begin work at 1 P.M. Given this new constraint, how many full-time English-speaking operators, full-time Spanish-speaking operators, and part-time operators should Lenny hire for each shift to minimize operating costs while answering all calls? (e) Lenny now has decided to investigate the option of hiring bilingual operators instead of monolingual operators. If all the operators are bilingual, how many operators should be working during each 2-hour shift to answer all phone calls? As in part (a), please provide an integer answer. (f) If all employees are bilingual, how many full-time and part-time employees should Lenny hire to begin on each shift to minimize operating costs while answering all calls? As in part (b), formulate a linear programming model to guide Lenny’s decision. (g) What is the maximum percentage increase in the hourly wage rate that Lenny can pay bilingual employees over monolingual employees without increasing the total operating costs? (h) What other features of the call center should Lenny explore to improve service or minimize operating costs?

4 Solving Linear Programming Problems: The Simplex Method We now are ready to begin studying the simplex method, a general procedure for solving linear programming problems. Developed by George Dantzig in 1947, it has proved to be a remarkably efficient method that is used routinely to solve huge problems on today’s computers. Except for its use on tiny problems, this method is always executed on a computer, and sophisticated software packages are widely available. Extensions and variations of the simplex method also are used to perform postoptimality analysis (including sensitivity analysis) on the model. This chapter describes and illustrates the main features of the simplex method. The first section introduces its general nature, including its geometric interpretation. The following three sections then develop the procedure for solving any linear programming model that is in our standard form (maximization, all functional constraints in form, and nonnegativity constraints on all variables) and has only nonnegative right-hand sides bi in the functional constraints. Certain details on resolving ties are deferred to Sec. 4.5, and Sec. 4.6 describes how to adapt the simplex method to other model forms. Next we discuss postoptimality analysis (Sec. 4.7), and describe the computer implementation of the simplex method (Sec. 4.8). Section 4.9 then introduces an alternative to the simplex method (the interior-point approach) for solving large linear programming problems.

4.1

THE ESSENCE OF THE SIMPLEX METHOD The simplex method is an algebraic procedure. However, its underlying concepts are geometric. Understanding these geometric concepts provides a strong intuitive feeling for how the simplex method operates and what makes it so efficient. Therefore, before delving into algebraic details, we focus in this section on the big picture from a geometric viewpoint. To illustrate the general geometric concepts, we shall use the Wyndor Glass Co. example presented in Sec. 3.1. (Sections 4.2 and 4.3 use the algebra of the simplex method to solve this same example.) Section 5.1 will elaborate further on these geometric concepts for larger problems. To refresh your memory, the model and graph for this example are repeated in Fig. 4.1. The five constraint boundaries and their points of intersection are highlighted in this figure because they are the keys to the analysis. Here, each constraint boundary is a line that forms the boundary of what is permitted by the corresponding constraint. The points 109

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x2

Maximize Z 3x1 5x2, subject to 4 x1 2x2 12 3x1 2x2 18 and x1 0, x2 0

x1 0 (0, 9) 3x1 2x2 18

(0, 6)

(2, 6)

(4, 6)

2x2 12

x1 4

Feasible region

FIGURE 4.1 Constraint boundaries and corner-point solutions for the Wyndor Glass Co. problem.

(4, 3)

x2 0

(0, 0) (4, 0)

(6, 0)

x1

of intersection are the corner-point solutions of the problem. The five that lie on the corners of the feasible region—(0, 0), (0, 6), (2, 6), (4, 3), and (4, 0)—are the corner-point feasible solutions (CPF solutions). [The other three—(0, 9), (4, 6), and (6, 0)—are called corner-point infeasible solutions.] In this example, each corner-point solution lies at the intersection of two constraint boundaries. (For a linear programming problem with n decision variables, each of its corner-point solutions lies at the intersection of n constraint boundaries.1) Certain pairs of the CPF solutions in Fig. 4.1 share a constraint boundary, and other pairs do not. It will be important to distinguish between these cases by using the following general definitions. For any linear programming problem with n decision variables, two CPF solutions are adjacent to each other if they share n 1 constraint boundaries. The two adjacent CPF solutions are connected by a line segment that lies on these same shared constraint boundaries. Such a line segment is referred to as an edge of the feasible region.

Since n 2 in the example, two of its CPF solutions are adjacent if they share one constraint boundary; for example, (0, 0) and (0, 6) are adjacent because they share the x1 0 constraint boundary. The feasible region in Fig. 4.1 has five edges, consisting of the five line segments forming the boundary of this region. Note that two edges emanate from each CPF solution. Thus, each CPF solution has two adjacent CPF solutions (each lying at the other end of one of the two edges), as enumerated in Table 4.1. (In each row 1

Although a corner-point solution is defined in terms of n constraint boundaries whose intersection gives this solution, it also is possible that one or more additional constraint boundaries pass through this same point.

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TABLE 4.1 Adjacent CPF solutions for each CPF solution of the Wyndor Glass Co. problem CPF Solution (0, (0, (2, (4, (4,

0) 6) 6) 3) 0)

Its Adjacent CPF Solutions (0, (2, (4, (4, (0,

6) 6) 3) 0) 0)

and and and and and

(4, (0, (0, (2, (4,

0) 0) 6) 6) 3)

of this table, the CPF solution in the first column is adjacent to each of the two CPF solutions in the second column, but the two CPF solutions in the second column are not adjacent to each other.) One reason for our interest in adjacent CPF solutions is the following general property about such solutions, which provides a very useful way of checking whether a CPF solution is an optimal solution. Optimality test: Consider any linear programming problem that possesses at least one optimal solution. If a CPF solution has no adjacent CPF solutions that are better (as measured by Z), then it must be an optimal solution. Thus, for the example, (2, 6) must be optimal simply because its Z 36 is larger than Z 30 for (0, 6) and Z 27 for (4, 3). (We will delve further into why this property holds in Sec. 5.1.) This optimality test is the one used by the simplex method for determining when an optimal solution has been reached. Now we are ready to apply the simplex method to the example. Solving the Example Here is an outline of what the simplex method does (from a geometric viewpoint) to solve the Wyndor Glass Co. problem. At each step, first the conclusion is stated and then the reason is given in parentheses. (Refer to Fig. 4.1 for a visualization.) Initialization: Choose (0, 0) as the initial CPF solution to examine. (This is a convenient choice because no calculations are required to identify this CPF solution.) Optimality Test: Conclude that (0, 0) is not an optimal solution. (Adjacent CPF solutions are better.) Iteration 1: Move to a better adjacent CPF solution, (0, 6), by performing the following three steps. 1. Considering the two edges of the feasible region that emanate from (0, 0), choose to move along the edge that leads up the x2 axis. (With an objective function of Z 3x1 5x2, moving up the x2 axis increases Z at a faster rate than moving along the x1 axis.) 2. Stop at the first new constraint boundary: 2x2 12. [Moving farther in the direction selected in step 1 leaves the feasible region; e.g., moving to the second new constraint boundary hit when moving in that direction gives (0, 9), which is a corner-point infeasible solution.]

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3. Solve for the intersection of the new set of constraint boundaries: (0, 6). (The equations for these constraint boundaries, x1 0 and 2x2 12, immediately yield this solution.) Optimality Test: Conclude that (0, 6) is not an optimal solution. (An adjacent CPF solution is better.) Iteration 2: Move to a better adjacent CPF solution, (2, 6), by performing the following three steps. 1. Considering the two edges of the feasible region that emanate from (0, 6), choose to move along the edge that leads to the right. (Moving along this edge increases Z, whereas backtracking to move back down the x2 axis decreases Z.) 2. Stop at the first new constraint boundary encountered when moving in that direction: 3x1 2x2 12. (Moving farther in the direction selected in step 1 leaves the feasible region.) 3. Solve for the intersection of the new set of constraint boundaries: (2, 6). (The equations for these constraint boundaries, 3x1 2x2 18 and 2x2 12, immediately yield this solution.) Optimality Test: Conclude that (2, 6) is an optimal solution, so stop. (None of the adjacent CPF solutions are better.) This sequence of CPF solutions examined is shown in Fig. 4.2, where each circled number identifies which iteration obtained that solution. Now let us look at the six key solution concepts of the simplex method that provide the rationale behind the above steps. (Keep in mind that these concepts also apply for solving problems with more than two decision variables where a graph like Fig. 4.2 is not available to help quickly find an optimal solution.) The Key Solution Concepts The first solution concept is based directly on the relationship between optimal solutions and CPF solutions given at the end of Sec. 3.2. FIGURE 4.2 This graph shows the sequence of CPF solutions (, , ) examined by the simplex method for the Wyndor Glass Co. problem. The optimal solution (2, 6) is found after just three solutions are examined.

x2 (0, 6)

Z 30 (2, 6) Z 36 1 2

Feasible region

(4, 3) Z 27

Z 12

0 (0, 0)

Z0

(4, 0)

x1

4.1 THE ESSENCE OF THE SIMPLEX METHOD

113

Solution concept 1: The simplex method focuses solely on CPF solutions. For any problem with at least one optimal solution, finding one requires only finding a best CPF solution.1 Since the number of feasible solutions generally is infinite, reducing the number of solutions that need to be examined to a small finite number ( just three in Fig. 4.2) is a tremendous simplification. The next solution concept defines the flow of the simplex method. Solution concept 2: The simplex method is an iterative algorithm (a systematic solution procedure that keeps repeating a fixed series of steps, called an iteration, until a desired result has been obtained) with the following structure. →

→ Initialization: Optimality test: → If no If yes ↓ Iteration:

Set up to start iterations, including finding an initial CPF solution. Is the current CPF solution optimal? Stop. Perform an iteration to find a better CPF solution.

When the example was solved, note how this flow diagram was followed through two iterations until an optimal solution was found. We next focus on how to get started. Solution concept 3: Whenever possible, the initialization of the simplex method chooses the origin (all decision variables equal to zero) to be the initial CPF solution. When there are too many decision variables to find an initial CPF solution graphically, this choice eliminates the need to use algebraic procedures to find and solve for an initial CPF solution. Choosing the origin commonly is possible when all the decision variables have nonnegativity constraints, because the intersection of these constraint boundaries yields the origin as a corner-point solution. This solution then is a CPF solution unless it is infeasible because it violates one or more of the functional constraints. If it is infeasible, special procedures described in Sec. 4.6 are needed to find the initial CPF solution. The next solution concept concerns the choice of a better CPF solution at each iteration. Solution concept 4: Given a CPF solution, it is much quicker computationally to gather information about its adjacent CPF solutions than about other CPF solutions. Therefore, each time the simplex method performs an iteration to move from the current CPF solution to a better one, it always chooses a CPF solution that is adjacent to the current one. No other CPF solutions are considered. Consequently, the entire path followed to eventually reach an optimal solution is along the edges of the feasible region. 1

The only restriction is that the problem must possess CPF solutions. This is ensured if the feasible region is bounded.

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The next focus is on which adjacent CPF solution to choose at each iteration. Solution concept 5: After the current CPF solution is identified, the simplex method examines each of the edges of the feasible region that emanate from this CPF solution. Each of these edges leads to an adjacent CPF solution at the other end, but the simplex method does not even take the time to solve for the adjacent CPF solution. Instead, it simply identifies the rate of improvement in Z that would be obtained by moving along the edge. Among the edges with a positive rate of improvement in Z, it then chooses to move along the one with the largest rate of improvement in Z. The iteration is completed by first solving for the adjacent CPF solution at the other end of this one edge and then relabeling this adjacent CPF solution as the current CPF solution for the optimality test and (if needed) the next iteration. At the first iteration of the example, moving from (0, 0) along the edge on the x1 axis would give a rate of improvement in Z of 3 (Z increases by 3 per unit increase in x1), whereas moving along the edge on the x2 axis would give a rate of improvement in Z of 5 (Z increases by 5 per unit increase in x2), so the decision is made to move along the latter edge. At the second iteration, the only edge emanating from (0, 6) that would yield a positive rate of improvement in Z is the edge leading to (2, 6), so the decision is made to move next along this edge. The final solution concept clarifies how the optimality test is performed efficiently. Solution concept 6: Solution concept 5 describes how the simplex method examines each of the edges of the feasible region that emanate from the current CPF solution. This examination of an edge leads to quickly identifying the rate of improvement in Z that would be obtained by moving along the edge toward the adjacent CPF solution at the other end. A positive rate of improvement in Z implies that the adjacent CPF solution is better than the current CPF solution, whereas a negative rate of improvement in Z implies that the adjacent CPF solution is worse. Therefore, the optimality test consists simply of checking whether any of the edges give a positive rate of improvement in Z. If none do, then the current CPF solution is optimal. In the example, moving along either edge from (2, 6) decreases Z. Since we want to maximize Z, this fact immediately gives the conclusion that (2, 6) is optimal.

4.2

SETTING UP THE SIMPLEX METHOD The preceding section stressed the geometric concepts that underlie the simplex method. However, this algorithm normally is run on a computer, which can follow only algebraic instructions. Therefore, it is necessary to translate the conceptually geometric procedure just described into a usable algebraic procedure. In this section, we introduce the algebraic language of the simplex method and relate it to the concepts of the preceding section. The algebraic procedure is based on solving systems of equations. Therefore, the first step in setting up the simplex method is to convert the functional inequality constraints to equivalent equality constraints. (The nonnegativity constraints are left as inequalities because they are treated separately.) This conversion is accomplished by introducing slack

4.2 SETTING UP THE SIMPLEX METHOD

115

variables. To illustrate, consider the first functional constraint in the Wyndor Glass Co. example of Sec. 3.1 x1 4. The slack variable for this constraint is defined to be x3 4 x1, which is the amount of slack in the left-hand side of the inequality. Thus, x1 x3 4. Given this equation, x1 4 if and only if 4 x1 x3 0. Therefore, the original constraint x1 4 is entirely equivalent to the pair of constraints x1 x3 4

and

x3 0.

Upon the introduction of slack variables for the other functional constraints, the original linear programming model for the example (shown below on the left) can now be replaced by the equivalent model (called the augmented form of the model) shown below on the right: Augmented Form of the Model1

Original Form of the Model Maximize

Z 3x1 5x2,

subject to

Z 3x1 5x2,

Maximize subject to

x1 2x2 4

(1)

3x1 2x2 12

(2)

2x2

3x1 2x2 18

(3)

3x1 2x2

and

x3

x1

4 x4

12 x5 18

and x1 0,

x2 0.

xj 0,

for j 1, 2, 3, 4, 5.

Although both forms of the model represent exactly the same problem, the new form is much more convenient for algebraic manipulation and for identification of CPF solutions. We call this the augmented form of the problem because the original form has been augmented by some supplementary variables needed to apply the simplex method. If a slack variable equals 0 in the current solution, then this solution lies on the constraint boundary for the corresponding functional constraint. A value greater than 0 means that the solution lies on the feasible side of this constraint boundary, whereas a value less than 0 means that the solution lies on the infeasible side of this constraint boundary. A demonstration of these properties is provided by the demonstration example in your OR Tutor entitled Interpretation of the Slack Variables. The terminology used in the preceding section (corner-point solutions, etc.) applies to the original form of the problem. We now introduce the corresponding terminology for the augmented form. An augmented solution is a solution for the original variables (the decision variables) that has been augmented by the corresponding values of the slack variables. 1

The slack variables are not shown in the objective function because the coefficients there are 0.

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For example, augmenting the solution (3, 2) in the example yields the augmented solution (3, 2, 1, 8, 5) because the corresponding values of the slack variables are x3 1, x4 8, and x5 5. A basic solution is an augmented corner-point solution. To illustrate, consider the corner-point infeasible solution (4, 6) in Fig. 4.1. Augmenting it with the resulting values of the slack variables x3 0, x4 0, and x5 6 yields the corresponding basic solution (4, 6, 0, 0, 6). The fact that corner-point solutions (and so basic solutions) can be either feasible or infeasible implies the following definition: A basic feasible (BF) solution is an augmented CPF solution. Thus, the CPF solution (0, 6) in the example is equivalent to the BF solution (0, 6, 4, 0, 6) for the problem in augmented form. The only difference between basic solutions and corner-point solutions (or between BF solutions and CPF solutions) is whether the values of the slack variables are included. For any basic solution, the corresponding corner-point solution is obtained simply by deleting the slack variables. Therefore, the geometric and algebraic relationships between these two solutions are very close, as described in Sec. 5.1. Because the terms basic solution and basic feasible solution are very important parts of the standard vocabulary of linear programming, we now need to clarify their algebraic properties. For the augmented form of the example, notice that the system of functional constraints has 5 variables and 3 equations, so Number of variables number of equations 5 3 2. This fact gives us 2 degrees of freedom in solving the system, since any two variables can be chosen to be set equal to any arbitrary value in order to solve the three equations in terms of the remaining three variables.1 The simplex method uses zero for this arbitrary value. Thus, two of the variables (called the nonbasic variables) are set equal to zero, and then the simultaneous solution of the three equations for the other three variables (called the basic variables) is a basic solution. These properties are described in the following general definitions. A basic solution has the following properties: 1. Each variable is designated as either a nonbasic variable or a basic variable. 2. The number of basic variables equals the number of functional constraints (now equations). Therefore, the number of nonbasic variables equals the total number of variables minus the number of functional constraints. 3. The nonbasic variables are set equal to zero. 4. The values of the basic variables are obtained as the simultaneous solution of the system of equations (functional constraints in augmented form). (The set of basic variables is often referred to as the basis.) 5. If the basic variables satisfy the nonnegativity constraints, the basic solution is a BF solution. 1

This method of determining the number of degrees of freedom for a system of equations is valid as long as the system does not include any redundant equations. This condition always holds for the system of equations formed from the functional constraints in the augmented form of a linear programming model.

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117

To illustrate these definitions, consider again the BF solution (0, 6, 4, 0, 6). This solution was obtained before by augmenting the CPF solution (0, 6). However, another way to obtain this same solution is to choose x1 and x4 to be the two nonbasic variables, and so the two variables are set equal to zero. The three equations then yield, respectively, x3 4, x2 6, and x5 6 as the solution for the three basic variables, as shown below (with the basic variables in bold type): (1) (2) (3)

x3

x1 2x2 3x1 2x2

4 x4 12 x5 18

x1 0 and x4 0 so x3 4 x2 6 x5 6

Because all three of these basic variables are nonnegative, this basic solution (0, 6, 4, 0, 6) is indeed a BF solution. Just as certain pairs of CPF solutions are adjacent, the corresponding pairs of BF solutions also are said to be adjacent. Here is an easy way to tell when two BF solutions are adjacent. Two BF solutions are adjacent if all but one of their nonbasic variables are the same. This implies that all but one of their basic variables also are the same, although perhaps with different numerical values.

Consequently, moving from the current BF solution to an adjacent one involves switching one variable from nonbasic to basic and vice versa for one other variable (and then adjusting the values of the basic variables to continue satisfying the system of equations). To illustrate adjacent BF solutions, consider one pair of adjacent CPF solutions in Fig. 4.1: (0, 0) and (0, 6). Their augmented solutions, (0, 0, 4, 12, 18) and (0, 6, 4, 0, 6), automatically are adjacent BF solutions. However, you do not need to look at Fig. 4.1 to draw this conclusion. Another signpost is that their nonbasic variables, (x1, x2) and (x1, x4), are the same with just the one exception—x2 has been replaced by x4. Consequently, moving from (0, 0, 4, 12, 18) to (0, 6, 4, 0, 6) involves switching x2 from nonbasic to basic and vice versa for x4. When we deal with the problem in augmented form, it is convenient to consider and manipulate the objective function equation at the same time as the new constraint equations. Therefore, before we start the simplex method, the problem needs to be rewritten once again in an equivalent way: Maximize

Z,

subject to (0) (1) (2) (3)

Z 3x1 5x2 0 x1 x3 4 2x2 x4 12 3x1 2x2 x5 18

and xj 0,

for j 1, 2, . . . , 5.

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It is just as if Eq. (0) actually were one of the original constraints; but because it already is in equality form, no slack variable is needed. While adding one more equation, we also have added one more unknown (Z) to the system of equations. Therefore, when using Eqs. (1) to (3) to obtain a basic solution as described above, we use Eq. (0) to solve for Z at the same time. Somewhat fortuitously, the model for the Wyndor Glass Co. problem fits our standard form, and all its functional constraints have nonnegative right-hand sides bi. If this had not been the case, then additional adjustments would have been needed at this point before the simplex method was applied. These details are deferred to Sec. 4.6, and we now focus on the simplex method itself.

4.3

THE ALGEBRA OF THE SIMPLEX METHOD We continue to use the prototype example of Sec. 3.1, as rewritten at the end of Sec. 4.2, for illustrative purposes. To start connecting the geometric and algebraic concepts of the simplex method, we begin by outlining side by side in Table 4.2 how the simplex method solves this example from both a geometric and an algebraic viewpoint. The geometric viewpoint (first presented in Sec. 4.1) is based on the original form of the model (no slack variables), so again refer to Fig. 4.1 for a visualization when you examine the second column of the table. Refer to the augmented form of the model presented at the end of Sec. 4.2 when you examine the third column of the table. We now fill in the details for each step of the third column of Table 4.2. Initialization The choice of x1 and x2 to be the nonbasic variables (the variables set equal to zero) for the initial BF solution is based on solution concept 3 in Sec. 4.1. This choice eliminates the work required to solve for the basic variables (x3, x4, x5) from the following system of equations (where the basic variables are shown in bold type): (1) (2) (3)

x1

x3

2x2 3x1 2x2

4 x4 12 x5 18

x1 0 and x2 0 so x3 4 x4 12 x5 18

Thus, the initial BF solution is (0, 0, 4, 12, 18). Notice that this solution can be read immediately because each equation has just one basic variable, which has a coefficient of 1, and this basic variable does not appear in any other equation. You will soon see that when the set of basic variables changes, the simplex method uses an algebraic procedure (Gaussian elimination) to convert the equations to this same convenient form for reading every subsequent BF solution as well. This form is called proper form from Gaussian elimination. Optimality Test The objective function is Z 3x1 5x2,

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119

TABLE 4.2 Geometric and algebraic interpretations of how the simplex method solves the Wyndor Glass Co. problem Method Sequence

Geometric Interpretation

Initialization Choose (0, 0) to be the initial CPF solution. Optimality test Iteration 1 Step 1

Step 2 Step 3

Optimality test Iteration 2 Step 1

Step 2 Step 3

Optimality test

Not optimal, because moving along either edge from (0, 0) increases Z. Move up the edge lying on the x2 axis. Stop when the first new constraint boundary (2x2 12) is reached. Find the intersection of the new pair of constraint boundaries: (0, 6) is the new CPF solution. Not optimal, because moving along the edge from (0, 6) to the right increases Z.

Algebraic Interpretation Choose x1 and x2 to be the nonbasic variables ( 0) for the initial BF solution: (0, 0, 4, 12, 18). Not optimal, because increasing either nonbasic variable (x1 or x2) increases Z. Increase x2 while adjusting other variable values to satisfy the system of equations. Stop when the first basic variable (x3, x4, or x5) drops to zero (x4). With x2 now a basic variable and x4 now a nonbasic variable, solve the system of equations: (0, 6, 4, 0, 6) is the new BF solution. Not optimal, because increasing one nonbasic variable (x1) increases Z.

Increase x1 while adjusting other variable values to satisfy the system of equations. Stop when the first new constraint Stop when the first basic variable (x2, boundary (3x1 2x2 18) is reached. x3, or x5) drops to zero (x5). Find the intersection of the new pair With x1 now a basic variable and x5 of constraint boundaries: (2, 6) is the now a nonbasic variable, solve the new CPF solution. system of equations: (2, 6, 2, 0, 0) is the new BF solution. (2, 6) is optimal, because moving (2, 6, 2, 0, 0) is optimal, because along either edge from (2, 6) decreases Z. increasing either nonbasic variable (x4 or x5) decreases Z. Move along this edge to the right.

so Z 0 for the initial BF solution. Because none of the basic variables (x3, x4, x5) have a nonzero coefficient in this objective function, the coefficient of each nonbasic variable (x1, x2) gives the rate of improvement in Z if that variable were to be increased from zero (while the values of the basic variables are adjusted to continue satisfying the system of equations).1 These rates of improvement (3 and 5) are positive. Therefore, based on solution concept 6 in Sec. 4.1, we conclude that (0, 0, 4, 12, 18) is not optimal. For each BF solution examined after subsequent iterations, at least one basic variable has a nonzero coefficient in the objective function. Therefore, the optimality test then will use the new Eq. (0) to rewrite the objective function in terms of just the nonbasic variables, as you will see later. 1 Note that this interpretation of the coefficients of the xj variables is based on these variables being on the righthand side, Z 3x1 5x2. When these variables are brought to the left-hand side for Eq. (0), Z 3x1 5x2 0, the nonzero coefficients change their signs.

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Determining the Direction of Movement (Step 1 of an Iteration) Increasing one nonbasic variable from zero (while adjusting the values of the basic variables to continue satisfying the system of equations) corresponds to moving along one edge emanating from the current CPF solution. Based on solution concepts 4 and 5 in Sec. 4.1, the choice of which nonbasic variable to increase is made as follows: Z 3x1 5x2 Increase x1? Rate of improvement in Z 3. Increase x2? Rate of improvement in Z 5. 5 3, so choose x2 to increase. As indicated next, we call x2 the entering basic variable for iteration 1. At any iteration of the simplex method, the purpose of step 1 is to choose one nonbasic variable to increase from zero (while the values of the basic variables are adjusted to continue satisfying the system of equations). Increasing this nonbasic variable from zero will convert it to a basic variable for the next BF solution. Therefore, this variable is called the entering basic variable for the current iteration (because it is entering the basis).

Determining Where to Stop (Step 2 of an Iteration) Step 2 addresses the question of how far to increase the entering basic variable x2 before stopping. Increasing x2 increases Z, so we want to go as far as possible without leaving the feasible region. The requirement to satisfy the functional constraints in augmented form (shown below) means that increasing x2 (while keeping the nonbasic variable x1 0) changes the values of some of the basic variables as shown on the right. (1) (2) (3)

x1 2x2 3x1 2x2

x3

4 x4 12 x5 18

x1 0, so x3 4 x4 12 2x2 x5 18 2x2.

The other requirement for feasibility is that all the variables be nonnegative. The nonbasic variables (including the entering basic variable) are nonnegative, but we need to check how far x2 can be increased without violating the nonnegativity constraints for the basic variables. x3 4 0

⇒ no upper bound on x2.

12 x4 12 2x2 0 ⇒ x2 6 minimum. 2 18 x5 18 2x2 0 ⇒ x2 9. 2 Thus, x2 can be increased just to 6, at which point x4 has dropped to 0. Increasing x2 beyond 6 would cause x4 to become negative, which would violate feasibility. These calculations are referred to as the minimum ratio test. The objective of this test is to determine which basic variable drops to zero first as the entering basic variable is increased. We can immediately rule out the basic variable in any equation where the coefficient of the entering basic variable is zero or negative, since such a basic variable would not decrease as the entering basic variable is increased. [This is what happened

4.3 THE ALGEBRA OF THE SIMPLEX METHOD

121

with x3 in Eq. (1) of the example.] However, for each equation where the coefficient of the entering basic variable is strictly positive ( 0), this test calculates the ratio of the right-hand side to the coefficient of the entering basic variable. The basic variable in the equation with the minimum ratio is the one that drops to zero first as the entering basic variable is increased. At any iteration of the simplex method, step 2 uses the minimum ratio test to determine which basic variable drops to zero first as the entering basic variable is increased. Decreasing this basic variable to zero will convert it to a nonbasic variable for the next BF solution. Therefore, this variable is called the leaving basic variable for the current iteration (because it is leaving the basis).

Thus, x4 is the leaving basic variable for iteration 1 of the example. Solving for the New BF Solution (Step 3 of an Iteration) Increasing x2 0 to x2 6 moves us from the initial BF solution on the left to the new BF solution on the right. Nonbasic variables: Basic variables:

Initial BF solution x1 0, x2 0 x3 4, x4 12, x5 18

New BF solution x1 0, x4 0 x3 ?, x2 6, x5 ?

The purpose of step 3 is to convert the system of equations to a more convenient form (proper form from Gaussian elimination) for conducting the optimality test and (if needed) the next iteration with this new BF solution. In the process, this form also will identify the values of x3 and x5 for the new solution. Here again is the complete original system of equations, where the new basic variables are shown in bold type (with Z playing the role of the basic variable in the objective function equation): (0) (1) (2) (3)

Z 3x1 5x2 x3 x1 2x2 x4 3x1 2x2 x5

0. 4. 12. 18.

Thus, x2 has replaced x4 as the basic variable in Eq. (2). To solve this system of equations for Z, x2, x3, and x5, we need to perform some elementary algebraic operations to reproduce the current pattern of coefficients of x4 (0, 0, 1, 0) as the new coefficients of x2. We can use either of two types of elementary algebraic operations: 1. Multiply (or divide) an equation by a nonzero constant. 2. Add (or subtract) a multiple of one equation to (or from) another equation. To prepare for performing these operations, note that the coefficients of x2 in the above system of equations are 5, 0, 2, and 3, respectively, whereas we want these coefficients to become 0, 0, 1, and 0, respectively. To turn the coefficient of 2 in Eq. (2) into 1, we use the first type of elementary algebraic operation by dividing Eq. (2) by 2 to obtain (2)

1 x2 x4 6. 2

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To turn the coefficients of 5 and 3 into zeros, we need to use the second type of elementary algebraic operation. In particular, we add 5 times this new Eq. (2) to Eq. (0), and subtract 2 times this new Eq. (2) from Eq. (3). The resulting complete new system of equations is (0)

Z 3x1

(1)

x1

(2) (3)

5 x4 2 x3 x2

3x1

30

4 1 x4 6 2 x4 x5 6.

Since x1 0 and x4 0, the equations in this form immediately yield the new BF solution, (x1, x2, x3, x4, x5) (0, 6, 4, 0, 6), which yields Z 30. This procedure for obtaining the simultaneous solution of a system of linear equations is called the Gauss-Jordan method of elimination, or Gaussian elimination for short.1 The key concept for this method is the use of elementary algebraic operations to reduce the original system of equations to proper form from Gaussian elimination, where each basic variable has been eliminated from all but one equation (its equation) and has a coefficient of 1 in that equation. Optimality Test for the New BF Solution The current Eq. (0) gives the value of the objective function in terms of just the current nonbasic variables 5 Z 30 3x1 x4. 2 Increasing either of these nonbasic variables from zero (while adjusting the values of the basic variables to continue satisfying the system of equations) would result in moving toward one of the two adjacent BF solutions. Because x1 has a positive coefficient, increasing x1 would lead to an adjacent BF solution that is better than the current BF solution, so the current solution is not optimal. Iteration 2 and the Resulting Optimal Solution Since Z 30 3x1 52 x4, Z can be increased by increasing x1, but not x4. Therefore, step 1 chooses x1 to be the entering basic variable. For step 2, the current system of equations yields the following conclusions about how far x1 can be increased (with x4 0): 4 x3 4 x1 0 ⇒ x1 4. 1 x2 6 0 ⇒ no upper bound on x1. 6 x5 6 3x1 0 ⇒ x1 2 3

minimum.

Therefore, the minimum ratio test indicates that x5 is the leaving basic variable. 1

Actually, there are some technical differences between the Gauss-Jordan method of elimination and Gaussian elimination, but we shall not make this distinction.

4.4 THE SIMPLEX METHOD IN TABULAR FORM

123

For step 3, with x1 replacing x5 as a basic variable, we perform elementary algebraic operations on the current system of equations to reproduce the current pattern of coefficients of x5 (0, 0, 0, 1) as the new coefficients of x1. This yields the following new system of equations: (0)

3 x4 x5 36 2 1 1 x3 x4 x5 2 3 3

Z

(1) (2) (3)

x2 x1

1 x4 6 2 1 1 x4 x5 2. 3 3

Therefore, the next BF solution is (x1, x2, x3, x4, x5) (2, 6, 2, 0, 0), yielding Z 36. To apply the optimality test to this new BF solution, we use the current Eq. (0) to express Z in terms of just the current nonbasic variables, 3 Z 36 x4 x5. 2 Increasing either x4 or x5 would decrease Z, so neither adjacent BF solution is as good as the current one. Therefore, based on solution concept 6 in Sec. 4.1, the current BF solution must be optimal. In terms of the original form of the problem (no slack variables), the optimal solution is x1 2, x2 6, which yields Z 3x1 5x2 36. To see another example of applying the simplex method, we recommend that you now view the demonstration entitled Simplex Method—Algebraic Form in your OR Tutor. This vivid demonstration simultaneously displays both the algebra and the geometry of the simplex method as it dynamically evolves step by step. Like the many other demonstration examples accompanying other sections of the book (including the next section), this computer demonstration highlights concepts that are difficult to convey on the printed page. To further help you learn the simplex method efficiently, your OR Courseware includes a procedure entitled Solve Interactively by the Simplex Method. This routine performs nearly all the calculations while you make the decisions step by step, thereby enabling you to focus on concepts rather than get bogged down in a lot of number crunching. Therefore, you probably will want to use this routine for your homework on this section. The software will help you get started by letting you know whenever you make a mistake on the first iteration of a problem. The next section includes a summary of the simplex method for a more convenient tabular form.

4.4

THE SIMPLEX METHOD IN TABULAR FORM The algebraic form of the simplex method presented in Sec. 4.3 may be the best one for learning the underlying logic of the algorithm. However, it is not the most convenient form for performing the required calculations. When you need to solve a problem by hand (or

124

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

interactively with your OR Courseware), we recommend the tabular form described in this section.1 The tabular form of the simplex method records only the essential information, namely, (1) the coefficients of the variables, (2) the constants on the right-hand sides of the equations, and (3) the basic variable appearing in each equation. This saves writing the symbols for the variables in each of the equations, but what is even more important is the fact that it permits highlighting the numbers involved in arithmetic calculations and recording the computations compactly. Table 4.3 compares the initial system of equations for the Wyndor Glass Co. problem in algebraic form (on the left) and in tabular form (on the right), where the table on the right is called a simplex tableau. The basic variable for each equation is shown in bold type on the left and in the first column of the simplex tableau on the right. [Although only the xj variables are basic or nonbasic, Z plays the role of the basic variable for Eq. (0).] All variables not listed in this basic variable column (x1, x2) automatically are nonbasic variables. After we set x1 0, x2 0, the right side column gives the resulting solution for the basic variables, so that the initial BF solution is (x1, x2, x3, x4, x5) (0, 0, 4, 12, 18) which yields Z 0. The tabular form of the simplex method uses a simplex tableau to compactly display the system of equations yielding the current BF solution. For this solution, each variable in the leftmost column equals the corresponding number in the rightmost column (and variables not listed equal zero). When the optimality test or an iteration is performed, the only relevant numbers are those to the right of the Z column. The term row refers to just a row of numbers to the right of the Z column (including the right side number), where row i corresponds to Eq. (i).

We summarize the tabular form of the simplex method below and, at the same time, briefly describe its application to the Wyndor Glass Co. problem. Keep in mind that the logic is identical to that for the algebraic form presented in the preceding section. Only the form for displaying both the current system of equations and the subsequent iteration has changed (plus we shall no longer bother to bring variables to the right-hand side of an equation before drawing our conclusions in the optimality test or in steps 1 and 2 of an iteration). 1

A form more convenient for automatic execution on a computer is presented in Sec. 5.2.

TABLE 4.3 Initial system of equations for the Wyndor Glass Co. problem (a) Algebraic Form

(b) Tabular Form Coefficient of:

(0) (1) (2) (3)

Z 3x1 5x2 x3 x4 x5 0 Z 3x1 5x2 x3 x4 x5 4 Z 3x1 2x2 x3 x4 x5 12 Z 3x1 2x2 x3 x4 x5 18

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

4.4 THE SIMPLEX METHOD IN TABULAR FORM

125

Summary of the Simplex Method (and Iteration 1 for the Example) Initialization. Introduce slack variables. Select the decision variables to be the initial nonbasic variables (set equal to zero) and the slack variables to be the initial basic variables. (See Sec. 4.6 for the necessary adjustments if the model is not in our standard form—maximization, only functional constraints, and all nonnegativity constraints— or if any bi values are negative.) For the Example: This selection yields the initial simplex tableau shown in Table 4.3b, so the initial BF solution is (0, 0, 4, 12, 18). Optimality Test. The current BF solution is optimal if and only if every coefficient in row 0 is nonnegative ( 0). If it is, stop; otherwise, go to an iteration to obtain the next BF solution, which involves changing one nonbasic variable to a basic variable (step 1) and vice versa (step 2) and then solving for the new solution (step 3). For the Example: Just as Z 3x1 5x2 indicates that increasing either x1 or x2 will increase Z, so the current BF solution is not optimal, the same conclusion is drawn from the equation Z 3x1 5x2 0. These coefficients of 3 and 5 are shown in row 0 of Table 4.3b. Iteration. Step 1: Determine the entering basic variable by selecting the variable (automatically a nonbasic variable) with the negative coefficient having the largest absolute value (i.e., the “most negative” coefficient) in Eq. (0). Put a box around the column below this coefficient, and call this the pivot column. For the Example: The most negative coefficient is 5 for x2 (5 3), so x2 is to be changed to a basic variable. (This change is indicated in Table 4.4 by the box around the x2 column below 5.) Step 2: Determine the leaving basic variable by applying the minimum ratio test. Minimum Ratio Test 1. 2. 3. 4.

Pick out each coefficient in the pivot column that is strictly positive ( 0). Divide each of these coefficients into the right side entry for the same row. Identify the row that has the smallest of these ratios. The basic variable for that row is the leaving basic variable, so replace that variable by the entering basic variable in the basic variable column of the next simplex tableau.

TABLE 4.4 Applying the minimum ratio test to determine the first leaving basic variable for the Wyndor Glass Co. problem Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3

(0) (1)

1 0

3 1

5 0

0 1

0 0

0 0

0 4

x4

(2)

0

0

2

0

1

0

12 12 6 minimum 2

x5

(3)

0

3

2

0

0

1

18 18 9 2

Ratio

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Put a box around this row and call it the pivot row. Also call the number that is in both boxes the pivot number. For the Example: The calculations for the minimum ratio test are shown to the right of Table 4.4. Thus, row 2 is the pivot row (see the box around this row in the first simplex tableau of Table 4.5), and x4 is the leaving basic variable. In the next simplex tableau (see the bottom of Table 4.5), x2 replaces x4 as the basic variable for row 2. Step 3: Solve for the new BF solution by using elementary row operations (multiply or divide a row by a nonzero constant; add or subtract a multiple of one row to another row) to construct a new simplex tableau in proper form from Gaussian elimination below the current one, and then return to the optimality test. The specific elementary row operations that need to be performed are listed below. 1. Divide the pivot row by the pivot number. Use this new pivot row in steps 2 and 3. 2. For each other row (including row 0) that has a negative coefficient in the pivot column, add to this row the product of the absolute value of this coefficient and the new pivot row. 3. For each other row that has a positive coefficient in the pivot column, subtract from this row the product of this coefficient and the new pivot row. For the Example: Since x2 is replacing x4 as a basic variable, we need to reproduce the first tableau’s pattern of coefficients in the column of x4 (0, 0, 1, 0) in the second tableau’s column of x2. To start, divide the pivot row (row 2) by the pivot number (2), which gives the new row 2 shown in Table 4.5. Next, we add to row 0 the product, 5 times the new row 2. Then we subtract from row 3 the product, 2 times the new row 2 (or equivalently, subtract from row 3 the old row 2). These calculations yield the new tableau shown in Table 4.6 for iteration 1. Thus, the new BF solution is (0, 6, 4, 0, 6), with Z 30. We next return to the optimality test to check if the new BF solution is optimal. Since the new row 0 still has a negative coefficient (3 for x1), the solution is not optimal, and so at least one more iteration is needed. TABLE 4.5 Simplex tableaux for the Wyndor Glass Co. problem after the first pivot row is divided by the first pivot number Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

0

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

1

Z x3 x2 x5

(0) (1) (2) (3)

1 0 0 0

0

1

0

1 2

0

6

Iteration

4.4 THE SIMPLEX METHOD IN TABULAR FORM

127

TABLE 4.6 First two simplex tableaux for the Wyndor Glass Co. problem Coefficient of: Iteration

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

Z

(0)

1

3

0

0

0

30

x3

(1)

0

1

0

1

0

4

x2

(2)

0

0

1

0

0

6

x5

(3)

0

3

0

0

1

6

0

1

5 2 0 1 2 1

Iteration 2 for the Example and the Resulting Optimal Solution The second iteration starts anew from the second tableau of Table 4.6 to find the next BF solution. Following the instructions for steps 1 and 2, we find x1 as the entering basic variable and x5 as the leaving basic variable, as shown in Table 4.7. For step 3, we start by dividing the pivot row (row 3) in Table 4.7 by the pivot number (3). Next, we add to row 0 the product, 3 times the new row 3. Then we subtract the new row 3 from row 1. We now have the set of tableaux shown in Table 4.8. Therefore, the new BF solution is (2, 6, 2, 0, 0), with Z 36. Going to the optimality test, we find that this solution is optimal because none of the coefficients in row 0 is negative, so the algorithm is finished. Consequently, the optimal solution for the Wyndor Glass Co. problem (before slack variables are introduced) is x1 2, x2 6. Now compare Table 4.8 with the work done in Sec. 4.3 to verify that these two forms of the simplex method really are equivalent. Then note how the algebraic form is superior for learning the logic behind the simplex method, but the tabular form organizes the TABLE 4.7 Steps 1 and 2 of iteration 2 for the Wyndor Glass Co. problem Coefficient of: Iteration

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z

(0)

1

3

0

0

5 2

0

30

x3

(1)

0

1

0

1

0

0

4

x2

(2)

0

0

1

0

1 2

0

6

x5

(3)

0

3

0

0

1

1

6

1

Ratio

4 4 1

6 2 minimum 3

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

TABLE 4.8 Complete set of simplex tableaux for the Wyndor Glass Co. problem Coefficient of: Iteration

0

1

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

Z

(0)

1

3

0

0

0

30

x3

(1)

0

1

0

1

0

4

x2

(2)

0

0

1

0

0

6

x5

(3)

0

3

0

0

5 2 0 1 2 1

1

6

Z

(0)

1

0

0

0

3 2

1

36

x3

(1)

0

0

0

1

1 3

1 3

2

x2

(2)

0

0

1

0

1 2

0

6

x1

(3)

0

1

0

0

1 3

1 3

2

2

work being done in a considerably more convenient and compact form. We generally use the tabular form from now on. An additional example of applying the simplex method in tabular form is available to you in the OR Tutor. See the demonstration entitled Simplex Method—Tabular Form.

4.5

TIE BREAKING IN THE SIMPLEX METHOD You may have noticed in the preceding two sections that we never said what to do if the various choice rules of the simplex method do not lead to a clear-cut decision, because of either ties or other similar ambiguities. We discuss these details now. Tie for the Entering Basic Variable Step 1 of each iteration chooses the nonbasic variable having the negative coefficient with the largest absolute value in the current Eq. (0) as the entering basic variable. Now suppose that two or more nonbasic variables are tied for having the largest negative coefficient (in absolute terms). For example, this would occur in the first iteration for the Wyndor Glass Co. problem if its objective function were changed to Z 3x1 3x2, so that the initial Eq. (0) became Z 3x1 3x2 0. How should this tie be broken? The answer is that the selection between these contenders may be made arbitrarily. The optimal solution will be reached eventually, regardless of the tied variable chosen, and there is no convenient method for predicting in advance which choice will lead there

4.5 TIE BREAKING IN THE SIMPLEX METHOD

129

sooner. In this example, the simplex method happens to reach the optimal solution (2, 6) in three iterations with x1 as the initial entering basic variable, versus two iterations if x2 is chosen. Tie for the Leaving Basic Variable—Degeneracy Now suppose that two or more basic variables tie for being the leaving basic variable in step 2 of an iteration. Does it matter which one is chosen? Theoretically it does, and in a very critical way, because of the following sequence of events that could occur. First, all the tied basic variables reach zero simultaneously as the entering basic variable is increased. Therefore, the one or ones not chosen to be the leaving basic variable also will have a value of zero in the new BF solution. (Note that basic variables with a value of zero are called degenerate, and the same term is applied to the corresponding BF solution.) Second, if one of these degenerate basic variables retains its value of zero until it is chosen at a subsequent iteration to be a leaving basic variable, the corresponding entering basic variable also must remain zero (since it cannot be increased without making the leaving basic variable negative), so the value of Z must remain unchanged. Third, if Z may remain the same rather than increase at each iteration, the simplex method may then go around in a loop, repeating the same sequence of solutions periodically rather than eventually increasing Z toward an optimal solution. In fact, examples have been artificially constructed so that they do become entrapped in just such a perpetual loop. Fortunately, although a perpetual loop is theoretically possible, it has rarely been known to occur in practical problems. If a loop were to occur, one could always get out of it by changing the choice of the leaving basic variable. Furthermore, special rules1 have been constructed for breaking ties so that such loops are always avoided. However, these rules frequently are ignored in actual application, and they will not be repeated here. For your purposes, just break this kind of tie arbitrarily and proceed without worrying about the degenerate basic variables that result. No Leaving Basic Variable—Unbounded Z In step 2 of an iteration, there is one other possible outcome that we have not yet discussed, namely, that no variable qualifies to be the leaving basic variable.2 This outcome would occur if the entering basic variable could be increased indefinitely without giving negative values to any of the current basic variables. In tabular form, this means that every coefficient in the pivot column (excluding row 0) is either negative or zero. As illustrated in Table 4.9, this situation arises in the example displayed in Fig. 3.6 on p. 36. In this example, the last two functional constraints of the Wyndor Glass Co. problem have been overlooked and so are not included in the model. Note in Fig. 3.6 how x2 can be increased indefinitely (thereby increasing Z indefinitely) without ever leaving the feasible region. Then note in Table 4.9 that x2 is the entering basic variable but the 1

See R. Bland, “New Finite Pivoting Rules for the Simplex Method,” Mathematics of Operations Research, 2: 103–107, 1977. 2 Note that the analogous case (no entering basic variable) cannot occur in step 1 of an iteration, because the optimality test would stop the algorithm first by indicating that an optimal solution had been reached.

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TABLE 4.9 Initial simplex tableau for the Wyndor Glass Co. problem without the last two functional constraints Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

Right Side

Ratio

Z x3

(0) (1)

1 0

3 1

5 0

0 1

0 4

None

With x1 0 and x2 increasing, x3 4 1x1 0x2 4 0.

only coefficient in the pivot column is zero. Because the minimum ratio test uses only coefficients that are greater than zero, there is no ratio to provide a leaving basic variable. The interpretation of a tableau like the one shown in Table 4.9 is that the constraints do not prevent the value of the objective function Z increasing indefinitely, so the simplex method would stop with the message that Z is unbounded. Because even linear programming has not discovered a way of making infinite profits, the real message for practical problems is that a mistake has been made! The model probably has been misformulated, either by omitting relevant constraints or by stating them incorrectly. Alternatively, a computational mistake may have occurred. Multiple Optimal Solutions We mentioned in Sec. 3.2 (under the definition of optimal solution) that a problem can have more than one optimal solution. This fact was illustrated in Fig. 3.5 by changing the objective function in the Wyndor Glass Co. problem to Z 3x1 2x2, so that every point on the line segment between (2, 6) and (4, 3) is optimal. Thus, all optimal solutions are a weighted average of these two optimal CPF solutions (x1, x2) w1(2, 6) w2(4, 3), where the weights w1 and w2 are numbers that satisfy the relationships w1 w2 1

and

w1 0,

w2 0.

For example, w1 13 and w2 23 give

1 2 2 8 (x1, x2) (2, 6) (4, 3) , 3 3 3 3

6 6 10 , 3 3 3

4

as one optimal solution. In general, any weighted average of two or more solutions (vectors) where the weights are nonnegative and sum to 1 is called a convex combination of these solutions. Thus, every optimal solution in the example is a convex combination of (2, 6) and (4, 3). This example is typical of problems with multiple optimal solutions. As indicated at the end of Sec. 3.2, any linear programming problem with multiple optimal solutions (and a bounded feasible region) has at least two CPF solutions that are optimal. Every optimal solution is a convex combination of these optimal CPF solutions. Consequently, in augmented form, every optimal solution is a convex combination of the optimal BF solutions.

(Problems 4.5-5 and 4.5-6 guide you through the reasoning behind this conclusion.)

4.5 TIE BREAKING IN THE SIMPLEX METHOD

131

The simplex method automatically stops after one optimal BF solution is found. However, for many applications of linear programming, there are intangible factors not incorporated into the model that can be used to make meaningful choices between alternative optimal solutions. In such cases, these other optimal solutions should be identified as well. As indicated above, this requires finding all the other optimal BF solutions, and then every optimal solution is a convex combination of the optimal BF solutions. After the simplex method finds one optimal BF solution, you can detect if there are any others and, if so, find them as follows: Whenever a problem has more than one optimal BF solution, at least one of the nonbasic variables has a coefficient of zero in the final row 0, so increasing any such variable will not change the value of Z. Therefore, these other optimal BF solutions can be identified (if desired) by performing additional iterations of the simplex method, each time choosing a nonbasic variable with a zero coefficient as the entering basic variable.1

To illustrate, consider again the case just mentioned, where the objective function in the Wyndor Glass Co. problem is changed to Z 3x1 2x2. The simplex method obtains the first three tableaux shown in Table 4.10 and stops with an optimal BF solution. How1

If such an iteration has no leaving basic variable, this indicates that the feasible region is unbounded and the entering basic variable can be increased indefinitely without changing the value of Z.

TABLE 4.10 Complete set of simplex tableaux to obtain all optimal BF solutions for the Wyndor Glass Co. problem with c2 2 Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

(0) (1) (2) (3)

1 0 0 0

3 1 0 3

2 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

0 4 12 18

No

0

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

2 0 2 2

3 1 0 3

0 0 1 0

0 0 0 1

12 4 12 6

No

1

Z x1 x4 x5 Z x1 x4

(0) (1) (2)

1 0 0

0 1 0

0 0 0

0 0 1

Yes

(3)

0

0

1

1 0 1 1 2

18 4 6

x2

0 1 3 3 2

Z

(0)

1

0

0

0

(1)

0

1

0

0

1 1 3

18

x1

0 1 3

x3

(2)

0

0

0

1

1 3

1 3

2

x2

(3)

0

0

1

0

1 2

0

6

Iteration

2

Extra

0

Solution Optimal?

3

2

Yes

132

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ever, because a nonbasic variable (x3) then has a zero coefficient in row 0, we perform one more iteration in Table 4.10 to identify the other optimal BF solution. Thus, the two optimal BF solutions are (4, 3, 0, 6, 0) and (2, 6, 2, 0, 0), each yielding Z 18. Notice that the last tableau also has a nonbasic variable (x4) with a zero coefficient in row 0. This situation is inevitable because the extra iteration does not change row 0, so this leaving basic variable necessarily retains its zero coefficient. Making x4 an entering basic variable now would only lead back to the third tableau. (Check this.) Therefore, these two are the only BF solutions that are optimal, and all other optimal solutions are a convex combination of these two. (x1, x2, x3, x4, x5) w1(2, 6, 2, 0, 0) w2(4, 3, 0, 6, 0), w1 w2 1, w1 0, w2 0.

4.6

ADAPTING TO OTHER MODEL FORMS Thus far we have presented the details of the simplex method under the assumptions that the problem is in our standard form (maximize Z subject to functional constraints in form and nonnegativity constraints on all variables) and that bi 0 for all i 1, 2, . . . , m. In this section we point out how to make the adjustments required for other legitimate forms of the linear programming model. You will see that all these adjustments can be made during the initialization, so the rest of the simplex method can then be applied just as you have learned it already. The only serious problem introduced by the other forms for functional constraints (the or forms, or having a negative right-hand side) lies in identifying an initial BF solution. Before, this initial solution was found very conveniently by letting the slack variables be the initial basic variables, so that each one just equals the nonnegative right-hand side of its equation. Now, something else must be done. The standard approach that is used for all these cases is the artificial-variable technique. This technique constructs a more convenient artificial problem by introducing a dummy variable (called an artificial variable) into each constraint that needs one. This new variable is introduced just for the purpose of being the initial basic variable for that equation. The usual nonnegativity constraints are placed on these variables, and the objective function also is modified to impose an exorbitant penalty on their having values larger than zero. The iterations of the simplex method then automatically force the artificial variables to disappear (become zero), one at a time, until they are all gone, after which the real problem is solved. To illustrate the artificial-variable technique, first we consider the case where the only nonstandard form in the problem is the presence of one or more equality constraints. Equality Constraints Any equality constraint ai1x1 ai2x2 ain xn bi actually is equivalent to a pair of inequality constraints: ai1x1 ai2x2 ain xn bi ai1x1 ai2x2 ain xn bi.

4.6 ADAPTING TO OTHER MODEL FORMS

133

However, rather than making this substitution and thereby increasing the number of constraints, it is more convenient to use the artificial-variable technique. We shall illustrate this technique with the following example. Example. Suppose that the Wyndor Glass Co. problem in Sec. 3.1 is modified to require that Plant 3 be used at full capacity. The only resulting change in the linear programming model is that the third constraint, 3x1 2x2 18, instead becomes an equality constraint 3x1 2x2 18, so that the complete model becomes the one shown in the upper right-hand corner of Fig. 4.3. This figure also shows in darker ink the feasible region which now consists of just the line segment connecting (2, 6) and (4, 3). After the slack variables still needed for the inequality constraints are introduced, the system of equations for the augmented form of the problem becomes (0) (1) (2) (3)

Z 3x1 5x2 0. x1 x3 4. 2x2 x4 12. 3x1 2x2 18.

Unfortunately, these equations do not have an obvious initial BF solution because there is no longer a slack variable to use as the initial basic variable for Eq. (3). It is necessary to find an initial BF solution to start the simplex method. This difficulty can be circumvented in the following way.

FIGURE 4.3 When the third functional constraint becomes an equality constraint, the feasible region for the Wyndor Glass Co. problem becomes the line segment between (2, 6) and (4, 3).

x2 10 Maximize subject to 8

6

and

Z 3x1 5x2, 4 x1 2x2 12 3x1 2x2 18 x1 0, x2 0

(2, 6)

4 (4, 3) 2

0

2

4

6

8

x1

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Obtaining an Initial BF Solution. The procedure is to construct an artificial problem that has the same optimal solution as the real problem by making two modifications of the real problem. 1. Apply the artificial-variable technique by introducing a nonnegative artificial variable (call it x5)1 into Eq. (3), just as if it were a slack variable (3)

3x1 2x2 x5 18.

2. Assign an overwhelming penalty to having x5 0 by changing the objective function Z 3x1 5x2 to Z 3x1 5x2 Mx5, where M symbolically represents a huge positive number. (This method of forcing x5 to be x5 0 in the optimal solution is called the Big M method.) Now find the optimal solution for the real problem by applying the simplex method to the artificial problem, starting with the following initial BF solution: Initial BF Solution Nonbasic variables: Basic variables:

x1 0, x3 4,

x2 0 x4 12,

x5 18.

Because x5 plays the role of the slack variable for the third constraint in the artificial problem, this constraint is equivalent to 3x1 2x2 18 (just as for the original Wyndor Glass Co. problem in Sec. 3.1). We show below the resulting artificial problem (before augmenting) next to the real problem. The Real Problem

The Artificial Problem Define x5 18 3x1 2x2.

Maximize Z 3x1 5x2,

Maximize

subject to

subject to

Z 3x1 5x2 M x5,

x1 2x2 4

(so

3x1 2x2 x5 4

3x1 2x2 12

(so

3x1 2x2 x5 12

3x1 2x2 18

(so

3x1 2x2 x5 18

(so

3x1 2x2 x5 18)

and x1 0,

x2 0.

and x1 0,

x2 0,

x5 0.

Therefore, just as in Sec. 3.1, the feasible region for (x1, x2) for the artificial problem is the one shown in Fig. 4.4. The only portion of this feasible region that coincides with the feasible region for the real problem is where x5 0 (so 3x1 2x2 18). Figure 4.4 also shows the order in which the simplex method examines the CPF solutions (or BF solutions after augmenting), where each circled number identifies which iteration obtained that solution. Note that the simplex method moves counterclockwise here 1

We shall always label the artificial variables by putting a bar over them.

4.6 ADAPTING TO OTHER MODEL FORMS

135

x2 Define x5 18 3x1 2x2. Maximize Z 3x1 5x2 Mx5, subject to x1 4 2x2 12 3x1 2x2 18 x1 0, x2 0, x5 0 and

Z 30 6M (2, 6) Z 36 3

(0, 6)

FIGURE 4.4 This graph shows the feasible region and the sequence of CPF solutions (, , , ) examined by the simplex method for the artificial problem that corresponds to the real problem of Fig. 4.3.

Feasible region

(4, 3)

1

0

(0, 0)

2

Z 0 18M

(4, 0)

Z 27

Z 12 6M x1

whereas it moved clockwise for the original Wyndor Glass Co. problem (see Fig. 4.2). The reason for this difference is the extra term Mxx5 in the objective function for the artificial problem. Before applying the simplex method and demonstrating that it follows the path shown in Fig. 4.4, the following preparatory step is needed. Converting Equation (0) to Proper Form. The system of equations after the artificial problem is augmented is (0) (1) (2) (3)

Z 3x1 5x2 Mx5 0 x3 4 x1 2x2 x4 12 3x1 2x2 x5 18

where the initial basic variables (x3, x4, x5) are shown in bold type. However, this system is not yet in proper form from Gaussian elimination because a basic variable x5 has a nonzero coefficient in Eq. (0). Recall that all basic variables must be algebraically eliminated from Eq. (0) before the simplex method can either apply the optimality test or find the entering basic variable. This elimination is necessary so that the negative of the coefficient of each nonbasic variable will give the rate at which Z would increase if that nonbasic variable were to be increased from 0 while adjusting the values of the basic variables accordingly. To algebraically eliminate x5 from Eq. (0), we need to subtract from Eq. (0) the product, M times Eq. (3).

New (0)

Z 3x1 5x2 Mx5 0 M(3x 1 2x2 Mx x5 18) Z (3M 3)x1 (2M 5)x2 18M.

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Application of the Simplex Method. This new Eq. (0) gives Z in terms of just the nonbasic variables (x1, x2), Z 18M (3M 3)x1 (2M 5)x2. Since 3M 3 2M 5 (remember that M represents a huge number), increasing x1 increases Z at a faster rate than increasing x2 does, so x1 is chosen as the entering basic variable. This leads to the move from (0, 0) to (4, 0) at iteration 1, shown in Fig. 4.4, thereby increasing Z by 4(3M 3). The quantities involving M never appear in the system of equations except for Eq. (0), so they need to be taken into account only in the optimality test and when an entering basic variable is determined. One way of dealing with these quantities is to assign some particular (huge) numerical value to M and use the resulting coefficients in Eq. (0) in the usual way. However, this approach may result in significant rounding errors that invalidate the optimality test. Therefore, it is better to do what we have just shown, namely, to express each coefficient in Eq. (0) as a linear function aM b of the symbolic quantity M by separately recording and updating the current numerical value of (1) the multiplicative factor a and (2) the additive term b. Because M is assumed to be so large that b always is negligible compared with M when a 0, the decisions in the optimality test and the choice of the entering basic variable are made by using just the multiplicative factors in the usual way, except for breaking ties with the additive factors. Using this approach on the example yields the simplex tableaux shown in Table 4.11. Note that the artificial variable x5 is a basic variable (xx5 0) in the first two tableaux and a nonbasic variable (xx5 0) in the last two. Therefore, the first two BF solutions for this artificial problem are infeasible for the real problem whereas the last two also are BF solutions for the real problem. This example involved only one equality constraint. If a linear programming model has more than one, each is handled in just the same way. (If the right-hand side is negative, multiply through both sides by 1 first.)

Negative Right-Hand Sides The technique mentioned in the preceding sentence for dealing with an equality constraint with a negative right-hand side (namely, multiply through both sides by 1) also works for any inequality constraint with a negative right-hand side. Multiplying through both sides of an inequality by 1 also reverses the direction of the inequality; i.e., changes to or vice versa. For example, doing this to the constraint x1 x2 1

(that is, x1 x2 1)

gives the equivalent constraint x1 x2 1

(that is, x2 1 x1)

but now the right-hand side is positive. Having nonnegative right-hand sides for all the functional constraints enables the simplex method to begin, because (after augmenting) these right-hand sides become the respective values of the initial basic variables, which must satisfy nonnegativity constraints.

4.6 ADAPTING TO OTHER MODEL FORMS

137

TABLE 4.11 Complete set of simplex tableaux for the problem shown in Fig. 4.4 Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x 5

Right Side

0

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

3M 3 1 0 3

2M 5 0 2 2

0 1 0 0

0 0 1 0

0 0 0 1

18M 4 12 18

1

Z x1 x4 x5

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

2M 5 0 2 2

3M 3 1 0 3

0 0 1 0

0 0 0 1

6M 12 4 12 6

Z

(0)

1

0

0

x1 x4

(1) (2)

0 0

1 0

0 0

x2

(3)

0

0

1

Z

(0)

1

0

x1

(1)

0

x3

(2)

x2

(3)

Iteration

2

9 2 1 3 3 2

0 1

0

0

3 2

M1

36

1

0

0

1 3

1 3

2

0

0

0

1

1 3

1 3

2

0

0

1

0

1 2

0

6

Extra

0

0

5 M 2 0 1 1 2

27 4 6 3

We next focus on how to augment constraints, such as x1 x2 1, with the help of the artificial-variable technique. Functional Constraints in Form To illustrate how the artificial-variable technique deals with functional constraints in form, we will use the model for designing Mary’s radiation therapy, as presented in Sec. 3.4. For your convenience, this model is repeated below, where we have placed a box around the constraint of special interest here. Radiation Therapy Example Minimize

Z 0.4x1 0.5x2,

subject to 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6 and x1 0,

x2 0.

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x2 27 15 Dots corner-point solutions Dark line segment feasible region Optimal solution (7.5, 4.5)

0.6x1 0.4x2 6 10

(6, 6) 5 (7.5, 4.5)

(8, 3)

FIGURE 4.5 Graphical display of the radiation therapy example and its corner-point solutions.

0.5x1 0.5x2 6

0.3x1 0.1x2 2.7

0

5

10

x1

The graphical solution for this example (originally presented in Fig. 3.12) is repeated here in a slightly different form in Fig. 4.5. The three lines in the figure, along with the two axes, constitute the five constraint boundaries of the problem. The dots lying at the intersection of a pair of constraint boundaries are the corner-point solutions. The only two corner-point feasible solutions are (6, 6) and (7.5, 4.5), and the feasible region is the line segment connecting these two points. The optimal solution is (x1, x2) (7.5, 4.5), with Z 5.25. We soon will show how the simplex method solves this problem by directly solving the corresponding artificial problem. However, first we must describe how to deal with the third constraint.

4.6 ADAPTING TO OTHER MODEL FORMS

139

Our approach involves introducing both a surplus variable x5 (defined as x5 0.6x1 0.4x2 6) and an artificial variable x6, as shown next.

0.6x1 0.4x2 6 0.6x1 0.4x2 x5 6 0.6x1 0.4x2 x5 x6 6

(x5 0) (x5 0, x6 0).

Here x5 is called a surplus variable because it subtracts the surplus of the left-hand side over the right-hand side to convert the inequality constraint to an equivalent equality constraint. Once this conversion is accomplished, the artificial variable is introduced just as for any equality constraint. After a slack variable x3 is introduced into the first constraint, an artificial variable x4 is introduced into the second constraint, and the Big M method is applied, so the complete artificial problem (in augmented form) is Z 0.4x1 0.5x2 Mxx4 Mxx6, 0.3x1 0.1x2 x3 2.7 0.5x1 0.5x2 x4 6 0.6x1 0.4x2 x5 x6 6 x2 0, x3 0, x5 0, x1 0, x4 0,

Minimize subject to

and

x6 0.

Note that the coefficients of the artificial variables in the objective function are M, instead of M, because we now are minimizing Z. Thus, even though x4 0 and/or x6 0 is possible for a feasible solution for the artificial problem, the huge unit penalty of M prevents this from occurring in an optimal solution. As usual, introducing artificial variables enlarges the feasible region. Compare below the original constraints for the real problem with the corresponding constraints on (x1, x2) for the artificial problem. Constraints on (x1, x2) for the Real Problem 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6 x1 0, x2 0

Constraints on (x1, x2) for the Artificial Problem 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 ( holds when x4 0) No such constraint (except when x6 0) x1 0, x2 0

Introducing the artificial variable x4 to play the role of a slack variable in the second constraint allows values of (x1, x2) below the 0.5x1 0.5x2 6 line in Fig. 4.5. Introducing x5 and x6 into the third constraint of the real problem (and moving these variables to the right-hand side) yields the equation 0.6x1 0.4x2 6 x5 x6. Because both x5 and x6 are constrained only to be nonnegative, their difference x5 x6 can be any positive or negative number. Therefore, 0.6x1 0.4x2 can have any value, which has the effect of eliminating the third constraint from the artificial problem and allowing points on either side of the 0.6x1 0.4x2 6 line in Fig. 4.5. (We keep the third constraint in the system of equations only because it will become relevant again later, after the Big M method forces x6 to be zero.) Consequently, the feasible region for the ar-

140

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

tificial problem is the entire polyhedron in Fig. 4.5 whose vertices are (0, 0), (9, 0), (7.5, 4.5), and (0, 12). Since the origin now is feasible for the artificial problem, the simplex method starts with (0, 0) as the initial CPF solution, i.e., with (x1, x2, x3, x4, x5, x6) (0, 0, 2.7, 6, 0, 6) as the initial BF solution. (Making the origin feasible as a convenient starting point for the simplex method is the whole point of creating the artificial problem.) We soon will trace the entire path followed by the simplex method from the origin to the optimal solution for both the artificial and real problems. But, first, how does the simplex method handle minimization? Minimization One straightforward way of minimizing Z with the simplex method is to exchange the roles of the positive and negative coefficients in row 0 for both the optimality test and step 1 of an iteration. However, rather than changing our instructions for the simplex method for this case, we present the following simple way of converting any minimization problem to an equivalent maximization problem: n

Minimizing

Z cj xj j1

is equivalent to n

maximizing

Z (cj)xj; j1

i.e., the two formulations yield the same optimal solution(s). The two formulations are equivalent because the smaller Z is, the larger Z is, so the solution that gives the smallest value of Z in the entire feasible region must also give the largest value of Z in this region. Therefore, in the radiation therapy example, we make the following change in the formulation:

Minimize Maximize

Z 0.4x1 0.5x2 Z 0.4x1 0.5x2.

After artificial variables x4 and x6 are introduced and then the Big M method is applied, the corresponding conversion is

Minimize Maximize

Z 0.4x1 0.5x2 Mxx4 Mxx6 Z 0.4x1 0.5x2 Mxx4 Mxx6.

Solving the Radiation Therapy Example We now are nearly ready to apply the simplex method to the radiation therapy example. By using the maximization form just obtained, the entire system of equations is now (0) (1)

Z 0.4x1 0.5x2 Mx 4 0.3x1 0.1x2 x3

Mx 6 0 2.7

4.6 ADAPTING TO OTHER MODEL FORMS

(2) (3)

0.5x1 0.5x2 0.6x1 0.4x2

x4

141

x5

6 x6 6.

The basic variables (x3, x4, x6) for the initial BF solution (for this artificial problem) are shown in bold type. Note that this system of equations is not yet in proper form from Gaussian elimination, as required by the simplex method, since the basic variables x4 and x6 still need to be algebraically eliminated from Eq. (0). Because x4 and x6 both have a coefficient of M, Eq. (0) needs to have subtracted from it both M times Eq. (2) and M times Eq. (3). The calculations for all the coefficients (and the right-hand sides) are summarized below, where the vectors are the relevant rows of the simplex tableau corresponding to the above system of equations. Row 0: M[0.4, M[0.5, M[0.6, New row 0 [1.1M 0.4,

0.5, 0.5, 0.4, 0.9M 0.5,

0, 0, 0, 0,

M, 1, 0, 0,

0, 0, 1, M,

M, 0, 1, 0,

0] 6] 6] 12M]

The resulting initial simplex tableau, ready to begin the simplex method, is shown at the top of Table 4.12. Applying the simplex method in just the usual way then yields the sequence of simplex tableaux shown in the rest of Table 4.12. For the optimality test and the selection of the entering basic variable at each iteration, the quantities involving M are treated just as discussed in connection with Table 4.11. Specifically, whenever M is present, only its multiplicative factor is used, unless there is a tie, in which case the tie is broken by using the corresponding additive terms. Just such a tie occurs in the last selection of an entering basic variable (see the next-to-last tableau), where the coefficients of x3 and x5 in row 0 both have the same multiplicative factor of 53. Comparing the additive terms, 161 73 leads to choosing x5 as the entering basic variable. Note in Table 4.12 the progression of values of the artificial variables x4 and x6 and of Z. We start with large values, x4 6 and x6 6, with Z 12M (Z 12M). The first iteration greatly reduces these values. The Big M method succeeds in driving x6 to zero (as a new nonbasic variable) at the second iteration and then in doing the same to x4 at the next iteration. With both x4 0 and x6 0, the basic solution given in the last tableau is guaranteed to be feasible for the real problem. Since it passes the optimality test, it also is optimal. Now see what the Big M method has done graphically in Fig. 4.6. The feasible region for the artificial problem initially has four CPF solutions—(0, 0), (9, 0), (0, 12), and (7.5, 4.5)—and then replaces the first three with two new CPF solutions—(8, 3), (6, 6)— after x6 decreases to x6 0 so that 0.6x1 0.4x2 6 becomes an additional constraint. (Note that the three replaced CPF solutions—(0, 0), (9, 0), and (0, 12)—actually were corner-point infeasible solutions for the real problem shown in Fig. 4.5.) Starting with the origin as the convenient initial CPF solution for the artificial problem, we move around the boundary to three other CPF solutions—(9, 0), (8, 3), and (7.5, 4.5). The last of these is the first one that also is feasible for the real problem. Fortuitously, this first feasible solution also is optimal, so no additional iterations are needed.

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4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

TABLE 4.12 The Big M method for the radiation therapy example Coefficient of: Iteration

0

Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Right Side

Z x3 x4 x6

(0) (1) (2) (3)

1 0 0 0

1.1M 0.4 0.3 0.5 0.6

0.9M 0.5 0.1 0.5 0.4

0.0 1.0 0.0 0.0

0.0 0.0 1.0 0.0

M 0 0 1

0 0 0 1

12M1 2.7 6.0 6.0

Z

(0)

1

0.0

16 11 M 30 30

11 4 M 3 3

0.0

M

0

2.1M 3.6

x1

(1)

0

1.0

1 3

10 3

0.0

0

0

9.0

x4

(2)

0

0.0

0

0

1.5

(3)

0

0.0

5 3 2

1.0

x6

1 3 0.2

0.0

1

1

0.6

Z

(0)

1

0.0

0.0

5 7 M 3 3

0.0

5 11 M 3 6

8 11 M 3 6

0.5M 4.7

x1

(1)

0

1.0

0.0

20 3

0.0

5 3

5 3

8.0

x4

(2)

0

0.0

0.0

1.0

(3)

0

0.0

1.0

0.0

5 3 5

5 3 5

0.5

x2

5 3 10.0

Z x1 x5 x2

(0) (1) (2) (3)

1 0 0 0

0.0 1.0 0.0 0.0

0.0 0.0 0.0 1.0

0.5 5.0 1.0 5.0

M 1.1 1.0 1 0.6 3.0

0 0 1 0

M 0 1 0

5.25 7.51 0.31 4.51

1

2

3

3.0

For other problems with artificial variables, it may be necessary to perform additional iterations to reach an optimal solution after the first feasible solution is obtained for the real problem. (This was the case for the example solved in Table 4.11.) Thus, the Big M method can be thought of as having two phases. In the first phase, all the artificial variables are driven to zero (because of the penalty of M per unit for being greater than zero) in order to reach an initial BF solution for the real problem. In the second phase, all the artificial variables are kept at zero (because of this same penalty) while the simplex method generates a sequence of BF solutions for the real problem that leads to an optimal solution. The two-phase method described next is a streamlined procedure for performing these two phases directly, without even introducing M explicitly. The Two-Phase Method For the radiation therapy example just solved in Table 4.12, recall its real objective function Real problem:

Minimize

Z 0.4x1 0.5x2.

4.6 ADAPTING TO OTHER MODEL FORMS

x2

Constraints for the artificial problem:

Z 6 1.2M

(0, 12)

143

0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 ( holds when x4 0) (0.6x1 0.4x2 6 when x6 0) x1 0, x2 0 (x4 0, x6 0)

Z 5.4 This dark line segment is the feasible region for the real problem (x4 0, x6 0).

(6, 6)

(7.5, 4.5) optimal 3

Z 5.25 (8, 3)

FIGURE 4.6 This graph shows the feasible region and the sequence of CPF solutions (, , , ) examined by the simplex method (with the Big M method) for the artificial problem that corresponds to the real problem of Fig. 4.5.

Z 4.7 0.5M

2

Feasible region for the artificial problem 0

Z 3.6 2.1M

1

(0, 0) (9, 0)

Z 0 12M

x1

However, the Big M method uses the following objective function (or its equivalent in maximization form) throughout the entire procedure: Big M method:

Minimize

Z 0.4x1 0.5x2 Mx4 Mx6.

Since the first two coefficients are negligible compared to M, the two-phase method is able to drop M by using the following two objective functions with completely different definitions of Z in turn. Two-phase method: Phase 1: Phase 2:

Minimize Minimize

Z x4 x6 Z 0.4x1 0.5x2

(until x4 0, x6 0). (with x4 0, x6 0).

The phase 1 objective function is obtained by dividing the Big M method objective function by M and then dropping the negligible terms. Since phase 1 concludes by obtaining

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a BF solution for the real problem (one where x4 0 and x6 0), this solution is then used as the initial BF solution for applying the simplex method to the real problem (with its real objective function) in phase 2. Before solving the example in this way, we summarize the general method. Summary of the Two-Phase Method. Initialization: Revise the constraints of the original problem by introducing artificial variables as needed to obtain an obvious initial BF solution for the artificial problem. Phase 1: The objective for this phase is to find a BF solution for the real problem. To do this, Minimize Z artificial variables, subject to revised constraints. The optimal solution obtained for this problem (with Z 0) will be a BF solution for the real problem. Phase 2: The objective for this phase is to find an optimal solution for the real problem. Since the artificial variables are not part of the real problem, these variables can now be dropped (they are all zero now anyway).1 Starting from the BF solution obtained at the end of phase 1, use the simplex method to solve the real problem. For the example, the problems to be solved by the simplex method in the respective phases are summarized below. Phase 1 Problem (Radiation Therapy Example): Minimize

Z x4 x6,

subject to 0.3x1 0.1x2 x3 2.7 0.5x1 0.5x2 x4 6 0.6x1 0.4x2 x5 x6 6 and x1 0,

x2 0,

x3 0,

x4 0,

x5 0,

x6 0.

Phase 2 Problem (Radiation Therapy Example): Minimize

Z 0.4x1 0.5x2,

subject to 0.3x1 0.1x2 x3 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 x5 6 and x1 0,

x2 0,

x3 0,

x5 0.

We are skipping over three other possibilities here: (1) artificial variables 0 (discussed in the next subsection), (2) artificial variables that are degenerate basic variables, and (3) retaining the artificial variables as nonbasic variables in phase 2 (and not allowing them to become basic) as an aid to subsequent postoptimality analysis. Your OR Courseware allows you to explore these possibilities. 1

4.6 ADAPTING TO OTHER MODEL FORMS

145

The only differences between these two problems are in the objective function and in the inclusion (phase 1) or exclusion (phase 2) of the artificial variables x4 and x6. Without the artificial variables, the phase 2 problem does not have an obvious initial BF solution. The sole purpose of solving the phase 1 problem is to obtain a BF solution with x4 0 and x6 0 so that this solution (without the artificial variables) can be used as the initial BF solution for phase 2. Table 4.13 shows the result of applying the simplex method to this phase 1 problem. [Row 0 in the initial tableau is obtained by converting Minimize Z x4 x6 to Maximize (Z) x4 x6 and then using elementary row operations to eliminate the basic variables x4 and x6 from Z x4 x6 0.] In the next-to-last tableau, there is a tie for the entering basic variable between x3 and x5, which is broken arbitrarily in favor of x3. The solution obtained at the end of phase 1, then, is (x1, x2, x3, x4, x5, x6) (6, 6, 0.3, 0, 0, 0) or, after x4 and x6 are dropped, (x1, x2, x3, x5) (6, 6, 0.3, 0). As claimed in the summary, this solution from phase 1 is indeed a BF solution for the real problem (the phase 2 problem) because it is the solution (after you set x5 0) to the system of equations consisting of the three functional constraints for the phase 2 problem. In fact, after deleting the x4 and x6 columns as well as row 0 for each iteration, Table TABLE 4.13 Phase 1 of the two-phase method for the radiation therapy example Coefficient of: Iteration

0

Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Right Side

Z x3 x4 x6

(0) (1) (2) (3)

1 0 0 0

1.1 0.3 0.5 0.6

0.9 0.1 0.5 0.4

00 01 00 00

0 0 1 0

1 0 0 1

0 0 0 1

12 2.7 6.0 6.0

Z

(0)

1

0.0

16 30

11 3

0

1

0

2.1

x1

(1)

0

1.0

1 3

10 3

0

0

0

9.0

x4

(2)

0

0.0

0

0

1.5

(3)

0

0.0

5 3 2

1

x6

1 3 0.2

0

1

1

0.6

Z

(0)

1

0.0

0.0

5 3

0

5 3

8 3

0.5

x1

(1)

0

1.0

0.0

20 3

0

5 3

5 3

8.0

x4

(2)

0

0.0

0.0

1

(3)

0

0.0

1.0

5 3 5

5 3 5

0.5

x2

5 3 10

Z x1

(0) (1)

1 0

0.0 1.0

0.0 0.0

00 00

0 5

1 5

0.0 6.0

x3

(2)

0

0.0

0.0

01

1

1

0.3

x2

(3)

0

0.0

1.0

00

1 4 3 5 6

5

5

6.0

1

2

3

0

3.0

146

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4.13 shows one way of using Gaussian elimination to solve this system of equations by reducing the system to the form displayed in the final tableau. Table 4.14 shows the preparations for beginning phase 2 after phase 1 is completed. Starting from the final tableau in Table 4.13, we drop the artificial variables (x4 and x6), substitute the phase 2 objective function (Z 0.4x1 0.5x2 in maximization form) into row 0, and then restore the proper form from Gaussian elimination (by algebraically eliminating the basic variables x1 and x2 from row 0). Thus, row 0 in the last tableau is obtained by performing the following elementary row operations in the next-to-last tableau: from row 0 subtract both the product, 0.4 times row 1, and the product, 0.5 times row 3. Except for the deletion of the two columns, note that rows 1 to 3 never change. The only adjustments occur in row 0 in order to replace the phase 1 objective function by the phase 2 objective function. The last tableau in Table 4.14 is the initial tableau for applying the simplex method to the phase 2 problem, as shown at the top of Table 4.15. Just one iteration then leads to the optimal solution shown in the second tableau: (x1, x2, x3, x5) (7.5, 4.5, 0, 0.3). This solution is the desired optimal solution for the real problem of interest rather than the artificial problem constructed for phase 1. Now we see what the two-phase method has done graphically in Fig. 4.7. Starting at the origin, phase 1 examines a total of four CPF solutions for the artificial problem. The first three actually were corner-point infeasible solutions for the real problem shown in Fig. 4.5. The fourth CPF solution, at (6, 6), is the first one that also is feasible for the real TABLE 4.14 Preparing to begin phase 2 for the radiation therapy example Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Right Side

Z x1

(0) (1)

1 0

00. 10.

0.0 0.0

0 0

0.0 5.0

1 5

0.0 6.0

x3

(2)

0

00.

0.0

1

1.0

1

0.3

x2

(3)

0

00.

1.0

0

1 4 3 5 6

5.0

5

6.0

Z x1

(0) (1)

1 0

00. 10.

0.0 0.0

0 0

0.0 5.0

0.0 6.0

x3 x2

(2) (3)

0 0

00. 00.

0.0 1.0

1 0

1.0 5.0

0.3 6.0

Substitute phase 2 objective function

Z x1 x3 x2

(0) (1) (2) (3)

1 0 0 0

0.4 10. 00. 00.

0.5 0.0 0.0 1.0

0 0 1 0

0.0 5.0 1.0 5.0

0.0 6.0 0.3 6.0

Restore proper form from Gaussian elimination

Z x1 x3 x2

(0) (1) (2) (3)

1 0 0 0

00. 10. 00. 00.

0.0 0.0 0.0 1.0

0 0 1 0

0.5 5.0 1.0 5.0

5.4 6.0 0.3 6.0

Final Phase 1 tableau

Drop x4 and x6

4.6 ADAPTING TO OTHER MODEL FORMS

147

TABLE 4.15 Phase 2 of the two-phase method for the radiation therapy example Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x5

Right Side

0

Z x1 x3 x2

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

0 0 0 1

0.0 0.0 1.0 0.0

0.5 5.0 1.0 5.0

5.40 6.00 0.30 6.00

1

Z x1 x5 x2

(0) (1) (2) (3)

1 0 0 0

0 1 0 0

0 0 0 1

0.5 5.0 1.0 5.0

0.0 0.0 1.0 0.0

5.25 7.50 0.30 4.50

Iteration

FIGURE 4.7 This graph shows the sequence of CPF solutions for phase 1 (, , , ) and then for phase 2 ( 0 , 1 ) when the two-phase method is applied to the radiation therapy example.

x2 (0, 12)

(6, 6) This dark line segment is the feasible region for the real problem (phase 2).

0 3

1 (7.5, 4.5) optimal Feasible region for the artificial problem (phase 1)

0

2

(8, 3)

1

(0, 0) (9, 0)

x1

148

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

problem, so it becomes the initial CPF solution for phase 2. One iteration in phase 2 leads to the optimal CPF solution at (7.5, 4.5). If the tie for the entering basic variable in the next-to-last tableau of Table 4.13 had been broken in the other way, then phase 1 would have gone directly from (8, 3) to (7.5, 4.5). After (7.5, 4.5) was used to set up the initial simplex tableau for phase 2, the optimality test would have revealed that this solution was optimal, so no iterations would be done. It is interesting to compare the Big M and two-phase methods. Begin with their objective functions. Big M Method: Minimize

Z 0.4x1 0.5x2 Mx4 Mxx6.

Two-Phase Method: Phase 1: Phase 2:

Minimize Minimize

Z x4 x6. Z 0.4x1 0.5x2.

Because the Mx4 and Mx6 terms dominate the 0.4x1 and 0.5x2 terms in the objective function for the Big M method, this objective function is essentially equivalent to the phase 1 objective function as long as x4 and/or x6 is greater than zero. Then, when both x4 0 and x6 0, the objective function for the Big M method becomes completely equivalent to the phase 2 objective function. Because of these virtual equivalencies in objective functions, the Big M and twophase methods generally have the same sequence of BF solutions. The one possible exception occurs when there is a tie for the entering basic variable in phase 1 of the twophase method, as happened in the third tableau of Table 4.13. Notice that the first three tableaux of Tables 4.12 and 4.13 are almost identical, with the only difference being that the multiplicative factors of M in Table 4.12 become the sole quantities in the corresponding spots in Table 4.13. Consequently, the additive terms that broke the tie for the entering basic variable in the third tableau of Table 4.12 were not present to break this same tie in Table 4.13. The result for this example was an extra iteration for the two-phase method. Generally, however, the advantage of having the additive factors is minimal. The two-phase method streamlines the Big M method by using only the multiplicative factors in phase 1 and by dropping the artificial variables in phase 2. (The Big M method could combine the multiplicative and additive factors by assigning an actual huge number to M, but this might create numerical instability problems.) For these reasons, the two-phase method is commonly used in computer codes. No Feasible Solutions So far in this section we have been concerned primarily with the fundamental problem of identifying an initial BF solution when an obvious one is not available. You have seen how the artificial-variable technique can be used to construct an artificial problem and obtain an initial BF solution for this artificial problem instead. Use of either the Big M method or the two-phase method then enables the simplex method to begin its pilgrim-

4.6 ADAPTING TO OTHER MODEL FORMS

149

age toward the BF solutions, and ultimately toward the optimal solution, for the real problem. However, you should be wary of a certain pitfall with this approach. There may be no obvious choice for the initial BF solution for the very good reason that there are no feasible solutions at all! Nevertheless, by constructing an artificial feasible solution, there is nothing to prevent the simplex method from proceeding as usual and ultimately reporting a supposedly optimal solution. Fortunately, the artificial-variable technique provides the following signpost to indicate when this has happened: If the original problem has no feasible solutions, then either the Big M method or phase 1 of the two-phase method yields a final solution that has at least one artificial variable greater than zero. Otherwise, they all equal zero.

To illustrate, let us change the first constraint in the radiation therapy example (see Fig. 4.5) as follows: 0.3x1 0.1x2 2.7

0.3x1 0.1x2 1.8,

so that the problem no longer has any feasible solutions. Applying the Big M method just as before (see Table 4.12) yields the tableaux shown in Table 4.16. (Phase 1 of the twophase method yields the same tableaux except that each expression involving M is replaced by just the multiplicative factor.) Hence, the Big M method normally would be indicating that the optimal solution is (3, 9, 0, 0, 0, 0.6). However, since an artificial variable x6 0.6 0, the real message here is that the problem has no feasible solutions.

TABLE 4.16 The Big M method for the revision of the radiation therapy example that has no feasible solutions Coefficient of: Iteration

0

Basic Variable

Eq.

Z

x1

x2

x3

x 4

x5

x 6

Z x3 x4 x6

(0) (1) (2) (3)

1 0 0 0

1.1M 0.4 0.3 0.5 0.6

0.9M 0.5 0.1 0.5 0.4

0 1 0 0

0.0 0.0 1.0 0.0

M 0 0 1

0 0 0 1

Z

(0)

1

0.0

16 11 M 30 30

11 4 M 3 3

0.0

M

0

5.4M 2.4

x1

(1)

0

1.0

1 3

10 3

0.0

0

0

6.0

x4

(2)

0

0.0

0

0

3.0

(3)

0

0.0

5 3 2

1.0

x6

1 3 0.2

0.0

1

1

2.4

Z x1 x2 x6

(0) (1) (2) (3)

1 0 0 0

0.0 1.0 0.0 0.0

0.0 0.0 1.0 0.0

M 0.5 5 5 1

1.6M 1.1 1.0 3.0 0.6

M 0 0 1

0 0 0 1

0.6M 5.7 3.0 9.0 0.6

1

2

Right Side 12M 1.8 6.0 6.0

150

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

Variables Allowed to Be Negative In most practical problems, negative values for the decision variables would have no physical meaning, so it is necessary to include nonnegativity constraints in the formulations of their linear programming models. However, this is not always the case. To illustrate, suppose that the Wyndor Glass Co. problem is changed so that product 1 already is in production, and the first decision variable x1 represents the increase in its production rate. Therefore, a negative value of x1 would indicate that product 1 is to be cut back by that amount. Such reductions might be desirable to allow a larger production rate for the new, more profitable product 2, so negative values should be allowed for x1 in the model. Since the procedure for determining the leaving basic variable requires that all the variables have nonnegativity constraints, any problem containing variables allowed to be negative must be converted to an equivalent problem involving only nonnegative variables before the simplex method is applied. Fortunately, this conversion can be done. The modification required for each variable depends upon whether it has a (negative) lower bound on the values allowed. Each of these two cases is now discussed. Variables with a Bound on the Negative Values Allowed. Consider any decision variable xj that is allowed to have negative values which satisfy a constraint of the form xj Lj, where Lj is some negative constant. This constraint can be converted to a nonnegativity constraint by making the change of variables x j xj Lj,

so

x j 0.

Thus, x j Lj would be substituted for xj throughout the model, so that the redefined decision variable x j cannot be negative. (This same technique can be used when Lj is positive to convert a functional constraint xj Lj to a nonnegativity constraint x j 0.) To illustrate, suppose that the current production rate for product 1 in the Wyndor Glass Co. problem is 10. With the definition of x1 just given, the complete model at this point is the same as that given in Sec. 3.1 except that the nonnegativity constraint x1 0 is replaced by x1 10. To obtain the equivalent model needed for the simplex method, this decision variable would be redefined as the total production rate of product 1 x j x1 10, which yields the changes in the objective function and constraints as shown: Z 3x1 5x2 3x1 2x2 4 3x1 2x2 12 3x1 2x2 18 x1 10, x2 0

Z 3(x 1 10) 5x2 3(x 1 10) 2x2 4 3(x 1 10) 2x2 12 3(x 1 10) 2x2 18 x 1 10 10, x2 0

Z 30 3x 1 5x2 2x 1 2x2 14 3x 1 2x2 12 3x 1 2x2 48 x 1 0, x2 0

4.6 ADAPTING TO OTHER MODEL FORMS

151

Variables with No Bound on the Negative Values Allowed. In the case where xj does not have a lower-bound constraint in the model formulated, another approach is required: xj is replaced throughout the model by the difference of two new nonnegative variables xj x j xj ,

where x j 0, xj 0.

Since x j and xj can have any nonnegative values, this difference xj xj can have any value (positive or negative), so it is a legitimate substitute for xj in the model. But after such substitutions, the simplex method can proceed with just nonnegative variables. The new variables x j and xj have a simple interpretation. As explained in the next paragraph, each BF solution for the new form of the model necessarily has the property that either x j 0 or xj 0 (or both). Therefore, at the optimal solution obtained by the simplex method (a BF solution),

0 x 0

x j x j

xj

j

if xj 0, otherwise; if xj 0, otherwise;

so that x j represents the positive part of the decision variable xj and xj its negative part (as suggested by the superscripts). For example, if xj 10, the above expressions give x j 10 and xj 0. This same value of xj xj xj 10 also would occur with larger values of xj and x j such that x x 10. Plotting these values of x and x on a two-dimensional graph gives a line j j j j with an endpoint at x 10, x 0 to avoid violating the nonnegativity constraints. This j j endpoint is the only corner-point solution on the line. Therefore, only this endpoint can be part of an overall CPF solution or BF solution involving all the variables of the model. This illustrates why each BF solution necessarily has either x j 0 or xj 0 (or both). To illustrate the use of the xj and xj , let us return to the example on the preceding page where x1 is redefined as the increase over the current production rate of 10 for product 1 in the Wyndor Glass Co. problem. However, now suppose that the x1 10 constraint was not included in the original model because it clearly would not change the optimal solution. (In some problems, certain variables do not need explicit lower-bound constraints because the functional constraints already prevent lower values.) Therefore, before the simplex method is applied, x1 would be replaced by the difference

x1 x1 x1,

where x1 0, x1 0,

as shown: Maximize subject to

Z 3x1 5x2, Z 3x1 5x2 4 2x2 12 3x1 2x2 18 x2 0 (only)

Maximize subject to

Z 3x1 3x1 5x2, Z 3x1 3x1 5x2 4 2x2 12 3x1 3x1 2x2 18 x1 0, x1 0, x2 0

152

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

From a computational viewpoint, this approach has the disadvantage that the new equivalent model to be used has more variables than the original model. In fact, if all the original variables lack lower-bound constraints, the new model will have twice as many variables. Fortunately, the approach can be modified slightly so that the number of variables is increased by only one, regardless of how many original variables need to be replaced. This modification is done by replacing each such variable xj by xj x j x,

where x j 0, x 0,

instead, where x is the same variable for all relevant j. The interpretation of x in this case is that x is the current value of the largest (in absolute terms) negative original variable, so that x j is the amount by which xj exceeds this value. Thus, the simplex method now can make some of the x j variables larger than zero even when x 0.

4.7

POSTOPTIMALITY ANALYSIS We stressed in Secs. 2.3, 2.4, and 2.5 that postoptimality analysis—the analysis done after an optimal solution is obtained for the initial version of the model—constitutes a very major and very important part of most operations research studies. The fact that postoptimality analysis is very important is particularly true for typical linear programming applications. In this section, we focus on the role of the simplex method in performing this analysis. Table 4.17 summarizes the typical steps in postoptimality analysis for linear programming studies. The rightmost column identifies some algorithmic techniques that involve the simplex method. These techniques are introduced briefly here with the technical details deferred to later chapters. Reoptimization As discussed in Sec. 3.7, linear programming models that arise in practice commonly are very large, with hundreds or thousands of functional constraints and decision variables. In such cases, many variations of the basic model may be of interest for considering different scenarios. Therefore, after having found an optimal solution for one version of a linear programming model, we frequently must solve again (often many times) for the soTABLE 4.17 Postoptimality analysis for linear programming Task

Purpose

Technique

Model debugging Model validation Final managerial decisions on resource allocations (the bi values) Evaluate estimates of model parameters Evaluate trade-offs between model parameters

Find errors and weaknesses in model Demonstrate validity of final model Make appropriate division of organizational resources between activities under study and other important activities Determine crucial estimates that may affect optimal solution for further study Determine best trade-off

Reoptimization See Sec. 2.4 Shadow prices

Sensitivity analysis Parametric linear programming

4.7 POSTOPTIMALITY ANALYSIS

153

lution of a slightly different version of the model. We nearly always have to solve again several times during the model debugging stage (described in Secs. 2.3 and 2.4), and we usually have to do so a large number of times during the later stages of postoptimality analysis as well. One approach is simply to reapply the simplex method from scratch for each new version of the model, even though each run may require hundreds or even thousands of iterations for large problems. However, a much more efficient approach is to reoptimize. Reoptimization involves deducing how changes in the model get carried along to the final simplex tableau (as described in Secs. 5.3 and 6.6). This revised tableau and the optimal solution for the prior model are then used as the initial tableau and the initial basic solution for solving the new model. If this solution is feasible for the new model, then the simplex method is applied in the usual way, starting from this initial BF solution. If the solution is not feasible, a related algorithm called the dual simplex method (described in Sec. 7.1) probably can be applied to find the new optimal solution,1 starting from this initial basic solution. The big advantage of this reoptimization technique over re-solving from scratch is that an optimal solution for the revised model probably is going to be much closer to the prior optimal solution than to an initial BF solution constructed in the usual way for the simplex method. Therefore, assuming that the model revisions were modest, only a few iterations should be required to reoptimize instead of the hundreds or thousands that may be required when you start from scratch. In fact, the optimal solutions for the prior and revised models are frequently the same, in which case the reoptimization technique requires only one application of the optimality test and no iterations. Shadow Prices Recall that linear programming problems often can be interpreted as allocating resources to activities. In particular, when the functional constraints are in form, we interpreted the bi (the right-hand sides) as the amounts of the respective resources being made available for the activities under consideration. In many cases, there may be some latitude in the amounts that will be made available. If so, the bi values used in the initial (validated) model actually may represent management’s tentative initial decision on how much of the organization’s resources will be provided to the activities considered in the model instead of to other important activities under the purview of management. From this broader perspective, some of the bi values can be increased in a revised model, but only if a sufficiently strong case can be made to management that this revision would be beneficial. Consequently, information on the economic contribution of the resources to the measure of performance (Z ) for the current study often would be extremely useful. The simplex method provides this information in the form of shadow prices for the respective resources. The shadow price for resource i (denoted by y*i ) measures the marginal value of this resource, i.e., the rate at which Z could be increased by (slightly) increasing the amount of 1

The one requirement for using the dual simplex method here is that the optimality test is still passed when applied to row 0 of the revised final tableau. If not, then still another algorithm called the primal-dual method can be used instead.

154

4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

this resource (bi) being made available.1,2 The simplex method identifies this shadow price by yi* coefficient of the ith slack variable in row 0 of the final simplex tableau.

To illustrate, for the Wyndor Glass Co. problem, Resource i production capacity of Plant i (i 1, 2, 3) being made available to the two new products under consideration, bi hours of production time per week being made available in Plant i for these new products. Providing a substantial amount of production time for the new products would require adjusting production times for the current products, so choosing the bi value is a difficult managerial decision. The tentative initial decision has been b1 4,

b2 12,

b3 18,

as reflected in the basic model considered in Sec. 3.1 and in this chapter. However, management now wishes to evaluate the effect of changing any of the bi values. The shadow prices for these three resources provide just the information that management needs. The final tableau in Table 4.8 (see p. 128) yields y*1 0 shadow price for resource 1, 3 y*2 shadow price for resource 2, 2 y*3 1 shadow price for resource 3. With just two decision variables, these numbers can be verified by checking graphically that individually increasing any bi by 1 indeed would increase the optimal value of Z by y*i . For example, Fig. 4.8 demonstrates this increase for resource 2 by reapplying the graphical method presented in Sec. 3.1. The optimal solution, (2, 6) with Z 36, changes to (53, 123) with Z 3712 when b2 is increased by 1 (from 12 to 13), so that 1 3 y*2 Z 37 36 . 2 2 Since Z is expressed in thousands of dollars of profit per week, y*2 32 indicates that adding 1 more hour of production time per week in Plant 2 for these two new products would increase their total profit by $1,500 per week. Should this actually be done? It depends on the marginal profitability of other products currently using this production time. If there is a current product that contributes less than $1,500 of weekly profit per hour of weekly production time in Plant 2, then some shift of production time to the new products would be worthwhile. We shall continue this story in Sec. 6.7, where the Wyndor OR team uses shadow prices as part of its sensitivity analysis of the model. 1

The increase in bi must be sufficiently small that the current set of basic variables remains optimal since this rate (marginal value) changes if the set of basic variables changes. 2 In the case of a functional constraint in or form, its shadow price is again defined as the rate at which Z could be increased by (slightly) increasing the value of bi, although the interpretation of bi now would normally be something other than the amount of a resource being made available.

4.7 POSTOPTIMALITY ANALYSIS

155

x2 3x1 2x2 18

Z 3x1 5x2

8 5 , 13 3 2

6 (2, 6) FIGURE 4.8 This graph shows that the shadow price is y2* 32 for resource 2 for the Wyndor Glass Co. problem. The two dots are the optimal solutions for b2 12 or b2 13, and plugging these solutions into the objective function reveals that increasing b2 by 1 increases Z by y2* 32.

5

13

Z 3 3 5 2 37 12 Z 3( 2 ) 5( 6 ) 36

2x2 13 2x2 12

Z

3 2

y*2

x1 4 4

2

0

2

4

6

x1

Figure 4.8 demonstrates that y*2 32 is the rate at which Z could be increased by increasing b2 slightly. However, it also demonstrates the common phenomenon that this interpretation holds only for a small increase in b2. Once b2 is increased beyond 18, the optimal solution stays at (0, 9) with no further increase in Z. (At that point, the set of basic variables in the optimal solution has changed, so a new final simplex tableau will be obtained with new shadow prices, including y*2 0.) Now note in Fig. 4.8 why y*1 0. Because the constraint on resource 1, x1 4, is not binding on the optimal solution (2, 6), there is a surplus of this resource. Therefore, increasing b1 beyond 4 cannot yield a new optimal solution with a larger value of Z. By contrast, the constraints on resources 2 and 3, 2x2 12 and 3x1 2x2 18, are binding constraints (constraints that hold with equality at the optimal solution). Because the limited supply of these resources (b2 12, b3 18) binds Z from being increased further, they have positive shadow prices. Economists refer to such resources as scarce goods, whereas resources available in surplus (such as resource 1) are free goods (resources with a zero shadow price). The kind of information provided by shadow prices clearly is valuable to management when it considers reallocations of resources within the organization. It also is very helpful when an increase in bi can be achieved only by going outside the organization to purchase more of the resource in the marketplace. For example, suppose that Z represents profit and that the unit profits of the activities (the cj values) include the costs (at regular prices) of all the resources consumed. Then a positive shadow price of y*i for resource i means that the total profit Z can be increased by y*i by purchasing 1 more unit of this resource at its regular price. Alternatively, if a premium price must be paid for the resource

156

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in the marketplace, then y*i represents the maximum premium (excess over the regular price) that would be worth paying.1 The theoretical foundation for shadow prices is provided by the duality theory described in Chap. 6. Sensitivity Analysis When discussing the certainty assumption for linear programming at the end of Sec. 3.3, we pointed out that the values used for the model parameters (the ai j, bi, and cj identified in Table 3.3) generally are just estimates of quantities whose true values will not become known until the linear programming study is implemented at some time in the future. A main purpose of sensitivity analysis is to identify the sensitive parameters (i.e., those that cannot be changed without changing the optimal solution). The sensitive parameters are the parameters that need to be estimated with special care to minimize the risk of obtaining an erroneous optimal solution. They also will need to be monitored particularly closely as the study is implemented. If it is discovered that the true value of a sensitive parameter differs from its estimated value in the model, this immediately signals a need to change the solution. How are the sensitive parameters identified? In the case of the bi, you have just seen that this information is given by the shadow prices provided by the simplex method. In particular, if y*i 0, then the optimal solution changes if bi is changed, so bi is a sensitive parameter. However, y*i 0 implies that the optimal solution is not sensitive to at least small changes in bi. Consequently, if the value used for bi is an estimate of the amount of the resource that will be available (rather than a managerial decision), then the bi values that need to be monitored more closely are those with positive shadow prices—especially those with large shadow prices. When there are just two variables, the sensitivity of the various parameters can be analyzed graphically. For example, in Fig. 4.9, c1 3 can be changed to any other value from 0 to 7.5 without the optimal solution changing from (2, 6). (The reason is that any value of c1 within this range keeps the slope of Z c1x1 5x2 between the slopes of the lines 2x2 12 and 3x1 2x2 18.) Similarly, if c2 5 is the only parameter changed, it can have any value greater than 2 without affecting the optimal solution. Hence, neither c1 nor c2 is a sensitive parameter. The easiest way to analyze the sensitivity of each of the aij parameters graphically is to check whether the corresponding constraint is binding at the optimal solution. Because x1 4 is not a binding constraint, any sufficiently small change in its coefficients (a11 1, a12 0) is not going to change the optimal solution, so these are not sensitive parameters. On the other hand, both 2x2 12 and 3x1 2x2 18 are binding constraints, so changing any one of their coefficients (a21 0, a22 2, a31 3, a32 2) is going to change the optimal solution, and therefore these are sensitive parameters. Typically, greater attention is given to performing sensitivity analysis on the bi and cj parameters than on the aij parameters. On real problems with hundreds or thousands of constraints and variables, the effect of changing one aij value is usually negligible, but 1

If the unit profits do not include the costs of the resources consumed, then y*i represents the maximum total unit price that would be worth paying to increase bi.

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157

x2 10

FIGURE 4.9 This graph demonstrates the sensitivity analysis of c1 and c2 for the Wyndor Glass Co. problem. Starting with the original objective function line [where c1 3, c2 5, and the optimal solution is (2, 6)], the other two lines show the extremes of how much the slope of the objective function line can change and still retain (2, 6) as an optimal solution. Thus, with c2 5, the allowable range for c1 is 0 c1 7.5. With c1 3, the allowable range for c2 is c2 2.

8 Z 36 3x1 5x2

Z 45 7.5x1 5x2 (or Z 18 3x1 2x2) (2, 6) optimal

Z 30 0x1 5x2

4

Feasible region

2

0

2

4

6

x1

changing one bi or cj value can have real impact. Furthermore, in many cases, the ai j values are determined by the technology being used (the aij values are sometimes called technological coefficients), so there may be relatively little (or no) uncertainty about their final values. This is fortunate, because there are far more aij parameters than bi and cj parameters for large problems. For problems with more than two (or possibly three) decision variables, you cannot analyze the sensitivity of the parameters graphically as was just done for the Wyndor Glass Co. problem. However, you can extract the same kind of information from the simplex method. Getting this information requires using the fundamental insight described in Sec. 5.3 to deduce the changes that get carried along to the final simplex tableau as a result of changing the value of a parameter in the original model. The rest of the procedure is described and illustrated in Secs. 6.6 and 6.7. Using Excel to Generate Sensitivity Analysis Information Sensitivity analysis normally is incorporated into software packages based on the simplex method. For example, the Excel Solver will generate sensitivity analysis information upon request. As was shown in Fig. 3.19 (see page 72), when the Solver gives the message that it has found a solution, it also gives on the right a list of three reports that can be provided. By selecting the second one (labeled “Sensitivity”) after solving the Wyndor Glass Co. problem, you will obtain the sensitivity report shown in Fig. 4.10. The upper table in this report provides sensitivity analysis information about the decision variables and their coefficients in the objective function. The lower table does the same for the functional constraints and their right-hand sides.

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FIGURE 4.10 The sensitivity report provided by the Excel Solver for the Wyndor Glass Co. problem.

Look first at the upper table in this figure. The “Final Value” column indicates the optimal solution. The next column gives the reduced costs. (We will not discuss these reduced costs now because the information they provide can also be gleaned from the rest of the upper table.) The next three columns provide the information needed to identify the allowable range to stay optimal for each coefficient cj in the objective function. For any cj, its allowable range to stay optimal is the range of values for this coefficient over which the current optimal solution remains optimal, assuming no change in the other coefficients.

The “Objective Coefficient” column gives the current value of each coefficient, and then the next two columns give the allowable increase and the allowable decrease from this value to remain within the allowable range. Therefore, 3 3 c1 3 4.5,

so

0 c1 7.5

is the allowable range for c1 over which the current optimal solution will stay optimal (assuming c2 5), just as was found graphically in Fig. 4.9. Similarly, since Excel uses 1E 30 (1030) to represent infinity, 5 3 c2 5 ,

so

2 c2

is the allowable range to stay optimal for c2. The fact that both the allowable increase and the allowable decrease are greater than zero for the coefficient of both decision variables provides another useful piece of information, as described below. When the upper table in the sensitivity report generated by the Excel Solver indicates that both the allowable increase and the allowable decrease are greater than zero for every objective coefficient, this is a signpost that the optimal solution in the “Final Value” column is the only optimal solution. Conversely, having any allowable increase or allowable decrease equal to zero is a signpost that there are multiple optimal solutions. Changing the corresponding coefficient a tiny amount beyond the zero allowed and re-solving provides another optimal CPF solution for the original model.

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159

Now consider the lower table in Fig. 4.10 that focuses on sensitivity analysis for the three functional constraints. The “Final Value” column gives the value of each constraint’s left-hand side for the optimal solution. The next two columns give the shadow price and the current value of the right-hand side (bi) for each constraint. When just one bi value is then changed, the last two columns give the allowable increase or allowable decrease in order to remain within its allowable range to stay feasible. For any bi, its allowable range to stay feasible is the range of values for this right-hand side over which the current optimal BF solution (with adjusted values1 for the basic variables) remains feasible, assuming no change in the other right-hand sides.

Thus, using the lower table in Fig. 4.10, combining the last two columns with the current values of the right-hand sides gives the following allowable ranges to stay feasible: 2 b1 6 b2 18 12 b3 24. This sensitivity report generated by the Excel Solver is typical of the sensitivity analysis information provided by linear programming software packages. You will see in Appendix 4.1 that LINDO provides essentially the same report. MPL/CPLEX does also when it is requested with the Solution File dialogue box. Once again, this information obtained algebraically also can be derived from graphical analysis for this two-variable problem. (See Prob. 4.7-1.) For example, when b2 is increased from 12 in Fig. 4.8, the originally optimal CPF solution at the intersection of two constraint boundaries 2x2 b2 and 3x1 2x2 18 will remain feasible (including x1 0) only for b2 18. The latter part of Chap. 6 will delve into this type of analysis more deeply. Parametric Linear Programming Sensitivity analysis involves changing one parameter at a time in the original model to check its effect on the optimal solution. By contrast, parametric linear programming (or parametric programming for short) involves the systematic study of how the optimal solution changes as many of the parameters change simultaneously over some range. This study can provide a very useful extension of sensitivity analysis, e.g., to check the effect of “correlated” parameters that change together due to exogenous factors such as the state of the economy. However, a more important application is the investigation of trade-offs in parameter values. For example, if the cj values represent the unit profits of the respective activities, it may be possible to increase some of the cj values at the expense of decreasing others by an appropriate shifting of personnel and equipment among activities. Similarly, if the bi values represent the amounts of the respective resources being made available, it may be possible to increase some of the bi values by agreeing to accept decreases in some of the others. The analysis of such possibilities is discussed and illustrated at the end of Sec. 6.7. 1

Since the values of the basic variables are obtained as the simultaneous solution of a system of equations (the functional constraints in augmented form), at least some of these values change if one of the right-hand sides changes. However, the adjusted values of the current set of basic variables still will satisfy the nonnegativity constraints, and so still will be feasible, as long as the new value of this right-hand side remains within its allowable range to stay feasible. If the adjusted basic solution is still feasible, it also will still be optimal. We shall elaborate further in Sec. 6.7.

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In some applications, the main purpose of the study is to determine the most appropriate trade-off between two basic factors, such as costs and benefits. The usual approach is to express one of these factors in the objective function (e.g., minimize total cost) and incorporate the other into the constraints (e.g., benefits minimum acceptable level), as was done for the Nori & Leets Co. air pollution problem in Sec. 3.4. Parametric linear programming then enables systematic investigation of what happens when the initial tentative decision on the trade-off (e.g., the minimum acceptable level for the benefits) is changed by improving one factor at the expense of the other. The algorithmic technique for parametric linear programming is a natural extension of that for sensitivity analysis, so it, too, is based on the simplex method. The procedure is described in Sec. 7.2.

4.8

COMPUTER IMPLEMENTATION If the electronic computer had never been invented, undoubtedly you would have never heard of linear programming and the simplex method. Even though it is possible to apply the simplex method by hand to solve tiny linear programming problems, the calculations involved are just too tedious to do this on a routine basis. However, the simplex method is ideally suited for execution on a computer. It is the computer revolution that has made possible the widespread application of linear programming in recent decades. Implementation of the Simplex Method Computer codes for the simplex method now are widely available for essentially all modern computer systems. These codes commonly are part of a sophisticated software package for mathematical programming that includes many of the procedures described in subsequent chapters (including those used for postoptimality analysis). These production computer codes do not closely follow either the algebraic form or the tabular form of the simplex method presented in Secs. 4.3 and 4.4. These forms can be streamlined considerably for computer implementation. Therefore, the codes use instead a matrix form (usually called the revised simplex method) that is especially well suited for the computer. This form accomplishes exactly the same things as the algebraic or tabular form, but it does this while computing and storing only the numbers that are actually needed for the current iteration; and then it carries along the essential data in a more compact form. The revised simplex method is described in Sec. 5.2. The simplex method is used routinely to solve surprisingly large linear programming problems. For example, powerful desktop computers (especially workstations) commonly are used to solve problems with many thousand functional constraints and a larger number of decision variables. We now are beginning to hear reports of successfully solved problems ranging into the hundreds of thousands of functional constraints and millions of decision variables.1 For certain special types of linear programming problems (such as the 1

Do not try this at home. Attacking such a massive problem requires an especially sophisticated linear programming system that uses the latest techniques for exploiting sparcity in the coefficient matrix as well as other special techniques (e.g., crashing techniques for quickly finding an advanced initial BF solution). When problems are re-solved periodically after minor updating of the data, much time often is saved by using (or modifying) the last optimal solution to provide the initial BF solution for the new run.

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161

transportation, assignment, and minimum cost flow problems to be described later in the book), even larger problems now can be solved by specialized versions of the simplex method. Several factors affect how long it will take to solve a linear programming problem by the general simplex method. The most important one is the number of ordinary functional constraints. In fact, computation time tends to be roughly proportional to the cube of this number, so that doubling this number may multiply the computation time by a factor of approximately 8. By contrast, the number of variables is a relatively minor factor.1 Thus, doubling the number of variables probably will not even double the computation time. A third factor of some importance is the density of the table of constraint coefficients (i.e., the proportion of the coefficients that are not zero), because this affects the computation time per iteration. (For large problems encountered in practice, it is common for the density to be under 5 percent, or even under 1 percent, and this much “sparcity” tends to greatly accelerate the simplex method.) One common rule of thumb for the number of iterations is that it tends to be roughly twice the number of functional constraints. With large linear programming problems, it is inevitable that some mistakes and faulty decisions will be made initially in formulating the model and inputting it into the computer. Therefore, as discussed in Sec. 2.4, a thorough process of testing and refining the model (model validation) is needed. The usual end product is not a single static model that is solved once by the simplex method. Instead, the OR team and management typically consider a long series of variations on a basic model (sometimes even thousands of variations) to examine different scenarios as part of postoptimality analysis. This entire process is greatly accelerated when it can be carried out interactively on a desktop computer. And, with the help of both mathematical programming modeling languages and improving computer technology, this now is becoming common practice. Until the mid-1980s, linear programming problems were solved almost exclusively on mainframe computers. Since then, there has been an explosion in the capability of doing linear programming on desktop computers, including personal computers as well as workstations. Workstations, including some with parallel processing capabilities, now are commonly used instead of mainframe computers to solve massive linear programming models. The fastest personal computers are not lagging far behind, although solving huge models usually requires additional memory. Linear Programming Software Featured in This Book A considerable number of excellent software packages for linear programming and its extensions now are available to fill a variety of needs. One that is widely regarded to be a particularly powerful package for solving massive problems is CPLEX, a product of ILOG, Inc., located in Silicon Valley. For more than a decade, CPLEX has helped to lead the way in solving larger and larger linear programming problems. An extensive research and development effort has enabled a series of upgrades with dramatic increases in efficiency. CPLEX 6.5 released in March 1999 provided another order-of-magnitude improvement. This software package has successfully solved real linear programming problems arising in industry with as many as 2 million functional constraints and a comparable number of 1

This statement assumes that the revised simplex method described in Sec. 5.2 is being used.

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decision variables! CPLEX 6.5 often uses the simplex method and its variants (such as the dual simplex method presented in Sec. 7.1) to solve these massive problems. In addition to the simplex method, CPLEX 6.5 also features some other powerful weapons for attacking linear programming problems. One is a lightning-fast algorithm that uses the interior-point approach introduced in the next section. This algorithm can solve some huge general linear programming problems that the simplex method cannot (and vice versa). Another feature is the network simplex method (described in Sec. 9.7) that can solve even larger special types of linear programming problems. CPLEX 6.5 also extends beyond linear programming by including state-of-the-art algorithms for integer programming (Chap. 12) and quadratic programming (Sec. 13.7). Because it often is used to solve really large problems, CPLEX normally is used in conjunction with a mathematical programming modeling language. As described in Sec. 3.7, modeling languages are designed for efficiently formulating large linear programming models (and related models) in a compact way, after which a solver is called upon to solve the model. Several of the prominent modeling languages support CPLEX as a solver. ILOG also has recently introduced its own modeling language, called OPL Studio, that can be used with CPLEX. (A trial version of OPL Studio is available at ILOG’s website, www.ilog.com.) As we mentioned in Sec. 3.7, the student version of CPLEX is included in your OR Courseware as the solver for the MPL modeling language. This version features the simplex method for solving linear programming problems. LINDO (short for Linear, INteractive, and Discrete Optimizer) is another prominent software package for linear programming and its extensions. A product of LINDO Systems based in Chicago, LINDO has an even longer history than CPLEX. Although not as powerful as CPLEX, the largest version of LINDO has solved problems with tens of thousands of functional constraints and hundreds of thousands of decision variables. Its longtime popularity is partially due to its ease of use. For relatively small (textbook-sized) problems, the model can be entered and solved in an intuitive straightforward manner, so LINDO provides a convenient tool for students to use. However, LINDO lacks some of the capabilities of modeling languages for dealing with large linear programming problems. For such problems, it may be more efficient to use the LINGO modeling language to formulate the model and then to call the solver it shares with LINDO to solve the model. You can download the student version of LINDO from the website, www.lindo.com. Appendix 4.1 provides an introduction to how to use LINDO. The CD-ROM also includes a LINDO tutorial, as well as LINDO formulations for all the examples in this book to which it can be applied. Spreadsheet-based solvers are becoming increasingly popular for linear programming and its extensions. Leading the way are the solvers produced by Frontline Systems for Microsoft Excel, Lotus 1-2-3, and Corel Quattro Pro. In addition to the basic solver shipped with these packages, two more powerful upgrades—Premium Solver and Premium Solver Plus—also are available. Because of the widespread use of spreadsheet packages such as Microsoft Excel today, these solvers are introducing large numbers of people to the potential of linear programming for the first time. For textbook-sized linear programming problems (and considerably larger problems as well), spreadsheets provide a convenient way to formulate and solve the model, as described in Sec. 3.6. The more powerful spreadsheet solvers can solve fairly large models with many thousand decision variables. How-

4.9 THE INTERIOR-POINT APPROACH TO SOLVING LINEAR PROGRAMMING PROBLEMS

163

ever, when the spreadsheet grows to an unwieldy size, a good modeling language and its solver may provide a more efficient approach to formulating and solving the model. Spreadsheets provide an excellent communication tool, especially when dealing with typical managers who are very comfortable with this format but not with the algebraic formulations of OR models. Therefore, optimization software packages and modeling languages now can commonly import and export data and results in a spreadsheet format. For example, the MPL modeling language now includes an enhancement (called the OptiMax 2000 Component Library) that enables the modeler to create the feel of a spreadsheet model for the user of the model while still using MPL to formulate the model very efficiently. (The student version of OptiMax 2000 is included in your OR Courseware.) Premium Solver is one of the Excel add-ins included on the CD-ROM. You can install this add-in to obtain a much better performance than with the standard Excel Solver. Consequently, all the software, tutorials, and examples packed on the CD-ROM are providing you with several attractive software options for linear programming. Available Software Options for Linear Programming. 1. Demonstration examples (in OR Tutor) and interactive routines for efficiently learning the simplex method. 2. Excel and its Premium Solver for formulating and solving linear programming models in a spreadsheet format. 3. MPL/CPLEX for efficiently formulating and solving large linear programming models. 4. LINGO and its solver (shared with LINDO) for an alternative way of efficiently formulating and solving large linear programming models. 5. LINDO for formulating and solving linear programming models in a straightforward way. Your instructor may specify which software to use. Whatever the choice, you will be gaining experience with the kind of state-of-the-art software that is used by OR professionals.

4.9

THE INTERIOR-POINT APPROACH TO SOLVING LINEAR PROGRAMMING PROBLEMS The most dramatic new development in operations research during the 1980s was the discovery of the interior-point approach to solving linear programming problems. This discovery was made in 1984 by a young mathematician at AT&T Bell Laboratories, Narendra Karmarkar, when he successfully developed a new algorithm for linear programming with this kind of approach. Although this particular algorithm experienced only mixed success in competing with the simplex method, the key solution concept described below appeared to have great potential for solving huge linear programming problems beyond the reach of the simplex method. Many top researchers subsequently worked on modifying Karmarkar’s algorithm to fully tap this potential. Much progress has been made (and continues to be made), and a number of powerful algorithms using the interior-point approach have been developed. Today, the more powerful software packages that are designed for solving really large linear programming problems (such as CPLEX) include at least one algorithm using the interior-point approach along with the simplex method. As research continues on these algorithms, their computer implementations continue to improve. This has spurred renewed research on the simplex method, and its computer implementations continue to improve as well (recall the dramatic advance by CPLEX 6.5

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cited in the preceding section). The competition between the two approaches for supremacy in solving huge problems is continuing. Now let us look at the key idea behind Karmarkar’s algorithm and its subsequent variants that use the interior-point approach. The Key Solution Concept Although radically different from the simplex method, Karmarkar’s algorithm does share a few of the same characteristics. It is an iterative algorithm. It gets started by identifying a feasible trial solution. At each iteration, it moves from the current trial solution to a better trial solution in the feasible region. It then continues this process until it reaches a trial solution that is (essentially) optimal. The big difference lies in the nature of these trial solutions. For the simplex method, the trial solutions are CPF solutions (or BF solutions after augmenting), so all movement is along edges on the boundary of the feasible region. For Karmarkar’s algorithm, the trial solutions are interior points, i.e., points inside the boundary of the feasible region. For this reason, Karmarkar’s algorithm and its variants are referred to as interior-point algorithms. To illustrate, Fig. 4.11 shows the path followed by the interior-point algorithm in your OR Courseware when it is applied to the Wyndor Glass Co. problem, starting from the

FIGURE 4.11 The curve from (1, 2) to (2, 6) shows a typical path followed by an interior-point algorithm, right through the interior of the feasible region for the Wyndor Glass Co. problem.

x2 (2, 6) optimal

6 (1.56, 5.5)

(1.38, 5)

4

2

0

(1.27, 4)

(1, 2)

2

4

x1

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TABLE 4.18 Output of interior-point algorithm in OR Courseware for Wyndor Glass Co. problem Iteration

x1

x2

Z

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 1.27298 1.37744 1.56291 1.80268 1.92134 1.96639 1.98385 1.99197 1.99599 1.99799 1.999 1.9995 1.99975 1.99987 1.99994

2 4 5 5.5 5.71816 5.82908 5.90595 5.95199 5.97594 5.98796 5.99398 5.99699 5.9985 5.99925 5.99962 5.99981

13 23.8189 29.1323 32.1887 33.9989 34.9094 35.429 35.7115 35.8556 35.9278 35.9639 35.9819 35.991 35.9955 35.9977 35.9989

initial trial solution (1, 2). Note how all the trial solutions (dots) shown on this path are inside the boundary of the feasible region as the path approaches the optimal solution (2, 6). (All the subsequent trial solutions not shown also are inside the boundary of the feasible region.) Contrast this path with the path followed by the simplex method around the boundary of the feasible region from (0, 0) to (0, 6) to (2, 6). Table 4.18 shows the actual output from your OR Courseware for this problem.1 (Try it yourself.) Note how the successive trial solutions keep getting closer and closer to the optimal solution, but never literally get there. However, the deviation becomes so infinitesimally small that the final trial solution can be taken to be the optimal solution for all practical purposes. Section 7.4 presents the details of the specific interior-point algorithm that is implemented in your OR Courseware. Comparison with the Simplex Method One meaningful way of comparing interior-point algorithms with the simplex method is to examine their theoretical properties regarding computational complexity. Karmarkar has proved that the original version of his algorithm is a polynomial time algorithm; i.e., the time required to solve any linear programming problem can be bounded above by a polynomial function of the size of the problem. Pathological counterexamples have been constructed to demonstrate that the simplex method does not possess this property, so it is an exponential time algorithm (i.e., the required time can be bounded above only by an exponential function of the problem size). This difference in worst-case performance 1 The routine is called Solve Automatically by the Interior-Point Algorithm. The option menu provides two choices for a certain parameter of the algorithm (defined in Sec. 7.4). The choice used here is the default value of 0.5.

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is noteworthy. However, it tells us nothing about their comparison in average performance on real problems, which is the more crucial issue. The two basic factors that determine the performance of an algorithm on a real problem are the average computer time per iteration and the number of iterations. Our next comparisons concern these factors. Interior-point algorithms are far more complicated than the simplex method. Considerably more extensive computations are required for each iteration to find the next trial solution. Therefore, the computer time per iteration for an interior-point algorithm is many times longer than that for the simplex method. For fairly small problems, the numbers of iterations needed by an interior-point algorithm and by the simplex method tend to be somewhat comparable. For example, on a problem with 10 functional constraints, roughly 20 iterations would be typical for either kind of algorithm. Consequently, on problems of similar size, the total computer time for an interior-point algorithm will tend to be many times longer than that for the simplex method. On the other hand, a key advantage of interior-point algorithms is that large problems do not require many more iterations than small problems. For example, a problem with 10,000 functional constraints probably will require well under 100 iterations. Even considering the very substantial computer time per iteration needed for a problem of this size, such a small number of iterations makes the problem quite tractable. By contrast, the simplex method might need 20,000 iterations and so might not finish within a reasonable amount of computer time. Therefore, interior-point algorithms often are faster than the simplex method for such huge problems. The reason for this very large difference in the number of iterations on huge problems is the difference in the paths followed. At each iteration, the simplex method moves from the current CPF solution to an adjacent CPF solution along an edge on the boundary of the feasible region. Huge problems have an astronomical number of CPF solutions. The path from the initial CPF solution to an optimal solution may be a very circuitous one around the boundary, taking only a small step each time to the next adjacent CPF solution, so a huge number of steps may be required to reach an optimal solution. By contrast, an interior-point algorithm bypasses all this by shooting through the interior of the feasible region toward an optimal solution. Adding more functional constraints adds more constraint boundaries to the feasible region, but has little effect on the number of trial solutions needed on this path through the interior. This makes it possible for interior-point algorithms to solve problems with a huge number of functional constraints. A final key comparison concerns the ability to perform the various kinds of postoptimality analysis described in Sec. 4.7. The simplex method and its extensions are very well suited to and are widely used for this kind of analysis. Unfortunately, the interiorpoint approach currently has limited capability in this area.1 Given the great importance of postoptimality analysis, this is a crucial drawback of interior-point algorithms. However, we point out next how the simplex method can be combined with the interior-point approach to overcome this drawback. 1

However, research aimed at increasing this capability continues to make progress. For example, see H. J. Greenberg, “Matrix Sensitivity Analysis from an Interior Solution of a Linear Program,” INFORMS Journal on Computing, 11: 316–327, 1999, and its references.

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The Complementary Roles of the Simplex Method and the Interior-Point Approach Ongoing research is continuing to provide substantial improvements in computer implementations of both the simplex method (including its variants) and interior-point algorithms. Therefore, any predictions about their future roles are risky. However, we do summarize below our current assessment of their complementary roles. The simplex method (and its variants) continues to be the standard algorithm for the routine use of linear programming. It continues to be the most efficient algorithm for problems with less than a few hundred functional constraints. It also is the most efficient for some (but not all) problems with up to several thousand functional constraints and nearly an unlimited number of decision variables, so most users are continuing to use the simplex method for such problems. However, as the number of functional constraints increases even further, it becomes increasingly likely that an interior-point approach will be the most efficient, so it often is now used instead. As the size grows into the tens of thousands of functional constraints, the interior-point approach may be the only one capable of solving the problem. However, this certainly is not always the case. As mentioned in the preceding section, the latest state-of-the-art software (CPLEX 6.5) is successfully using the simplex method and its variants to solve some truly massive problems with hundreds of thousands, or even millions of functional constraints and decision variables. These generalizations about how the interior-point approach and the simplex method should compare for various problem sizes will not hold across the board. The specific software packages and computer equipment being used have a major impact. The comparison also is affected considerably by the specific type of linear programming problem being solved. As time goes on, we should learn much more about how to identify specific types which are better suited for one kind of algorithm. One of the by-products of the emergence of the interior-point approach has been a major renewal of efforts to improve the efficiency of computer implementations of the simplex method. As we indicated, impressive progress has been made in recent years, and more lies ahead. At the same time, ongoing research and development of the interior-point approach will further increase its power, and perhaps at a faster rate than for the simplex method. Improving computer technology, such as massive parallel processing (a huge number of computer units operating in parallel on different parts of the same problem), also will substantially increase the size of problem that either kind of algorithm can solve. However, it now appears that the interior-point approach has much greater potential to take advantage of parallel processing than the simplex method does. As discussed earlier, a key disadvantage of the interior-point approach is its limited capability for performing postoptimality analysis. To overcome this drawback, researchers have been developing procedures for switching over to the simplex method after an interior-point algorithm has finished. Recall that the trial solutions obtained by an interior-point algorithm keep getting closer and closer to an optimal solution (the best CPF solution), but never quite get there. Therefore, a switching procedure requires identifying a CPF solution (or BF solution after augmenting) that is very close to the final trial solution. For example, by looking at Fig. 4.11, it is easy to see that the final trial solution in Table 4.18 is very near the CPF solution (2, 6). Unfortunately, on problems with thou-

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sands of decision variables (so no graph is available), identifying a nearby CPF (or BF) solution is a very challenging and time-consuming task. However, good progress has been made in developing procedures to do this. Once this nearby BF solution has been found, the optimality test for the simplex method is applied to check whether this actually is the optimal BF solution. If it is not optimal, some iterations of the simplex method are conducted to move from this BF solution to an optimal solution. Generally, only a very few iterations (perhaps one) are needed because the interior-point algorithm has brought us so close to an optimal solution. Therefore, these iterations should be done quite quickly, even on problems that are too huge to be solved from scratch. After an optimal solution is actually reached, the simplex method and its variants are applied to help perform postoptimality analysis. Because of the difficulties involved in applying a switching procedure (including the extra computer time), some practitioners prefer to just use the simplex method from the outset. This makes good sense when you only occasionally encounter problems that are large enough for an interior-point algorithm to be modestly faster (before switching) than the simplex method. This modest speed-up would not justify both the extra computer time for a switching procedure and the high cost of acquiring (and learning to use) a software package based on the interior-point approach. However, for organizations which frequently must deal with extremely large linear programming problems, acquiring a state-of-the-art software package of this kind (including a switching procedure) probably is worthwhile. For sufficiently huge problems, the only available way of solving them may be with such a package. Applications of huge linear programming models sometimes lead to savings of millions of dollars. Just one such application can pay many times over for a state-of-the-art software package based on the interior-point approach plus switching over to the simplex method at the end.

4.10

CONCLUSIONS The simplex method is an efficient and reliable algorithm for solving linear programming problems. It also provides the basis for performing the various parts of postoptimality analysis very efficiently. Although it has a useful geometric interpretation, the simplex method is an algebraic procedure. At each iteration, it moves from the current BF solution to a better, adjacent BF solution by choosing both an entering basic variable and a leaving basic variable and then using Gaussian elimination to solve a system of linear equations. When the current solution has no adjacent BF solution that is better, the current solution is optimal and the algorithm stops. We presented the full algebraic form of the simplex method to convey its logic, and then we streamlined the method to a more convenient tabular form. To set up for starting the simplex method, it is sometimes necessary to use artificial variables to obtain an initial BF solution for an artificial problem. If so, either the Big M method or the two-phase method is used to ensure that the simplex method obtains an optimal solution for the real problem. Computer implementations of the simplex method and its variants have become so powerful that they now are frequently used to solve linear programming problems with

APPENDIX 4.1

AN INTRODUCTION TO USING LINDO

169

many thousand functional constraints and decision variables, and occasionally vastly larger problems. Interior-point algorithms also provide a powerful tool for solving very large problems.

APPENDIX 4.1

AN INTRODUCTION TO USING LINDO The LINDO software is designed to be easy to learn and to use, especially for small problems of the size you will encounter in this book. In addition to linear programming, it also can be used to solve both integer programming problems (Chap. 12) and quadratic programming problems (Sec. 13.7). Our focus in this appendix is on its use for linear programming. LINDO allows you to enter a model in a straightforward algebraic way. For example, here is a nice way of entering the LINDO model for the Wyndor Glass Co. example introduced in Sec. 3.1. ! Wyndor Glass Co. Problem. LINDO model ! X1 batches of product 1 per week ! X2 batches of product 2 per week ! Profit, in 1000 of dollars MAX Profit) 3 X1 5 X2 Subject to ! Production time Plant1) X1 4 Plant2) 2 X2 12 Plant3) 3 X1 2 X2 18 END In addition to the basic model, this formulation includes several clarifying comments, where each comment is indicated by starting with an exclamation point. Thus, the first three lines give the title and the definitions of the decision variables. The decision variables can be either lowercase or uppercase, but uppercase usually is used so the variables won’t be dwarfed by the following “subscripts.” Another option is to use a suggestive word (or abbreviation of a word), such as the name of the product being produced, to represent the decision variable throughout the model, provided the word does not exceed eight letters. The fifth line of the LINDO formulation indicates that the objective of the model is to maximize the objective function, 3x1 5x2. The word Profit followed by a parenthesis clarifies that this quantity being maximized is profit. The comment on the fourth line further clarifies that the objective function is expressed in units of thousands of dollars. The number 1000 in this comment does not have the usual comma in front of the last three digits because LINDO does not accept commas. (It also does not accept parentheses in algebraic expressions.) The comment on the seventh line points out that the following constraints are on the production times being used. The next three lines start by giving a name (followed by a parenthesis) for each of the functional constraints. These constraints are written in the usual way except for the inequality signs. Because many keyboards do not include and signs, LINDO interprets either or as and either or as . (On systems that include and signs, LINDO will not recognize them.) The end of the constraints is signified by the word END. No nonnegativity constraints are stated because LINDO automatically assumes that all the variables have these constraints. If, say, x1 had not had a nonnegativity constraint, this would have to be indicated by typing FREE X1 on the next line below END.

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To solve this model in the Windows version of LINDO, either select the Solve command from the Solve menu or press the Solve button on the toolbar. On a platform other than Windows, simply type GO followed by a return at the colon prompt. Figure A4.1 shows the resulting solution report delivered by LINDO. Both the top line and bottom line in this figure indicate that an optimal solution was found at iteration 2 of the simplex method. Next comes the value of the objective function for this solution. Below this, we have the values of x1 and x2 for the optimal solution. The column to the right of these values gives the reduced costs. We have not discussed reduced costs in this chapter because the information they provide can also be gleaned from the allowable range to stay optimal for the coefficients in the objective function, and these allowable ranges also are readily available (as you will see in the next figure). When the variable is a basic variable in the optimal solution (as for both variables in the Wyndor problem), its reduced cost automatically is 0. When the variable is a nonbasic variable, its reduced cost provides some interesting information. This variable is 0 because its coefficient in the objective function is too small (when maximizing the objective function) or too large (when minimizing) to justify undertaking the activity represented by the variable. The reduced cost indicates how much this coefficient can be increased (when maximizing) or decreased (when minimizing) before the optimal solution would change and this variable would become a basic variable. However, recall that this same information already is available from the allowable range to stay optimal for the coefficient of this variable in the objective function. The reduced cost (for a nonbasic variable) is just the allowable increase (when maximizing) from the current value of this coefficient to remain within its allowable range to stay optimal or the allowable decrease (when minimizing). Below the variable values and reduced costs in Fig. A4.1, we next have information about the three functional constraints. The Slack or Surplus column gives the difference between the two sides of each constraint. The Dual Prices column gives, by another name, the shadow prices discussed in Sec. 4.7 for these constraints.1 (This alternate name comes from the fact found in Sec. 6.1 that these shadow prices are just the optimal values of the dual variables introduced in Chap. 6.) When LINDO provides you with this solution report, it also asks you whether you want to do range (sensitivity) analysis. Answering yes (by pressing the Y key) provides you with the additional range report shown in Fig. A4.2. This report is identical to the last three columns of the 1

However, beware that LINDO uses a different sign convention from the common one adopted here (see the second footnote for the definition of shadow price in Sec. 4.7), so that for minimization problems, its shadow prices (dual prices) are the negative of ours.

FIGURE A4.1 The solution report provided by LINDO for the Wyndor Glass Co. problem.

LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE Profit) 36.00000 VARIABLE X1 X2 ROW Plant1) Plant2) Plant3)

VALUE 2.000000 6.000000

REDUCED COST .000000 .000000

SLACK OR SURPLUS 2.000000 .000000 .000000

NO. ITERATIONS= 2

DUAL PRICES .000000 1.500000 1.000000

SELECTED REFERENCES

171

RANGES IN WHICH THE BASIS IS UNCHANGED: VARIABLE X1 X2 ROW FIGURE A4.2 The range report provided by LINDO for the Wyndor Glass Co. problem.

Plant1 Plant2 Plant3

OBJ COEFFICIENT RANGES CURRENT ALLOWABLE COEF INCREASE 3.000000 4.500000 5.000000 INFINITY RIGHTHAND SIDE RANGES CURRENT ALLOWABLE RHS INCREASE 4.000000 INFINITY 12.000000 6.000000 18.000000 6.000000

ALLOWABLE DECREASE 3.000000 3.000000 ALLOWABLE DECREASE 2.000000 6.000000 6.000000

tables in the sensitivity report generated by the Excel Solver, as shown earlier in Fig. 4.10. Thus, as already discussed in Sec. 4.7, the first two rows of this range report indicate that the allowable range to stay optimal for each coefficient in the objective function (assuming no other change in the model) is 0 c1 7.5 2 c2 Similarly, the last three rows indicate that the allowable range to stay feasible for each right-hand side (assuming no other change in the model) is 2 b1 6 b2 18 12 b3 24 To print your results with the Windows version of LINDO, you simply need to use the Print command to send the contents of the active window to the printer. If you are running LINDO on a platform other than Windows, you can use the DIVERT command (followed by the file name) to send screen output to a file, which can then print from either the operating system or a word processing package. These are the basics for getting started with LINDO. The LINDO tutorial on the CD-ROM also provides some additional details. The LINGO/LINDO files on the CD-ROM for various chapters show the LINDO formulations for numerous examples. In addition, LINDO includes a Help menu to provide guidance. These resources should enable you to apply LINDO to any linear programming problem you will encounter in this book. (We will discuss applications to other problem types in Chaps. 12 and 13.) For more advanced applications, the LINDO User’s Manual (Selected Reference 4 for this chapter) might be needed.

SELECTED REFERENCES 1. Bazaraa, M. S., J. J. Jarvis, and H. D. Sherali: Linear Programming and Network Flows, 2d ed., Wiley, New York, 1990. 2. Calvert, J. E., and W. L. Voxman: Linear Programming, Harcourt Brace Jovanovich, Orlando, FL, 1989.

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4 SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

3. Dantzig, G.B., and M.N. Thapa: Linear Programming 1: Introduction, Springer, New York, 1997. 4. LINDO User’s Manual, LINDO Systems, Inc., Chicago, IL, e-mail: [email protected], 1999. 5. Vanderbei, R. J.: Linear Programming: Foundations and Extensions, Kluwer Academic Publishers, Boston, MA, 1996.

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE Demonstration Examples in OR Tutor: Interpretation of the Slack Variables Simplex Method—Algebraic Form Simplex Method—Tabular Form

Interactive Routines: Enter or Revise a General Linear Programming Model Set Up for the Simplex Method—Interactive Only Solve Interactively by the Simplex Method

An Automatic Routine: Solve Automatically by the Interior-Point Algorithm

An Excel Add-In: Premium Solver

Files (Chapter 3) for Solving the Wyndor and Radiation Therapy Examples: Excel File LINGO/LINDO File MPL/CPLEX File

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The corresponding demonstration example listed above may be helpful. I: We suggest that you use the corresponding interactive routine listed above (the printout records your work). C: Use the computer with any of the software options available to you (or as instructed by your instructor) to solve the problem automatically. (See Sec. 4.8 for a listing of the options featured in this book and on the CD-ROM.) An asterisk on the problem number indicates that at least a partial answer is given in the back of the book.

4.1-1. Consider the following problem. Z x1 2x2,

Maximize subject to 2 x2 2 x1 x2 3

x1

and x1 0,

x2 0.

(a) Plot the feasible region and circle all the CPF solutions.

CHAPTER 4 PROBLEMS

173

(b) For each CPF solution, identify the pair of constraint boundary equations that it satisfies. (c) For each CPF solution, use this pair of constraint boundary equations to solve algebraically for the values of x1 and x2 at the corner point. (d) For each CPF solution, identify its adjacent CPF solutions. (e) For each pair of adjacent CPF solutions, identify the constraint boundary they share by giving its equation.

The objective is to maximize the total profit from the two activities. The unit profit for activity 1 is $1,000 and the unit profit for activity 2 is $2,000. (a) Calculate the total profit for each CPF solution. Use this information to find an optimal solution. (b) Use the solution concepts of the simplex method given in Sec. 4.1 to identify the sequence of CPF solutions that would be examined by the simplex method to reach an optimal solution.

4.1-2. Consider the following problem.

4.1-4.* Consider the linear programming model (given in the back of the book) that was formulated for Prob. 3.2-3. (a) Use graphical analysis to identify all the corner-point solutions for this model. Label each as either feasible or infeasible. (b) Calculate the value of the objective function for each of the CPF solutions. Use this information to identify an optimal solution. (c) Use the solution concepts of the simplex method given in Sec. 4.1 to identify which sequence of CPF solutions might be examined by the simplex method to reach an optimal solution. (Hint: There are two alternative sequences to be identified for this particular model.)

Z 3x1 2x2,

Maximize subject to 2x1 x2 6 x1 2x2 6 and x1 0,

x2 0.

(a) Use the graphical method to solve this problem. Circle all the corner points on the graph. (b) For each CPF solution, identify the pair of constraint boundary equations it satisfies. (c) For each CPF solution, identify its adjacent CPF solutions. (d) Calculate Z for each CPF solution. Use this information to identify an optimal solution. (e) Describe graphically what the simplex method does step by step to solve the problem. 4.1-3. A certain linear programming model involving two activities has the feasible region shown below.

subject to x1 3x2 8 x1 x2 4 and x1 0,

Maximize

(0, 6 23 )

x2 0.

Z 3x1 2x2,

subject to

6

x1 3x2 4 x1 3x2 15 2x1 x2 10

(5, 5)

and

(6, 4)

4

x1 0,

Feasible region

Maximize (8, 0) 2

x2 0.

4.1-7. Describe graphically what the simplex method does step by step to solve the following problem.

2

0

Z x1 2x2,

Maximize

4.1-6. Repeat Prob. 4.1-4 for the following problem.

8

Level of Activity 2

4.1-5. Repeat Prob. 4.1-4 for the following problem.

4 6 Level of Activity 1

8

Z 2x1 3x2,

subject to 3x1 x2 1 4x1 2x2 20

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4x1 x2 10 x1 2x2 5 and x1 0,

x2 0.

4.1-8. Describe graphically what the simplex method does step by step to solve the following problem. Minimize

Z 5x1 7x2,

subject to 2x1 3x2 42 3x1 4x2 60 x1 x2 18 and x1 0,

x2 0.

(b) For each CPF solution, identify the corresponding BF solution by calculating the values of the slack variables. For each BF solution, use the values of the variables to identify the nonbasic variables and the basic variables. (c) For each BF solution, demonstrate (by plugging in the solution) that, after the nonbasic variables are set equal to zero, this BF solution also is the simultaneous solution of the system of equations obtained in part (a). 4.2-2. Reconsider the model in Prob. 4.1-5. Follow the instructions of Prob. 4.2-1 for parts (a), (b), and (c). (d) Repeat part (b) for the corner-point infeasible solutions and the corresponding basic infeasible solutions. (e) Repeat part (c) for the basic infeasible solutions. 4.2-3. Follow the instructions of Prob. 4.2-1 for the model in Prob. 4.1-6. 4.3-1. Work through the simplex method (in algebraic form) step by step to solve the model in Prob. 4.1-4.

D,I

4.1-9. Label each of the following statements about linear programming problems as true or false, and then justify your answer. (a) For minimization problems, if the objective function evaluated at a CPF solution is no larger than its value at every adjacent CPF solution, then that solution is optimal. (b) Only CPF solutions can be optimal, so the number of optimal solutions cannot exceed the number of CPF solutions. (c) If multiple optimal solutions exist, then an optimal CPF solution may have an adjacent CPF solution that also is optimal (the same value of Z). 4.1-10. The following statements give inaccurate paraphrases of the six solution concepts presented in Sec. 4.1. In each case, explain what is wrong with the statement. (a) The best CPF solution always is an optimal solution. (b) An iteration of the simplex method checks whether the current CPF solution is optimal and, if not, moves to a new CPF solution. (c) Although any CPF solution can be chosen to be the initial CPF solution, the simplex method always chooses the origin. (d) When the simplex method is ready to choose a new CPF solution to move to from the current CPF solution, it only considers adjacent CPF solutions because one of them is likely to be an optimal solution. (e) To choose the new CPF solution to move to from the current CPF solution, the simplex method identifies all the adjacent CPF solutions and determines which one gives the largest rate of improvement in the value of the objective function. 4.2-1. Reconsider the model in Prob. 4.1-4. (a) Introduce slack variables in order to write the functional constraints in augmented form.

4.3-2. Reconsider the model in Prob. 4.1-5. (a) Work through the simplex method (in algebraic form) by hand to solve this model. D,I (b) Repeat part (a) with the corresponding interactive routine in your OR Tutor. C (c) Verify the optimal solution you obtained by using a software package based on the simplex method. 4.3-3. Follow the instructions of Prob. 4.3-2 for the model in Prob. 4.1-6. 4.3-4.* Work through the simplex method (in algebraic form) step by step to solve the following problem.

D,I

Maximize

Z 4x1 3x2 6x3,

subject to 3x1 x2 3x3 30 2x1 2x2 3x3 40 and x1 0,

x2 0,

x3 0.

4.3-5. Work through the simplex method (in algebraic form) step by step to solve the following problem.

D,I

Maximize

Z x1 2x2 4x3,

subject to 3x1 x2 5x3 10 x1 4x2 x3 8 2x1 4x2 2x3 7

CHAPTER 4 PROBLEMS

and x1 0,

x2 0,

x3 0.

4.3-6. Work through the simplex method (in algebraic form) step by step to solve the following problem.

D,I

Maximize

Z x1 2x2 2x3,

subject to 5x1 2x2 3x3 15 x1 4x2 2x3 12 2x1 4x2 x3 8 and x1 0,

x2 0,

x3 0.

4.3-7. Consider the following problem. Maximize

Z 5x1 3x2 4x3,

4.4-2. Repeat Prob. 4.3-2, using the tabular form of the simplex method.

D,I,C

and x2 0,

x3 0.

You are given the information that the nonzero variables in the optimal solution are x2 and x3. (a) Describe how you can use this information to adapt the simplex method to solve this problem in the minimum possible number of iterations (when you start from the usual initial BF solution). Do not actually perform any iterations. (b) Use the procedure developed in part (a) to solve this problem by hand. (Do not use your OR Courseware.) 4.3-8. Consider the following problem. Z 2x1 4x2 3x3,

subject to x1 3x2 2x3 30 x1 x2 x3 24 3x1 5x2 3x3 60 and x1 0,

4.3-9. Label each of the following statements as true or false, and then justify your answer by referring to specific statements (with page citations) in the chapter. (a) The simplex method’s rule for choosing the entering basic variable is used because it always leads to the best adjacent BF solution (largest Z). (b) The simplex method’s minimum ratio rule for choosing the leaving basic variable is used because making another choice with a larger ratio would yield a basic solution that is not feasible. (c) When the simplex method solves for the next BF solution, elementary algebraic operations are used to eliminate each nonbasic variable from all but one equation (its equation) and to give it a coefficient of 1 in that one equation. D,I

2x1 x2 x3 20 3x1 x2 2x3 30

Maximize

(b) Use the procedure developed in part (a) to solve this problem by hand. (Do not use your OR Courseware.)

4.4-1. Repeat Prob. 4.3-1, using the tabular form of the simplex method.

subject to

x1 0,

175

x2 0,

x3 0.

You are given the information that x1 0, x2 0, and x3 0 in the optimal solution. (a) Describe how you can use this information to adapt the simplex method to solve this problem in the minimum possible number of iterations (when you start from the usual initial BF solution). Do not actually perform any iterations.

4.4-3. Repeat Prob. 4.3-3, using the tabular form of the simplex method.

D,I,C

4.4-4. Consider the following problem. Maximize

Z 2x1 x2,

subject to x1 x2 40 4x1 x2 100 and x1 0,

x2 0.

(a) Solve this problem graphically in a freehand manner. Also identify all the CPF solutions. (b) Now repeat part (a) when using a ruler to draw the graph carefully. D (c) Use hand calculations to solve this problem by the simplex method in algebraic form. D,I (d) Now use your OR Courseware to solve this problem interactively by the simplex method in algebraic form. D (e) Use hand calculations to solve this problem by the simplex method in tabular form. D,I (f) Now use your OR Courseware to solve this problem interactively by the simplex method in tabular form. C (g) Use a software package based on the simplex method to solve the problem.

176

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4.4-5. Repeat Prob. 4.4-4 for the following problem. subject to

x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step in algebraic form to solve this problem. D,I (b) Work through the simplex method step by step in tabular form to solve the problem. C (c) Use a computer package based on the simplex method to solve the problem. D,I

x1 2x2 30 x1 x2 20 and x1 0,

and

Z 2x1 3x2,

Maximize

x2 0.

4.4-6. Consider the following problem. Maximize

Z 2x1 4x2 3x3,

4.4-9. Work through the simplex method step by step (in tabular form) to solve the following problem.

D,I

subject to

Maximize

3x1 4x2 2x3 60 2x1 x2 2x3 40 x1 3x2 2x3 80

subject to 3x1 x2 x3 6 x1 x2 2x3 1 x1 x2 x3 2

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step in algebraic form. D,I (b) Work through the simplex method step by step in tabular form. C (c) Use a software package based on the simplex method to solve the problem. D,I

4.4-7. Consider the following problem. Maximize

Z 3x1 5x2 6x3,

and x1 0,

x2 0,

x3 0.

4.4-10. Work through the simplex method step by step to solve the following problem.

D,I

Maximize

Z x1 x2 2x3,

subject to x1 2x2 x3 20 2x1 4x2 2x3 60 2x1 3x2 x3 50

subject to 2x1 x2 x3 x1 2x2 x3 x1 x2 2x3 x1 x2 x3

Z 2x1 x2 x3,

4 4 4 3

and x1 0,

x2 0,

x3 0.

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step in algebraic form. D,I (b) Work through the simplex method in tabular form. C (c) Use a computer package based on the simplex method to solve the problem. D,I

4.4-8. Consider the following problem. Maximize

Z 2x1 x2 x3,

subject to x1 x2 3x3 4 2x1 x2 3x3 10 x1 x2 x3 7

4.5-1. Consider the following statements about linear programming and the simplex method. Label each statement as true or false, and then justify your answer. (a) In a particular iteration of the simplex method, if there is a tie for which variable should be the leaving basic variable, then the next BF solution must have at least one basic variable equal to zero. (b) If there is no leaving basic variable at some iteration, then the problem has no feasible solutions. (c) If at least one of the basic variables has a coefficient of zero in row 0 of the final tableau, then the problem has multiple optimal solutions. (d) If the problem has multiple optimal solutions, then the problem must have a bounded feasible region.

CHAPTER 4 PROBLEMS

4.5-2. Suppose that the following constraints have been provided for a linear programming model with decision variables x1 and x2. x1 3x2 30 3x1 x2 30 and x1 0,

x2 0.

(a) Demonstrate graphically that the feasible region is unbounded. (b) If the objective is to maximize Z x1 x2, does the model have an optimal solution? If so, find it. If not, explain why not. (c) Repeat part (b) when the objective is to maximize Z x1 x2. (d) For objective functions where this model has no optimal solution, does this mean that there are no good solutions according to the model? Explain. What probably went wrong when formulating the model? D,I (e) Select an objective function for which this model has no optimal solution. Then work through the simplex method step by step to demonstrate that Z is unbounded. C (f) For the objective function selected in part (e), use a software package based on the simplex method to determine that Z is unbounded. 4.5-3. Follow the instructions of Prob. 4.5-2 when the constraints are the following: 2x1 x2 20 x1 2x2 20 and

(a) Show that any convex combination of any set of feasible solutions must be a feasible solution (so that any convex combination of CPF solutions must be feasible). (b) Use the result quoted in part (a) to show that any convex combination of BF solutions must be a feasible solution. 4.5-6. Using the facts given in Prob. 4.5-5, show that the following statements must be true for any linear programming problem that has a bounded feasible region and multiple optimal solutions: (a) Every convex combination of the optimal BF solutions must be optimal. (b) No other feasible solution can be optimal. 4.5-7. Consider a two-variable linear programming problem whose CPF solutions are (0, 0), (6, 0), (6, 3), (3, 3), and (0, 2). (See Prob. 3.2-2 for a graph of the feasible region.) (a) Use the graph of the feasible region to identify all the constraints for the model. (b) For each pair of adjacent CPF solutions, give an example of an objective function such that all the points on the line segment between these two corner points are multiple optimal solutions. (c) Now suppose that the objective function is Z x1 2x2. Use the graphical method to find all the optimal solutions. D,I (d) For the objective function in part (c), work through the simplex method step by step to find all the optimal BF solutions. Then write an algebraic expression that identifies all the optimal solutions. 4.5-8. Consider the following problem.

D,I

x1 0,

x2 0.

Maximize

Z 5x1 x2 3x3 4x4,

subject to x1 2x2 4x3 3x4 20 4x1 6x2 5x3 4x4 40 2x1 3x2 3x3 8x4 50

subject to x1 x2 3 x3 x4 2 and xj 0,

for j 1, 2, 3, 4.

Work through the simplex method step by step to find all the optimal BF solutions.

and x1 0,

Z x1 x2 x3 x4,

Maximize

4.5-4. Consider the following problem.

D,I

177

x2 0,

x3 0,

x4 0.

Work through the simplex method step by step to demonstrate that Z is unbounded. 4.5-5. A basic property of any linear programming problem with a bounded feasible region is that every feasible solution can be expressed as a convex combination of the CPF solutions (perhaps in more than one way). Similarly, for the augmented form of the problem, every feasible solution can be expressed as a convex combination of the BF solutions.

4.6-1.* Consider the following problem. Z 2x1 3x2,

Maximize subject to x1 2x2 4 x1 x2 3 and x1 0,

x2 0.

178

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

(a) Solve this problem graphically. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (c) Continue from part (b) to work through the simplex method step by step to solve the problem. 4.6-2. Consider the following problem. Maximize

Z 4x1 2x2 3x3 5x4,

subject to 2x1 3x2 4x3 2x4 300 8x1 x2 x3 5x4 300 and xj 0,

for j 1, 2, 3, 4.

(a) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (b) Work through the simplex method step by step to solve the problem. (c) Using the two-phase method, construct the complete first simplex tableau for phase 1 and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (d) Work through phase 1 step by step. (e) Construct the complete first simplex tableau for phase 2. I (f) Work through phase 2 step by step to solve the problem. (g) Compare the sequence of BF solutions obtained in part (b) with that in parts (d) and ( f ). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (h) Use a software package based on the simplex method to solve the problem. 4.6-3. Consider the following problem. Minimize

Z 3x1 2x2,

4.6-4.* Consider the following problem. Z 2x1 3x2 x3,

Minimize subject to

x1 4x2 2x3 8 3x1 2x2 2x3 6 and x1 0,

x2 0,

x3 0.

(a) Reformulate this problem to fit our standard form for a linear programming model presented in Sec. 3.2. I (b) Using the Big M method, work through the simplex method step by step to solve the problem. I (c) Using the two-phase method, work through the simplex method step by step to solve the problem. (d) Compare the sequence of BF solutions obtained in parts (b) and (c). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (e) Use a software package based on the simplex method to solve the problem. 4.6-5. For the Big M method, explain why the simplex method never would choose an artificial variable to be an entering basic variable once all the artificial variables are nonbasic. 4.6-6. Consider the following problem. Z 90x1 70x2,

Maximize subject to 2x1 x2 2 x1 x2 2 and

subject to 2x1 x2 10 3x1 2x2 6 x1 x2 6 and x1 0,

I

initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. (c) Work through the simplex method step by step to solve the problem.

x2 0.

(a) Solve this problem graphically. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding

x1 0,

x2 0.

(a) Demonstrate graphically that this problem has no feasible solutions. C (b) Use a computer package based on the simplex method to determine that the problem has no feasible solutions. I (c) Using the Big M method, work through the simplex method step by step to demonstrate that the problem has no feasible solutions. I (d) Repeat part (c) when using phase 1 of the two-phase method.

CHAPTER 4 PROBLEMS

4.6-7. Follow the instructions of Prob. 4.6-6 for the following problem. Minimize

Z 5,000x1 7,000x2,

subject to 2x1 x2 1 x1 2x2 1 and x1 0,

(a) Using the two-phase method, work through phase 1 step by step. C (b) Use a software package based on the simplex method to formulate and solve the phase 1 problem. I (c) Work through phase 2 step by step to solve the original problem. C (d) Use a computer code based on the simplex method to solve the original problem. I

4.6-10.* Consider the following problem. x2 0.

Minimize

4.6-8. Consider the following problem. Maximize

179

subject to

Z 2x1 5x2 3x3,

2x1 x2 3x3 60 3x1 3x2 5x3 120

subject to x1 2x2 x3 20 2x1 4x2 x3 50

and x1 0,

and x1 0,

x3 0.

4.6-9. Consider the following problem. Z 2x1 x2 3x3,

subject to

x3 0.

(a) Using the Big M method, work through the simplex method step by step to solve the problem. I (b) Using the two-phase method, work through the simplex method step by step to solve the problem. (c) Compare the sequence of BF solutions obtained in parts (a) and (b). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (d) Use a software package based on the simplex method to solve the problem. 4.6-11. Follow the instructions of Prob. 4.6-10 for the following problem. Minimize

Z 3x1 2x2 7x3,

subject to x1 x2 x3 10 2x1 x2 x3 10 and x1 0,

x2 0,

x3 0.

4.6-12. Follow the instructions of Prob. 4.6-10 for the following problem. Minimize

Z 3x1 2x2 x3,

subject to

5x1 2x2 7x3 420 3x1 2x2 5x3 280

x1 x2 x3 7 3x1 x2 x3 10

and x1 0,

x2 0,

I

x2 0,

(a) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (b) Work through the simplex method step by step to solve the problem. I (c) Using the two-phase method, construct the complete first simplex tableau for phase 1 and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (d) Work through phase 1 step by step. (e) Construct the complete first simplex tableau for phase 2. I (f) Work through phase 2 step by step to solve the problem. (g) Compare the sequence of BF solutions obtained in part (b) with that in parts (d) and ( f ). Which of these solutions are feasible only for the artificial problem obtained by introducing artificial variables and which are actually feasible for the real problem? C (h) Use a software package based on the simplex method to solve the problem.

Minimize

Z 3x1 2x2 4x3,

and x2 0,

x3 0.

x1 0,

x2 0,

x3 0.

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SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4.6-13. Label each of the following statements as true or false, and then justify your answer. (a) When a linear programming model has an equality constraint, an artificial variable is introduced into this constraint in order to start the simplex method with an obvious initial basic solution that is feasible for the original model. (b) When an artificial problem is created by introducing artificial variables and using the Big M method, if all artificial variables in an optimal solution for the artificial problem are equal to zero, then the real problem has no feasible solutions. (c) The two-phase method is commonly used in practice because it usually requires fewer iterations to reach an optimal solution than the Big M method does. 4.6-14. Consider the following problem. Maximize Z x1 4x2 2x3, subject to 4x1 x2 x3 5 x1 x2 2x3 10 and x3 0 x2 0, (no nonnegativity constraint for x1). (a) Reformulate this problem so all variables have nonnegativity constraints. D,I (b) Work through the simplex method step by step to solve the problem. C (c) Use a software package based on the simplex method to solve the problem. 4.6-15.* Consider the following problem. Maximize Z x1 4x2, subject to 3x1 x2 6 x1 2x2 4 x1 2x2 3 (no lower bound constraint for x1). (a) Solve this problem graphically. (b) Reformulate this problem so that it has only two functional constraints and all variables have nonnegativity constraints. D,I (c) Work through the simplex method step by step to solve the problem. 4.6-16. Consider the following problem. Maximize Z x1 2x2 x3, subject to 3x2 x3 120 x1 x2 4x3 80 3x1 x2 2x3 100 (no nonnegativity constraints).

(a) Reformulate this problem so that all variables have nonnegativity constraints. D,I (b) Work through the simplex method step by step to solve the problem. C (c) Use a computer package based on the simplex method to solve the problem. 4.6-17. This chapter has described the simplex method as applied to linear programming problems where the objective function is to be maximized. Section 4.6 then described how to convert a minimization problem to an equivalent maximization problem for applying the simplex method. Another option with minimization problems is to make a few modifications in the instructions for the simplex method given in the chapter in order to apply the algorithm directly. (a) Describe what these modifications would need to be. (b) Using the Big M method, apply the modified algorithm developed in part (a) to solve the following problem directly by hand. (Do not use your OR Courseware.) Minimize

Z 3x1 8x2 5x3,

subject to 3x1 3x2 4x3 70 3x1 5x2 2x3 70 and x1 0,

x2 0,

x3 0.

4.6-18. Consider the following problem. Maximize

Z 2x1 x2 4x3 3x4,

subject to x1 x2 3x3 2x4 4 x1 x2 x3 x4 1 2x1 x2 x3 x4 2 x1 2x2 x3 2x4 2 and x2 0,

x3 0,

x4 0

(no nonnegativity constraint for x1). (a) Reformulate this problem to fit our standard form for a linear programming model presented in Sec. 3.2. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. (c) Using the two-phase method, construct row 0 of the first simplex tableau for phase 1. C (d) Use a computer package based on the simplex method to solve the problem.

CHAPTER 4 PROBLEMS

I

4.6-19. Consider the following problem. Maximize

and x2 0,

x3 0.

Work through the simplex method step by step to demonstrate that this problem does not possess any feasible solutions. 4.7-1. Refer to Fig. 4.10 and the resulting allowable range to stay feasible for the respective right-hand sides of the Wyndor Glass Co. problem given in Sec. 3.1. Use graphical analysis to demonstrate that each given allowable range is correct. 4.7-2. Reconsider the model in Prob. 4.1-5. Interpret the right-hand side of the respective functional constraints as the amount available of the respective resources. (a) Use graphical analysis as in Fig. 4.8 to determine the shadow prices for the respective resources. (b) Use graphical analysis to perform sensitivity analysis on this model. In particular, check each parameter of the model to determine whether it is a sensitive parameter (a parameter whose value cannot be changed without changing the optimal solution) by examining the graph that identifies the optimal solution. (c) Use graphical analysis as in Fig. 4.9 to determine the allowable range for each cj value (coefficient of xj in the objective function) over which the current optimal solution will remain optimal. (d) Changing just one bi value (the right-hand side of functional constraint i) will shift the corresponding constraint boundary. If the current optimal CPF solution lies on this constraint boundary, this CPF solution also will shift. Use graphical analysis to determine the allowable range for each bi value over which this CPF solution will remain feasible. C (e) Verify your answers in parts (a), (c), and (d) by using a computer package based on the simplex method to solve the problem and then to generate sensitivity analysis information. 4.7-3. Repeat Prob. 4.7-2 for the model in Prob. 4.1-6. 4.7-4. You are given the following linear programming problem. Z 4x1 2x2,

subject to 2x1 3x2 16 x1 3x2 17 x1 3x2 5

x1 0,

x2 0.

(a) Solve this problem graphically. (b) Use graphical analysis to find the shadow prices for the resources. (c) Determine how many additional units of resource 1 would be needed to increase the optimal value of Z by 15.

x1 x2 2x3 20 15x1 6x2 5x3 50 x1 3x2 5x3 30

Maximize

and

Z 4x1 5x2 3x3,

subject to

x1 0,

181

(resource 1) (resource 2) (resource 3)

4.7-5. Consider the following problem. Maximize

Z x1 7x2 3x3,

subject to 2x1 x2 x3 4 4x1 3x2 x3 2 3x1 2x2 x3 3

(resource 1) (resource 2) (resource 3)

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the three resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient, and the allowable range to stay feasible for each right-hand side. D,I

4.7-6.* Consider the following problem. Maximize

Z 2x1 2x2 3x3,

subject to x1 x2 x3 4 2x1 x2 x3 2 x1 x2 3x3 12

(resource 1) (resource 2) (resource 3)

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the three resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient and the allowable range to stay feasible for each right-hand side. D,I

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4

SOLVING LINEAR PROGRAMMING PROBLEMS: THE SIMPLEX METHOD

4.7-7. Consider the following problem. Maximize

and x1 0,

Z 2x1 4x2 x3,

subject to

x2 0,

x3 0,

x4 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the two resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient, and the allowable range to stay feasible for each right-hand side. D,I

2x1 3x2 x3 30 2x1 x2 x3 10 4x1 2x2 2x3 40

(resource 1) (resource 2) (resource 3)

and x1 0,

x2 0,

x3 0.

(a) Work through the simplex method step by step to solve the problem. (b) Identify the shadow prices for the three resources and describe their significance. C (c) Use a software package based on the simplex method to solve the problem and then to generate sensitivity information. Use this information to identify the shadow price for each resource, the allowable range to stay optimal for each objective function coefficient, and the allowable range to stay feasible for each right-hand side. D,I

4.7-8. Consider the following problem. Maximize

4.9.1. Use the interior-point algorithm in your OR Courseware to solve the model in Prob. 4.1-4. Choose 0.5 from the Option menu, use (x1, x2) (0.1, 0.4) as the initial trial solution, and run 15 iterations. Draw a graph of the feasible region, and then plot the trajectory of the trial solutions through this feasible region. 4.9-2. Repeat Prob. 4.9-1 for the model in Prob. 4.1-5. 4.9-3. Repeat Prob. 4.9-1 for the model in Prob. 4.1-6.

Z 5x1 4x2 x3 3x4,

subject to 3x1 2x2 3x3 x4 24 3x1 3x2 x3 3x4 36

CASE 4.1

(resource 1) (resource 2)

FABRICS AND FALL FASHIONS From the tenth floor of her office building, Katherine Rally watches the swarms of New Yorkers fight their way through the streets infested with yellow cabs and the sidewalks littered with hot dog stands. On this sweltering July day, she pays particular attention to the fashions worn by the various women and wonders what they will choose to wear in the fall. Her thoughts are not simply random musings; they are critical to her work since she owns and manages TrendLines, an elite women’s clothing company. Today is an especially important day because she must meet with Ted Lawson, the production manager, to decide upon next month’s production plan for the fall line. Specifically, she must determine the quantity of each clothing item she should produce given the plant’s production capacity, limited resources, and demand forecasts. Accurate planning for next month’s production is critical to fall sales since the items produced next month will appear in stores during September, and women generally buy the majority of the fall fashions when they first appear in September. She turns back to her sprawling glass desk and looks at the numerous papers covering it. Her eyes roam across the clothing patterns designed almost six months ago,

CASE 4.1

FABRICS AND FALL FASHIONS

183

the lists of materials requirements for each pattern, and the lists of demand forecasts for each pattern determined by customer surveys at fashion shows. She remembers the hectic and sometimes nightmarish days of designing the fall line and presenting it at fashion shows in New York, Milan, and Paris. Ultimately, she paid her team of six designers a total of $860,000 for their work on her fall line. With the cost of hiring runway models, hair stylists, and makeup artists, sewing and fitting clothes, building the set, choreographing and rehearsing the show, and renting the conference hall, each of the three fashion shows cost her an additional $2,700,000. She studies the clothing patterns and material requirements. Her fall line consists of both professional and casual fashions. She determined the prices for each clothing item by taking into account the quality and cost of material, the cost of labor and machining, the demand for the item, and the prestige of the TrendLines brand name. The fall professional fashions include:

Clothing Item

Materials Requirements

Price

Labor and Machine Cost

Tailored wool slacks

3 yards of wool 2 yards of acetate for lining 1.5 yards of cashmere 1.5 yards of silk 0.5 yard of silk 2 yards of rayon 1.5 yards of acetate for lining 2.5 yards of wool 1.5 yards of acetate for lining

$300

$160

$450 $180 $120 $270

$150 $100 $ 60 $120

$320

$140

Cashmere sweater Silk blouse Silk camisole Tailored skirt Wool blazer

The fall casual fashions include:

Clothing Item

Materials Requirements

Price

Labor and Machine Cost

Velvet pants

3 yards of velvet 2 yards of acetate for lining 1.5 yards of cotton 0.5 yard of cotton 1.5 yards of velvet 1.5 yards of rayon

$350

$175

$130 $ 75 $200 $120

$ 60 $ 40 $160 $ 90

Cotton sweater Cotton miniskirt Velvet shirt Button-down blouse

She knows that for the next month, she has ordered 45,000 yards of wool, 28,000 yards of acetate, 9,000 yards of cashmere, 18,000 yards of silk, 30,000 yards of rayon, 20,000 yards of velvet, and 30,000 yards of cotton for production. The prices of the materials are listed on the next page.

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Material

Price per yard

Wool Acetate Cashmere Silk Rayon Velvet Cotton

$ 9.00 $ 1.50 $60.00 $13.00 $ 2.25 $12.00 $ 2.50

Any material that is not used in production can be sent back to the textile wholesaler for a full refund, although scrap material cannot be sent back to the wholesaler. She knows that the production of both the silk blouse and cotton sweater leaves leftover scraps of material. Specifically, for the production of one silk blouse or one cotton sweater, 2 yards of silk and cotton, respectively, are needed. From these 2 yards, 1.5 yards are used for the silk blouse or the cotton sweater and 0.5 yard is left as scrap material. She does not want to waste the material, so she plans to use the rectangular scrap of silk or cotton to produce a silk camisole or cotton miniskirt, respectively. Therefore, whenever a silk blouse is produced, a silk camisole is also produced. Likewise, whenever a cotton sweater is produced, a cotton miniskirt is also produced. Note that it is possible to produce a silk camisole without producing a silk blouse and a cotton miniskirt without producing a cotton sweater. The demand forecasts indicate that some items have limited demand. Specifically, because the velvet pants and velvet shirts are fashion fads, TrendLines has forecasted that it can sell only 5,500 pairs of velvet pants and 6,000 velvet shirts. TrendLines does not want to produce more than the forecasted demand because once the pants and shirts go out of style, the company cannot sell them. TrendLines can produce less than the forecasted demand, however, since the company is not required to meet the demand. The cashmere sweater also has limited demand because it is quite expensive, and TrendLines knows it can sell at most 4,000 cashmere sweaters. The silk blouses and camisoles have limited demand because many women think silk is too hard to care for, and TrendLines projects that it can sell at most 12,000 silk blouses and 15,000 silk camisoles. The demand forecasts also indicate that the wool slacks, tailored skirts, and wool blazers have a great demand because they are basic items needed in every professional wardrobe. Specifically, the demand for wool slacks is 7,000 pairs of slacks, and the demand for wool blazers is 5,000 blazers. Katherine wants to meet at least 60 percent of the demand for these two items in order to maintain her loyal customer base and not lose business in the future. Although the demand for tailored skirts could not be estimated, Katherine feels she should make at least 2,800 of them. (a) Ted is trying to convince Katherine not to produce any velvet shirts since the demand for this fashion fad is quite low. He argues that this fashion fad alone accounts for $500,000 of the fixed design and other costs. The net contribution (price of clothing item materials cost labor cost) from selling the fashion fad should cover these fixed costs. Each velvet shirt generates a net contribution of $22. He argues that given the net contribution,

CASE 4.2

NEW FRONTIERS

185

even satisfying the maximum demand will not yield a profit. What do you think of Ted’s argument? (b) Formulate and solve a linear programming problem to maximize profit given the production, resource, and demand constraints.

Before she makes her final decision, Katherine plans to explore the following questions independently except where otherwise indicated. (c) The textile wholesaler informs Katherine that the velvet cannot be sent back because the demand forecasts show that the demand for velvet will decrease in the future. Katherine can therefore get no refund for the velvet. How does this fact change the production plan? (d) What is an intuitive economic explanation for the difference between the solutions found in parts (b) and (c)? (e) The sewing staff encounters difficulties sewing the arms and lining into the wool blazers since the blazer pattern has an awkward shape and the heavy wool material is difficult to cut and sew. The increased labor time to sew a wool blazer increases the labor and machine cost for each blazer by $80. Given this new cost, how many of each clothing item should TrendLines produce to maximize profit? (f) The textile wholesaler informs Katherine that since another textile customer canceled his order, she can obtain an extra 10,000 yards of acetate. How many of each clothing item should TrendLines now produce to maximize profit? (g) TrendLines assumes that it can sell every item that was not sold during September and October in a big sale in November at 60 percent of the original price. Therefore, it can sell all items in unlimited quantity during the November sale. (The previously mentioned upper limits on demand concern only the sales during September and October.) What should the new production plan be to maximize profit?

CASE 4.2

NEW FRONTIERS Rob Richman, president of AmeriBank, takes off his glasses, rubs his eyes in exhaustion, and squints at the clock in his study. It reads 3 A.M. For the last several hours, Rob has been poring over AmeriBank’s financial statements from the last three quarters of operation. AmeriBank, a medium-sized bank with branches throughout the United States, is headed for dire economic straits. The bank, which provides transaction, savings, and investment and loan services, has been experiencing a steady decline in its net income over the past year, and trends show that the decline will continue. The bank is simply losing customers to nonbank and foreign bank competitors. AmeriBank is not alone in its struggle to stay out of the red. From his daily industry readings, Rob knows that many American banks have been suffering significant losses because of increasing competition from nonbank and foreign bank competitors offering services typically in the domain of American banks. Because the nonbank and foreign bank competitors specialize in particular services, they are able to better capture the market for those services by offering less expensive, more efficient, more convenient services. For example, large corporations now turn to foreign banks and commercial paper offerings for loans, and affluent Americans now turn to money-market funds for investment. Banks face the daunting challenge of distinguishing themselves from nonbank and foreign bank competitors.

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4

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Rob has concluded that one strategy for distinguishing AmeriBank from its competitors is to improve services that nonbank and foreign bank competitors do not readily provide: transaction services. He has decided that a more convenient transaction method must logically succeed the automatic teller machine, and he believes that electronic banking over the Internet allows this convenient transaction method. Over the Internet, customers are able to perform transactions on their desktop computers either at home or at work. The explosion of the Internet means that many potential customers understand and use the World Wide Web. He therefore feels that if AmeriBank offers Web banking (as the practice of Internet banking is commonly called), the bank will attract many new customers. Before Rob undertakes the project to make Web banking possible, however, he needs to understand the market for Web banking and the services AmeriBank should provide over the Internet. For example, should the bank only allow customers to access account balances and historical transaction information over the Internet, or should the bank develop a strategy to allow customers to make deposits and withdrawals over the Internet? Should the bank try to recapture a portion of the investment market by continuously running stock prices and allowing customers to make stock transactions over the Internet for a minimal fee? Because AmeriBank is not in the business of performing surveys, Rob has decided to outsource the survey project to a professional survey company. He has opened the project up for bidding by several survey companies and will award the project to the company which is willing to perform the survey for the least cost. Sophisticated Surveys is one of three survey companies competing for the project. Rob provided each survey company with a list of survey requirements to ensure that AmeriBank receives the needed information for planning the Web banking project. Because different age groups require different services, AmeriBank is interested in surveying four different age groups. The first group encompasses customers who are 18 to 25 years old. The bank assumes that this age group has limited yearly income and performs minimal transactions. The second group encompasses customers who are 26 to 40 years old. This age group has significant sources of income, performs many transactions, requires numerous loans for new houses and cars, and invests in various securities. The third group encompasses customers who are 41 to 50 years old. These customers typically have the same level of income and perform the same number of transactions as the second age group, but the bank assumes that these customers are less likely to use Web banking since they have not become as comfortable with the explosion of computers or the Internet. Finally, the fourth group encompasses customers who are 51 years of age and over. These customers commonly crave security and require continuous information on retirement funds. The banks believes that it is highly unlikely that customers in this age group will use Web banking, but the bank desires to learn the needs of this age group for the future. AmeriBank wants to interview 2,000 customers with at least 20 percent from the first age group, at least 27.5 percent from the second age group, at least 15 percent from the third age group, and at least 15 percent from the fourth age group. Rob understands that the Internet is a recent phenomenon and that some customers may not have heard of the World Wide Web. He therefore wants to ensure that the sur-

CASE 4.2

NEW FRONTIERS

187

vey includes a mix of customers who know the Internet well and those that have less exposure to the Internet. To ensure that AmeriBank obtains the correct mix, he wants to interview at least 15 percent of customers from the Silicon Valley where Internet use is high, at least 35 percent of customers from big cities where Internet use is medium, and at least 20 percent of customers from small towns where Internet use is low. Sophisticated Surveys has performed an initial analysis of these survey requirements to determine the cost of surveying different populations. The costs per person surveyed are listed in the following table:

Age Group Region Silicon Valley Big cities Small towns

18 to 25

26 to 40

41 to 50

51 and over

$4.75 $5.25 $6.50

$6.50 $5.75 $7.50

$6.50 $6.25 $7.50

$5.00 $6.25 $7.25

Sophisticated Surveys explores the following options cumulatively. (a) Formulate a linear programming model to minimize costs while meeting all survey constraints imposed by AmeriBank. (b) If the profit margin for Sophisticated Surveys is 15 percent of cost, what bid will they submit? (c) After submitting its bid, Sophisticated Surveys is informed that it has the lowest cost but that AmeriBank does not like the solution. Specifically, Rob feels that the selected survey population is not representative enough of the banking customer population. Rob wants at least 50 people of each age group surveyed in each region. What is the new bid made by Sophisticated Surveys? (d) Rob feels that Sophisticated Survey oversampled the 18- to 25-year-old population and the Silicon Valley population. He imposes a new constraint that no more than 600 individuals can be surveyed from the 18- to 25-year-old population and no more than 650 individuals can be surveyed from the Silicon Valley population. What is the new bid? (e) When Sophisticated Surveys calculated the cost of reaching and surveying particular individuals, the company thought that reaching individuals in young populations would be easiest. In a recently completed survey, however, Sophisticated Surveys learned that this assumption was wrong. The new costs for surveying the 18- to 25-year-old population are listed below. Region survey cost per person Silicon Valley Big cities Small towns

$6.50 $6.75 $7.00

Given the new costs, what is the new bid?

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(f) To ensure the desired sampling of individuals, Rob imposes even stricter requirements. He fixes the exact percentage of people that should be surveyed from each population. The requirements are listed below: Population percentage of people surveyed 18 26 41 51

to 25 to 40 to 50 and over

25% 35% 20% 20%

Silicon Valley Big cities Small towns

20% 50% 30%

By how much would these new requirements increase the cost of surveying for Sophisticated Surveys? Given the 15 percent profit margin, what would Sophisticated Surveys bid?

CASE 4.3

ASSIGNING STUDENTS TO SCHOOLS The Springfield school board has made the decision to close one of its middle schools (sixth, seventh, and eighth grades) at the end of this school year and reassign all of next year’s middle school students to the three remaining middle schools. The school district provides bussing for all middle school students who must travel more than approximately a mile, so the school board wants a plan for reassigning the students that will minimize the total bussing cost. The annual cost per student of bussing from each of the six residential areas of the city to each of the schools is shown in the following table (along with other basic data for next year), where 0 indicates that bussing is not needed and a dash indicates an infeasible assignment.

No. of Area Students 1 2 3 4 5 6

450 600 550 350 500 450

Percentage Percentage Percentage Bussing Cost per Student in 6th in 7th in 8th Grade Grade Grade School 1 School 2 School 3 32 37 30 28 39 34

38 28 32 40 34 28

30 35 38 32 27 38 School capacity:

$300 — $600 $200 0 $500

0 $400 $300 $500 — $300

$700 $500 $200 — $400 0

900

1,100

1,000

The school board also has imposed the restriction that each grade must constitute between 30 and 36 percent of each school’s population. The above table shows the percentage of each area’s middle school population for next year that falls into each of

CASE 4.3

ASSIGNING STUDENTS TO SCHOOLS

189

the three grades. The school attendance zone boundaries can be drawn so as to split any given area among more than one school, but assume that the percentages shown in the table will continue to hold for any partial assignment of an area to a school. You have been hired as an operations research consultant to assist the school board in determining how many students in each area should be assigned to each school. (a) Formulate a linear programming model for this problem. (b) Solve the model. (c) What is your resulting recommendation to the school board?

After seeing your recommendation, the school board expresses concern about all the splitting of residential areas among multiple schools. They indicate that they “would like to keep each neighborhood together.” (d) Adjust your recommendation as well as you can to enable each area to be assigned to just one school. (Adding this restriction may force you to fudge on some other constraints.) How much does this increase the total bussing cost? (This line of analysis will be pursued more rigorously in Case 12.4.)

The school board is considering eliminating some bussing to reduce costs. Option 1 is to eliminate bussing only for students traveling 1 to 1.5 miles, where the cost per student is given in the table as $200. Option 2 is to also eliminate bussing for students traveling 1.5 to 2 miles, where the estimated cost per student is $300. (e) Revise the model from part (a) to fit Option 1, and solve. Compare these results with those from part (c), including the reduction in total bussing cost. (f) Repeat part (e) for Option 2.

The school board now needs to choose among the three alternative bussing plans (the current one or Option 1 or Option 2). One important factor is bussing costs. However, the school board also wants to place equal weight on a second factor: the inconvenience and safety problems caused by forcing students to travel by foot or bicycle a substantial distance (more than a mile, and especially more than 1.5 miles). Therefore, they want to choose a plan that provides the best trade-off between these two factors. (g) Use your results from parts (c), (e), and ( f ) to summarize the key information related to these two factors that the school board needs to make this decision. (h) Which decision do you think should be made? Why?

Note: This case will be continued in later chapters (Cases 6.3 and 12.4), so we suggest that you save your analysis, including your basic model.

5 The Theory of the Simplex Method Chapter 4 introduced the basic mechanics of the simplex method. Now we shall delve a little more deeply into this algorithm by examining some of its underlying theory. The first section further develops the general geometric and algebraic properties that form the foundation of the simplex method. We then describe the matrix form of the simplex method (called the revised simplex method), which streamlines the procedure considerably for computer implementation. Next we present a fundamental insight about a property of the simplex method that enables us to deduce how changes that are made in the original model get carried along to the final simplex tableau. This insight will provide the key to the important topics of Chap. 6 (duality theory and sensitivity analysis).

5.1

FOUNDATIONS OF THE SIMPLEX METHOD Section 4.1 introduced corner-point feasible (CPF) solutions and the key role they play in the simplex method. These geometric concepts were related to the algebra of the simplex method in Secs. 4.2 and 4.3. However, all this was done in the context of the Wyndor Glass Co. problem, which has only two decision variables and so has a straightforward geometric interpretation. How do these concepts generalize to higher dimensions when we deal with larger problems? We address this question in this section. We begin by introducing some basic terminology for any linear programming problem with n decision variables. While we are doing this, you may find it helpful to refer to Fig. 5.1 (which repeats Fig. 4.1) to interpret these definitions in two dimensions (n 2). Terminology It may seem intuitively clear that optimal solutions for any linear programming problem must lie on the boundary of the feasible region, and in fact this is a general property. Because boundary is a geometric concept, our initial definitions clarify how the boundary of the feasible region is identified algebraically. The constraint boundary equation for any constraint is obtained by replacing its , , or sign by an sign.

Consequently, the form of a constraint boundary equation is ai1x1 ai2 x2 ain xn bi for functional constraints and xj 0 for nonnegativity constraints. Each such 190

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

Maximize Z 3x1 5x2, subject to 4 x1 2x2 12 2x1 3x2 18 and x1 0, x2 0

x1 0 (0, 9) 3x1 2x2 18

(0, 6)

(2, 6)

191

(4, 6)

2x2 12

x1 4

Feasible region FIGURE 5.1 Constraint boundaries, constraint boundary equations, and corner-point solutions for the Wyndor Glass Co. problem.

(4, 3)

(0, 0) (4, 0)

(6, 0)

x2 0

equation defines a “flat” geometric shape (called a hyperplane) in n-dimensional space, analogous to the line in two-dimensional space and the plane in three-dimensional space. This hyperplane forms the constraint boundary for the corresponding constraint. When the constraint has either a or a sign, this constraint boundary separates the points that satisfy the constraint (all the points on one side up to and including the constraint boundary) from the points that violate the constraint (all those on the other side of the constraint boundary). When the constraint has an sign, only the points on the constraint boundary satisfy the constraint. For example, the Wyndor Glass Co. problem has five constraints (three functional constraints and two nonnegativity constraints), so it has the five constraint boundary equations shown in Fig. 5.1. Because n 2, the hyperplanes defined by these constraint boundary equations are simply lines. Therefore, the constraint boundaries for the five constraints are the five lines shown in Fig. 5.1. The boundary of the feasible region contains just those feasible solutions that satisfy one or more of the constraint boundary equations.

Geometrically, any point on the boundary of the feasible region lies on one or more of the hyperplanes defined by the respective constraint boundary equations. Thus, in Fig. 5.1, the boundary consists of the five darker line segments. Next, we give a general definition of CPF solution in n-dimensional space. A corner-point feasible (CPF) solution is a feasible solution that does not lie on any line segment1 connecting two other feasible solutions. 1

An algebraic expression for a line segment is given in Appendix 2.

192

5

THE THEORY OF THE SIMPLEX METHOD

As this definition implies, a feasible solution that does lie on a line segment connecting two other feasible solutions is not a CPF solution. To illustrate when n 2, consider Fig. 5.1. The point (2, 3) is not a CPF solution, because it lies on various such line segments, e.g., the line segment connecting (0, 3) and (4, 3). Similarly, (0, 3) is not a CPF solution, because it lies on the line segment connecting (0, 0) and (0, 6). However, (0, 0) is a CPF solution, because it is impossible to find two other feasible solutions that lie on completely opposite sides of (0, 0). (Try it.) When the number of decision variables n is greater than 2 or 3, this definition for CPF solution is not a very convenient one for identifying such solutions. Therefore, it will prove most helpful to interpret these solutions algebraically. For the Wyndor Glass Co. example, each CPF solution in Fig. 5.1 lies at the intersection of two (n 2) constraint lines; i.e., it is the simultaneous solution of a system of two constraint boundary equations. This situation is summarized in Table 5.1, where defining equations refer to the constraint boundary equations that yield (define) the indicated CPF solution. For any linear programming problem with n decision variables, each CPF solution lies at the intersection of n constraint boundaries; i.e., it is the simultaneous solution of a system of n constraint boundary equations.

However, this is not to say that every set of n constraint boundary equations chosen from the n m constraints (n nonnegativity and m functional constraints) yields a CPF solution. In particular, the simultaneous solution of such a system of equations might violate one or more of the other m constraints not chosen, in which case it is a corner-point infeasible solution. The example has three such solutions, as summarized in Table 5.2. (Check to see why they are infeasible.) Furthermore, a system of n constraint boundary equations might have no solution at all. This occurs twice in the example, with the pairs of equations (1) x1 0 and x1 4 and (2) x2 0 and 2x2 12. Such systems are of no interest to us. The final possibility (which never occurs in the example) is that a system of n constraint boundary equations has multiple solutions because of redundant equations. You need not be concerned with this case either, because the simplex method circumvents its difficulties. TABLE 5.1 Defining equations for each CPF solution for the Wyndor Glass Co. problem CPF Solution

Defining Equations

(0, 0)

x1 0 x2 0

(0, 6)

x1 0 2x2 12

(2, 6)

2x2 12 3x1 2x2 18

(4, 3)

3x1 2x2 18 x1 4

(4, 0)

x1 4 x2 0

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

193

TABLE 5.2 Defining equations for each corner-point infeasible solution for the Wyndor Glass Co. problem Corner-Point Infeasible Solution

Defining Equations

(0, 9)

x1 0 3x1 2x2 18

(4, 6)

2x2 12 x1 4

(6, 0)

3x1 2x2 18 x2 0

To summarize for the example, with five constraints and two variables, there are 10 pairs of constraint boundary equations. Five of these pairs became defining equations for CPF solutions (Table 5.1), three became defining equations for corner-point infeasible solutions (Table 5.2), and each of the final two pairs had no solution. Adjacent CPF Solutions Section 4.1 introduced adjacent CPF solutions and their role in solving linear programming problems. We now elaborate. Recall from Chap. 4 that (when we ignore slack, surplus, and artificial variables) each iteration of the simplex method moves from the current CPF solution to an adjacent one. What is the path followed in this process? What really is meant by adjacent CPF solution? First we address these questions from a geometric viewpoint, and then we turn to algebraic interpretations. These questions are easy to answer when n 2. In this case, the boundary of the feasible region consists of several connected line segments forming a polygon, as shown in Fig. 5.1 by the five darker line segments. These line segments are the edges of the feasible region. Emanating from each CPF solution are two such edges leading to an adjacent CPF solution at the other end. (Note in Fig. 5.1 how each CPF solution has two adjacent ones.) The path followed in an iteration is to move along one of these edges from one end to the other. In Fig. 5.1, the first iteration involves moving along the edge from (0, 0) to (0, 6), and then the next iteration moves along the edge from (0, 6) to (2, 6). As Table 5.1 illustrates, each of these moves to an adjacent CPF solution involves just one change in the set of defining equations (constraint boundaries on which the solution lies). When n 3, the answers are slightly more complicated. To help you visualize what is going on, Fig. 5.2 shows a three-dimensional drawing of a typical feasible region when n 3, where the dots are the CPF solutions. This feasible region is a polyhedron rather than the polygon we had with n 2 (Fig. 5.1), because the constraint boundaries now are planes rather than lines. The faces of the polyhedron form the boundary of the feasible region, where each face is the portion of a constraint boundary that satisfies the other constraints as well. Note that each CPF solution lies at the intersection of three constraint boundaries (sometimes including some of the x1 0, x2 0, and x3 0 constraint boundaries for the nonnegativity

194

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THE THEORY OF THE SIMPLEX METHOD

Constraints

x3

x1 4 x2 4 x1 x2 6 x1 2x3 4 x1 0, x2 0, x3 0

(4, 0, 4)

(4, 2, 4)

(0, 0, 2)

(4, 0, 0)

(2, 4, 3)

x1

(0, 0, 0) (0, 4, 2) (4, 2, 0) FIGURE 5.2 Feasible region and CPF solutions for a three-variable linear programming problem.

x2

(0, 4, 0)

(2, 4, 0)

constraints), and the solution also satisfies the other constraints. Such intersections that do not satisfy one or more of the other constraints yield corner-point infeasible solutions instead. The darker line segment in Fig. 5.2 depicts the path of the simplex method on a typical iteration. The point (2, 4, 3) is the current CPF solution to begin the iteration, and the point (4, 2, 4) will be the new CPF solution at the end of the iteration. The point (2, 4, 3) lies at the intersection of the x2 4, x1 x2 6, and x1 2x3 4 constraint boundaries, so these three equations are the defining equations for this CPF solution. If the x2 4 defining equation were removed, the intersection of the other two constraint boundaries (planes) would form a line. One segment of this line, shown as the dark line segment from (2, 4, 3) to (4, 2, 4) in Fig. 5.2, lies on the boundary of the feasible region, whereas the rest of the line is infeasible. This line segment is an edge of the feasible region, and its endpoints (2, 4, 3) and (4, 2, 4) are adjacent CPF solutions. For n 3, all the edges of the feasible region are formed in this way as the feasible segment of the line lying at the intersection of two constraint boundaries, and the two endpoints of an edge are adjacent CPF solutions. In Fig. 5.2 there are 15 edges of the feasible region, and so there are 15 pairs of adjacent CPF solutions. For the current CPF solution (2, 4, 3), there are three ways to remove one of its three defining equations to obtain an intersection of the other two constraint boundaries, so there are three edges emanating from (2, 4, 3). These edges lead to (4, 2, 4), (0, 4, 2), and (2, 4, 0), so these are the CPF solutions that are adjacent to (2, 4, 3). For the next iteration, the simplex method chooses one of these three edges, say, the darker line segment in Fig. 5.2, and then moves along this edge away from (2, 4, 3) until it reaches the first new constraint boundary, x1 4, at its other endpoint. [We cannot continue farther along this line to the next constraint boundary, x2 0, because this leads

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

195

to a corner-point infeasible solution—(6, 0, 5).] The intersection of this first new constraint boundary with the two constraint boundaries forming the edge yields the new CPF solution (4, 2, 4). When n 3, these same concepts generalize to higher dimensions, except the constraint boundaries now are hyperplanes instead of planes. Let us summarize. Consider any linear programming problem with n decision variables and a bounded feasible region. A CPF solution lies at the intersection of n constraint boundaries (and satisfies the other constraints as well). An edge of the feasible region is a feasible line segment that lies at the intersection of n 1 constraint boundaries, where each endpoint lies on one additional constraint boundary (so that these endpoints are CPF solutions). Two CPF solutions are adjacent if the line segment connecting them is an edge of the feasible region. Emanating from each CPF solution are n such edges, each one leading to one of the n adjacent CPF solutions. Each iteration of the simplex method moves from the current CPF solution to an adjacent one by moving along one of these n edges.

When you shift from a geometric viewpoint to an algebraic one, intersection of constraint boundaries changes to simultaneous solution of constraint boundary equations. The n constraint boundary equations yielding (defining) a CPF solution are its defining equations, where deleting one of these equations yields a line whose feasible segment is an edge of the feasible region. We next analyze some key properties of CPF solutions and then describe the implications of all these concepts for interpreting the simplex method. However, while the above summary is fresh in your mind, let us give you a preview of its implications. When the simplex method chooses an entering basic variable, the geometric interpretation is that it is choosing one of the edges emanating from the current CPF solution to move along. Increasing this variable from zero (and simultaneously changing the values of the other basic variables accordingly) corresponds to moving along this edge. Having one of the basic variables (the leaving basic variable) decrease so far that it reaches zero corresponds to reaching the first new constraint boundary at the other end of this edge of the feasible region. Properties of CPF Solutions We now focus on three key properties of CPF solutions that hold for any linear programming problem that has feasible solutions and a bounded feasible region. Property 1: (a) If there is exactly one optimal solution, then it must be a CPF solution. (b) If there are multiple optimal solutions (and a bounded feasible region), then at least two must be adjacent CPF solutions. Property 1 is a rather intuitive one from a geometric viewpoint. First consider Case (a), which is illustrated by the Wyndor Glass Co. problem (see Fig. 5.1) where the one optimal solution (2, 6) is indeed a CPF solution. Note that there is nothing special about this example that led to this result. For any problem having just one optimal solution, it always is possible to keep raising the objective function line (hyperplane) until it just touches one point (the optimal solution) at a corner of the feasible region. We now give an algebraic proof for this case. Proof of Case (a) of Property 1: We set up a proof by contradiction by assuming that there is exactly one optimal solution and that it is not a CPF solution.

196

5 THE THEORY OF THE SIMPLEX METHOD

We then show below that this assumption leads to a contradiction and so cannot be true. (The solution assumed to be optimal will be denoted by x*, and its objective function value by Z*.) Recall the definition of CPF solution (a feasible solution that does not lie on any line segment connecting two other feasible solutions). Since we have assumed that the optimal solution x* is not a CPF solution, this implies that there must be two other feasible solutions such that the line segment connecting them contains the optimal solution. Let the vectors x and x denote these two other feasible solutions, and let Z 1 and Z 2 denote their respective objective function values. Like each other point on the line segment connecting x and x , x* x (1 )x for some value of such that 0 1. Thus, Z* Z2 (1 )Z1. Since the weights and 1 add to 1, the only possibilities for how Z*, Z1, and Z2 compare are (1) Z* Z1 Z2, (2) Z1 Z* Z2, and (3) Z1 Z* Z2. The first possibility implies that x and x also are optimal, which contradicts the assumption that there is exactly one optimal solution. Both the latter possibilities contradict the assumption that x* (not a CPF solution) is optimal. The resulting conclusion is that it is impossible to have a single optimal solution that is not a CPF solution. Now consider Case (b), which was demonstrated in Sec. 3.2 under the definition of optimal solution by changing the objective function in the example to Z 3x1 2x2 (see Fig. 3.5 on page 35). What then happens when you are solving graphically is that the objective function line keeps getting raised until it contains the line segment connecting the two CPF solutions (2, 6) and (4, 3). The same thing would happen in higher dimensions except that an objective function hyperplane would keep getting raised until it contained the line segment(s) connecting two (or more) adjacent CPF solutions. As a consequence, all optimal solutions can be obtained as weighted averages of optimal CPF solutions. (This situation is described further in Probs. 4.5-5 and 4.5-6.) The real significance of Property 1 is that it greatly simplifies the search for an optimal solution because now only CPF solutions need to be considered. The magnitude of this simplification is emphasized in Property 2. Property 2: There are only a finite number of CPF solutions. This property certainly holds in Figs. 5.1 and 5.2, where there are just 5 and 10 CPF solutions, respectively. To see why the number is finite in general, recall that each CPF solution is the simultaneous solution of a system of n out of the m n constraint boundary equations. The number of different combinations of m n equations taken n at a time is mn

(m n)!

, n m!n! which is a finite number. This number, in turn, in an upper bound on the number of CPF solutions. In Fig. 5.1, m 3 and n 2, so there are 10 different systems of two equa-

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

197

tions, but only half of them yield CPF solutions. In Fig. 5.2, m 4 and n 3, which gives 35 different systems of three equations, but only 10 yield CPF solutions. Property 2 suggests that, in principle, an optimal solution can be obtained by exhaustive enumeration; i.e., find and compare all the finite number of CPF solutions. Unfortunately, there are finite numbers, and then there are finite numbers that (for all practical purposes) might as well be infinite. For example, a rather small linear programming problem with only m 50 and n 50 would have 100!/(50!)2 1029 systems of equations to be solved! By contrast, the simplex method would need to examine only approximately 100 CPF solutions for a problem of this size. This tremendous savings can be obtained because of the optimality test given in Sec. 4.1 and restated here as Property 3. Property 3: If a CPF solution has no adjacent CPF solutions that are better (as measured by Z), then there are no better CPF solutions anywhere. Therefore, such a CPF solution is guaranteed to be an optimal solution (by Property 1), assuming only that the problem possesses at least one optimal solution (guaranteed if the problem possesses feasible solutions and a bounded feasible region). To illustrate Property 3, consider Fig. 5.1 for the Wyndor Glass Co. example. For the CPF solution (2, 6), its adjacent CPF solutions are (0, 6) and (4, 3), and neither has a better value of Z than (2, 6) does. This outcome implies that none of the other CPF solutions—(0, 0) and (4, 0)—can be better than (2, 6), so (2, 6) must be optimal. By contrast, Fig. 5.3 shows a feasible region that can never occur for a linear programming problem but that does violate Property 3. The problem shown is identical to the Wyndor Glass Co. example (including the same objective function) except for the en-

FIGURE 5.3 Modification of the Wyndor Glass Co. problem that violates both linear programming and Property 3 for CPF solutions in linear programming.

x2

6

(0, 6)

(2, 6)

( 83 , 5) (4, 5) Z 36 3x1 5x2

4

2

(4, 0) (0, 0)

2

4

x1

198

5

THE THEORY OF THE SIMPLEX METHOD

largement of the feasible region to the right of ( 83 , 5). Consequently, the adjacent CPF solutions for (2, 6) now are (0, 6) and ( 83 , 5), and again neither is better than (2, 6). However, another CPF solution (4, 5) now is better than (2, 6), thereby violating Property 3. The reason is that the boundary of the feasible region goes down from (2, 6) to ( 83 , 5) and then “bends outward” to (4, 5), beyond the objective function line passing through (2, 6). The key point is that the kind of situation illustrated in Fig. 5.3 can never occur in linear programming. The feasible region in Fig. 5.3 implies that the 2x2 12 and 3x1 2x2 18 constraints apply for 0 x1 83 . However, under the condition that 83 x1 4, the 3x1 2x2 18 constraint is dropped and replaced by x2 5. Such “conditional constraints” just are not allowed in linear programming. The basic reason that Property 3 holds for any linear programming problem is that the feasible region always has the property of being a convex set, as defined in Appendix 2 and illustrated in several figures there. For two-variable linear programming problems, this convex property means that the angle inside the feasible region at every CPF solution is less than 180°. This property is illustrated in Fig. 5.1, where the angles at (0, 0), (0, 6), and (4, 0) are 90° and those at (2, 6) and (4, 3) are between 90° and 180°. By contrast, the feasible region in Fig. 5.3 is not a convex set, because the angle at ( 83 , 5) is more than 180°. This is the kind of “bending outward” at an angle greater than 180° that can never occur in linear programming. In higher dimensions, the same intuitive notion of “never bending outward” continues to apply. To clarify the significance of a convex feasible region, consider the objective function hyperplane that passes through a CPF solution that has no adjacent CPF solutions that are better. [In the original Wyndor Glass Co. example, this hyperplane is the objective function line passing through (2, 6).] All these adjacent solutions [(0, 6) and (4, 3) in the example] must lie either on the hyperplane or on the unfavorable side (as measured by Z) of the hyperplane. The feasible region being convex means that its boundary cannot “bend outward” beyond an adjacent CPF solution to give another CPF solution that lies on the favorable side of the hyperplane. So Property 3 holds. Extensions to the Augmented Form of the Problem For any linear programming problem in our standard form (including functional constraints in form), the appearance of the functional constraints after slack variables are introduced is as follows: (1) a11x1 a12x2 a1n xn xn1 (2) a21x1 a22x2 a2n xn xn2

b1 b2

............................................................

(m) am1x1 am2x2 amn xn

xnm bm,

where xn1, xn2, . . . , xnm are the slack variables. For other linear programming problems, Sec. 4.6 described how essentially this same appearance (proper form from Gaussian elimination) can be obtained by introducing artificial variables, etc. Thus, the original solutions (x1, x2, . . . , xn) now are augmented by the corresponding values of the slack or artificial variables (xn1, xn2, . . . , xnm) and perhaps some surplus variables as well. This augmentation led in Sec. 4.2 to defining basic solutions as augmented corner-point solutions and basic feasible solutions (BF solutions) as augmented CPF so-

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

199

lutions. Consequently, the preceding three properties of CPF solutions also hold for BF solutions. Now let us clarify the algebraic relationships between basic solutions and corner-point solutions. Recall that each corner-point solution is the simultaneous solution of a system of n constraint boundary equations, which we called its defining equations. The key question is: How do we tell whether a particular constraint boundary equation is one of the defining equations when the problem is in augmented form? The answer, fortunately, is a simple one. Each constraint has an indicating variable that completely indicates (by whether its value is zero) whether that constraint’s boundary equation is satisfied by the current solution. A summary appears in Table 5.3. For the type of constraint in each row of the table, note that the corresponding constraint boundary equation (fourth column) is satisfied if and only if this constraint’s indicating variable (fifth column) equals zero. In the last row (functional constraint in form), the indicating variable xni xsi actually is the difference between the artificial variable xni and the surplus variable xsi . Thus, whenever a constraint boundary equation is one of the defining equations for a corner-point solution, its indicating variable has a value of zero in the augmented form of the problem. Each such indicating variable is called a nonbasic variable for the corresponding basic solution. The resulting conclusions and terminology (already introduced in Sec. 4.2) are summarized next. Each basic solution has m basic variables, and the rest of the variables are nonbasic variables set equal to zero. (The number of nonbasic variables equals n plus the number of surplus variables.) The values of the basic variables are given by the simultaneous solution of the system of m equations for the problem in augmented form (after the nonbasic variables are set to zero). This basic solution is the augmented corner-point solution whose n defining equations are those indicated by the nonbasic variables. In particular, whenever an indicating variable in the fifth column of Table 5.3 is a nonbasic variable, the constraint boundary equation in the fourth column is a defining equation for the corner-point solution. (For functional constraints in form, at least one of the two supplementary variables xni and xsi always is a nonbasic variable, but the constraint boundary equation becomes a defining equation only if both of these variables are nonbasic variables.)

TABLE 5.3 Indicating variables for constraint boundary equations* Type of Constraint

Form of Constraint

Nonnegativity

xj 0 n

Functional ()

aijxj bi

j1 n

Functional ()

aijxj bi j1

Functional ()

aijxj bi j1

n

Constraint in Augmented Form xj 0 n

Constraint Boundary Equation

Indicating Variable

xj 0

xj

n

aijxj xni bi

aijxj bi

j1

xni

j1

n

n

aijxj xni bi j1

aijxj bi j1

n

aijxj xni xs bi j1 i

*Indicating variable 0 ⇒ constraint boundary equation satisfied; indicating variable 0 ⇒ constraint boundary equation violated.

n

aijxj bi j1

xni xni xsi

200

5

THE THEORY OF THE SIMPLEX METHOD

Now consider the basic feasible solutions. Note that the only requirements for a solution to be feasible in the augmented form of the problem are that it satisfy the system of equations and that all the variables be nonnegative. A BF solution is a basic solution where all m basic variables are nonnegative ( 0). A BF solution is said to be degenerate if any of these m variables equals zero.

Thus, it is possible for a variable to be zero and still not be a nonbasic variable for the current BF solution. (This case corresponds to a CPF solution that satisfies another constraint boundary equation in addition to its n defining equations.) Therefore, it is necessary to keep track of which is the current set of nonbasic variables (or the current set of basic variables) rather than to rely upon their zero values. We noted earlier that not every system of n constraint boundary equations yields a corner-point solution, because either the system has no solution or it has multiple solutions. For analogous reasons, not every set of n nonbasic variables yields a basic solution. However, these cases are avoided by the simplex method. To illustrate these definitions, consider the Wyndor Glass Co. example once more. Its constraint boundary equations and indicating variables are shown in Table 5.4. Augmenting each of the CPF solutions (see Table 5.1) yields the BF solutions listed in Table 5.5. This table places adjacent BF solutions next to each other, except for the pair consisting of the first and last solutions listed. Notice that in each case the nonbasic variables necessarily are the indicating variables for the defining equations. Thus, adjacent BF solutions differ by having just one different nonbasic variable. Also notice that each BF solution is the simultaneous solution of the system of equations for the problem in augmented form (see Table 5.4) when the nonbasic variables are set equal to zero. Similarly, the three corner-point infeasible solutions (see Table 5.2) yield the three basic infeasible solutions shown in Table 5.6. The other two sets of nonbasic variables, (1) x1 and x3 and (2) x2 and x4, do not yield a basic solution, because setting either pair of variables equal to zero leads to having no solution for the system of Eqs. (1) to (3) given in Table 5.4. This conclusion parallels the observation we made early in this section that the corresponding sets of constraint boundary equations do not yield a solution.

TABLE 5.4 Indicating variables for the constraint boundary equations of the Wyndor Glass Co. problem* Constraint

Constraint in Augmented Form

Constraint Boundary Equation

Indicating Variable

x1 0 x2 0 x1 4 2x2 12 3x1 x2 18

x1 0 x2 0 (1) 2x1 2x2 x3x3x3 24 (2) 3x1 2x2 x3x4x3 12 (3) 3x1 2x2 x3x3x5 18

x1 0 x2 0 x1 4 2x2 12 3x1 2x2 18

x1 x2 x3 x4 x5

*Indicating variable 0 ⇒ constraint boundary equation satisfied; indicating variable 0 ⇒ constraint boundary equation violated.

5.1 FOUNDATIONS OF THE SIMPLEX METHOD

201

TABLE 5.5 BF solutions for the Wyndor Glass Co. problem CPF Solution

Defining Equations

(0, 0)

x1 0 x2 0

(0, 6)

BF Solution

Nonbasic Variables

(0, 0, 4, 12, 18)

x1 x2

x1 0 2x2 12

(0, 6, 4, 0, 6)

x1 x4

(2, 6)

2x2 12 3x1 2x2 18

(2, 6, 2, 0, 0)

x4 x5

(4, 3)

3x1 2x2 18 x1 4

(4, 3, 0, 6, 0)

x5 x3

(4, 0)

x1 4 x2 0

(4, 0, 0, 12, 6)

x3 x2

The simplex method starts at a BF solution and then iteratively moves to a better adjacent BF solution until an optimal solution is reached. At each iteration, how is the adjacent BF solution reached? For the original form of the problem, recall that an adjacent CPF solution is reached from the current one by (1) deleting one constraint boundary (defining equation) from the set of n constraint boundaries defining the current solution, (2) moving away from the current solution in the feasible direction along the intersection of the remaining n 1 constraint boundaries (an edge of the feasible region), and (3) stopping when the first new constraint boundary (defining equation) is reached. Equivalently, in our new terminology, the simplex method reaches an adjacent BF solution from the current one by (1) deleting one variable (the entering basic variable) from the set of n nonbasic variables defining the current solution, (2) moving away from the current solution by increasing this one variable from zero (and adjusting the other basic variables to still satisfy the system of equations) while keeping the remaining n 1 nonbasic variables at zero, and (3) stopping when the first of the basic variables (the leaving basic variable) reaches a value of zero (its constraint boundary). With either interpretation, the choice among the n alternatives in step 1 is made by selecting the one that would give the best rate of improvement in Z (per unit increase in the entering basic variable) during step 2. TABLE 5.6 Basic infeasible solutions for the Wyndor Glass Co. problem Corner-Point Infeasible Solution

Defining Equations

Basic Infeasible Solution

Nonbasic Variables

(0, 9)

x1 0 3x1 2x2 18

(0, 9, 4, 6, 0)

x1 x5

(4, 6)

2x2 12 x1 4

(4, 6, 0, 0, 6)

x4 x3

(6, 0)

3x1 2x2 18 x2 0

(6, 0, 2, 12, 0)

x5 x2

202

5

THE THEORY OF THE SIMPLEX METHOD

TABLE 5.7 Sequence of solutions obtained by the simplex method for the Wyndor Glass Co. problem Iteration

CPF Solution

Defining Equations

0

(0, 0)

x1 0 x2 0

1

(0, 6)

2

(2, 6)

BF Solution

Nonbasic Variables

Functional Constraints in Augmented Form

(0, 0, 4, 12, 18)

x1 0 x2 0

x1 2x2 x3 4 2x2 x4 12 3x1 2x2 x5 18

x1 0 2x2 12

(0, 6, 4, 0, 6)

x1 0 x4 0

x1 2x2 x3 4 2x2 x4 12 3x1 2x2 x5 18

2x2 12 3x1 2x2 18

(2, 6, 2, 0, 0)

x4 0 x5 0

x1 2x2 x3 4 2x2 x4 12 3x1 2x2 x5 18

Table 5.7 illustrates the close correspondence between these geometric and algebraic interpretations of the simplex method. Using the results already presented in Secs. 4.3 and 4.4, the fourth column summarizes the sequence of BF solutions found for the Wyndor Glass Co. problem, and the second column shows the corresponding CPF solutions. In the third column, note how each iteration results in deleting one constraint boundary (defining equation) and substituting a new one to obtain the new CPF solution. Similarly, note in the fifth column how each iteration results in deleting one nonbasic variable and substituting a new one to obtain the new BF solution. Furthermore, the nonbasic variables being deleted and added are the indicating variables for the defining equations being deleted and added in the third column. The last column displays the initial system of equations [excluding Eq. (0)] for the augmented form of the problem, with the current basic variables shown in bold type. In each case, note how setting the nonbasic variables equal to zero and then solving this system of equations for the basic variables must yield the same solution for (x1, x2) as the corresponding pair of defining equations in the third column.

5.2

THE REVISED SIMPLEX METHOD The simplex method as described in Chap. 4 (hereafter called the original simplex method ) is a straightforward algebraic procedure. However, this way of executing the algorithm (in either algebraic or tabular form) is not the most efficient computational procedure for computers because it computes and stores many numbers that are not needed at the current iteration and that may not even become relevant for decision making at subsequent iterations. The only pieces of information relevant at each iteration are the coefficients of the nonbasic variables in Eq. (0), the coefficients of the entering basic variable in the other equations, and the right-hand sides of the equations. It would be very useful to have a procedure that could obtain this information efficiently without computing and storing all the other coefficients. As mentioned in Sec. 4.8, these considerations motivated the development of the revised simplex method. This method was designed to accomplish exactly the same things as the original simplex method, but in a way that is more efficient for execution on a computer. Thus, it is a streamlined version of the original procedure. It computes and stores

5.2 THE REVISED SIMPLEX METHOD

203

only the information that is currently needed, and it carries along the essential data in a more compact form. The revised simplex method explicitly uses matrix manipulations, so it is necessary to describe the problem in matrix notation. (See Appendix 4 for a review of matrices.) To help you distinguish between matrices, vectors, and scalars, we consistently use BOLDFACE CAPITAL letters to represent matrices, boldface lowercase letters to represent vectors, and italicized letters in ordinary print to represent scalars. We also use a boldface zero (0) to denote a null vector (a vector whose elements all are zero) in either column or row form (which one should be clear from the context), whereas a zero in ordinary print (0) continues to represent the number zero. Using matrices, our standard form for the general linear programming model given in Sec. 3.2 becomes Z cx,

Maximize subject to Ax b

x 0,

and

where c is the row vector c [c1, c2, . . . , cn], x, b, and 0 are the column vectors such that x1 x x 2 , xn

b1 b b 2 , bm

0 0 0 , 0

and A is the matrix a11 a12 … a1n a a22 … a2n A 21 . ……………………… am1 am2 … amn To obtain the augmented form of the problem, introduce the column vector of slack variables xn1 x xs n2 xnm so that the constraints become [A, I]

x b x

s

and

x 0, x

s

204

5

THE THEORY OF THE SIMPLEX METHOD

where I is the m m identity matrix, and the null vector 0 now has n m elements. (We comment at the end of the section about how to deal with problems that are not in our standard form.) Solving for a Basic Feasible Solution Recall that the general approach of the simplex method is to obtain a sequence of improving BF solutions until an optimal solution is reached. One of the key features of the revised simplex method involves the way in which it solves for each new BF solution after identifying its basic and nonbasic variables. Given these variables, the resulting basic solution is the solution of the m equations [A, I]

x b, x

s

in which the n nonbasic variables from the n m elements of

x x

s

are set equal to zero. Eliminating these n variables by equating them to zero leaves a set of m equations in m unknowns (the basic variables). This set of equations can be denoted by BxB b, where the vector of basic variables xB1 x xB B2 xBm is obtained by eliminating the nonbasic variables from

x , x

s

and the basis matrix B11 B12 … B1m B B22 … B2m 21 B ………………………… Bm1 Bm2 … Bmm is obtained by eliminating the columns corresponding to coefficients of nonbasic variables from [A, I]. (In addition, the elements of xB and, therefore, the columns of B may be placed in a different order when the simplex method is executed.) The simplex method introduces only basic variables such that B is nonsingular, so that B1 always will exist. Therefore, to solve BxB b, both sides are premultiplied by B1: B1BxB B1b.

5.2 THE REVISED SIMPLEX METHOD

205

Since B1B I, the desired solution for the basic variables is xB B1b. Let cB be the vector whose elements are the objective function coefficients (including zeros for slack variables) for the corresponding elements of xB. The value of the objective function for this basic solution is then Z cBxB cBB1b. Example. To illustrate this method of solving for a BF solution, consider again the Wyndor Glass Co. problem presented in Sec. 3.1 and solved by the original simplex method in Table 4.8. In this case, c [3, 5],

1 [A, I] 0 3

0 2 2

1 0 0

0 1 0

0 0 , 1

4 b 12 , 18

x x 1 , x2

x3 xs x4 . x5

Referring to Table 4.8, we see that the sequence of BF solutions obtained by the simplex method (original or revised) is the following: Iteration 0 x3 xB x4 , x 5

1 B 0 0

cB [0, 0, 0],

0 1 0

0 0 B1, 1

so

1 x3 x4 0 0 x5

0 1 0

4 0 4 0 12 12 , 1 18 18

4 Z [0, 0, 0] 12 0. 18

so

Iteration 1 x3 xB x2 , x5

1 B 0 0

1 x3 x2 0 0 x5

0

0 2 2

0 0 , 1

1 B1 0 0

so 4 0 4 0 12 6 , 1 18 6

1 2

1

cB [0, 5, 0],

so

4 Z [0, 5, 0] 6 30. 6

0 1 2

1

0 0 , 1

206

5

THE THEORY OF THE SIMPLEX METHOD

Iteration 2 x3 xB x2 , x1

1 B 0 0

1 x3 x2 0 0 x1

1 3 1 2 1 3

0 2 2

1 0 , 3

1 1 B 0 0

1 3 1 2 1 3

13 0 , 1 3

so

cB [0, 5, 3],

1 3 4 2 0 12 6 , 1 2 3 18 2 Z [0, 5, 3] 6 36. 2

so

Matrix Form of the Current Set of Equations The last preliminary before we summarize the revised simplex method is to show the matrix form of the set of equations appearing in the simplex tableau for any iteration of the original simplex method. For the original set of equations, the matrix form is

1 0

Z c 0 0 . x A I b xs

This set of equations also is exhibited in the first simplex tableau of Table 5.8. The algebraic operations performed by the simplex method (multiply an equation by a constant and add a multiple of one equation to another equation) are expressed in maTABLE 5.8 Initial and later simplex tableaux in matrix form Coefficient of: Basic Variable

Eq.

Z

Original Variables

Slack Variables

Right Side

0

Z xB

(0) (1, 2, . . . , m)

1 0

c A

0 I

0 b

Any

Z xB

(0) (1, 2, . . . m)

1 0

cBB1A c B1 A

cBB1 B1

Iteration

cBB1b B1b

5.2 THE REVISED SIMPLEX METHOD

207

trix form by premultiplying both sides of the original set of equations by the appropriate matrix. This matrix would have the same elements as the identity matrix, except that each multiple for an algebraic operation would go into the spot needed to have the matrix multiplication perform this operation. Even after a series of algebraic operations over several iterations, we still can deduce what this matrix must be (symbolically) for the entire series by using what we already know about the right-hand sides of the new set of equations. In particular, after any iteration, xB B1b and Z cBB1b, so the right-hand sides of the new set of equations have become cBB1b cBB1 0 . B1 b B1b

Z 1 xB 0

Because we perform the same series of algebraic operations on both sides of the original set of operations, we use this same matrix that premultiplies the original right-hand side to premultiply the original left-hand side. Consequently, since 1 cBB1 1 0 B1 0

cBB1A c cBB1 , B1A B1

c 0 1 A I 0

the desired matrix form of the set of equations after any iteration is

1 0

cBB1A c B1A

Z cBB1 cBB1b . x 1 B B1b x s

The second simplex tableau of Table 5.8 also exhibits this same set of equations. Example. To illustrate this matrix form for the current set of equations, we will show how it yields the final set of equations resulting from iteration 2 for the Wyndor Glass Co. problem. Using the B1 and cB given for iteration 2 at the end of the preceding subsection, we have 1 B A 0 0

1 3 1 2 1 3

13 1 0 0 1 3 3

1 3 1 2 1 3

13 0 [0, 32 , 1], 1 3

1

1

cBB

1 [0, 5, 3] 0 0

0 cBB1A c [0, 5, 3] 0 1

0 0 2 0 2 1

0 1 [3, 5] [0, 0]. 0

0 1 , 0

208

5

THE THEORY OF THE SIMPLEX METHOD

Also, by using the values of xB B1b and Z cBB1b calculated at the end of the preceding subsection, these results give the following set of equations: 1 0 0 0

0 0 0 1

0 0 1 0

0 1 0 0

3 2 1 3 1 2 1 3

Z 1 x1 1 3 x2 0 x3 1 x4 3 x5

36 2 , 6 2

as shown in the final simplex tableau in Table 4.8.

The Overall Procedure There are two key implications from the matrix form of the current set of equations shown at the bottom of Table 5.8. The first is that only B1 needs to be derived to be able to calculate all the numbers in the simplex tableau from the original parameters (A, b, cB) of the problem. (This implication is the essence of the fundamental insight described in the next section.) The second is that any one of these numbers can be obtained individually, usually by performing only a vector multiplication (one row times one column) instead of a complete matrix multiplication. Therefore, the required numbers to perform an iteration of the simplex method can be obtained as needed without expending the computational effort to obtain all the numbers. These two key implications are incorporated into the following summary of the overall procedure. Summary of the Revised Simplex Method. 1. Initialization: Same as for the original simplex method. 2. Iteration: Step 1 Determine the entering basic variable: Same as for the original simplex method. Step 2 Determine the leaving basic variable: Same as for the original simplex method, except calculate only the numbers required to do this [the coefficients of the entering basic variable in every equation but Eq. (0), and then, for each strictly positive coefficient, the right-hand side of that equation].1 Step 3 Determine the new BF solution: Derive B1 and set xB B1b. 3. Optimality test: Same as for the original simplex method, except calculate only the numbers required to do this test, i.e., the coefficients of the nonbasic variables in Eq. (0). In step 3 of an iteration, B1 could be derived each time by using a standard computer routine for inverting a matrix. However, since B (and therefore B1) changes so little from one iteration to the next, it is much more efficient to derive the new B1 (denote it by B1 new) from the B1 at the preceding iteration (denote it by B1 old). (For the initial BF solution, 1

Because the value of xB is the entire vector of right-hand sides except for Eq. (0), the relevant right-hand sides need not be calculated here if xB was calculated in step 3 of the preceding iteration.

5.2 THE REVISED SIMPLEX METHOD

209

B I B1.) One method for doing this derivation is based directly upon the interpretation of the elements of B1 [the coefficients of the slack variables in the current Eqs. (1), (2), . . . , (m)] presented in the next section, as well as upon the procedure used by the original simplex method to obtain the new set of equations from the preceding set. To describe this method formally, let xk entering basic variable, a ik coefficient of xk in current Eq. (i), for i 1, 2, . . . , m (calculated in step 2 of an iteration), r number of equation containing the leaving basic variable. Recall that the new set of equations [excluding Eq. (0)] can be obtained from the preceding set by subtracting a ik /a rk times Eq. (r) from Eq. (i), for all i 1, 2, . . . , m except i r, and then dividing Eq. (r) by a rk. Therefore, the element in row i and column j of B1 new is (B1 new)ij

a ik 1 (B1 old)rj old)ij a r (B k 1 (B1 ) a rk old rj

if i r, if i r.

These formulas are expressed in matrix notation as 1 B1 new EB old,

where matrix E is an identity matrix except that its rth column is replaced by the vector 1 2 , m

where

a i k a i rk 1 a rk

if i r, if i r.

Thus, E [U1, U2, . . . , Ur1, , Ur1, . . . , Um], where the m elements of each of the Ui column vectors are 0 except for a 1 in the ith position. Example. We shall illustrate the revised simplex method by applying it to the Wyndor Glass Co. problem. The initial basic variables are the slack variables x3 xB x4 . x5 Iteration 1 Because the initial B1 I, no calculations are needed to obtain the numbers required to identify the entering basic variable x2 (c2 5 3 c1) and the leaving basic variable x4 (a12 0, b2/a22 1 2 2 1 2 8 b3/a32, so r 2). Thus, the new set of basic variables is x3 xB x2 . x5

210

5

THE THEORY OF THE SIMPLEX METHOD

To obtain the new B1, a12 0 a22 1 1 , 2 a22 a32 1 a22 so

B

1

1 0 0

0 1 0 0 1 0

0 1 2

1

1 0 0 0 1 0

0 1 0

0 1 2

1

0 0 , 1

so that 1 xB 0 0

4 0 4 0 12 6 . 1 18 6

0 1 2

1

To test whether this solution is optimal, we calculate the coefficients of the nonbasic variables (x1 and x4) in Eq. (0). Performing only the relevant parts of the matrix multiplications, we obtain 1 cBB1A c [0, 5, 0] 0 0 1

cBB

— [0, 5, 0] — —

0 1 0 0 1 3

0 1 2

1 0 1 2

1

— — [3, —] [3, —], —

— — [—, 52 , —], —

so the coefficients of x1 and x4 are 3 and 52 , respectively. Since x1 has a negative coefficient, this solution is not optimal. Iteration 2 Using these coefficients of the nonbasic variables in Eq. (0), since only x1 has a negative coefficient, we begin the next iteration by identifying x1 as the entering basic variable. To determine the leaving basic variable, we must calculate the other coefficients of x1: 1 B1A 0 0

0 1 2

1

0 1 0 0 1 3

1 — — 0 — 3

— — . —

5.2 THE REVISED SIMPLEX METHOD

211

By using the right side column for the current BF solution (the value of xB) just given for iteration 1, the ratios 4/1 6/3 indicate that x5 is the leaving basic variable, so the new set of basic variables is x3 xB x2 x1

a 11 1 a 31 3 a 21 0 . a 31 1 1 3 a 31

with

Therefore, the new B1 is 1 B1 0 0

0 1 3 1 1 0 0 1 0 3 0

0 1 2

1

1 0 0 0 1 0

1 3 1 2 1 3

1 3 0 , 1 3

so that 1 xB 0 0

1 3 1 2 1 3

2 1 3 4 0 12 6 . 1 2 3 18

Applying the optimality test, we find that the coefficients of the nonbasic variables (x4 and x5) in Eq. (0) are — cBB1 [0, 5, 3] — —

1 3 1 2 1 3

1 3 0 [—, 32 , 1]. 1 3

Because both coefficients ( and 1) are nonnegative, the current solution (x1 2, x2 6, x3 2, x4 0, x5 0) is optimal and the procedure terminates. 3 2

General Observations The preceding discussion was limited to the case of linear programming problems fitting our standard form given in Sec. 3.2. However, the modifications for other forms are relatively straightforward. The initialization would be conducted just as it would for the original simplex method (see Sec. 4.6). When this step involves introducing artificial variables to obtain an initial BF solution (and thereby to obtain an identity matrix as the initial basis matrix), these variables are included among the m elements of xs. Let us summarize the advantages of the revised simplex method over the original simplex method. One advantage is that the number of arithmetic computations may be reduced. This is especially true when the A matrix contains a large number of zero elements (which is usually the case for the large problems arising in practice). The amount of information that must be stored at each iteration is less, sometimes considerably so. The revised simplex method also permits the control of the rounding errors inevitably generated

212

5 THE THEORY OF THE SIMPLEX METHOD

by computers. This control can be exercised by periodically obtaining the current B1 by directly inverting B. Furthermore, some of the postoptimality analysis problems discussed in Sec. 4.7 can be handled more conveniently with the revised simplex method. For all these reasons, the revised simplex method is usually preferable to the original simplex method for computer execution.

5.3

A FUNDAMENTAL INSIGHT We shall now focus on a property of the simplex method (in any form) that has been revealed by the revised simplex method in the preceding section.1 This fundamental insight provides the key to both duality theory and sensitivity analysis (Chap. 6), two very important parts of linear programming. The insight involves the coefficients of the slack variables and the information they give. It is a direct result of the initialization, where the ith slack variable xni is given a coefficient of 1 in Eq. (i) and a coefficient of 0 in every other equation [including Eq. (0)] for i 1, 2, . . . , m, as shown by the null vector 0 and the identity matrix I in the slack variables column for iteration 0 in Table 5.8. (For most of this section, we are assuming that the problem is in our standard form, with bi 0 for all i 1, 2, . . . , m, so that no additional adjustments are needed in the initialization.) The other key factor is that subsequent iterations change the initial equations only by 1. Multiplying (or dividing) an entire equation by a nonzero constant 2. Adding (or subtracting) a multiple of one entire equation to another entire equation As already described in the preceding section, a sequence of these kinds of elementary algebraic operations is equivalent to premultiplying the initial simplex tableau by some matrix. (See Appendix 4 for a review of matrices.) The consequence can be summarized as follows. Verbal description of fundamental insight: After any iteration, the coefficients of the slack variables in each equation immediately reveal how that equation has been obtained from the initial equations. As one example of the importance of this insight, recall from Table 5.8 that the matrix formula for the optimal solution obtained by the simplex method is xB B1b, where xB is the vector of basic variables, B1 is the matrix of coefficients of slack variables for rows 1 to m of the final tableau, and b is the vector of original right-hand sides (resource availabilities). (We soon will denote this particular B1 by S*.) Postoptimality analysis normally includes an investigation of possible changes in b. By using this formula, you can see exactly how the optimal BF solution changes (or whether it becomes infeasible because of negative variables) as a function of b. You do not have to reapply the simplex method over and over for each new b, because the coefficients of the slack 1

However, since some instructors do not cover the preceding section, we have written this section in a way that can be understood without first reading Sec. 5.2. It is helpful to take a brief look at the matrix notation introduced at the beginning of Sec. 5.2, including the resulting key equation, xB B1b.

5.3 A FUNDAMENTAL INSIGHT

213

variables tell all! In a similar fashion, this fundamental insight provides a tremendous computational saving for the rest of sensitivity analysis as well. To spell out the how and the why of this insight, let us look again at the Wyndor Glass Co. example. (The OR Tutor also includes another demonstration example.) Example. Table 5.9 shows the relevant portion of the simplex tableau for demonstrating this fundamental insight. Light lines have been drawn around the coefficients of the slack variables in all the tableaux in this table because these are the crucial coefficients for applying the insight. To avoid clutter, we then identify the pivot row and pivot column by a single box around the pivot number only. Iteration 1 To demonstrate the fundamental insight, our focus is on the algebraic operations performed by the simplex method while using Gaussian elimination to obtain the new BF solution. If we do all the algebraic operations with the old row 2 (the pivot row) rather than the new one, then the algebraic operations spelled out in Chap. 4 for iteration 1 are New New New New

row row row row

0 old row 0 ( 52 )(old 1 old row 1 (0)(old 2 ( 12 )(old 3 old row 3 (1)(old

row row row row

2), 2), 2), 2).

TABLE 5.9 Simplex tableaux without leftmost columns for the Wyndor Glass Co. problem Coefficient of: Iteration

x1

x2

x3

0

3 1 0 3

5 0 2 2

0 1 0 0

3

0

0

1

0

1

0

1

0

3

0

0

0

0

0

3 2

0

0

1

1 3

0

1

0

1

0

0

1

2

x4 0 0 1 0 5 2 0 1 2 1

x5

Right Side

0 0 0 1

0 4 12 18

0

30

0

4

0

6

1

6

1

36

1

3

2

1 2

0

6

1 3

1 3

2

214

5

THE THEORY OF THE SIMPLEX METHOD

Ignoring row 0 for the moment, we see that these algebraic operations amount to premultiplying rows 1 to 3 of the initial tableau by the matrix 1 0 0

0 1 2

1

0 0 . 1

Rows 1 to 3 of the initial tableau are 1 Old rows 1–3 0 3

0 2 2

1 0 0

0 1 0

4 12 , 18

0 0 1

where the third, fourth, and fifth columns (the coefficients of the slack variables) form an identity matrix. Therefore, 1 New rows 1–3 0 0 1 0 3

0 1 0

1 0 0

0 1 2

1 0 1 2

1

0 1 0 0 1 3 0 0 1

0 2 2

1 0 0

0 1 0

4 12 18

0 0 1

4 6 . 6

Note how the first matrix is reproduced exactly in the box below it as the coefficients of the slack variables in rows 1 to 3 of the new tableau, because the coefficients of the slack variables in rows 1 to 3 of the initial tableau form an identity matrix. Thus, just as stated in the verbal description of the fundamental insight, the coefficients of the slack variables in the new tableau do indeed provide a record of the algebraic operations performed. This insight is not much to get excited about after just one iteration, since you can readily see from the initial tableau what the algebraic operations had to be, but it becomes invaluable after all the iterations are completed. For row 0, the algebraic operation performed amounts to the following matrix calculations, where now our focus is on the vector [0, 52, 0] that premultiplies rows 1 to 3 of the initial tableau. New row 0 [3,

5 0,

0,

0 0] [0,

[3,

0,

0,

5 2

, 5 2

,

1 0] 0 3

0 2 2

0,

30].

1 0 0

0 1 0

0 0 1

4 12 18

Note how this vector is reproduced exactly in the box below it as the coefficients of the slack variables in row 0 of the new tableau, just as was claimed in the statement of the fundamental insight. (Once again, the reason is the identity matrix for the coefficients of the slack variables in rows 1 to 3 of the initial tableau, along with the zeros for these coefficients in row 0 of the initial tableau.)

5.3 A FUNDAMENTAL INSIGHT

215

Iteration 2 The algebraic operations performed on the second tableau of Table 5.9 for iteration 2 are New New New New

row row row row

0 old row 0 (1)(old row 3), 1 old row 1 (1 3 )(old row 3), 2 old row 2 (0)(old row 3), 3 (1 3 )(old row 3).

Ignoring row 0 for the moment, we see that these operations amount to premultiplying rows 1 to 3 of this tableau by the matrix 1 0 0

0 1 3 1 0 . 1 0 3

Writing this second tableau as the matrix product shown for iteration 1 (namely, the corresponding matrix times rows 1 to 3 of the initial tableau) then yields 1 Final rows 1–3 0 0

0 0 1

0 1 3 1 0 0 1 1 0 3 0 1 0 0 0 1 0

1 0 0

1

0 1 0 0 1 3

0 2 2

1 0 0

0 1 0

0 0 1

4 12 18

1 3 1 2 1 3

1 3 1 0 0 1 3 3

0 2 2

1 0 0

0 1 0

0 0 1

4 12 18

1 3 1 2 1 3

13 0

0 1 2

1 3

2 6 . 2

The first two matrices shown on the first line of these calculations summarize the algebraic operations of the second and first iterations, respectively. Their product, shown as the first matrix on the second line, then combines the algebraic operations of the two iterations. Note how this matrix is reproduced exactly in the box below it as the coefficients of the slack variables in rows 1 to 3 of the new (final) tableau shown on the third line. What this portion of the tableau reveals is how the entire final tableau (except row 0) has been obtained from the initial tableau, namely, Final row 1 (1)(initial row 1) ( 13 )(initial row 2) ( 13 )(initial row 3), Final row 2 (0)(initial row 1) (1 2 )(initial row 2) (0)(initial row 3), Final row 3 (0)(initial row 1) (1 3 )(initial row 2) (1 3 )(initial row 3). To see why these multipliers of the initial rows are correct, you would have to trace through all the algebraic operations of both iterations. For example, why does final row 1 include ( 13 )(initial row 2), even though a multiple of row 2 has never been added directly to row 1? The reason is that initial row 2 was subtracted from initial row 3 in iteration 1, and then ( 13 )(old row 3) was subtracted from old row 1 in iteration 2.

216

5

THE THEORY OF THE SIMPLEX METHOD

However, there is no need for you to trace through. Even when the simplex method has gone through hundreds or thousands of iterations, the coefficients of the slack variables in the final tableau will reveal how this tableau has been obtained from the initial tableau. Furthermore, the same algebraic operations would give these same coefficients even if the values of some of the parameters in the original model (initial tableau) were changed, so these coefficients also reveal how the rest of the final tableau changes with changes in the initial tableau. To complete this story for row 0, the fundamental insight reveals that the entire final row 0 can be calculated from the initial tableau by using just the coefficients of the slack variables in the final row 0—[0, 32, 1]. This calculation is shown below, where the first vector is row 0 of the initial tableau and the matrix is rows 1 to 3 of the initial tableau. Final row 0 [3,

5 0,

0,

0 0] [0, [0,

0,

0,

3 2

, 3 2

,

1 1] 0 3

0 2 2

1,

36].

1 0 0

0 1 0

0 0 1

4 12 18

Note again how the vector premultiplying rows 1 to 3 of the initial tableau is reproduced exactly as the coefficients of the slack variables in the final row 0. These quantities must be identical because of the coefficients of the slack variables in the initial tableau (an identity matrix below a null vector). This conclusion is the row 0 part of the fundamental insight. Mathematical Summary Because its primary applications involve the final tableau, we shall now give a general mathematical expression for the fundamental insight just in terms of this tableau, using matrix notation. If you have not read Sec. 5.2, you now need to know that the parameters of the model are given by the matrix A aij and the vectors b bi and c cj, as displayed at the beginning of that section. The only other notation needed is summarized and illustrated in Table 5.10. Notice how vector t (representing row 0) and matrix T (representing the other rows) together correspond to the rows of the initial tableau in Table 5.9, whereas vector t* and matrix T* together correspond to the rows of the final tableau in Table 5.9. This table also shows these vectors and matrices partitioned into three parts: the coefficients of the original variables, the coefficients of the slack variables (our focus), and the right-hand side. Once again, the notation distinguishes between parts of the initial tableau and the final tableau by using an asterisk only in the latter case. For the coefficients of the slack variables (the middle part) in the initial tableau of Table 5.10, notice the null vector 0 in row 0 and the identity matrix I below, which provide the keys for the fundamental insight. The vector and matrix in the same location of the final tableau, y* and S*, then play a prominent role in the equations for the fundamental insight. A and b in the initial tableau turn into A* and b* in the final tableau. For row 0 of the final tableau, the coefficients of the decision variables are z* c (so the vector z* is what has been added to the vector of initial coefficients, c), and the right-hand side Z* denotes the optimal value of Z.

5.3 A FUNDAMENTAL INSIGHT

217

TABLE 5.10 General notation for initial and final simplex tableaux in matrix form, illustrated by the Wyndor Glass Co. problem Initial Tableau Row 0:

t [3, 5

Other rows:

T

Combined:

T A

1 0 3

t

0, 0, 0

0 2 2

1 0 0

0 1 0

0 0 1

c

0 I

0 . b

0] [c 0 0]. 4 12 [A I b]. 18

Final Tableau Row 0:

t* [0, 0

1 13 2 1 0 0 6 [A* S* b*]. 2 1 1 0 3 2 3

Other rows:

0 T* 0 1

0 1 0

Combined:

T*

z* c A*

t*

36] [z* c y* Z*].

0, 3 2 , 1 1 3

y* S*

Z* b* .

It is helpful at this point to look back at Table 5.8 in Sec. 5.2 and compare it with Table 5.10. (If you haven’t previously studied Sec. 5.2, you will need to read the definition of the basis matrix B and the vectors xB and cB given early in that section before looking at Table 5.8.) The notation for the components of the initial simplex tableau is the same in the two tables. The lower part of Table 5.8 shows any later simplex tableau in matrix form, whereas the lower part of Table 5.10 gives the final tableau in matrix form. Note that the matrix B1 in Table 5.8 is in the same location as S* in Table 5.10. Thus, S* B1 when B is the basis matrix for the optimal solution found by the simplex method. Referring to Table 5.10 again, suppose now that you are given the initial tableau, t and T, and just y* and S* from the final tableau. How can this information alone be used to calculate the rest of the final tableau? The answer is provided by Table 5.8. This table includes some information that is not directly relevant to our current discussion, namely, how y* and S* themselves can be calculated (y* cBB1 and S* B1) by knowing the set of basic variables and so the basis matrix B for the optimal solution found by the simplex method. However, the lower part of this table also shows how the rest of the final tableau can be obtained from the coefficients of the slack variables, which is summarized as follows. Fundamental Insight (1) t* t y*T [y*A c y* y*b]. (2) T* S*T [S*A S* S*b].

218

5

THE THEORY OF THE SIMPLEX METHOD

Thus, by knowing the parameters of the model in the initial tableau (c, A, and b) and only the coefficients of the slack variables in the final tableau (y* and S*), these equations enable calculating all the other numbers in the final tableau. We already used these two equations when dealing with iteration 2 for the Wyndor Glass Co. problem in the preceding subsection. In particular, the right-hand side of the expression for final row 0 for iteration 2 is just t y*T, and the second line of the expression for final rows 1 to 3 is just S*T. Now let us summarize the mathematical logic behind the two equations for the fundamental insight. To derive Eq. (2), recall that the entire sequence of algebraic operations performed by the simplex method (excluding those involving row 0) is equivalent to premultiplying T by some matrix, call it M. Therefore, T* MT, but now we need to identify M. By writing out the component parts of T and T*, this equation becomes [A* S* b*] M [A I b] ↑ [MA M Mb]. ↑ Because the middle (or any other) component of these equal matrices must be the same, it follows that M S*, so Eq. (2) is a valid equation. Equation (1) is derived in a similar fashion by noting that the entire sequence of algebraic operations involving row 0 amounts to adding some linear combination of the rows in T to t, which is equivalent to adding to t some vector times T. Denoting this vector by v, we thereby have t* t vT, but v still needs to be identified. Writing out the component parts of t and t* yields [z* c y* Z*] [c 0 0] v [A I b] [c vA v vb]. ↑ ↑ Equating the middle component of these equal vectors gives v y*, which validates Eq. (1). Adapting to Other Model Forms Thus far, the fundamental insight has been described under the assumption that the original model is in our standard form, described in Sec. 3.2. However, the above mathematical logic now reveals just what adjustments are needed for other forms of the original model. The key is the identity matrix I in the initial tableau, which turns into S* in the final tableau. If some artificial variables must be introduced into the initial tableau to serve as initial basic variables, then it is the set of columns (appropriately ordered) for all the initial basic variables (both slack and artificial) that forms I in this tableau. (The columns for any surplus variables are extraneous.) The same columns in the final tableau provide S* for the T* S*T equation and y* for the t* t y*T equation. If M’s were introduced into the

5.3 A FUNDAMENTAL INSIGHT

219

preliminary row 0 as coefficients for artificial variables, then the t for the t* t y*T equation is the row 0 for the initial tableau after these nonzero coefficients for basic variables are algebraically eliminated. (Alternatively, the preliminary row 0 can be used for t, but then these M’s must be subtracted from the final row 0 to give y*.) (See Prob. 5.3-11.) Applications The fundamental insight has a variety of important applications in linear programming. One of these applications involves the revised simplex method. As described in the preceding section (see Table 5.8), this method used B1 and the initial tableau to calculate all the relevant numbers in the current tableau for every iteration. It goes even further than the fundamental insight by using B1 to calculate y* itself as y* cBB1. Another application involves the interpretation of the shadow prices ( y1*, y2*, . . . , y*m) described in Sec. 4.7. The fundamental insight reveals that Z* (the value of Z for the optimal solution) is m

Z* y*b yi*bi, i1

so, e.g., 3 Z* 0b1 b2 b3 2 for the Wyndor Glass Co. problem. This equation immediately yields the interpretation for the yi* values given in Sec. 4.7. Another group of extremely important applications involves various postoptimality tasks (reoptimization technique, sensitivity analysis, parametric linear programming— described in Sec. 4.7) that investigate the effect of making one or more changes in the original model. In particular, suppose that the simplex method already has been applied to obtain an optimal solution (as well as y* and S*) for the original model, and then these changes are made. If exactly the same sequence of algebraic operations were to be applied to the revised initial tableau, what would be the resulting changes in the final tableau? Because y* and S* don’t change, the fundamental insight reveals the answer immediately. For example, consider the change from b2 12 to b2 13 as illustrated in Fig. 4.8 for the Wyndor Glass Co. problem. It is not necessary to solve for the new optimal solution (x1, x2) ( 53 , 123 ) because the values of the basic variables in the final tableau (b*) are immediately revealed by the fundamental insight: x3 x2 b* S*b x1

73 13 4 0 13 1 23 . 5 1 3 3 18 There is an even easier way to make this calculation. Since the only change is in the second component of b (b2 1), which gets premultiplied by only the second column of S*, the change in b* can be calculated as simply 1 3 b* 12 b2 1 3

1 0 0

13 1 2 , 1 3

1 3

1 2 1 3

220

5

THE THEORY OF THE SIMPLEX METHOD

so the original values of the basic variables in the final tableau (x3 2, x2 6, x1 2) now become x3 2 1 3 17 3 1 13 x2 6 2 2 . 1 5 x1 2 3 3 (If any of these new values were negative, and thus infeasible, then the reoptimization technique described in Sec. 4.7 would be applied, starting from this revised final tableau.) Applying incremental analysis to the preceding equation for Z* also immediately yields 3 3 Z* b2 . 2 2 The fundamental insight can be applied to investigating other kinds of changes in the original model in a very similar fashion; it is the crux of the sensitivity analysis procedure described in the latter part of Chap. 6. You also will see in the next chapter that the fundamental insight plays a key role in the very useful duality theory for linear programming.

5.4

CONCLUSIONS Although the simplex method is an algebraic procedure, it is based on some fairly simple geometric concepts. These concepts enable one to use the algorithm to examine only a relatively small number of BF solutions before reaching and identifying an optimal solution. Chapter 4 describes how elementary algebraic operations are used to execute the algebraic form of the simplex method, and then how the tableau form of the simplex method uses the equivalent elementary row operations in the same way. Studying the simplex method in these forms is a good way of getting started in learning its basic concepts. However, these forms of the simplex method do not provide the most efficient form for execution on a computer. Matrix operations are a faster way of combining and executing elementary algebraic operations or row operations. Therefore, by using the matrix form of the simplex method, the revised simplex method provides an effective way of adapting the simplex method for computer implementation. The final simplex tableau includes complete information on how it can be algebraically reconstructed directly from the initial simplex tableau. This fundamental insight has some very important applications, especially for postoptimality analysis.

SELECTED REFERENCES 1. Bazaraa, M. S., J. J. Jarvis, and H. D. Sherali: Linear Programming and Network Flows, 2d ed., Wiley, New York, 1990. 2. Dantzig, G. B., and M. N. Thapa: Linear Programming 1: Introduction, Springer, New York, 1997. 3. Schriver, A: Theory of Linear and Integer Programming, Wiley, New York, 1986. 4. Vanderbei, R. J.: Linear Programming: Foundations and Extensions, Kluwer Academic Publishers, Boston, MA, 1996.

CHAPTER 5 PROBLEMS

221

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE A Demonstration Example in OR Tutor: Fundamental Insight

Interactive Routines: Enter or Revise a General Linear Programming Model Set Up for the Simplex Method—Interactive Only Solve Interactively by the Simplex Method

Files (Chapter 3) for Solving the Wyndor Example: Excel File LINGO/LINDO File MPL/CPLEX File

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The demonstration example listed above may be helpful. I: You can check some of your work by using the interactive routines listed above for the original simplex method. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 5.1-1.* Consider the following problem. Z 3x1 2x2,

Maximize subject to 2x1 x2 6 x1 2x2 6

5.1-2. Repeat Prob. 5.1-1 for the model in Prob. 3.1-5.

and x1 0,

(d) Do the following for each set of two defining equations from part (b): Identify the indicating variable for each defining equation. Display the set of equations from part (c) after deleting these two indicating (nonbasic) variables. Then use the latter set of equations to solve for the two remaining variables (the basic variables). Compare the resulting basic solution to the corresponding basic solution obtained in part (c). (e) Without executing the simplex method, use its geometric interpretation (and the objective function) to identify the path (sequence of CPF solutions) it would follow to reach the optimal solution. For each of these CPF solutions in turn, identify the following decisions being made for the next iteration: (i) which defining equation is being deleted and which is being added; (ii) which indicating variable is being deleted (the entering basic variable) and which is being added (the leaving basic variable).

x2 0.

(a) Solve this problem graphically. Identify the CPF solutions by circling them on the graph. (b) Identify all the sets of two defining equations for this problem. For each set, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or corner-point infeasible solution. (c) Introduce slack variables in order to write the functional constraints in augmented form. Use these slack variables to identify the basic solution that corresponds to each corner-point solution found in part (b).

5.1-3. Consider the following problem. Maximize

Z 2x1 3x2,

subject to 3x1 x2 1 4x1 2x2 20 4x1 x2 10 x1 2x2 5 and x1 0,

x2 0.

222

5

THE THEORY OF THE SIMPLEX METHOD

(a) Solve this problem graphically. Identify the CPF solutions by circling them on the graph. (b) Develop a table giving each of the CPF solutions and the corresponding defining equations, BF solution, and nonbasic variables. Calculate Z for each of these solutions, and use just this information to identify the optimal solution. (c) Develop the corresponding table for the corner-point infeasible solutions, etc. Also identify the sets of defining equations and nonbasic variables that do not yield a solution.

x1 x2 15 2x1 x2 90 2x1 x2 30

3x1 x2 x3 60 x1 x2 2x3 10 x1 x2 x3 20

and x1 0,

and x2 0,

x3 0.

After slack variables are introduced and then one complete iteration of the simplex method is performed, the following simplex tableau is obtained.

Coefficient of: Iteration

1

Z x1 2x2,

subject to

Z 2x1 x2 x3,

subject to

x1 0,

5.1-7. Consider the following problem. Minimize

5.1-4. Consider the following problem. Maximize

(a) Identify the 10 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or cornerpoint infeasible solution. (b) For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables. (Compare with Table 6.9.)

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

x6

Right Side

Z x4 x1 x6

(0) (1) (2) (3)

1 0 0 0

0 0 1 0

1 4 1 2

3 5 2 3

0 1 0 0

2 3 1 1

0 0 0 1

20 30 10 10

(a) Identify the CPF solution obtained at iteration 1. (b) Identify the constraint boundary equations that define this CPF solution. 5.1-5. Consider the three-variable linear programming problem shown in Fig. 5.2. (a) Construct a table like Table 5.1, giving the set of defining equations for each CPF solution. (b) What are the defining equations for the corner-point infeasible solution (6, 0, 5)? (c) Identify one of the systems of three constraint boundary equations that yields neither a CPF solution nor a corner-point infeasible solution. Explain why this occurs for this system. 5.1-6. Consider the linear programming problem given in Table 6.1 as the dual problem for the Wyndor Glass Co. example.

x2 0.

(a) Solve this problem graphically. (b) Develop a table giving each of the CPF solutions and the corresponding defining equations, BF solution, and nonbasic variables. 5.1-8. Reconsider the model in Problem 4.6-3. (a) Identify the 10 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or a corner-point infeasible solution. (b) For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables. 5.1-9. Reconsider the model in Prob. 3.1-4. (a) Identify the 15 sets of defining equations for this problem. For each one, solve (if a solution exists) for the corresponding corner-point solution, and classify it as a CPF solution or a corner-point infeasible solution. (b) For each corner-point solution, give the corresponding basic solution and its set of nonbasic variables. 5.1-10. Each of the following statements is true under most circumstances, but not always. In each case, indicate when the statement will not be true and why. (a) The best CPF solution is an optimal solution. (b) An optimal solution is a CPF solution. (c) A CPF solution is the only optimal solution if none of its adjacent CPF solutions are better (as measured by the value of the objective function). 5.1-11. Consider the original form (before augmenting) of a linear programming problem with n decision variables (each with a nonnegativity constraint) and m functional constraints. Label each of the following statements as true or false, and then justify your

CHAPTER 5 PROBLEMS

answer with specific references (including page citations) to material in the chapter. (a) If a feasible solution is optimal, it must be a CPF solution. (b) The number of CPF solutions is at least (m n)! . m!n! (c) If a CPF solution has adjacent CPF solutions that are better (as measured by Z), then one of these adjacent CPF solutions must be an optimal solution. 5.1-12. Label each of the following statements about linear programming problems as true or false, and then justify your answer. (a) If a feasible solution is optimal but not a CPF solution, then infinitely many optimal solutions exist. (b) If the value of the objective function is equal at two different feasible points x* and x**, then all points on the line segment connecting x* and x** are feasible and Z has the same value at all those points. (c) If the problem has n variables (before augmenting), then the simultaneous solution of any set of n constraint boundary equations is a CPF solution. 5.1-13. Consider the augmented form of linear programming problems that have feasible solutions and a bounded feasible region. Label each of the following statements as true or false, and then justify your answer by referring to specific statements (with page citations) in the chapter. (a) There must be at least one optimal solution. (b) An optimal solution must be a BF solution. (c) The number of BF solutions is finite. 5.1-14.* Reconsider the model in Prob. 4.6-10. Now you are given the information that the basic variables in the optimal solution are x2 and x3. Use this information to identify a system of three constraint boundary equations whose simultaneous solution must be this optimal solution. Then solve this system of equations to obtain this solution. 5.1-15. Reconsider Prob. 4.3-7. Now use the given information and the theory of the simplex method to identify a system of three constraint boundary equations (in x1, x2, x3) whose simultaneous solution must be the optimal solution, without applying the simplex method. Solve this system of equations to find the optimal solution. 5.1-16. Reconsider Prob. 4.3-8. Using the given information and the theory of the simplex method, analyze the constraints of the problem in order to identify a system of three constraint boundary equations whose simultaneous solution must be the optimal solution (not augmented). Then solve this system of equations to obtain this solution.

223

5.1-17. Consider the following problem. Maximize

Z 2x1 2x2 3x3,

subject to 2x1 x2 2x3 4 x1 x2 x3 3 and x1 0,

x2 0,

x3 0.

Let x4 and x5 be the slack variables for the respective functional constraints. Starting with these two variables as the basic variables for the initial BF solution, you now are given the information that the simplex method proceeds as follows to obtain the optimal solution in two iterations: (1) In iteration 1, the entering basic variable is x3 and the leaving basic variable is x4; (2) in iteration 2, the entering basic variable is x2 and the leaving basic variable is x5. (a) Develop a three-dimensional drawing of the feasible region for this problem, and show the path followed by the simplex method. (b) Give a geometric interpretation of why the simplex method followed this path. (c) For each of the two edges of the feasible region traversed by the simplex method, give the equation of each of the two constraint boundaries on which it lies, and then give the equation of the additional constraint boundary at each endpoint. (d) Identify the set of defining equations for each of the three CPF solutions (including the initial one) obtained by the simplex method. Use the defining equations to solve for these solutions. (e) For each CPF solution obtained in part (d ), give the corresponding BF solution and its set of nonbasic variables. Explain how these nonbasic variables identify the defining equations obtained in part (d ). 5.1-18. Consider the following problem. Maximize

Z 3x1 4x2 2x3,

subject to x1 x2 x3 20 x1 2x2 x3 30 and x1 0,

x2 0,

x3 0.

Let x4 and x5 be the slack variables for the respective functional constraints. Starting with these two variables as the basic variables for the initial BF solution, you now are given the information that the simplex method proceeds as follows to obtain the optimal solution in two iterations: (1) In iteration 1, the entering basic vari-

224

5

THE THEORY OF THE SIMPLEX METHOD

able is x2 and the leaving basic variable is x5; (2) in iteration 2, the entering basic variable is x1 and the leaving basic variable is x4. Follow the instructions of Prob. 5.1-17 for this situation. 5.1-19. By inspecting Fig. 5.2, explain why Property 1b for CPF solutions holds for this problem if it has the following objective function. (a) Maximize Z x3. (b) Maximize Z x1 2x3. 5.1-20. Consider the three-variable linear programming problem shown in Fig. 5.2. (a) Explain in geometric terms why the set of solutions satisfying any individual constraint is a convex set, as defined in Appendix 2. (b) Use the conclusion in part (a) to explain why the entire feasible region (the set of solutions that simultaneously satisfies every constraint) is a convex set. 5.1-21. Suppose that the three-variable linear programming problem given in Fig. 5.2 has the objective function Maximize

Z 3x1 4x2 3x3.

method as it goes through one iteration in moving from (2, 4, 3) to (4, 2, 4). (You are given the information that it is moving along this line segment.) (a) What is the entering basic variable? (b) What is the leaving basic variable? (c) What is the new BF solution? 5.1-24. Consider a two-variable mathematical programming problem that has the feasible region shown on the graph, where the six dots correspond to CPF solutions. The problem has a linear objective function, and the two dashed lines are objective function lines passing through the optimal solution (4, 5) and the secondbest CPF solution (2, 5). Note that the nonoptimal solution (2, 5) is better than both of its adjacent CPF solutions, which violates Property 3 in Sec. 5.1 for CPF solutions in linear programming. Demonstrate that this problem cannot be a linear programming problem by constructing the feasible region that would result if the six line segments on the boundary were constraint boundaries for linear programming constraints. x2

Without using the algebra of the simplex method, apply just its geometric reasoning (including choosing the edge giving the maximum rate of increase of Z) to determine and explain the path it would follow in Fig. 5.2 from the origin to the optimal solution. 5.1-22. Consider the three-variable linear programming problem shown in Fig. 5.2. (a) Construct a table like Table 5.4, giving the indicating variable for each constraint boundary equation and original constraint. (b) For the CPF solution (2, 4, 3) and its three adjacent CPF solutions (4, 2, 4), (0, 4, 2), and (2, 4, 0), construct a table like Table 5.5, showing the corresponding defining equations, BF solution, and nonbasic variables. (c) Use the sets of defining equations from part (b) to demonstrate that (4, 2, 4), (0, 4, 2), and (2, 4, 0) are indeed adjacent to (2, 4, 3), but that none of these three CPF solutions are adjacent to each other. Then use the sets of nonbasic variables from part (b) to demonstrate the same thing. 5.1-23. The formula for the line passing through (2, 4, 3) and (4, 2, 4) in Fig. 5.2 can be written as (2, 4, 3) [(4, 2, 4) (2, 4, 3)] (2, 4, 3) (2, 2, 1), where 0 1 for just the line segment between these points. After augmenting with the slack variables x4, x5, x6, x7 for the respective functional constraints, this formula becomes (2, 4, 3, 2, 0, 0, 0) (2, 2, 1, 2, 2, 0, 0). Use this formula directly to answer each of the following questions, and thereby relate the algebra and geometry of the simplex

(2, 5)

5

(4, 5)

4 3 2 1

0

1

2

3

4

x1

5.2-1. Consider the following problem. Maximize

Z 8x1 4x2 6x3 3x4 9x5,

subject to x1 2x2 3x3 3x4 x5 180 4x1 3x2 2x3 x4 x5 270 x1 3x2 2x3 x4 3x5 180

(resource 1) (resource 2) (resource 3)

CHAPTER 5 PROBLEMS

and

225

and

xj 0,

j 1, . . . , 5.

You are given the facts that the basic variables in the optimal solution are x3, x1, and x5 and that 1 3 1 0 1 11 3 2 4 1 1 6 9 3 . 27 0 1 3 2 3 10

x1 0,

Coefficient of: Basic Variable

Eq.

Z

Z

(0)

x2 x6 x3

(1) (2) (3)

5.2-2.* Work through the revised simplex method step by step to solve the following problem.

I

Z 5x1 8x2 7x3 4x4 6x5,

subject to 2x1 3x2 3x3 2x4 2x5 20 3x1 5x2 4x3 2x4 4x5 30 and xj 0,

j 1, 2, 3, 4, 5.

5.2-3. Work through the revised simplex method step by step to solve the model given in Prob. 4.3-4.

I

5.2-4. Reconsider Prob. 5.1-1. For the sequence of CPF solutions identified in part (e), construct the basis matrix B for each of the corresponding BF solutions. For each one, invert B manually, use this B1 to calculate the current solution, and then perform the next iteration (or demonstrate that the current solution is optimal). 5.2-5. Work through the revised simplex method step by step to solve the model given in Prob. 4.1-5.

I

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

(a) Use the given information to identify the optimal solution. (b) Use the given information to identify the shadow prices for the three resources.

Maximize

x2 0,

x1

x2

x3

x4

x5

x6

1

1

1

0

0 0 0

1 0 1

3 1 2

0 1 0

Right Side

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau. D

5.3-2. Consider the following problem. Z 4x1 3x2 x3 2x4,

Maximize subject to

4x1 2x2 x3 x4 5 3x1 x2 2x3 x4 4 and x1 0,

x2 0,

x3 0,

x4 0.

Let x5 and x6 denote the slack variables for the respective constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

5.2-6. Work through the revised simplex method step by step to solve the model given in Prob. 4.7-6.

I

5.2-7. Work through the revised simplex method step by step to solve each of the following models: (a) Model given in Prob. 3.1-5. (b) Model given in Prob. 4.7-8.

I

D

5.3-1.* Consider the following problem. Maximize

Z x1 x2 2x3,

subject to 2x1 2x2 3x3 5 x1 x2 x3 3 x1 x2 x3 2

Coefficient of: Basic Variable

Eq.

Z

Z

(0)

x2 x4

(1) (2)

x1

x2

x3

x4

x5

x6

1

1

1

0 0

1 1

1 2

Right Side

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau.

226

D

5

THE THEORY OF THE SIMPLEX METHOD

5.3-3. Consider the following problem. Coefficient of:

Z 6x1 x2 2x3,

Maximize

Basic Variable

Eq.

Z

Z

(0)

x4

(1)

x1

x2

x3

x4

x5

x6

1

0

3 2

1 2

0

1

1

2

1 2 1 2

1 2 1 2

subject to 1 2x1 2x2 x3 2 2 3 4x1 2x2 x3 3 2 1 2x1 2x2 x3 1 2 and x1 0,

x2 0,

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

Coefficient of: Eq.

Z

Z

(0)

x5 x3 x1

(1) (2) (3)

x1

x2

x3

x4

x5

x6

1

2

0

2

0 0 0

1 2 1

1 0 0

2 4 1

Right Side

(2)

0

0

x2

(3)

0

0

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau. D

Basic Variable

x3

Right Side

5.3-5. Consider the following problem. Z 20x1 6x2 8x3,

Maximize subject to

8x1 2x2 3x3 200 4x1 3x2 3x3 100 2x1 3x2 x3 50 2x1 3x2 x3 20 and x1 0,

Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. D

5.3-4. Consider the following problem. Maximize

Z x1 x2 2x3,

subject to x1 x2 3x3 15 2x1 x2 x3 2 x1 x2 x3 4 and x1 0,

x2 0,

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After the simplex method is applied, a portion of the final simplex tableau is as follows:

x2 0,

x3 0.

Let x4, x5, x6, and x7 denote the slack variables for the first through fourth constraints, respectively. Suppose that after some number of iterations of the simplex method, a portion of the current simplex tableau is as follows: Coefficient of: Basic Variable

Eq.

Z

Z

(0)

1

x1

(1)

0

x2

(2)

0

x6

(3)

0

x7

(4)

0

x1

x2

x3

x4

x5

x6

x7

9 4

1 2

0

0

0

0

0

0

1

0

0

1

3 1 16 8 1 1 4 2 3 1 8 4 0 0

Right Side

CHAPTER 5 PROBLEMS

(a) Use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the current simplex tableau. Show your calculations. (b) Indicate which of these missing numbers would be generated by the revised simplex method in order to perform the next iteration. (c) Identify the defining equations of the CPF solution corresponding to the BF solution in the current simplex tableau. 5.3-6. You are using the simplex method to solve the following linear programming problem. D

Z 6x1 5x2 x3 4x4,

Maximize

227

Now suppose that your boss has inserted her best estimate of the values of c1, c2, c3, and b without informing you and then has run the simplex method. You are given the resulting final simplex tableau below (where x4 and x5 are the slack variables for the respective functional constraints), but you are unable to read the value of Z*. Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Z

(0)

1

7 10

0

0

3 5

4 5

x2

(1)

0

1

0

x3

(2)

0

0

1

3 5 1 5

1 5 2 5

subject to 3x1 2x2 3x3 x4 120 3x1 3x2 x3 3x4 180 and x1 0,

x2 0,

x3 0,

x4 0.

You have obtained the following final simplex tableau where x5 and x6 are the slack variables for the respective constraints. Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Z

(0)

1

0

1 4

0

1 2

3 4

5 4

x1

(1)

0

1

x3

(2)

0

0

5 6 1 2

1 12 1 4

1 4 1 4

11 12 1 4

0 1

x5

x6

Right Side Z*

b*1 b*2

Use the fundamental insight presented in Sec. 5.3 to identify Z*, b*1, and b*2. Show your calculations. D

5.3-7. Consider the following problem. Maximize Z c1x1 c2x2 c3x3,

subject to x1 2x2 x3 b 2x1 x2 3x3 2b and x1 0,

x2 0,

x3 0.

Note that values have not been assigned to the coefficients in the objective function (c1, c2, c3), and that the only specification for the right-hand side of the functional constraints is that the second one (2b) be twice as large as the first (b).

1 5 3 5

Right Side Z*

1 3

(a) Use the fundamental insight presented in Sec. 5.3 to identify the value of (c1, c2, c3) that was used. (b) Use the fundamental insight presented in Sec. 5.3 to identify the value of b that was used. (c) Calculate the value of Z* in two ways, where one way uses your results from part (a) and the other way uses your result from part (b). Show your two methods for finding Z*. 5.3-8. For iteration 2 of the example in Sec. 5.3, the following expression was shown: Final row 0 [3,

5 0, [0,

0,

0 0]

3 2

1 1] 0 3

,

0 2 2

1 0 0

0 1 0

0 0 1

4 12 . 18

Derive this expression by combining the algebraic operations (in matrix form) for iterations 1 and 2 that affect row 0. 5.3-9. Most of the description of the fundamental insight presented in Sec. 5.3 assumes that the problem is in our standard form. Now consider each of the following other forms, where the additional adjustments in the initialization step are those presented in Sec. 4.6, including the use of artificial variables and the Big M method where appropriate. Describe the resulting adjustments in the fundamental insight. (a) Equality constraints (b) Functional constraints in form (c) Negative right-hand sides (d) Variables allowed to be negative (with no lower bound) 5.3-10. Reconsider the model in Prob. 4.6-6. Use artificial variables and the Big M method to construct the complete first sim-

228

5

THE THEORY OF THE SIMPLEX METHOD

plex tableau for the simplex method, and then identify the columns that will contains S* for applying the fundamental insight in the final tableau. Explain why these are the appropriate columns. 5.3-11. Consider the following problem. Z 2x1 3x2 2x3,

Minimize subject to

x1 4x2 2x3 8 3x1 2x2 2x3 6 and

(c) When you apply the t* t vT equation, another option is to use t [2, 3, 2, 0, M, 0, M, 0], which is the preliminary row 0 before the algebraic elimination of the nonzero coefficients of the initial basic variables x5 and x7. Repeat part (b) for this equation with this new t. After you derive the new v, show that this equation yields the same final row 0 for this problem as the equation derived in part (b). (d) Identify the defining equations of the CPF solution corresponding to the optimal BF solution in the final simplex tableau. 5.3-12. Consider the following problem.

x1 0,

x2 0,

x3 0.

Let x4 and x6 be the surplus variables for the first and second constraints, respectively. Let x5 and x7 be the corresponding artificial variables. After you make the adjustments described in Sec. 4.6 for this model form when using the Big M method, the initial simplex tableau ready to apply the simplex method is as follows:

Z

x1

x2

x3

x4 x 5 x6 x 7

Z

(0) 1 4M 2 6M 3 2M 2 M

x5 x7

(1) 0 (2) 0

1 3

4 2

2 0

subject to x1 3x2 2x3 20 x1 5x2 2x3 10 and x1 0,

Coefficient of: Basic Variable Eq.

0

M

Right Side

0 14M

1 1 0 0 0 0 1 1

8 6

After you apply the simplex method, a portion of the final simplex tableau is as follows:

Z

Z

(0) 1

x2 x1

(1) 0 (2) 0

x1 x2 x3 x4

x 5 M 0.5

0.3 0.2

x6

x 7

Eq.

Z

Z

(0)

x1 x5

(1) (2)

M 0.5

(a) Based on the above tableaux, use the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Examine the mathematical logic presented in Sec. 5.3 to validate the fundamental insight (see the T* MT and t* t vT equations and the subsequent derivations of M and v). This logic assumes that the original model fits our standard form, whereas the current problem does not fit this form. Show how, with minor adjustments, this same logic applies to the current problem when t is row 0 and T is rows 1 and 2 in the initial simplex tableau given above. Derive M and v for this problem.

x3 0.

Coefficient of: Basic Variable

Right Side

0.1 0.4

x2 0,

Let x4 be the artificial variable for the first constraint. Let x5 and x6 be the surplus variable and artificial variable, respectively, for the second constraint. You are now given the information that a portion of the final simplex tableau is as follows:

Coefficient of: Basic Variable Eq.

Z 2x1 4x2 3x3,

Maximize

x1

x2

x3

x 4

x5

x 6

1

M2

0

M

0 0

1 1

0 1

0 1

Right Side

(a) Extend the fundamental insight presented in Sec. 5.3 to identify the missing numbers in the final simplex tableau. Show your calculations. (b) Identify the defining equations of the CPF solution corresponding to the optimal solution in the final simplex tableau. 5.3-13. Consider the following problem. Maximize

Z 3x1 7x2 2x3,

subject to 2x1 2x2 x3 10 3x1 x2 x3 20 and x1 0,

x2 0,

x3 0.

CHAPTER 5 PROBLEMS

You are given the fact that the basic variables in the optimal solution are x1 and x3. (a) Introduce slack variables, and then use the given information to find the optimal solution directly by Gaussian elimination. (b) Extend the work in part (a) to find the shadow prices. (c) Use the given information to identify the defining equations of the optimal CPF solution, and then solve these equations to obtain the optimal solution.

229

(d) Construct the basis matrix B for the optimal BF solution, invert B manually, and then use this B1 to solve for the optimal solution and the shadow prices y*. Then apply the optimality test for the revised simplex method to verify that this solution is optimal. (e) Given B1 and y* from part (d ), use the fundamental insight presented in Sec. 5.3 to construct the complete final simplex tableau.

6 Duality Theory and Sensitivity Analysis

One of the most important discoveries in the early development of linear programming was the concept of duality and its many important ramifications. This discovery revealed that every linear programming problem has associated with it another linear programming problem called the dual. The relationships between the dual problem and the original problem (called the primal) prove to be extremely useful in a variety of ways. For example, you soon will see that the shadow prices described in Sec. 4.7 actually are provided by the optimal solution for the dual problem. We shall describe many other valuable applications of duality theory in this chapter as well. One of the key uses of duality theory lies in the interpretation and implementation of sensitivity analysis. As we already mentioned in Secs. 2.3, 3.3, and 4.7, sensitivity analysis is a very important part of almost every linear programming study. Because most of the parameter values used in the original model are just estimates of future conditions, the effect on the optimal solution if other conditions prevail instead needs to be investigated. Furthermore, certain parameter values (such as resource amounts) may represent managerial decisions, in which case the choice of the parameter values may be the main issue to be studied, which can be done through sensitivity analysis. For greater clarity, the first three sections discuss duality theory under the assumption that the primal linear programming problem is in our standard form (but with no restriction that the bi values need to be positive). Other forms are then discussed in Sec. 6.4. We begin the chapter by introducing the essence of duality theory and its applications. We then describe the economic interpretation of the dual problem (Sec. 6.2) and delve deeper into the relationships between the primal and dual problems (Sec. 6.3). Section 6.5 focuses on the role of duality theory in sensitivity analysis. The basic procedure for sensitivity analysis (which is based on the fundamental insight of Sec. 5.3) is summarized in Sec. 6.6 and illustrated in Sec. 6.7. 230

6.1 THE ESSENCE OF DUALITY THEORY

6.1

231

THE ESSENCE OF DUALITY THEORY Given our standard form for the primal problem at the left (perhaps after conversion from another form), its dual problem has the form shown to the right. Primal Problem

Dual Problem m

n

Maximize

Z c j x j,

Minimize

W bi yi, i1

j1

subject to

subject to m

n

aij x j bi, j1

for i 1, 2, . . . , m

aij yi cj, i1

for j 1, 2, . . . , n

and

and xj 0,

yi 0,

for j 1, 2, . . . , n.

for i 1, 2, . . . , m.

Thus, the dual problem uses exactly the same parameters as the primal problem, but in different locations. To highlight the comparison, now look at these same two problems in matrix notation (as introduced at the beginning of Sec. 5.2), where c and y [y1, y2, . . . , ym] are row vectors but b and x are column vectors. Primal Problem Maximize

Z cx,

subject to

Dual Problem Minimize

W yb,

subject to yA c

Ax b and

and x 0.

y 0.

To illustrate, the primal and dual problems for the Wyndor Glass Co. example of Sec. 3.1 are shown in Table 6.1 in both algebraic and matrix form. The primal-dual table for linear programming (Table 6.2) also helps to highlight the correspondence between the two problems. It shows all the linear programming parameters (the aij, bi, and cj) and how they are used to construct the two problems. All the headings for the primal problem are horizontal, whereas the headings for the dual problem are read by turning the book sideways. For the primal problem, each column (except the Right Side column) gives the coefficients of a single variable in the respective constraints and then in the objective function, whereas each row (except the bottom one) gives the parameters for a single contraint. For the dual problem, each row (except the Right Side row) gives the coefficients of a single variable in the respective constraints and then in the objective function, whereas each column (except the rightmost one) gives the parameters for a single constraint. In addition, the Right Side column gives the right-hand sides for the primal problem and the objective function coefficients for the dual problem, whereas the bottom row gives the objective function coefficients for the primal problem and the righthand sides for the dual problem.

232

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.1 Primal and dual problems for the Wyndor Glass Co. example Primal Problem in Algebraic Form

Dual Problem in Algebraic Form

Z 3x1 5x2,

Maximize subject to

W 4y1 12y2 18y3,

Minimize subject to

3x1 2x2 4

y12y2 3y3 3

3x1 2x2 12

2y2 2y3 5

3x1 2x2 18 and

and

x2 0.

x1 0,

y1 0,

Primal Problem in Matrix Form Z [3, 5]

Maximize

x1

x , 2

0 x1 2 x2 2

Minimize

4 12 18

0

x 0. 2

4 W [y1,y2,y3] 12 18

subject to [y1, y2, y3]

and x1

y3 0.

Dual Problem in Matrix Form

subject to

1 0 3

y2 0,

1 0 3

0 2 [3, 5] 2

and [y1, y2, y3] [0, 0, 0].

Consequently, (1) the parameters for a constraint in either problem are the coefficients of a variable in the other problem and (2) the coefficients for the objective function of either problem are the right sides for the other problem. Thus, there is a direct correspondence between these entities in the two problems, as summarized in Table 6.3. These correspondences are a key to some of the applications of duality theory, including sensitivity analysis. Origin of the Dual Problem Duality theory is based directly on the fundamental insight (particularly with regard to row 0) presented in Sec. 5.3. To see why, we continue to use the notation introduced in Table 5.10 for row 0 of the final tableau, except for replacing Z* by W* and dropping the asterisks from z* and y* when referring to any tableau. Thus, at any given iteration of the simplex method for the primal problem, the current numbers in row 0 are denoted as shown in the (partial) tableau given in Table 6.4. For the coefficients of x1, x2, . . . , xn, recall that z (z1, z2, . . . , zn) denotes the vector that the simplex method added to the vector of initial coefficients, c, in the process of reaching the current tableau. (Do not confuse z with the value of the objective function Z.) Similarly, since the initial coefficients of xn1, xn2, . . . , xnm in row 0 all are 0, y (y1, y2, . . . , ym) denotes the vector that the simplex method has added to these coefficients. Also recall [see Eq. (1) in the

6.1 THE ESSENCE OF DUALITY THEORY

233

TABLE 6.2 Primal-dual table for linear programming, illustrated by the Wyndor Glass Co. example (a) General Case Primal Problem Coefficient of:

…

xn

a11 a21

ym

a12 a22

VI c1

VI c2

… …

Right Side b1 b2

bm

Coefficients for Objective Function (Minimize)

x2

… a1n … a2n ……………………………… … am1 am2 amn

y1 y2

Right Side

Dual Problem

Coefficient of:

x1

VI cn

Coefficients for Objective Function (Maximize) (b) Wyndor Glass Co. Example

y1 y2 y3

x1

x2

1 0 3

0 2 2

VI 3

VI 5

4 12 18

“Mathematical Summary” subsection of Sec. 5.3] that the fundamental insight led to the following relationships between these quantities and the parameters of the original model: m

W yb bi yi , i1

m

z yA,

so

zj aij yi ,

for j 1, 2, . . . , n.

i1

To illustrate these relationships with the Wyndor example, the first equation gives W 4y1 12y2 18y3, which is just the objective function for the dual problem shown TABLE 6.3 Correspondence between entities in primal and dual problems One Problem

Other Problem

Constraint i ←→ Variable i Objective function ←→ Right sides

234

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.4 Notation for entries in row 0 of a simplex tableau Coefficient of: Iteration

Basic Variable

Eq.

Z

x1

x2

…

xn

xn1

xn2

…

xnm

Right Side

Any

Z

(0)

1

z1 c1

z2 c2

…

zn cn

y1

y2

…

ym

W

in the upper right-hand box of Table 6.1. The second set of equations give z1 y1 3y3 and z2 2y2 2y3, which are the left-hand sides of the functional constraints for this dual problem. Thus, by subtracting the right-hand sides of these constraints (c1 3 and c2 5), (z1 c1) and (z2 c2) can be interpreted as being the surplus variables for these functional constraints. The remaining key is to express what the simplex method tries to accomplish (according to the optimality test) in terms of these symbols. Specifically, it seeks a set of basic variables, and the corresponding BF solution, such that all coefficients in row 0 are nonnegative. It then stops with this optimal solution. Using the notation in Table 6.4, this goal is expressed symbolically as follows: Condition for Optimality: for j 1, 2, . . . , n, zj cj 0 yi 0 for i 1, 2, . . . , m. After we substitute the preceding expression for zj, the condition for optimality says that the simplex method can be interpreted as seeking values for y1, y2, . . . , ym such that m

W biyi, i1

subject to m

aijyi cj, i1

for j 1, 2, . . . , n

and yi 0,

for i 1, 2, . . . , m.

But, except for lacking an objective for W, this problem is precisely the dual problem! To complete the formulation, let us now explore what the missing objective should be. Since W is just the current value of Z, and since the objective for the primal problem is to maximize Z, a natural first reaction is that W should be maximized also. However, this is not correct for the following rather subtle reason: The only feasible solutions for this new problem are those that satisfy the condition for optimality for the primal problem. Therefore, it is only the optimal solution for the primal problem that corresponds to a feasible solution for this new problem. As a consequence, the optimal value of Z in the primal problem is the minimum feasible value of W in the new problem, so W should be minimized. (The full justification for this conclusion is provided by the relationships we develop in Sec. 6.3.) Adding this objective of minimizing W gives the complete dual problem.

6.1 THE ESSENCE OF DUALITY THEORY

235

Consequently, the dual problem may be viewed as a restatement in linear programming terms of the goal of the simplex method, namely, to reach a solution for the primal problem that satisfies the optimality test. Before this goal has been reached, the corresponding y in row 0 (coefficients of slack variables) of the current tableau must be infeasible for the dual problem. However, after the goal is reached, the corresponding y must be an optimal solution (labeled y*) for the dual problem, because it is a feasible solution that attains the minimum feasible value of W. This optimal solution (y1*, y2*, . . . , *) provides for the primal problem the shadow prices that were described in Sec. 4.7. ym Furthermore, this optimal W is just the optimal value of Z, so the optimal objective function values are equal for the two problems. This fact also implies that cx yb for any x and y that are feasible for the primal and dual problems, respectively. To illustrate, the left-hand side of Table 6.5 shows row 0 for the respective iterations when the simplex method is applied to the Wyndor Glass Co. example. In each case, row 0 is partitioned into three parts: the coefficients of the decision variables (x1, x2), the coefficients of the slack variables (x3, x4, x5), and the right-hand side (value of Z). Since the coefficients of the slack variables give the corresponding values of the dual variables (y1, y2, y3), each row 0 identifies a corresponding solution for the dual problem, as shown in the y1, y2, and y3 columns of Table 6.5. To interpret the next two columns, recall that (z1 c1) and (z2 c2) are the surplus variables for the functional constraints in the dual problem, so the full dual problem after augmenting with these surplus variables is W 4y1 12y2 18y3,

Minimize subject to y1

3y3 (z1 c1) 3 2y2 2y3 (z2 c2) 5

and y1 0,

y2 0,

y3 0.

Therefore, by using the numbers in the y1, y2, and y3 columns, the values of these surplus variables can be calculated as z1 c1 y1 3y3 3, z2 c2 2y2 2y3 5.

TABLE 6.5 Row 0 and corresponding dual solution for each iteration for the Wyndor Glass Co. example Primal Problem Iteration

Dual Problem

Row 0

0

[3,

5

0,

1

[3,

0

0,

2

[0,

0

0,

0, 5 , 2 3 , 2

y1

y2

y3

z1 c1

z2 c2

0 5 2 3 2

0

3

5

0

0

3

0

30

1

0

0

36

0

0]

0

0

30]

0

1

36]

0

W

236

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Thus, a negative value for either surplus variable indicates that the corresponding constraint is violated. Also included in the rightmost column of the table is the calculated value of the dual objective function W 4y1 12y2 18y3. As displayed in Table 6.4, all these quantities to the right of row 0 in Table 6.5 already are identified by row 0 without requiring any new calculations. In particular, note in Table 6.5 how each number obtained for the dual problem already appears in row 0 in the spot indicated by Table 6.4. For the initial row 0, Table 6.5 shows that the corresponding dual solution (y1, y2, y3) (0, 0, 0) is infeasible because both surplus variables are negative. The first iteration succeeds in eliminating one of these negative values, but not the other. After two iterations, the optimality test is satisfied for the primal problem because all the dual variables and surplus variables are nonnegative. This dual solution (y1*, y2*, y3*) (0, 32, 1) is optimal (as could be verified by applying the simplex method directly to the dual problem), so the optimal value of Z and W is Z* 36 W*. Summary of Primal-Dual Relationships Now let us summarize the newly discovered key relationships between the primal and dual problems. Weak duality property: If x is a feasible solution for the primal problem and y is a feasible solution for the dual problem, then cx yb. For example, for the Wyndor Glass Co. problem, one feasible solution is x1 3, x2 3, which yields Z cx 24, and one feasible solution for the dual problem is y1 1, y2 1, y3 2, which yields a larger objective function value W yb 52. These are just sample feasible solutions for the two problems. For any such pair of feasible solutions, this inequality must hold because the maximum feasible value of Z cx (36) equals the minimum feasible value of the dual objective function W yb, which is our next property. Strong duality property: If x* is an optimal solution for the primal problem and y* is an optimal solution for the dual problem, then cx* y*b. Thus, these two properties imply that cx yb for feasible solutions if one or both of them are not optimal for their respective problems, whereas equality holds when both are optimal. The weak duality property describes the relationship between any pair of solutions for the primal and dual problems where both solutions are feasible for their respective problems. At each iteration, the simplex method finds a specific pair of solutions for the two problems, where the primal solution is feasible but the dual solution is not feasible (except at the final iteration). Our next property describes this situation and the relationship between this pair of solutions. Complementary solutions property: At each iteration, the simplex method simultaneously identifies a CPF solution x for the primal problem and a complementary solution y for the dual problem (found in row 0, the coefficients of the slack variables), where cx yb.

6.1 THE ESSENCE OF DUALITY THEORY

237

If x is not optimal for the primal problem, then y is not feasible for the dual problem. To illustrate, after one iteration for the Wyndor Glass Co. problem, x1 0, x2 6, and y1 0, y2 52, y3 0, with cx 30 yb. This x is feasible for the primal problem, but this y is not feasible for the dual problem (since it violates the constraint, y1 3y3 3). The complementary solutions property also holds at the final iteration of the simplex method, where an optimal solution is found for the primal problem. However, more can be said about the complementary solution y in this case, as presented in the next property. Complementary optimal solutions property: At the final iteration, the simplex method simultaneously identifies an optimal solution x* for the primal problem and a complementary optimal solution y* for the dual problem (found in row 0, the coefficients of the slack variables), where cx* y*b. The y*i are the shadow prices for the primal problem. For the example, the final iteration yields x1* 2, x2* 6, and y1* 0, y2* 32, y3* 1, with cx* 36 y*b. We shall take a closer look at some of these properties in Sec. 6.3. There you will see that the complementary solutions property can be extended considerably further. In particular, after slack and surplus variables are introduced to augment the respective problems, every basic solution in the primal problem has a complementary basic solution in the dual problem. We already have noted that the simplex method identifies the values of the surplus variables for the dual problem as zj cj in Table 6.4. This result then leads to an additional complementary slackness property that relates the basic variables in one problem to the nonbasic variables in the other (Tables 6.7 and 6.8), but more about that later. In Sec. 6.4, after describing how to construct the dual problem when the primal problem is not in our standard form, we discuss another very useful property, which is summarized as follows: Symmetry property: For any primal problem and its dual problem, all relationships between them must be symmetric because the dual of this dual problem is this primal problem. Therefore, all the preceding properties hold regardless of which of the two problems is labeled as the primal problem. (The direction of the inequality for the weak duality property does require that the primal problem be expressed or reexpressed in maximization form and the dual problem in minimization form.) Consequently, the simplex method can be applied to either problem, and it simultaneously will identify complementary solutions (ultimately a complementary optimal solution) for the other problem. So far, we have focused on the relationships between feasible or optimal solutions in the primal problem and corresponding solutions in the dual problem. However, it is possible that the primal (or dual) problem either has no feasible solutions or has feasible solutions but no optimal solution (because the objective function is unbounded). Our final property summarizes the primal-dual relationships under all these possibilities.

238

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Duality theorem: The following are the only possible relationships between the primal and dual problems. 1. If one problem has feasible solutions and a bounded objective function (and so has an optimal solution), then so does the other problem, so both the weak and strong duality properties are applicable. 2. If one problem has feasible solutions and an unbounded objective function (and so no optimal solution), then the other problem has no feasible solutions. 3. If one problem has no feasible solutions, then the other problem has either no feasible solutions or an unbounded objective function. Applications As we have just implied, one important application of duality theory is that the dual problem can be solved directly by the simplex method in order to identify an optimal solution for the primal problem. We discussed in Sec. 4.8 that the number of functional constraints affects the computational effort of the simplex method far more than the number of variables does. If m n, so that the dual problem has fewer functional constraints (n) than the primal problem (m), then applying the simplex method directly to the dual problem instead of the primal problem probably will achieve a substantial reduction in computational effort. The weak and strong duality properties describe key relationships between the primal and dual problems. One useful application is for evaluating a proposed solution for the primal problem. For example, suppose that x is a feasible solution that has been proposed for implementation and that a feasible solution y has been found by inspection for the dual problem such that cx yb. In this case, x must be optimal without the simplex method even being applied! Even if cx yb, then yb still provides an upper bound on the optimal value of Z, so if yb cx is small, intangible factors favoring x may lead to its selection without further ado. One of the key applications of the complementary solutions property is its use in the dual simplex method presented in Sec. 7.1. This algorithm operates on the primal problem exactly as if the simplex method were being applied simultaneously to the dual problem, which can be done because of this property. Because the roles of row 0 and the right side in the simplex tableau have been reversed, the dual simplex method requires that row 0 begin and remain nonnegative while the right side begins with some negative values (subsequent iterations strive to reach a nonnegative right side). Consequently, this algorithm occasionally is used because it is more convenient to set up the initial tableau in this form than in the form required by the simplex method. Furthermore, it frequently is used for reoptimization (discussed in Sec. 4.7), because changes in the original model lead to the revised final tableau fitting this form. This situation is common for certain types of sensitivity analysis, as you will see later in the chapter. In general terms, duality theory plays a central role in sensitivity analysis. This role is the topic of Sec. 6.5. Another important application is its use in the economic interpretation of the dual problem and the resulting insights for analyzing the primal problem. You already have seen one example when we discussed shadow prices in Sec. 4.7. The next section describes how this interpretation extends to the entire dual problem and then to the simplex method.

6.2 ECONOMIC INTERPRETATION OF DUALITY

6.2

239

ECONOMIC INTERPRETATION OF DUALITY The economic interpretation of duality is based directly upon the typical interpretation for the primal problem (linear programming problem in our standard form) presented in Sec. 3.2. To refresh your memory, we have summarized this interpretation of the primal problem in Table 6.6. Interpretation of the Dual Problem To see how this interpretation of the primal problem leads to an economic interpretation for the dual problem,1 note in Table 6.4 that W is the value of Z (total profit) at the current iteration. Because W b1y1 b2 y2 … bm ym , each bi yi can thereby be interpreted as the current contribution to profit by having bi units of resource i available for the primal problem. Thus, The dual variable yi is interpreted as the contribution to profit per unit of resource i (i 1, 2, . . . , m), when the current set of basic variables is used to obtain the primal solution.

In other words, the yi values (or y*i values in the optimal solution) are just the shadow prices discussed in Sec. 4.7. For example, when iteration 2 of the simplex method finds the optimal solution for the Wyndor problem, it also finds the optimal values of the dual variables (as shown in the bottom row of Table 6.5) to be y1* 0, y2* 32, and y3* 1. These are precisely the shadow prices found in Sec. 4.7 for this problem through graphical analysis. Recall that the resources for the Wyndor problem are the production capacities of the three plants being made available to the two new products under consideration, so that bi is the number of hours of production time per week being made available in Plant i for these new products, where i 1, 2, 3. As discussed in Sec. 4.7, the shadow prices indicate that individually increasing any bi by 1 would increase the optimal value of the objective function (total weekly profit in units of thousands of dollars) by y*i . Thus, y*i can be interpreted as the contribution to profit per unit of resource i when using the optimal solution.

1

Actually, several slightly different interpretations have been proposed. The one presented here seems to us to be the most useful because it also directly interprets what the simplex method does in the primal problem.

TABLE 6.6 Economic interpretation of the primal problem Quantity xj cj Z bi aij

Interpretation Level of activity j (j 1, 2, . . . , n) Unit profit from activity j Total profit from all activities Amount of resource i available (i 1, 2, . . . , m) Amount of resource i consumed by each unit of activity j

240

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

This interpretation of the dual variables leads to our interpretation of the overall dual problem. Specifically, since each unit of activity j in the primal problem consumes aij units of resource i,

m i1 ai j yi is interpreted as the current contribution to profit of the mix of resources that would be consumed if 1 unit of activity j were used ( j 1, 2, . . . , n).

For the Wyndor problem, 1 unit of activity j corresponds to producing 1 batch of product j per week, where j 1, 2. The mix of resources consumed by producing 1 batch of product 1 is 1 hour of production time in Plant 1 and 3 hours in Plant 3. The corresponding mix per batch of product 2 is 2 hours each in Plants 2 and 3. Thus, y1 3y3 and 2y2 2y3 are interpreted as the current contributions to profit (in thousands of dollars per week) of these respective mixes of resources per batch produced per week of the respective products. For each activity j, this same mix of resources (and more) probably can be used in other ways as well, but no alternative use should be considered if it is less profitable than 1 unit of activity j. Since cj is interpreted as the unit profit from activity j, each functional constraint in the dual problem is interpreted as follows:

m i1 aij yi cj says that the actual contribution to profit of the above mix of resources must be at least as much as if they were used by 1 unit of activity j; otherwise, we would not be making the best possible use of these resources.

For the Wyndor problem, the unit profits (in thousands of dollars per week) are c1 3 and c2 5, so the dual functional constraints with this interpretation are y1 y3 3 and 2y2 2y3 5. Similarly, the interpretation of the nonnegativity constraints is the following: yi 0 says that the contribution to profit of resource i (i 1, 2, . . . , m) must be nonnegative: otherwise, it would be better not to use this resource at all.

The objective m

Minimize

W bi yi i1

can be viewed as minimizing the total implicit value of the resources consumed by the activities. For the Wyndor problem, the total implicit value (in thousands of dollars per week) of the resources consumed by the two products is W 4y1 12y2 18y3. This interpretation can be sharpened somewhat by differentiating between basic and nonbasic variables in the primal problem for any given BF solution (x1, x2, . . . , xnm). Recall that the basic variables (the only variables whose values can be nonzero) always have a coefficient of zero in row 0. Therefore, referring again to Table 6.4 and the accompanying equation for zj, we see that m

aij yi cj, i1 yi 0,

if xj 0

( j 1, 2, . . . , n),

if xni 0

(i 1, 2, . . . , m).

(This is one version of the complementary slackness property discussed in the next section.) The economic interpretation of the first statement is that whenever an activity j op-

6.2 ECONOMIC INTERPRETATION OF DUALITY

241

erates at a strictly positive level (xj 0), the marginal value of the resources it consumes must equal (as opposed to exceeding) the unit profit from this activity. The second statement implies that the marginal value of resource i is zero (yi 0) whenever the supply of this resource is not exhausted by the activities (xni 0). In economic terminology, such a resource is a “free good”; the price of goods that are oversupplied must drop to zero by the law of supply and demand. This fact is what justifies interpreting the objective for the dual problem as minimizing the total implicit value of the resources consumed, rather than the resources allocated. To illustrate these two statements, consider the optimal BF solution (2, 6, 2, 0, 0) for the Wyndor problem. The basic variables are x1, x2, and x3, so their coefficients in row 0 are zero, as shown in the bottom row of Table 6.5. This bottom row also gives the corresponding dual solution: y1* 0, y2* 32, y3* 1, with surplus variables (z1* c1) 0 and (z2* c2) 0. Since x1 0 and x2 0, both these surplus variables and direct calculations indicate that y1* 3y3* c1 3 and 2y2* 2y3* c2 5. Therefore, the value of the resources consumed per batch of the respective products produced does indeed equal the respective unit profits. The slack variable for the constraint on the amount of Plant 1 capacity used is x3 0, so the marginal value of adding any Plant 1 capacity would be zero (y1* 0). Interpretation of the Simplex Method The interpretation of the dual problem also provides an economic interpretation of what the simplex method does in the primal problem. The goal of the simplex method is to find how to use the available resources in the most profitable feasible way. To attain this goal, we must reach a BF solution that satisfies all the requirements on profitable use of the resources (the constraints of the dual problem). These requirements comprise the condition for optimality for the algorithm. For any given BF solution, the requirements (dual constraints) associated with the basic variables are automatically satisfied (with equality). However, those associated with nonbasic variables may or may not be satisfied. In particular, if an original variable xj is nonbasic so that activity j is not used, then the current contribution to profit of the resources that would be required to undertake each unit of activity j m

aij yi

i1

may be smaller than, larger than, or equal to the unit profit cj obtainable from the activity. If it is smaller, so that zj cj 0 in row 0 of the simplex tableau, then these resources can be used more profitably by initiating this activity. If it is larger (zj cj 0), then these resources already are being assigned elsewhere in a more profitable way, so they should not be diverted to activity j. If zj cj 0, there would be no change in profitability by initiating activity j. Similarly, if a slack variable xni is nonbasic so that the total allocation bi of resource i is being used, then yi is the current contribution to profit of this resource on a marginal basis. Hence, if yi 0, profit can be increased by cutting back on the use of this resource (i.e., increasing xni). If yi 0, it is worthwhile to continue fully using this resource, whereas this decision does not affect profitability if yi 0.

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DUALITY THEORY AND SENSITIVITY ANALYSIS

Therefore, what the simplex method does is to examine all the nonbasic variables in the current BF solution to see which ones can provide a more profitable use of the resources by being increased. If none can, so that no feasible shifts or reductions in the current proposed use of the resources can increase profit, then the current solution must be optimal. If one or more can, the simplex method selects the variable that, if increased by 1, would improve the profitability of the use of the resources the most. It then actually increases this variable (the entering basic variable) as much as it can until the marginal values of the resources change. This increase results in a new BF solution with a new row 0 (dual solution), and the whole process is repeated. The economic interpretation of the dual problem considerably expands our ability to analyze the primal problem. However, you already have seen in Sec. 6.1 that this interpretation is just one ramification of the relationships between the two problems. In the next section, we delve into these relationships more deeply.

6.3

PRIMAL-DUAL RELATIONSHIPS Because the dual problem is a linear programming problem, it also has corner-point solutions. Furthermore, by using the augmented form of the problem, we can express these corner-point solutions as basic solutions. Because the functional constraints have the form, this augmented form is obtained by subtracting the surplus (rather than adding the slack) from the left-hand side of each constraint j ( j 1, 2, . . . , n).1 This surplus is m

zj cj aijyi cj ,

for j 1, 2, . . . , n.

i1

Thus, zjcj plays the role of the surplus variable for constraint j (or its slack variable if the constraint is multiplied through by 1). Therefore, augmenting each corner-point solution (y1, y2, . . . , ym) yields a basic solution (y1, y2, . . . , ym , z1 c1, z2 c2, . . . , zn cn) by using this expression for zj cj. Since the augmented form of the dual problem has n functional constraints and n m variables, each basic solution has n basic variables and m nonbasic variables. (Note how m and n reverse their previous roles here because, as Table 6.3 indicates, dual constraints correspond to primal variables and dual variables correspond to primal constraints.) Complementary Basic Solutions One of the important relationships between the primal and dual problems is a direct correspondence between their basic solutions. The key to this correspondence is row 0 of the simplex tableau for the primal basic solution, such as shown in Table 6.4 or 6.5. Such a row 0 can be obtained for any primal basic solution, feasible or not, by using the formulas given in the bottom part of Table 5.8. Note again in Tables 6.4 and 6.5 how a complete solution for the dual problem (including the surplus variables) can be read directly from row 0. Thus, because of its coefficient in 1

You might wonder why we do not also introduce artificial variables into these constraints as discussed in Sec. 4.6. The reason is that these variables have no purpose other than to change the feasible region temporarily as a convenience in starting the simplex method. We are not interested now in applying the simplex method to the dual problem, and we do not want to change its feasible region.

6.3 PRIMAL-DUAL RELATIONSHIPS

243

TABLE 6.7 Association between variables in primal and dual problems Primal Variable

Associated Dual Variable

Any problem

(Decision variable) xj (Slack variable) xni

zj cj (surplus variable) j 1, 2, . . . , n yi (decision variable) i 1, 2, . . . , m

Wyndor problem

Decision variables: Decision variables: Slack variables: Decision variables: Decision variables:

z1 c1 (surplus variables) z2 c2 y1 (decision variables) y2 y3

x1 x2 x3 x4 x5

row 0, each variable in the primal problem has an associated variable in the dual problem, as summarized in Table 6.7, first for any problem and then for the Wyndor problem. A key insight here is that the dual solution read from row 0 must also be a basic solution! The reason is that the m basic variables for the primal problem are required to have a coefficient of zero in row 0, which thereby requires the m associated dual variables to be zero, i.e., nonbasic variables for the dual problem. The values of the remaining n (basic) variables then will be the simultaneous solution to the system of equations given at the beginning of this section. In matrix form, this system of equations is z c yA c, and the fundamental insight of Sec. 5.3 actually identifies its solution for z c and y as being the corresponding entries in row 0. Because of the symmetry property quoted in Sec. 6.1 (and the direct association between variables shown in Table 6.7), the correspondence between basic solutions in the primal and dual problems is a symmetric one. Furthermore, a pair of complementary basic solutions has the same objective function value, shown as W in Table 6.4. Let us now summarize our conclusions about the correspondence between primal and dual basic solutions, where the first property extends the complementary solutions property of Sec. 6.1 to the augmented forms of the two problems and then to any basic solution (feasible or not) in the primal problem. Complementary basic solutions property: Each basic solution in the primal problem has a complementary basic solution in the dual problem, where their respective objective function values (Z and W) are equal. Given row 0 of the simplex tableau for the primal basic solution, the complementary dual basic solution (y, z c) is found as shown in Table 6.4. The next property shows how to identify the basic and nonbasic variables in this complementary basic solution. Complementary slackness property: Given the association between variables in Table 6.7, the variables in the primal basic solution and the complementary dual basic solution satisfy the complementary slackness relationship shown in Table 6.8. Furthermore, this relationship is a symmetric one, so that these two basic solutions are complementary to each other. The reason for using the name complementary slackness for this latter property is that it says (in part) that for each pair of associated variables, if one of them has slack in its

244

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DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.8 Complementary slackness relationship for complementary basic solutions Primal Variable

Associated Dual Variable

Basic Nonbasic

Nonbasic Basic

(m variables) (n variables)

nonnegativity constraint (a basic variable 0), then the other one must have no slack (a nonbasic variable 0). We mentioned in Sec. 6.2 that this property has a useful economic interpretation for linear programming problems. Example. To illustrate these two properties, again consider the Wyndor Glass Co. problem of Sec. 3.1. All eight of its basic solutions (five feasible and three infeasible) are shown in Table 6.9. Thus, its dual problem (see Table 6.1) also must have eight basic solutions, each complementary to one of these primal solutions, as shown in Table 6.9. The three BF solutions obtained by the simplex method for the primal problem are the first, fifth, and sixth primal solutions shown in Table 6.9. You already saw in Table 6.5 how the complementary basic solutions for the dual problem can be read directly from row 0, starting with the coefficients of the slack variables and then the original variables. The other dual basic solutions also could be identified in this way by constructing row 0 for each of the other primal basic solutions, using the formulas given in the bottom part of Table 5.8. Alternatively, for each primal basic solution, the complementary slackness property can be used to identify the basic and nonbasic variables for the complementary dual basic solution, so that the system of equations given at the beginning of the section can be TABLE 6.9 Complementary basic solutions for the Wyndor Glass Co. example Primal Problem

Dual Problem ZW

No.

Basic Solution

Feasible?

Feasible?

1 2 3

(0, 0, 4, 12, 18) (4, 0, 0, 12, 6) (6, 0, 2, 12, 0)

Yes Yes No

0 12 18

No No No

4

(4, 3, 0, 6, 0)

Yes

27

No

5

(0, 6, 4, 0, 6)

Yes

30

No

6

(2, 6, 2, 0, 0)

Yes

36

Yes

7

(4, 6, 0, 0, 6)

No

42

Yes

8

(0, 9, 4, 6, 0)

No

45

Yes

Basic Solution (0, 0, 0, 3, 5) (3, 0, 0, 0, 5) (0, 0, 1, 0, 3) 9 5 , 0, , 0, 0 2 2 5 0, , 0, 3, 0 2 3 0, , 1, 0, 0 2 5 3, , 0, 0, 0 2 5 9 0, 0, , , 0 2 2

6.3 PRIMAL-DUAL RELATIONSHIPS

245

solved directly to obtain this complementary solution. For example, consider the next-tolast primal basic solution in Table 6.9, (4, 6, 0, 0, 6). Note that x1, x2, and x5 are basic variables, since these variables are not equal to 0. Table 6.7 indicates that the associated dual variables are (z1 c1), (z2 c2), and y3. Table 6.8 specifies that these associated dual variables are nonbasic variables in the complementary basic solution, so z1 c1 0,

z2 c2 0,

y3 0.

Consequently, the augmented form of the functional constraints in the dual problem, y1

3y3 (z1 c1) 3 2y2 2y3 (z2 c2) 5,

reduce to y1

003 2y2 0 0 5,

so that y1 3 and y2 52. Combining these values with the values of 0 for the nonbasic variables gives the basic solution (3, 52, 0, 0, 0), shown in the rightmost column and nextto-last row of Table 6.9. Note that this dual solution is feasible for the dual problem because all five variables satisfy the nonnegativity constraints. Finally, notice that Table 6.9 demonstrates that (0, 32, 1, 0, 0) is the optimal solution for the dual problem, because it is the basic feasible solution with minimal W (36).

Relationships between Complementary Basic Solutions We now turn our attention to the relationships between complementary basic solutions, beginning with their feasibility relationships. The middle columns in Table 6.9 provide some valuable clues. For the pairs of complementary solutions, notice how the yes or no answers on feasibility also satisfy a complementary relationship in most cases. In particular, with one exception, whenever one solution is feasible, the other is not. (It also is possible for neither solution to be feasible, as happened with the third pair.) The one exception is the sixth pair, where the primal solution is known to be optimal. The explanation is suggested by the Z W column. Because the sixth dual solution also is optimal (by the complementary optimal solutions property), with W 36, the first five dual solutions cannot be feasible because W 36 (remember that the dual problem objective is to minimize W). By the same token, the last two primal solutions cannot be feasible because Z 36. This explanation is further supported by the strong duality property that optimal primal and dual solutions have Z W. Next, let us state the extension of the complementary optimal solutions property of Sec. 6.1 for the augmented forms of the two problems. Complementary optimal basic solutions property: Each optimal basic solution in the primal problem has a complementary optimal basic solution in the dual problem, where their respective objective function values (Z and W) are equal. Given row 0 of the simplex tableau for the optimal primal solution, the complementary optimal dual solution (y*, z* c) is found as shown in Table 6.4.

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DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.10 Classification of basic solutions Satisfies Condition for Optimality?

Feasible?

Yes

No

Yes

Optimal

Suboptimal

No

Superoptimal

Neither feasible nor superoptimal

To review the reasoning behind this property, note that the dual solution (y*, z* c) must be feasible for the dual problem because the condition for optimality for the primal problem requires that all these dual variables (including surplus variables) be nonnegative. Since this solution is feasible, it must be optimal for the dual problem by the weak duality property (since W Z, so y*b cx* where x* is optimal for the primal problem). Basic solutions can be classified according to whether they satisfy each of two conditions. One is the condition for feasibility, namely, whether all the variables (including slack variables) in the augmented solution are nonnegative. The other is the condition for optimality, namely, whether all the coefficients in row 0 (i.e., all the variables in the complementary basic solution) are nonnegative. Our names for the different types of basic solutions are summarized in Table 6.10. For example, in Table 6.9, primal basic solutions 1, 2, 4, and 5 are suboptimal, 6 is optimal, 7 and 8 are superoptimal, and 3 is neither feasible nor superoptimal. Given these definitions, the general relationships between complementary basic solutions are summarized in Table 6.11. The resulting range of possible (common) values for the objective functions (Z W) for the first three pairs given in Table 6.11 (the last pair can have any value) is shown in Fig. 6.1. Thus, while the simplex method is dealing directly with suboptimal basic solutions and working toward optimality in the primal problem, it is simultaneously dealing indirectly with complementary superoptimal solutions and working toward feasibility in the dual problem. Conversely, it sometimes is more convenient (or necessary) to work directly with superoptimal basic solutions and to move toward feasibility in the primal problem, which is the purpose of the dual simplex method described in Sec. 7.1. The third and fourth columns of Table 6.11 introduce two other common terms that are used to describe a pair of complementary basic solutions. The two solutions are said to be primal feasible if the primal basic solution is feasible, whereas they are called dual feasible if the complementary dual basic solution is feasible for the dual problem. Using TABLE 6.11 Relationships between complementary basic solutions Both Basic Solutions Primal Basic Solution

Complementary Dual Basic Solution

Suboptimal Optimal Superoptimal Neither feasible nor superoptimal

Superoptimal Optimal Suboptimal Neither feasible nor superoptimal

Primal Feasible?

Dual Feasible?

Yes Yes No No

No Yes Yes No

6.4 ADAPTING TO OTHER PRIMAL FORMS

Primal problem

Dual problem

n

cj xj Z

j1

Superoptimal

247

m

W

bi yi

i 1

Suboptimal

(optimal) Z*

W* (optimal)

Suboptimal

Superoptimal

FIGURE 6.1 Range of possible values of Z W for certain types of complementary basic solutions.

this terminology, the simplex method deals with primal feasible solutions and strives toward achieving dual feasibility as well. When this is achieved, the two complementary basic solutions are optimal for their respective problems. These relationships prove very useful, particularly in sensitivity analysis, as you will see later in the chapter.

6.4

ADAPTING TO OTHER PRIMAL FORMS Thus far it has been assumed that the model for the primal problem is in our standard form. However, we indicated at the beginning of the chapter that any linear programming problem, whether in our standard form or not, possesses a dual problem. Therefore, this section focuses on how the dual problem changes for other primal forms. Each nonstandard form was discussed in Sec. 4.6, and we pointed out how it is possible to convert each one to an equivalent standard form if so desired. These conversions are summarized in Table 6.12. Hence, you always have the option of converting any model to our standard form and then constructing its dual problem in the usual way. To illustrate, we do this for our standard dual problem (it must have a dual also) in Table 6.13. Note that what we end up with is just our standard primal problem! Since any pair of primal and dual problems can be converted to these forms, this fact implies that the dual of the dual problem always is the primal problem. Therefore, for any primal problem and its dual problem, all relationships between them must be symmetric. This is just the symmetry property already stated in Sec. 6.1 (without proof), but now Table 6.13 demonstrates why it holds.

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.12 Conversions to standard form for linear programming models Nonstandard Form

Equivalent Standard Form

Minimize

Maximize

Z

n

(Z)

n

aij xj bi

aij xj bi

j1

j1

n

n

aij xj bi j1

aij xj bi j1

xj unconstrained in sign

x j

x j ,

n

and

aij xj bi j1

x j

0,

x j 0

One consequence of the symmetry property is that all the statements made earlier in the chapter about the relationships of the dual problem to the primal problem also hold in reverse. Another consequence is that it is immaterial which problem is called the primal and which is called the dual. In practice, you might see a linear programming problem fitting our standard form being referred to as the dual problem. The convention is that the model formulated to fit the actual problem is called the primal problem, regardless of its form. Our illustration of how to construct the dual problem for a nonstandard primal problem did not involve either equality constraints or variables unconstrained in sign. Actually, for these two forms, a shortcut is available. It is possible to show (see Probs. 6.4-7 and 6.4-2a) that an equality constraint in the primal problem should be treated just like a constraint in TABLE 6.13 Constructing the dual of the dual problem Dual Problem Minimize

Converted to Standard Form

W yb,

Maximize

(W) yb,

subject to

subject to yA c

→

yA c and

and

y 0.

Converted to Standard Form

Its Dual Problem

Maximize

→

y 0.

Z cx,

Minimize

Ax b

(Z) cx,

subject to

subject to

→

248

Ax b and

and x 0.

x 0.

6.4 ADAPTING TO OTHER PRIMAL FORMS

249

constructing the dual problem except that the nonnegativity constraint for the corresponding dual variable should be deleted (i.e., this variable is unconstrained in sign). By the symmetry property, deleting a nonnegativity constraint in the primal problem affects the dual problem only by changing the corresponding inequality constraint to an equality constraint. Another shortcut involves functional constraints in form for a maximization problem. The straightforward (but longer) approach would begin by converting each such constraint to form n

n

aij xj bi → j1 aij xj bi. j1 Constructing the dual problem in the usual way then gives aij as the coefficient of yi in functional constraint j (which has form) and a coefficient of bi in the objective function (which is to be minimized), where yi also has a nonnegativity constraint yi 0. Now suppose we define a new variable yi yi. The changes caused by expressing the dual problem in terms of yi instead of yi are that (1) the coefficients of the variable become ai j for functional constraint j and bi for the objective function and (2) the constraint on the variable becomes yi 0 (a nonpositivity constraint). The shortcut is to use yi instead of yi as a dual variable so that the parameters in the original constraint (aij and bi) immediately become the coefficients of this variable in the dual problem. Here is a useful mnemonic device for remembering what the forms of dual constraints should be. With a maximization problem, it might seem sensible for a functional constraint to be in form, slightly odd to be in form, and somewhat bizarre to be in form. Similarly, for a minimization problem, it might seem sensible to be in form, slightly odd to be in form, and somewhat bizarre to be in form. For the constraint on an individual variable in either kind of problem, it might seem sensible to have a nonnegativity constraint, somewhat odd to have no constraint (so the variable is unconstrained in sign), and quite bizarre for the variable to be restricted to be less than or equal to zero. Now recall the correspondence between entities in the primal and dual problems indicated in Table 6.3; namely, functional constraint i in one problem corresponds to variable i in the other problem, and vice versa. The sensible-odd-bizarre method, or SOB method for short, says that the form of a functional constraint or the constraint on a variable in the dual problem should be sensible, odd, or bizarre, depending on whether the form for the corresponding entity in the primal problem is sensible, odd, or bizarre. Here is a summary. The SOB Method for Determining the Form of Constraints in the Dual.1 1. Formulate the primal problem in either maximization form or minimization form, and then the dual problem automatically will be in the other form. 2. Label the different forms of functional constraints and of constraints on individual variables in the primal problem as being sensible, odd, or bizarre according to Table 6.14. 1

This particular mnemonic device (and a related one) for remembering what the forms of dual constraints should be has been suggested by Arthur T. Benjamin, a mathematics professor at Harvey Mudd College. An interesting and wonderfully bizarre fact about Professor Benjamin himself is that he is one of the world’s great human calculators who can perform such feats as quickly multiplying six-digit numbers in his head.

250

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.14 Corresponding primal-dual forms Label

Primal Problem (or Dual Problem)

Dual Problem (or Primal Problem)

Maximize

Minimize

Z (or W)

W (or Z)

Sensible Odd Bizarre

Constraint i: form form form

Variable yi (or xi): yi 0 Unconstrained yi 0

Sensible Odd Bizarre

Variable xj (or yj): Constraint j: xj 0 ←→ form Unconstrained ←→ form xj 0 ←→ form

←→ ←→ ←→

The labeling of the functional constraints depends on whether the problem is a maximization problem (use the second column) or a minimization problem (use the third column). 3. For each constraint on an individual variable in the dual problem, use the form that has the same label as for the functional constraint in the primal problem that corresponds to this dual variable (as indicated by Table 6.3). 4. For each functional constraint in the dual problem, use the form that has the same label as for the constraint on the corresponding individual variable in the primal problem (as indicated by Table 6.3). The arrows between the second and third columns of Table 6.14 spell out the correspondence between the forms of constraints in the primal and dual. Note that the correspondence always is between a functional constraint in one problem and a constraint on an individual variable in the other problem. Since the primal problem can be either a maximization or minimization problem, where the dual then will be of the opposite type, the second column of the table gives the form for whichever is the maximization problem and the third column gives the form for the other problem (a minimization problem). To illustrate, consider the radiation therapy example presented in Sec. 3.4. (Its model is shown on p. 46.) To show the conversion in both directions in Table 6.14, we begin with the maximization form of this model as the primal problem, before using the (original) minimization form. The primal problem in maximization form is shown on the left side of Table 6.15. By using the second column of Table 6.14 to represent this problem, the arrows in this table indicate the form of the dual problem in the third column. These same arrows are used in Table 6.15 to show the resulting dual problem. (Because of these arrows, we have placed the functional constraints last in the dual problem rather than in their usual top position.) Beside each constraint in both problems, we have inserted (in parentheses) an S, O, or B to label the form as sensible, odd, or bizarre. As prescribed by the SOB method, the label for each dual constraint always is the same as for the corresponding primal constraint.

6.4 ADAPTING TO OTHER PRIMAL FORMS

251

TABLE 6.15 One primal-dual form for the radiation therapy example Primal Problem Maximize

Dual Problem

Z 0.4x1 0.5x2,

(S) (O) (B)

Minimize

W 2.7y1 6y2 6y3,

subject to

subject to 0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6

←→ ←→ ←→

y1 0 y2 unconstrained in sign y3 0

(S) (O) (B)

and

and x1 0 x2 0

(S) (S)

←→ ←→

0.3y1 0.5y2 0.6y3 0.4 0.1y1 0.5y2 0.4y3 0.5

(S) (S)

However, there was no need (other than for illustrative purposes) to convert the primal problem to maximization form. Using the original minimization form, the equivalent primal problem is shown on the left side of Table 6.16. Now we use the third column of Table 6.14 to represent this primal problem, where the arrows indicate the form of the dual problem in the second column. These same arrows in Table 6.16 show the resulting dual problem on the right side. Again, the labels on the constraints show the application of the SOB method. Just as the primal problems in Tables 6.15 and 6.16 are equivalent, the two dual problems also are completely equivalent. The key to recognizing this equivalency lies in the fact that the variables in each version of the dual problem are the negative of those in the other version (y1 y1, y2 y2, y3 y3). Therefore, for each version, if the variables in the other version are used instead, and if both the objective function and the constraints are multiplied through by 1, then the other version is obtained. (Problem 6.4-5 asks you to verify this.) If the simplex method is to be applied to either a primal or a dual problem that has any variables constrained to be nonpositive (for example, y3 0 in the dual problem of Table 6.15), this variable may be replaced by its nonnegative counterpart (for example, y3 y3). TABLE 6.16 The other primal-dual form for the radiation therapy example Primal Problem Minimize

Z 0.4x1 0.5x2,

0.3x1 0.1x2 2.7 0.5x1 0.5x2 6 0.6x1 0.4x2 6

W 2.7y1 6y2 6y3,

←→ ←→ ←→

y1 0 y2 unconstrained in sign y3 0

(B) (O) (S)

and

and (S) (S)

Maximize subject to

subject to (B) (O) (S)

Dual Problem

x1 0 x2 0

←→ ←→

0.3y1 0.5y2 0.6y3 0.4 0.1y1 0.5y2 0.4y3 0.6

(S) (S)

252

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

When artificial variables are used to help the simplex method solve a primal problem, the duality interpretation of row 0 of the simplex tableau is the following: Since artificial variables play the role of slack variables, their coefficients in row 0 now provide the values of the corresponding dual variables in the complementary basic solution for the dual problem. Since artificial variables are used to replace the real problem with a more convenient artificial problem, this dual problem actually is the dual of the artificial problem. However, after all the artificial variables become nonbasic, we are back to the real primal and dual problems. With the two-phase method, the artificial variables would need to be retained in phase 2 in order to read off the complete dual solution from row 0. With the Big M method, since M has been added initially to the coefficient of each artificial variable in row 0, the current value of each corresponding dual variable is the current coefficient of this artificial variable minus M. For example, look at row 0 in the final simplex tableau for the radiation therapy example, given at the bottom of Table 4.12 on p. 142. After M is subtracted from the coefficients of the artificial variables x4 and x6, the optimal solution for the corresponding dual problem given in Table 6.15 is read from the coefficients of x3, x4, and x6 as (y1, y2, y3) (0.5, 1.1, 0). As usual, the surplus variables for the two functional constraints are read from the coefficients of x1 and x2 as z1 c1 0 and z2 c2 0.

6.5

THE ROLE OF DUALITY THEORY IN SENSITIVITY ANALYSIS As described further in the next two sections, sensitivity analysis basically involves investigating the effect on the optimal solution of making changes in the values of the model parameters aij , bi, and cj. However, changing parameter values in the primal problem also changes the corresponding values in the dual problem. Therefore, you have your choice of which problem to use to investigate each change. Because of the primal-dual relationships presented in Secs. 6.1 and 6.3 (especially the complementary basic solutions property), it is easy to move back and forth between the two problems as desired. In some cases, it is more convenient to analyze the dual problem directly in order to determine the complementary effect on the primal problem. We begin by considering two such cases. Changes in the Coefficients of a Nonbasic Variable Suppose that the changes made in the original model occur in the coefficients of a variable that was nonbasic in the original optimal solution. What is the effect of these changes on this solution? Is it still feasible? Is it still optimal? Because the variable involved is nonbasic (value of zero), changing its coefficients cannot affect the feasibility of the solution. Therefore, the open question in this case is whether it is still optimal. As Tables 6.10 and 6.11 indicate, an equivalent question is whether the complementary basic solution for the dual problem is still feasible after these changes are made. Since these changes affect the dual problem by changing only one constraint, this question can be answered simply by checking whether this complementary basic solution still satisfies this revised constraint. We shall illustrate this case in the corresponding subsection of Sec. 6.7 after developing a relevant example.

6.5 THE ROLE OF DUALITY THEORY IN SENSITIVITY ANALYSIS

253

Introduction of a New Variable As indicated in Table 6.6, the decision variables in the model typically represent the levels of the various activities under consideration. In some situations, these activities were selected from a larger group of possible activities, where the remaining activities were not included in the original model because they seemed less attractive. Or perhaps these other activities did not come to light until after the original model was formulated and solved. Either way, the key question is whether any of these previously unconsidered activities are sufficiently worthwhile to warrant initiation. In other words, would adding any of these activities to the model change the original optimal solution? Adding another activity amounts to introducing a new variable, with the appropriate coefficients in the functional constraints and objective function, into the model. The only resulting change in the dual problem is to add a new constraint (see Table 6.3). After these changes are made, would the original optimal solution, along with the new variable equal to zero (nonbasic), still be optimal for the primal problem? As for the preceding case, an equivalent question is whether the complementary basic solution for the dual problem is still feasible. And, as before, this question can be answered simply by checking whether this complementary basic solution satisfies one constraint, which in this case is the new constraint for the dual problem. To illustrate, suppose for the Wyndor Glass Co. problem of Sec. 3.1 that a possible third new product now is being considered for inclusion in the product line. Letting xnew represent the production rate for this product, we show the resulting revised model as follows: Maximize

Z 3x1 5x2 4xnew,

subject to x1 2x2 2xnew 4 3x1 2x2 3xnew 12 3x1 2x2 xnew 18 and x1 0,

x2 0,

xnew 0.

After we introduced slack variables, the original optimal solution for this problem without xnew (given by Table 4.8) was (x1, x2, x3, x4, x5) (2, 6, 2, 0, 0). Is this solution, along with xnew 0, still optimal? To answer this question, we need to check the complementary basic solution for the dual problem. As indicated by the complementary optimal basic solutions property in Sec. 6.3, this solution is given in row 0 of the final simplex tableau for the primal problem, using the locations shown in Table 6.4 and illustrated in Table 6.5. Therefore, as given in both the bottom row of Table 6.5 and the sixth row of Table 6.9, the solution is

3 (y1, y2, y3, z1 c1, z2 c2) 0, , 1, 0, 0 . 2 (Alternatively, this complementary basic solution can be derived in the way that was illustrated in Sec. 6.3 for the complementary basic solution in the next-to-last row of Table 6.9.)

254

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Since this solution was optimal for the original dual problem, it certainly satisfies the original dual constraints shown in Table 6.1. But does it satisfy this new dual constraint? 2y1 3y2 y3 4 Plugging in this solution, we see that

3 2(0) 3 (1) 4 2 is satisfied, so this dual solution is still feasible (and thus still optimal). Consequently, the original primal solution (2, 6, 2, 0, 0), along with xnew 0, is still optimal, so this third possible new product should not be added to the product line. This approach also makes it very easy to conduct sensitivity analysis on the coefficients of the new variable added to the primal problem. By simply checking the new dual constraint, you can immediately see how far any of these parameter values can be changed before they affect the feasibility of the dual solution and so the optimality of the primal solution. Other Applications Already we have discussed two other key applications of duality theory to sensitivity analysis, namely, shadow prices and the dual simplex method. As described in Secs. 4.7 and 6.2, the optimal dual solution (y1*, y2*, . . . , ym*) provides the shadow prices for the respective resources that indicate how Z would change if (small) changes were made in the bi (the resource amounts). The resulting analysis will be illustrated in some detail in Sec. 6.7. In more general terms, the economic interpretation of the dual problem and of the simplex method presented in Sec. 6.2 provides some useful insights for sensitivity analysis. When we investigate the effect of changing the bi or the aij values (for basic variables), the original optimal solution may become a superoptimal basic solution (as defined in Table 6.10) instead. If we then want to reoptimize to identify the new optimal solution, the dual simplex method (discussed at the end of Secs. 6.1 and 6.3) should be applied, starting from this basic solution. We mentioned in Sec. 6.1 that sometimes it is more efficient to solve the dual problem directly by the simplex method in order to identify an optimal solution for the primal problem. When the solution has been found in this way, sensitivity analysis for the primal problem then is conducted by applying the procedure described in the next two sections directly to the dual problem and then inferring the complementary effects on the primal problem (e.g., see Table 6.11). This approach to sensitivity analysis is relatively straightforward because of the close primal-dual relationships described in Secs. 6.1 and 6.3. (See Prob. 6.6-3.)

6.6

THE ESSENCE OF SENSITIVITY ANALYSIS The work of the operations research team usually is not even nearly done when the simplex method has been successfully applied to identify an optimal solution for the model. As we pointed out at the end of Sec. 3.3, one assumption of linear programming is that all the parameters of the model (aij, bi, and cj) are known constants. Actually, the parameter values used in the model normally are just estimates based on a prediction of future conditions. The data obtained to develop these estimates often are rather crude or non-

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

255

existent, so that the parameters in the original formulation may represent little more than quick rules of thumb provided by harassed line personnel. The data may even represent deliberate overestimates or underestimates to protect the interests of the estimators. Thus, the successful manager and operations research staff will maintain a healthy skepticism about the original numbers coming out of the computer and will view them in many cases as only a starting point for further analysis of the problem. An “optimal” solution is optimal only with respect to the specific model being used to represent the real problem, and such a solution becomes a reliable guide for action only after it has been verified as performing well for other reasonable representations of the problem. Furthermore, the model parameters (particularly bi) sometimes are set as a result of managerial policy decisions (e.g., the amount of certain resources to be made available to the activities), and these decisions should be reviewed after their potential consequences are recognized. For these reasons it is important to perform sensitivity analysis to investigate the effect on the optimal solution provided by the simplex method if the parameters take on other possible values. Usually there will be some parameters that can be assigned any reasonable value without the optimality of this solution being affected. However, there may also be parameters with likely alternative values that would yield a new optimal solution. This situation is particularly serious if the original solution would then have a substantially inferior value of the objective function, or perhaps even be infeasible! Therefore, one main purpose of sensitivity analysis is to identify the sensitive parameters (i.e., the parameters whose values cannot be changed without changing the optimal solution). For certain parameters that are not categorized as sensitive, it is also very helpful to determine the range of values of the parameter over which the optimal solution will remain unchanged. (We call this range of values the allowable range to stay optimal.) In some cases, changing a parameter value can affect the feasibility of the optimal BF solution. For such parameters, it is useful to determine the range of values over which the optimal BF solution (with adjusted values for the basic variables) will remain feasible. (We call this range of values the allowable range to stay feasible.) In the next section, we will describe the specific procedures for obtaining this kind of information. Such information is invaluable in two ways. First, it identifies the more important parameters, so that special care can be taken to estimate them closely and to select a solution that performs well for most of their likely values. Second, it identifies the parameters that will need to be monitored particularly closely as the study is implemented. If it is discovered that the true value of a parameter lies outside its allowable range, this immediately signals a need to change the solution. For small problems, it would be straightforward to check the effect of a variety of changes in parameter values simply by reapplying the simplex method each time to see if the optimal solution changes. This is particularly convenient when using a spreadsheet formulation. Once the Solver has been set up to obtain an optimal solution, all you have to do is make any desired change on the spreadsheet and then click on the Solve button again. However, for larger problems of the size typically encountered in practice, sensitivity analysis would require an exorbitant computational effort if it were necessary to reapply the simplex method from the beginning to investigate each new change in a parameter value. Fortunately, the fundamental insight discussed in Sec. 5.3 virtually eliminates computational effort. The basic idea is that the fundamental insight immediately reveals

256

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

just how any changes in the original model would change the numbers in the final simplex tableau (assuming that the same sequence of algebraic operations originally performed by the simplex method were to be duplicated ). Therefore, after making a few simple calculations to revise this tableau, we can check easily whether the original optimal BF solution is now nonoptimal (or infeasible). If so, this solution would be used as the initial basic solution to restart the simplex method (or dual simplex method) to find the new optimal solution, if desired. If the changes in the model are not major, only a very few iterations should be required to reach the new optimal solution from this “advanced” initial basic solution. To describe this procedure more specifically, consider the following situation. The simplex method already has been used to obtain an optimal solution for a linear programming model with specified values for the bi , cj , and aij parameters. To initiate sensitivity analysis, at least one of the parameters is changed. After the changes are made, let bi , cj , and aij denote the values of the various parameters. Thus, in matrix notation, bb ,

c c,

AA ,

for the revised model. The first step is to revise the final simplex tableau to reflect these changes. Continuing to use the notation presented in Table 5.10, as well as the accompanying formulas for the fundamental insight [(1) t* t y*T and (2) T* S*T], we see that the revised final tableau is calculated from y* and S* (which have not changed) and the new initial tableau, as shown in Table 6.17. Example (Variation 1 of the Wyndor Model). To illustrate, suppose that the first revision in the model for the Wyndor Glass Co. problem of Sec. 3.1 is the one shown in Table 6.18. Thus, the changes from the original model are c1 3 4, a31 3 2, and b2 12 24. Figure 6.2 shows the graphical effect of these changes. For the original model, the simplex method already has identified the optimal CPF solution as (2, 6), lying at the intersection of the two constraint boundaries, shown as dashed lines 2x2 12 and 3x1 2x2 18. Now the revision of the model has shifted both of these constraint boundaries as shown by the dark lines 2x2 24 and 2x1 2x2 18. Consequently, the previous TABLE 6.17 Revised final simplex tableau resulting from changes in original model Coefficient of: Eq.

Z

Original Variables

Slack Variables

Right Side

(0)

1

c

0

0

(1, 2, . . . , m)

0

A

I

b

(0)

1

z* c y*A c

y*

Z* y*b

(1, 2, . . . , m)

0

A* S*A

S*

b* S*b

New initial tableau

Revised final tableau

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

257

TABLE 6.18 The original model and the first revised model (variation 1) for conducting sensitivity analysis on the Wyndor Glass Co. model Original Model Maximize

Revised Model

Z [3, 5]

x , x1

Maximize

Z [4, 5]

0 x1 2 x2 2

1 0 2

4 12 18

0 x1 2 x2 2

4 24 18

and

and

x 0.

x 0.

FIGURE 6.2 Shift of the final corner-point solution from (2, 6) to (3, 12) for Variation 1 of the Wyndor Glass Co. model where c1 3 4, a31 3 2, and b2 12 24.

x1

subject to

subject to

1 0 3

x , 2

2

x2 x1 0

(3, 12)

2x2 24

10 x1 4

(0, 9) optimal 8

6

2x2 12

(2, 6)

4 2x1 2x2 18 2

3x1 2x2 18 x2 0

0

2

4

6

8

x1

258

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

CPF solution (2, 6) now shifts to the new intersection (3, 12), which is a corner-point infeasible solution for the revised model. The procedure described in the preceding paragraphs finds this shift algebraically (in augmented form). Furthermore, it does so in a manner that is very efficient even for huge problems where graphical analysis is impossible. To carry out this procedure, we begin by displaying the parameters of the revised model in matrix form: c [4, 5],

1 A 0 2

0 2 , 2

4 b 24 . 18

The resulting new initial simplex tableau is shown at the top of Table 6.19. Below this tableau is the original final tableau (as first given in Table 4.8). We have drawn dark boxes around the portions of this final tableau that the changes in the model definitely do not change, namely, the coefficients of the slack variables in both row 0 (y*) and the rest of the rows (S*). Thus, y* [0, , 1], 3 2

1 S* 0 0

1 3 1 2 1 3

13 0 . 1 3

TABLE 6.19 Obtaining the revised final simplex tableau for Variation 1 of the Wyndor Glass Co. model Coefficient of:

New initial tableau

Final tableau for original model

Basic Variable

Eq.

Z

x1

x2

x3

Z x3 x4 x5

(0) (1) (2) (3)

1 0 0 0

4 1 0 2

5 0 2 2

0 1 0 0

Z

(0)

1

0

0

0

x3

(1)

0

0

0

1

x2

(2)

0

0

1

0

x1

(3)

0

1

0

0

Z

(0)

1

2

0

0

x3

(1)

0

1 3

0

1

x2

(2)

0

0

1

0

x1

(3)

0

2 3

0

0

Revised final tableau

x4 0 0 1 0

3 2 1 3 1 2 1 3 3 2 1 3 1 2 1 3

x5

Right Side

0 0 0 1

0 4 24 18

1

36

1 3

2

0

6

1 3

2

1

54

1 3

6

0

12

1 3

2

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

259

These coefficients of the slack variables necessarily are unchanged with the same algebraic operations originally performed by the simplex method because the coefficients of these same variables in the initial tableau are unchanged. However, because other portions of the initial tableau have changed, there will be changes in the rest of the final tableau as well. Using the formulas in Table 6.17, we calculate the revised numbers in the rest of the final tableau as follows: 1 z* c [0, 32, 1] 0 2

0 2 [4, 5] [2, 0], 2

1 1 13 1 3 1 0 0 A* 0 2 1 1 0 3 3 2

13 0 2 0 2 2 3

4 Z* [0, 32, 1] 24 54, 18

0 1 , 0

1 6 1 13 4 3 1 0 24 12 . b* 0 2 1 1 2 0 3 3 18 The resulting revised final tableau is shown at the bottom of Table 6.19. Actually, we can substantially streamline these calculations for obtaining the revised final tableau. Because none of the coefficients of x2 changed in the original model (tableau), none of them can change in the final tableau, so we can delete their calculation. Several other original parameters (a11, a21, b1, b3) also were not changed, so another shortcut is to calculate only the incremental changes in the final tableau in terms of the incremental changes in the initial tableau, ignoring those terms in the vector or matrix multiplication that involve zero change in the initial tableau. In particular, the only incremental changes in the initial tableau are c1 1, a31 1, and b2 12, so these are the only terms that need be considered. This streamlined approach is shown below, where a zero or dash appears in each spot where no calculation is needed.

0 (z* c) y* A c [0, 32, 1] 0 1

— — [1, —] [2, —]. —

0 Z* y* b [0, , 1] 12 18. 0 3 2

1 A* S* A 0 0

1 3 1 2 1 3

13 0 0 0 1 3 1

13 — — 0 1 — 3

— — . —

1 1 4 13 0 3 1 0 12 6 . b* S* b 0 2 1 1 0 3 4 3 0 Adding these increments to the original quantities in the final tableau (middle of Table 6.19) then yields the revised final tableau (bottom of Table 6.19).

260

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

This incremental analysis also provides a useful general insight, namely, that changes in the final tableau must be proportional to each change in the initial tableau. We illustrate in the next section how this property enables us to use linear interpolation or extrapolation to determine the range of values for a given parameter over which the final basic solution remains both feasible and optimal. After obtaining the revised final simplex tableau, we next convert the tableau to proper form from Gaussian elimination (as needed). In particular, the basic variable for row i must have a coefficient of 1 in that row and a coefficient of 0 in every other row (including row 0) for the tableau to be in the proper form for identifying and evaluating the current basic solution. Therefore, if the changes have violated this requirement (which can occur only if the original constraint coefficients of a basic variable have been changed), further changes must be made to restore this form. This restoration is done by using Gaussian elimination, i.e., by successively applying step 3 of an iteration for the simplex method (see Chap. 4) as if each violating basic variable were an entering basic variable. Note that these algebraic operations may also cause further changes in the right side column, so that the current basic solution can be read from this column only when the proper form from Gaussian elimination has been fully restored. For the example, the revised final simplex tableau shown in the top half of Table 6.20 is not in proper form from Gaussian elimination because of the column for the basic variable x1. Specifically, the coefficient of x1 in its row (row 3) is 23 instead of 1, and it has nonzero coefficients (2 and 13) in rows 0 and 1. To restore proper form, row 3 is multiplied by 32; then 2 times this new row 3 is added to row 0 and 13 times new row 3 is subtracted from row 1. This yields the proper form from Gaussian elimination shown in

TABLE 6.20 Converting the revised final simplex tableau to proper form from Gaussian elimination for Variation 1 of the Wyndor Glass Co. model Coefficient of:

Revised final tableau

Converted to proper form

Basic Variable

Eq.

Z

x1

x2

Z

(0)

1

2

0

0

x3

(1)

0

1 3

0

1

x2

(2)

0

0

1

0

x1

(3)

0

2 3

0

0

Z

(0)

1

0

0

0

x3

(1)

0

0

0

1

x2

(2)

0

0

1

0

x1

(3)

0

1

0

0

x3

x4 3 2 1 3 1 2 1 3 1 2 1 2 1 2 1 2

x5

Right Side

1

54

1 3

6

0

12

1 3

2

2

48

1 2

7

0

12

1 2

3

6.6 THE ESSENCE OF SENSITIVITY ANALYSIS

261

the bottom half of Table 6.20, which now can be used to identify the new values for the current (previously optimal) basic solution: (x1, x2, x3, x4, x5) (3, 12, 7, 0, 0). Because x1 is negative, this basic solution no longer is feasible. However, it is superoptimal (as defined in Table 6.10), and so dual feasible, because all the coefficients in row 0 still are nonnegative. Therefore, the dual simplex method can be used to reoptimize (if desired), by starting from this basic solution. (The sensitivity analysis routine in the OR Courseware includes this option.) Referring to Fig. 6.2 (and ignoring slack variables), the dual simplex method uses just one iteration to move from the corner-point solution (3, 12) to the optimal CPF solution (0, 9). (It is often useful in sensitivity analysis to identify the solutions that are optimal for some set of likely values of the model parameters and then to determine which of these solutions most consistently performs well for the various likely parameter values.) If the basic solution (3, 12, 7, 0, 0) had been neither primal feasible nor dual feasible (i.e., if the tableau had negative entries in both the right side column and row 0), artificial variables could have been introduced to convert the tableau to the proper form for an initial simplex tableau.1 The General Procedure. When one is testing to see how sensitive the original optimal solution is to the various parameters of the model, the common approach is to check each parameter (or at least cj and bi) individually. In addition to finding allowable ranges as described in the next section, this check might include changing the value of the parameter from its initial estimate to other possibilities in the range of likely values (including the endpoints of this range). Then some combinations of simultaneous changes of parameter values (such as changing an entire functional constraint) may be investigated. Each time one (or more) of the parameters is changed, the procedure described and illustrated here would be applied. Let us now summarize this procedure. Summary of Procedure for Sensitivity Analysis 1. Revision of model: Make the desired change or changes in the model to be investigated next. 2. Revision of final tableau: Use the fundamental insight (as summarized by the formulas on the bottom of Table 6.17) to determine the resulting changes in the final simplex tableau. (See Table 6.19 for an illustration.) 3. Conversion to proper form from Gaussian elimination: Convert this tableau to the proper form for identifying and evaluating the current basic solution by applying (as necessary) Gaussian elimination. (See Table 6.20 for an illustration.) 4. Feasibility test: Test this solution for feasibility by checking whether all its basic variable values in the right-side column of the tableau still are nonnegative. 5. Optimality test: Test this solution for optimality (if feasible) by checking whether all its nonbasic variable coefficients in row 0 of the tableau still are nonnegative. 6. Reoptimization: If this solution fails either test, the new optimal solution can be obtained (if desired) by using the current tableau as the initial simplex tableau (and making any necessary conversions) for the simplex method or dual simplex method. 1

There also exists a primal-dual algorithm that can be directly applied to such a simplex tableau without any conversion.

262

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

The interactive routine entitled sensitivity analysis in the OR Courseware will enable you to efficiently practice applying this procedure. In addition, a demonstration in OR Tutor (also entitled sensitivity analysis) provides you with another example. In the next section, we shall discuss and illustrate the application of this procedure to each of the major categories of revisions in the original model. This discussion will involve, in part, expanding upon the example introduced in this section for investigating changes in the Wyndor Glass Co. model. In fact, we shall begin by individually checking each of the preceding changes. At the same time, we shall integrate some of the applications of duality theory to sensitivity analysis discussed in Sec. 6.5.

6.7

APPLYING SENSITIVITY ANALYSIS Sensitivity analysis often begins with the investigation of changes in the values of bi, the amount of resource i (i 1, 2, . . . , m) being made available for the activities under consideration. The reason is that there generally is more flexibility in setting and adjusting these values than there is for the other parameters of the model. As already discussed in Secs. 4.7 and 6.2, the economic interpretation of the dual variables (the yi) as shadow prices is extremely useful for deciding which changes should be considered. Case 1—Changes in bi Suppose that the only changes in the current model are that one or more of the bi parameters (i 1, 2, . . . , m) has been changed. In this case, the only resulting changes in the final simplex tableau are in the right-side column. Consequently, the tableau still will be in proper form from Gaussian elimination and all the nonbasic variable coefficients in row 0 still will be nonnegative. Therefore, both the conversion to proper form from Gaussian elimination and the optimality test steps of the general procedure can be skipped. After revising the right-side column of the tableau, the only question will be whether all the basic variable values in this column still are nonnegative (the feasibility test). As shown in Table 6.17, when the vector of the bi values is changed from b to b, the formulas for calculating the new right-side column in the final tableau are Right side of final row 0: Right side of final rows 1, 2, . . . , m:

Z* y*b , b* S*b .

(See the bottom of Table 6.17 for the location of the unchanged vector y* and matrix S* in the final tableau.) Example (Variation 2 of the Wyndor Model). Sensitivity analysis is begun for the original Wyndor Glass Co. problem of Sec. 3.1 by examining the optimal values of the yi dual variables ( y1* 0, y2* 32, y3* 1). These shadow prices give the marginal value of each resource i for the activities (two new products) under consideration, where marginal value is expressed in the units of Z (thousands of dollars of profit per week). As discussed in Sec. 4.7 (see Fig. 4.8), the total profit from these activities can be increased $1,500 per week ( y2* times $1,000 per week) for each additional unit of resource 2 (hour of production time per week in Plant 2) that is made available. This increase in profit holds for relatively small changes that do not affect the feasibility of the current basic solution (and so do not affect the yi* values).

6.7 APPLYING SENSITIVITY ANALYSIS

263

Consequently, the OR team has investigated the marginal profitability from the other current uses of this resource to determine if any are less than $1,500 per week. This investigation reveals that one old product is far less profitable. The production rate for this product already has been reduced to the minimum amount that would justify its marketing expenses. However, it can be discontinued altogether, which would provide an additional 12 units of resource 2 for the new products. Thus, the next step is to determine the profit that could be obtained from the new products if this shift were made. This shift changes b2 from 12 to 24 in the linear programming model. Figure 6.3 shows the graphical effect of this change, including the shift in the final corner-point solution from (2, 6) to (2, 12). (Note that this figure differs from Fig. 6.2, which depicts Variation 1 of the Wyndor model, because the constraint 3x1 2x2 18 has not been changed here.) Thus, for Variation 2 of the Wyndor model, the only revision in the original model is the following change in the vector of the bi values: 4 b 12 → b 18

4 24 . 18

so only b2 has a new value.

FIGURE 6.3 Feasible region for Variation 2 of the Wyndor Glass Co. model where b2 12 → 24.

x2 14

x1 0

(2, 12) 2x2 24

10 x1 4

(0, 9) optimal 8 (2, 6)

2x2 12

6

Z 45 3x1 5x2

4 Feasible region

3x1 2x2 18

2

x2 0 0

2

4

6

8

x1

264

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Analysis of Variation 2. When the fundamental insight (Table 6.17) is applied, the effect of this change in b2 on the original final simplex tableau (middle of Table 6.19) is that the entries in the right-side column change to the following values: 4 Z* y*b [0, , 1] 24 54, 18 3 2

1 b* S*b 0 0

1 3 1 2 1 3

6 13 4 0 24 12 , 1 2 3 18

x3 6 so x2 12 . x1 2

Equivalently, because the only change in the original model is b2 24 12 12, incremental analysis can be used to calculate these same values more quickly. Incremental analysis involves calculating just the increments in the tableau values caused by the change (or changes) in the original model, and then adding these increments to the original values. In this case, the increments in Z* and b* are b1 Z* y* b y* b2 y* b3

0 12 , 0 b1 b* S* b S* b2 S* b3

0 12 . 0

Therefore, using the second component of y* and the second column of S*, the only calculations needed are 3 Z* (12) 18, 2 1 b1* (12) 4, 3 1 b2* (12) 6, 2 1 b3* (12) 4, 3

so Z* 36 18 54, so b1* 2 4 6, so b2* 6 6 12, so b3* 2 4 2,

where the original values of these quantities are obtained from the right-side column in the original final tableau (middle of Table 6.19). The resulting revised final tableau corresponds completely to this original final tableau except for replacing the right-side column with these new values. Therefore, the current (previously optimal) basic solution has become (x1, x2, x3, x4, x5) (2, 12, 6, 0, 0), which fails the feasibility test because of the negative value. The dual simplex method now can be applied, starting with this revised simplex tableau, to find the new optimal so-

6.7 APPLYING SENSITIVITY ANALYSIS

265

TABLE 6.21 Data for Variation 2 of the Wyndor Glass Co. model Final Simplex Tableau after Reoptimization Coefficient of: Basic Variable

Eq.

Z

Z

(0)

1

Model Parameters c1 3, a11 1, a21 0, a31 3,

c2 5 a12 0, a22 2, a32 2,

(n 2) b1 4 b2 24 b3 18

x3

(1)

0

x2

(2)

0

x4

(3)

0

x1 9 2 1 3 2 3

x2

x3

x4

0

0

0

0

1

0

1

0

0

0

0

1

x5 5 2 0 1 2 1

Right Side 45 4 9 6

lution. This method leads in just one iteration to the new final simplex tableau shown in Table 6.21. (Alternatively, the simplex method could be applied from the beginning, which also would lead to this final tableau in just one iteration in this case.) This tableau indicates that the new optimal solution is (x1, x2, x3, x4, x5) (0, 9, 4, 6, 0), with Z 45, thereby providing an increase in profit from the new products of 9 units ($9,000 per week) over the previous Z 36. The fact that x4 6 indicates that 6 of the 12 additional units of resource 2 are unused by this solution. Based on the results with b2 24, the relatively unprofitable old product will be discontinued and the unused 6 units of resource 2 will be saved for some future use. Since y3* still is positive, a similar study is made of the possibility of changing the allocation of resource 3, but the resulting decision is to retain the current allocation. Therefore, the current linear programming model at this point (Variation 2) has the parameter values and optimal solution shown in Table 6.21. This model will be used as the starting point for investigating other types of changes in the model later in this section. However, before turning to these other cases, let us take a broader look at the current case. The Allowable Range to Stay Feasible. Although b2 12 proved to be too large an increase in b2 to retain feasibility (and so optimality) with the basic solution where x1, x2, and x3 are the basic variables (middle of Table 6.19), the above incremental analysis shows immediately just how large an increase is feasible. In particular, note that 1 b1* 2 b2, 3 1 b2* 6 b2, 2 1 b3* 2 b2, 3

266

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

where these three quantities are the values of x3, x2, and x1, respectively, for this basic solution. The solution remains feasible, and so optimal, as long as all three quantities remain nonnegative. 1 1 2 b2 0 ⇒ b2 2 ⇒ b2 6, 3 3 1 1 6 b2 0 ⇒ b2 6 ⇒ b2 12, 2 2 1 1 2 b2 0 ⇒ 2 b2 3 3

⇒ b2

6.

Therefore, since b2 12 b2, the solution remains feasible only if 6 b2 6,

that is,

6 b2 18.

(Verify this graphically in Fig. 6.3.) As introduced in Sec. 4.7, this range of values for b2 is referred to as its allowable range to stay feasible. For any bi, recall from Sec. 4.7 that its allowable range to stay feasible is the range of values over which the current optimal BF solution1 (with adjusted values for the basic variables) remains feasible. Thus, the shadow price for bi remains valid for evaluating the effect on Z of changing bi only as long as bi remains within this allowable range. (It is assumed that the change in this one bi value is the only change in the model.) The adjusted values for the basic variables are obtained from the formula b* S*b . The calculation of the allowable range to stay feasible then is based on finding the range of values of bi such that b* 0. Many linear programming software packages use this same technique for automatically generating the allowable range to stay feasible for each bi. (A similar technique, discussed under Cases 2a and 3, also is used to generate an allowable range to stay optimal for each cj.) In Chap. 4, we showed the corresponding output for the Excel Solver and LINDO in Figs. 4.10 and 4.13, respectively. Table 6.22 summarizes this same output with respect to the bi for the original Wyndor Glass Co. model. For example, both the allowable increase and allowable decrease for b2 are 6, that is, 6 b2 6. The above analysis shows how these quantities were calculated. 1

When there is more than one optimal BF solution for the current model (before changing bi), we are referring here to the one obtained by the simplex method.

TABLE 6.22 Typical software output for sensitivity analysis of the right-hand sides for the original Wyndor Glass Co. model Constraint Plant 1 Plant 2 Plant 3

Shadow Price

Current RHS

Allowable Increase

Allowable Decrease

0.0 1.5 1.0

4 12 18

6 6

2 6 6

6.7 APPLYING SENSITIVITY ANALYSIS

267

Analyzing Simultaneous Changes in Right-Hand Sides. When multiple bi values are changed simultaneously, the formula b* S*b can again be used to see how the righthand sides change in the final tableau. If all these right-hand sides still are nonnegative, the feasibility test will indicate that the revised solution provided by this tableau still is feasible. Since row 0 has not changed, being feasible implies that this solution also is optimal. Although this approach works fine for checking the effect of a specific set of changes in the bi, it does not give much insight into how far the bi can be simultaneously changed from their original values before the revised solution will no longer be feasible. As part of postoptimality analysis, the management of an organization often is interested in investigating the effect of various changes in policy decisions (e.g., the amounts of resources being made available to the activities under consideration) that determine the right-hand sides. Rather than considering just one specific set of changes, management may want to explore directions of changes where some right-hand sides increase while others decrease. Shadow prices are invaluable for this kind of exploration. However, shadow prices remain valid for evaluating the effect of such changes on Z only within certain ranges of changes. For each bi, the allowable range to stay feasible gives this range if none of the other bi are changing at the same time. What do these allowable ranges become when some of the bi are changing simultaneously? A partial answer to this question is provided by the following 100 percent rule, which combines the allowable changes (increase or decrease) for the individual bi that are given by the last two columns of a table like Table 6.22. The 100 Percent Rule for Simultaneous Changes in Right-Hand Sides: The shadow prices remain valid for predicting the effect of simultaneously changing the right-hand sides of some of the functional constraints as long as the changes are not too large. To check whether the changes are small enough, calculate for each change the percentage of the allowable change (increase or decrease) for that right-hand side to remain within its allowable range to stay feasible. If the sum of the percentage changes does not exceed 100 percent, the shadow prices definitely will still be valid. (If the sum does exceed 100 percent, then we cannot be sure.) Example (Variation 3 of the Wyndor Model). To illustrate this rule, consider Variation 3 of the Wyndor Glass Co. model, which revises the original model by changing the right-hand side vector as follows: 4 b 12 b 18

4 15 . 15

The calculations for the 100 percent rule in this case are b2: 12 15. b3: 18 15.

15 12 Percentage of allowable increase 100 50% 6 18 15 Percentage of allowable decrease 100 50% 6 Sum 100%

Since the sum of 100 percent barely does not exceed 100 percent, the shadow prices definitely are valid for predicting the effect of these changes on Z. In particular, since

268

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

the shadow prices of b2 and b3 are 1.5 and 1, respectively, the resulting change in Z would be Z 1.5(3) 1(3) 1.5, so Z* would increase from 36 to 37.5. Figure 6.4 shows the feasible region for this revised model. (The dashed lines show the original locations of the revised constraint boundary lines.) The optimal solution now is the CPF solution (0, 7.5), which gives Z 3x1 5x2 0 5(7.5) 37.5, just as predicted by the shadow prices. However, note what would happen if either b2 were further increased above 15 or b3 were further decreased below 15, so that the sum of the percentages of allowable changes would exceed 100 percent. This would cause the previously optimal corner-point solution to slide to the left of the x2 axis (x1 0), so this infeasible solution would no longer be optimal. Consequently, the old shadow prices would no longer be valid for predicting the new value of Z*.

FIGURE 6.4 Feasible region for Variation 3 of the Wyndor Glass Co. model where b2 12 15 and b3 18 15.

x2

8 (0, 7.5) optimal

2x2 15

6 x1 4 4 Feasible region 2 3x1 2x2 15

0

2

4

6

8

x1

6.7 APPLYING SENSITIVITY ANALYSIS

269

Case 2a—Changes in the Coefficients of a Nonbasic Variable Consider a particular variable xj (fixed j) that is a nonbasic variable in the optimal solution shown by the final simplex tableau. In Case 2a, the only change in the current model is that one or more of the coefficients of this variable—cj , a1j , a2j , . . . , amj —have been changed. Thus, letting cj and aij denote the new values of these parameters, with Aj (column j of matrix A ) as the vector containing the aij , we have cj → cj ,

Aj → A j

for the revised model. As described at the beginning of Sec. 6.5, duality theory provides a very convenient way of checking these changes. In particular, if the complementary basic solution y* in the dual problem still satisfies the single dual constraint that has changed, then the original optimal solution in the primal problem remains optimal as is. Conversely, if y* violates this dual constraint, then this primal solution is no longer optimal. If the optimal solution has changed and you wish to find the new one, you can do so rather easily. Simply apply the fundamental insight to revise the xj column (the only one that has changed) in the final simplex tableau. Specifically, the formulas in Table 6.17 reduce to the following: Coefficient of xj in final row 0: Coefficient of xj in final rows 1 to m:

z j* cj y*A j cj , Aj* S*A j.

With the current basic solution no longer optimal, the new value of zj* cj now will be the one negative coefficient in row 0, so restart the simplex method with xj as the initial entering basic variable. Note that this procedure is a streamlined version of the general procedure summarized at the end of Sec. 6.6. Steps 3 and 4 (conversion to proper form from Gaussian elimination and the feasibility test) have been deleted as irrelevant, because the only column being changed in the revision of the final tableau (before reoptimization) is for the nonbasic variable xj. Step 5 (optimality test) has been replaced by a quicker test of optimality to be performed right after step 1 (revision of model). It is only if this test reveals that the optimal solution has changed, and you wish to find the new one, that steps 2 and 6 (revision of final tableau and reoptimization) are needed. Example (Variation 4 of the Wyndor Model). Since x1 is nonbasic in the current optimal solution (see Table 6.21) for Variation 2 of the Wyndor Glass Co. model, the next step in its sensitivity analysis is to check whether any reasonable changes in the estimates of the coefficients of x1 could still make it advisable to introduce product 1. The set of changes that goes as far as realistically possible to make product 1 more attractive would be to reset c1 4 and a31 2. Rather than exploring each of these changes independently (as is often done in sensitivity analysis), we will consider them together. Thus, the changes under consideration are c1 3 → c1 4,

1 A1 0 → A1 3

1 0 . 2

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6

DUALITY THEORY AND SENSITIVITY ANALYSIS

These two changes in Variation 2 give us Variation 4 of the Wyndor model. Variation 4 actually is equivalent to Variation 1 considered in Sec. 6.6 and depicted in Fig. 6.2, since Variation 1 combined these two changes with the change in the original Wyndor model (b2 12 24) that gave Variation 2. However, the key difference from the treatment of Variation 1 in Sec. 6.6 is that the analysis of Variation 4 treats Variation 2 as being the original model, so our starting point is the final simplex tableau given in Table 6.21 where x1 now is a nonbasic variable. The change in a31 revises the feasible region from that shown in Fig. 6.3 to the corresponding region in Fig. 6.5. The change in c1 revises the objective function from Z 3x1 5x2 to Z 4x1 5x2. Figure 6.5 shows that the optimal objective function line Z 45 4x1 5x2 still passes through the current optimal solution (0, 9), so this solution remains optimal after these changes in a31 and c1. To use duality theory to draw this same conclusion, observe that the changes in c1 and a31 lead to a single revised constraint for the dual problem, namely, the constraint that a11y1 a21y2 a31y3 c1. Both this revised constraint and the current y* (coefficients of the slack variables in row 0 of Table 6.21) are shown below. 5 y2* 0, y3* , y1* 0, 2 y1 3y3 3 → y1 2y3 4, 5 0 2 4. 2 Since y* still satisfies the revised constraint, the current primal solution (Table 6.21) is still optimal. Because this solution is still optimal, there is no need to revise the xj column in the final tableau (step 2). Nevertheless, we do so below for illustrative purposes. 1 5 * 1 c1 [0, 0, 2] 0 4 1. z 1 c1 y*A 2

1 1 0 0 1 1 0 A1* S*A 0 1 0 1 . 2 1 1 2 0 2 The fact that z1* c1 0 again confirms the optimality of the current solution. Since z1* c1 is the surplus variable for the revised constraint in the dual problem, this way of testing for optimality is equivalent to the one used above. This completes the analysis of the effect of changing the current model (Variation 2) to Variation 4. Because any larger changes in the original estimates of the coefficients of x1 would be unrealistic, the OR team concludes that these coefficients are insensitive parameters in the current model. Therefore, they will be kept fixed at their best estimates shown in Table 6.21—c1 3 and a31 3—for the remainder of the sensitivity analysis. The Allowable Range to Stay Optimal. We have just described and illustrated how to analyze simultaneous changes in the coefficients of a nonbasic variable xj. It is common practice in sensitivity analysis to also focus on the effect of changing just one param-

6.7 APPLYING SENSITIVITY ANALYSIS

271

x2 12

2x2 24

x1 4

10 (0, 9) optimal 8

6

4

FIGURE 6.5 Feasible region for Variation 4 of the Wyndor model where Variation 2 (Fig. 6.3) has been revised so a31 3 2 and c1 3 4.

Feasible region

Z 45 4x1 5x2

2

0

2x1 2x2 18

2

4

6

8

10 x1

eter, cj. As introduced in Sec. 4.7, this involves streamlining the above approach to find the allowable range to stay optimal for cj. For any cj, recall from Sec. 4.7 that its allowable range to stay optimal is the range of values over which the current optimal solution (as obtained by the simplex method for the current model before cj is changed) remains optimal. (It is assumed that the change in this one cj is the only change in the current model.) When xj is a nonbasic variable for this solution, the solution remains optimal as long as z*j cj 0, where z*j y*Aj is a constant unaffected by any change in the value of cj. Therefore, the allowable range to stay optimal for cj can be calculated as cj y*Aj. For example, consider the current model (Variation 2) for the Wyndor Glass Co. problem summarized on the left side of Table 6.21, where the current optimal solution (with c1 3) is given on the right side. When considering only the decision variables, x1 and x2, this optimal solution is (x1, x2) = (0, 9), as displayed in Fig. 6.3. When just c1 is changed, this solution remains optimal as long as 1 c1 y*A1 [0, 0, 52] 0 712, 3 1 so c1 72 is the allowable range to stay optimal.

272

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

An alternative to performing this vector multiplication is to note in Table 6.21 that z1* c1 92 (the coefficient of x1 in row 0) when c1 3, so z1* 3 92 712. Since z1* y*A1, this immediately yields the same allowable range. Figure 6.3 provides graphical insight into why c1 712 is the allowable range. At c1 712, the objective function becomes Z 7.5x1 5x2 2.5(3x1 2x2), so the optimal objective line will lie on top of the constraint boundary line 3x1 2x2 18 shown in the figure. Thus, at this endpoint of the allowable range, we have multiple optimal solutions consisting of the line segment between (0, 9) and (4, 3). If c1 were to be increased any further (c1 712 ), only (4, 3) would be optimal. Consequently, we need c1 712 for (0, 9) to remain optimal. For any nonbasic decision variable xj , the value of z*j cj sometimes is referred to as the reduced cost for xj , because it is the minimum amount by which the unit cost of activity j would have to be reduced to make it worthwhile to undertake activity j (increase xj from zero). Interpreting cj as the unit profit of activity j (so reducing the unit cost increases cj by the same amount), the value of z*j cj thereby is the maximum allowable increase in cj to keep the current BF solution optimal. The sensitivity analysis information generated by linear programming software packages normally includes both the reduced cost and the allowable range to stay optimal for each coefficient in the objective function (along with the types of information displayed in Table 6.22). This was illustrated in Figs. 4.10, 4.12, and 4.13 for the Excel Solver and LINDO. Table 6.23 displays this information in a typical form for our current model (Variation 2 of the Wyndor Glass Co. model). The last three columns are used to calculate the allowable range to stay optimal for each coefficient, so these allowable ranges are c1 3 4.5 7.5, c2 5 3 2. As was discussed in Sec. 4.7, if any of the allowable increases or decreases had turned out to be zero, this would have been a signpost that the optimal solution given in the table is only one of multiple optimal solutions. In this case, changing the corresponding coefficient a tiny amount beyond the zero allowed and re-solving would provide another optimal CPF solution for the original model. Thus far, we have described how to calculate the type of information in Table 6.23 for only nonbasic variables. For a basic variable like x2, the reduced cost automatically is 0. We will discuss how to obtain the allowable range to stay optimal for cj when xj is a basic variable under Case 3.

TABLE 6.23 Typical software output for sensitivity analysis of the objective function coefficients for Variation 2 of the Wyndor Glass Co. model Variable

Value

Reduced Cost

Current Coefficient

Allowable Increase

Allowable Decrease

x1 x2

0 9

4.5 0.0

3 5

4.5

3

6.7 APPLYING SENSITIVITY ANALYSIS

273

Analyzing Simultaneous Changes in Objective Function Coefficients. Regardless of whether xj is a basic or nonbasic variable, the allowable range to stay optimal for cj is valid only if this objective function coefficient is the only one being changed. However, when simultaneous changes are made in the coefficients of the objective function, a 100 percent rule is available for checking whether the original solution must still be optimal. Much like the 100 percent rule for simultaneous changes in right-hand sides, this 100 percent rule combines the allowable changes (increase or decrease) for the individual cj that are given by the last two columns of a table like Table 6.23, as described below. The 100 Percent Rule for Simultaneous Changes in Objective Function Coefficients: If simultaneous changes are made in the coefficients of the objective function, calculate for each change the percentage of the allowable change (increase or decrease) for that coefficient to remain within its allowable range to stay optimal. If the sum of the percentage changes does not exceed 100 percent, the original optimal solution definitely will still be optimal. (If the sum does exceed 100 percent, then we cannot be sure.) Using Table 6.23 (and referring to Fig. 6.3 for visualization), this 100 percent rule says that (0, 9) will remain optimal for Variation 2 of the Wyndor Glass Co. model even if we simultaneously increase c1 from 3 and decrease c2 from 5 as long as these changes are not too large. For example, if c1 is increased by 1.5 (3313 percent of the allowable change), then c2 can be decreased by as much as 2 (6623 percent of the allowable change). Similarly, if c1 is increased by 3 (6623 percent of the allowable change), then c2 can only be decreased by as much as 1 (3313 percent of the allowable change). These maximum changes revise the objective function to either Z 4.5x1 3x2 or Z 6x1 4x2, which causes the optimal objective function line in Fig. 6.3 to rotate clockwise until it coincides with the constraint boundary equation 3x1 2x2 18. In general, when objective function coefficients change in the same direction, it is possible for the percentages of allowable changes to sum to more than 100 percent without changing the optimal solution. We will give an example at the end of the discussion of Case 3. Case 2b—Introduction of a New Variable After solving for the optimal solution, we may discover that the linear programming formulation did not consider all the attractive alternative activities. Considering a new activity requires introducing a new variable with the appropriate coefficients into the objective function and constraints of the current model—which is Case 2b. The convenient way to deal with this case is to treat it just as if it were Case 2a! This is done by pretending that the new variable xj actually was in the original model with all its coefficients equal to zero (so that they still are zero in the final simplex tableau) and that xj is a nonbasic variable in the current BF solution. Therefore, if we change these zero coefficients to their actual values for the new variable, the procedure (including any reoptimization) does indeed become identical to that for Case 2a. In particular, all you have to do to check whether the current solution still is optimal is to check whether the complementary basic solution y* satisfies the one new

274

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

dual constraint that corresponds to the new variable in the primal problem. We already have described this approach and then illustrated it for the Wyndor Glass Co. problem in Sec. 6.5. Case 3—Changes in the Coefficients of a Basic Variable Now suppose that the variable xj (fixed j) under consideration is a basic variable in the optimal solution shown by the final simplex tableau. Case 3 assumes that the only changes in the current model are made to the coefficients of this variable. Case 3 differs from Case 2a because of the requirement that a simplex tableau be in proper form from Gaussian elimination. This requirement allows the column for a nonbasic variable to be anything, so it does not affect Case 2a. However, for Case 3, the basic variable xj must have a coefficient of 1 in its row of the simplex tableau and a coefficient of 0 in every other row (including row 0). Therefore, after the changes in the xj column of the final simplex tableau have been calculated,1 it probably will be necessary to apply Gaussian elimination to restore this form, as illustrated in Table 6.20. In turn, this step probably will change the value of the current basic solution and may make it either infeasible or nonoptimal (so reoptimization may be needed). Consequently, all the steps of the overall procedure summarized at the end of Sec. 6.6 are required for Case 3. Before Gaussian elimination is applied, the formulas for revising the xj column are the same as for Case 2a, as summarized below. Coefficient of xj in final row 0: Coefficient of xj in final rows 1 to m:

z j* cj y*A j cj. A*j S*A j.

Example (Variation 5 of the Wyndor Model). Because x2 is a basic variable in Table 6.21 for Variation 2 of the Wyndor Glass Co. model, sensitivity analysis of its coefficients fits Case 3. Given the current optimal solution (x1 0, x2 9), product 2 is the only new product that should be introduced, and its production rate should be relatively large. Therefore, the key question now is whether the initial estimates that led to the coefficients of x2 in the current model (Variation 2) could have overestimated the attractiveness of product 2 so much as to invalidate this conclusion. This question can be tested by checking the most pessimistic set of reasonable estimates for these coefficients, which turns out to be c2 3, a22 3, and a32 4. Consequently, the changes to be investigated (Variation 5 of the Wyndor model) are c2 5 → c2 3,

0 A2 2 → A 2 2

0 3 . 4

The graphical effect of these changes is that the feasible region changes from the one shown in Fig. 6.3 to the one in Fig. 6.6. The optimal solution in Fig. 6.3 is (x1, x2) (0, 9), which is the corner-point solution lying at the intersection of the x1 0 and 3x1 2x2 18 constraint boundaries. With the revision of the constraints, the corre1

For the relatively sophisticated reader, we should point out a possible pitfall for Case 3 that would be discovered at this point. Specifically, the changes in the initial tableau can destroy the linear independence of the columns of coefficients of basic variables. This event occurs only if the unit coefficient of the basic variable xj in the final tableau has been changed to zero at this point, in which case more extensive simplex method calculations must be used for Case 3.

6.7 APPLYING SENSITIVITY ANALYSIS

275

sponding corner-point solution in Fig. 6.6 is (0, 92 ). However, this solution no longer is optimal, because the revised objective function of Z 3x1 3x2 now yields a new optimal solution of (x1, x2) (4, 32 ). Analysis of Variation 5. Now let us see how we draw these same conclusions algebraically. Because the only changes in the model are in the coefficients of x2, the only resulting changes in the final simplex tableau (Table 6.21) are in the x2 column. Therefore, the above formulas are used to recompute just this column. 0 z2 c2 y*A 2 c2 [0, 0, 52] 3 3 7. 4 1 A2* S*A 2 0 0

FIGURE 6.6 Feasible region for Variation 5 of the Wyndor model where Variation 2 (Fig. 6.3) has been revised so c2 5 3, a22 2 3, and a32 2 4.

0 0 1

x2 x1 0

2x2 24

12

10 x1 4

(0, 9)

3x2 24

8

6

(0, 92)

3x1 2x2 18

4

3x1 4x2 18 3 (4, 2 )optimal

2 Feasible region

x2 0 0

2

4

6

8

x1

0 0 0 1 3 2 . 2 1 4 1

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6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.24 Sensitivity analysis procedure applied to Variation 5 of the Wyndor Glass Co. model Coefficient of:

Revised final tableau

Converted to proper form

New final tableau after reoptimization (only one iteration of the simplex method needed in this case)

Basic Variable

Eq.

Z

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4

(3)

0

x1 9 2 1 3 2 3 3 4 1 3 4 9 4

x2

x3

x4

7

0

0

0

1

0

2

0

0

1

0

1

Z

(0)

1

x3

(1)

0

0

0

0

0

1

0

x2

(2)

0

1

0

0

x4

(3)

0

0

0

1

Z

(0)

1

0

0

x1

(1)

0

1

0

x2

(2)

0

0

1

x4

(3)

0

0

0

3 4 1 3 4 9 4

0 0 0 1

x5 5 2 0 1 2

Right Side 45 4 9

1

6

3 4 0 1 4 3 4

27 2 4 9 2 21 2

3 4 0 1 4 3 4

33 2 4 3 2 39 2

(Equivalently, incremental analysis with c2 2, a22 1, and a32 2 can be used in the same way to obtain this column.) The resulting revised final tableau is shown at the top of Table 6.24. Note that the new coefficients of the basic variable x2 do not have the required values, so the conversion to proper form from Gaussian elimination must be applied next. This step involves dividing row 2 by 2, subtracting 7 times the new row 2 from row 0, and adding the new row 2 to row 3. The resulting second tableau in Table 6.24 gives the new value of the current basic solution, namely, x3 4, x2 92, x4 221 (x1 0, x5 0). Since all these variables are nonnegative, the solution is still feasible. However, because of the negative coefficient of x1 in row 0, we know that it is no longer optimal. Therefore, the simplex method would be applied to this tableau, with this solution as the initial BF solution, to find the new optimal solution. The initial entering basic variable is x1, with x3 as the leaving basic variable. Just one iteration is needed in this case to reach the new optimal solution x1 4, x2 32, x4 329 (x3 0, x5 0), as shown in the last tableau of Table 6.24. All this analysis suggests that c2, a22, and a32 are relatively sensitive parameters. However, additional data for estimating them more closely can be obtained only by conducting a pilot run. Therefore, the OR team recommends that production of product 2 be ini-

6.7 APPLYING SENSITIVITY ANALYSIS

277

tiated immediately on a small scale (x2 32) and that this experience be used to guide the decision on whether the remaining production capacity should be allocated to product 2 or product 1. The Allowable Range to Stay Optimal. For Case 2a, we described how to find the allowable range to stay optimal for any cj such that xj is a nonbasic variable for the current optimal solution (before cj is changed). When xj is a basic variable instead, the procedure is somewhat more involved because of the need to convert to proper form from Gaussian elimination before testing for optimality. To illustrate the procedure, consider Variation 5 of the Wyndor Glass Co. model (with c2 3, a22 3, a23 4) that is graphed in Fig. 6.6 and solved in Table 6.24. Since x2 is a basic variable for the optimal solution (with c2 3) given at the bottom of this table, the steps needed to find the allowable range to stay optimal for c2 are the following: 1. Since x2 is a basic variable, note that its coefficient in the new final row 0 (see the bottom tableau in Table 6.24) is automatically z2* c2 0 before c2 is changed from its current value of 3. 2. Now increment c2 3 by c2 (so c2 3 c2). This changes the coefficient noted in step 1 to z2* c2 c2, which changes row 0 to

3 3 33 Row 0 0, c2, , 0, . 4 4 2 3. With this coefficient now not zero, we must perform elementary row operations to restore proper form from Gaussian elimination. In particular, add to row 0 the product, c2 times row 2, to obtain the new row 0, as shown below.

0, c , 34, c 0, 34 c 3 1 0, c , c , 0, c 4 4 2

2

2

2

2

2

33 2 3 c2 2

3 3 3 1 33 3 New row 0 0, 0, c2, 0, c2 c2 4 4 4 4 2 2

4. Using this new row 0, solve for the range of values of c2 that keeps the coefficients of the nonbasic variables (x3 and x5) nonnegative. 3 3 c2 0 4 4 3 1 c2 0 4 4

3 3 ⇒ c2 4 4 1 3 ⇒ c2 4 4

⇒ c2 1. ⇒ c2 3.

Thus, the range of values is 3 c2 1. 5. Since c2 3 c2, add 3 to this range of values, which yields 0 c2 4 as the allowable range to stay optimal for c2.

278

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

With just two decision variables, this allowable range can be verified graphically by using Fig. 6.6 with an objective function of Z 3x1 c2 x2. With the current value of c2 3, the optimal solution is (4, 23). When c2 is increased, this solution remains optimal only for c2 4. For c2 4, (0, 29) becomes optimal (with a tie at c2 4), because of the constraint boundary 3x1 4x2 18. When c2 is decreased instead, (4, 23) remains optimal only for c2 0. For c2 0, (4, 0) becomes optimal because of the constraint boundary x1 4. In a similar manner, the allowable range to stay optimal for c1 (with c2 fixed at 3) can be derived either algebraically or graphically to be c1 94. (Problem 6.7-13 asks you to verify this both ways.) Thus, the allowable decrease for c1 from its current value of 3 is only 34. However, it is possible to decrease c1 by a larger amount without changing the optimal solution if c2 also decreases sufficiently. For example, suppose that both c1 and c2 are decreased by 1 from their current value of 3, so that the objective function changes from Z 3x1 3x2 to Z 2x1 2x2. According to the 100 percent rule for simultaneous changes in objective function coefficients, the percentages of allowable changes are 13313 percent and 3313 percent, respectively, which sum to far over 100 percent. However, the slope of the objective function line has not changed at all, so (4, 32) still is optimal. Case 4—Introduction of a New Constraint In this case, a new constraint must be introduced to the model after it has already been solved. This case may occur because the constraint was overlooked initially or because new considerations have arisen since the model was formulated. Another possibility is that the constraint was deleted purposely to decrease computational effort because it appeared to be less restrictive than other constraints already in the model, but now this impression needs to be checked with the optimal solution actually obtained. To see if the current optimal solution would be affected by a new constraint, all you have to do is to check directly whether the optimal solution satisfies the constraint. If it does, then it would still be the best feasible solution (i.e., the optimal solution), even if the constraint were added to the model. The reason is that a new constraint can only eliminate some previously feasible solutions without adding any new ones. If the new constraint does eliminate the current optimal solution, and if you want to find the new solution, then introduce this constraint into the final simplex tableau (as an additional row) just as if this were the initial tableau, where the usual additional variable (slack variable or artificial variable) is designated to be the basic variable for this new row. Because the new row probably will have nonzero coefficients for some of the other basic variables, the conversion to proper form from Gaussian elimination is applied next, and then the reoptimization step is applied in the usual way. Just as for some of the preceding cases, this procedure for Case 4 is a streamlined version of the general procedure summarized at the end of Sec. 6.6. The only question to be addressed for this case is whether the previously optimal solution still is feasible, so step 5 (optimality test) has been deleted. Step 4 (feasibility test) has been replaced by a much quicker test of feasibility (does the previously optimal solution satisfy the new constraint?) to be performed right after step 1 (revision of model). It is only if this test provides a negative answer, and you wish to reoptimize, that steps 2, 3, and 6 are used (revision of final tableau, conversion to proper form from Gaussian elimination, and reoptimization).

6.7 APPLYING SENSITIVITY ANALYSIS

279

Example (Variation 6 of the Wyndor Model). To illustrate this case, we consider Variation 6 of the Wyndor Glass Co. model, which simply introduces the new constraint 2x1 3x2 24 into the Variation 2 model given in Table 6.21. The graphical effect is shown in Fig. 6.7. The previous optimal solution (0, 9) violates the new constraint, so the optimal solution changes to (0, 8). To analyze this example algebraically, note that (0, 9) yields 2x1 3x2 27 24, so this previous optimal solution is no longer feasible. To find the new optimal solution, add the new constraint to the current final simplex tableau as just described, with the slack variable x6 as its initial basic variable. This step yields the first tableau shown in Table 6.25. The conversion to proper form from Gaussian elimination then requires subtracting from the new row the product, 3 times row 2, which identifies the current basic solution x3 4, x2 9, x4 6, x6 3 (x1 0, x5 0), as shown in the second tableau. Applying the dual simplex method (described in Sec. 7.1) to this tableau then leads in just one iteration (more are sometimes needed) to the new optimal solution in the last tableau of Table 6.25.

FIGURE 6.7 Feasible region for Variation 6 of the Wyndor model where Variation 2 (Fig. 6.3) has been revised by adding the new constraint, 2x1 3x2 24.

x2

x1 0

14

2x2 24

12

10 x1 4

(0, 9) 8

(0, 8) optimal

6

2x1 3x2 24

4

2

Feasible region 3x1 2x2 18

0

2

4

6

8

x2 0 10

12

14

x1

280

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

TABLE 6.25 Sensitivity analysis procedure applied to Variation 6 of the Wyndor Glass Co. model Coefficient of:

Revised final tableau

Converted to proper form

New final tableau after reoptimization (only one iteration of dual simplex method needed in this case)

Basic Variable

Eq.

Z

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4 x6

(3) New

0 0

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4

(3)

0

x6

New

0

Z

(0)

1

x3

(1)

0

x2

(2)

0

x4

(3)

0

x5

New

0

x1 9 2 1 3 2 3 2 9 2 1 3 2 3 5 2 1 3 1 2 3 4 3 5 3

x2

x3

x4

0

0

0

0

1

0

x5 5 2 0 1 2 1 0

1

0

0

0 3

0 0

1 0

0

0

0

0

1

0

1

0

0

0

0

1

0

0

0

0

0

0

0

0

1

0

0

1

0

0

0

0

0

1

0

0

0

0

1

5 2 0 1 2 1 3 2

x6

Right Side

0

45

0

4

0

9

0 1

6 24

0

45

0

4

0

9

0

6

1

3

5 3 0 1 3 2 3 2 3

40 4 8 8 2

Systematic Sensitivity Analysis—Parametric Programming So far we have described how to test specific changes in the model parameters. Another common approach to sensitivity analysis is to vary one or more parameters continuously over some interval(s) to see when the optimal solution changes. For example, with Variation 2 of the Wyndor Glass Co. model, rather than beginning by testing the specific change from b2 12 to b2 24, we might instead set b2 12 and then vary continuously from 0 to 12 (the maximum value of interest). The geometric interpretation in Fig. 6.3 is that the 2x2 12 constraint line is being shifted upward to 2x2 12 , with being increased from 0 to 12. The result is that the original optimal CPF solution (2, 6) shifts up the 3x1 2x2 18 constraint line toward (2, 12). This corner-point solution remains optimal as long as it is still feasible (x1 0), after which (0, 9) becomes the optimal solution. The algebraic calculations of the effect of having b2 are directly analogous to those for the Case 1 example where b2 12. In particular, we use the expressions for Z* and b* given for Case 1,

6.7 APPLYING SENSITIVITY ANALYSIS

281

Z* y*b b* S*b where b now is 4 12 b 18 and where y* and S* are given in the boxes in the middle tableau in Table 6.19. These equations indicate that the optimal solution is 3 Z* 36 2 1 x3 2 3 1 x2 6 2 1 x1 2 3

(x4 0, x5 0)

for small enough that this solution still is feasible, i.e., for 6. For 6, the dual simplex method (described in Sec. 7.1) yields the tableau shown in Table 6.21 except for the value of x4. Thus, Z 45, x3 4, x2 9 (along with x1 0, x5 0), and the expression for b* yields x4 b3* 0(4) 1(12 ) 1(18) 6 . This information can then be used (along with other data not incorporated into the model on the effect of increasing b2) to decide whether to retain the original optimal solution and, if not, how much to increase b2. In a similar way, we can investigate the effect on the optimal solution of varying several parameters simultaneously. When we vary just the bi parameters, we express the new value bi in terms of the original value bi as follows: bi bi i,

for i 1, 2, . . . , m,

where the i values are input constants specifying the desired rate of increase (positive or negative) of the corresponding right-hand side as is increased. For example, suppose that it is possible to shift some of the production of a current Wyndor Glass Co. product from Plant 2 to Plant 3, thereby increasing b2 by decreasing b3. Also suppose that b3 decreases twice as fast as b2 increases. Then b2 12 b3 18 2, where the (nonnegative) value of measures the amount of production shifted. (Thus, 1 0, 2 1, and 3 2 in this case.) In Fig. 6.3, the geometric interpretation is that as is increased from 0, the 2x2 12 constraint line is being pushed up to 2x2 12 (ignore the 2x2 24 line) and simultaneously the 3x1 2x2 18 constraint line is being

282

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

pushed down to 3x1 2x2 18 2. The original optimal CPF solution (2, 6) lies at the intersection of the 2x2 12 and 3x1 2x2 18 lines, so shifting these lines causes this corner-point solution to shift. However, with the objective function of Z 3x1 5x2, this corner-point solution will remain optimal as long as it is still feasible (x1 0). An algebraic investigation of simultaneously changing b2 and b3 in this way again involves using the formulas for Case 1 (treating as representing an unknown number) to calculate the resulting changes in the final tableau (middle of Table 6.19), namely, 4 [0, , 1] 12 36 12, Z* y*b 18 2 3 2

1 b* S*b 0 0

1 3 1 2 1 3

2 13 4 0 12 6 12 . 1 2 3 18 2

Therefore, the optimal solution becomes 1 Z* 36 2 x3 2

(x4 0,

1 x2 6 2 x1 2

x5 0)

for small enough that this solution still is feasible, i.e., for 2. (Check this conclusion in Fig. 6.3.) However, the fact that Z decreases as increases from 0 indicates that the best choice for is 0, so none of the possible shifting of production should be done. The approach to varying several cj parameters simultaneously is similar. In this case, we express the new value cj in terms of the original value of cj as cj cj j,

for j 1, 2, . . . , n,

where the j are input constants specifying the desired rate of increase (positive or negative) of cj as is increased. To illustrate this case, reconsider the sensitivity analysis of c1 and c2 for the Wyndor Glass Co. problem that was performed earlier in this section. Starting with Variation 2 of the Wyndor model presented in Table 6.21 and Fig. 6.3, we separately considered the effect of changing c1 from 3 to 4 (its most optimistic estimate) and c2 from 5 to 3 (its most pessimistic estimate). Now we can simultaneously consider both changes, as well as various intermediate cases with smaller changes, by setting c1 3

and

c2 5 2,

where the value of measures the fraction of the maximum possible change that is made. The result is to replace the original objective function Z 3x1 5x2 by a function of Z() (3 )x1 (5 2)x2,

6.7 APPLYING SENSITIVITY ANALYSIS

283

so the optimization now can be performed for any desired (fixed) value of between 0 and 1. By checking the effect as increases from 0 to 1, we can determine just when and how the optimal solution changes as the error in the original estimates of these parameters increases. Considering these changes simultaneously is especially appropriate if there are factors that cause the parameters to change together. Are the two products competitive in some sense, so that a larger-than-expected unit profit for one implies a smaller-thanexpected unit profit for the other? Are they both affected by some exogenous factor, such as the advertising emphasis of a competitor? Is it possible to simultaneously change both unit profits through appropriate shifting of personnel and equipment? In the feasible region shown in Fig. 6.3, the geometric interpretation of changing the objective function from Z 3x1 5x2 to Z() (3 )x1 (5 2)x2 is that we are changing the slope of the original objective function line (Z 45 3x1 5x2) that passes through the optimal solution (0, 9). If is increased enough, this slope will change sufficiently that the optimal solution will switch from (0, 9) to another CPF solution (4, 3). (Check graphically whether this occurs for 1.) The algebraic procedure for dealing simultaneously with these two changes ( c1 and c2 2) is shown in Table 6.26. Although the changes now are expressed in terms of rather than specific numerical amounts, is treated just as an unknown number. The table displays just the relevant rows of the tableaux involved (row 0 and the row for the basic variable x2). The first tableau shown is just the final tableau for the current version of the model (before c1 and c2 are changed) as given in Table 6.21. Refer to the formulas in Table 6.17. The only changes in the revised final tableau shown next are that c1 and c2 are subtracted from the row 0 coefficients of x1 and x2, respectively. To convert this tableau to proper form from Gaussian elimination, we subtract 2 times row 2 from row 0, which yields the last tableau shown. The expressions in terms of for the coeffiTABLE 6.26 Dealing with c1 and c2 2 for Variation 2 of the Wyndor model as given in Table 6.21 Coefficient of: Basic Variable

Eq.

Z

Z

(0)

1

x2

(2)

0

Z()

(0)

1

x2

(2)

0

Z()

(0)

1

x2

(2)

0

Final tableau

Revised final tableau when c1 and c2 2

Converted to proper form

x1 9 2 3 2 9 2 3 2 9 4 2 3 2

x2

x3

x4

0

0

0

1

0

0

2

0

0

1

0

0

0

0

0

1

0

0

x5 5 2 1 2 5 2 1 2 5 2 1 2

Right Side 45 9

45 9 45 18 9

284

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

cients of nonbasic variables x1 and x5 in row 0 of this tableau show that the current BF solution remains optimal for 98. Because 1 is the maximum realistic value of , this indicates that c1 and c2 together are insensitive parameters with respect to the Variation 2 model in Table 6.21. There is no need to try to estimate these parameters more closely unless other parameters change (as occurred for Variation 5 of the Wyndor model). As we discussed in Sec. 4.7, this way of continuously varying several parameters simultaneously is referred to as parametric linear programming. Section 7.2 presents the complete parametric linear programming procedure (including identifying new optimal solutions for larger values of ) when just the cj parameters are being varied and then when just the bi parameters are being varied. Some linear programming software packages also include routines for varying just the coefficients of a single variable or just the parameters of a single constraint. In addition to the other applications discussed in Sec. 4.7, these procedures provide a convenient way of conducting sensitivity analysis systematically.

6.8

CONCLUSIONS Every linear programming problem has associated with it a dual linear programming problem. There are a number of very useful relationships between the original (primal) problem and its dual problem that enhance our ability to analyze the primal problem. For example, the economic interpretation of the dual problem gives shadow prices that measure the marginal value of the resources in the primal problem and provides an interpretation of the simplex method. Because the simplex method can be applied directly to either problem in order to solve both of them simultaneously, considerable computational effort sometimes can be saved by dealing directly with the dual problem. Duality theory, including the dual simplex method for working with superoptimal basic solutions, also plays a major role in sensitivity analysis. The values used for the parameters of a linear programming model generally are just estimates. Therefore, sensitivity analysis needs to be performed to investigate what happens if these estimates are wrong. The fundamental insight of Sec. 5.3 provides the key to performing this investigation efficiently. The general objectives are to identify the sensitive parameters that affect the optimal solution, to try to estimate these sensitive parameters more closely, and then to select a solution that remains good over the range of likely values of the sensitive parameters. This analysis is a very important part of most linear programming studies.

SELECTED REFERENCES 1. Bazaraa, M. S., J. J. Jarvis, and H. D. Sherali: Linear Programming and Network Flows, 2d ed., Wiley, New York, 1990. 2. Dantzig, G. B., and M. N. Thapa: Linear Programming 1: Introduction, Springer, New York, 1997. 3. Hillier, F. S., M. S. Hillier, and G. J. Lieberman: Introduction to Management Science: A Modeling and Case Studies Approach with Spreadsheets, Irwin/McGraw-Hill, Burr Ridge, IL, 2000, chap. 4. 4. Vanderbei, R. J.: Linear Programming: Foundations and Extensions, Kluwer Academic Publishers, Boston, MA, 1996.

CHAPTER 6 PROBLEMS

285

LEARNING AIDS FOR THIS CHAPTER IN YOUR OR COURSEWARE A Demonstration Example in OR Tutor: Sensitivity Analysis

Interactive Routines: Enter or Revise a General Linear Programming Model Solve Interactively by the Simplex Method Sensitivity Analysis

An Excel Add-In: Premium Solver

Files (Chapter 3) for Solving the Wyndor Example: Excel File LINGO/LINDO File MPL/CPLEX File

See Appendix 1 for documentation of the software.

PROBLEMS The symbols to the left of some of the problems (or their parts) have the following meaning: D: The demonstration example listed above may be helpful. I: We suggest that you use the corresponding interactive routine listed above (the printout records your work). C: Use the computer with any of the software options available to you (or as instructed by your instructor) to solve the problem automatically. An asterisk on the problem number indicates that at least a partial answer is given in the back of the book. 6.1-1. Construct the primal-dual table and the dual problem for each of the following linear programming models fitting our standard form. (a) Model in Prob. 4.1-6 (b) Model in Prob. 4.7-8 6.1-2.* Construct the dual problem for each of the following linear programming models fitting our standard form. (a) Model in Prob. 3.1-5 (b) Model in Prob. 4.7-6

6.1-3. Consider the linear programming model in Prob. 4.5-4. (a) Construct the primal-dual table and the dual problem for this model. (b) What does the fact that Z is unbounded for this model imply about its dual problem? 6.1-4. For each of the following linear programming models, give your recommendation on which is the more efficient way (probably) to obtain an optimal solution: by applying the simplex method directly to this primal problem or by applying the simplex method directly to the dual problem instead. Explain. (a) Maximize Z 10x1 4x2 7x3, subject to 3x1 x1 5x1 x1 2x1

x2 2x2 x2 x2 x2

2x3 3x3 2x3 x3 x3

25 25 40 90 20

and x1 0,

x2 0,

x3 0.

286

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Z 2x1 5x2 3x3 4x4 x5,

(b) Maximize

Maximize

subject to x1 3x2 2x3 3x4 x5 6 4x1 6x2 5x3 7x4 x5 15 and xj 0,

for j 1, 2, 3, 4, 5.

6.1-5. Consider the following problem. Maximize

Z x1 2x2 x3,

subject to x1 x2 2x3 12 x1 x2 x3 1 and x1 0,

x2 0,

x3 0.

(a) Construct the dual problem. (b) Use duality theory to show that the optimal solution for the primal problem has Z 0. 6.1-6. Consider the following problem. Maximize

6.1-8. Consider the following problem.

Z 2x1 6x2 9x3,

subject to x1x1 x3 3 x1x2 2x3 5

(resource 1) (resource 2)

Z x1 2x2,

subject to x1 x2 2 4x1 x2 4 and x1 0,

x2 0.

(a) Demonstrate graphically that this problem has no feasible solutions. (b) Construct the dual problem. (c) Demonstrate graphically that the dual problem has an unbounded objective function. 6.1-9. Construct and graph a primal problem with two decision variables and two functional constraints that has feasible solutions and an unbounded objective function. Then construct the dual problem and demonstrate graphically that it has no feasible solutions. 6.1-10. Construct a pair of primal and dual problems, each with two decision variables and two functional constraints, such that both problems have no feasible solutions. Demonstrate this property graphically. 6.1-11. Construct a pair of primal and dual problems, each with two decision variables and two functional constraints, such that the primal problem has no feasible solutions and the dual problem has an unbounded objective function.

and x1 0,

x2 0,

x3 0.

(a) Construct the dual problem for this primal problem. (b) Solve the dual problem graphically. Use this solution to identify the shadow prices for the resources in the primal problem. C (c) Confirm your results from part (b) by solving the primal problem automatically by the simplex method and then identifying the shadow prices. 6.1-7. Follow the instructions of Prob. 6.1-6 for the following problem. Maximize

Z x1 3x2 2x3,

subject to 2x1 2x2 2x3 6 2x1 x2 2x3 4

(resource 1) (resource 2)

and x1 0,

x2 0,

x3 0.

6.1-12. Use the weak duality property to prove that if both the primal and the dual problem have feasible solutions, then both must have an optimal solution. 6.1-13. Consider the primal and dual problems in our standard form presented in matrix notation at the beginning of Sec. 6.1. Use only this definition of the dual problem for a primal problem in this form to prove each of the following results. (a) The weak duality property presented in Sec. 6.1. (b) If the primal problem has an unbounded feasible region that permits increasing Z indefinitely, then the dual problem has no feasible solutions. 6.1-14. Consider the primal and dual problems in our standard form presented in matrix notation at the beginning of Sec. 6.1. Let y* denote the optimal solution for this dual problem. Suppose that b is then replaced by b . Let x denote the optimal solution for the new primal problem. Prove that cx y*b .

CHAPTER 6 PROBLEMS

287

6.1-15. For any linear programming problem in our standard form and its dual problem, label each of the following statements as true or false and then justify your answer. (a) The sum of the number of functional constraints and the number of variables (before augmenting) is the same for both the primal and the dual problems. (b) At each iteration, the simplex method simultaneously identifies a CPF solution for the primal problem and a CPF solution for the dual problem such that their objective function values are the same. (c) If the primal problem has an unbounded objective function, then the optimal value of the objective function for the dual problem must be zero.

6.3-3. Consider the primal and dual problems for the Wyndor Glass Co. example given in Table 6.1. Using Tables 5.5, 5.6, 6.8, and 6.9, construct a new table showing the eight sets of nonbasic variables for the primal problem in column 1, the corresponding sets of associated variables for the dual problem in column 2, and the set of nonbasic variables for each complementary basic solution in the dual problem in column 3. Explain why this table demonstrates the complementary slackness property for this example.

6.2-1. Consider the simplex tableaux for the Wyndor Glass Co. problem given in Table 4.8. For each tableau, give the economic interpretation of the following items: (a) Each of the coefficients of the slack variables (x3, x4, x5) in row 0 (b) Each of the coefficients of the decision variables (x1, x2) in row 0 (c) The resulting choice for the entering basic variable (or the decision to stop after the final tableau)

6.3-5. Consider the following problem.

6.3-1.* Consider the following problem. Maximize

Z 6x1 8x2,

subject to 5x1 2x2 20 x1 2x2 10 and x1 0,

x2 0.

(a) Construct the dual problem for this primal problem. (b) Solve both the primal problem and the dual problem graphically. Identify the CPF solutions and corner-point infeasible solutions for both problems. Calculate the objective function values for all these solutions. (c) Use the information obtained in part (b) to construct a table listing the complementary basic solutions for these problems. (Use the same column headings as for Table 6.9.) I (d) Work through the simplex method step by step to solve the primal problem. After each iteration (including iteration 0), identify the BF solution for this problem and the complementary basic solution for the dual problem. Also identify the corresponding corner-point solutions. 6.3-2. Consider the model with two functional constraints and two variables given in Prob. 4.1-5. Follow the instructions of Prob. 6.3-1 for this model.

6.3-4. Suppose that a primal problem has a degenerate BF solution (one or more basic variables equal to zero) as its optimal solution. What does this degeneracy imply about the dual problem? Why? Is the converse also true?

Z 2x1 4x2,

Maximize subject to x1 x2 1 and x1 0,

x2 0.

(a) Construct the dual problem, and then find its optimal solution by inspection. (b) Use the complementary slackness property and the optimal solution for the dual problem to find the optimal solution for the primal problem. (c) Suppose that c1, the coefficient of x1 in the primal objective function, actually can have any value in the model. For what values of c1 does the dual problem have no feasible solutions? For these values, what does duality theory then imply about the primal problem? 6.3-6. Consider the following problem. Maximize

Z 2x1 7x2 4x3,

subject to x1 2x2 x3 10 3x1 3x2 2x3 10 and x1 0,

x2 0,

x3 0.

(a) Construct the dual problem for this primal problem. (b) Use the dual problem to demonstrate that the optimal value of Z for the primal problem cannot exceed 25. (c) It has been conjectured that x2 and x3 should be the basic variables for the optimal solution of the primal problem. Directly derive this basic solution (and Z) by using Gaussian elimination. Simultaneously derive and identify the complementary ba-

288

I

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

sic solution for the dual problem by using Eq. (0) for the primal problem. Then draw your conclusions about whether these two basic solutions are optimal for their respective problems. (d) Solve the dual problem graphically. Use this solution to identify the basic variables and the nonbasic variables for the optimal solution of the primal problem. Directly derive this solution, using Gaussian elimination.

6.3-7.* Reconsider the model of Prob. 6.1-4b. (a) Construct its dual problem. (b) Solve this dual problem graphically. (c) Use the result from part (b) to identify the nonbasic variables and basic variables for the optimal BF solution for the primal problem. (d) Use the results from part (c) to obtain the optimal solution for the primal problem directly by using Gaussian elimination to solve for its basic variables, starting from the initial system of equations [excluding Eq. (0)] constructed for the simplex method and setting the nonbasic variables to zero. (e) Use the results from part (c) to identify the defining equations (see Sec. 5.1) for the optimal CPF solution for the primal problem, and then use these equations to find this solution. 6.3-8. Consider the model given in Prob. 5.3-13. (a) Construct the dual problem. (b) Use the given information about the basic variables in the optimal primal solution to identify the nonbasic variables and basic variables for the optimal dual solution. (c) Use the results from part (b) to identify the defining equations (see Sec. 5.1) for the optimal CPF solution for the dual problem, and then use these equations to find this solution. (d) Solve the dual problem graphically to verify your results from part (c). 6.3-9. Consider the model given in Prob. 3.1-4. (a) Construct the dual problem for this model. (b) Use the fact that (x1, x2) (13, 5) is optimal for the primal problem to identify the nonbasic variables and basic variables for the optimal BF solution for the dual problem. (c) Identify this optimal solution for the dual problem by directly deriving Eq. (0) corresponding to the optimal primal solution identified in part (b). Derive this equation by using Gaussian elimination. (d) Use the results from part (b) to identify the defining equations (see Sec. 5.1) for the optimal CPF solution for the dual problem. Verify your optimal dual solution from part (c) by checking to see that it satisfies this system of equations. 6.3-10. Suppose that you also want information about the dual problem when you apply the revised simplex method (see Sec. 5.2) to the primal problem in our standard form.

(a) How would you identify the optimal solution for the dual problem? (b) After obtaining the BF solution at each iteration, how would you identify the complementary basic solution in the dual problem? 6.4-1. Consider the following problem. Maximize

Z x1 x2,

subject to x1 2x2 10 2x1 x2 2 and x2 0

(x1 unconstrained in sign).

(a) Use the SOB method to construct the dual problem. (b) Use Table 6.12 to convert the primal problem to our standard form given at the beginning of Sec. 6.1, and construct the corresponding dual problem. Then show that this dual problem is equivalent to the one obtained in part (a). 6.4-2. Consider the primal and dual problems in our standard form presented in matrix notation at the beginning of Sec. 6.1. Use only this definition of the dual problem for a primal problem in this form to prove each of the following results. (a) If the functional constraints for the primal problem Ax b are changed to Ax b, the only resulting change in the dual problem is to delete the nonnegativity constraints, y 0. (Hint: The constraints Ax b are equivalent to the set of constraints Ax b and Ax b.) (b) If the functional constraints for the primal problem Ax b are changed to Ax b, the only resulting change in the dual problem is that the nonnegativity constraints y 0 are replaced by nonpositivity constraints y 0, where the current dual variables are interpreted as the negative of the original dual variables. (Hint: The constraints Ax b are equivalent to Ax b.) (c) If the nonnegativity constraints for the primal problem x 0 are deleted, the only resulting change in the dual problem is to replace the functional constraints yA c by yA c. (Hint: A variable unconstrained in sign can be replaced by the difference of two nonnegative variables.) 6.4-3.* Construct the dual problem for the linear programming problem given in Prob. 4.6-4. 6.4-4. Consider the following problem. Minimize

Z x1 2x2,

subject to 2x1 x2 1 2x1 2x2 1

CHAPTER 6 PROBLEMS

and x1 0,

289

and x2 0.

(a) Construct the dual problem. (b) Use graphical analysis of the dual problem to determine whether the primal problem has feasible solutions and, if so, whether its objective function is bounded.

x1 0,

x2 0.

(a) Demonstrate graphically that this problem has an unbounded objective function. (b) Construct the dual problem. (c) Demonstrate graphically that the dual problem has no feasible solutions.

6.4-5. Consider the two versions of the dual problem for the radiation therapy example that are given in Tables 6.15 and 6.16. Review in Sec. 6.4 the general discussion of why these two versions are completely equivalent. Then fill in the details to verify this equivalency by proceeding step by step to convert the version in Table 6.15 to equivalent forms until the version in Table 6.16 is obtained.

6.5-1. Consider the model of Prob. 6.7-1. Use duality theory directly to determine whether the current basic solution remains optimal after each of the following independent changes. (a) The change in part (e) of Prob. 6.7-1 (b) The change in part (g) of Prob. 6.7-1

6.4-6. For each of the following linear programming models, use the SOB method to construct its dual problem. (a) Model in Prob. 4.6-3 (b) Model in Prob. 4.6-8 (c) Model in Prob. 4.6-18

6.5-2. Consider the model of Prob. 6.7-3. Use duality theory directly to determine whether the current basic solution remains optimal after each of the following independent changes. (a) The change in part (c) of Prob. 6.7-3 (b) The change in part ( f ) of Prob. 6.7-3

6.4-7. Consider the model with equality constraints given in Prob. 4.6-2. (a) Construct its dual problem. (b) Demonstrate that the answer in part (a) is correct (i.e., equality constraints yield dual variables without nonnegativity constraints) by first converting the primal problem to our standard form (see Table 6.12), then constructing its dual problem, and next converting this dual problem to the form obtained in part (a).

6.5-3. Consider the model of Prob. 6.7-4. Use duality theory directly to determine whether the current basic solution remains optimal after each of the following independent changes. (a) The change in part (b) of Prob. 6.7-4 (b) The change in part (d ) of Prob. 6.7-4

6.4-8.* Consider the model without nonnegativity constraints given in Prob. 4.6-16. (a) Construct its dual problem. (b) Demonstrate that the answer in part (a) is correct (i.e., variables without nonnegativity constraints yield equality constraints in the dual problem) by first converting the primal problem to our standard form (see Table 6.12), then constructing its dual problem, and finally converting this dual problem to the form obtained in part (a). 6.4-9. Consider the dual problem for the Wyndor Glass Co. example given in Table 6.1. Demonstrate that its dual problem is the primal problem given in Table 6.1 by going through the conversion steps given in Table 6.13. 6.4-10. Consider the following problem. Minimize

6.5-4. Reconsider part (d) of Prob. 6.7-6. Use duality theory directly to determine whether the original optimal solution is still optimal. 6.6-1.* Consider the following problem. Maximize subject to 6x1 3x2 5x3 25 3x1 4x2 5x3 20 and x1 0,

x1 2x2 2 x1 x2 4

x2 0,

x3 0.

The corresponding final set of equations yielding the optimal solution is (0) (1)

Z x1 3x2,

subject to

Z 3x1 x2 4x3,

(2)

Z

2x2

1 3 x4 x5 17 5 5

1 x1 x2 3

1 1 5 x4 x5 3 3 3

1 2 x2 x3 x4 x5 3. 5 5

(a) Identify the optimal solution from this set of equations. (b) Construct the dual problem.

290

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

(c) Identify the optimal solution for the dual problem from the final set of equations. Verify this solution by solving the dual problem graphically. (d) Suppose that the original problem is changed to Maximize

subject to 4y1 2y1 y1 y1

subject to

x2 0,

3y2 4 y2 3 2y2 1 y2 2

y1 0,

x3 0.

Use duality theory to determine whether the previous optimal solution is still optimal. (e) Use the fundamental insight presented in Sec. 5.3 to identify the new coefficients of x2 in the final set of equations after it has been adjusted for the changes in the original problem given in part (d ). (f) Now suppose that the only change in the original problem is that a new variable xnew has been introduced into the model as follows: Maximize

and

and x1 0,

W 5y1 4y2,

Minimize

Z 3x1 3x2 4x3,

6x1 2x2 5x3 25 3x1 3x2 5x3 20

6.6-3. Consider the following problem.

D,I

y2 0.

Because this primal problem has more functional constraints than variables, suppose that the simplex method has been applied directly to its dual problem. If we let x5 and x6 denote the slack variables for this dual problem, the resulting final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

x6

Right Side

Z x2 x4

(0) (1) (2)

1 0 0

3 1 2

0 1 0

2 1 3

0 0 1

1 1 1

1 1 2

9 1 3

Z 3x1 x2 4x3 2xnew,

subject to 6x1 3x2 5x3 3xnew 25 3x1 4x2 5x3 2xnew 20 and x1 0,

x2 0,

x3 0,

xnew 0.

Use duality theory to determine whether the previous optimal solution, along with xnew 0, is still optimal. (g) Use the fundamental insight presented in Sec. 5.3 to identify the coefficients of xnew as a nonbasic variable in the final set of equations resulting from the introduction of xnew into the original model as shown in part ( f ). 6.6-2. Reconsider the model of Prob. 6.6-1. You are now to conduct sensitivity analysis by independently investigating each of the following six changes in the original model. For each change, use the sensitivity analysis procedure to revise the given final set of equations (in tableau form) and convert it to proper form from Gaussian elimination. Then test this solution for feasibility and for optimality. (Do not reoptimize.) (a) Change the right-hand side of constraint 1 to b1 15. (b) Change the right-hand side of constraint 2 to b2 5. (c) Change the coefficient of x2 in the objective function to c2 4. (d) Change the coefficient of x3 in the objective function to c3 3. (e) Change the coefficient of x2 in constraint 2 to a22 1. (f) Change the coefficient of x1 in constraint 1 to a11 10. D,I

For each of the following independent changes in the original primal model, you now are to conduct sensitivity analysis by directly investigating the effect on the dual problem and then inferring the complementary effect on the primal problem. For each change, apply the procedure for sensitivity analysis summarized at the end of Sec. 6.6 to the dual problem (do not reoptimize), and then give your conclusions as to whether the current basic solution for the primal problem still is feasible and whether it still is optimal. Then check your conclusions by a direct graphical analysis of the primal problem. (a) Change the objective function to W 3y1 5y2. (b) Change the right-hand sides of the functional constraints to 3, 5, 2, and 3, respectively. (c) Change the first constraint to 2y1 4y2 7. (d) Change the second constraint to 5y1 2y2 10. 6.7-1.* Consider the following problem.

D,I

Maximize

Z 5x1 5x2 13x3,

subject to x1 x2 3x3 20 12x1 4x2 10x3 90 and xj 0

( j 1, 2, 3).

CHAPTER 6 PROBLEMS

If we let x4 and x5 be the slack variables for the respective constraints, the simplex method yields the following final set of equations: (0) (1) (2)

Z

2x3 5x4 100. x1 x2 3x3 x4 20. 16x1 2x3 4x4 x5 = 10.

Now you are to conduct sensitivity analysis by independently investigating each of the following nine changes in the original model. For each change, use the sensitivity analysis procedure to revise this set of equations (in tableau form) and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. (Do not reoptimize.) (a) Change the right-hand side of constraint 1 to b1 30. (b) Change the right-hand side of constraint 2 to b2 70. (c) Change the right-hand sides to

b1 10 . 100 b2

(d) Change the coefficient of x3 in the objective function to c3 8.

291

6.7-2.* Reconsider the model of Prob. 6.7-1. Suppose that we now want to apply parametric linear programming analysis to this problem. Specifically, the right-hand sides of the functional constraints are changed to 20 2

(for constraint 1)

and 90

(for constraint 2),

where can be assigned any positive or negative values. Express the basic solution (and Z) corresponding to the original optimal solution as a function of . Determine the lower and upper bounds on before this solution would become infeasible. 6.7-3. Consider the following problem.

D,I

Z 2x1 x2 x3,

Maximize subject to

3x1 x2 x3 60 x1 x2 2x3 10 x1 x2 x3 20 and x1 0,

x2 0,

x3 0.

Let x4, x5, and x6 denote the slack variables for the respective constraints. After we apply the simplex method, the final simplex tableau is

(e) Change the coefficients of x1 to

c1 2 a11 0 . a21 5 (f) Change the coefficients of x2 to

c2 6 a12 2 . a22 5 (g) Introduce a new variable x6 with coefficients

c6 10 a16 3 . a26 5 (h) Introduce a new constraint 2x1 3x2 5x3 50. (Denote its slack variable by x6.) (i) Change constraint 2 to 10x1 5x2 10x3 100.

Coefficient of: Basic Variable

Eq.

Z

x1

x2

Z

(0)

1

0

0

x4

(1)

0

0

0

x1

(2)

0

1

0

x2

(3)

0

0

1

x3 3 2 1 1 2 3 2

x4 0 1 0 0

x5

x6

3 2 1 1 2 1 2

1 2 2 1 2 1 2

Right Side 25 10 15 5

Now you are to conduct sensitivity analysis by independently investigating each of the following six changes in the original model. For each change, use the sensitivity analysis procedure to revise this final tableau and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. If either test fails, reoptimize to find a new optimal solution.

292

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

(a) Change the right-hand sides

b1 60 b2 10 b3 20

from

to

b1 70 b2 20 . b3 10

form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. If either test fails, reoptimize to find a new optimal solution. (a) Change the right-hand sides to

(b) Change the coefficients of x1

c1 2 a11 3 a 1 21 a31 1

from

b 30. b1

c1 1 a11 2 a 2 . 21 a31 0

to

(b) Change the coefficients of x3 to

c3 2 a13 3 . a23 2

(c) Change the coefficients of x3

c3 1 a13 1 a 2 23 a33 1

from

to

c3 2 a13 3 a 1 . 23 a33 2

(d) Change the objective function to Z 3x1 2x2 3x3. (e) Introduce a new constraint 3x1 2x2 x3 30. (Denote its slack variable by x7.) (f) Introduce a new variable x8 with coefficients

c8 1 a18 2 a 1 . 28 a38 2

Maximize

(c) Change the coefficients of x1 to

c1 4 a11 3 . a21 2 (d) Introduce a new variable x6 with coefficients

c6 3 a16 1 . a26 2 (e) Change the objective function to Z x1 5x2 2x3. (f) Introduce a new constraint 3x1 2x2 3x3 25. (g) Change constraint 2 to x1 2x2 2x3 35.

6.7-4. Consider the following problem.

D,I

Z 2x1 7x2 3x3,

subject to

6.7-5. Reconsider the model of Prob. 6.7-4. Suppose that we now want to apply parametric linear programming analysis to this problem. Specifically, the right-hand sides of the functional constraints are changed to 30 3

x1 3x2 4x3 30 x1 4x2 x3 10

10

x1 0,

x2 0,

(for constraint 1)

and

and x3 0.

By letting x4 and x5 be the slack variables for the respective constraints, the simplex method yields the following final set of equations: (0) (1) (2)

20

2

x2 x3 2x5 20, x2 5x3 x4 x5 20, x1 4x2 x3 x5 10.

Z

Now you are to conduct sensitivity analysis by independently investigating each of the following seven changes in the original model. For each change, use the sensitivity analysis procedure to revise this set of equations (in tableau form) and convert it to proper

(for constraint 2),

where can be assigned any positive or negative values. Express the basic solution (and Z) corresponding to the original optimal solution as a function of . Determine the lower and upper bounds on before this solution would become infeasible. D,I

6.7-6. Consider the following problem. Maximize

Z 2x1 x2 x3,

subject to 3x1 2x2 2x3 15 x1 x2 x3 3 x1 x2 x3 4

CHAPTER 6 PROBLEMS

and x1 0,

x2 0,

x3 0.

If we let x4, x5, and x6 be the slack variables for the respective constraints, the simplex method yields the following final set of equations: (0) (1) (2) (3)

2x3 x4 x5 18, x2 5x3 x4 3x5 24, 2x3 x5 x6 7, x1 4x3 x4 2x5 21.

Z

Now you are to conduct sensitivity analysis by independently investigating each of the following eight changes in the original model. For each change, use the sensitivity analysis procedure to revise this set of equations (in tableau form) and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. If either test fails, reoptimize to find a new optimal solution. (a) Change the right-hand sides to

b1 10 b2 4 . b3 2

of type A, but the vendor providing these subassemblies would only be able to increase its supply rate from the current 2,000 per day to a maximum of 3,000 per day. Each toy requires only one subassembly of type B, but the vendor providing these subassemblies would be unable to increase its supply rate above the current level of 1,000 per day. Because no other vendors currently are available to provide these subassemblies, management is considering initiating a new production process internally that would simultaneously produce an equal number of subassemblies of the two types to supplement the supply from the two vendors. It is estimated that the company’s cost for producing one subassembly of each type would be $2.50 more than the cost of purchasing these subassemblies from the two vendors. Management wants to determine both the production rate of the toy and the production rate of each pair of subassemblies (one A and one B) that would maximize the total profit. The following table summarizes the data for the problem. Resource Usage per Unit of Each Activity

Resource

(b) Change the coefficient of x3 in the objective function to c3 2. (c) Change the coefficient of x1 in the objective function to c1 3. (d) Change the coefficients of x3 to

c3 4 a13 3 a 2 . 23 a33 1 (e) Change the coefficients of x1 and x2 to

c1 1 a11 1 a 2 21 a31 3

293

and

c2 2 a12 2 a 3 , 22 a32 2

respectively. (f) Change the objective function to Z 5x1 x2 3x3. (g) Change constraint 1 to 2x1 x2 4x3 12. (h) Introduce a new constraint 2x1 x2 3x3 60. 6.7-7. One of the products of the G. A. Tanner Company is a special kind of toy that provides an estimated unit profit of $3. Because of a large demand for this toy, management would like to increase its production rate from the current level of 1,000 per day. However, a limited supply of two subassemblies (A and B) from vendors makes this difficult. Each toy requires two subassemblies

Produce Produce Amount of Resource Toys Subassemblies Available

Subassembly A Subassembly B

$2 $1

.00$1 .00$1

Unit profit

$3

$2.50

3,000 1,000

(a) Formulate a linear programming model for this problem and use the graphical method to obtain its optimal solution. C (b) Use a software package based on the simplex method to solve for an optimal solution. C (c) Since the stated unit profits for the two activities are only estimates, management wants to know how much each of these estimates can be off before the optimal solution would change. Begin exploring this question for the first activity (producing toys) by using the same software package to resolve for an optimal solution and total profit as the unit profit for this activity increases in 50-cent increments from $2.00 to $4.00. What conclusion can be drawn about how much the estimate of this unit profit can differ in each direction from its original value of $3.00 before the optimal solution would change? C (d) Repeat part (c) for the second activity (producing subassemblies) by re-solving as the unit profit for this activity increases in 50-cent increments from $3.50 to $1.50 (with the unit profit for the first activity fixed at $3). C (e) Use the same software package to generate the usual output (as in Table 6.23) for sensitivity analysis of the unit profits.

294

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Use this output to obtain the allowable range to stay optimal for each unit profit. (f) Use graphical analysis to verify the allowable ranges obtained in part (e). (g) For each of the 16 combinations of unit profits considered in parts (c) and (d ) where both unit profits differ from their original estimates, use the 100 percent rule for simultaneous changes in objective function coefficients to determine if the original optimal solution must still be optimal. (h) For each of the combinations of unit profits considered in part (g) where it was found that the original optimal solution is not guaranteed to still be optimal, use graphical analysis to determine whether this solution is still optimal. 6.7-8. Reconsider Prob. 6.7-7. After further negotiations with each vendor, management of the G. A. Tanner Co. has learned that either of them would be willing to consider increasing their supply of their respective subassemblies over the previously stated maxima (3,000 subassemblies of type A per day and 1,000 of type B per day) if the company would pay a small premium over the regular price for the extra subassemblies. The size of the premium for each type of subassembly remains to be negotiated. The demand for the toy being produced is sufficiently high that 2,500 per day could be sold if the supply of subassemblies could be increased enough to support this production rate. Assume that the original estimates of unit profits given in Prob. 6.7-7 are accurate. (a) Formulate a linear programming model for this problem with the original maximum supply levels and the additional constraint that no more than 2,500 toys should be produced per day. Then use the graphical method to obtain its optimal solution. C (b) Use a software package based on the simplex method to solve for an optimal solution. C (c) Without considering the premium, use the same software package to determine the shadow price for the subassembly A constraint by solving the model again after increasing the maximum supply by 1. Use this shadow price to determine the maximum premium that the company should be willing to pay for each subassembly of this type. C (d) Repeat part (c) for the subassembly B constraint. C (e) Estimate how much the maximum supply of subassemblies of type A could be increased before the shadow price (and the corresponding premium) found in part (c) would no longer be valid by using the same software package to resolve for an optimal solution and the total profit (excluding the premium) as the maximum supply increases in increments of 100 from 3,000 to 4,000. C (f) Repeat part (e) for subassemblies of type B by re-solving as the maximum supply increases in increments of 100 from 1,000 to 2,000.

(g) Use the same software package to generate the usual output (as in Table 6.23) for sensitivity analysis of the supplies being made available of the subassemblies. Use this output to obtain the allowable range to stay feasible for each subassembly supply. (h) Use graphical analysis to verify the allowable ranges obtained in part (g). (i) For each of the four combinations where the maximum supply of subassembly A is either 3,500 or 4,000 and the maximum supply of subassembly B is either 1,500 or 2,000, use the 100 percent rule for simultaneous changes in right-hand sides to determine whether the original shadow prices definitely will still be valid. (j) For each of the combinations considered in part (i) where it was found that the original shadow prices are not guaranteed to still be valid, use graphical analysis to determine whether these shadow prices actually are still valid for predicting the effect of changing the right-hand sides. C

6.7-9 Consider the Distribution Unlimited Co. problem presented in Sec. 3.4 and summarized in Fig. 3.13. Although Fig. 3.13 gives estimated unit costs for shipping through the various shipping lanes, there actually is some uncertainty about what these unit costs will turn out to be. Therefore, before adopting the optimal solution given at the end of Sec. 3.4, management wants additional information about the effect of inaccuracies in estimating these unit costs. Use a computer package based on the simplex method to generate sensitivity analysis information preparatory to addressing the following questions. (a) Which of the unit shipping costs given in Fig. 3.13 has the smallest margin for error without invalidating the optimal solution given in Sec. 3.4? Where should the greatest effort be placed in estimating the unit shipping costs? (b) What is the allowable range to stay optimal for each of the unit shipping costs? (c) How should these allowable ranges be interpreted to management? (d) If the estimates change for more than one of the unit shipping costs, how can you use the generated sensitivity analysis information to determine whether the optimal solution might change?

C

6.7-10. Consider the Union Airways problem presented in Sec. 3.4, including the data given in Table 3.19. Management is about to begin negotiations on a new contract with the union that represents the company’s customer service agents. This might result in some small changes in the daily costs per agent given in Table 3.19 for the various shifts. Several possible changes listed below are being considered separately. In each case, management would like to know whether the change might

C

CHAPTER 6 PROBLEMS

result in the original optimal solution (given in Sec. 3.4) no longer being optimal. Answer this question in parts (a) to (e) by using a software package based on the simplex method to generate sensitivity analysis information. If the optimal solution might change, use the software package to re-solve for the optimal solution. (a) The daily cost per agent for Shift 2 changes from $160 to $165. (b) The daily cost per agent for Shift 4 changes from $180 to $170. (c) The changes in parts (a) and (b) both occur. (d) The daily cost per agent increases by $4 for shifts 2, 4, and 5, but decreases by $4 for shifts 1 and 3. (e) The daily cost per agent increases by 2 percent for each shift. 6.7-11. Consider the following problem.

6.7-13. Consider Variation 5 of the Wyndor Glass Co. model (see Fig. 6.6 and Table 6.24), where the changes in the parameter values given in Table 6.21 are c2 3, a22 3, and a32 4. Verify both algebraically and graphically that the allowable range to stay optimal for c1 is c1 94. 6.7-14. Consider the following problem. Z 3x1 x2 2x3,

Maximize subject to

x1 x2 2x3 20 2x1 x2 x3 10 and

Z c1x1 c2x2,

Maximize

295

x1 0,

subject to

x2 0,

x3 0.

Let x4 and x5 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is

2x1 x2 b1 x1 x2 b2 and x1 0,

x2 0.

Coefficient of:

Let x3 and x4 denote the slack variables for the respective functional constraints. When c1 3, c2 2, b1 30, and b2 10, the simplex method yields the following final simplex tableau.

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Right Side

Z x2 x1

(0) (1) (2)

1 0 0

0 0 1

0 1 0

1 1 1

1 2 1

40 10 20

(a) Use graphical analysis to determine the allowable range to stay optimal for c1 and c2. (b) Use algebraic analysis to derive and verify your answers in part (a). (c) Use graphical analysis to determine the allowable range to stay feasible for b1 and b2. (d) Use algebraic analysis to derive and verify your answers in part (c) C (e) Use a software package based on the simplex method to find these allowable ranges. 6.7-12. Consider Variation 5 of the Wyndor Glass Co. model (see Fig. 6.6 and Table 6.24), where the changes in the parameter values given in Table 6.21 are c2 3, a22 3, and a32 4. Use the formula b* S*b to find the allowable range to stay feasible for each bi. Then interpret each allowable range graphically.

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x3 x2

(0) (1) (2)

1 0 0

8 3 5

0 0 1

0 1 0

3 1 1

4 1 2

100 30 40

(a) Perform sensitivity analysis to determine which of the 11 parameters of the model are sensitive parameters in the sense that any change in just that parameter’s value will change the optimal solution. (b) Use algebraic analysis to find the allowable range to stay optimal for each cj. (c) Use algebraic analysis to find the allowable range to stay feasible for each bi. C (d) Use a software package based on the simplex method to find these allowable ranges. 6.7-15. For the problem given in Table 6.21, find the allowable range to stay optimal for c2. Show your work algebraically, using the tableau given in Table 6.21. Then justify your answer from a geometric viewpoint, referring to Fig. 6.3. 6.7-16.* For the original Wyndor Glass Co. problem, use the last tableau in Table 4.8 to do the following. (a) Find the allowable range to stay feasible for each bi. (b) Find the allowable range to stay optimal for c1 and c2. C (c) Use a software package based on the simplex method to find these allowable ranges.

296

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

6.7-17. For Variation 6 of the Wyndor Glass Co. model presented in Sec. 6.7, use the last tableau in Table 6.25 to do the following. (a) Find the allowable range to stay feasible for each bi. (b) Find the allowable range to stay optimal for c1 and c2. C (c) Use a software package based on the simplex method to find these allowable ranges.

gallons of cream left in its inventory. The linear programming formulation for this problem is shown below in algebraic form. Let

C gallons of chocolate ice cream produced, V gallons of vanilla ice cream produced, B gallons of banana ice cream produced.

Maximize 6.7-18. Ken and Larry, Inc., supplies its ice cream parlors with three flavors of ice cream: chocolate, vanilla, and banana. Because of extremely hot weather and a high demand for its products, the company has run short of its supply of ingredients: milk, sugar, and cream. Hence, they will not be able to fill all the orders received from their retail outlets, the ice cream parlors. Owing to these circumstances, the company has decided to choose the amount of each flavor to produce that will maximize total profit, given the constraints on supply of the basic ingredients. The chocolate, vanilla, and banana flavors generate, respectively, $1.00, $0.90, and $0.95 of profit per gallon sold. The company has only 200 gallons of milk, 150 pounds of sugar, and 60

profit 1.00 C 0.90 V 0.95 B,

subject to Milk: Sugar: Cream:

0.45 C 0.50 V 0.40 B 200 gallons 0.50 C 0.40 V 0.40 B 150 pounds 0.10 C 0.15 V 0.20 B 60 gallons

and C 0,

V 0,

B 0.

This problem was solved using the Excel Solver. The spreadsheet (already solved) and the sensitivity report are shown below. [Note: The numbers in the sensitivity report for the milk constraint are missing on purpose, since you will be asked to fill in these numbers in part ( f ).]

CHAPTER 6 PROBLEMS

For each of the following parts, answer the question as specifically and completely as is possible without solving the problem again on the Excel Solver. Note: Each part is independent (i.e., any change made to the model in one part does not apply to any other parts). (a) What is the optimal solution and total profit? (b) Suppose the profit per gallon of banana changes to $1.00. Will the optimal solution change, and what can be said about the effect on total profit? (c) Suppose the profit per gallon of banana changes to 92 cents. Will the optimal solution change, and what can be said about the effect on total profit? (d) Suppose the company discovers that 3 gallons of cream have gone sour and so must be thrown out. Will the optimal solution change, and what can be said about the effect on total profit? (e) Suppose the company has the opportunity to buy an additional 15 pounds of sugar at a total cost of $15. Should they? Explain. (f) Fill in all the sensitivity report information for the milk constraint, given just the optimal solution for the problem. Explain how you were able to deduce each number. 6.7-19. David, LaDeana, and Lydia are the sole partners and workers in a company which produces fine clocks. David and LaDeana each are available to work a maximum of 40 hours per week at the company, while Lydia is available to work a maximum of 20 hours per week. The company makes two different types of clocks: a grandfather clock and a wall clock. To make a clock, David (a mechanical engineer) assembles the inside mechanical parts of the clock while LaDeana (a woodworker) produces the hand-carved wood casings. Lydia is responsible for taking orders and shipping the clocks. The amount of time required for each of these tasks is shown below.

Time Required Task Assemble clock mechanism Carve wood casing Shipping

Grandfather Clock

Wall Clock

6 hours 8 hours 3 hours

4 hours 4 hours 3 hours

Each grandfather clock built and shipped yields a profit of $300, while each wall clock yields a profit of $200. The three partners now want to determine how many clocks of each type should be produced per week to maximize the total profit.

297

(a) Formulate a linear programming model for this problem. (b) Use the graphical method to solve the model. C (c) Use a software package based on the simplex method to solve the model. C (d) Use this same software package to generate sensitivity analysis information. (e) Use this sensitivity analysis information to determine whether the optimal solution must remain optimal if the estimate of the unit profit for grandfather clocks is changed from $300 to $375 (with no other changes in the model). (f) Repeat part (e) if, in addition to this change in the unit profit for grandfather clocks, the estimated unit profit for wall clocks also changes from $200 to $175. (g) Use graphical analysis to verify your answers in parts (e) and ( f ). (h) To increase the total profit, the three partners have agreed that one of them will slightly increase the maximum number of hours available to work per week. The choice of which one will be based on which one would increase the total profit the most. Use the sensitivity analysis information to make this choice. (Assume no change in the original estimates of the unit profits.) (i) Explain why one of the shadow prices is equal to zero. (j) Can the shadow prices given in the sensitivity analysis information be validly used to determine the effect if Lydia were to change her maximum number of hours available to work per week from 20 to 25? If so, what would be the increase in the total profit? (k) Repeat part ( j) if, in addition to the change for Lydia, David also were to change his maximum number of hours available to work per week from 40 to 35. (l) Use graphical analysis to verify your answer in part (k). 6.7-20. Consider the Union Airways problem presented in Sec. 3.4, including the data given in Table 3.19. Management now is considering increasing the level of service provided to customers by increasing one or more of the numbers in the rightmost column of Table 3.19 for the minimum number of agents needed in the various time periods. To guide them in making this decision, they would like to know what impact this change would have on total cost. Use a software package based on the simplex method to generate sensitivity analysis information in preparation for addressing the following questions. (a) Which of the numbers in the rightmost column of Table 3.19 can be increased without increasing total cost? In each case, indicate how much it can be increased (if it is the only one being changed) without increasing total cost. (b) For each of the other numbers, how much would the total cost increase per increase of 1 in the number? For each answer, in-

C

298

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

dicate how much the number can be increased (if it is the only one being changed) before the answer is no longer valid. (c) Do your answers in part (b) definitely remain valid if all the numbers considered in part (b) are simultaneously increased by 1? (d) Do your answers in part (b) definitely remain valid if all 10 numbers are simultaneously increased by 1? (e) How far can all 10 numbers be simultaneously increased by the same amount before your answers in part (b) may no longer be valid? 6.7-21. Consider the following problem. Z 2x1 5x2,

Maximize subject to x1 2x2 10 x1 3x2 12 and x1 0,

x2 0.

Let x3 and x4 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Right Side

Z x1 x2

(0) (1) (2)

1 0 0

0 1 0

0 0 1

1 3 1

1 2 1

22 6 2

While doing postoptimality analysis, you learn that all four bi and cj values used in the original model just given are accurate only to within 50 percent. In other words, their ranges of likely values are 5 b1 15, 6 b2 18, 1 c1 3, and 2.5 c2 7.5. Your job now is to perform sensitivity analysis to determine for each parameter individually (assuming the other three parameters equal their values in the original model) whether this uncertainty might affect either the feasibility or the optimality of the above basic solution (perhaps with new values for the basic variables). Specifically, determine the allowable range to stay feasible for each bi and the allowable range to stay optimal for each cj. Then, for each parameter and its range of likely values, indicate which part of this range lies within the allowable range and which parts correspond to values for which the current basic solution will no longer be both feasible and optimal. (a) Perform this sensitivity analysis graphically on the original model.

(b) Now perform this sensitivity analysis as described and illustrated in Sec. 6.7 for b1 and c1. (c) Repeat part (b) for b2. (d) Repeat part (b) for c2. 6.7-22. Reconsider Prob. 6.7-21. Now use a software package based on the simplex method to generate sensitivity analysis information preparatory to doing parts (a) and (c) below. C (a) Suppose that the estimates for c1 and c2 are correct but the estimates for both b1 and b2 are incorrect. Consider the following four cases where the true values of b1 and b2 differ from their estimates by the same percentage: (1) both b1 and b2 are smaller than their estimates, (2) both b1 and b2 are larger than their estimates, (3) b1 is smaller and b2 is larger than their estimates, and (4) b1 is larger and b2 is smaller than their estimates. For each of these cases, use the 100 percent rule for simultaneous changes in right-hand sides to determine how large the percentage error can be while guaranteeing that the original shadow prices still will be valid. (b) For each of the four cases considered in part (a), start with the final simplex tableau given in Prob. 6.7-21 and use algebraic analysis based on the fundamental insight presented in Sec. 5.3 to determine how large the percentage error can be without invalidating the original shadow prices. C (c) Suppose that the estimates for b1 and b2 are correct but the estimates for both c1 and c2 are incorrect. Consider the following four cases where the true values of c1 and c2 differ from their estimates by the same percentage: (1) both c1 and c2 are smaller than their estimates, (2) both c1 and c2 are larger than their estimates, (3) c1 is smaller and c2 is larger than their estimates, and (4) c1 is larger and c2 is smaller than their estimates. For each of these cases, use the 100 percent rule for simultaneous changes in objective function coefficients to determine how large the percentage error can be while guaranteeing that the original optimal solution must still be optimal. (d) For each of the four cases considered in part (c), start with the final simplex tableau given in Prob. 6.7-21 and use algebraic analysis based on the fundamental insight presented in Sec. 5.3 to determine how large the percentage error can be without invalidating the original optimal solution. 6.7-23. Consider the following problem. Maximize

Z 3x1 4x2 8x3,

subject to 2x1 3x2 5x3 9 x1 2x2 3x3 5 and x1 0,

x2 0,

x3 0.

CHAPTER 6 PROBLEMS

Let x4 and x5 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

1 0 0

0 1 0

1 1 1

0 0 1

1 3 1

1 5 2

14 2 1

While doing postoptimality analysis, you learn that some of the parameter values used in the original model just given are just rough estimates, where the range of likely values in each case is within 50 percent of the value used here. For each of these following parameters, perform sensitivity analysis to determine whether this uncertainty might affect either the feasibility or the optimality of the above basic solution. Specifically, for each parameter, determine the allowable range of values for which the current basic solution (perhaps with new values for the basic variables) will remain both feasible and optimal. Then, for each parameter and its range of likely values, indicate which part of this range lies within the allowable range and which parts correspond to values for which the current basic solution will no longer be both feasible and optimal. (a) Parameter b2 (b) Parameter c2 (c) Parameter a22 (d) Parameter c3 (e) Parameter a12 (f) Parameter b1 6.7-24. Consider Variation 5 of the Wyndor Glass Co. model presented in Sec. 6.7, where c2 3, a22 3, a32 4, and where the other parameters are given in Table 6.21. Starting from the resulting final tableau given at the bottom of Table 6.24, construct a table like Table 6.26 to perform parametric linear programming analysis, where c1 3

and

c2 3 2.

How far can be increased above 0 before the current basic solution is no longer optimal? 6.7-25. Reconsider the model of Prob. 6.7-6. Suppose that you now have the option of making trade-offs in the profitability of the first two activities, whereby the objective function coefficient of x1 can be increased by any amount by simultaneously decreasing the objective function coefficient of x2 by the same amount. Thus, the alternative choices of the objective function are Z() (2 )x1 (1 )x2 x3, where any nonnegative value of can be chosen.

299

Construct a table like Table 6.26 to perform parametric linear programming analysis on this problem. Determine the upper bound on before the original optimal solution would become nonoptimal. Then determine the best choice of over this range. 6.7-26. Consider the following parametric linear programming problem. Z() (10 4)x1 (4 )x2 (7 )x3,

Maximize subject to

3x1 x2 2x3 7 2x1 x2 3x3 5

(resource 1), (resource 2),

and x1 0,

x2 0,

x3 0,

where can be assigned any positive or negative values. Let x4 and x5 be the slack variables for the respective constraints. After we apply the simplex method with 0, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x2

(0) (1) (2)

1 0 0

0 1 0

0 0 1

3 1 5

2 1 2

2 1 3

24 2 1

(a) Determine the range of values of over which the above BF solution will remain optimal. Then find the best choice of within this range. (b) Given that is within the range of values found in part (a), find the allowable range to stay feasible for b1 (the available amount of resource 1). Then do the same for b2 (the available amount of resource 2). (c) Given that is within the range of values found in part (a), identify the shadow prices (as a function of ) for the two resources. Use this information to determine how the optimal value of the objective function would change (as a function of ) if the available amount of resource 1 were decreased by 1 and the available amount of resource 2 simultaneously were increased by 1. (d) Construct the dual of this parametric linear programming problem. Set 0 and solve this dual problem graphically to find the corresponding shadow prices for the two resources of the primal problem. Then find these shadow prices as a function of [within the range of values found in part (a)] by algebraically solving for this same optimal CPF solution for the dual problem as a function of .

300

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

6.7-27. Consider the following parametric linear programming problem. Z() 2x1 4x2 5x3,

Maximize

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x 5

x 6

Right Side

Z x2 x4

(0) (1) (2)

1 0 0

1 3 1

0 1 0

1 2 2

0 0 1

M 0 1

M2 1 1

20 10 5

subject to x1 3x2 2x3 5 x1 2x2 3x3 6 2 and x1 0,

x2 0,

x3 0,

where can be assigned any positive or negative values. Let x4 and x5 be the slack variables for the respective functional constraints. After we apply the simplex method with 0, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

0 1 2

0 1 0

1 5 1

0 0 1

1 3 1

1 2 1

11 3 1

(a) Use the fundamental insight (Sec. 5.3) to revise this tableau to reflect the inclusion of the parameter in the original model. Show the complete tableau needed to apply the feasibility test and the optimality test for any value of . Express the corresponding basic solution (and Z) as a function of . (b) Determine the range of nonnegative values of over which this basic solution is feasible. (c) Determine the range of nonnegative values of over which this basic solution is both feasible and optimal. Determine the best choice of over this range. 6.7-29. Consider the following problem. Z 10x1 4x2,

Maximize subject to 3x1 x2 30 2x1 x2 25 and (a) Express the BF solution (and Z) given in this tableau as a function of . Determine the lower and upper bounds on before this optimal solution would become infeasible. Then determine the best choice of between these bounds. (b) Given that is between the bounds found in part (a), determine the allowable range to stay optimal for c1 (the coefficient of x1 in the objective function). 6.7-28. Consider the following parametric linear programming problem, where the parameter must be nonnegative: Maximize

Z() (5 2)x1 (2 )x2 (3 )x3,

subject to 4x1 x2 2x3 5 5 3x1 x2 2x3 10 10 and x1 0,

x2 0,

x3 0.

Let x4 be the surplus variable for the first functional constraint, and let x5 and x6 be the artificial variables for the respective functional constraints. After we apply the simplex method with the Big M method and with 0, the final simplex tableau is

x1 0,

x2 0.

Let x3 and x4 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

Right Side

Z x2 x1

(0) (1) (2)

1 0 0

0 0 1

0 1 0

2 2 1

2 3 1

110 15 5

Now suppose that both of the following changes are made simultaneously in the original model: 1. The first constraint is changed to 4x1 x2 40. 2. Parametric programming is introduced to change the objective function to the alternative choices of Z() (10 2)x1 (4 )x2, where any nonnegative value of can be chosen.

CHAPTER 6 PROBLEMS

301

nology, so 0 1. Given , the coefficients of x1 in the model become

(a) Construct the resulting revised final tableau (as a function of ), and then convert this tableau to proper form from Gaussian elimination. Use this tableau to identify the new optimal solution that applies for either 0 or sufficiently small values of . (b) What is the upper bound on before this optimal solution would become nonoptimal? (c) Over the range of from zero to this upper bound, which choice of gives the largest value of the objective function?

c1 9 9 a11 2 . a21 5 Construct the resulting revised final tableau (as a function of ), and convert this tableau to proper form from Gaussian elimination. Use this tableau to identify the current basic solution as a function of . Over the allowable values of 0 1, give the range of values of for which this solution is both feasible and optimal. What is the best choice of within this range?

6.7-30. Consider the following problem. Z 9x1 8x2 5x3,

Maximize subject to

6.7-31. Consider the following problem.

2x1 3x2 x3 4 5x1 4x2 3x3 11

subject to

and x1 0,

x2 0,

2x1 2x2 x3 5 3x1 x2 x3 10

x3 0.

Let x4 and x5 denote the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

1 0 0

0 1 0

2 5 7

0 0 1

2 3 5

1 1 2

19 1 2

D,I

Z 3x1 5x2 2x3,

Maximize

(a) Suppose that a new technology has become available for conducting the first activity considered in this problem. If the new technology were adopted to replace the existing one, the coefficients of x1 in the model would change from

c1 9 a11 2 a21 5

to

c1 18 a11 3 . a21 6

Use the sensitivity analysis procedure to investigate the potential effect and desirability of adopting the new technology. Specifically, assuming it were adopted, construct the resulting revised final tableau, convert this tableau to proper form from Gaussian elimination, and then reoptimize (if necessary) to find the new optimal solution. (b) Now suppose that you have the option of mixing the old and new technologies for conducting the first activity. Let denote the fraction of the technology used that is from the new tech-

and x1 0,

x2 0,

x3 0.

Let x4 and x5 be the slack variables for the respective functional constraints. After we apply the simplex method, the final simplex tableau is

Coefficient of: Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Z x1 x3

(0) (1) (2)

1 0 0

0 1 0

20 3 8

0 0 1

9 1 3

7 1 2

115 15 35

Parametric linear programming analysis now is to be applied simultaneously to the objective function and right-hand sides, where the model in terms of the new parameter is the following: Maximize

Z() (3 2)x1 (5 )x2 (2 )x3,

subject to 2x1 2x2 x3 5 6 3x1 x2 x3 10 8 and x1 0,

x2 0,

x3 0.

Construct the resulting revised final tableau (as a function of ), and convert this tableau to proper form from Gaussian elimination.

302

6 DUALITY THEORY AND SENSITIVITY ANALYSIS

Use this tableau to identify the current basic solution as a function of . For 0, give the range of values of for which this solution is both feasible and optimal. What is the best choice of within this range?

(d) If the unit profit is below this breakeven point, how much can the old product’s production rate be decreased (assuming its previous rate was larger than this decrease) before the final BF solution would become infeasible?

6.7-32. Consider the Wyndor Glass Co. problem described in Sec. 3.1. Suppose that, in addition to considering the introduction of two new products, management now is considering changing the production rate of a certain old product that is still profitable. Refer to Table 3.1. The number of production hours per week used per unit production rate of this old product is 1, 4, and 3 for Plants 1, 2, and 3, respectively. Therefore, if we let denote the change (positive or negative) in the production rate of this old product, the right-hand sides of the three functional constraints in Sec. 3.1 become 4 , 12 4, and 18 3, respectively. Thus, choosing a negative value of would free additional capacity for producing more of the two new products, whereas a positive value would have the opposite effect. (a) Use a parametric linear programming formulation to determine the effect of different choices of on the optimal solution for the product mix of the two new products given in the final tableau of Table 4.8. In particular, use the fundamental insight of Sec. 5.3 to obtain expressions for Z and the basic variables x3, x2, and x1 in terms of , assuming that is sufficiently close to zero that this “final” basic solution still is feasible and thus optimal for the given value of . (b) Now consider the broader question of the choice of along with the product mix for the two new products. What is the breakeven unit profit for the old product (in comparison with the two new products) below which its production rate should be decreased ( 0) in favor of the new products and above which its production rate should be increased ( 0)? (c) If the unit profit is above this breakeven point, how much can the old product’s production rate be increased before the final BF solution would become infeasible?

6.7-33. Consider the following problem. Maximize

Z 2x1 x2 3x3,

subject to x1 x2 x3 3 x1 2x2 x3 1 x1 2x2 x3 2 and x1 0,

x2 0,

x3 0.

Suppose that the Big M method (see Sec. 4.6) is used to obtain the initial (artificial) BF solution. Let x4 be the artificial slack variable for the first constraint, x5 the surplus variable for the second constraint, x6 the artificial variable for the second constraint, and x7 the slack variable for the third constraint. The corresponding final set of equations yielding the optimal solution is (0) (1) (2) (3)

Z 5x2 (M 2)x4 Mx6 x7 8, x1 x2 x4 x7 1, 2x2 x3 x7 2, 3x2 x4 x5 x6 2.

Suppose that the original objective function is changed to Z 2x1 3x2 4x3 and that the original third constraint is changed to 2x2 x3 1. Use the sensitivity analysis procedure to revise the final set of equations (in tableau form) and convert it to proper form from Gaussian elimination for identifying and evaluating the current basic solution. Then test this solution for feasibility and for optimality. (Do not reoptimize.)

CASE 6.1 CONTROLLING AIR POLLUTION Refer to Sec. 3.4 (subsection entitled “Controlling Air Pollution”) for the Nori & Leets Co. problem. After the OR team obtained an optimal solution, we mentioned that the team then conducted sensitivity analysis. We now continue this story by having you retrace the steps taken by the OR team, after we provide some additional background. The values of the various parameters in the original formulation of the model are given in Tables 3.12, 3.13, and 3.14. Since the company does not have much prior experience with the pollution abatement methods under consideration, the cost estimates given in Table 3.14 are fairly rough, and each one could easily be off by as much as 10 percent in either direction. There also is some uncertainty about the parameter val-

CASE 6.1 CONTROLLING AIR POLLUTION

303

ues given in Table 3.13, but less so than for Table 3.14. By contrast, the values in Table 3.12 are policy standards, and so are prescribed constants. However, there still is considerable debate about where to set these policy standards on the required reductions in the emission rates of the various pollutants. The numbers in Table 3.12 actually are preliminary values tentatively agreed upon before learning what the total cost would be to meet these standards. Both the city and company officials agree that the final decision on these policy standards should be based on the tradeoff between costs and benefits. With this in mind, the city has concluded that each 10 percent increase in the policy standards over the current values (all the numbers in Table 3.12) would be worth $3.5 million to the city. Therefore, the city has agreed to reduce the company’s tax payments to the city by $3.5 million for each 10 percent reduction in the policy standards (up to 50 percent) that is accepted by the company. Finally, there has been some debate about the relative values of the policy standards for the three pollutants. As indicated in Table 3.12, the required reduction for particulates now is less than half of that for either sulfur oxides or hydrocarbons. Some have argued for decreasing this disparity. Others contend that an even greater disparity is justified because sulfur oxides and hydrocarbons cause considerably more damage than particulates. Agreement has been reached that this issue will be reexamined after information is obtained about which trade-offs in policy standards (increasing one while decreasing another) are available without increasing the total cost. (a) Use any available linear programming software to solve the model for this problem as formulated in Sec. 3.4. In addition to the optimal solution, obtain the additional output provided for performing postoptimality analysis (e.g., the Sensitivity Report when using Excel). This output provides the basis for the following steps. (b) Ignoring the constraints with no uncertainty about their parameter values (namely, xj 1 for j 1, 2, . . . , 6), identify the parameters of the model that should be classified as sensitive parameters. (Hint: See the subsection “Sensitivity Analysis” in Sec. 4.7.) Make a resulting recommendation about which parameters should be estimated more closely, if possible. (c) Analyze the effect of an inaccuracy in estimating each cost parameter given in Table 3.14. If the true value is 10 percent less than the estimated value, would this alter the optimal solution? Would it change if the true value were 10 percent more than the estimated value? Make a resulting recommendation about where to focus further work in estimating the cost parameters more closely. (d) Consider the case where your model has been converted to maximization form before applying the simplex method. Use Table 6.14 to construct the corresponding dual problem, and use the output from applying the simplex method to the primal problem to identify an optimal solution for this dual problem. If the primal problem had been left in minimization form, how would this affect the form of the dual problem and the sign of the optimal dual variables? (e) For each pollutant, use your results from part (d) to specify the rate at which the total cost of an optimal solution would change with any small change in the required reduction in the annual emission rate of the pollutant. Also specify how much this required reduction can be changed (up or down) without affecting the rate of change in the total cost. (f) For each unit change in the policy standard for particulates given in Table 3.12, determine the change in the opposite direction for sulfur oxides that would keep the total cost of an optimal solution unchanged. Repeat this for hydrocarbons instead of sulfur oxides. Then do

304

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

it for a simultaneous and equal change for both sulfur oxides and hydrocarbons in the opposite direction from particulates. (g) Letting denote the percentage increase in all the policy standards given in Table 3.12, formulate the problem of analyzing the effect of simultaneous proportional increases in these standards as a parametric linear programming problem. Then use your results from part (e) to determine the rate at which the total cost of an optimal solution would increase with a small increase in from zero. (h) Use the simplex method to find an optimal solution for the parametric linear programming problem formulated in part (g) for each 10, 20, 30, 40, 50. Considering the tax incentive offered by the city, use these results to determine which value of (including the option of 0) should be chosen to minimize the company’s total cost of both pollution abatement and taxes. (i) For the value of chosen in part (h), repeat parts (e) and ( f ) so that the decision makers can make a final decision on the relative values of the policy standards for the three pollutants.

CASE 6.2

FARM MANAGEMENT The Ploughman family owns and operates a 640-acre farm that has been in the family for several generations. The Ploughmans always have had to work hard to make a decent living from the farm and have had to endure some occasional difficult years. Stories about earlier generations overcoming hardships due to droughts, floods, etc., are an important part of the family history. However, the Ploughmans enjoy their selfreliant lifestyle and gain considerable satisfaction from continuing the family tradition of successfully living off the land during an era when many family farms are being abandoned or taken over by large agricultural corporations. John Ploughman is the current manager of the farm while his wife Eunice runs the house and manages the farm’s finances. John’s father, Grandpa Ploughman, lives with them and still puts in many hours working on the farm. John and Eunice’s older children, Frank, Phyllis, and Carl, also are given heavy chores before and after school. The entire famiy can produce a total of 4,000 person-hours worth of labor during the winter and spring months and 4,500 person-hours during the summer and fall. If any of these person-hours are not needed, Frank, Phyllis, and Carl will use them to work on a neighboring farm for $5 per hour during the winter and spring months and $5.50 per hour during the summer and fall. The farm supports two types of livestock: dairy cows and laying hens, as well as three crops: soybeans, corn, and wheat. (All three are cash crops, but the corn also is a feed crop for the cows and the wheat also is used for chicken feed.) The crops are harvested during the late summer and fall. During the winter months, John, Eunice, and Grandpa make a decision about the mix of livestock and crops for the coming year. Currently, the family has just completed a particularly successful harvest which has provided an investment fund of $20,000 that can be used to purchase more livestock. (Other money is available for ongoing expenses, including the next planting of crops.) The family currently has 30 cows valued at $35,000 and 2,000 hens valued at $5,000. They wish to keep all this livestock and perhaps purchase more. Each new cow would cost $1,500, and each new hen would cost $3.

CASE 6.2

FARM MANAGEMENT

305

Over a year’s time, the value of a herd of cows will decrease by about 10 percent and the value of a flock of hens will decrease by about 25 percent due to aging. Each cow will require 2 acres of land for grazing and 10 person-hours of work per month, while producing a net annual cash income of $850 for the family. The corresponding figures for each hen are: no significant acreage, 0.05 person-hour per month, and an annual net cash income of $4.25. The chicken house can accommodate a maximum of 5,000 hens, and the size of the barn limits the herd to a maximum of 42 cows. For each acre planted in each of the three crops, the following table gives the number of person-hours of work that will be required during the first and second halves of the year, as well as a rough estimate of the crop’s net value (in either income or savings in purchasing feed for the livestock).

Data per acre planted

Winter and spring, person-hours Summer and fall, person-hours Net value

Soybeans

Corn

Wheat

1.0 1.4 $70

0.9 1.2 $60

0.6 0.7 $40

To provide much of the feed for the livestock, John wants to plant at least 1 acre of corn for each cow in the coming year’s herd and at least 0.05 acre of wheat for each hen in the coming year’s flock. John, Eunice, and Grandpa now are discussing how much acreage should be planted in each of the crops and how many cows and hens to have for the coming year. Their objective is to maximize the family’s monetary worth at the end of the coming year (the sum of the net income from the livestock for the coming year plus the net value of the crops for the coming year plus what remains from the investment fund plus the value of the livestock at the end of the coming year plus any income from working on a neighboring farm, minus living expenses of $40,000 for the year). (a) Identify verbally the components of a linear programming model for this problem. (b) Formulate this model. (Either an algebraic or a spreadsheet formulation is acceptable.) (c) Obtain an optimal solution and generate the additional output provided for performing postoptimality analysis (e.g., the Sensitivity Report when using Excel). What does the model predict regarding the family’s monetary worth at the end of the coming year? (d) Find the allowable range to stay optimal for the net value per acre planted for each of the three crops.

The above estimates of the net value per acre planted in each of the three crops assumes good weather conditions. Adverse weather conditions would harm the crops and greatly reduce the resulting value. The scenarios particularly feared by the family are a drought, a flood, an early frost, both a drought and an early frost, and both a flood and an early frost. The estimated net values for the year under these scenarios are shown on the next page.

306

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

Net Value per Acre Planted Scenario Drought Flood Early frost Drought and early frost Flood and early frost

Soybeans

Corn

Wheat

$10 $15 $50 $15 $10

$15 $20 $40 $20 $10

0 $10 $30 $10 $ 5

(e) Find an optimal solution under each scenario after making the necessary adjustments to the linear programming model formulated in part (b). In each case, what is the prediction regarding the family’s monetary worth at the end of the year? (f) For the optimal solution obtained under each of the six scenarios [including the good weather scenario considered in parts (a) to (d )], calculate what the family’s monetary worth would be at the end of the year if each of the other five scenarios occur instead. In your judgment, which solution provides the best balance between yielding a large monetary worth under good weather conditions and avoiding an overly small monetary worth under adverse weather conditions.

Grandpa has researched what the weather conditions were in past years as far back as weather records have been kept, and obtained the following data. Scenario Good weather Drought Flood Early frost Drought and early frost Flood and early frost

Frequency 40% 20% 10% 15% 10% 5%

With these data, the family has decided to use the following approach to making its planting and livestock decisions. Rather than the optimistic approach of assuming that good weather conditions will prevail [as done in parts (a) to (d)], the average net value under all weather conditions will be used for each crop (weighting the net values under the various scenarios by the frequencies in the above table). (g) Modify the linear programming model formulated in part (b) to fit this new approach. (h) Repeat part (c) for this modified model. (i) Use a shadow price obtained in part (h) to analyze whether it would be worthwhile for the family to obtain a bank loan with a 10 percent interest rate to purchase more livestock now beyond what can be obtained with the $20,000 from the investment fund. (j) For each of the three crops, use the postoptimality analysis information obtained in part (h) to identify how much latitude for error is available in estimating the net value per acre planted for that crop without changing the optimal solution. Which two net values need to be estimated most carefully? If both estimates are incorrect simultaneously, how close do the estimates need to be to guarantee that the optimal solution will not change?

CASE 6.3

ASSIGNING STUDENTS TO SCHOOLS (REVISITED)

307

This problem illustrates a kind of situation that is frequently faced by various kinds of organizations. To describe the situation in general terms, an organization faces an uncertain future where any one of a number of scenarios may unfold. Which one will occur depends on conditions that are outside the control of the organization. The organization needs to choose the levels of various activities, but the unit contribution of each activity to the overall measure of performance is greatly affected by which scenario unfolds. Under these circumstances, what is the best mix of activities? (k) Think about specific situations outside of farm management that fit this description. Describe one.

CASE 6.3

ASSIGNING STUDENTS TO SCHOOLS (REVISITED) Reconsider Case 4.3. The Springfield School Board still has the policy of providing bussing for all middle school students who must travel more than approximately 1 mile. Another current policy is to allow splitting residential areas among multiple schools if this will reduce the total bussing cost. (This latter policy will be reversed in Case 12.4.) However, before adopting a bussing plan based on parts (a) and (b) of Case 4.3, the school board now wants to conduct some postoptimality analysis. (a) If you have not already done so for parts (a) and (b) of Case 4.3, formulate and solve a linear programming model for this problem. (Either an algebraic or a spreadsheet formulation is acceptable.) (b) Generate a sensitivity analysis report with the same software package as used in part (a).

One concern of the school board is the ongoing road construction in area 6. These construction projects have been delaying traffic considerably and are likely to affect the cost of bussing students from area 6, perhaps increasing them as much as 10 percent. (c) Use the report from part (b) to check how much the bussing cost from area 6 to school 1 can increase (assuming no change in the costs for the other schools) before the current optimal solution would no longer be optimal. If the allowable increase is less than 10 percent, re-solve to find the new optimal solution with a 10 percent increase. (d) Repeat part (c) for school 2 (assuming no change in the costs for the other schools). (e) Now assume that the bussing cost from area 6 would increase by the same percentage for all the schools. Use the report from part (b) to determine how large this percentage can be before the current optimal solution might no longer be optimal. If the allowable increase is less than 10 percent, re-solve to find the new optimal solution with a 10 percent increase.

The school board has the option of adding portable classrooms to increase the capacity of one or more of the middle schools for a few years. However, this is a costly move that the board would consider only if it would significantly decrease bussing costs. Each portable classroom holds 20 students and has a leasing cost of $2,500 per year. To analyze this option, the school board decides to assume that the road construction in area 6 will wind down without significantly increasing the bussing costs from that area.

308

6

DUALITY THEORY AND SENSITIVITY ANALYSIS

(f) For each school, use the corresponding shadow price from the report obtained in part (b) to determine whether it would be worthwhile to add any portable classrooms. (g) For each school where it is worthwhile to add any portable classrooms, use the report from part (b) to determine how many could be added before the shadow price would no longer be valid (assuming this is the only school receiving portable classrooms). (h) If it would be worthwhile to add portable classrooms to more than one school, use the report from part (b) to determine the combinations of the number to add for which the shadow prices definitely would still be valid. Then use the shadow prices to determine which of these combinations is best in terms of minimizing the total cost of bussing students and leasing portable classrooms. Re-solve to find the corresponding optimal solution for assigning students to schools. (i) If part (h) was applicable, modify the best combination of portable classrooms found there by adding one more to the school with the most favorable shadow price. Find the corresponding optimal solution for assigning students to schools and generate the corresponding sensitivity analysis report. Use this information to assess whether the plan developed in part (h) is the best one available for minimizing the total cost of bussing students and leasing portable classrooms. If not, find the best plan.

7 Other Algorithms for Linear Programming The key to the extremely widespread use of linear programming is the availability of an exceptionally efficient algorithm—the simplex method—that will routinely solve the largesize problems that typically arise in practice. However, the simplex method is only part of the arsenal of algorithms regularly used by linear programming practitioners. We now turn to these other algorithms. This chapter focuses first on three particularly important algorithms that are, in fact, variants of the simplex method. In particular, the next three sections present the dual simplex method (a modification particularly useful for sensitivity analysis), parametric linear programming (an extension for systematic sensitivity analysis), and the upper bound technique (a streamlined version of the simplex method for dealing with variables having upper bounds). Section 4.9 introduced another algorithmic approach to linear programming—a type of algorithm that moves through the interior of the feasible region. We describe this interior-point approach further in Sec. 7.4. We next introduce linear goal programming where, rather than having a single objective (maximize or minimize Z) as for linear programming, the problem instead has several goals toward which we must strive simultaneously. Certain formulation techniques enable converting a linear goal programming problem back into a linear programming problem so that solution procedures based on the simplex method can still be used. Section 7.5 describes these techniques and procedures.

7.1

THE DUAL SIMPLEX METHOD The dual simplex method is based on the duality theory presented in the first part of Chap. 6. To describe the basic idea behind this method, it is helpful to use some terminology introduced in Tables 6.10 and 6.11 of Sec. 6.3 for describing any pair of complementary basic solutions in the primal and dual problems. In particular, recall that both solutions are said to be primal feasible if the primal basic solution is feasible, whereas they are called dual feasible if the complementary dual basic solution is feasible for the dual problem. Also recall (as indicated on the right side of Table 6.11) that each complementary basic solution is optimal for its problem only if it is both primal feasible and dual feasible. The dual simplex method can be thought of as the mirror image of the simplex method. The simplex method deals directly with basic solutions in the primal problem that are primal feasible but not dual feasible. It then moves toward an optimal solution by striving 309

310

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

to achieve dual feasibility as well (the optimality test for the simplex method). By contrast, the dual simplex method deals with basic solutions in the primal problem that are dual feasible but not primal feasible. It then moves toward an optimal solution by striving to achieve primal feasibility as well. Furthermore, the dual simplex method deals with a problem as if the simplex method were being applied simultaneously to its dual problem. If we make their initial basic solutions complementary, the two methods move in complete sequence, obtaining complementary basic solutions with each iteration. The dual simplex method is very useful in certain special types of situations. Ordinarily it is easier to find an initial basic solution that is feasible than one that is dual feasible. However, it is occasionally necessary to introduce many artificial variables to construct an initial BF solution artificially. In such cases it may be easier to begin with a dual feasible basic solution and use the dual simplex method. Furthermore, fewer iterations may be required when it is not necessary to drive many artificial variables to zero. As we mentioned several times in Chap. 6 as well as in Sec. 4.7, another important primary application of the dual simplex method is its use in conjunction with sensitivity analysis. Suppose that an optimal solution has been obtained by the simplex method but that it becomes necessary (or of interest for sensitivity analysis) to make minor changes in the model. If the formerly optimal basic solution is no longer primal feasible (but still satisfies the optimality test), you can immediately apply the dual simplex method by starting with this dual feasible basic solution. Applying the dual simplex method in this way usually leads to the new optimal solution much more quickly than would solving the new problem from the beginning with the simplex method. The dual simplex method also can be useful in solving huge linear programming problems from scratch because it is such an efficient algorithm. Recent computational experience with the latest versions of CPLEX indicates that the dual simplex method often is more efficient than the simplex method for solving particularly massive problems encountered in practice. The rules for the dual simplex method are very similar to those for the simplex method. In fact, once the methods are started, the only difference between them is in the criteria used for selecting the entering and leaving basic variables and for stopping the algorithm. To start the dual simplex method (for a maximization problem), we must have all the coefficients in Eq. (0) nonnegative (so that the basic solution is dual feasible). The basic solutions will be infeasible (except for the last one) only because some of the variables are negative. The method continues to decrease the value of the objective function, always retaining nonnegative coefficients in Eq. (0), until all the variables are nonnegative. Such a basic solution is feasible (it satisfies all the equations) and is, therefore, optimal by the simplex method criterion of nonnegative coefficients in Eq. (0). The details of the dual simplex method are summarized next. Summary of the Dual Simplex Method. 1. Initialization: After converting any functional constraints in form to form (by multiplying through both sides by 1), introduce slack variables as needed to construct a set of equations describing the problem. Find a basic solution such that the coefficients in Eq. (0) are zero for basic variables and nonnegative for nonbasic variables (so the solution is optimal if it is feasible). Go to the feasibility test.

7.1 THE DUAL SIMPLEX METHOD

311

2. Feasibility test: Check to see whether all the basic variables are nonnegative. If they are, then this solution is feasible, and therefore optimal, so stop. Otherwise, go to an iteration. 3. Iteration: Step 1 Determine the leaving basic variable: Select the negative basic variable that has the largest absolute value. Step 2 Determine the entering basic variable: Select the nonbasic variable whose coefficient in Eq. (0) reaches zero first as an increasing multiple of the equation containing the leaving basic variable is added to Eq. (0). This selection is made by checking the nonbasic variables with negative coefficients in that equation (the one containing the leaving basic variable) and selecting the one with the smallest absolute value of the ratio of the Eq. (0) coefficient to the coefficient in that equation. Step 3 Determine the new basic solution: Starting from the current set of equations, solve for the basic variables in terms of the nonbasic variables by Gaussian elimination. When we set the nonbasic variables equal to zero, each basic variable (and Z) equals the new right-hand side of the one equation in which it appears (with a coefficient of 1). Return to the feasibility test. To fully understand the dual simplex method, you must realize that the method proceeds just as if the simplex method were being applied to the complementary basic solutions in the dual problem. (In fact, this interpretation was the motivation for constructing the method as it is.) Step 1 of an iteration, determining the leaving basic variable, is equivalent to determining the entering basic variable in the dual problem. The negative variable with the largest absolute value corresponds to the negative coefficient with the largest absolute value in Eq. (0) of the dual problem (see Table 6.3). Step 2, determining the entering basic variable, is equivalent to determining the leaving basic variable in the dual problem. The coefficient in Eq. (0) that reaches zero first corresponds to the variable in the dual problem that reaches zero first. The two criteria for stopping the algorithm are also complementary. We shall now illustrate the dual simplex method by applying it to the dual problem for the Wyndor Glass Co. (see Table 6.1). Normally this method is applied directly to the problem of concern (a primal problem). However, we have chosen this problem because you have already seen the simplex method applied to its dual problem (namely, the primal problem1) in Table 4.8 so you can compare the two. To facilitate the comparison, we shall continue to denote the decision variables in the problem being solved by yi rather than xj. In maximization form, the problem to be solved is Maximize

Z 4y1 12y2 18y3,

subject to y1

3y3 3 2y2 2y3 5

and y1 0, 1

y2 0,

y3 0.

Recall that the symmetry property in Sec. 6.1 points out that the dual of a dual problem is the original primal problem.

312

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

TABLE 7.1 Dual simplex method applied to the Wyndor Glass Co. dual problem Coefficient of: Iteration

0

1

2

Basic Variable

Eq.

Z

y1

y2

y3

Z y4 y5

(0) (1) (2)

1 0 0

4 1 0

12 0 2

18 3 2

0 1 0

0 0 1

0 3 5

Z y4

(0) (1)

1 0

4 1

0 0

6 3

0 1

y2

(2)

0

0

1

1

0

6 0 1 2

30 3 5 2

2 1 3 1 3

0

0

36

1

0

1

1

0

2 1 3 1 3

6

0

1 2

3 2

Z

(0)

1

y3

(1)

0

y2

(2)

0

y4

y5

Right Side

Since negative right-hand sides are now allowed, we do not need to introduce artificial variables to be the initial basic variables. Instead, we simply convert the functional constraints to form and introduce slack variables to play this role. The resulting initial set of equations is that shown for iteration 0 in Table 7.1. Notice that all the coefficients in Eq. (0) are nonnegative, so the solution is optimal if it is feasible. The initial basic solution is y1 0, y2 0, y3 0, y4 3, y5 5, with Z 0, which is not feasible because of the negative values. The leaving basic variable is y5 (5 3), and the entering basic variable is y2 (12/2 18/2), which leads to the second set of equations, labeled as iteration 1 in Table 7.1. The corresponding basic solution is y1 0, y2 5 2 , y3 0, y4 3, y5 0, with Z 30, which is not feasible. The next leaving basic variable is y4, and the entering basic variable is y3 (6/3 4/1), which leads to the final set of equations in Table 7.1. The corresponding basic solution is y1 0, y2 3 2 , y3 1, y4 0, y5 0, with Z 36, which is feasible and therefore optimal. Notice that the optimal solution for the dual of this problem1 is x*1 2, x*2 6, x*3 2, x*4 0, x*5 0, as was obtained in Table 4.8 by the simplex method. We suggest that you now trace through Tables 7.1 and 4.8 simultaneously and compare the complementary steps for the two mirror-image methods.

7.2

PARAMETRIC LINEAR PROGRAMMING At the end of Sec. 6.7 we described parametric linear programming and its use for conducting sensitivity analysis systematically by gradually changing various model parameters simultaneously. We shall now present the algorithmic procedure, first for the case where the cj parameters are being changed and then where the bi parameters are varied. 1 The complementary optimal basic solutions property presented in Sec. 6.3 indicates how to read the optimal solution for the dual problem from row 0 of the final simplex tableau for the primal problem. This same conclusion holds regardless of whether the simplex method or the dual simplex method is used to obtain the final tableau.

7.2 PARAMETRIC LINEAR PROGRAMMING

313

Systematic Changes in the cj Parameters For the case where the cj parameters are being changed, the objective function of the ordinary linear programming model n

Z cj xj j1

is replaced by n

Z( ) (cj j )xj, j1

where the j are given input constants representing the relative rates at which the coefficients are to be changed. Therefore, gradually increasing from zero changes the coefficients at these relative rates. The values assigned to the j may represent interesting simultaneous changes of the cj for systematic sensitivity analysis of the effect of increasing the magnitude of these changes. They may also be based on how the coefficients (e.g., unit profits) would change together with respect to some factor measured by . This factor might be uncontrollable, e.g., the state of the economy. However, it may also be under the control of the decision maker, e.g., the amount of personnel and equipment to shift from some of the activities to others. For any given value of , the optimal solution of the corresponding linear programming problem can be obtained by the simplex method. This solution may have been obtained already for the original problem where 0. However, the objective is to find the optimal solution of the modified linear programming problem [maximize Z( ) subject to the original constraints] as a function of . Therefore, in the solution procedure you need to be able to determine when and how the optimal solution changes (if it does) as increases from zero to any specified positive number. Figure 7.1 illustrates how Z*( ), the objective function value for the optimal solution (given ), changes as increases. In fact, Z*( ) always has this piecewise linear and con-

FIGURE 7.1 The objective function value for an optimal solution as a function of for parametric linear programming with systematic changes in the cj parameters.

Z* ( )

0

1

2

314

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

vex1 form (see Prob. 7.2-7). The corresponding optimal solution changes (as increases) just at the values of where the slope of the Z*( ) function changes. Thus, Fig. 7.1 depicts a problem where three different solutions are optimal for different values of , the first for 0 1, the second for 1 2, and the third for 2. Because the value of each xj remains the same within each of these intervals for , the value of Z*( ) varies with only because the coefficients of the xj are changing as a linear function of

. The solution procedure is based directly upon the sensitivity analysis procedure for investigating changes in the cj parameters (Cases 2a and 3, Sec. 6.7). As described in the last subsection of Sec. 6.7, the only basic difference with parametric linear programming is that the changes now are expressed in terms of rather than as specific numbers. To illustrate, suppose that 1 2 and 2 1 for the original Wyndor Glass Co. problem presented in Sec. 3.1, so that Z( ) (3 2 )x1 (5 )x2. Beginning with the final simplex tableau for 0 (Table 4.8), we see that its Eq. (0) (0)

3 Z x4 x5 36 2

would first have these changes from the original ( 0) coefficients added into it on the left-hand side: (0)

3 Z 2 x1 x2 x4 x5 36. 2

Because both x1 and x2 are basic variables [appearing in Eqs. (3) and (2), respectively], they both need to be eliminated algebraically from Eq. (0): 3 Z 2 x1 x2 x4 x5 36 2 2 times Eq. (3) times Eq. (2) (0)

3 7 2 Z x4 1 x5 36 2 . 2 6 3

The optimality test says that the current BF solution will remain optimal as long as these coefficients of the nonbasic variables remain nonnegative: 3 7 0, 2 6

9 for 0 , 7

2 1 0, 3

for all 0.

Therefore, after is increased past 9 7 , x4 would need to be the entering basic variable for another iteration of the simplex method to find the new optimal solution. Then would be increased further until another coefficient goes negative, and so on until has been increased as far as desired. This entire procedure is now summarized, and the example is completed in Table 7.2. 1

See Appendix 2 for a definition and discussion of convex functions.

7.2 PARAMETRIC LINEAR PROGRAMMING

315

TABLE 7.2 The cj parametric linear programming procedure applied to the Wyndor Glass Co. example Coefficient of: Range of

Basic Variable

Eq.

Z

x1

x2

x3

x4

x5

Right Side

Optimal Solution

Z( )

(0)

1

0

0

0

9 7

6

3 2

3

36 2

x4 0

1 3 1 2 1 3

1 3

2

x3 2

0

6

x2 6

1 3

2

x1 2

5

2

27 5

x3 0

x5 0 9 0 7

x3

(1)

0

0

0

1

x2

(2)

0

0

1

0

x1

(3)

0

1

0

0

Z( )

(0)

1

0

0

9 7

2

0

x5 0 9 5 7

3

x4

(1)

0

0

0

x2

(2)

0

0

1

x1

(3)

0

1

0

3 2 1

Z( )

(0)

1

0

5

x4 x5 x1

(1) (2) (3)

0 0 0

0 0 1

2 2 0

5

1

1

6

x4 6

3

x2 3

4

x1 4 x2 0 x3 0 x4 12 x5 6 x1 4

0

1 2 0

3 2

0

0

12 8

0 3 1

1 0 0

0 1 0

12 6 4

0

Summary of the Parametric Linear Programming Procedure for Systematic Changes in the cj Parameters. 1. Solve the problem with 0 by the simplex method. 2. Use the sensitivity analysis procedure (Cases 2a and 3, Sec. 6.7) to introduce the cj j changes into Eq. (0). 3. Increase until one of the nonbasic variables has its coefficient in Eq. (0) go negative (or until has been increased as far as desired). 4. Use this variable as the entering basic variable for an iteration of the simplex method to find the new optimal solution. Return to step 3. Systematic Changes in the bi Parameters For the case where the bi parameters change systematically, the one modification made in the original linear programming model is that bi is replaced by bi i , for i 1, 2, . . . , m, where the i are given input constants. Thus, the problem becomes n

Maximize

Z( ) cj xj, j1

316

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

subject to n

aij xj bi i

for i 1, 2, . . . , m

j1

and xj 0

for j 1, 2, . . . , n.

The goal is to identify the optimal solution as a function of . With this formulation, the corresponding objective function value Z*( ) always has the piecewise linear and concave1 form shown in Fig. 7.2. (See Prob. 7.2-8.) The set of basic variables in the optimal solution still changes (as increases) only where the slope of Z*( ) changes. However, in contrast to the preceding case, the values of these variables now change as a (linear) function of between the slope changes. The reason is that increasing changes the right-hand sides in the initial set of equations, which then causes changes in the right-hand sides in the final set of equations, i.e., in the values of the final set of basic variables. Figure 7.2 depicts a problem with three sets of basic variables that are optimal for different values of , the first for 0 1, the second for 1 2, and the third for 2. Within each of these intervals of , the value of Z*( ) varies with

despite the fixed coefficients cj because the xj values are changing. The following solution procedure summary is very similar to that just presented for systematic changes in the cj parameters. The reason is that changing the bi values is equivalent to changing the coefficients in the objective function of the dual model. Therefore, the procedure for the primal problem is exactly complementary to applying simultaneously the procedure for systematic changes in the cj parameters to the dual problem. Consequently, the dual simplex method (see Sec. 7.1) now would be used to obtain each new optimal solution, and the applicable sensitivity analysis case (see Sec. 6.7) now is Case 1, but these differences are the only major differences.

1

See Appendix 2 for a definition and discussion of concave functions.

FIGURE 7.2 The objective function value for an optimal solution as a function of for parametric linear programming with systematic changes in the bi parameters.

Z* ( )

0

1

2

7.3 THE UPPER BOUND TECHNIQUE

317

Summary of the Parametric Linear Programming Procedure for Systematic Changes in the bi Parameters. 1. Solve the problem with 0 by the simplex method. 2. Use the sensitivity analysis procedure (Case 1, Sec. 6.7) to introduce the bi i

changes to the right side column. 3. Increase until one of the basic variables has its value in the right side column go negative (or until has been increased as far as desired). 4. Use this variable as the leaving basic variable for an iteration of the dual simplex method to find the new optimal solution. Return to step 3. To illustrate this procedure in a way that demonstrates its duality relationship with the procedure for systematic changes in the cj parameters, we now apply it to the dual problem for the Wyndor Glass Co. (see Table 6.1). In particular, suppose that 1 2 and 2 1 so that the functional constraints become y1 3y3 3 2

2y2 2y3 5

or or

y1

3y3 3 2

2y2 2y3 5 .

Thus, the dual of this problem is just the example considered in Table 7.2. This problem with 0 has already been solved in Table 7.1, so we begin with the final simplex tableau given there. Using the sensitivity analysis procedure for Case 1, Sec. 6.7, we find that the entries in the right side column of the tableau change to the values given below. [2, 6] Z* y*b

3 2

5 36 2 ,

2 1 1 0 3 3 2

3 . b* S*b 3 1 5

1 7 2 6 3 2

Therefore, the two basic variables in this tableau 3 2

y3 3

and

9 7

y2 6

remain nonnegative for 0 9 7 . Increasing past 9 7 requires making y2 a leaving basic variable for another iteration of the dual simplex method, and so on, as summarized in Table 7.3. We suggest that you now trace through Tables 7.2 and 7.3 simultaneously to note the duality relationship between the two procedures.

7.3

THE UPPER BOUND TECHNIQUE It is fairly common in linear programming problems for some of or all the individual xj variables to have upper bound constraints xj uj, where uj is a positive constant representing the maximum feasible value of xj. We pointed out in Sec. 4.8 that the most important determinant of computation time for the simplex

318

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

TABLE 7.3 The bi parametric linear programming procedure applied to the dual of the Wyndor Glass Co. example Coefficient of: Range of 9 0 7

9 5 7

5

Basic Variable

Eq.

Z

y1

Z( )

(0)

1

y3

(1)

0

y2

(2)

0

Z( )

(0)

1

y2

y3

2 1 3 1 3

0

0

0

1

1

0

0

6

0

y4 2 1 3 1 3 4

y3

(1)

0

0

1

1

0

y1

(2)

0

1

3

0

1

Right Side

Optimal Solution

1 2

36 2

3 2

3 9 7

6

y1 y4 y5 0 3 2

y3 3 9 7

y2 6

y5 6 0

3

27 5

y2 y4 y5 0

1 2 3 2

5

2 9 7

2

5

y3 2 9 7

y1 2 y2 y3 y4 0

Z( )

(0)

1

0

12

6

4

0

12 8

y5 y1

(1) (2)

0 0

0 1

2 0

2 3

0 1

1 0

5

3 2

y5 5

y1 3 2

method is the number of functional constraints, whereas the number of nonnegativity constraints is relatively unimportant. Therefore, having a large number of upper bound constraints among the functional constraints greatly increases the computational effort required. The upper bound technique avoids this increased effort by removing the upper bound constraints from the functional constraints and treating them separately, essentially like nonnegativity constraints. Removing the upper bound constraints in this way causes no problems as long as none of the variables gets increased over its upper bound. The only time the simplex method increases some of the variables is when the entering basic variable is increased to obtain a new BF solution. Therefore, the upper bound technique simply applies the simplex method in the usual way to the remainder of the problem (i.e., without the upper bound constraints) but with the one additional restriction that each new BF solution must satisfy the upper bound constraints in addition to the usual lower bound (nonnegativity) constraints. To implement this idea, note that a decision variable xj with an upper bound constraint xj uj can always be replaced by xj uj yj, where yj would then be the decision variable. In other words, you have a choice between letting the decision variable be the amount above zero (xj) or the amount below uj (yj uj xj). (We shall refer to xj and yj as complementary decision variables.) Because 0 xj uj

7.3 THE UPPER BOUND TECHNIQUE

319

it also follows that 0 yj uj. Thus, at any point during the simplex method, you can either 1. Use xj, where 0 xj uj, or 2. Replace xj by uj yj , where 0 yj uj. The upper bound technique uses the following rule to make this choice: Rule: Begin with choice 1. Whenever xj 0, use choice 1, so xj is nonbasic. Whenever xj uj , use choice 2, so yj 0 is nonbasic. Switch choices only when the other extreme value of xj is reached. Therefore, whenever a basic variable reaches its upper bound, you should switch choices and use its complementary decision variable as the new nonbasic variable (the leaving basic variable) for identifying the new BF solution. Thus, the one substantive modification being made in the simplex method is in the rule for selecting the leaving basic variable. Recall that the simplex method selects as the leaving basic variable the one that would be the first to become infeasible by going negative as the entering basic variable is increased. The modification now made is to select instead the variable that would be the first to become infeasible in any way, either by going negative or by going over the upper bound, as the entering basic variable is increased. (Notice that one possibility is that the entering basic variable may become infeasible first by going over its upper bound, so that its complementary decision variable becomes the leaving basic variable.) If the leaving basic variable reaches zero, then proceed as usual with the simplex method. However, if it reaches its upper bound instead, then switch choices and make its complementary decision variable the leaving basic variable. To illustrate, consider this problem: Maximize

Z 2x1 x2 2x3,

subject to 4x1 x2 12 2x1 x3 4 and 0 x1 4,

0 x2 15,

0 x3 6.

Thus, all three variables have upper bound constraints (u1 4, u2 15, u3 6). The two equality constraints are already in proper form from Gaussian elimination for identifying the initial BF solution (x1 0, x2 12, x3 4), and none of the variables in this solution exceeds its upper bound, so x2 and x3 can be used as the initial basic variables without artificial variables being introduced. However, these variables then need to be eliminated algebraically from the objective function to obtain the initial Eq. (0), as follows:

(0)

Z 2( (2x1 x2 2x3 0 Z 2( (4x1 x2 2x3 12) Z 2( (2x1 x2 x3 4) Z 2( (2x1 x2 2x3 20.

320

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

TABLE 7.4 Equations and calculations for the initial leaving basic variable in the example for the upper bound technique Initial Set of Equations

Maximum Feasible Value of x1

(0) Z 2x1 x2 x3 20

x1 4 (since u1 4) 12 x1 3 4 64 x1 1 minimum (because u3 6) 2

(1) Z 4x1 x2 x3 12 (2) Z 2x1 x2 x3 4

To start the first iteration, this initial Eq. (0) indicates that the initial entering basic variable is x1. Since the upper bound constraints are not to be included, the entire initial set of equations and the corresponding calculations for selecting the leaving basic variables are those shown in Table 7.4. The second column shows how much the entering basic variable x1 can be increased from zero before some basic variable (including x1) becomes infeasible. The maximum value given next to Eq. (0) is just the upper bound constraint for x1. For Eq. (1), since the coefficient of x1 is positive, increasing x1 to 3 decreases the basic variable in this equation (x2) from 12 to its lower bound of zero. For Eq. (2), since the coefficient of x1 is negative, increasing x1 to 1 increases the basic variable in this equation (x3) from 4 to its upper bound of 6. Because Eq. (2) has the smallest maximum feasible value of x1 in Table 7.4, the basic variable in this equation (x3) provides the leaving basic variable. However, because x3 reached its upper bound, replace x3 by 6 y3, so that y3 0 becomes the new nonbasic variable for the next BF solution and x1 becomes the new basic variable in Eq. (2). This replacement leads to the following changes in this equation: (2)

2x1 x3 4 → 2x1 6 y3 4 → 2x1 y3 2 1 → x1 y3 1 2

Therefore, after we eliminate x1 algebraically from the other equations, the second complete set of equations becomes (0) (1) (2)

Zx2x2 y3 22 Zx2x2 2y3 8 1 Zx1x2 y3 1. 2

The resulting BF solution is x1 1, x2 8, y3 0. By the optimality test, it also is an optimal solution, so x1 1, x2 8, x3 6 y3 6 is the desired solution for the original problem.

7.4

AN INTERIOR-POINT ALGORITHM In Sec. 4.9 we discussed a dramatic development in linear programming that occurred in 1984, the invention by Narendra Karmarkar of AT&T Bell Laboratories of a powerful algorithm for solving huge linear programming problems with an approach very different

7.4 AN INTERIOR-POINT ALGORITHM

321

from the simplex method. We now introduce the nature of Karmarkar’s approach by describing a relatively elementary variant (the “affine” or “affine-scaling” variant) of his algorithm.1 (Your OR Courseware also includes this variant under the title, Solve Automatically by the Interior-Point Algorithm.) Throughout this section we shall focus on Karmarkar’s main ideas on an intuitive level while avoiding mathematical details. In particular, we shall bypass certain details that are needed for the full implementation of the algorithm (e.g., how to find an initial feasible trial solution) but are not central to a basic conceptual understanding. The ideas to be described can be summarized as follows: Concept 1: Shoot through the interior of the feasible region toward an optimal solution. Concept 2: Move in a direction that improves the objective function value at the fastest possible rate. Concept 3: Transform the feasible region to place the current trial solution near its center, thereby enabling a large improvement when concept 2 is implemented. To illustrate these ideas throughout the section, we shall use the following example: Maximize

Z x1 2x2,

subject to x1 x2 8 and x1 0,

x2 0.

This problem is depicted graphically in Fig. 7.3, where the optimal solution is seen to be (x1, x2) (0, 8) with Z 16. The Relevance of the Gradient for Concepts 1 and 2 The algorithm begins with an initial trial solution that (like all subsequent trial solutions) lies in the interior of the feasible region, i.e., inside the boundary of the feasible region. Thus, for the example, the solution must not lie on any of the three lines (x1 0, x2 0, x1 x2 8) that form the boundary of this region in Fig. 7.3. (A trial solution that lies on the boundary cannot be used because this would lead to the undefined mathematical operation of division by zero at one point in the algorithm.) We have arbitrarily chosen (x1, x2) (2, 2) to be the initial trial solution. To begin implementing concepts 1 and 2, note in Fig. 7.3 that the direction of movement from (2, 2) that increases Z at the fastest possible rate is perpendicular to (and toward) the objective function line Z 16 x1 2x2. We have shown this direction by the arrow from (2, 2) to (3, 4). Using vector addition, we have (3, 4) (2, 2) (1, 2), 1

The basic approach for this variant actually was proposed in 1967 by a Russian mathematician I. I. Dikin and then rediscovered soon after the appearance of Karmarkar’s work by a number of researchers, including E. R. Barnes, T. M. Cavalier, and A. L. Soyster. Also see R. J. Vanderbei, M. S. Meketon, and B. A. Freedman, “A Modification of Karmarkar’s Linear Programming Algorithm,” Algorithmica, 1(4) (Special Issue on New Approaches to Linear Programming): 395–407, 1986.

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OTHER ALGORITHMS FOR LINEAR PROGRAMMING

x2 8

(0, 8) optimal

Z 16 x1 2x2

6

4

(3, 4)

(2, 2)

2

FIGURE 7.3 Example for the interior-point algorithm.

2

0

4

6

8

x1

where the vector (1, 2) is the gradient of the objective function. (We will discuss gradients further in Sec. 13.5 in the broader context of nonlinear programming, where algorithms similar to Karmarkar’s have long been used.) The components of (1, 2) are just the coefficients in the objective function. Thus, with one subsequent modification, the gradient (1, 2) defines the ideal direction to which to move, where the question of the distance to move will be considered later. The algorithm actually operates on linear programming problems after they have been rewritten in augmented form. Letting x3 be the slack variable for the functional constraint of the example, we see that this form is Maximize

Z x1 2x2,

subject to x1 x2 x3 8 and x1 0,

x2 0,

x3 0.

In matrix notation (slightly different from Chap. 5 because the slack variable now is incorporated into the notation), the augmented form can be written in general as Maximize subject to Ax b

Z cTx,

7.4 AN INTERIOR-POINT ALGORITHM

323

and x 0, where 1 c 2 , 0

x1 x x2 , x3

A [1,

1,

1],

b [8],

0 0 0 0

for the example. Note that cT [1, 2, 0] now is the gradient of the objective function. The augmented form of the example is depicted graphically in Fig. 7.4. The feasible region now consists of the triangle with vertices (8, 0, 0), (0, 8, 0), and (0, 0, 8). Points in the interior of this feasible region are those where x1 0, x2 0, and x3 0. Each of these three xj 0 conditions has the effect of forcing (x1, x2) away from one of the three lines forming the boundary of the feasible region in Fig. 7.3. Using the Projected Gradient to Implement Concepts 1 and 2 In augmented form, the initial trial solution for the example is (x1, x2, x3) (2, 2, 4). Adding the gradient (1, 2, 0) leads to (3, 4, 4) (2, 2, 4) (1, 2, 0). However, now there is a complication. The algorithm cannot move from (2, 2, 4) toward (3, 4, 4), because (3, 4, 4) is infeasible! When x1 3 and x2 4, then x3 8 x1 x2 1 instead of 4. The point (3, 4, 4) lies on the near side as you look down on the feasible triangle in Fig. 7.4. Therefore, to remain feasible, the algorithm (indirectly) projects the point (3, 4, 4) down onto the feasible triangle by dropping a line that is perpendicular to this triangle. A vector from (0, 0, 0) to (1, 1, 1) is perpendicular to this triangle, so the perpendicular line through (3, 4, 4) is given by the equation (x1, x2, x3) (3, 4, 4) (1, 1, 1), where is a scalar. Since the triangle satisfies the equation x1 x2 x3 8, this perpendicular line intersects the triangle at (2, 3, 3). Because (2, 3, 3) (2, 2, 4) (0, 1, 1), the projected gradient of the objective function (the gradient projected onto the feasible region) is (0, 1, 1). It is this projected gradient that defines the direction of movement for the algorithm, as shown by the arrow in Fig. 7.4. A formula is available for computing the projected gradient directly. By defining the projection matrix P as P I AT(AAT)1A, the projected gradient (in column form) is cp Pc.

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x3 8

(2, 2, 4)

(3, 3, 4)

(2, 3, 3) 0

8 x1

FIGURE 7.4 Example in augmented form for the interior-point algorithm.

8 x2

(0, 8, 0) optimal

Thus, for the example,

1 P 0 0

0 1 0

1 0 0 1 [1 1 1

1

1 1] 1 1

1 0 0

0 1 0

1 0 1 0 1 [1 3 1 1

1

1]

1 0 0

0 1 0

1 0 1 0 1 3 1 1

23 1 1 1 3 1 1 3

1 1 1

1

[1

1 3 2 3 1 3

1

1]

1 3 1 3 , 2 3

7.4 AN INTERIOR-POINT ALGORITHM

325

so 23 cp 13 1 3

1 3 2 3 1 3

0 1 3 1 1 3 2 1 . 2 1 3 0

Moving from (2, 2, 4) in the direction of the projected gradient (0, 1, 1) involves increasing from zero in the formula 2 x 2 4cp 4

2 0 2 4 1 , 4 1

where the coefficient 4 is used simply to give an upper bound of 1 for to maintain feasibility (all xj 0). Note that increasing to 1 would cause x3 to decrease to x3 4 4(1)(1) 0, where 1 yields x3 0. Thus, measures the fraction used of the distance that could be moved before the feasible region is left. How large should be made for moving to the next trial solution? Because the increase in Z is proportional to , a value close to the upper bound of 1 is good for giving a relatively large step toward optimality on the current iteration. However, the problem with a value too close to 1 is that the next trial solution then is jammed against a constraint boundary, thereby making it difficult to take large improving steps during subsequent iterations. Therefore, it is very helpful for trial solutions to be near the center of the feasible region (or at least near the center of the portion of the feasible region in the vicinity of an optimal solution), and not too close to any constraint boundary. With this in mind, Karmarkar has stated for his algorithm that a value as large as 0.25 should be “safe.” In practice, much larger values (for example, 0.9) sometimes are used. For the purposes of this example (and the problems at the end of the chapter), we have chosen 0.5. (Your OR Courseware uses 0.5 as the default value, but also has 0.9 available.) A Centering Scheme for Implementing Concept 3 We now have just one more step to complete the description of the algorithm, namely, a special scheme for transforming the feasible region to place the current trial solution near its center. We have just described the benefit of having the trial solution near the center, but another important benefit of this centering scheme is that it keeps turning the direction of the projected gradient to point more nearly toward an optimal solution as the algorithm converges toward this solution. The basic idea of the centering scheme is straightforward—simply change the scale (units) for each of the variables so that the trial solution becomes equidistant from the constraint boundaries in the new coordinate system. (Karmarkar’s original algorithm uses a more sophisticated centering scheme.) For the example, there are three constraint boundaries in Fig. 7.3, each one corresponding to a zero value for one of the three variables of the problem in augmented form, namely, x1 0, x2 0, and x3 0. In Fig. 7.4, see how these three constraint boundaries intersect the Ax b (x1 x2 x3 8) plane to form the boundary of the feasible re-

326

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OTHER ALGORITHMS FOR LINEAR PROGRAMMING

gion. The initial trial solution is (x1, x2, x3) (2, 2, 4), so this solution is 2 units away from the x1 0 and x2 0 constraint boundaries and 4 units away from the x3 0 constraint boundary, when the units of the respective variables are used. However, whatever these units are in each case, they are quite arbitrary and can be changed as desired without changing the problem. Therefore, let us rescale the variables as follows: 1 ~x x , 1 2

2 ~x x , 2 2

~x x 3 3 4

in order to make the current trial solution of (x1, x2, x3) (2, 2, 4) become ~ , ~x , ~x ) (1, 1, 1). (x 1

2

3

~ for x , 2x ~ for x , and 4x ~ for x ), the problem In these new coordinates (substituting 2x 1 1 2 2 3 3 becomes ~ 4x ~, Maximize Z 2x 1

2

subject to ~ 2x ~x 4x ~ 8 2x 1 2 3 and ~x 0, 1

~x 0, 2

~x 0, 3

as depicted graphically in Fig. 7.5. Note that the trial solution (1, 1, 1) in Fig. 7.5 is equidistant from the three constraint boundaries ~x 1 0, ~x 2 0, ~x 3 0. For each subsequent iteration as well, the problem is rescaled again to achieve this same property, so that the current trial solution always is (1, 1, 1) in the current coordinates.

~ x3

FIGURE 7.5 Example after rescaling for iteration 1.

2

(1, 1, 1) 0

( 54, 74, 12) 4 ~ x2

(0, 4, 0) optimal

4

~ x1

7.4 AN INTERIOR-POINT ALGORITHM

327

Summary and Illustration of the Algorithm Now let us summarize and illustrate the algorithm by going through the first iteration for the example, then giving a summary of the general procedure, and finally applying this summary to a second iteration. Iteration 1. Given the initial trial solution (x1, x2, x3) (2, 2, 4), let D be the corresponding diagonal matrix such that x Dx~, so that 2 D 0 0

0 2 0

0 0 . 4

The rescaled variables then are the components of 1 0 0 x 1 2 2 x1 1 x ~ x D1x 0 0 x2 2 . 2 2 x x3 1 3 0 0 4 4 In these new coordinates, A and c have become Ã AD [1

1

2 1] 0 0

2 ~c Dc 0 0

0 2 0

0 0 4

0 0 [2 4

0 2 0

2

4],

1 2 2 4 . 0 0

Therefore, the projection matrix is P I ÃT(ÃÃT)1Ã

1 P 0 0

0 1 0

2 0 0 2 [2 1 4

1 P 0 0

0 1 0

4 4 0 0 21 4 4 4 1 8 8

2

2 4] 2 4

1

56 8 8 1 6 1 16 3

[2

2

4]

1 6 13 5 13 , 6 1 1 3 3

so that the projected gradient is 56 cp Pc~ 16 1 3

1 6 5 6

1 3

1 1 3 2 1 3 4 3 . 1 2 3 0

Define v as the absolute value of the negative component of cp having the largest absolute value, so that v 2 2 in this case. Consequently, in the current coordinates, the

328

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

~ , ~x , ~x ) (1, 1, 1) to the next algorithm now moves from the current trial solution (x 1 2 3 trial solution 5 4 1 1 1 7 0.5 ~ x 1 cp 1 3 , v 2 4 1 1 2 1 2 as shown in Fig. 7.5. (The definition of v has been chosen to make the smallest component of ~ x equal to zero when 1 in this equation for the next trial solution.) In the original coordinates, this solution is x1 2 0 0 5 4 5 2 7 7 x2 Dx~ 0 2 0 4 2 . 1 x3 0 0 4 2 2 This completes the iteration, and this new solution will be used to start the next iteration. These steps can be summarized as follows for any iteration. Summary of the Interior-Point Algorithm. 1. Given the current trial solution (x1, x2, . . . , xn), set 0

0 x1 0 0

0 0 x2 D 0 0 x3

0

0

xn 0 0 2. Calculate Ã AD and ~ c Dc. 3. Calculate P I ÃT(ÃÃT)1Ã and cp Pc~. 4. Identify the negative component of cp having the largest absolute value, and set v equal to this absolute value. Then calculate 1 1 ~ x cp, v 1 where is a selected constant between 0 and 1 (for example, 0.5). 5. Calculate x Dx~ as the trial solution for the next iteration (step 1). (If this trial solution is virtually unchanged from the preceding one, then the algorithm has virtually converged to an optimal solution, so stop.) Now let us apply this summary to iteration 2 for the example. Iteration 2. Step 1: Given the current trial solution (x1, x2, x3) (5 2 , 72 , 2), set 52 0 0 D 0 7 2 0 . 0 0 2

7.4 AN INTERIOR-POINT ALGORITHM

329

(Note that the rescaled variables are ~x 1 2 5 x1 2 5 0 0 x1 ~ 2 2 1 x 2 D x 0 7 0 x2 7 x2 , ~ 1 1 x3 2 x3 0 0 2 x3 so that the BF solutions in these new coordinates are 8 1 56 ~ x D1 0 0 , 0 0

0 0 1 ~ x D 8 176 , 0 0

and 0 0 1 ~ x D 0 0 , 8 4 as depicted in Fig. 7.6.) Step 2: Ã AD [ , , 2] 5 7 2 2

and

52 ~c Dc 7 . 0

~ x3

FIGURE 7.6 Example after rescaling for iteration 2.

4

2

(1, 1, 1) 0 1 16 7 3

~ x2

2 16 (0.83, 1.40, 0.5) 5

(0, 167 , 0) optimal

4

~ x1

330

7

OTHER ALGORITHMS FOR LINEAR PROGRAMMING

Step 3: 1138 P 178 2 9

178 41 90 14 45

29 1445 37 45

and

1 11 2 cp 1630 3 . 4 1 15

Step 4: 4 11 5 1 11 2, so v 4 115 and 1 11 2 2372 38 1 0.83 461 0.5 1 3 3 ~ x 1 41 60 328 1.40 . 1 41 15 1 5 2 1 0.50 Step 5: 1 6356 65 2.08 3227 ~ x Dx 656 4.92 1 1.00 is the trial solution for iteration 3. Since there is little to be learned by repeating these calculations for additional iterations, we shall stop here. However, we do show in Fig. 7.7 the reconfigured feasible region after rescaling based on the trial solution just obtained for iteration 3. As always, the ~ , ~x , ~x ) (1, 1, 1), equidistant from the ~x rescaling has placed the trial solution at (x 1 2 3 1 ~ ~ 0, x 2 0, and x 3 0 constraint boundaries. Note in Figs. 7.5, 7.6, and 7.7 how the sequence of iterations and rescaling have the effect of “sliding” the optimal solution toward (1, 1, 1) while the other BF solutions tend to slide away. Eventually, after enough itera~ , ~x , ~x ) (0, 1, 0) after rescaling, while tions, the optimal solution will lie very near (x 1 2 3 the other two BF solutions will be very far from the origin on the ~x 1 and ~x 3 axes. Step 5 of that iteration then will yield a solution in the original coordinates very near the optimal solution of (x1, x2, x3) (0, 8, 0). Figure 7.8 shows the progress of the algorithm in the original x1 x2 coordinate system before the problem is augmented. The three points—(x1, x2) (2, 2), (2.5, 3.5), and (2.08, 4.92)—are the trial solutions for initiating iterations 1, 2, and 3, respectively. We then have drawn a smooth curve through and beyond these points to show the trajectory of the algorithm in subsequent iterations as it approaches (x1, x2) (0, 8). The functional constraint for this particular example happened to be an inequality constraint. However, equality constraints cause no difficulty for the algorithm, since it deals with the constraints only after any necessary augmenting has been done to convert them to equality form (Ax b) anyway. To illustrate, suppose that the only change in the example is that the constraint x1 x2 8 is changed to x1 x2 8. Thus, the feasible region in Fig. 7.3 changes to just the line segment between (8, 0) and (0, 8). Given an initial feasible trial solution in the interior (x1 0 and x2 0) of this line segment—say, (x1, x2) (4, 4)—the algorithm can proceed just as presented in the five-step summary with just the two variables and A [1, 1]. For each iteration, the projected gradient points along this line segment in the direction of (0, 8). With 12 , iteration 1 leads from (4, 4)

7.4 AN INTERIOR-POINT ALGORITHM

331

~ x3 8

(1, 1, 1) 0 3.85 1.63

FIGURE 7.7 Example after rescaling for iteration 3.

~ x1

(0, 1.63, 0) optimal

~ x2

to (2, 6), iteration 2 leads from (2, 6) to (1, 7), etc. (Problem 7.4-3 asks you to verify these results.) Although either version of the example has only one functional constraint, having more than one leads to just one change in the procedure as already illustrated (other than more extensive calculations). Having a single functional constraint in the example meant that A had only a single row, so the (ÃÃT)1 term in step 3 only involved taking the reciprocal of the number obtained from the vector product ÃÃT. Multiple functional constraints mean that A has multiple rows, so then the (ÃÃT)1 term involves finding the inverse of the matrix obtained from the matrix product ÃÃT. To conclude, we need to add a comment to place the algorithm into better perspective. For our extremely small example, the algorithm requires relatively extensive calcul