(1,k)-configuration facets for the generalized assignment problem - Elsie Sterbin Gottlieb and M. R. Rao

Mathematical Programming 46 (1990) 53-60 North-Holland

53

(1, k)-CONFIGURATION FACETS FOR THE GENERALIZED ASSIGNMENT PROBLEM Elsie S t e r b i n G O T T L I E B * Baruch College, The City University of New York, NY, USA M.R. R A O * * Stern School of Business, New York University, NY, USA Received 21 August 1986 Revised manuscript received April 1988 A class of facet defining inequalities for the generalized assignment problem is derived. These inequalities are based upon multiple knapsack constraints and are derived from (1, k)configuration inequalities. Key words: Generalized assignment problem, knapsack problem, special ordered sets, integer polytope, facets, (1, k)-configurations.

1. Introduction In [3] we d e r i v e various classes o f valid i n e q u a l i t i e s for the g e n e r a l i z e d a s s i g n m e n t p r o b l e m ( G A P ) . M o r e restrictive c o n d i t i o n s n e e d to be i m p o s e d in o r d e r to t r a n s f o r m these valid inequalities into facet defining inequalities. These facet defining inequalities are p a r t i c u l a r l y i m p o r t a n t since t h e y are essential for defining the c o n v e x hull o f 0-1 solutions. In this p a p e r we derive a class o f facets b a s e d u p o n the (1, k ) - c o n f i g u r a t i o n i n e q u a l i t i e s d i s c u s s e d in [3]. W e e m p l o y the s a m e n o t a t i o n as in [3] a n d refer the r e a d e r to S e c t i o n 3 a n d Section 4.3 o f that p a p e r . A d d i t i o n a l classes o f facet defining i n e q u a l i t i e s , w h i c h are s p e c i a l i z a t i o n s o f o t h e r v a l i d i n e q u a l i t i e s d e r i v e d in [3], m a y b e f o u n d in [2] a n d [4].

2. (1, k)-Configuration facets In this s e c t i o n we derive a class o f facets o f K wYH , w h e r e I wI = w a n d IHf = h, b a s e d u p o n the j o i n t valid i n e q u a l i t i e s in [3, S e c t i o n 4.3]. A s s u m e that N ' u { z } , w h e r e ]N'[ = n', is a (1, k ) - c o n f i g u r a t i o n for s o m e a r b i t r a r y k n a p s a c k p. F o r specific r, k ~< r <~ n', t h e r e exist ("~') sets T(r) ~_ N ' , ] T(r)] = r, which are a r b i t r a r i l y n u m b e r e d * Partial financial support under NSF grant #CCR-8812736. ** Partial financial support under NSF grant #DMS-8606188.

54

E.S. Gottlieb, M . R . R a o / (1, k )-Configuration f a c e t s

and denoted as T~', t = 1. . . . , (~'). Recall that in [3, Section 4.3] the index f ( r , t), i c W \ p , is such that a~(~,t) = min{ai~, aidi(r,t)},

where aidi(r,t): min{aij [j c T7}.

Let aig,(r,t) =max{aiz, aia,(~,t)},

gi(r, t ) # f ( r , t).

For ease of notation we denote d~(r, t ) , f ( r , t) and g~(r, t) as d~,f and g~, respectively. The conditions of the following theorem determine for each knapsack i6 W \ p individual subsets Nic_ N \ ( N ' u { z } ) , Ri, and R*. Theorem 2.1. Suppose (i) N ' u {z} is a (1, k)-configuration for knapsack p, (ii) for some r, k<~ r<~ n', and some t, 1 <~ t<~ (~r'), i c W \ p implies there exists a set N~ ~ N \ ( N ' w {z}) such that {f} u N~ is an ( n~ + 1, ri)-cover for knapsack i, (iii) there exist w - 1 sets Ri c N~ such that IR~[= n and R~ c~ Rk = O, i, k c W \ p , (iv) a~; + aig, -{-~j~g* ao <~ bi, i c W\p, where R* = Ri\ei and alei = max{aij [j c Ri}; then (r-k+l)Xp~+

Y. xpj+ jE T~t

Z

Y.

xij <~r+

i~ W \ p j ~ Ni(z,r,t)

Z

r~,

(1)

i~ W \ p

where

S,(z, r, t) = N; u {z} u {d,}, is a facet of KYw~,

H={z}uN'u{

Uw\ Ni}, y = n ' + l + iG

p

2

(ni+2).

i~ W \ p

Proof. In this proof we require the construction of various solutions which are demonstrated using Example 2.1 given after the proof. The conditions are such that the graph of the support contains at least w - 1 cycles of length four. Each cycle contains node p c W, one node i ~ W\p, the corresponding node di c N and node zcN. We first show that (1) is a facet of K~wH,

H={z}uT:u

I U

N/}, y = r + l +

[.ic W \ p

~

(n/+2).

i~ W \ p

Subsequently, we will employ a lifting procedure to show that (1) is a facet of

K~,_,,

H={zIu

~,

U

N,y=n'+l+

iE W \ p

Z

(n~+2).

i~ W \ p

It follows from [3, Proposition 4.3] that (1) is a valid inequality for KwH, Y H = { z } u T , r u { U i ~ w \ p N i } . Assume that (1) is not a facet of Kwh. Y Then there exists a stronger inequality

~p~X,z+ Y, ~pjx,j+ Z j e Tt

Z

i~ W \ p j e Ni(z,r,t)

~ox,j<~ro,

(2)

E.S. Gottlieb, M.R. Rao /

(1,

k)-Configuration facets

55

which is a facet of K Y n . Define: F = {x c R y Ix satisfies (1) as equality, xo ~ {0, 1}} n K Y H , G = {x c NY Ix satisfies (2) as equality, xii ~ {0, 1}} c~ KYH. The assumption that (1) is not a facet implies F ~ (3. We now show that the conditions o f the t h e o r e m imply that (2) is a scalar multiple o f (1); hence, it must be the case that F = G and (1) is a facet. Since the p r o o f relies on the construction o f various solutions to PYH satisfying (1) as equality, we define solution c o m p o n e n t s for an individual k n a p s a c k i ~ W \ p :

f xif~ = 1, x~(a): ~ r / - 1 of the r~ variables xo, j c R~, equal one, /lx/j = 0 , otherwise. There exist (r,~l) = r~ such solution c o m p o n e n t s ; however, for ease o f notation all c o m p o n e n t s are denoted as x ~°). xi(2);

jxii=l, / tx~j=O,

j~Ri, otherwise.

Xt(3): •

ix~j=l,

jc{Z,g~}uR~,

tx!i = 0 ,

otherwise.

/x!j = 1,

j c { g i } w R '~ t ,

tx~j=O,

otherwise.

xi(4) ~

The c o m p o n e n t s x ~¢~) and x ~¢-~) are valid for k n a p s a c k i by condition (ii); the c o m p o n e n t s x ~¢s) and x ~¢4) are valid for k n a p s a c k i by condition (iv). For k n a p s a c k p we define the following solution c o m p o n e n t s , which are valid by condition (i):

/

Xp~ = 0 ,

xP(°): l x p j = l, I xpj=O,

j c TT, otherwise,

and

f Xpz = 1, xP(~): Ixt~i = 1, for k - 1 variables j c T~, ( xf~j = O,

otherwise.

We consider two cases d e p e n d i n g u p o n whether or not Rf = Ni for all i ~ W \ p .

Case A. Ri = Ni for all i c W \ p Due to the structure o f the facet defining inequality, it is easy to verify that the SOS constraints for j c ~_Ji~ w\p R~ u T~\{Ui~ w\p di} are satisfied since these nodes are not on any cycle. Hence, the following constructions are designed to satisfy the SOS constraints for each j c {z} w { U ~ w\p di}.

56

E . S . Gottlieb, M . R . R a o / (1, k ) - C o n f i g u r a t i o n f a c e t s

Construction L This construction produces r + 1 solutions. The solutions for this construction are displayed in Table 1. (1) Let Cr be an r by r submatrix, the rows of which are selected from a complete ( k - 1 ) - c l u t t e r [1] on r elements such that the rank is r. Let each column of C, correspond to a different variable xpj, j c T~,. Construct r solutions such that xp~ = 1 and the variables xpj, j ~ T~, take the values of a row in Cr, a different row being selected for each solution; this is valid due to condition (i). In each of the r solutions, knapsack i ~ W \ p is assigned the solution component x i(2). Since in each component i¢ W\p, xo=O, j ~ { z , di}, all of the SOS constraints are satisfied. Since ~i~w\p ri variables equal one for knapsack i c W \ p and, in addition,

(r-k+l)xp~+ ~ xpj--r,

(3)

j~ T r

the r solutions are in F and, consequently, in G. Since Cr has rank r and k - 1 elements equal one in each row, it follows that

~pj =1rp, j c T ~ . Consequently,

~pz+(k-1)~+

Z

Y~ ~ / j = ~ o .

(4)

ic W \ p j ~ R i

(2) Construct one solution such that knapsack p is assigned solution component x p(°), and each knapsack i c W \ p is assigned the solution component x ~(2). By construction, the SOS constraints are satisfied. Since r + ) ~ w \ p r~ variables equal one, the solution is in F. Consequently,

r~p + Z

Z ~'o= 7ro,

(5)

i~ l,V\p j ~ R i

which together with (4) implies that

~rpz = ( r - k + l )Trp.

(6)

Construction II. For each knapsack q c W\p, rq + 1 solutions are constructed as follows: Table 1 Solutions x c F for Example 2.1, Case A, Construction 1 i: j:

p=l z

1

2 2

3

z

2

3 4

5

z

3

6

7

1 1 0 0 1 0 1 0 1 0 0 1

0 0 1 1 0 0 1 1 0 0 1 1

0 0 1 1 1 0 0 1 1 1 0 0 1 1 1

0 1 1 1

0 0 1 1

0 0 1 1 1

E.S. Gottlieb, M.R. Rao / (1, k)-Configuration facets

57

(1) Knapsack q is assigned the solution component x q°), each knapsack i~ W \ { p , q} is assigned the solution component x i(2~, and knapsack p is assigned the solution component x p(°) if fq = z and x p(I) if fq = dq. Note that it is always possible to make such an assignment for knapsack p without violating the SOS constraints since k - 1 < r. Since there exist rq distinct solution components x q(n, we can construct rq different solutions to P Y H . (2) Construct one solution such that knapsack q is assigned the solution component x q(4), knapsack p is assigned the solution component x p(°~ if fq = dq and x p(1) iffq = z. Each knapsack i c W \ { p , q} is assigned the solution c o m p o n e n t x ~(2). For a given q, all of the SOS constraints are satisfied since in each of the rq + 1 solutions xij = O, i c W \ { p , q}, j E {z, d~}, and by construction Xpz + Xqz = 1 and Xpdq + X q d q = 1. In each of t h e ~ , q c w \ p (rq+ |) solutions ~ W \ p r~ variables equal one for knapsacks i e W \ p and (3) is satisfied; hence the solutions are in F. The solutions for this construction are displayed in Table 2. The solutions of Construction II(1) imply that 77-qj= 3Tq,

q C W \ p , j ~ Rq.

(7)

rirri= fro.

(8)

Thus, (5) becomes rrrt, +

E i~ W \ p

Since these solutions imply r~rp+Tr% + ( r q - l ) T r q +

~

riTri--Tro,

q6 W\p,

(9)

i~ ~z\{ p,q}

it follows that ~lTqfq = 3"i'q,

q C

(lo)

W\p.

Table 2 Solutions x c F for E x a m p l e 2.1, C a s e A, C o n s t r u c t i o n 11 i: j:

p=l z

1

2 2

3

z

2

3 4

5

z

3

6

7

8

0 1 1 l 0 1 1 1 1 1 0 0

1 0 0 1 1 0 1 0 0 1 1 0

0 0 1 1 1 0 0 1 1 1 0 0 1 1 1

q=2,f?= z

1 1 1 0

0 0 0 0

0 0 0 1

q=3,~=

1 1 1 1

0 0 0 1

0 0 0 1

0 0 0 0

1 1 1 1

1 1 1 1

1 1 1 0

0 1 1 1

1 0 1 1

1 1 0 0

d3

58

E.S. Gottlieb, M . R . R a o / (1, k)-Configuration f a c e t s

F r o m Construction II(2) it follows that rTrp + 7rqgq + ( rq -- 1 ) 7rq +

riTri = ~o,

q ~ W\p,

ic W\{p,q}

(11)

which, in turn, implies that 7rqgq = 7ro,

(12)

q ~ W\p.

Construction III. This construction p r o d u c e s w - 1 solutions, one for each q W \ p . K n a p s a c k q is assigned the solution c o m p o n e n t xq~3); for k n a p s a c k p, Xpz = Xpd,, = 0, and the remaining r - 1 variables equal one; each k n a p s a c k i ~ W \ { p , q} is assigned the solution c o m p o n e n t x i~2). In each solution, constructed for a particular q, x~j = O, i c W \ q , j c {z, di}; hence, all o f the SOS constraints are satisfied. These solutions are in F since rq + 1 + r - 1 + Zi~ w\{p,q} ri : r + ~i~ W\p ri" The solutions for this construction are displayed in Table 3. D u e to (7), (10) and (12), for each q ~ W \ p there exists a solution such that (r-1)Trp+(rq+l)Trq+

riTri = 7ro,

Z

q C W\p.

(13)

i E W \ { p,q}

This, together with (8), implies that ~q = Crp,

q c W\p.

Since r,rrp +

Z

riTTp = 77-0,

ieW\p

it follows that 7r0=Trp{r+

~ i~ W \ p

r~}

and (2) is a scalar multiple o f (1). C a s e B. Ri c Ni f o r s o m e i ~ W \ p For this case we require the following construction in addition to I, II and III. Construction IV. This construction defines Z i ~ w \ p ( n i - r i ) additional solutions, one for each k ~ N i \ R ~ , i c W \ p . For knapsack q ~ W \ p , Xqk = 1 and Xqj = 1 for any

Table 3 Solutions x c F for Example 2.1, Case A, Construction III i: j:

p=l z 1 2 3

2 z245

0

1

0

1

1

1

1

0

0

0

1

1

1

q=2

0

1

1

0

0

0

1

1

1

1

1

1

0

q=3

3 z3678

E.S. Gottlieb, M . R . R a o / (1, k)-Configuration f a c e t s

59

r q - 1 variables, j c Rq; Xqj = 0, otherwise. I f k ~ Ui~ w\(p,q~ R~, each knapsack i W \ { p , q} is assigned the solution component x ~2~ and knapsack p can be assigned any solution component such that (3) is satisfied. On the other hand, it may be the case that k c U ~ wxlp, q~ Ri. Note that since the sets R~ are disjoint, k can be in only one such set, say k ~ R~. Each knapsack i c W \ { p , q, s} is assigned the solution component x ~(2). For knapsack s, x~j = 1, j c R~\k; x~j~= 1; x~j = 0, otherwise. Knapsack p is assigned the solution component x p~°) iff~ = z and X p(1) such that Xps~= 0 iff~ = d~. It follows from the construction that all of the SOS constraints are satisfied. All solutions are in F since (3) is satisfied and for knapsacks i6 W \ p , Y.~w\p r~ variables equal one. It is easy to verify that these solutions imply that ~rqk=~rp,

k6Nq\Rq,

q~ W \ p ,

(14)

and the desired result follows. The solutions for this construction are displayed in Table 4. Table 4 S o l u t i o n s x e F f o r E x a m p l e 2.1, C a s e B, C o n s t r u c t i o n IV i: j:

p=|

2

z

1

2

3

z

2

0

1

1

l

0

0 0

1 1

1 1

1 1

1 0

3

4

5

9

z

3

4

6

7

0

1

0 0

0 1

8

10

0

1

0

0

0

1

1

1

0

q=2, k=9

1 1

0 0

0 0

0 0

1 0

1 1

1 1

0 0

0 1

q=3, k=4, s=2 q=3, k=10

We now show that in either Case A or B the sequential lifting procedure [5] applied to the inequality (1) yields 7rpj = 0 for each xpi , j c N ' \ T~. For any j c N ' \ T~, the following solution to the lifting problem is feasible: knapsack p has solution component x p(°), knapsacks i~ W \ p have solution component x "2). The objective function value equals r + ~ i ~ w\p r~, which is the right-hand side of (1); hence, this solution is also optimal. Consequently, ~5,i = 0 for each j c N ' \ T~ and (1) is a facet of KWH,

H={z}u

u

U

N~ , y = n ' + l +

ic W \ p

2

(n~+2).

[]

ic W \ p

Example 2.1. Case A. Suppose that the conditions of Theorem 2.1 are satisfied with: W={1,2,3}, w=3, p=l; N={z,l,2,...,10}, N'={1,2,3}, n'=3, k=2, r=3; N 2 = R 2 = { 4 , 5 } , e2=5, r2=2, f z = z , g2=d2=2; N3=R3={6,7,8}, e3=8, r3=3, f3 = d3 = 3, g3 = z. The facet defining inequality of KYwH, where y = 13, is

2x,,z+ Z x~j+ Z j~- N'

Z

xq~<8.

(15)

ic W \ p jc Ni(z,r,t )

The graph of the support is shown in Figure 1. Case B. This is identical to Case A except that now N2 = {4, 5, 9} and N3 = {4, 6, 7, 8, 10}; hence, N 2 \ R 2 -- {9) and N3\R3 = {4, 10}. The facet defining inequality of K Y w,, where y = 16, is still (15).

60

E.S. Gottlieb, M.R. R a o / ( 1 , k)-Configurationfacets nodes i e W

nodes J e N Z

1

2

1

4

2

5 -6

8 ..

",.9

"~ ~0 Fig. 1. Graph of support of the facet defining inequality (15) for Example 2.1; Case A: Graph without dotted lines; Case B: Graph including dotted lines.

References [1] J. Edmonds and D.R. Fulkerson, "Bottleneck extrema," Journal of Combinatorial Theory 8 (1970) 299-306. [2] E.S. Gottlieb, "Contributions to the polyhedral approach to generalized assignment problems," Ph.D. Thesis, Graduate School of Business Administration, New York University (New York, 1985). [3] E.S. Gottlieb and M.R. Rao, "The generalized assignment problem: valid inequalities and facets," Mathematical Programming 46 (1990) 31-52, this issue. [4] E.S. Gottlieb and M.R. Rao, "The generalized assignment problem II: three classes of facets," Technical Report #8622-Stat, Baruch College, The City University of New York (New York, 1986). [5] M.W. Padberg, "A note on zero-one programming," Operations Research 23 (1975) 833-837.