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Chapter 2. Some Valid Inequalities for the Set Partitioning Problem

Chapter 2. Some Valid Inequalities for the Set Partitioning Problem

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E. Balm



14



I



min{cx A X = e, x,



=0



or 1,; E N }



where A = (a,-)is an m x n matrix of 0's and l's, e is an rn-vector of l's,

N = (1,. . ., n } . We will denote by a, the j t h column of A , and assume that A has no

zero row and n o zero column. Also, we will write M = (1,. . ., m}.

The convex hull and the dimensions of a set S, and the vertex set of a polytope T,

will be denoted by conv S, dim S and vert T respectively.

Denoting by "conv" the convex hull, we will call



1



P = conv{x E R" Ax



=



e, xi



=0



or 1,;E N }



the setpartitioning polytope, and denote the linear programming relaxation of P by

LP={xER"IAx=e,xsO}.

Clearly, vert P = P n (0, l}",

We will also refer to



p = conv{x E R" I Ax



=se, x, = 0 or l , j E N } ,



the sef packing polytope associated with P.

Whenever P # 0, we have

dim P =G dim LP = n - r(A )

where r(A) is the rank of A.

An inequality

7rx



s



rro



(1)



satisfied by all x E P is called valid for P. A valid inequality (1) such that

rrx



=



570



(1')



for exactly k + 1 affinely independent points x E P, 0 k s dimP, defines a

k-dimensional face of P and will itself be called a face (though since dim P < n, a

given face can be defined by more than one inequality). If k < d i m P , the face is

proper, otherwise it is improper. In the latter case, the hyperplane defined by (1')

contains all of P, and is called singular.

A valid inequality (1) is a cut, or cutting plane, if it is violated by some x E LP \ P.

A face of P, whether proper or not, may or may not be a cutting plane. If

dim P = dimLP, then the affine hull of P is the same as that of LP; hence any

hyperplane which contains all of P, also contains all of LP, and therefore n o

improper face of P is a cutting plane. If dim P < dim LP, then improper faces of P

may also be cutting planes.

Proper faces of maximal dimension are called facets. Evidently, P has faces

(hence facets) if and only if dimP 2 1, which implies n > r(A). If dim P = dimLP,



Some oalid inequalities



15



then the facets of P are of dimension n - r ( A )- 1, i.e., each facet contains exactly

n - r ( A ) affinely independent points of P. Since 0 P, these affinely independent

points are linearly independent vectors.

A valid inequality (1) is maximal if for any k E N and any T ; > T k there exists

x E P such that

T:x*



+ j E N2

-{k)



TjX,



>To.



This notion is the same as that of a minimal inequality (see Gomory and Johnson

[12]; and, more recently Jeroslow [13]), except that here we find it more convenient

to consider inequalities of the form S rather than 3 , in order to have a

nonnegative righthand side.

The following is an outline of the content of this paper.

We start (Section 2) with a class of homogeneous canonical inequalities that we

call elementary, since all the subsequent inequalities can be built up from these first

ones by various composition rules. The elementary inequalities, together with the

0-1 condition and the constraints Ax S e, imply the constraints Ax 3 e ; but they

also cut off fractional points satisfying Ax = e, x 3 0 . We discuss the conditions

under which a given elementary inequality is (a) a cutting plane, (b) maximal, (c) a

facet or an improper face of P.

When a given elementary inequality is not maximal, it can be strengthened. In

Section 3 we discuss two systematic strengthening procedures for these inequalities.

In Section 4 we show that each elementary inequality is equivalent on LP to a set

packing inequality and to each of several set covering inequalities. The first one of

these equivalences suggests a graph-theoretical interpretation. We introduce a

“strong” intersection graph of the matrix A defining P, and show that a set packing

inequality is valid for P if and only if it corresponds to a complete subgraph of the

strong intersection graph of A ; and it is maximal if and only if this complete

subgraph is a clique.

The next two sections deal with composite inequalities, obtained by certain rules

from the elementary inequalities. These composite inequalities have the following

property. Given an integer basic solution to the system Ax = e, x a 0 , and a set S of

nonbasic variables, none of which can be pivoted into the basis with a value of 1

without making the solution infeasible, there exists a composite inequality which

can be used as a primal all-integer cut to pivot into the basis any of the variables in S

without losing feasibility.

Finally, in Section 7 we introduce a class of inequalities which are satisfied by

every feasible integer solution better than a given one, and which can be

strengthened to a desired degree by performing implicit enumeration on certain

subproblems. We then discuss a hybrid primal cutting plane/implicit enumeration

algorithm based on these results.

Throughout the paper, the statements are illustrated on numerical examples.



E. Balas



16



2. Elementary inequalities

We shall denote

M

k



={i E



1



M



a,k



=



I},



I



N, = { k E N atk= l},



1



N,k = { j E N, a,ak = o),



G

k



= M\



Mk,



k E N,



Is,= N , N,,i E M,

iE



G k ,



k E N.



N,, is the index set of those columns a, orthogonal to ak and such that a,, = 1.

Since alk= 0 (as a result of i E G k ) , x k = I implies that at least one of the variables

x,, j E N , k , must be one.

Valid inequalities of the form



where Q C N i k , for some i E G k , will be called elementary. They play a central role

as building blocks for all the inequalities discussed in this paper. These elementary

inequalities are canonical in the sense of [4] (i.e., they have coefficients equal to 0, 1

or - l), hence each of them is parallel to a (n - 1 Q 1 - 1)-dimensional face of the

unit cube.



Remark 2.1.



The slack of an elementary inequality is a 0-1 variable.



Proof. Since Q C NikC Ni for some i E M, the sum of the variables indexed by Q

cannot exceed 1.

Proposition 2.1.



For every k



EN



and i E



a,, the inequality



is satisfied by all x E P.

Proof. From the definition of Nik,for every x E vert P, XI, = 1 implies x, = 1 for at

least one j E Nik.But this is precisely the condition expressed by (2); thus (2) is

satisfied by all x E v e r t P, hence by all x E P. 0

Remark 2.2.



The number of distinct inequalities ( 2 ) is at most



c k E N (M



k



I.



Proof. There is one inequality (2) for every zero entry of the matrix A, but some

of these inequalities may be identical.

The converse of Proposition 2.1 is not true in general, i.e., a 0-1 point satisfying

all inequalities (2) need not be in P, as one can easily see from the counterexample

offered by R such that X, = 1, V j E N. However, a weaker converse property holds.



Some valid inequalities



17



Proposition 2.2. A n y x E (0, l}", x # 0, which satisfies A x s e and all the inequalities (2), also satisfies Ax 3 e.

Proof. Let X E (0, l}", X # 0, be such that AX s e, AX# e. Then there exists i E M

such that X, = 0, V j E N,.

Further, since X # 0, there exists k E fit such that Xk = 1.

Therefore X violates the inequality



since Nik C Ni. 0



Corollary 2.2.1. Every nonzero vertex of P not contained in P is cut off by some

inequality ( 2 ) ; and every inequality (2) cuts off some x E p \ P.

Proof. Every x E p \ P violates A x 2 e ; hence if it is a nonzero vertex of P,

according to Proposition 2.2 it violates some inequality (2). On the other hand,

every inequality (2) cuts off the point X € p defined by ?k = 1, Xj =o,

VjEN\{k}. 0

Proposition 2.3.



For k E N, i E M kand Q



is valid if and only if x E vert P and



xk



Nik, the inequality



= 1 implies xi = 0, vj E Nik\ Q.



Proof. Necessity: if x E vert P and X I , = x, = 1 for some j E Nik\ Q, then from

Remark 2.1, x, = 0, v j E Q (since Q C N g k ) and

,

x violates (3).

Sufficiency: if x E vert P and X k = 1 implies x, = 0, v j E Nik \ 0, then (3) is valid

because (2) is valid. 0

Next we illustrate the elementary inequalities on a numerical example.



Example 2.1. Consider the numerical example of [5], i.e., the set partitioning

polytope with coefficient matrix A (where the blanks are zeroes):

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



1



1



1 1 1

1 1 1



1



1



1



1 1

1



For k = 1 , MI = {3,4,5}; N,, = {3,12}, N4,= {3,4}, N51= {3,4,5,12}, and the

inequalities (2) are



E. Balas



18

XI



- x3 - x * *c 0



x1-



XI



x3



- x4 =s 0



- x.1-



xs- x,* c 0,



x4-



where the last inequality is dominated by each of the other two and hence is

redundant. Further, x1 = 1 implies x4 = x l Z= 0 for any feasible partition (this can be

seen by inspection; systematic checking of such implications is discussed in Section

3) and therefore each of the sets N31,N41 and N51can be replaced by Q = {3}, and

each of the above inequalities can be replaced by

XI



- x3 =s 0.



In the next section we discuss procedures for strengthening elementary inequalities of the type (3) (which subsumes (2)) by systematically reducing the size of

the sets Q subject to the condition of Proposition 2.3.

As mentioned in Section 1, a valid inequality may o r may not be a cut, i.e., may

or may not be violated by some x E LP, P.

Proposition 2.4. The (valid) inequality (3) is a cut i f and only if there exists no

8 E R" satisfying

-1



8ai a



8e 3 0 .



1

0



j=k,

jEQ,

j E N , Q,



(4)



(5)



Proof. According to a classical result (see, for instance, [20, Theorem 1.4.4]), (3) is

a consequence of the system Ax = e, x 3 0 if and only if there exists 8 E R"

satisfying (4)and (5). If (3) is a consequence of Ax = e, x 3 0, it is clearly not a cut.

Conversely, if (3) is not a cut, then it is satisfied by all x E LP, hence a consequence

of Ax = e, x 3 0 . 0



Next we address the question of when a given elementary inequality is

undominated, i.e., maximal. First, if for some j E N, x E P implies x j = 0, then

clearly the coefficient of x j can be made arbitrarily large without invalidating the

given inequality. Therefore, without loss of generality, we can exclude this

degenerate case from our statement.

Proposition 2.5. Assume that the inequality (3), where Q C Ni,for some k E N ,

i E M,,is valid; and for every j E N there exists x E vert P such that xj = 1. Then ( 3 )

is maximal i f and only if

(i) for every j E Q there exists x E vert P such that x j = x k = 1;

(ii) for every j E # .. { k } there exists x E vert P such that x, = 1 and x k 2 x h ,

V h E Q.



Some ualid inequalities



19



Proof. This is a specialization of the statement that a valid inequalityrx s rofor a

0-1 polytope T C R" is maximal if and only if for every j E N there exists x E T

such that xJ = 1 and r x = ro. 0



If a valid inequality is not maximal, then at least one of its coefficients can be

increased without cutting off any x E P. In the case of an arbitrary polytope, this is

all we know, and it is not true in general that more than one coefficient can be

increased without invalidating the inequality. In the case of elementary inequalities

for P, however, one can say more.

Corollary 2.5.1. Assume that for every j E N there exists x E vert P such that

xj = 1. Let (3) be a valid, but not maximal inequality, with Q C Nikfor some k E N,

i E M k , and let S1, S2 be the sets of those j E N for which conditions (i) and (ii),

respectively, are violated. Then all x E P satisfy the inequality



and the inequalities



for every T 5



\



{ k } such that ahaJ# 0, V h ,j E T.



Proof. The validity of (6) follows from Proposition 2.3 and the definition of S , .

To prove the validity of (7), let x E vert P be such that XI, = 1. Then xi = 0,

V j E Sz n Ni\ {k} (hence V j E Sz n T ) , since otherwise from the definition of Sz,

x,, > XI, = 1 for some h E Q, which is impossible. Further, from (3), xi = 1 for some

j E Q. Hence (7) holds for all x E P such that xk = 1.

Now let XI, = 0. From the definition of T, xj = 1 for at most one E 7';and from

the definition of S2, xj = 1 for some j E Sz f l T implies xk < xh for some h E 0,

i.e.,

0

x h = 1 for at least one h E Q. Hence (7) also holds for all x E P such that xk = 0.



Clearly, if for some S ' C N the nondegeneracy assumption of Proposition 2.5

(and Corollary 2.5.1) is violated for all j E S', then the coefficient of each xi, j E S',

can be made arbitrarily large, in addition to the changes in the coefficients of xi,

j E S = S , U S2, justified by the Corollary.

From the above Corollary, nonmaximal elementary inequalities can be

strengthened, provided we know S. In the following sections we give several

procedures for identifying subsets of S.

Next we turn t o the question of when a maximal elementary inequality is a face of

maximal dimension, i.e., a facet or an improper face of P. This question is of

interest since P is the intersection of the halfspaces defining its facets and

improper faces. The next proposition gives a sufficient condition for an elementary

inequality to be a facet or an improper face of P.



E. Balas



20



Proposition 2.6. Suppose (3) is a maximal (valid) inequality for P, with Q C N,,

for some k E N, i E f i k . Let N ' = N \ Q U {k}, and



P N . = P n { x E R " ) x=, O , V j E Q U { k } } .

Then dimP 2 dimP,. + q, where q = 1 Q 1.

If dimP = d i m P N 8 +4, then ( 3 ) is an improper face of P.

If dimP = dimP,. + q + 1, then (3) is either a facet, or an improper face of P.

Proof. Let d = dimP, d ' = dimPNs.Since (3) is maximal, for every j E Q there

exists x' Evert P such that x : = x i = 1. Also, since Q C N,, xl, = 0, V h E Q \ { j }

for each of these q points x ' . With each point x', j = 1,. . ., q, we associate a row

vector y' E R", obtained by permuting the components of x' so that x i comes first,

and the components indexed by Q come next.

Further, let z E R"", j = 1,. . ., d ' + 1, be a maximal set of affinely independent

vertices of PN,, and let yq" E R", j = 1,. . ., d ' + 1 be row vectors of the form

y"' = ( O , z ' ) , where 0 has q components. Clearly, each yq+' is, modulo the

permutation of components, a vertex of P. Then the matrix Y whose rows are the

vectors y ' , i = 1, . . . , q + d ' + 1, is of the form



x=



I x*

[ ;+-]

XI



where X I is the q x ( q + 1) matrix



(tne blanks stand for zeroes), Z is the ( d ' + 1) x ( n - q - 1) matrix whose rows are

the vectors ti,j = 1,. . ., d ' + 1, 0 is the ( d ' + 1) x ( q + 1) zero matrix, and X , is a

q x ( n - q - 1) matrix of zeroes and ones.

Since X and Z are of full row rank, so is Y ; and since Y has q + d ' + 1 rows, it

follows that P contains at least q + d ' + 1 affinely independent points; hence

d 2 d ' + q.

If d = d ' + q, then the d ' + q + 1 rows of Y define a maximum-cardinality set of

affinely independent points of P; and since each of these points satisfies (3) with

equality, the same is true of every other point of P. Hence in this case (3) is an

improper face of P.

If d = d ' + + 1, then there exists a point x ' E P which, together with the

d ' + q + 1points corresponding to the rows of Y , forms an affinely independent set.

If x ' also satisfies ( 3 ) with equality, then (3) is an improper face of P; otherwise (3) is

a facet of P.



Some valid inequalities



In example 2.1, the inequalities (2) for k = 1 and i = 3,4, i.e., the



Example 2.2.

inequalities

XI



21



- x3 - X I 2 s 0,



XI



- x3 - xq c 0



are cutting planes, since each of them cuts off the fractional point 2 defined by

XI = Xz = Xs = ffs = 1, Xj = 0 otherwise; but they are not maximal, since the

conditions of Proposition 2.5 are violated for j = 9,12 in the case of the first

inequality and j = 4 in the case of the second one. Therefore, x 1 - x3 S 0 and

x I - x 3 + x s - x I 2s 0 are both valid (Corollary 2.5.1). T h e inequality x 1 - x3 s 0 is

maximal, since the assumption and conditions of Proposition 2.5 are satisfied. It is

also a facet of P, since the dimensionality condition of Proposition 2.6 is satisfied

and the point X defined by X, = XI4 = XIS = 1, Xj = 0 otherwise, does not lie on



a,



XI



- x3 = 0.



On the other hand, if P’ is the set partitioning polytope obtained from P by

removing the last column of A, then x 1- x 3 s 0 is an improper face of P’ since

x E P’ implies x , - x 3 = 0.



3. Strengthening procedures



An inequality r r ‘ x s rro is called stronger than rrx G rro, if .rr: 3 rrj for all j , and

for at least one j .

In this section we discuss two procedures for replacing a valid elementary

inequality which is not maximal, with a stronger valid elementary inequality. The

first procedure uses information from the other elementary inequalities in which xk

has a positive coefficient; the second one uses information from the elementary

inequalities in which x, has a positive coefficient for some j E Q.

rr; > T



For some k E N, let the index sets Qi C Nik, i E Mk, be such that



Proposition 3.1.

the inequalities



are satisfied by all x E P. For each j E



S.t



oh,define



Uhe,,&



. j E dh



and for i E Mk, let



1



T, = { j E Q, Q G ) Q~

~ for some /I E



Then the inequalities



Mk}.



E. Balas



22



are satisfied by all x E P.

Proof. From the definition of the sets



for all



j



QU),



x E P with x j = 1 implies



E Q,, i E GI,.Therefore, if j E T,, then x E P with x, = 1 implies



for some h E f i e ; which implies XI, = 0, since (3‘) holds for z = h.

Hence x E P and xe = 1 implies xi = 0, V j E T,. Therefore, if the system (3’) is

satisfied by all x E P, then so is the system (8). 0

Proposition 3.1 can be used to strengthen the inequalities (2) by replacing the sets

N,, with Qi = Nik., T,. It can then again be applied t o the strengthened inequalities,

and so on, until n o further strengthening is possible on the basis of this proposition

alone.

Applying the proposition to an inequality of the system (3’) consists of identifying

the set T . This can be done by bit manipulation and the use of logical “and” and

I.

logical “or”. The number of operations required is bounded by 1 QiX lak



I



Example 3.1. Consider again the set partitioning polytope defined by the matrix

of Example 2.1, and let us use Theorem 3.1 to strengthen the inequality

XI



- x3- x * 2 c 0



a,



associated with N31. For k = 1,

= {3,4,5}, and N31

= (3,121,

NS1= {3,4,5,12}. Setting Qh = N h l , h = 3,4,5, we have

Q(3) = {4,5,121,



= {3,4}



and



Q(l2) = {3,4,51,



and we find that Q(12) 3 Q4. Hence T3= {12}, and the above inequality can be

replaced by

x1-



x3 5z



0.



Since {3} is contained in each of N4]and NS1,the inequalities associated with

these two sets can both be replaced by x 1 - x3 =s0.

A second application of Proposition 3.1 brings n o further improvement.

For k = 2, a2

= {1,5}, N,, = {13,14}, Nsz = {5,13} and none of the two corresponding inequalities can be strengthened via Proposition 3.1.

= {1,2,3,4}; N l s = {1,6,8,9,14}, N 2 S = {1,2,11,15}, N3s= {2,6,8},

For k = 5, a5

N4,= {2,8,9,11}. Using Proposition 3.1 to strengthen the inequality



Some valid inequalities



associated with N , 5 ,we set Qh = Nh5,h



=



23



1,2,3,4,



Q(1) = {2,6,8,9,11,14,15}, Q(6) = {1,2,8,9,14), Q(8) = {1,2,6,9,11,14},

Q(9) = {1,2,6,8,11,14), Q(14) = {1,6,8,9),



and we find that

Q(1) IIQ3,



Q(9) 3 0



3 ,



and hence TI = {1,9} and the inequality associated with N I 5can be replaced b y

xs - x(j- xs - x,4 s 0.



When Proposition 3.1 is used to strengthen all rather than just one of the

elementary inequalities in which a certain variable x k has a positive coefficient, it is

convenient to work with the set

Qo=



U Q,



,€Mk



and instead of forming the sets T, by looking at each

the set

T=



u



j-E Q,, i E Mk,



form directly



T,



i=G&



= { j E Qol Q(j)> Qh for some h



€Gk)



by looking once at each j E Qo, and then use Qi \ T in place of Q i , T, in (8).

The number of operations required for applying Proposition 3.1 once to all

elementary inequalities in which xk has a positive coefficient is then bounded by

I a O I x l G k ( .



Next we discuss a second procedure for strengthening elementary inequalities.

Proposition 3.2.

inequalities



Let the index sets QikC Nik,i E Gk., k E N, be such that the



are satisfied by all x E P.



E. Balas



24



Proof. Let k E N , i E M k , and 1 E u # k . Then there exists h E 6%n M lsuch that

Qhkn Qbl= 0,and therefore adding the two elementary inequalities corresponding

to Qhkand Qhfrespectively yields



Since QhkU QhlC Nh,adding equation h of Ax



e to the kist inequality yields



s 1.



f



xk



=



I ENh\QhkUQhl



But then xk = 1 implies x1 = 0 and therefore (3”) can be replaced by (9). 0

If Proposition 3.2 is applied to several elementary inequalities, then repeated

applications may yield additional improvements like in the case of Proposition 3.1.

Applying Proposition 3.2 to an inequality (3”)consists of identifying the set Utk.

Again, this can be done by bit manipulation and use of logical “and” and logical

“or”. The number of operations required for each j E Q,k is bounded by

n

G, 1, hence the total number of operations is bounded by I Qa 1 X I Qk 1, like in the

case of Proposition 3.1.



Example 3.2.



Consider the set partitioning polytope defined by the matrix B



1 2 3

1



1



2



1

1



4



1



For k = 1,



6



7



8



9



10



1



1

1



3



5



4 5



1



1



1

1



1



1



1

1



1



G, = {3,4,5,6,7}, and



N ~=I {2,5,7},



N4i = {2,6,8},



N61 = {3,4,6},



N71



NSI= {3,5,8},



= {4,5,7}.



An attempt to apply Proposition 3.1 fails to strengthen any of the inequalities

associated with k = 1. On the other hand, Proposition 3.2 can be fruitfully applied

to replace both N,, and N4, with smaller sets, after applying Proposition 3.1 to

k = 2. We have A?f, = {1,2,5,6,7}, A?, f l M 2 = {5,6,7} and N5z= (3, lo}, Nh2= {3,4},

N72= (4). Applying Proposition 3.1 we find that Q(3) 3 N72;thus T5= T6= {3}, and

the sets N5*, N6*, can be replaced by Ns2\ Ts = (10) and N6*\ Th= (4) respectively.

Now writing Q,,= N , , and Q,Z= N,, \ T, for i = 5,6,7, we can apply Proposition

3.2 since



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Chapter 2. Some Valid Inequalities for the Set Partitioning Problem

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