Minimal inequalities for an infiniterelaxation of integer programs

# Minimal inequalities for an infinite relaxation of integer programs

Amitabh Basu
Carnegie Mellon University, abasu1@andrew.cmu.edu

Michele Conforti

Gérard Cornuéjols
Carnegie Mellon University and Université d’Aix-Marseille
gc0v@andrew.cmu.edu

Giacomo Zambelli
Supported by NSF grant CMMI0653419, ONR grant N00014-03-1-0188 and ANR grant BLAN06-1-138894.
April 17, 2009, revised November 19, 2009
###### Abstract

We show that maximal -free convex sets are polyhedra when is the set of integral points in some rational polyhedron of . This result extends a theorem of Lovász characterizing maximal lattice-free convex sets. Our theorem has implications in integer programming. In particular, we show that maximal -free convex sets are in one-to-one correspondence with minimal inequalities.

## 1 Introduction

Consider a mixed integer linear program, and the optimal tableau of the linear programming relaxation. We select rows of the tableau, relative to basic integer variables . Let denote the nonbasic variables. Let be the value of in the basic solution associated with the tableau, , and suppose . The tableau restricted to these rows is of the form

 x=f+m∑j=1rjsj,x≥0 integral,s≥0, % and sj∈Z,j∈I, (1)

where , , and denotes the set of integer nonbasic variables.

An important question in integer programming is to derive valid inequalities for (1), cutting off the current infeasible solution , . We will consider a simplified model where the integrality conditions are relaxed on all nonbasic variables. On the other hand, we can present our results in a more general context, where the constraints , , are replaced by constraints , where is the set of integral points in some given rational polyhedron such that , i.e. contains affinely independent points. Recall that a polyhedron is rational if the matrix and vector have rational entries.

So we study the following model, introduced by Johnson [8].

 x=f+m∑j=1rjsj,x∈S,s≥0, (2)

where . Note that every inequality cutting off the point can be expressed in terms of the nonbasic variables only, and can therefore be written in the form .

In this paper we are interested in “formulas” for deriving such inequalities. More formally, we are interested in functions such that the inequality

 m∑j=1ψ(rj)sj≥1

is valid for (2) for every choice of and vectors . We refer to such functions as valid functions (with respect to and ). Note that, if is a valid function and is a function such that , then is also valid, and the inequality is implied by . Therefore we only need to investigate (pointwise) minimal valid functions.

Andersen, Louveaux, Weismantel, Wolsey [1] characterize minimal valid functions for the case , . Borozan and Cornuéjols [6] extend this result to for any . These papers and a result of Zambelli [11] show a one-to-one correspondence between minimal valid functions and maximal lattice-free convex sets with in the interior. These results have been further generalized in [4]. Minimal valid functions for the case are intersection cuts [2].

Our interest in model (2) arose from a recent paper of Dey and Wolsey [7]. They introduce the notion of -free convex set as a convex set without points of in its interior, and show the connection between valid functions and -free convex sets with in their interior.

A class of valid functions can be defined as follows. A function is positively homogeneous if for every and every , and it is subadditive if for all . A function is sublinear if it is positively homogeneous and subadditive. It is easy to observe that sublinear functions are also convex.

Assume that is a sublinear function such that the set

 Bψ={x∈Rn|ψ(x−f)≤1} (3)

is -free. Note that is closed and convex because is convex. Since is positively homogeneous, , thus is in the interior of . We claim that is a valid function. Indeed, given any solution to (2), we have

 m∑j=1ψ(rj)¯sj≥ψ(m∑j=1rj¯sj)=ψ(¯x−f)≥1,

where the first inequality follows from sublinearity and the last one follows from the fact that is not in the interior of .

Dey and Wolsey [7] show that every minimal valid function is sublinear and is an -free convex set with in its interior. In this paper, we prove that if is a minimal valid function, then is a maximal -free convex set.

In Section 2, we show that maximal -free convex sets are polyhedra. Therefore a maximal -free convex set containing in its interior can be uniquely written in the form . Let be the function defined by

 ψB(r)=maxi=1,…,kair,∀r∈Rn. (4)

It is easy to observe that the above function is sublinear and . In Section 3 we will prove that every minimal valid function is of the form for some maximal -free convex set containing in its interior. Conversely, if is a maximal -free convex set containing in its interior, then is a minimal valid function.

## 2 Maximal S-free convex sets

Let be the set of integral points in some rational polyhedron of . We say that is an -free convex set if is convex and does not contain any point of in its interior. We say that is a maximal -free convex set if it is an -free convex set and it is not properly contained in any -free convex set. It follows from Zorn’s lemma that every -free convex set is contained in a maximal -free convex set.

When , an -free convex set is called a lattice-free convex set. The following theorem of Lovász characterizes maximal lattice-free convex sets. A linear subspace or cone in is rational if it can be generated by rational vectors, i.e. vectors with rational coordinates.

###### Theorem 1.

(Lovász [9]) A set is a maximal lattice-free convex set if and only if one of the following holds:

• is a polyhedron of the form where is a polytope, is a rational linear space, , does not contain any integral point in its interior and there is an integral point in the relative interior of each facet of ;

• is a hyperplane of that is not rational.

Lovász only gives a sketch of the proof. A complete proof can be found in [4]. The next theorem is an extension of Lovász’ theorem to maximal -free convex sets.

Given a convex set , we denote by its recession cone and by its lineality space. Given a set , we denote by the linear space generated by . Given a -dimensional linear space and a subset of , we say that is a lattice of if there exists a linear bijection such that .

###### Theorem 2.

Let be the set of integral points in some rational polyhedron of such that . A set is a maximal -free convex set if and only if one of the following holds:

• is a polyhedron such that has nonempty interior, does not contain any point of in its interior and there is a point of in the relative interior of each of its facets.

• is a half-space of such that has empty interior and the boundary of is a supporting hyperplane of .

• is a hyperplane of such that is not rational.

Furthermore, if (i) holds, the recession cone of is rational and it is contained in the lineality space of .

We illustrate case (i) of the theorem in the plane in Figure 1. The question of the polyhedrality of maximal -free convex sets was raised by Dey and Wolsey [7]. They proved that this is the case for a maximal -free convex set , under the assumptions that has nonempty interior and that the recession cone of is finitely generated and rational. Theorem 2 settles the question in general.

To prove Theorem 2 we will need the following lemmas. The first one is proved in [4] and is an easy consequence of Dirichlet’s theorem.

###### Lemma 3.

Let and . For every and , there exists an integral point at distance less than from the half line

###### Lemma 4.

Let be an -free convex set such that has nonempty interior. For every , is -free.

###### Proof.

Let and . Suppose by contradiction that there exists . We show that . If not, , which implies that there is a hyperplane separating the line and , a contradiction. Thus there exists such that , i.e. there exists such that contains the open ball of radius centered at . Since , it follows that . Since , by Lemma 3 there exists at distance less than from the half line . Thus , hence . Note that the half-line is in , since and . Since is a rational polyhedron, for sufficiently small every integral point at distance at most from is in . Therefore , a contradiction. ∎

###### Proof of Theorem 2.

The proof of the “if” part is standard, and it is similar to the proof for the lattice-free case (see [4]). We show the “only if” part. Let be a maximal -free convex set. If , then is contained in some affine hyperplane . Since has empty interior, is -free, thus by maximality of . Next we show that is not rational. Suppose not. Then the linear subspace is rational. Therefore the projection of onto is a lattice of (see, for example, Barvinok [3] p 284 problem 3). The projection of onto is a subset of . Let be the projection of onto . Then is the projection of onto . Since is a hyperplane, . This implies that is bounded : otherwise there is an unbounded direction in and so for some . Since , this would imply that which is a contradiction. Fix . Since is a lattice and , there is a finite number of points at distance less than from the bounded set in . It follows that there exists such that every point of has distance at least from . Let . The set is -free by the choice of , but strictly contains , contradicting the maximality of . Therefore holds when . Hence we may assume . If has empty interior, then there exists a hyperplane separating and which is supporting for . By maximality of case follows.

Therefore we may assume that has nonempty interior. We show that satisfies .

###### Claim 1.

There exists a rational polyhedron such that:
,
The set is lattice-free,
For every facet of , is a facet of ,
For every facet of , contains an integral point in its relative interior.

Since is a rational polyhedron, there exist integral and such that . The set satisfies . The set is lattice-free since is -free and does not contain any point in , thus also satisfies . Let be the system containing all inequalities of that define facets of . Let . Then satisfies . See Figure 2 for an illustration.

It will be more convenient to write as intersection of the half-spaces defining the facets of , . We construct a sequence of rational polyhedra such that satisfies , , and such that satisfies . Given , we construct as follows. Let , where is the set of half spaces defining facets of . Let be a half-space in defining a facet of that does not contain an integral point in its relative interior; if no such exists, then satisfies and we are done. If does not contain any integral point in its interior, let . Otherwise, since is rational, among all integral points in the interior of there exists one, say , at minimum distance from . Let be the half-space containing with on its boundary. Let . Observe that defines a facet of since is in the interior of and it is on the boundary of . So are satisfied and has fewer facets that violate than .

Let be a maximal lattice-free convex set containing the set defined in Claim 1. As remarked earlier, such a set exists. By Theorem 1, is a polyhedron with an integral point in the relative interior of each of its facets. Let be a hyperplane that defines a facet of . Since is a facet of with an integral point in its relative interior, it follows that defines a facet of . This implies that . Therefore we can write as

 T=P∩k⋂i=1Hi, (5)

where are halfspaces. Let , .

###### Claim 2.

is a polyhedron.

We first show that, for , . Consider . Since and is contained in , . Since and , it follows that . Hence because .

Thus, for , there exists a hyperplane separating and . Hence there exists a halfspace such that and is disjoint from the interior of . We claim that the set is -free. Indeed, let . Then is not interior of . Since and , is in the interior of . Hence, by (5), there exists such that is not in the interior of . Thus . By construction, is not in the interior of , hence is not in the interior of . Thus is an -free convex set containing . Since is maximal, .

###### Claim 3.

.

Let . We show . By Lemma 4 applied to , is lattice-free. We observe that is -free. If not, let . Since , , hence , a contradiction. Hence, by maximality of , . Thus . Suppose that . Then there exists a facet of that is not parallel to . By construction, is a facet of containing an integral point in its relative interior. The point is then in the interior of , a contradiction.

###### Claim 4.

is rational.

Consider the maximal lattice-free convex set containing considered earlier. By Theorem 1, , and is rational. Clearly . Hence, if the claim does not hold, there exists a rational vector . By (5), . Since , . Hence . We will show that is -free, contradicting the maximality of . Suppose there exists . Since , . Therefore . Since , then where the last equality follows from . This contradicts the fact that is lattice-free.

By Lemma 4 and by the maximality of , .

###### Claim 5.

is rational.

Since and are both rational, we only need to show . The “” direction follows from . For the other direction, note that, since , we have , hence .

###### Claim 6.

Every facet of contains a point of in its relative interior.

Let be the linear space generated by . By Claim 5, is rational. Let , , be the projections of , , , respectively, onto . Since is rational, is a lattice of and . Also, is a maximal -free convex set of , since for any -free set of , is -free. Note that implies that is bounded. Otherwise there is an unbounded direction in and so for some . Since , this would imply that which is a contradiction. Let . Given , let . The polyhedron is a polytope since it has the same recession cone as . The polytope contains points of in its interior by the maximality of . Since is a lattice of , has a finite number of points in , hence there exists one minimizing , say . By construction, the polyhedron does not contain any point of in its interior and contains . By the maximality of , hence contains in its relative interior, and contains a point of in its relative interior. ∎

###### Corollary 5.

For every maximal -free convex set there exists a maximal lattice-free convex set such that, for every facet of , is a facet of .

###### Proof.

Let be defined as in Claim 1 in the proof of Theorem 2. It follows from the proof that is a maximal lattice-free convex set with the desired properties. ∎

## 3 Minimal valid functions

In this section we study minimal valid functions. We find it convenient to state our results in terms of an infinite model introduced by Dey and Wolsey [7].

Throughout this section, is a set of integral points in some rational polyhedron of such that , and is a point in . Let be the set of all infinite dimensional vectors such that

 f+∑r∈Rnrsr∈S sr≥0,r∈Rn (6) s has finite support

where has finite support means that is zero for all but a finite number of .

A function is valid (with respect to and ) if the linear inequality

 ∑r∈Rnψ(r)sr≥1 (7)

is satisfied by every . Note that this definition coincides with the one we gave in the introduction.

Given two functions we say that dominates if for all . A valid function is minimal if there is no valid function that dominates .

###### Theorem 6.

For every minimal valid function , there exists a maximal -free convex set with in its interior such that dominates . Furthermore, if is a maximal -free convex set containing in its interior, then is a minimal valid function.

We will need the following lemma.

###### Lemma 7.

Every valid function is dominated by a sublinear valid function.

###### Sketch of proof..

Given a valid function , define the following function . For all , let . Following the proof of Lemma 18 in [4] one can show that is a valid sublinear function that dominates . ∎

Given a valid sublinear function , the set is closed, convex, and contains in its interior. Since is a valid function, is -free. Indeed the interior of is . Its boundary is , and its recession cone is .

Before proving Theorem 6, we need the following general theorem about sublinear functions. Let be a closed, convex set in with the origin in its interior. The polar of is the set . Clearly is closed and convex, and since , it is well known that is bounded. In particular, is a compact set. Also, since , . Let

 ^K={y∈K∗|∃x∈K such that xy=1}. (8)

Note that is contained in the relative boundary of . Let be defined by

 ρK(r)=supy∈^Kry,for all r∈Rn. (9)

It is easy to show that is sublinear.

###### Theorem 8 (Basu et al. [5]).

Let be a closed convex set containing the origin in its interior. Then . Furthermore, for every sublinear function such that , we have for every .

###### Remark 9.

Let be a polyhedron containing the origin in its interior. Let such that . Then .

###### Proof.

The polar of is (see Theorem 9.1 in Schrijver [10]). Furthermore, is the union of all the facets of that do not contain the origin, therefore

 ρK(r)=supy∈^Kyr=maxi=1,…,tair

for all . ∎

###### Remark 10.

Let be a closed -free convex set in with in its interior, and let . Then is a valid function.

• Proof: Let . Then is in , therefore because is -free. By Theorem 8, . Thus

 1≤ρK(∑r∈Rnrsr)≤∑r∈RnρK(r)sr,

where the second inequality follows from the sublinearity of .

###### Lemma 11.

Let be a closed -free convex set containing in its interior, and let . There exists a maximal -free convex set such that for .

###### Proof.

Since is an -free convex set, it is contained in some maximal -free convex set . The set satisfies one of the statements (i)-(iii) of Theorem 2. By assumption, and is in the interior of . Since , is a full dimensional polyhedron, thus . This implies that , hence case (i) applies.

Thus is a polyhedron and is rational. Let us choose such that the dimension of is largest possible.

Since is a polyhedron containing in its interior, there exists and such that , , and . Without loss of generality, we may assume that where denotes the th row of , . By our assumption, . Therefore , since for all . Furthermore , since .

Let . Note that . By our choice of , is not -free for any , otherwise would be contained in a maximal -free convex set whose lineality space contains , a contradiction.

Let . Since , is a rational space. Note that , implying that for .

We observe next that we may assume that is bounded. Indeed, let , , be the projections onto of , , and , respectively. Since is a rational space, is a lattice of and . Note that is bounded, since . If we are given a maximal -free convex set in such that and for , then is the set . Since contains a point of in the relative interior of each of its facets, contains a point of in the relative interior of each of its facets, thus is a maximal -free convex set.

Thus we assume that is bounded, so . If all facets of contain a point of in their relative interior, then is a maximal -free convex set, thus the statement of the lemma holds. Otherwise we describe a procedure that replaces one of the inequalities defining a facet of without any point of in its relative interior with an inequality which is a convex combination of the inequalities of , such that the new polyhedron thus obtained is -free and has one fewer facet without points of in its relative interior. More formally, suppose the facet of defined by does not contain any point of in its relative interior. Given , let

 P(λ)={x∈Rn|[λd1+(1−λ)d2](x−f)≤1,di(x−f)≤1i=2,…,t}.

Note that and is obtained from by removing the inequality . Furthermore, given , we have .

Let be generators of . Note that, since is bounded, for every there exists such that . Let be the generators of satisfying

 d1rj>0 dirj≤0 i=2,…,t.

Note that, if no such generators exist, then is bounded. Otherwise is bounded if and only if, for

 [λd1+(1−λ)d2]rj>0.

This is the case if and only if , where

 λ∗=maxj=1,…,h−d2rj(d1−d2)rj.

Let be one of the vectors attaining the maximum in the previous equation. Then .

Note that is not -free otherwise is -free by Lemma 4, and so is , a contradiction.

Thus contains a point of in its interior. That is, there exists a point such that and for . Since is -free, . Thus there exists such that . Note that, since is bounded, there is a finite number of points of in the interior of . So we may choose such that is maximum. Thus is -free and is in the relative interior of the facet of defined by