Minimal inequalities for an infinite relaxation of integer programs

TL;DR

This paper proves that maximal S-free convex sets are polyhedra when S is the set of integral points in a rational polyhedron, extending Lovász's theorem and linking these sets to minimal inequalities in integer programming.

Contribution

It extends Lovász's theorem to a broader class of sets and establishes a correspondence between maximal S-free convex sets and minimal inequalities.

Findings

Maximal S-free convex sets are polyhedra for S as integral points in rational polyhedra.

Established a one-to-one correspondence between maximal S-free convex sets and minimal inequalities.

Extended classical results to new settings in integer programming.

Abstract

We show that maximal $S$-free convex sets are polyhedra when $S$ is the set of integral points in some rational polyhedron of $\mathbb{R}^n$. This result extends a theorem of Lov\'asz characterizing maximal lattice-free convex sets. Our theorem has implications in integer programming. In particular, we show that maximal $S$-free convex sets are in one-to-one correspondence with minimal inequalities.

Equations

$$x=f+\sum _{j=1}^m{r}^j{s}_j,x\ge 0\text{mbox}integral,s\ge 0,\text{mbox}and{s}_j\in \mathbb{Z},j\in I,$$

$$x=f+\sum _{j=1}^m{r}^j{s}_j,x\in S,s\ge 0,$$

$$\sum _{j=1}^m\psi (r^j)s_j\ge 1$$

$$B_{\psi }=\{x\in {\mathbb{R}}^n|\psi (x-f)\le 1\}$$

$$\sum _{j=1}^m\psi (r^j){\bar{s}}_j\ge \psi (\sum _{j=1}^m{r}^j{\bar{s}}_j)=\psi (\bar{x}-f)\ge 1,$$

$${\psi }_B(r)=\underset{i=1,\dots ,k}{\max }{a}_{i}r,\forall r\in {\mathbb{R}}^n.$$

$$T=P\cap \bigcap _{i=1}^k{H}_i,$$

$$f+\sum _{r\in {\mathbb{R}}^n}r{s}_r\in S$$

$$s_r\ge 0,r\in {\mathbb{R}}^n$$

$$s\text{mbox}hasfinitesupport$$

$$\sum _{r\in {\mathbb{R}}^n}\psi (r)s_r\ge 1$$

$$\hat{K}=\{y\in K^{\ast }|\exists x\in K\text{mbox}suchthatxy=1\}.$$

$${\rho }_K(r)=\underset{y\in \hat{K}}{\sup }ry,\text{mbox}forallr\in {\mathbb{R}}^n.$$

$${\rho }_K(r)=\underset{y\in \hat{K}}{\sup }yr=\underset{i=1,\dots ,t}{\max }{a}_{i}r$$

$$1\le {\rho }_K(\sum _{r\in {\mathbb{R}}^n}r{s}_r)\le \sum _{r\in {\mathbb{R}}^n}{\rho }_K(r)s_r,$$

$$P(\lambda )=\{x\in {\mathbb{R}}^n|[\lambda d_1+(1-\lambda )d_2](x-f)\le 1,d_i(x-f)\le 1i=2,\dots ,t\}.$$

$$d_1{r}_j>0$$

$$d_i{r}_j\le 0$$

$$[\lambda d_1+(1-\lambda )d_2]r_j>0.$$

$${\lambda }^{\ast }=\underset{j=1,\dots ,h}{\max }\frac{-d_2{r}_j}{(d_1-d_2)r_j}.$$

$$\psi (r)\ge {\rho }_K(r)=\underset{y\in \hat{K}}{\sup }yr=\underset{y\in \mathbf{cl}(\mathrm{conv}(\hat{K}))}{\max }yr\ge \underset{i=1,\dots ,k}{\max }{a}_{i}r={\psi }_B(r).$$

Full text

1 Introduction

Consider a mixed integer linear program, and the optimal tableau of the linear programming relaxation. We select $n$ rows of the tableau, relative to $n$ basic integer variables $x_{1},\ldots,x_{n}$. Let $s_{1},\ldots,s_{m}$ denote the nonbasic variables. Let $f_{i}\geq 0$ be the value of $x_{i}$ in the basic solution associated with the tableau, $i=1,\ldots,n$, and suppose $f\notin\mathbb{Z}^{n}$. The tableau restricted to these $n$ rows is of the form

[TABLE]

where $r^{j}\in\mathbb{R}^{n}$, $j=1,\ldots,m$, and $I$ 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 $x=f$, $s=0$. 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 $x\geq 0$, $x\in\mathbb{Z}^{n}$, are replaced by constraints $x\in S$, where $S$ is the set of integral points in some given rational polyhedron such that $\dim(S)=n$, i.e. $S$ contains $n+1$ affinely independent points. Recall that a polyhedron $Ax\leq b$ is rational if the matrix $A$ and vector $b$ have rational entries.

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

[TABLE]

where $f\in\mathrm{conv}(S)\setminus\mathbb{Z}^{n}$. Note that every inequality cutting off the point $(f,0)$ can be expressed in terms of the nonbasic variables $s$ only, and can therefore be written in the form $\sum_{j=1}^{m}\alpha_{j}s_{j}\geq 1$.

In this paper we are interested in “formulas” for deriving such inequalities. More formally, we are interested in functions $\psi\,:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that the inequality

[TABLE]

is valid for (2) for every choice of $m$ and vectors $r^{1},\ldots,r^{m}\in\mathbb{R}^{n}$. We refer to such functions $\psi$ as valid functions (with respect to $f$ and $S$). Note that, if $\psi$ is a valid function and $\psi^{\prime}$ is a function such that $\psi\leq\psi^{\prime}$, then $\psi^{\prime}$ is also valid, and the inequality $\sum_{j=1}^{m}\psi^{\prime}(r^{j})s_{j}\geq 1$ is implied by $\sum_{j=1}^{m}\psi(r^{j})s_{j}\geq 1$. Therefore we only need to investigate (pointwise) minimal valid functions.

Andersen, Louveaux, Weismantel, Wolsey [1] characterize minimal valid functions for the case $n=2$, $S=\mathbb{Z}^{2}$. Borozan and Cornuéjols [6] extend this result to $S=\mathbb{Z}^{n}$ for any $n$. 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 $f$ in the interior. These results have been further generalized in [4]. Minimal valid functions for the case $S=\mathbb{Z}^{n}$ are intersection cuts [2].

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

A class of valid functions can be defined as follows. A function $\psi$ is positively homogeneous if $\psi(\lambda r)=\lambda(\psi r)$ for every $r\in\mathbb{R}^{n}$ and every $\lambda\geq 0$, and it is subadditive if $\psi(r)+\psi(r^{\prime})\geq\psi(r+r^{\prime})$ for all $r,r^{\prime}\in\mathbb{R}^{n}$. A function $\psi$ is sublinear if it is positively homogeneous and subadditive. It is easy to observe that sublinear functions are also convex.

Assume that $\psi$ is a sublinear function such that the set

[TABLE]

is $S$-free. Note that $B_{\psi}$ is closed and convex because $\psi$ is convex. Since $\psi$ is positively homogeneous, $\psi(0)=0$, thus $f$ is in the interior of $B_{\psi}$. We claim that $\psi$ is a valid function. Indeed, given any solution $(\bar{x},\bar{s})$ to (2), we have

[TABLE]

where the first inequality follows from sublinearity and the last one follows from the fact that $\bar{x}$ is not in the interior of $B_{\psi}$.

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

In Section 2, we show that maximal $S$-free convex sets are polyhedra. Therefore a maximal $S$-free convex set $B\subseteq\mathbb{R}^{n}$ containing $f$ in its interior can be uniquely written in the form $B=\{x\in\mathbb{R}^{n}:\;a_{i}(x-f)\leq 1,\;i=1,\ldots,k\}$. Let $\psi_{B}:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be the function defined by

[TABLE]

It is easy to observe that the above function is sublinear and $B=\{x\in\mathbb{R}^{n}\,|\,\psi_{B}(x-f)\leq 1\}$. In Section 3 we will prove that every minimal valid function is of the form $\psi_{B}$ for some maximal $S$-free convex set $B$ containing $f$ in its interior. Conversely, if $B$ is a maximal $S$-free convex set containing $f$ in its interior, then $\psi_{B}$ is a minimal valid function.

2 Maximal $S$-free convex sets

Let $S\subseteq\mathbb{Z}^{n}$ be the set of integral points in some rational polyhedron of $\mathbb{R}^{n}$. We say that $B\subset\mathbb{R}^{n}$ is an $S$-free convex set if $B$ is convex and does not contain any point of $S$ in its interior. We say that $B$ is a maximal $S$-free convex set if it is an $S$-free convex set and it is not properly contained in any $S$-free convex set. It follows from Zorn’s lemma that every $S$-free convex set is contained in a maximal $S$-free convex set.

When $S=\mathbb{Z}^{n}$, an $S$-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 $\mathbb{R}^{n}$ is rational if it can be generated by rational vectors, i.e. vectors with rational coordinates.

Theorem 1**.**

(Lovász [9])* A set $B\subset\mathbb{R}^{n}$ is a maximal lattice-free convex set if and only if one of the following holds:*

• (i)

$B$* is a polyhedron of the form $B=P+L$ where $P$ is a polytope, $L$ is a rational linear space, $\dim(B)=\dim(P)+\dim(L)=n$, $B$ does not contain any integral point in its interior and there is an integral point in the relative interior of each facet of $B$;*

• (ii)

$B$* is a hyperplane of $\mathbb{R}^{n}$ 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 $S$-free convex sets.

Given a convex set $K\subset\mathbb{R}^{n}$, we denote by $\mathrm{rec}(K)$ its recession cone and by $\mathrm{lin}(K)$ its lineality space. Given a set $X\subseteq\mathbb{R}^{n}$, we denote by $\langle X\rangle$ the linear space generated by $X$. Given a $k$-dimensional linear space $V$ and a subset $\Lambda$ of $V$, we say that $\Lambda$ is a lattice of $V$ if there exists a linear bijection $f:\,\mathbb{R}^{k}\rightarrow V$ such that $\Lambda=f(\mathbb{Z}^{k})$.

Theorem 2**.**

Let $S$ be the set of integral points in some rational polyhedron of $\mathbb{R}^{n}$ such that $\dim(S)=n$. A set $B\subset\mathbb{R}^{n}$ is a maximal $S$-free convex set if and only if one of the following holds:

• $(i)$

$B$* is a polyhedron such that $B\cap\mathrm{conv}(S)$ has nonempty interior, $B$ does not contain any point of $S$ in its interior and there is a point of $S$ in the relative interior of each of its facets.*

• $(ii)$

$B$* is a half-space of $\mathbb{R}^{n}$ such that $B\cap\mathrm{conv}(S)$ has empty interior and the boundary of $B$ is a supporting hyperplane of $\mathrm{conv}(S)$.*

• $(iii)$

$B$* is a hyperplane of $\mathbb{R}^{n}$ such that $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$ is not rational.*

Furthermore, if (i) holds, the recession cone of $B\cap\mathrm{conv}(S)$ is rational and it is contained in the lineality space of $B$.

We illustrate case (i) of the theorem in the plane in Figure 1. The question of the polyhedrality of maximal $S$-free convex sets was raised by Dey and Wolsey [7]. They proved that this is the case for a maximal $S$-free convex set $B$, under the assumptions that $B\cap\mathrm{conv}(S)$ has nonempty interior and that the recession cone of $B\cap\mathrm{conv}(S)$ 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 $y\in\mathbb{Z}^{n}$ and $r\in\mathbb{R}^{n}$. For every $\varepsilon>0$ and $\bar{\lambda}\geq 0$, there exists an integral point at distance less than $\varepsilon$ from the half line $\{y+\lambda r\,|\,\lambda\geq\bar{\lambda}\}.$

Lemma 4**.**

Let $B$ be an $S$-free convex set such that $B\cap\mathrm{conv}(S)$ has nonempty interior. For every $r\in\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$, $B+\langle r\rangle$ is $S$-free.

Proof.

Let $C=\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$ and $r\in C\setminus\{0\}$. Suppose by contradiction that there exists $y\in S\cap\mathrm{\bf int}(B+\langle r\rangle)$. We show that $y\in\mathrm{\bf int}(B)+\langle r\rangle$. If not, $(y+\langle r\rangle)\cap\mathrm{\bf int}(B)=\emptyset$, which implies that there is a hyperplane $H$ separating the line $y+\langle r\rangle$ and $B+\langle r\rangle$, a contradiction. Thus there exists $\bar{\lambda}$ such that $\bar{y}=y+\bar{\lambda}r\in\mathrm{\bf int}(B)$, i.e. there exists $\varepsilon>0$ such that $B$ contains the open ball $B_{\varepsilon}(\bar{y})$ of radius $\varepsilon$ centered at $\bar{y}$. Since $r\in C\subseteq\mathrm{rec}(B)$, it follows that $B_{\varepsilon}(\bar{y})+\{\lambda r\,|\,\lambda\geq 0\}\subset B$. Since $y\in\mathbb{Z}^{n}$, by Lemma 3 there exists $z\in\mathbb{Z}^{n}$ at distance less than $\varepsilon$ from the half line $\{y+\lambda r\,|\,\lambda\geq\bar{\lambda}\}$. Thus $z\in B_{\varepsilon}(\bar{y})+\{\lambda r\,|\,\lambda\geq 0\}$, hence $z\in\mathrm{\bf int}(B)$. Note that the half-line $\{y+\lambda r\,|\,\lambda\geq\bar{\lambda}\}$ is in $\mathrm{conv}(S)$, since $y\in S$ and $r\in\mathrm{rec}(\mathrm{conv}(S))$. Since $\mathrm{conv}(S)$ is a rational polyhedron, for $\varepsilon>0$ sufficiently small every integral point at distance at most $\varepsilon$ from $\mathrm{conv}(S)$ is in $\mathrm{conv}(S)$. Therefore $z\in S$, 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 $B$ be a maximal $S$-free convex set. If $\dim(B)<n$, then $B$ is contained in some affine hyperplane $K$. Since $K$ has empty interior, $K$ is $S$-free, thus $B=K$ by maximality of $B$. Next we show that $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$ is not rational. Suppose not. Then the linear subspace $L=\langle\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))\rangle$ is rational. Therefore the projection $\Lambda$ of $\mathbb{Z}^{n}$ onto $L^{\bot}$ is a lattice of $L^{\bot}$ (see, for example, Barvinok [3] p 284 problem 3). The projection $S^{\prime}$ of $S$ onto $L^{\bot}$ is a subset of $\Lambda$. Let $B^{\prime}$ be the projection of $B$ onto $L^{\bot}$. Then $B^{\prime}\cap\mathrm{conv}(S^{\prime})$ is the projection of $B\cap\mathrm{conv}(S)$ onto $L^{\bot}$. Since $B$ is a hyperplane, $\mathrm{lin}(B)=\mathrm{rec}(B)$. This implies that $B^{\prime}\cap\mathrm{conv}(S^{\prime})$ is bounded : otherwise there is an unbounded direction $d\in L^{\bot}$ in $\mathrm{rec}(B^{\prime})\cap\mathrm{rec}(\mathrm{conv}(S^{\prime}))$ and so $d+l\in\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$ for some $l\in L$. Since $\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))=\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$, this would imply that $d\in L$ which is a contradiction. Fix $\delta>0$. Since $\Lambda$ is a lattice and $S^{\prime}\subseteq\Lambda$, there is a finite number of points at distance less than $\delta$ from the bounded set $B^{\prime}\cap\mathrm{conv}(S^{\prime})$ in $L^{\bot}$. It follows that there exists $\varepsilon>0$ such that every point of $S^{\prime}$ has distance at least $\varepsilon$ from $B^{\prime}\cap\mathrm{conv}(S^{\prime})$. Let $B^{\prime\prime}=\{v+w\,|\,v\in B,\,w\in L^{\bot},\|w\|\leq\varepsilon\}$. The set $B^{\prime\prime}$ is $S$-free by the choice of $\varepsilon$, but $B^{\prime\prime}$ strictly contains $B$, contradicting the maximality of $B$. Therefore $(iii)$ holds when $\dim(B)<n$. Hence we may assume $\dim(B)=n$. If $B\cap\mathrm{conv}(S)$ has empty interior, then there exists a hyperplane separating $B$ and $\mathrm{conv}(S)$ which is supporting for $\mathrm{conv}(S)$. By maximality of $B$ case $(ii)$ follows.

Therefore we may assume that $B\cap\mathrm{conv}(S)$ has nonempty interior. We show that $B$ satisfies $(i)$.

Claim 1**.**

*There exists a rational polyhedron $P$ such that:

$i)$ $\mathrm{conv}(S)\subset\mathrm{\bf int}(P)$,

$ii)$ The set $K=B\cap P$ is lattice-free,

$iii)$ For every facet $F$ of $P$, $F\cap K$ is a facet of $K$,

$iv)$ For every facet $F$ of $P$, $F\cap K$ contains an integral point in its relative interior.*

Since $\mathrm{conv}(S)$ is a rational polyhedron, there exist integral $A$ and $b$ such that $\mathrm{conv}(S)=\{x\in\mathbb{R}^{n}\,|\,Ax\leq b\}$. The set $P^{\prime}=\{x\in\mathbb{R}^{n}\,|\,Ax\leq b+\frac{1}{2}\mathbf{1}\}$ satisfies $i)$. The set $B\cap P^{\prime}$ is lattice-free since $B$ is $S$-free and $P^{\prime}$ does not contain any point in $\mathbb{Z}^{n}\setminus S$, thus $P^{\prime}$ also satisfies $ii)$. Let $\bar{A}x\leq\bar{b}$ be the system containing all inequalities of $Ax\leq b+\frac{1}{2}\mathbf{1}$ that define facets of $B\cap P^{\prime}$. Let $P_{0}=\{x\in\mathbb{R}^{n}\,|\,\bar{A}x\leq\bar{b}\}$. Then $P_{0}$ satisfies $i),ii),iii)$. See Figure 2 for an illustration.

It will be more convenient to write $P_{0}$ as intersection of the half-spaces defining the facets of $P_{0}$, $P_{0}=\cap_{H\in\mathcal{F}_{0}}H$. We construct a sequence of rational polyhedra $P_{0}\subset P_{1}\subset\ldots\subset P_{t}$ such that $P_{i}$ satisfies $i),ii),iii)$, $i=1,\ldots,t$, and such that $P_{t}$ satisfies $iv)$. Given $P_{i}$, we construct $P_{i+1}$ as follows. Let $P_{i}=\cap_{H\in\mathcal{F}_{i}}H$, where $\mathcal{F}_{i}$ is the set of half spaces defining facets of $P_{i}$. Let $\bar{H}$ be a half-space in $\mathcal{F}_{i}$ defining a facet of $B\cap P_{i}$ that does not contain an integral point in its relative interior; if no such $\bar{H}$ exists, then $P_{i}$ satisfies $iv)$ and we are done. If $B\cap\bigcap_{H\in\mathcal{F}_{i}\setminus\{\bar{H}\}}H$ does not contain any integral point in its interior, let $\mathcal{F}_{i+1}=\mathcal{F}_{i}\setminus\{\bar{H}\}$. Otherwise, since $P_{i}$ is rational, among all integral points in the interior of $B\cap\bigcap_{H\in\mathcal{F}_{i}\setminus\{\bar{H}\}}H$ there exists one, say $\bar{x}$, at minimum distance from $\bar{H}$. Let $H^{\prime}$ be the half-space containing $\bar{H}$ with $\bar{x}$ on its boundary. Let $\mathcal{F}_{i+1}=\mathcal{F}_{i}\setminus\{\bar{H}\}\cap\{H^{\prime}\}$. Observe that $H^{\prime}$ defines a facet of $P_{i+1}$ since $\bar{x}$ is in the interior of $B\cap\bigcap_{H\in\mathcal{F}_{i+1}\setminus\{H^{\prime}\}}H$ and it is on the boundary of $H^{\prime}$. So $i),ii),iii)$ are satisfied and $P_{i+1}$ has fewer facets that violate $iv)$ than $P_{i}$. $\diamond$

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

[TABLE]

where $H_{i}$ are halfspaces. Let $\bar{H}_{i}=\mathbb{R}^{n}\setminus\mathrm{\bf int}(H_{i})$, $i=1,\ldots,k$.

Claim 2**.**

$B$* is a polyhedron.*

We first show that, for $i=1,\ldots,k$, $\mathrm{\bf int}(B)\cap(\bar{H}_{i}\cap\mathrm{conv}(S))=\emptyset$. Consider $y\in\mathrm{\bf int}(B)\cap\bar{H}_{i}$. Since $y\in\bar{H}_{i}$ and $K$ is contained in $T$, $y\notin\mathrm{\bf int}(K)$. Since $K=B\cap P$ and $y\in\mathrm{\bf int}(B)\setminus\mathrm{\bf int}(K)$, it follows that $y\notin\mathrm{\bf int}(P)$. Hence $y\notin\mathrm{conv}(S)$ because $\mathrm{conv}(S)\subseteq\mathrm{\bf int}(P)$.

Thus, for $i=1,\ldots,k$, there exists a hyperplane separating $B$ and $\bar{H}_{i}\cap\mathrm{conv}(S)$. Hence there exists a halfspace $K_{i}$ such that $B\subset K_{i}$ and $\bar{H}_{i}\cap\mathrm{conv}(S)$ is disjoint from the interior of $K_{i}$. We claim that the set $B^{\prime}=\cap_{i=1}^{k}K_{i}$ is $S$-free. Indeed, let $y\in S$. Then $y$ is not interior of $T$. Since $y\in\mathrm{conv}(S)$ and $\mathrm{conv}(S)\subseteq\mathrm{\bf int}(P)$, $y$ is in the interior of $P$. Hence, by (5), there exists $i\in\{1,\ldots,k\}$ such that $y$ is not in the interior of $H_{i}$. Thus $y\in\bar{H}_{i}\cap\mathrm{conv}(S)$. By construction, $y$ is not in the interior of $K_{i}$, hence $y$ is not in the interior of $B^{\prime}$. Thus $B^{\prime}$ is an $S$-free convex set containing $B$. Since $B$ is maximal, $B^{\prime}=B$. $\diamond$

Claim 3**.**

$\mathrm{lin}(K)=\mathrm{rec}(K)$.

Let $r\in\mathrm{rec}(K)$. We show $-r\in\mathrm{rec}(K)$. By Lemma 4 applied to $\mathbb{Z}^{n}$, $K+\langle r\rangle$ is lattice-free. We observe that $B+\langle r\rangle$ is $S$-free. If not, let $y\in S\cap\mathrm{\bf int}(B+\langle r\rangle)$. Since $S\subseteq\mathrm{\bf int}(P)$, $y\in\mathrm{\bf int}(P+\langle r\rangle)$, hence $y\in\mathrm{\bf int}(K+\langle r\rangle)$, a contradiction. Hence, by maximality of $B$, $B=B+\langle r\rangle$. Thus $-r\in\mathrm{rec}(B)$. Suppose that $-r\notin\mathrm{rec}(P)$. Then there exists a facet $F$ of $P$ that is not parallel to $r$. By construction, $F\cap K$ is a facet of $K$ containing an integral point $\bar{x}$ in its relative interior. The point $\bar{x}$ is then in the interior of $K+\langle r\rangle$, a contradiction. $\diamond$

Claim 4**.**

$\mathrm{lin}(K)$* is rational.*

Consider the maximal lattice-free convex set $T$ containing $K$ considered earlier. By Theorem 1, $\mathrm{lin}(T)=\mathrm{rec}(T)$, and $\mathrm{lin}(T)$ is rational. Clearly $\mathrm{lin}(T)\supseteq\mathrm{lin}(K)$. Hence, if the claim does not hold, there exists a rational vector $r\in\mathrm{lin}(T)\setminus\mathrm{lin}(K)$. By (5), $r\in\mathrm{lin}(P)$. Since $K=B\cap P$, $r\notin\mathrm{lin}(B)$. Hence $B\subset B+\langle r\rangle$. We will show that $B+\langle r\rangle$ is $S$-free, contradicting the maximality of $B$. Suppose there exists $y\in S\cap\mathrm{\bf int}(B+\langle r\rangle)$. Since $\mathrm{conv}(S)\subseteq\mathrm{\bf int}(P)$, $y\in\mathrm{\bf int}(P)\subseteq\mathrm{\bf int}(P)+\langle r\rangle$. Therefore $y\in\mathrm{\bf int}(B\cap P)+\langle r\rangle$. Since $B\cap P\subseteq T$, then $y\in\mathrm{\bf int}(T)+\langle r\rangle=\mathrm{\bf int}(T)$ where the last equality follows from $r\in\mathrm{lin}(T)$. This contradicts the fact that $T$ is lattice-free. $\diamond$

By Lemma 4 and by the maximality of $B$, $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))=\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$.

Claim 5**.**

$\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$* is rational.*

Since $\mathrm{lin}(K)$ and $\mathrm{rec}(\mathrm{conv}(S))$ are both rational, we only need to show $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))=\mathrm{lin}(K)\cap\mathrm{rec}(\mathrm{conv}(S))$. The “$\supseteq$” direction follows from $B\supseteq K$. For the other direction, note that, since $\mathrm{conv}(S)\subseteq P$, we have $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))\subseteq\mathrm{lin}(B)\cap\mathrm{rec}(P)=\mathrm{lin}(B\cap P)=\mathrm{lin}(K)$, hence $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))\subseteq\mathrm{lin}(K)\cap\mathrm{rec}(\mathrm{conv}(S))$. $\diamond$

Claim 6**.**

Every facet of $B$ contains a point of $S$ in its relative interior.

Let $L$ be the linear space generated by $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$. By Claim 5, $L$ is rational. Let $B^{\prime}$, $S^{\prime}$, $\Lambda$ be the projections of $B$, $S$, $\mathbb{Z}^{n}$, respectively, onto $L^{\bot}$. Since $L$ is rational, $\Lambda$ is a lattice of $L^{\bot}$ and $S^{\prime}=\mathrm{conv}(S^{\prime})\cap\Lambda$. Also, $B^{\prime}$ is a maximal $S^{\prime}$-free convex set of $L^{\bot}$, since for any $S^{\prime}$-free set $D$ of $L^{\bot}$, $D+L$ is $S$-free. Note that $\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))=\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$ implies that $B^{\prime}\cap\mathrm{conv}(S^{\prime})$ is bounded. Otherwise there is an unbounded direction $d\in L^{\bot}$ in $\mathrm{rec}(B^{\prime})\cap\mathrm{rec}(\mathrm{conv}(S^{\prime}))$ and so $d+l\in\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$ for some $l\in L$. Since $\mathrm{rec}(B)\cap\mathrm{rec}(\mathrm{conv}(S))=\mathrm{lin}(B)\cap\mathrm{rec}(\mathrm{conv}(S))$, this would imply that $d\in L$ which is a contradiction. Let $B^{\prime}=\{x\in L^{\bot}\,|\,\alpha_{i}x\leq\beta_{i},\,i=1,\ldots,t\}$. Given $\varepsilon>0$, let $\bar{B}=\{x\in L^{\bot}\,|\,\alpha_{i}x\leq\beta_{i},\,i=1,\ldots,t-1,\,\alpha_{t}x\leq\beta_{t}+\varepsilon\}$. The polyhedron $\mathrm{conv}(S^{\prime})\cap\bar{B}$ is a polytope since it has the same recession cone as $\mathrm{conv}(S^{\prime})\cap B^{\prime}$. The polytope $\mathrm{conv}(S^{\prime})\cap\bar{B}$ contains points of $S^{\prime}$ in its interior by the maximality of $B^{\prime}$. Since $\Lambda$ is a lattice of $L^{\bot}$, $\mathrm{\bf int}(\mathrm{conv}(S^{\prime})\cap\bar{B})$ has a finite number of points in $S^{\prime}$, hence there exists one minimizing $\alpha_{t}x$, say $z$. By construction, the polyhedron $B^{\prime\prime}=\{x\in L^{\bot}\,|\,\alpha_{i}x\leq\beta_{i},\,i=1,\ldots,t-1,\,\alpha_{t}x\leq\alpha_{t}z\}$ does not contain any point of $S^{\prime}$ in its interior and contains $B^{\prime}$. By the maximality of $B^{\prime}$, $B^{\prime}=B^{\prime\prime}$ hence $B^{\prime}$ contains $z$ in its relative interior, and $B$ contains a point of $S$ in its relative interior. ∎

Corollary 5**.**

For every maximal $S$-free convex set $B$ there exists a maximal lattice-free convex set $K$ such that, for every facet $F$ of $B$, $F\cap K$ is a facet of $K$.

Proof.

Let $K$ be defined as in Claim 1 in the proof of Theorem 2. It follows from the proof that $K$ 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, $S\subseteq\mathbb{Z}^{n}$ is a set of integral points in some rational polyhedron of $\mathbb{R}^{n}$ such that $\dim(S)=n$, and $f$ is a point in $\mathrm{conv}(S)\setminus\mathbb{Z}^{n}$. Let $R_{f,S}$ be the set of all infinite dimensional vectors $s=(s_{r})_{r\in\mathbb{R}^{n}}$ such that

[TABLE]

where $s$ has finite support means that $s_{r}$ is zero for all but a finite number of $r\in\mathbb{R}^{n}$.

A function $\psi\,:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ is valid (with respect to $f$ and $S$) if the linear inequality

[TABLE]

is satisfied by every $s\in R_{f,S}$. Note that this definition coincides with the one we gave in the introduction.

Given two functions $\psi,\psi^{\prime}$ we say that $\psi^{\prime}$ dominates $\psi$ if $\psi^{\prime}(r)\leq\psi(r)$ for all $r\in\mathbb{R}^{n}$. A valid function $\psi$ is minimal if there is no valid function $\psi^{\prime}\neq\psi$ that dominates $\psi$.

Theorem 6**.**

For every minimal valid function $\psi$, there exists a maximal $S$-free convex set $B$ with $f$ in its interior such that $\psi_{B}$ dominates $\psi$. Furthermore, if $B$ is a maximal $S$-free convex set containing $f$ in its interior, then $\psi_{B}$ 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 $\psi$, define the following function $\bar{\psi}$. For all $\bar{r}\in\mathbb{R}^{n}$, let $\bar{\psi}(\bar{r})=\inf\{\sum_{r\in\mathbb{R}^{n}}\psi(r)s_{r}\,|\,\sum_{r\in\mathbb{R}^{n}}rs_{r}=\bar{r},\,s\geq 0\mbox{ with finite support}\}$. Following the proof of Lemma 18 in [4] one can show that $\bar{\psi}$ is a valid sublinear function that dominates $\psi$. ∎

Given a valid sublinear function $\psi$, the set $B_{\psi}=\{x\in\mathbb{R}^{n}\,|\,\psi(x-f)\leq 1\}$ is closed, convex, and contains $f$ in its interior. Since $\psi$ is a valid function, $B_{\psi}$ is $S$-free. Indeed the interior of $B_{\psi}$ is $\mathrm{\bf int}(B_{\psi})=\{x\in\mathbb{R}^{n}\,:\,\psi(x-f)<1\}$. Its boundary is $\mathrm{\bf bd}(B_{\psi})=\{x\in\mathbb{R}^{n}\,:\,\psi(x-f)=1\}$, and its recession cone is $\mathrm{rec}(B_{\psi})=\{x\in\mathbb{R}^{n}\,:\,\psi(x-f)\leq 0\}$.

Before proving Theorem 6, we need the following general theorem about sublinear functions. Let $K$ be a closed, convex set in $\mathbb{R}^{n}$ with the origin in its interior. The polar of $K$ is the set $K^{*}=\{y\in\mathbb{R}^{n}\,|\,ry\leq 1\mbox{ for all }r\in K\}$. Clearly $K^{*}$ is closed and convex, and since $0\in\mathrm{\bf int}(K)$, it is well known that $K^{*}$ is bounded. In particular, $K^{*}$ is a compact set. Also, since $0\in K$, $K^{**}=K$. Let

[TABLE]

Note that $\hat{K}$ is contained in the relative boundary of $K^{*}$. Let $\rho_{K}:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ be defined by

[TABLE]

It is easy to show that $\rho_{K}$ is sublinear.

Theorem 8** (Basu et al. [5]).**

Let $K\subset\mathbb{R}^{n}$ be a closed convex set containing the origin in its interior. Then $K=\{r\in\mathbb{R}^{n}\,|\,\rho_{K}(r)\leq 1\}$. Furthermore, for every sublinear function $\sigma$ such that $K=\{r\,|\,\sigma(r)\leq 1\}$, we have $\rho_{K}(r)\leq\sigma(r)$ for every $r\in\mathbb{R}^{n}$.

Remark 9**.**

Let $K\subset\mathbb{R}^{n}$ be a polyhedron containing the origin in its interior. Let $a_{1},\ldots,a_{t}\in\mathbb{R}^{n}$ such that $K=\{r\in\mathbb{R}^{n}\,|\,a_{i}r\leq 1,\,i=1,\ldots,t\}$. Then $\rho_{K}(r)=\max_{i=1,\ldots,t}a_{i}r$.

Proof.

The polar of $K$ is $K^{*}=\mathrm{conv}\{0,a_{1},\ldots,a_{t}\}$ (see Theorem 9.1 in Schrijver [10]). Furthermore, $\hat{K}$ is the union of all the facets of $K^{*}$ that do not contain the origin, therefore

[TABLE]

for all $r\in\mathbb{R}^{n}$. ∎

Remark 10**.**

Let $B$ be a closed $S$-free convex set in $\mathbb{R}^{n}$ with $f$ in its interior, and let $K=B-f$. Then $\rho_{K}$ is a valid function.

• Proof: Let $s\in R_{f,S}$. Then $x=f+\sum_{r\in\mathbb{R}^{n}}rs_{r}$ is in $S$, therefore $x\notin\mathrm{\bf int}(B)$ because $B$ is $S$-free. By Theorem 8, $\rho_{K}(x-f)\geq 1$. Thus

[TABLE]

where the second inequality follows from the sublinearity of $\rho_{K}$. $\Box$

Lemma 11**.**

Let $C$ be a closed $S$-free convex set containing $f$ in its interior, and let $K=C-f$. There exists a maximal $S$-free convex set $B=\{x\in\mathbb{R}^{n}\,|\,a_{i}(x-f)\leq 1,\,\,i=1,\ldots,k\}$ such that $a_{i}\in\mathrm{\bf cl}(\mathrm{conv}(\hat{K}))$ for $i=1,\ldots,k$.

Proof.

Since $C$ is an $S$-free convex set, it is contained in some maximal $S$-free convex set $T$. The set $T$ satisfies one of the statements (i)-(iii) of Theorem 2. By assumption, $f\in\mathrm{conv}(S)$ and $f$ is in the interior of $C$. Since $\dim(S)=n$, $\mathrm{conv}(S)$ is a full dimensional polyhedron, thus $\mathrm{\bf int}(C\cap\mathrm{conv}(S))\neq\emptyset$. This implies that $\mathrm{\bf int}(T\cap\mathrm{conv}(S))\neq\emptyset$, hence case (i) applies.

Thus $T$ is a polyhedron and $\mathrm{rec}(T\cap\mathrm{conv}(S))=\mathrm{lin}(T)\cap\mathrm{rec}(\mathrm{conv}(S))$ is rational. Let us choose $T$ such that the dimension of $\mathrm{lin}(T)$ is largest possible.

Since $T$ is a polyhedron containing $f$ in its interior, there exists $D\in\mathbb{R}^{t\times q}$ and $b\in\mathbb{R}^{t}$ such that $b_{i}>0$, $i=1,\ldots,t$, and $T=\{x\in\mathbb{R}^{n}\,|\,D(x-f)\leq b\}$. Without loss of generality, we may assume that $\sup_{x\in C}d_{i}(x-f)=1$ where $d_{i}$ denotes the $i$th row of $D$, $i=1,\ldots,t$. By our assumption, $\sup_{r\in K}d_{i}r=1$. Therefore $d_{i}\in K^{*}$, since $d_{i}r\leq 1$ for all $r\in K$. Furthermore $d_{i}\in\mathrm{\bf cl}(\hat{K})$, since $\sup_{r\in K}d_{i}r=1$.

Let $P=\{x\in\mathbb{R}^{n}\,|\,D(x-f)\leq e\}$. Note that $\mathrm{lin}(P)=\mathrm{lin}(T)$. By our choice of $T$, $P+\langle r\rangle$ is not $S$-free for any $r\in\mathrm{rec}(\mathrm{conv}(S))\setminus\mathrm{lin}(P)$, otherwise $P$ would be contained in a maximal $S$-free convex set whose lineality space contains $\mathrm{lin}(T)+\langle r\rangle$, a contradiction.

Let $L=\langle\mathrm{rec}(P\cap\mathrm{conv}(S))\rangle$. Since $\mathrm{lin}(P)=\mathrm{lin}(T)$, $L$ is a rational space. Note that $L\subseteq\mathrm{lin}(P)$, implying that $d_{i}\in L^{\bot}$ for $i=1,\ldots,t$.

We observe next that we may assume that $P\cap\mathrm{conv}(S)$ is bounded. Indeed, let $\bar{P}$, $\bar{S}$, $\Lambda$ be the projections onto $L^{\bot}$ of $P$, $S$, and $\mathbb{Z}^{n}$, respectively. Since $L$ is a rational space, $\Lambda$ is a lattice of $L^{\bot}$ and $\bar{S}=\mathrm{conv}(\bar{S})\cap\Lambda$. Note that $\bar{P}\cap\mathrm{conv}(\bar{S})$ is bounded, since $L\supseteq\mathrm{rec}(P\cap\mathrm{conv}(S))$. If we are given a maximal $\bar{S}$-free convex set $\bar{B}$ in $L^{\bot}$ such that $\bar{B}=\{x\in L^{\bot}\,|\,a_{i}(x-f)\leq 1,\,i=1,\ldots,h\}$ and $a_{i}\in\mathrm{conv}\{d_{1},\ldots,d_{t}\}$ for $i=1,\ldots,h$, then $B=\bar{B}+L$ is the set $B=\{x\in\mathbb{R}^{n}\,|\,a_{i}(x-f)\leq 1,\,i=1,\ldots,h\}$. Since $\bar{B}$ contains a point of $\bar{S}$ in the relative interior of each of its facets, $B$ contains a point of $S$ in the relative interior of each of its facets, thus $B$ is a maximal $S$-free convex set.

Thus we assume that $P\cap\mathrm{conv}(S)$ is bounded, so $\dim(L)=0$. If all facets of $P$ contain a point of $S$ in their relative interior, then $P$ is a maximal $S$-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 $P$ without any point of $S$ in its relative interior with an inequality which is a convex combination of the inequalities of $D(x-f)\leq e$, such that the new polyhedron thus obtained is $S$-free and has one fewer facet without points of $S$ in its relative interior. More formally, suppose the facet of $P$ defined by $d_{1}(x-f)\leq 1$ does not contain any point of $S$ in its relative interior. Given $\lambda\in[0,1]$, let

[TABLE]

Note that $P(1)=P$ and $P(0)$ is obtained from $P$ by removing the inequality $d_{1}(x-f)\leq 1$. Furthermore, given $0\leq\lambda^{\prime}\leq\lambda^{\prime\prime}\leq 1$, we have $P(\lambda^{\prime})\supseteq P(\lambda^{\prime\prime})$.

Let $r_{1},\ldots,r_{m}$ be generators of $\mathrm{rec}(\mathrm{conv}(S))$. Note that, since $P\cap\mathrm{conv}(S)$ is bounded, for every $j=1,\ldots,m$ there exists $i\in\{1,\ldots,t\}$ such that $d_{i}r_{j}>0$. Let $r_{1},\ldots,r_{h}$ be the generators of $\mathrm{rec}(\mathrm{conv}(S))$ satisfying

[TABLE]

Note that, if no such generators exist, then $P(0)\cap\mathrm{conv}(S)$ is bounded. Otherwise $P(\lambda)\cap\mathrm{conv}(S)$ is bounded if and only if, for $j=1,\ldots,h$

[TABLE]

This is the case if and only if $\lambda>\lambda^{*}$, where

[TABLE]

Let $r^{*}$ be one of the vectors $r_{1},\ldots,r_{h}$ attaining the maximum in the previous equation. Then $r^{*}\in\mathrm{rec}(P(\lambda^{*}\cap\mathrm{conv}(S)$.

Note that $P(\lambda^{*})$ is not $S$-free otherwise $P(\lambda^{*})+\langle r^{*}\rangle$ is $S$-free by Lemma 4, and so is $P+\langle r^{*}\rangle$, a contradiction.

Thus $P(\lambda^{*})$ contains a point of $S$ in its interior. That is, there exists a point $\bar{x}\in S$ such that $[\lambda^{*}d_{1}+(1-\lambda^{*})d_{2}](\bar{x}-f)<1$ and $d_{i}(\bar{x}-f)<1$ for $i=2,\ldots,t$. Since $P$ is $S$-free, $d_{1}(\bar{x}-f)>1$. Thus there exists $\bar{\lambda}>\lambda^{*}$ such that $[\bar{\lambda}d_{1}+(1-\bar{\lambda})d_{2}](\bar{x}-f)=1$. Note that, since $P(\bar{\lambda})\cap\mathrm{conv}(S)$ is bounded, there is a finite number of points of $S$ in the interior of $P(\bar{\lambda})$. So we may choose $\bar{x}$ such that $\bar{\lambda}$ is maximum. Thus $P(\bar{\lambda})$ is $S$-free and $\bar{x}$ is in the relative interior of the facet of $P(\bar{\lambda})$ defined by $[\bar{\lambda}d_{1}+(1-\bar{\lambda})d_{2}](x-f)\leq 1$.

Note that, for $i=2,\ldots,t$, if $d_{i}(x-f)\leq 1$ defines a facet of $P$ with a point of $S$ in its relative interior, then it also defines a facet of $P(\bar{\lambda})$ with a point of $S$ in its relative interior, because $P\subset P(\bar{\lambda})$. Thus repeating the above construction at most $t$ times, we obtain a set $B$ satisfying the lemma. ∎

Remark 12**.**

Let $C$ and $K$ be as in Lemma 11. Given any maximal $S$-free convex set $B=\{x\in\mathbb{R}^{n}\,|\,a_{i}(x-f)\leq 1,\,\,i=1,\ldots,k\}$ containing $C$, then $a_{1},\ldots,a_{k}\in K^{*}$. If $\mathrm{rec}(C)$ is not full dimensional, then the origin is not an extreme point of $K^{*}$. Since all extreme points of $K^{*}$ are contained in $\{0\}\cup\hat{K}$, in this case $\mathrm{\bf cl}(\mathrm{conv}(\hat{K}))=K^{*}$. Therefore, when $\mathrm{rec}(C)$ is not full dimensional, every maximal $S$-free convex set containing $C$ satisfies the statement of Lemma 11.

Proof of Theorem 6.

We first show that any valid function is dominated by a function of the form $\psi_{B}$, for some maximal $S$-free convex set $B$ containing $f$ in its interior.

Let $\psi$ be a valid function. By Lemma 7, we may assume that $\psi$ is sublinear. Let $K=\{r\in\mathbb{R}^{n}\,|\,\psi(r)\leq 1\}$, and let $\hat{K}$ be defined as in (8). Note that $K=B_{\psi}-f$. Thus, by Remark 10, $\sum_{r\in\mathbb{R}^{n}}\rho_{K}(r)s_{r}\geq 1$ is valid for $R_{f,S}$. Since $\psi$ is sublinear, it follows from Theorem 8 that $\rho_{K}(r)\leq\psi(r)$ for every $r\in\mathbb{R}^{n}$.

By Lemma 11, there exists a maximal $S$-free convex set $B=\{x\in\mathbb{R}^{n}\,|\,a_{i}(x-f)\leq 1,\,\,i=1,\ldots,k\}$ such that $a_{i}\in\mathrm{\bf cl}(\mathrm{conv}(\hat{K}))$ for $i=1,\ldots,k$.

Then

[TABLE]

This shows that $\psi_{B}$ dominates $\psi$ for all $r\in\mathbb{R}^{n}$.

To complete the proof of the theorem, we need to show that, given a maximal $S$-free convex set $B$, the function $\psi_{B}$ is minimal. Consider any valid function $\psi$ dominating $\psi_{B}$. Then $B_{\psi}\supseteq B$ and $B_{\psi}$ is $S$-free. By maximality of $B$, $B=B_{\psi}$. By Theorem 8 and Remark 9, $\psi_{B}(r)\leq\psi(r)$ for all $r\in\mathbb{R}^{n}$, proving $\psi=\psi_{B}$.

∎

Example 13**.**

We illustrate the ideas behind the proof in the following two-dimensional example. Consider $f=(\frac{1}{4},\frac{1}{2})$ and $S=\{(x_{1},x_{2})\,|\,x_{1}\geq 0)\}$. See Figure 3. Then the function $\psi(r)=\max\{4r_{1}+8r_{2},4r_{1}-8r_{2}\}$ is a valid linear inequality for $R_{f,S}$. The corresponding $B_{\psi}$ is $\{(x_{1},x_{2})\,|\,4(x_{1}-\frac{1}{4})+8(x_{2}-\frac{1}{2})\leq 1,4(x_{1}-\frac{1}{4})-8(x_{2}-\frac{1}{2})\leq 1\}$. Note that $B_{\psi}$ is not a maximal $S$-free convex set and it corresponds to $C$ in Lemma 11. Following the procedure outlined in the proof, we obtain the maximal $S$-free convex set $B=\{(x_{1},x_{2})\,|\,4(x_{1}-\frac{1}{4})+4(x_{2}-\frac{1}{2})\leq 1,4(x_{1}-\frac{1}{4})-4(x_{2}-\frac{1}{2})\leq 1\}$. Then, $\psi_{B}(r)=\max\{4r_{1}+4r_{2},4r_{1}-4r_{2}\}$ and $\psi_{B}$ dominates $\psi$.

Remark 14**.**

Note that $\psi$ is nonnegative if and only if $\mathrm{rec}(B_{\psi})$ is not full-dimensional. It follows from Remark 12 that, for every maximal $S$-free convex set $B$ containing $B_{\psi}$, we have $\psi_{B}(r)\leq\psi(r)$ for every $r\in\mathbb{R}^{n}$ when $\psi$ is nonnegative.

A statement similar to the one of Theorem 6 was shown by Borozan-Cornuéjols [6] for a model similar to (3) when $S=\mathbb{Z}^{n}$ and the vectors $s$ are elements of $\mathbb{R}^{\mathbb{Q}^{n}}$. In this case, it is easy to show that, for every valid inequality $\sum_{r\in\mathbb{Q}^{n}}\psi(r)s_{r}\geq 1$, the function $\psi:\,\mathbb{Q}^{n}\rightarrow\mathbb{R}$ is nonnegative. Remark 14 explains why in this context it is much easier to prove that minimal inequalities arise from maximal lattice-free convex sets.