Unique lifting of integer variables in minimal inequalities

Transcription

1 Unique liting o integer variables in minimal inequalities Amitabh Basu 1, Manoel Campêlo 2,6, Michele Conorti 3,8, Gérard Cornuéjols 4,7, Giacomo Zambelli 5,8 December 6, 2011 Abstract This paper contributes to the theory o cutting planes or mixed integer linear programs (MILPs. Minimal valid inequalities are well understood or a relaxation o an MILP in tableau orm where all the nonbasic variables are continuous; they are derived using the gauge unction o maximal lattice-ree convex sets. In this paper we study liting unctions or the nonbasic integer variables starting rom such minimal valid inequalities. We characterize precisely when the lited coeicient is equal to the coeicient o the corresponding continuous variable in every minimal liting (This result irst appeared in the proceedings o IPCO The answer is a nonconvex region that can be obtained as a inite union o convex polyhedra. We then establish a necessary and suicient condition or the uniqueness o the liting unction. Keywords: mixed integer linear programming, minimal valid inequality, liting Mathematics Subject Classiication (2000: 90C11 1 Introduction In the context o mixed integer linear programming, there has been a renewed interest recently in the study o cutting planes that cut o a basic solution o the linear programming relaxation. More precisely, consider a mixed integer linear set in which the variables are partitioned into a basic set B and a nonbasic set N, and K B N indexes the integer variables: x i = i j N a ijx j or i B x 0 x k Z or k K. (1 1 Department o Mathematics, University o Caliornia, Davis, CA abasu@math.ucdavis.edu 2 Departamento de Estatística e Matemática Aplicada, Universidade Federal do Ceará, Brazil. mcampelo@lia.uc.br 3 Dipartimento di Matematica Pura e Applicata, Universitá di Padova, Via Trieste 63, Padova, Italy. conorti@math.unipd.it 4 Tepper School o Business, Carnegie Mellon University, Pittsburgh, PA gc0v@andrew.cmu.edu 5 London School o Economics and Political Sciences, Houghton Street, London WC2A 2AE. G.Zambelli@lse.ac.uk 6 Supported by CNPq Brazil. 7 Supported by NSF grant CMMI and ONR grant N Supported by the Progetto di Eccellenza o the Fondazione Cassa Risparmio di Padova e Rovigo. 1

2 Let X be the relaxation o (1 obtained by dropping the nonnegativity restriction on all the basic variables x i, i B. The convex hull o X is the corner polyhedron introduced by Gomory [16] (see also [17]. Note that, or any i B \ K, the equation x i = i j N a ijx j can be removed rom the ormulation o X since it just deines variable x i. Thereore, throughout the paper, we will assume B K, i.e. all basic variables are integer. Andersen, Louveaux, Weismantel and Wolsey [2] studied the corner polyhedron when B = 2 and B = K, i.e. all nonbasic variables are continuous. They give a complete characterization o the corner polyhedron using intersection cuts (Balas [3] arising rom splits, triangles and quadrilaterals. This very elegant result has been extended to B > 2 and B = K by showing a correspondence between minimal valid inequalities and maximal lattice-ree convex sets [8], [11]. These results and their extensions [9], [15] are best described in an ininite model, which we motivate next. 1.1 The Ininite Model A classical amily o cutting planes or (1 is that o Gomory mixed integer cuts. For a given row i B o the tableau, the Gomory mixed integer cut is o the orm j N\K ψ(a ijx j + j N K π(a ijx j 1 where ψ and π are unctions given by simple ormulas. A nice eature o the Gomory mixed integer cut is that, or ixed i, the same unctions ψ, π are used or any possible choice o the a ij s in (1. It is well known that the Gomory mixed integer cuts are also valid or X. More generally, let a j be the vector with entries a ij, i B; we are interested in pairs (ψ, π o unctions such that the inequality j N\K ψ(aj x j + j N K π(aj x j 1 is valid or X or any possible number o nonbasic variables and any choice o the nonbasic coeicients a ij. Since we are interested in nonredundant inequalities, we can assume that the unction (ψ, π is pointwise minimal. While a general characterization o minimal valid unctions seems hopeless (see or example [7], when N K = the minimal valid unctions ψ are well understood in terms o maximal lattice-ree convex sets, as already mentioned. Starting rom such a minimal valid unction ψ, an interesting question is how to generate a unction π such that (ψ, π is valid and minimal. Recent papers [13], [14] study when such a unction π is unique. Here we prove two theorems that generalize and uniy results rom these two papers. In order to ormalize the concept o valid unction (ψ, π, we introduce the ollowing ininite model. In the setting below, we also allow urther linear constraints on the basic variables. Let S be the set o integral points in some rational polyhedron in R n such that dim(s = n (or example, S could be the set o nonnegative integer points. Let R n \ S. Consider the ollowing ininite relaxation o (1, introduced in [15]. x = + rs r + ry r, r R n r R n x S, s r R +, r R n, y r Z +, r R n, s, y have inite support where the nonbasic continuous variables have been renamed s and the nonbasic integer variables have been renamed y, and where an ininite dimensional vector has inite support i it (2 2

3 has a inite number o nonzero entries. Given two unctions ψ, π : R n R, (ψ, π is said to be valid or (2 i the inequality r R n ψ(rs r + r R n π(ry r 1 holds or every (x, s, y satisying (2. We also consider the ininite model where we only have continuous nonbasic variables. x = + r R n rs r x S, s r R +, r R n, s has inite support. A unction ψ : R n R is said to be valid or (3 i the inequality r R n ψ(rs r 1 holds or every (x, s satisying (3. Given a valid unction ψ or (3, a unction π is a liting o ψ i (ψ, π is valid or (2. One is interested only in (pointwise minimal valid unctions, since non-minimal ones are implied by some minimal valid unction. I ψ is a minimal valid unction or (3 and π is a liting o ψ such that (ψ, π is a minimal valid unction or (2 then we say that π is a minimal liting o ψ. It can be shown, using Zorn s Lemma, that or every liting π o ψ there exists some minimal liting π o ψ such that π π. 1.2 Sequence Independent Liting and Unique Liting Functions While minimal valid unctions or (3 have a simple characterization [9], minimal valid unctions or (2 are not well understood. A general idea to derive minimal valid unctions or (2 is to start rom some minimal valid unction ψ or (3, and construct a minimal liting π o ψ. While there is no general technique to compute such a minimal liting π, it is known that there exists a region R ψ, containing the origin in its interior, where ψ coincides with π or any minimal liting π. This latter act was proved by Dey and Wolsey [14] or the case o S = Z 2, and by Conorti, Cornuéjols and Zambelli [13] or the general case. In the latter paper, the authors describe the set R ψ in the case where ψ is deined by a simplicial maximal lattice-ree polytope. In this paper we give a precise description o the region R ψ (Theorem 4 in general (This result irst appeared in the proceedings o IPCO 2010 [6]. The importance o the region R ψ comes rom the act that or any ray r R ψ, the minimal liting coeicient π(r is unique, i.e. it is the same or every minimal liting. Thus, we get sequence independent liting coeicients or the rays r in R ψ. Moreover, these coeicients can be computed directly rom the unction ψ or which we have more direct tools. These ideas are related to the results o Balas and Jeroslow [4]. Furthermore, it is remarked in [13] that, i π is a minimal liting o ψ, then π(r = π(r or every r, r R n such that r r Z n lin(conv(s. Thereore the coeicients o any minimal liting π are uniquely determined in the region R ψ +(Z n lin(conv(s (throughout this paper, we use + to denote the Minkowski sum o two sets. In particular, whenever R n can be covered by translating R ψ by integer vectors in lin(conv(s, the unction ψ has a unique minimal liting π. As mentioned above, i ψ has a unique minimal liting, we can compute the best possible coeicients or all the integer variables in our problem, in a sequence independent manner. Thus, it is very useul to recognize the minimal valid unctions ψ with unique minimal litings. The second main result in this paper (Theorem 5 is to show that, or the case when S = Z n, the covering property is in act a necessary and suicient (3 3

4 condition or the uniqueness o minimal litings : i R ψ +Z n R n, then there are at least two distinct minimal litings or ψ. Theorem 5 thus converts the question o recognizing minimal valid unctions ψ that have a unique liting unction to the geometric question o covering R n by lattice translates o the region R ψ. This equivalence is utilized by Basu, Cornuéjols and Köppe to study the unique liting properties o certain amilies o minimal valid unctions [10]. 2 Overview o the Main Results Let S be the set o integral points in some rational polyhedron in R n such that dim(s = n (where dim(s denotes the dimension o the aine hull o S, and let R n \ S. To state our main results, we need to explain the characterization o minimal valid unctions or (3. Given a polyhedron P, we denote by rec(p and lin(p the recession cone and the lineality space o P. 2.1 Minimal Valid Functions and Maximal Lattice-Free Convex Sets We say that a convex set B R n is S-ree i B does not contain any point o S in its interior. When S = Z n, S-ree convex sets are called lattice-ree convex sets. A set B is a maximal S-ree convex set i it is an S-ree convex set that is not properly contained in any S-ree convex set. It was proved in [9] that maximal S-ree convex sets are polyhedra containing a point o S in the relative interior o each acet. The ollowing characterization o maximal S-ree convex sets and the subsequent remark will be needed in the proos, but the reader can skip them or now. Theorem 1. [9] A ull-dimensional convex set B is a maximal S-ree convex set i and only i B is a polyhedron that does not contain any point o S in its interior and each acet o B contains a point o S in its relative interior. Furthermore i B conv(s has nonempty interior, then lin(b contains rec(b conv(s implying that rec(b conv(s = lin(b conv(s, and lin(b conv(s is a rational subspace. Remark 2. The proo o Theorem 1 in [9] implies the ollowing. Given a maximal S-ree convex set B, there exists δ > 0 such that no point o S \ B is at a distance less than δ rom B. Given an S-ree polyhedron B R n containing in its interior, B can be uniquely written in the orm B = {x R n : a i (x 1, i I}, (4 where I is a inite set o indices in one-to-one correspondence with the acets o B. Let ψ B : R n R be the unction deined by ψ B (r = max i I a ir, r R n. (5 Theorem 3. [9] I B is a maximal S-ree convex set containing in its interior, then ψ B is a minimal valid unction or (3. Conversely, let ψ be a minimal valid unction or (3. Then the set B ψ := {x R n ψ(x 1} 4

5 is a maximal S-ree convex set containing in its interior, and ψ = ψ Bψ. It ollows easily rom the ormula in (5 and Theorem 3 that minimal valid unctions or (3 are sublinear. In particular, ψ is subadditive, i.e., ψ(r 1 + ψ(r 2 ψ(r 1 + r 2 or all r 1, r 2 R n (see also [9]. 2.2 Minimal Liting Functions This paper has two main contributions, which we state next. Given a minimal valid unction ψ or (3, B ψ deined in Theorem 3 is a maximal S-ree convex set containing in its interior. Following (4-(5, it can be uniquely written as B ψ = {x R n a i (x 1, i I} and so ψ(r = max i I a i r. Given x R n, let R(x := {r R n ψ(r + ψ(x r = ψ(x }. We deine R ψ := {r R n π(r = ψ(r or all minimal litings π o ψ}. Theorem 4. Let ψ be a minimal valid unction or (3. Then R ψ = x S B ψ R(x. Figure 1 illustrates the region R ψ or several examples. The second contribution is to give a necessary and suicient condition or the existence o a unique minimal liting unction in the case S = Z n. Theorem 5. Let S = Z n, and let ψ be a minimal valid unction or (3. There exists a unique minimal liting π o ψ i and only i R ψ + Z n = R n. The proos o Theorems 4 and 5 are given in Sections 4 and 5, respectively. We conclude this section with two propositions, used in the proo o Theorem 5, that are o independent interest. The irst gives a geometric description and properties o the regions R(x and R ψ, while the second (Proposition 7 states that all minimal litings are continuous unctions. Proposition 6. Let ψ be a minimal valid unction or (3 and let B ψ = {x R n : a i (x 1, i I}. Let L ψ := {r R n a i r = a j r or all i, j I}. Then i For all x R n, R(x is a polyhedron, namely R(x = {r R n a i r + a j (x r ψ(x or all i, j I}. ii rec(r(x = lin(r(x = L ψ or every x R n. iii R(x = R(x or every x, x R n such that x x L ψ. iv R ψ is a inite union o polyhedra. v L ψ is contained in the interior o R ψ. 5

6 x 1 R(x 1 x 3 R(x 3 R(x 2 x 2 B ψ B ψ R(x 1 x 1 x 2 R(x 2 R(x 6 x 6 R(x 5 R(x 4 x 3 x 4 x 5 (a A maximal Z 2 -ree triangle with three integer points R(x 3 (b A maximal Z 2 -ree triangle with integer vertices x 3 R(x 3 x 1 x 2 R(x 2 R(x 1 x 1 x 2 R(x 2 R(x 1 B ψ B ψ l 1 l (c A wedge (d A truncated wedge Figure 1: Regions R(x or some maximal S-ree convex sets B in the plane and or x S B. The thick dark line indicates the boundary o B ψ. For a particular x, the dark gray regions denote R(x translated by. The jagged lines in a region indicate that it extends to ininity. For example, in Figure 1(c, R(x 1 is the strip between lines l 1 and l. Figure 1(b shows an example where R(x is ull-dimensional or x 2, x 4, x 6, but is not ull-dimensional or x 1, x 3, x 5. 6

7 Proo. i Given x R n, let h I such that ψ(x = a h (x. Then, or all r R n, ψ(x = a h (x = a h r+a h (x r max i I a ir+max a i(x r = ψ(r+ψ(x r. i I It ollows rom the above and rom the deinition o R(x that r R(x i and only i max i I a i r+max i I a i (x r ψ(x. That is, r R(x i and only i a i r+a j (x r ψ(x or all i, j I. ii Let x R n. By i, a vector r R n belongs to rec(r(x i and only i a i r a j r 0 or all i, j I. The latter condition is veriied i and only i, or all i, j I, r satisies a i r a j r 0 and a j r a i r 0, that is, i and only i r L ψ. This shows that lin(r(x = rec(r(x = L ψ. iii I x x L ψ, then, or all i I, a i (x x = α or some constant α. For all r R n, it ollows that a i (x r = a i (x x +a i (x r = α+a i (x r, i I. Choosing r = 0, we also have that a i (x = α+a i (x or all i I and so ψ(x = ψ(x. Thereore, or all i, j I, a i r + a j (x r ψ(x i and only i a i r + a j (x r ψ(x. By i, this implies that r R(x i and only i r R(x. iv Let B := B ψ and let L := rec(b conv(s. By Theorem 1, L is a rational subspace contained in lin(b. Since lin(b = {r R n a i r = 0 or all i I}, we have that L lin(b L ψ. Let P be the projection o B onto the orthogonal complement L o L, and let Q be the projection o conv(s onto L. Since L = rec(b conv(s, it ollows that P Q is a polytope. Given two elements x, x S whose orthogonal projections onto L coincide, it ollows that x x L L ψ, and thereore by iii R(x = R(x. It ollows that the number o sets R(x, x S B, is at most the cardinality o the orthogonal projection S o S B onto L. Let Λ be the orthogonal projection o Z n onto L. Since L is a rational space, it ollows that Λ is a lattice (see or example [5]. In particular, there exists ε > 0 such that y z ε or all y, z Λ. Since P Q is a polytope, it ollows that P Q Λ is inite. Since S P Q Λ, it ollows that S is a inite set. We conclude that the amily o polyhedra R(x, x S B, has a inite number o elements, thus R ψ = x S B R(x is the union o a inite number o polyhedra. v By ii, or every r R ψ, {r} + L ψ is contained in R ψ. It was proved in [13] that R ψ contains the origin in its interior. That is, there exists ε > 0 such that r R ψ or all r such that r < ε. It then ollows that {r r < ε} + L ψ R ψ, hence L ψ is contained in the interior o R ψ. We observe that Proposition 6 i implies that, or every x R n, the region R(x is the intersection o a polyhedral cone C with the translation o C by x. More ormally, let k I such that ψ(x = a k (x, and let C := {r R n (a i a k r 0, i I}. We claim that R(x = C ((x C, where (x C := {r x r C}. Observe that (x C = {r (a i a k (x r 0, i I}. To show R(x C ((x C, note that, or all i I, the inequalities (a i a k r 0 and (a i a k (x r 0 are equivalent to the inequalities a i r +a k (x r ψ(x and a k r +a i (x r ψ(x, respectively. To show C ((x C R(x, note that, or all i, j I, the inequality a i r a j (x r ψ(x is the sum o the inequalities (a i a k r 0 and (a j a k (x r 0. 7

8 Proposition 7. Let ψ be a minimal valid unction or (3. Every minimal liting or ψ is a continuous unction. Proo. Let π be a minimal liting o ψ. It means that the unction (ψ, π is a minimal valid unction or (2. It is known that, i (ψ, π is a minimal valid unction or (2, then π is subadditive. The latter act was shown by Johnson [18] or the case S = Z n, but the proo or the general case is identical (see also [12] or a proo. To prove that π is continuous, we need to show that, given r R n, or every δ > 0 there exists ε > 0 such that π(r π( r < δ or all r R n satisying r r < ε. Since ψ is a continuous unction, ψ(0 = 0 and ψ coincides with π in some open ball containing the origin, it ollows that, or every δ > 0, there exists ε > 0 such that, or all r R n satisying r r < ε, we have ψ(r r < δ, π(r r = ψ(r r, and π( r r = ψ( r r. Hence, or all r R n satisying r r < ε, π(r π( r max{π( r r, π(r r} = max{ψ(r r, ψ( r r} < δ, where the irst inequality ollows rom the subadditivity o π, since π(r π( r + π(r r and π( r π(r + π( r r. 3 Minimum liting coeicient o a single variable Given r R n, we consider the set o solutions to x = + r R n rs r + r y r x S s 0 (6 y r 0, y r Z s has inite support. Given a minimal valid unction ψ or (3 and scalar λ, we say that the inequality r R n ψ(rs r+ λy r 1 is valid or (6 i it holds or every (x, s, y r satisying (6. We denote by π (r the minimum value o λ or which r R n ψ(rs r + λy r 1 is valid or (6. We observe that, or any liting π o ψ, we have π (r π(r. Indeed, r R n ψ(rs r + π(r y r 1 is valid or (6, since, or any ( s, ȳ r satisying (6, the vector ( s, ȳ, deined by ȳ r = 0 or all r R n \ {r }, satisies (2. It is easy to show (see [13] that, i ψ is a minimal valid unction or (3 and π is a minimal liting o ψ, then π ψ. So, or every minimal valid unction ψ or (3 and every minimal liting π o ψ, we have the ollowing relation, π (r π(r ψ(r or all r R n. In general π is not a liting o ψ, but i it is, then the above relation implies that it is the unique minimal liting o ψ. 8

9 Lemma 8. For any r R n such that π (r = ψ(r, we have π(r = ψ(r or every minimal liting π o ψ. Conversely, i π (r < ψ(r or some r R n, then there exists some minimal liting π o ψ such that π(r = π (r < ψ(r. Proo. The irst part ollows rom π π ψ. For the second part, given r R n such that π (r < ψ(r, we can deine a unction π : R n R by π (r = π (r and π (r = ψ(r or all r R n, r r. Since ψ is valid or (3, it ollows by the deinition o π (r that π is a liting o ψ. As observed in the introduction, there exists a minimal liting π such that π π. Since π π and π (r = π (r, it ollows that π(r = π (r. Next we present a geometric characterization, introduced in [13], o the unction π. Let r R n. Given a maximal S-ree convex set B = {x R n a i (x 1, i I}, or any λ R, we deine the set B(λ, r R n+1 as ollows ( x B(λ, r = { R n+1 a i (x + (λ a i r x n+1 1, i I}. (7 x n+1 Observe that B(λ, r (R n {0} = B {0}. For λ = 0, the vector ( r 1 belongs to the lineality space o B(λ, r, thus B(0, r = (B {0} + L where L is the line o R n+1 in the direction ( ( r 1. For λ 0, the point 0 + λ 1 ( r 1 satisies at equality all the inequalities a i (x + (λ a i r x n+1 1, i I, thereore B(λ, r has a unique minimal ace F (i.e., it is the translation o a polyhedral cone, and ( 0 + λ 1 ( r 1 F. The ollowing theorem, has been proved in [13]. Theorem 9. Let r R n. Given a maximal S-ree convex set B with in its interior, let ψ = ψ B. Given λ R, the inequality r R n ψ(rs r + λy r 1 is valid or (6 i and only i B(λ, r is S Z + -ree. In particular, by the above theorem, π (r is the smallest possible λ such that B(λ, r is S Z + -ree. Observe that, as λ gets smaller, the set B(λ, r (R n R + becomes larger, so or λ suiciently small B(λ, r will not be S Z + -ree. Intuitively, there will be a blocking point in S Z + that will not belong to B(λ, r as long as λ > π (r, it will be on the boundary o B(λ, r or λ = π (r, and it will be in the interior or λ < π (r. This act is proven in Theorem 11 below, and it will play a central role in the proos o Theorem 4 and 5. The next lemma will be needed in the proo o Theorem 11 and o Theorem 5. Lemma 10. Let r R n, and let B be a maximal S-ree convex set with in its interior. For every λ such that B(λ, r is S Z + -ree, B(λ, r is maximal S Z + -ree. Proo. Since B is a maximal S-ree convex set, then by Theorem 1 each acet o B contains a point x o S in its relative interior. Thereore the corresponding acet o B(λ, r contains the point ( x 0 in its relative interior. I B(λ, r is S Z + -ree, by Theorem 1 it is a maximal S Z + -ree convex set. Theorem 11. Consider r R n and λ R. We have λ = π (r i and only i B(λ, r is S Z + -ree and contains a point ( x S Z+ such that > 0. 9

10 Proo. We irst prove that, i B(λ, r is S Z + -ree and contains a point ( x S Z+ such that > 0, then λ = π (r. Since B(λ, r is S Z + -ree, Theorem 9 and the deinition o π imply that λ π (r. We claim that ( x is in the interior o B(λ ϵ, r or every ϵ > 0. Indeed, or all i I, a i ( x + (λ ϵ a i r = a i ( x + (λ a i r ϵ < 1, where the inequality holds because ( x B(λ, r implies that a i (x + (λ a i r x n+1 1 and > 0 implies ϵ > 0. This proves our claim. Since ( x is in the interior o B(λ ϵ, r or every ϵ > 0, Theorem 9 and the deinition o π imply that λ ϵ < π (r. Hence λ = π (r. We now prove the converse. Let λ = π (r. The deinition o π and Theorem 9 imply ( that B(λ, r is S Z + -ree. It only remains to show that B(λ, r contains a point x S Z+ such that > 0. Note that, or all λ R, B(λ, r (R n {0} = B {0}. We consider two possible cases: either there exists ( r r n+1 rec(b(λ, r conv(s Z + such that r n+1 > 0, or rec(b(λ, r conv(s Z + = rec(b conv(s {0}. Case 1. There exists ( r r n+1 rec(b(λ, r conv(s Z + such that r n+1 > 0. Note that rec(conv(s Z + = rec(conv(s R +, thus r rec(conv(s. By Lemma 10, B(λ, r is a maximal S Z + -ree convex set, thus by Theorem 1, rec(b(λ, r conv(s Z + is rational. Hence, we can choose ( r r n+1 integral. Since B is a maximal S-ree convex set, there exists x S B. Let ( ( x = ( x 0 + r r n+1. Since r rec(conv(s and x S, it ollows that x conv(s. Since x Z n, we conclude that x S. Furthermore, = r n+1, thus Z and > 0. Finally, since ( x 0 B(λ, r and ( r r n+1 rec(b(λ, r, we conclude that ( x B(π (r, r. Case 2. rec(b(λ, r conv(s Z + = rec(b conv(s {0}. Claim. ε > 0 such that rec(b(λ ε, r conv(s Z + = rec(b conv(s {0}. Since conv(s is a polyhedron, conv(s = {x R n Cx d} or some matrix (C, d. By assumption, there is no vector ( r 1 in rec(b(λ, r conv(s Z +. Thus the system a i r + (λ a i r 0, i I Cr 0 is ineasible. By Farkas Lemma, there exist scalars µ i 0, i I and a nonnegative vector γ such that µ i a i + γc = 0 λ ( i I i I µ i ( i I µ i a i r > 0. 10

11 This implies that there exists some ε > 0 small enough such that µ i a i + γc = 0 i I (λ ε( µ i ( µ i a i r > 0, i I i I thus the system a i r + (λ ε a i r 0, i I Cr 0 is ineasible. This implies that rec(b(λ ε, r conv(s Z + = rec(b conv(s {0}. This concludes the proo o the claim. By the previous claim, there exists ε such that rec(b(λ ε, r conv(s Z + = rec(b conv(s {0}. This implies that there exists a scalar M such that M or every point ( x B(λ ε, r (S Z +. Remark 2 and Lemma 10 imply that there exists δ > 0 such that, or every ( x (S Z + \B(λ, r, there exists h I such that a h ( x +(λ a h r 1+δ. Choose ε > 0 such that ε ε and εm δ. Since π (r = λ, Theorem 9 implies that B(λ ε, r has a point ( x S Z+ in its interior. Thus a i ( x + (λ ε a i r < 1, i I. We show that ( x is also in B(λ, r. Suppose not. Then, by our choice o δ, there exists h I such that a h ( x + (λ a h r 1 + δ. It ollows rom the deinition o B(λ, r given in (7 that B(λ ε, r (R n R + B(λ ε, r (R n R + because ε ε; thereore M. Hence 1+δ a h ( x +(λ a h r a h ( x +(λ ε a h r +εm < 1+εM 1+δ, a contradiction. Hence ( x is in B(λ, r. Since B is S-ree and B(λ ε, r (R n {0} = B {0}, it ollows that B(λ ε, r does not contain any point o S {0} in its interior. Thus > 0. 4 Proo o Theorem 4 The proo o Theorem 4 will ollow easily rom the next proposition. Proposition 12. Let ψ be a minimal valid unction or (3, and let r R n. The ollowing are equivalent. i π (r = ψ(r. ii There exists x S B ψ such that ψ(r + ψ(x r = ψ(x = 1. 11

12 Proo. We irst prove ii implies i. The vector (s, y r deined by y r = 1, s x r = 1, s r = 0 or all r x r, is a solution o (6. Since r R n ψ(rs r + π (r y r 1 is valid or (6, it ollows that thus π (r = ψ(r. 1 π (r + ψ(x r ψ(r + ψ(x r = 1, We now prove i implies ii. By Theorem 11, B(π (r, r contains a point ( x x n+1 S Z + such that x n+1 > 0. Since π (r = ψ(r = max i I a i r, the coeicients o x n+1 in the inequalities (7 deining B(π(r, r are all nonnegative. This shows that B(π (r, r contains the point ( x 1 and thereore ai (x +ψ(r a i r 1 or all i I. This implies that max i I a i (x r + ψ(r 1. Hence we have that ψ(x r + ψ(r 1. Conversely, we also have 1 ψ(x ψ(x r + ψ(r, where the irst inequality ollows rom the act that x S and that, by Theorem 3, B ψ is S-ree, while the second inequality ollows rom the act that ψ is subadditive. We conclude that ψ(r + ψ(x r = ψ(x = 1. Proo o Theorem 4. By Lemma 8, R ψ = {r R n : π (r = ψ(r}. By Proposition 12, given r R n, π (r = ψ(r i and only i r R(x or some x S B ψ. Thus R ψ = x S Bψ R(x. 5 Proo o Theorem 5 Let ψ be a minimal valid unction or (3 and let B := B ψ = {x R n a i (x 1, i I}. In this section we assume S = Z n. Under this assumption, Theorem 1 shows that: B is a maximal lattice-ree convex set. rec(b = lin(b = {r R n a i r = 0 or all i I} and rec(b is a rational subspace. ψ(r = max i I a i r 0 or every r R n. Let L ψ := {r R n a i r = a j r or all i, j I}. To prove Theorem 5, we need the ollowing three lemmas. Lemma 13. L ψ = lin(b. Furthermore, R ψ + Z n is a closed set. Proo. Let L := lin(b. It ollows rom Theorem 1 that L is a rational linear subspace and rec(b = L. Given r L ψ, it ollows that a i r = a j r or all i, j I, which implies that a i r = α or all i I or some constant α that only depends on r. I α 0 then r rec(b = L, while i α 0 then r rec(b = L, thus r L. The converse inclusion L L ψ is trivial. Next we show that R ψ + Z n is a closed set. It ollows rom Proposition 6 that R ψ is the union o a inite number o polyhedra, R 1,..., R k, and that rec(r i = L or i = 1,..., k. Let R i be the orthogonal projection o R i onto L. It ollows that R i is a polytope, so in particular R i is compact. Furthermore, R i = R i + L. Let Λ be the orthogonal projection o Z n onto L. Since L is rational, Λ is a lattice, so in particular it is closed. Note that R i + Z n = ( R i + Λ + L. Since the Minkowski sum o a compact set and a closed set is closed (see e.g. [1] Lemma 5.3 (4, it ollows that R i + Λ is closed. Since the Minkowski sum o a closed set and a linear subspace is closed, it ollows that ( R i + Λ + L is closed. 12

13 Lemma 14. Let H := R n \(R ψ +Z n. I H, then there exists r R ψ such that r belongs to the closure o H. Any such vector satisies ψ( r > 0. Proo. By Lemma 13, R ψ + Z n is closed. It ollows that, i R ψ + Z n R n, then there exists a point r R ψ + Z n such that r belongs to the closure o H. Since r R ψ + Z n, there exists r R ψ and w Z n such that r = r + w. We show that the point r belongs to the closure o H. I not, then there exists an open ball C centered at r contained in the interior o R ψ, thus C + w is an open ball centered at r contained in the interior o R ψ + Z n, contradicting the act that r is in the closure o H. Finally, given r on the boundary o R ψ, Proposition 6 v and the act that L ψ = lin(b imply that r / lin(b, thus ψ( r > 0. Lemma 15. Let r R n. I π (r > 0, then rec(b(π (r, r (R n R + = rec(b {0}. Proo. Let λ = π (r > 0. Let ( r r n+1 R n R + such that ( r r n+1 rec(b(λ, r. I r n+1 = 0, it ollows that r rec(b, thus ( r r n+1 rec(b {0}. So suppose by contradiction that r n+1 > 0. We may assume without loss o generality that r n+1 = 1. Since B(λ, r is a maximal Z n Z + -ree convex set, it ollows rom Theorem 1 that a i r + (λ a i r = 0 or all i I. Since λ > 0, the point ˆr := r r λ satisies a iˆr = 1 or all i I. It ollows that ˆr / rec(b while ˆr rec(b, contradicting the act that rec(b = lin(b. Proo o Theorem 5. As already mentioned in the introduction, it is shown in [13] that i R ψ + Z n = R n then π is the unique minimal valid liting o ψ. We show the converse, that is, we show that i R ψ + Z n does not cover all o R n, then there exist at least two distinct minimal litings o ψ. We irst observe that there exist two distinct minimal litings o ψ i and only i π is not a liting or ψ. Indeed, i π is a liting or ψ, then π is the unique minimal liting or ψ, since any other liting π satisies π π. Conversely, suppose that π is not a liting or ψ. Given any minimal liting π o ψ, since π is not a valid liting it ollows that there exists r R n such that π (r < π(r. By Lemma 8, there exists a minimal liting π such that π (r = π (r. Thus π and π are distinct minimal litings o ψ. Thus, we only need to show that, i R ψ + Z n R n, then π is not a liting. Suppose by contradiction that R ψ + Z n R n but π is a liting o ψ. It ollows that π is a minimal liting or ψ, thus by Proposition 7, π is a continuous unction. Let H = R n \ (R ψ + Z n. We will show the ollowing. Claim 1. There exists p H such that B(π (p, p contains a point ( x 1 with x Z n. Beore proving Claim 1, we use it to conclude the proo o Theorem 5. By Claim 1, there exists p R n \ (R ψ + Z n such that B(π (p, p contains a point ( x 1 with x Z n. It ollows that a i ( x + (π (p a i p 1 or all i I. Moreover, by Theorem 9, B(π (p, p is S Z + -ree, thus there exists h I such that a h ( x + (π (p a h p = 1. This shows that a i ( x p 1 π (p or all i I and a h ( x p = 1 π (p. Since ψ( x p = max i I a i ( x p = a h ( x p, we have that ψ( x p = 1 π (p. 13

14 Since π is a liting, π (p + π ( x p 1 and since π ( x p ψ( x p, this shows that π ( x p = ψ( x p. It ollows that x p R ψ. In particular, x p R( x or some x Z n B. By deinition o R( x, x ( x p R( x, that is, x x + p R( x. This implies that p R( x + Z n, a contradiction. The remainder o the proo is devoted to showing Claim 1. By Lemma 14, the closure o H contains a point r R ψ and such a vector satisies π ( r = ψ( r > 0. Since, by Proposition 7, π is continuous, there exists ε > 0 such that, or every r R n satisying r r ε, π (r > 0. Let µ = min{π (r r r ε}. Note that µ is well deined since π is continuous and {r R n r r ε} is compact. Furthermore, by the choice o ε, µ > 0. Let M = µ 1. Claim 2. For every r R n satisying r r < ε and every ( x x n+1 B(π (r, r, we have x n+1 M. Let r R n such that r r < ε. By Lemma 15, max{x n+1 ( x x n+1 B(π (r, r} is bounded. The above is a linear program, thus the set o its optimal solutions contains a minimal ace o B(π (r, r. Since the point ( r π (r( 1 satisies all inequalities ai (x + (π (r a i rx n+1 1 at equality, it ollows that B(π (r, r has a unique minimal ace and that ( r ( π (r( 1 is in it. It ollows that r π (r( 1 is an optimal solution, thus max{x n+1 ( x x n+1 B(π (r, r} = 1 π (r M. This concludes the proo o Claim 2. By Remark 2 and Lemma 10, there exists δ > 0 such that, or every ( x (Z n Z + \ B(ψ( r, r, a i ( x + (ψ( r a i r 1 + δ, or some i I. Since π is a continuous unction, π ( r = ψ( r, and r is in the closure o H, there exists p H such that p r < ε and (ψ( r a i r (π (p a i p < δ M or all i I. By Theorem 11 and Claim 2, B(π (p, p contains a point ( x Z n Z + such that 0 < M. We conclude by showing that ( x 1 is in B(π (p, p, thus proving Claim 1. First we show that ( x 1 is in B(ψ( r, r. Indeed, or all i I, a i ( x + (ψ( r a i r a i ( x + (ψ( r a i r < a i ( x + (π (p a i p + δ M 1 + δ, (8 where the irst inequality ollows rom the acts that ψ( r a i r and 1, while the last inequality ollows rom the acts that ( x B(π (p, p and M. By our choice o δ, the strict inequality in (8 shows that ( x 1 is in B(ψ( r, r. In particular, since ψ( r ai r or all i I, it ollows that a i ( x 1 or all i I. We inally show that a i ( x + π (p a i p 1 or all i I, implying that ( x 1 is in B(π (p, p. Indeed, i π (p a i p 0, then a i ( x + π (p a i p 1 + π (p a i p 1, while i π (p a i p > 0, then a i ( x + π (p a i p a i ( x + (π (p a i p 1. 14

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