Minimal Valid Inequalities for Integer Constraints

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1 Minimal Valid Inequalities for Integer Constraints Valentin Borozan LIF, Faculté des Sciences de Luminy, Université de Marseille, France and Gérard Cornuéjols Tepper School of Business, Carnegie Mellon University, Pittsburgh, PA 1513 and LIF, Faculté des Sciences de Luminy, Université de Marseille, France July 007 Dedicated to George Nemhauser for his 70th birthday Abstract In this note we consider an infinite relaxation of mixed integer linear programs. We show that any minimal valid inequality for this infinite relaxation arises from a nonnegative, piecewise linear, convex and homogeneous function. 1 Introduction Consider an integer program (IP min cx Ax = b x j Z for j = 1,..., p, x j 0 for j = 1,..., n, where p n, the matrix A Q m n, the row vector c Q n, the column vector b Q m are data, and x R n is a column vector of variables. We assume that A has full row rank m. A common approach to solve (IP is to first solve the linear programming relaxation (LP obtained by ignoring the integrality restrictions on x. A basic optimal solution of (LP is of the form x i = f i + j J rj x j for i B, (1 where B and J denote the sets of basic and nonbasic variables respectively. We have f 0. If f i Z for all i B {1,..., p}, then the basic solution x i = f i for all i B and x i = 0 Supported by NSF grant DMI , ONR grant N and ANR grant BLAN

2 otherwise, is an optimal solution of the integer program. On the other hand, if f i Z for some i B {1,..., p}, the above basic solution is not feasible to (IP and one may want to generate one or several cutting planes that cut it off while being satisfied by all the feasible solutions to (IP. Different strategies have been proposed for generating cutting planes. For example, Balas [] introduced intersection cuts obtained by intersecting the rays f + αr j, α 0, with a convex set whose interior contains f but no integral point. Most general purpose cutting planes used in state-of-the-art solvers are obtained by generating a linear combination of the original constraints Ax = b, and by applying integrality arguments to the resulting equation (Gomory s Mixed Integer Cuts, MIR inequalities and split cuts are examples. Recently, there has been interest in cutting planes that cannot be deduced from a single equation, but can be deduced by integrality arguments involving two equations (Dey and Richard [5], Andersen, Louveaux, Weismantel and Wolsey [1]. Let us consider the following relaxation of the integer program (IP starting from its equivalent form (1. We drop the nonnegativity restriction on all basic variables x i, i B, and we drop the integrality restriction on all the nonbasic variables x j, j J. Furthermore, we drop the constraints x i = f i + j J rj x j for i B {p + 1,..., n}. We are left with a system of the form x = f + k j=1 rj s j x Z q s 0 ( where we now denote by s the nonbasic variables and by x the remaining basic variables. We will keep this notation in the remainder of the paper. This relaxation of (IP is denoted by R f (r 1,..., r k where f, r 1,..., r k Q q. Such a relaxation was considered in [1] for the case q =. Gomory and Johnson [8] suggested relaxing the k-dimensional space of variables s = (s 1,..., s k to an infinite-dimensional space, where the variables s r are defined for any r Q q. We get the infinite relaxation R f x = f + finite rs r x Z q s 0 (3 where finite means that s r > 0 for a finite number of r Q q, i.e. the vector s has finite support. A feasible solution of the infinite relaxation R f is a vector (x, s with finite support that satisfies the three conditions (3. We say that an inequality is valid for R f if it is satisfied by all its feasible solutions. Note that conv(r f (r 1,..., r k is the face of conv(r f obtained by setting s r = 0 for all r Q q \ {r 1,..., r k }. Any valid inequality for (3 yields a valid inequality for ( by simply restricting it to the space r 1,..., r k. In the remainder, we assume f Z q. Thus the basic solution x = f, s = 0 is not feasible for R f. In this paper, we study valid inequalities for R f that cut off this infeasible basic solution. These inequalities can be stated in terms of the variables s only:

3 finite ψ(rs r 1. Our main interest is in minimal valid inequalities for R f, namely valid inequalities finite ψ(rs r 1 such that there is no valid inequality finite ψ (rs r 1 where ψ ψ and ψ (r < ψ(r for at least one r Q q. We show that, for a minimal valid inequality, the function ψ is nonnegative, piecewise linear, convex and homogeneous (namely ψ(λr = λψ(r for any scalar λ Q + and r Q q. The function ψ is not always continuous or finite. However, when it is finite, the piecewise linear function ψ has at most q pieces. Minimal Valid Inequalities We consider the infinite integer programming problem R f defined by (3 where we assume f Z q. Note that R f since x = 0, s r = 1 for r = f and s r = 0 otherwise, is a feasible solution of (3. Any valid inequality for R f that cuts off the infeasible solution s = 0 can be written as finite ψ f (rs r 1. (4 We say that the function ψ f : Q q R {+ } is valid if the corresponding inequality (4 is satisfied by every feasible solution of R f, i.e. by every s with finite support such that (x, s R f. We assume that there exists at least one feasible solution of R f such that finite ψ f (rs r < + (otherwise the function ψ f is uninteresting. When ψ f (r = + and s r = 0, we define ψ f (rs r = 0. Equivalently, finite is computed by summing only over the finitely many vectors r such that s r > 0. To simplify notation, we write ψ instead of ψ f in the remainder. Lemma.1. If the function ψ is valid, then ψ 0. p 1 D Proof. Suppose ψ( r < 0 for some r =... where p 1,..., p q Z and D Z + is a p q D common denominator. Let ( x, s be a feasible solution in R f such that finite ψ(r s r < +. Let ( x, s be defined by s r := s r + MD where M is a positive integer, s r := s r for r r, and x := f + r s r. The point ( x, s is a feasible solution in R f since x = x+md r Z q. We have ψ(r s r = ψ(r sr + ψ( rmd. By making the positive integer M large, we can make ψ( rmd as negative as we want. Therefore ψ(r s r < 1, contradicting the fact that ( x, s is feasible. A valid function ψ is minimal if there is no valid function ψ such that ψ ψ and ψ (r < ψ(r for at least one r Q q. Lemma.. If ψ is a minimal valid function, then ψ(0 = 0. Proof. If ( x, s is a feasible solution in R f, then so is ( x, s defined by s r := s r for r 0, and s 0 = 0. Therefore, if ψ is valid, then ψ defined by ψ (r = ψ(r for r 0 and ψ (0 = 0 is also valid. Since ψ is minimal, it follows that ψ(0 = 0. 3

4 A function g is subadditive if g(a + g(b g(a + b. Lemma.3. If ψ is a minimal valid function, then ψ is subadditive. Proof. Let r 1, r Q q. Define the function ψ as follows. { ψ(r ψ (r := 1 + ψ(r if r = r 1 + r ψ(r if r r 1 + r. We will show that ψ is valid. Consider any ( x, s R f. Define ( x, s as follows s r 1 + s r 1 +r if r = r1 s s r := r + s r 1 +r if r = r 0 if r = r 1 + r s r otherwise. Using the definitions of ψ and s, it is easy to verify that ψ (r s r = r r ψ(r s r. (5 Furthermore we have x = f + r s r = f + r s r. Since x Z q and s 0, this implies that ( x, s R f. Since ψ is valid, this implies r ψ(r s r 1. Therefore, by (5, ψ (r s r 1. Thus ψ is valid. Since ψ is minimal, we get ψ(r 1 + ψ(r ψ(r 1 + r. Lemma.4. If ψ is a minimal valid function, then ψ is homogeneous. Proof. Let r Q q and λ Q +. We will show ψ(λ r = λψ( r. This holds when λ = 0 so assume now λ > 0. Define the function ψ as follows. { 1 ψ (r := λψ(λ r if r = r ψ(r otherwise. We will show that ψ is valid. Consider any ( x, s R f. Define ( x, s as follows s λ r + 1 λ s r if r = λ r s r := 0 if r = r s r otherwise. Using the definition of ψ and s, it is easy to verify that ψ (r s r = ψ(r s r. (6 Furthermore we have x = f + r s r = f + r s r. Since x Z q and s 0, this implies that ( x, s R f. Since ψ is valid, this implies r ψ(r s r 1. Therefore, by (6, ψ (r s r 1. Thus ψ is valid. Since ψ is minimal, we get ψ(λ r λψ( r. By reversing the roles of r and λ r, we get ψ( r 1 λ ψ(λ r. Therefore we have equality. 4 r

5 Corollary.5. If ψ is a minimal valid function, then ψ is convex. Proof. Let r 1, r Q q and 0 < t < 1 rational. Then, by Lemmas.3 and.4, tψ(r 1 + (1 tψ(r = ψ(tr 1 + ψ((1 tr ψ(tr 1 + (1 tr. Example.6. In R q, suppose 0 < f i < 1 for some i = 1,..., q. Define ψ as follows. ψ(r := { ri 1 f i if r i 0 r i f i if r i 0. This inequality is a Gomory mixed integer cut [7] obtained from row i of R f. Equivalently, it is a simple split inequality [4] obtained from the disjunction x i 0 or x i 1. Example.7. In R q, suppose f i = 1 for i = 1,..., q. Define ψ(r := q ( r r q. This inequality is an intersection cut [] obtained from the octahedron centered at f with vertices f ± q ei, where e i denotes the ith unit vector. This octahedron has q facets, each of which contains a 0,1 point in its center. As one would expect, a minimal valid inequality may be implied by a linear combination of other minimal valid inequalities. This is the case here. Indeed, the above intersection cut from the octahedron is implied by the q split cuts ψ i (r := r i for i = 1,..., q since ψ = q i=1 1 q ψi and finite ψi (rs r 1 for i = 1,..., q imply finite ψ(rs r 1. A minimal valid function ψ may not be continuous nor finite, as shown by the following examples. ( 1 Example.8. In R, let f :=. Define ψ as follows. 0 ( r1 ψ r ( 1 Example.9. In R, let f := 1 := r if r > 0 r 1 if r = 0 + if r < 0. It is easy to verify that this function ( ( is valid. In ( particular ψ(x f 1 for all x Z 0 1 i and equality holds when x =, and for i Z. The minimality of ψ ( ( M for r 0 can be verified directly. For r < 0, convexity implies 1 ψ r1 ψ r 0 ( ( ( 1 M ψ r1 r1 r1. Thus ψ M r r r. When M goes to +, this implies ψ = r +.. Define ψ as follows. ( r1 ψ r := 3 r if r 1 < r < r 1 and r > 0 r if r = r 1 or r 1, and r 0 + otherwise. 5

6 3 Maximal lattice-free convex sets For a convex function ψ : Q q R {+ }, define B ψ := {x Q q : ψ(x f 1}. (7 We are interested in the properties of B ψ when ψ is a minimal valid function. For the function ψ of Example.8, B ψ contains all the rational points in the band 0 < x 1 and in the segment x = 0, 1 x 1 1 but the half-lines x = 0, x 1 > 1 are not in B ψ. Define the boundary of B ψ to be the set {x Q q : ψ(x f = 1}. In Example.8, the boundary of B ψ consists of the line x = 1 and the two points (0, 0 and (1, 0. Lemma 3.1. Let ψ be a minimal valid function for R f. Then B ψ is a convex subset of Q q that contains no integral point x Z q except possibly on its boundary. Furthermore f B ψ and, if ψ is finite, then B ψ contains f in its interior. Proof. The fact that B ψ is convex follows from Corollary.5. Let x Z q. Then ( x, s R f where s x f = 1 and s r = 0 otherwise. Since ψ is valid, ψ( x f 1. If, in addition, x B ψ, this implies ψ( x f = 1. f B ψ since ψ(0 = 0. Now assume ψ(r < + for all r Q q. For each unit vector e i, the homogeneity of ψ implies the existence of λ i > 0 such that ψ(λ i e i 1. Similarly, for i = 1,..., q, choose any µ i > 0 such that ψ(µ i ( e i 1. Then the convex hull of the q points f + λ i e i, f µ i e i is contained in B ψ by convexity and it contains f in its interior. Thus f is in the interior of B ψ. Lemma 3.. Let B be any closed convex set in R q with nonempty interior and no integral point in its interior. Then the following function ψ is valid for R f whenever f Q q is in the interior of B: Set ψ(r = 0 for any r in the recession cone of B, ψ(r = 1 for any r Q q such that the point f + r is on the boundary of B, all other values of ψ being defined by homogeneity. Furthermore B ψ = B Q q. Proof. The convexity of B and homogeneity of ψ imply that ψ is subadditive: Let a, b Q q. Suppose first that neither a nor b is in the recession cone of B. Set α > 0 to be the scalar such that ψ(αa = 1. Similarly set β > 0 such that ψ(βb = 1. Then f + αa, f + βb B. By convexity of B, for any 0 λ 1 we have Set λ := β α+β λψ(αa + (1 λψ(βb = 1 ψ(λαa + (1 λβb (8 in (8. We get, by homogeneity of ψ ψ(a + ψ(b ψ(a + b. (9 If both a, b are in the recession cone of B, then a + b also is and (9 holds. So we may assume that b is in the recession cone of B but not a. Choose α > 0 such that ψ(αa = 1. Then αa B and, since b is in the recession cone of B, we also have αa + αb B. Thus ψ(α(a + b 1. Now (9 holds since ψ(αa + ψ(αb = 1 and ψ is homogeneous. 6

7 Suppose ψ is not valid. Then there exists ( x, s R f such that ψ(r s r < 1. Since x = f + r s r, the subadditivity and homogeneity of ψ imply ψ( x f = ψ( r s r ψ(r s r < 1. Thus x Z q is in B ψ but not on its boundary, a contradiction. Lemma 3.3. Let ψ and ψ be two valid functions. Then ψ ψ if and only if B ψ B ψ. Proof. Immediate from the definition of B ψ. Theorem 3.4. If ψ is a finite minimal valid function for R f, then ψ is a continuous nonegative homogeneous convex piecewise linear function with at most q pieces. Proof. Let ψ be a finite minimal valid function and let B ψ be the corresponding convex set as defined in (7. By Lemma 3.1, B ψ is convex, contains f in its interior and has integral points only on its boundary, and by Lemmas 3. and 3.3 it is inclusion maximal with these properties. By a theorem of Doignon [6], Bell [3] and Scarf [11], every maximal convex set in R q with no integral point in its interior is a polyhedron with at most q facets. The proof of the upper bound on the number of facets is simple and elegant: By maximality, each facet F contains an integral point x F in its relative interior. If there are more than q facets, two integral points x F and x F must be identical modulo. Then their middle point 1 (xf + x F is integral and interior, contradiction. Consider a facet F of B ψ. We have ψ(x f = 1 for all x F. By homogeneity (Lemma.4, ψ is linear in the cone {r Q q : r = λ(x f with λ 0, x F }. Since the union of these cones over all facets of B ψ covers Q q, the function ψ is piecewise linear with at most q pieces. integral point integral point integral point f integral point r 0 1 r 1 x-space B ψ ψ Figure 1: A maximal lattice-free convex set and corresponding function ψ in R Corollary 3.5. Let ψ is a finite minimal valid function for R f. Then B ψ is a polyhedron with at most q facets, each facet contains an integral point in its relative interior, and B ψ contains f but no integral point in its interior. 7

8 Conversely, any polyhedron B Q q with nonempty interior, no integral point in its interior, but an integral point in the relative interior of each facet corresponds to a finite minimal valid function ψ such that B ψ = B. To see this, as in Lemma 3., choose f Q q in the interior of B and let ψ(x f = 1 for x on the facets of B. The other values of ψ(r are defined by homogeneity. The function so defined is minimal because B ψ is a maximal convex set with no integral point in its interior (Lemma 3.3. As stated in the proof of Theorem 3.4 above, a maximal convex set in R q with no integral point in its interior is a polyhedron with at most q facets. Lovász [9] points out that the unbounded case reduces to the bounded case as follows. Let b 1,..., b q be any basis of the integer lattice. Let U 1 (L 1 be the linear subspace (lattice spanned by b 1,..., b k and U the linear subspace spanned by b k+1,..., b q. Let K 1 be a maximal convex set in U 1 free of the lattice L 1 (By a lattice-free convex set S, we mean that S does not contain lattice points in its interior. Then the set K 1 + U, called cylinder above K 1, is a maximal lattice-free convex set for the whole space. Conversely, every unbounded maximal lattice-free convex set is a cylinder above a bounded maximal lattice-free convex set in some lattice subspace. Theorem 3.6. If ψ is a minimal valid function for R f, then ψ is a nonnegative homogeneous convex piecewise linear function. Proof. By Theorem 3.4, it suffices to consider the case when ψ is not finite everywhere. Let B ψ be the corresponding convex set as defined in (7. Let B ψ be the closure of B ψ. Bψ is a lattice-free convex set and by Lemma 3.3, it is maximal. Therefore, by [6], [3] and [11], Bψ is a full-dimensional polyhedron with at most q facets. By Lemma 3.1, f B ψ. If f were in the interior of B ψ, then ψ would be finite by Lemma 3., contradicting our assumption. Therefore the point f lies in one of the faces of B ψ. In each facet of B ψ containing f, we can apply induction to conclude that ψ is piecewise linear in that subspace. In the directions r such that the ray f + λr, λ 0, goes through the interior of B ψ, we get a piecewise linear function since ψ equals 1 on the encountered facets and ψ is homogeneous in each of the corresponding cones. Finally, in the directions r such that the ray f +λr, λ 0, only touches B ψ in f, we have ψ(r = +. References [1] K. Andersen, Q. Louveaux, R. Weismantel and L. Wolsey, Cutting Planes from Two Rows of a Simplex Tableau, Proceedings of IPCO XII, Ithaca, New York ( [] E. Balas, Intersection Cuts - A New Type of Cutting Planes for Integer Programming, Operations Research 19 ( [3] D.E. Bell, A Theorem Concerning the Integer Lattice, Studies in Applied Mathematics 56 ( [4] W. Cook, R. Kannan and A. Schrijver, Chvátal closures for mixed integer programming problems, Mathematical Programming 47 ( [5] S.S. Dey and J.-P. P. Richard, Facets for the Two-Dimensional Infinite Group Problems, technical report (007. 8

9 [6] J.-P. Doignon, Convexity in Cristallographical Lattices, Journal of Geometry 3 ( [7] R.E. Gomory, An Algorithm for Integer Solutions to Linear Programs, Recent Advances in Mathematical Programming, R.L. Graves and P. Wolfe eds., McGraw-Hill, New York ( [8] R.E. Gomory and E.L. Johnson, Some Continuous Functions Related to Corner Polyhedra, Part I, Mathematical Programming 3 ( [9] L. Lovász, Geometry of Numbers and Integer Programming, Mathematical Programming: Recent Developments and Applications, M. Iri and K. Tanabe eds., Kluwer ( [10] G.L. Nemhauser and L.A. Wolsey, Integer and Combinatorial Optimization, Wiley (1988. [11] H.E. Scarf, An Observation on the Structure of Production Sets with Indivisibilities, Proceedings of the National Academy of Sciences of the United States of America 74 (