Minimal Valid Inequalities for Integer Constraints Valentin Borozan LIF, Facult´e des Sciences de Luminy, Universit´e de Marseille, France
[email protected] and G´erard Cornu´ejols ∗ Tepper School of Business, Carnegie Mellon University, Pittsburgh, PA 15213 and LIF, Facult´e des Sciences de Luminy, Universit´e de Marseille, France
[email protected] July 2007 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 xj ∈ Z for j = 1, . . . , p, xj ≥ 0 for j = 1, . . . , n, where p ≤ n, the matrix A ∈ Qm×n , the row vector c ∈ Qn , the column vector b ∈ Qm are data, and x ∈ Rn 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
xi = fi +
P j∈J
r j xj
for i ∈ B,
(1)
where B and J denote the sets of basic and nonbasic variables respectively. We have f ≥ 0. If fi ∈ Z for all i ∈ B ∩ {1, . . . , p}, then the basic solution xi = fi for all i ∈ B and xi = 0 ∗
Supported by NSF grant DMI-0352885, ONR grant N00014-97-1-0196 and ANR grant BLAN06-1-138894.
1
otherwise, is an optimal solution of the integer program. On the other hand, if fi 6∈ 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 [2] introduced intersection cuts obtained by intersecting the rays f + αrj , 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 xi , i ∈ B, and we drop the integrality restriction P on all the nonbasic variables xj , j ∈ J. Furthermore, we drop the constraints xi = fi + j∈J rj xj for i ∈ B ∩ {p + 1, . . . , n}. We are left with a system of the form
x = f+ x ∈ Zq s ≥ 0
Pk
j=1 r
js
j
(2)
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 Rf (r1 , . . . , rk ) where f, r1 , . . . , rk ∈ Qq . Such a relaxation was considered in [1] for the case q = 2. Gomory and Johnson [8] suggested relaxing the k-dimensional space of variables s = (s1 , . . . , sk ) to an infinite-dimensional space, where the variables sr are defined for any r ∈ Qq . We get the infinite relaxation Rf x = f+ x ∈ Zq s ≥ 0
P
f inite rsr
(3)
P where f inite means that sr > 0 for a finite number of r ∈ Qq , i.e. the vector s has finite support. A feasible solution of the infinite relaxation Rf is a vector (x, s) with finite support that satisfies the three conditions (3). We say that an inequality is valid for Rf if it is satisfied by all its feasible solutions. Note that conv(Rf (r1 , . . . , rk )) is the face of conv(Rf ) obtained by setting sr = 0 for all r ∈ Qq \ {r1 , . . . , rk }. Any valid inequality for (3) yields a valid inequality for (2) by simply restricting it to the space r1 , . . . , rk . In the remainder, we assume f 6∈ Zq . Thus the basic solution x = f , s = 0 is not feasible for Rf . In this paper, we study valid inequalities for Rf that cut off this infeasible basic solution. These inequalities can be stated in terms of the variables s only: 2
P
f inite ψ(r)sr
≥ 1.
P Our main interest is in minimal valid inequalities for R , namely valid inequalities f f inite P ψ(r)sr ≥ 1 such that there is no valid inequality f inite ψ 0 (r)sr ≥ 1 where ψ 0 ≤ ψ and ψ 0 (r) < ψ(r) for at least one r ∈ Qq . 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 ∈ Qq ). The function ψ is not always continuous or finite. However, when it is finite, the piecewise linear function ψ has at most 2q pieces.
2
Minimal Valid Inequalities
We consider the infinite integer programming problem Rf defined by (3) where we assume f 6∈ Zq . Note that Rf 6= ∅ since x = 0, sr = 1 for r = −f and sr = 0 otherwise, is a feasible solution of (3). Any valid inequality for Rf that cuts off the infeasible solution s = 0 can be written as P
f inite ψf (r)sr
≥ 1.
(4)
We say that the function ψf : Qq → R ∪ {+∞} is valid if the corresponding inequality (4) is satisfied by every feasible solution of Rf , i.e. by every s with finite support such that (x, s) ∈ Rf . P We assume that there exists at least one feasible solution of Rf such that f inite ψf (r)sr < +∞ (otherwise the function ψ Pf is uninteresting). When ψf (r) = +∞ and sr = 0, we define ψf (r)sr = 0. Equivalently, f inite is computed by summing only over the finitely many vectors r such that sr > 0. To simplify notation, we write ψ instead of ψf in the remainder. Lemma 2.1. If the function ψ is valid, then ψ ≥ 0. p1 D
Proof. Suppose ψ(˜ r) < 0 for some r˜ = . . . where p1 , . . . , pq ∈ Z and D ∈ Z+ is a pq D
common denominator. P Let (¯ x, s¯) be a feasible solution in Rf such that f inite ψ(r)¯ sr < +∞. Let (˜ x, s˜) P be defined by s˜r˜ := s¯r˜ + M D where M is a positive integer, s˜r := s¯r for r 6= r˜, and x ˜ := f +P r˜ sr . q x, s˜) is a feasible solution in Rf since x ˜=x ¯ +M D˜ r ∈ Z . We have ψ(r)˜ sr = P The point (˜ ψ(r)¯ sr + ψ(˜ r)M D. By making the positive integer M large, we can make ψ(˜ r )M D as P negative as we want. Therefore ψ(r)˜ sr < 1, contradicting the fact that (˜ x, s˜) is feasible. A valid function ψ is minimal if there is no valid function ψ 0 such that ψ 0 ≤ ψ and ψ 0 (r) < ψ(r) for at least one r ∈ Qq . Lemma 2.2. If ψ is a minimal valid function, then ψ(0) = 0. Proof. If (¯ x, s¯) is a feasible solution in Rf , then so is (¯ x, s˜) defined by s˜r := s¯r for r 6= 0, and s˜0 = 0. Therefore, if ψ is valid, then ψ 0 defined by ψ 0 (r) = ψ(r) for r 6= 0 and ψ 0 (0) = 0 is also valid. Since ψ is minimal, it follows that ψ(0) = 0. 3
A function g is subadditive if g(a) + g(b) ≥ g(a + b). Lemma 2.3. If ψ is a minimal valid function, then ψ is subadditive. Proof. Let r1 , r2 ∈ Qq . Define the function ψ 0 as follows. ½ ψ(r1 ) + ψ(r2 ) if r = r1 + r2 0 ψ (r) := ψ(r) if r 6= r1 + r2 . We will show that ψ 0 is valid. Consider any (¯ x, s¯) ∈ Rf . Define (¯ x, s˜) as follows 1 s¯ 1 + s¯r1 +r2 if r = r r s¯r2 + s¯r1 +r2 if r = r2 s˜r := if r = r1 + r2 0 s¯r otherwise. Using the definitions of ψ 0 and s˜, it is easy to verify that X
ψ 0 (r)¯ sr =
r
X
ψ(r)˜ sr .
(5)
r
P P Furthermore we have x ¯ = f + r¯ sr = f + r˜ sr . Since x ¯ ∈ Zq and s˜ ≥ 0, this implies that (¯ x, s˜) ∈ Rf . P P 0 Since ψ is valid, this implies r ψ(r)˜ sr ≥ 1. Therefore, by (5), ψ (r)¯ sr ≥ 1. Thus ψ 0 is valid. Since ψ is minimal, we get ψ(r1 ) + ψ(r2 ) ≥ ψ(r1 + r2 ). Lemma 2.4. If ψ is a minimal valid function, then ψ is homogeneous. Proof. Let r˜ ∈ Qq 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 0 as follows. ½ 1 r) if r = r˜ λ ψ(λ˜ ψ 0 (r) := ψ(r) otherwise. 0 We will show that ψ is valid. Consider any (¯ x, s¯) ∈ Rf . Define (¯ x, s˜) as follows 1 r s¯λ˜r + λ s¯r˜ if r = λ˜ 0 if r = r˜ s˜r := s¯r otherwise. Using the definition of ψ 0 and s˜, it is easy to verify that X
ψ 0 (r)¯ sr =
X r
ψ(r)˜ sr .
(6)
P P Furthermore we have x ¯ = f + r¯ sr = f + r˜ sr . Since x ¯ ∈ Zq and s˜ ≥ 0, this implies that (¯ x, s˜) ∈ Rf . P P 0 Since ψ is valid, this implies r ψ(r)˜ sr ≥ 1. Therefore, by (6), ψ (r)¯ sr ≥ 1. Thus ψ 0 is valid. Since ψ is minimal, we get ψ(λ˜ r) ≥ λψ(˜ r). By reversing the roles of r˜ and λ˜ r, we get 1 r). ψ(˜ r) ≥ λ ψ(λ˜ Therefore we have equality. 4
Corollary 2.5. If ψ is a minimal valid function, then ψ is convex. Proof. Let r1 , r2 ∈ Qq and 0 < t < 1 rational. Then, by Lemmas 2.3 and 2.4, tψ(r1 ) + (1 − t)ψ(r2 ) = ψ(tr1 ) + ψ((1 − t)r2 ) ≥ ψ(tr1 + (1 − t)r2 ).
Example 2.6. In Rq , suppose 0 < fi < 1 for some i = 1, . . . , q. Define ψ as follows. ( ri 1−fi if ri ≥ 0 ψ(r) := −ri if ri ≤ 0. fi This inequality is a Gomory mixed integer cut [7] obtained from row i of Rf . Equivalently, it is a simple split inequality [4] obtained from the disjunction xi ≤ 0 or xi ≥ 1. Example 2.7. In Rq , suppose fi = 12 for i = 1, . . . , q. Define ψ(r) := 2q (|r1 | + . . . + |rq |). This inequality is an intersection cut [2] obtained from the octahedron centered at f with vertices f ± 2q ei , where ei denotes the ith unit vector. This octahedron has 2q 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 octahedron is implied by the q split cuts ψ i (r) P := 2|ri | for i = 1, . . . , q since Pq the P 1 i i ψ = i=1 q ψ and f inite ψ (r)sr ≥ 1 for i = 1, . . . , q imply f inite ψ(r)sr ≥ 1. A minimal valid function ψ may not be continuous nor finite, as shown by the following examples. µ 1 ¶ 2 2 Example 2.8. In R , let f := . Define ψ as follows. 0 µ ¶ if r2 > 0 r2 r1 2|r1 | if r2 = 0 ψ := r2 +∞ if r2 < 0. It is easy to verify that this function is valid. ψ(x − f ) ≥ 1 for all x ∈ Z2 µ ¶ µ ¶ In µparticular ¶ 0 1 i and equality holds when x = , and for i ∈ Z. The minimality of ψ 0 0 1 µ ¶ µ M ¶ r1 1 2 for r2 ≥ 0 can be verified directly. For r2 < 0, convexity implies 2 ψ ≥ψ − r2 0 µ ¶ µ ¶ µ ¶ M − r1 r1 r1 1 . Thus ψ ≥ M − |r2 |. When M goes to +∞, this implies ψ = 2ψ −r2 r2 r2 +∞. µ 1 ¶ 2 2 Example 2.9. In R , let f := . Define ψ as follows. 1 2 2 µ ¶ 3 r2 if − r1 < r2 < r1 and r2 > 0 r1 2r2 if r2 = r1 or − r1 , and r2 ≥ 0 ψ := r2 +∞ otherwise.
5
3
Maximal lattice-free convex sets
For a convex function ψ : Qq → R ∪ {+∞}, define Bψ := {x ∈ Qq : ψ(x − f ) ≤ 1}.
(7)
We are interested in the properties of Bψ when ψ is a minimal valid function. For the function ψ of Example 2.8, Bψ contains all the rational points in the band 0 < x2 ≤ 1 and in the segment x2 = 0, −1 ≤ x1 ≤ 1 but the half-lines x2 = 0, |x1 | > 1 are not in Bψ . Define the boundary of Bψ to be the set {x ∈ Qq : ψ(x − f ) = 1}. In Example 2.8, the boundary of Bψ consists of the line x2 = 1 and the two points (0, 0) and (1, 0). Lemma 3.1. Let ψ be a minimal valid function for Rf . Then Bψ is a convex subset of Qq that contains no integral point x ¯ ∈ Zq except possibly on its boundary. Furthermore f ∈ Bψ and, if ψ is finite, then Bψ contains f in its interior. Proof. The fact that Bψ is convex follows from Corollary 2.5. Let x ¯ ∈ Zq . Then (¯ x, s¯) ∈ Rf where s¯x¯−f = 1 and s¯r = 0 otherwise. Since ψ is valid, ψ(¯ x − f ) ≥ 1. If, in addition, x ¯ ∈ Bψ , this implies ψ(¯ x − f ) = 1. f ∈ Bψ since ψ(0) = 0. Now assume ψ(r) < +∞ for all r ∈ Qq . For each unit vector ei , the homogeneity of ψ implies the existence of λi > 0 such that ψ(λi ei ) ≤ 1. Similarly, for i = 1, . . . , q, choose any µi > 0 such that ψ(µi (−ei )) ≤ 1. Then the convex hull of the 2q points f + λi ei , f − µi ei is contained in Bψ by convexity and it contains f in its interior. Thus f is in the interior of Bψ . Lemma 3.2. Let B be any closed convex set in Rq with nonempty interior and no integral point in its interior. Then the following function ψ is valid for Rf whenever f ∈ Qq 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 ∈ Qq such that the point f + r is on the boundary of B, all other values of ψ being defined by homogeneity. Furthermore Bψ = B ∩ Qq . Proof. The convexity of B and homogeneity of ψ imply that ψ is subadditive: Let a, b ∈ Qq . 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 λψ(αa) + (1 − λ)ψ(βb) = 1 ≥ ψ(λαa + (1 − λ)βb) Set λ :=
β α+β
(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
Suppose x, s¯) ∈ Rf such that P ψ is not valid. Then there exists (¯ x ¯ = f + r¯ sr , the subadditivity and homogeneity of ψ imply X X ψ(¯ x − f ) = ψ( r¯ sr ) ≤ ψ(r)¯ sr < 1.
P
ψ(r)¯ sr < 1. Since
Thus x ¯ ∈ Zq is in Bψ but not on its boundary, a contradiction. Lemma 3.3. Let ψ and ψ 0 be two valid functions. Then ψ ≤ ψ 0 if and only if Bψ0 ⊆ Bψ . Proof. Immediate from the definition of Bψ . Theorem 3.4. If ψ is a finite minimal valid function for Rf , then ψ is a continuous nonegative homogeneous convex piecewise linear function with at most 2q pieces. Proof. Let ψ be a finite minimal valid function and let Bψ be the corresponding convex set as defined in (7). By Lemma 3.1, Bψ is convex, contains f in its interior and has integral points only on its boundary, and by Lemmas 3.2 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 Rq with no integral point in its interior is a polyhedron with at most 2q 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 xF in its relative interior. If there are more than 2q facets, two 0 0 integral points xF and xF must be identical modulo 2. Then their middle point 12 (xF + xF ) is integral and interior, contradiction. Consider a facet F of Bψ . We have ψ(x − f ) = 1 for all x ∈ F . By homogeneity (Lemma 2.4), ψ is linear in the cone {r ∈ Qq : r = λ(x − f ) with λ ≥ 0, x ∈ F }. Since the union of these cones over all facets of Bψ covers Qq , the function ψ is piecewise linear with at most 2q pieces.
integral point integral point
x-space
r2
integral point
2 1
f
integral point
r1
0
Bψ
ψ
Figure 1: A maximal lattice-free convex set and corresponding function ψ in R2 Corollary 3.5. Let ψ is a finite minimal valid function for Rf . Then Bψ is a polyhedron with at most 2q facets, each facet contains an integral point in its relative interior, and Bψ contains f but no integral point in its interior. 7
Conversely, any polyhedron B ∈ Qq 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 Bψ = B. To see this, as in Lemma 3.2, choose f ∈ Qq 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 Bψ 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 Rq with no integral point in its interior is a polyhedron with at most 2q facets. Lov´ asz [9] points out that the unbounded case reduces to the bounded case as follows. Let b1 , . . . , bq be any basis of the integer lattice. Let U1 (L1 ) be the linear subspace (lattice) spanned by b1 , . . . , bk and U2 the linear subspace spanned by bk+1 , . . . , bq . Let K1 be a maximal convex set in U1 free of the lattice L1 (By a lattice-free convex set S, we mean that S does not contain lattice points in its interior). Then the set K1 + U2 , called cylinder above K1 , 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 Rf , 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 ¯ψ be the closure of Bψ . B ¯ψ is a Bψ be the corresponding convex set as defined in (7). Let B ¯ψ lattice-free convex set and by Lemma 3.3, it is maximal. Therefore, by [6], [3] and [11], B q is a full-dimensional polyhedron with at most 2 facets. By Lemma 3.1, f ∈ Bψ . If f were in the interior of Bψ , then ψ would be finite by Lemma 3.2, contradicting our assumption. ¯ψ . In each facet of B ¯ψ containing f , we can Therefore the point f lies in one of the faces of B apply induction to conclude that ψ is piecewise linear in that subspace. In the directions r ¯ψ , we get a piecewise linear such that the ray f + λr, λ ≥ 0, goes through the interior of B 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 ¯ψ in f , we have ψ(r) = +∞. B
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