Multi-Row Cuts in Integer Programming. Tepper School of Business Carnegie Mellon University, Pittsburgh

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1 Multi-Row Cuts in Integer Programming Gérard Cornuéjols Tepper School o Business Carnegie Mellon University, Pittsburgh March 2011

2 Mixed Integer Linear Programming min s.t. cx Ax = b x j Z or j = 1,..., p x j 0 or j = 1,..., n. Common approach to solving MILP: First solve the LP relaxation. Basic optimal solution: x i = i + j N r j x j or i B. I i Z or some i B {1,..., p}, add cutting planes: Gomory 1963 Mixed Integer Cuts, Marchand and Wolsey 2001 MIR inequalities, Balas, Ceria and Cornuéjols 1993 lit-and-project cuts, or instance, are used in commercial codes.

3 Reerences This talk Borozan and Cornuéjols MOR 2009 Basu, Conorti, Cornuéjols and Zambelli SIDMA 2010 Basu, Conorti, Campelo, Cornuéjols and Zambelli IPCO 2010 Basu, Cornuéjols and Margot working paper 2010 Basu, Cornuéjols and Köppe working paper 2011 Related work Corner polyhedron Gomory LAA 1969 Gomory and Johnson MP 1972 Intersection cuts Balas OR 1971 The work that motivated me Andersen, Louveaux, Weismantel and Wolsey IPCO 2007 Dey and Richard MOR 2008 Dey and Wolsey IPCO 2008, SIOPT 2010

4 Corner Polyhedron Gomory 1969 Relax nonnegativity on basic variables x j. In our work, we make a urther relaxation, as suggested by Andersen, Louveaux, Weismantel and Wolsey 2007 Relax integrality on nonbasic variables. r 1 x = + k j=1 r j s j x Z q s 0 Example ) r 2 x1 Feasible set {( x 2 ( x1 x 2 Z 2 : ) = + r 1 s 1 + r 2 s 2 where s 1 0, s 2 0}

5 Formulas or Deriving Cutting Planes x = + k j=1 r j s j x Z q s 0 Every inequality cutting o the point ( x, s) = (, 0) can be expressed in terms o the nonbasic variables s only, in the orm k j=1 α js j 1. We are interested in ormulas or deriving such inequalities. More ormally, we are interested in unctions ψ : R q R such that the inequality k ψ(r j )s j 1 j=1 is valid or every choice o k and vectors r 1,..., r k R q. Such unctions ψ will be called valid unctions with respect to.

6 Intersection Cuts Balas 1971 Assume Z q. Want to cut o the basic solution s = 0, x =. r 2 S r 1 Any convex set S with int(s) with no integer point in int(s).

7 Intersection Cuts Balas 1971 Assume Z q. Want to cut o the basic solution s = 0, x =. r 2 S r 1 intersection cut Any convex set S with int(s) with no integer point in int(s). The gauge o S, i.e. ψ(r) = in {λ > 0 : 1 λ r S } is a valid unction. Intersection cut: ψ(r 1 )s 1 + ψ(r 2 )s 2 1.

8 Minimal Valid Functions Our main interest is in minimal valid unctions ψ : R q R, i.e. there is no valid unction ψ ψ where ψ (r) < ψ(r) or at least one r R q. r 2 S r 1 Bigger convex set

9 Minimal Valid Functions Our main interest is in minimal valid unctions ψ : R q R, i.e. there is no valid unction ψ ψ where ψ (r) < ψ(r) or at least one r R q. r 2 S r 1 intersection cut Bigger convex set Better cut: ψ(r 1 )s 1 + ψ(r 2 )s 2 1.

10 Theorem Borozan and Cornuéjols MOR 2009 On Q q (extension to R q due to Basu, Conorti, Cornuéjols, Zambelli) Let R q \ Z q. I ψ : R q R is a minimal valid unction, then ψ is nonnegative piecewise linear positively homogeneous and convex. Furthermore B ψ := {x R q : ψ(x ) 1} is a maximal Z q -ree convex set containing in its interior. Conversely, or any maximal Z q -ree convex set B containing in its interior, the gauge o B is a minimal valid unction ψ. DEFINITION A convex set is Z q -ree i it does not have any integral point in its interior. However, it may have integral points on its boundary.

11 Maximal Z q -Free Convex Sets...are polyhedra Lovász 1989 Z q -ree convex set contains no integral point in its interior Maximal: each edge contains an integral point in its relative interior.

12 Maximal Z q -Free Convex Sets...are polyhedra Lovász 1989 Z q -ree convex set contains no integral point in its interior Maximal: each edge contains an integral point in its relative interior.

13 Maximal Z q -Free Convex Sets...are polyhedra Lovász 1989 Z q -ree convex set contains no integral point in its interior Maximal: each edge contains an integral point in its relative interior. In the plane: it is a strip, a triangle or a quadrilateral.

14 Maximal Z q -Free Sets in the Plane Split, triangles and quadrilaterals generate split, triangle and quadrilateral inequalities ψ(r)s r 1, where the unction ψ is the gauge o S..

15 Maximal Z q -Free Sets in the Plane Split, triangles and quadrilaterals generate split, triangle and quadrilateral inequalities ψ(r)s r 1, where the unction ψ is the gauge o S. I S = {x R q : a i (x ) 1, i = 1,..., t}, then ψ = max i=1,...,t a i r.

16 Lovász Theorem THEOREM i and only i A set K R q is a maximal Z q -ree convex set either K is a polyhedron o the orm K = P + L where P is a polytope, L is a rational linear space, dim(p) + dim(l) = p, K does not contain any point o Z q in its interior and there is a point o Z q in the relative interior o each acet o K. or K is an irrational hyperplane. irrational hyperplane cylinder

17 Generalization o the Lovász and Borozan-Cornuéjols theorems Here, we consider a system o the orm x = + k j=1 r j s j x S s 0 where S = P Z q or some rational polyhedron P R q. This model has been studied in the 70s by Glover 1974, Balas 1972 and Johnson 1981, and recently by Dey and Wolsey 2009, and Günlük and Fukusawa Basu, Conorti, Cornuéjols and Zambelli generalize the Lovász and Borozan-Cornuéjols theorems to such systems SIDMA 2010.

18 Integer Liting Dey-Wolsey 2010 QUESTION: How should we deal with INTEGER nonbasic variables?

19 Integer Liting Here, we consider a system o the orm x = + k j=1 r j s j + l i=1 ρi y i x Z q s 0 y Z l. We are interested in unctions ψ : R q R and φ : R q R such that the inequality k ψ(r j )s j + j=1 l φ(ρ i )y i 1 i=1 is valid or every choice o integers k, l and vectors r 1,..., r k R q and ρ 1,..., ρ l R q.

20 Integer Liting Basu, Campelo, Conorti, Cornuéjols, Zambelli IPCO 2010 Starting rom a minimal valid unction ψ : R q R, what can we say about a minimal liting unction φ? Clearly, φ ψ. Are there regions R where we can guarantee that φ(r) = ψ(r) or all r R? THEOREM Let ψ be minimal. φ(r) = ψ(r) or r R = t R(x t) where the union is taken over all integral points x t on the boundary o the maximal Z q -ree convex set B ψ deining ψ. Conversely, i r R, there exists a minimal liting φ where φ(r) < ψ(r). x3 R(x3) R(x2) Bψ R(x1) x1 x2 THEOREM A minimal unction ψ has a unique minimal liting φ i and only i R + Z q covers R q.

21 Body with a Unique Liting Characterizing when the integer liting is unique. Example: Split inequalities, Gomory Mixed Integer Cuts. Another example:

22 Bodies with a Unique Liting Basu, Cornuéjols, Köppe 2011 THEOREM Let B be a maximal lattice-ree simplicial polytope in R n. Then B is either a body with a unique liting or all int(b), or a body with multiple litings or all int(b). THEOREM Let be a simplex in R n such that it is a maximal lattice-ree convex body and each acet o has exactly one integer point in its relative interior. Then is a body with a unique liting or all int(b) i and only i all the vertices o are integral, i.e., is an aine unimodular transormation o conv{0, ne 1,..., ne n }

23 Bodies with a Unique Liting Basu, Cornuéjols, Köppe 2011 THEOREM Let R n+1 be a maximal lattice-ree 2-partitionable simplex with hyperplanes H 1, H 2 such that H 1 deines a acet o and this is the only acet o with more than one lattice point in its relative interior. Then is a body with a unique liting or all int(b) i and only i H 2 is an aine unimodular transormation o conv{0, ne 1,..., ne n } THEOREM Let B R n be a maximal lattice-ree simplicial polytope and let int(b). Then the volume o the region R where the liting is unique is an aine unction o the coordinates o.

24 Split Inequalities Cook-Kannan-Schrijver 1990 Widely used in commercial solvers. QUESTION: Can we generate any intersection cut using a sequence o split inequalities?

25 Split Inequalities Cook-Kannan-Schrijver 1990 P := {x R n : Ax b} S := P (Z p R n p ). For π Z n such that π p+1 =... = π n = 0 and π 0 Z, deine πx π 0 πx π split inequality Π 1 := P {x : πx π 0 } Π 2 := P {x : πx π 0 + 1} Π 1 P Π 2 We call cx c 0 a split inequality i there exists (π, π 0 ) Z p Z such that cx c 0 is valid or Π 1 Π 2.

26 Split Inequalities Cook-Kannan-Schrijver 1990 P := {x R n : Ax b} S := P (Z p R n p ). For π Z n such that π p+1 =... = π n = 0 and π 0 Z, deine πx π 0 πx π split inequality Π 1 := P {x : πx π 0 } Π 2 := P {x : πx π 0 + 1} Π 1 P Π 2 We call cx c 0 a split inequality i there exists (π, π 0 ) Z p Z such that cx c 0 is valid or Π 1 Π 2. The split closure is the intersection o all split inequalities. THEOREM Cook, Kannan, Schrijver 1990 The split closure is a polyhedron.

27 Split Rank Cook-Kannan-Schrijver 1990 P := {x R n : Ax b} S := P (Z p R n p ). Let P 0 = P. For k 1, let P k denote the split closure o P k 1. Let αx β be a valid inequality or conv(s). The smallest k such that αx β is valid or P k is called the split rank o αx β, i such an integer k exists.

28 Split Rank Cook-Kannan-Schrijver 1990 Let P 0 = P. For k 1, let P k denote the split closure o P k 1. Let αx β be a valid inequality or conv(s). The smallest k such that αx β is valid or P k is called the split rank o αx β, i such an integer k exists. In the mixed integer case, inequalities may have ininite split rank, i.e. there is no inite k such that αx β is valid or P k, as shown by the ollowing example. O C B A P is a simplex with vertices O = (0, 0, 0), A = (2, 0, 0), B = (0, 2, 0) and C = ( 1 2, 1 2, 1 2 ). S := P (Z 2 R). Thus conv(s) = P {y 0}.

29 Split Rank Cook-Kannan-Schrijver 1990 In the mixed integer case, inequalities may have ininite split rank, i.e. there is no inite k such that αx β is valid or P k, as shown by the ollowing example. O C B A P is a simplex with vertices O = (0, 0, 0), A = (2, 0, 0), B = (0, 2, 0) and C = ( 1 2, 1 2, 1 2 ). S := P (Z 2 R). Thus conv(s) = P {y 0}. Consider a simplex P with vertices O, A, B and C = ( 1 2, 1 2, t) with t > 0. Let C 1 = C, let C 2 be the point on the edge AC with coordinate x 1 = 1 and C 3 the point on BC with x 2 = 1. Observe that no split disjunction removes all three points C 1, C 2, C 3. Thus ( 1 2, 1 2, t 3 ) P1. By induction, ( 1 2, 1 2, t 3 k ) P k. Thereore y 0 has ininite split rank.

30 The Andersen-Louveaux-Weismantel-Wolsey Model The Cook-Kannan-Schrijver example can be written as x 1 y, x 2 y, x 1 + x 2 + 2y 2. Introducing nonnegative slack variables, and eliminating y, we get Note that y 0 s 1 + s 2 + s 3 2 Remember the Andersen, Louveaux, Weismantel, Wolsey 2007 model: x 1 = s s s 3 x 2 = s s s 3 x i Z or i = 1, 2 s j 0 or j = 1, 2, 3. x 1 = 1 + n j=1 r j 1 s j x 2 = 2 + n j=1 r j 2 s j x i Z or i = 1, 2 s j 0 or j = 1,..., n.

31 The Cook-Kannan-Schrijver Example Continued (0,1) r 2 r 1 r 3 (0,0) (1,0) x 1 = s s s 3 x 2 = s s s 3 x i Z or i = 1, 2 s j 0 or j = 1, 2, 3. Recall: Inequality with ininite split rank is s 1 + s 2 + s 3 2 This is the intersection cut associated with the triangle (0,1) r 2 r 1 r 3 (0,0) (1,0)

32 The Dey-Louveaux Theorem 2009 Andersen, Louveaux, Weismantel, Wolsey 2007 model in R 2 : x 1 = 1 + n j=1 r j 1 s j x 2 = 2 + n j=1 r j 2 s j x i Z or i = 1, 2 s j 0 or j = 1,..., n. THEOREM Every intersection cut has a inite split rank, except or those generated rom a maximal Z 2 -ree triangle with integral vertices and rays pointing to the corners. (0,1) r 2 r 1 r 3 (0,0) (1,0)

33 A Property o the Triangles that Generate Intersection Cuts with Ininite Split Rank (0,1) r 2 (0,1) r 2 r 1 r 3 (0,0) (1,0) r 1 r 3 (0,0) (1,0) Not all integral points can it on the two parallel lines o a split. IMPRECISE DEFINITION I every integral point o K R q lies on the two parallel hyperplanes o a split, we say that K has the 2-hyperplane property.

34 Intersection Cuts with Finite Split Rank THEOREM Basu, Cornuéjols, Margot 2010 Let K be a rational lattice-ree polytope in R q containing in its interior and having rays going into its corners. The intersection cut arising rom K has inite split rank i and only i K has the 2-hyperplane property. PRECISE DEFINITION A set S o points in R q is 2-partitionable i either S 1 or there exists a partition o S into nonempty sets S 1, S 2 and a split such that S 1 is contained in one o its boundary hyperplanes and S 2 is contained in the other. A polytope is 2-partitionable i its integer points are 2-partitionable. Let K I be the convex hull o the integer points in K. We say that K has the 2-hyperplane property i every ace o K I that is not contained in a acet o K is 2-partitionable.

35 Idea o Proo I K does not have the 2-hyperplane property, it is not too hard to show that the intersection cut arising rom K has ininite split rank. The diicult part o the theorem is to show i K has the 2-hyperplane property, then the intersection cut arising rom K has inite split rank. Our proo is by induction on the dimension q. We deine the notions o intersecting split and englobing split, and we show that the theorem holds when there is a sequence o intersecting splits ollowed by an englobing split. We use Chvátal cuts to reduce K to K I. The theorem is proved by replacing each o the Chvátal cuts by a inite collection o intersecting splits or enlarged polytopes, and using the 2-hyperplane property or proving that a inal englobing split exists.

36 Thank you Papers available on