LP Relaxations of Mixed Integer Programs

Transcription

1 LP Relaxations of Mixed Integer Programs John E. Mitchell Department of Mathematical Sciences RPI, Troy, NY USA February 2015 Mitchell LP Relaxations 1 / 29

2 LP Relaxations LP relaxations We want to solve an integer program by solving a sequence of linear programs. We can relax the integrality constraints to give an LP relaxation. Let S := {x Z n : Ax b, x 0} = Z n P where P := {x R n : Ax b, x 0}. Since we restrict x 0, the polyhedron P has extreme points if it is nonempty. Mitchell LP Relaxations 2 / 29

3 LP Relaxations Convex hull of S Theorem The convex hull of S is a polyhedron. The proof is in Nemhauser and Wolsey. It is important to note that the extreme rays of conv(s) are the extreme rays of P. Mitchell LP Relaxations 3 / 29

4 LP Relaxations Equivalent LP formulation Our integer program of interest is min{c T x : x S} (IP). We relate this to the linear program min{c T x : x conv(s)} (CIP). We don t know the constraints of the linear program (CIP) explicitly. Mitchell LP Relaxations 4 / 29

5 LP Relaxations (IP) and (CIP) S c T x = z Mitchell LP Relaxations 5 / 29

6 LP Relaxations (IP) and (CIP) S conv(s) c T x = z Mitchell LP Relaxations 5 / 29

7 Value of (IP) and (CIP) LP Relaxations Theorem For any c R n, we have: 1 The objective value of (IP) is unbounded below if and only if the objective value of (CIP) is unbounded below. 2 If (CIP) has a bounded optimal value then it has an optimal solution that is an optimal solution to (IP). This optimal solution is an extreme point of conv(s). 3 If x 0 is an optimal solution to (IP) then it is an optimal solution to (CIP). The proof is in the text. It relies on the fact that the minimizer of a concave function over a convex set is at an extreme point. Mitchell LP Relaxations 6 / 29

8 Value of (IP) and (CIP) LP Relaxations Theorem For any c R n, we have: 1 The objective value of (IP) is unbounded below if and only if the objective value of (CIP) is unbounded below. 2 If (CIP) has a bounded optimal value then it has an optimal solution that is an optimal solution to (IP). This optimal solution is an extreme point of conv(s). 3 If x 0 is an optimal solution to (IP) then it is an optimal solution to (CIP). The proof is in the text. It relies on the fact that the minimizer of a concave function over a convex set is at an extreme point. Mitchell LP Relaxations 6 / 29

9 Value of (IP) and (CIP) LP Relaxations Theorem For any c R n, we have: 1 The objective value of (IP) is unbounded below if and only if the objective value of (CIP) is unbounded below. 2 If (CIP) has a bounded optimal value then it has an optimal solution that is an optimal solution to (IP). This optimal solution is an extreme point of conv(s). 3 If x 0 is an optimal solution to (IP) then it is an optimal solution to (CIP). The proof is in the text. It relies on the fact that the minimizer of a concave function over a convex set is at an extreme point. Mitchell LP Relaxations 6 / 29

10 Value of (IP) and (CIP) LP Relaxations Theorem For any c R n, we have: 1 The objective value of (IP) is unbounded below if and only if the objective value of (CIP) is unbounded below. 2 If (CIP) has a bounded optimal value then it has an optimal solution that is an optimal solution to (IP). This optimal solution is an extreme point of conv(s). 3 If x 0 is an optimal solution to (IP) then it is an optimal solution to (CIP). The proof is in the text. It relies on the fact that the minimizer of a concave function over a convex set is at an extreme point. Mitchell LP Relaxations 6 / 29

11 LP Relaxations Possible cases for an integer linear program Corollary (IP) is either unbounded or infeasible or has an optimal solution. Mitchell LP Relaxations 7 / 29

12 Set up LP relaxation LP Relaxations So in principle, could solve (IP) by solving the linear program (CIP). But we don t know a polyhedral description of conv(s). Look at the LP relaxation min {c T x : Ax b, x 0} = min {c T x : x P}. (LP) Mitchell LP Relaxations 8 / 29

13 LP Relaxations Compare (IP) to LP relaxation We can relate (LP) to (IP) in the following theorem. The proof exploits the fact that S P. Theorem Let z LP and z IP be the optimal values of (LP) and (IP) respectively. 1 If P = then S =. 2 If z LP is finite then either S = or z IP is finite. If S then z IP z LP. 3 If x LP is optimal for (LP) and if x LP Z n then x LP is optimal for (IP). Mitchell LP Relaxations 9 / 29

14 LP Relaxations Compare (IP) to LP relaxation We can relate (LP) to (IP) in the following theorem. The proof exploits the fact that S P. Theorem Let z LP and z IP be the optimal values of (LP) and (IP) respectively. 1 If P = then S =. 2 If z LP is finite then either S = or z IP is finite. If S then z IP z LP. 3 If x LP is optimal for (LP) and if x LP Z n then x LP is optimal for (IP). Mitchell LP Relaxations 9 / 29

15 LP Relaxations Compare (IP) to LP relaxation We can relate (LP) to (IP) in the following theorem. The proof exploits the fact that S P. Theorem Let z LP and z IP be the optimal values of (LP) and (IP) respectively. 1 If P = then S =. 2 If z LP is finite then either S = or z IP is finite. If S then z IP z LP. 3 If x LP is optimal for (LP) and if x LP Z n then x LP is optimal for (IP). Mitchell LP Relaxations 9 / 29

16 LP Relaxations Compare (IP) to LP relaxation We can relate (LP) to (IP) in the following theorem. The proof exploits the fact that S P. Theorem Let z LP and z IP be the optimal values of (LP) and (IP) respectively. 1 If P = then S =. 2 If z LP is finite then either S = or z IP is finite. If S then z IP z LP. 3 If x LP is optimal for (LP) and if x LP Z n then x LP is optimal for (IP). Mitchell LP Relaxations 9 / 29

17 LP Relaxations Cutting plane algorithm illustration Solve linear program relaxation {x R n + : Ax b} Soln to LP relaxation conv(s) Mitchell LP Relaxations 10 / 29

18 LP Relaxations Cutting plane algorithm illustration Solve linear program relaxation {x R n + : Ax b} Cutting plane a1 T x = b 1 Soln to LP relaxation conv(s) Mitchell LP Relaxations 10 / 29

19 LP Relaxations Cutting plane algorithm illustration Solve linear program relaxation {x R n + : Ax b, a1 T x b 1} Soln to LP relaxation conv(s) Mitchell LP Relaxations 10 / 29

20 LP Relaxations Cutting plane algorithm illustration Solve linear program relaxation {x R n + : Ax b, a T 1 x b 1} Cutting plane a 2 T x = b 2 Soln to LP relaxation conv(s) Mitchell LP Relaxations 10 / 29

21 LP Relaxations Cutting plane algorithm illustration Solve linear program relaxation {y R n + : Ax b, a1 T x b 1, a2 T x b 2} Solution to LP relaxation conv(s) Mitchell LP Relaxations 10 / 29

22 LP Relaxations Cutting plane algorithm illustration Solve linear program relaxation {x R n + : Ax b, a1 T x b 1, a2 T x b 2} Solution to LP relaxation conv(s) Solution is integral, so optimal to IP Mitchell LP Relaxations 10 / 29

23 LP Relaxations A cutting plane approach An integer program has many LP relaxations. The aim of a cutting plane approach is to add linear constraints to the original LP relaxation in order to tighten up the LP relaxation. These linear constraints are satisfied by every point in S but they might not be satisfied by every point in P. Adding these constraints gives an improved LP relaxation of the integer program. We would like to improve the relaxation so that it resembles conv(s) in some sense, at least in the neighborhood of an optimal solution to (IP). Mitchell LP Relaxations 11 / 29

24 Polyhedral description of conv(s) We would like to get a polyhedral description of conv(s). We d like as compact a description as possible. We first set up some notation. Given a matrix A R m n and a vector b R m, we construct the polyhedron Q := {x R n : Ax b}. Note: We have changed notation from Section 1. In particular, any nonnegativity constraints are included in the system Ax b. Mitchell LP Relaxations 12 / 29

25 Example 0 Full-dimensional example Q := x R 2 : x 1 + x 2 6 x 2 6 x 1 x 2 2 2x 1 + x 2 10 x 1 0 x 2 0 Mitchell LP Relaxations 13 / 29

26 Example 1 First example with implicit equality in R 2 Q := x R 2 : x 1 + x 2 3 x 1 + x 2 3 3x 1 + x 2 3 x 1 0 x 2 0 Mitchell LP Relaxations 14 / 29

27 Example 2 Second example with implicit equality in R 2 Q := x R 2 : 2x 1 + x 2 4 x 1 + x 2 4 x 1 0 x 2 0 Mitchell LP Relaxations 15 / 29

28 Implicit equality constraints We construct three sets of indices: M := {1,..., m} M = := {i M : A T i x = b i x Q} M := M \ M = = {i M : A T i x > b i for some x Q}. We also use the notation A =, A, b = and b in an analogous way. Mitchell LP Relaxations 16 / 29

29 Example 0 Full-dimensional example Q := x R 2 : x 1 + x 2 6 M x 2 6 M x 1 x 2 2 M 2x 1 + x 2 10 M x 1 0 M x 2 0 M Mitchell LP Relaxations 17 / 29

30 Example 1 First example with implicit equality in R 2 Q := x R 2 : x 1 + x 2 3 M = x 1 + x 2 3 M = 3x 1 + x 2 3 M x 1 0 M x 2 0 M Mitchell LP Relaxations 18 / 29

31 Example 2 Second example with implicit equality in R 2 Q := x R 2 : 2x 1 + x 2 4 M = x 1 + x 2 4 M = x 1 0 M = x 2 0 M Mitchell LP Relaxations 19 / 29

32 Valid inequalities Describing Polyhedra by Facets Definition The inequality π T x π 0 is a valid inequality for Q if it is satisfied by all points in Q. We also represent the inequality using the pair (π, π 0 ). Definition Let π T x π 0 be a valid inequality for Q. Let H = {x R n : π T x = π 0 } so F := Q H is a face of Q. Then (π, π 0 ) represents F. If F then (π, π 0 ) supports Q. We extend the notation M =, M to the face F using the notation M = F, M F. Mitchell LP Relaxations 20 / 29

33 Valid inequalities Describing Polyhedra by Facets Definition The inequality π T x π 0 is a valid inequality for Q if it is satisfied by all points in Q. We also represent the inequality using the pair (π, π 0 ). Definition Let π T x π 0 be a valid inequality for Q. Let H = {x R n : π T x = π 0 } so F := Q H is a face of Q. Then (π, π 0 ) represents F. If F then (π, π 0 ) supports Q. We extend the notation M =, M to the face F using the notation M = F, M F. Mitchell LP Relaxations 20 / 29

34 Impicit equalities for faces Proposition If F is a nonempty face of Q then F is a polyhedron and F = {x R n : ai T x = b i i M F =, at i x b i i M F }, where M F = M= and M F M. Mitchell LP Relaxations 21 / 29

35 Facets Proposition If F is a facet of Q then there exists some inequality a T k x b k with k M that represents F. Proof. Now dim(f) = dim(p)-1, so rank(a = F, b= F ) = rank(a=, b = ) + 1. Therefore, we only need one inequality in M that is in M F =, and this inequality represents the facet. Mitchell LP Relaxations 22 / 29

36 Facets Proposition If F is a facet of Q then there exists some inequality a T k x b k with k M that represents F. Proof. Now dim(f) = dim(p)-1, so rank(a = F, b= F ) = rank(a=, b = ) + 1. Therefore, we only need one inequality in M that is in M F =, and this inequality represents the facet. Mitchell LP Relaxations 22 / 29

37 One inequality per facet Proposition For each facet F of Q, one of the inequalities representing F is necessary in the description of Q. Proposition Every inequality a T i x b i for i M that represents a face of Q of dimension less than dim(q) 1 is irrelevant to the description of Q. Thus, later we will emphasize finding facets of the convex hull of S. Mitchell LP Relaxations 23 / 29

38 One inequality per facet Proposition For each facet F of Q, one of the inequalities representing F is necessary in the description of Q. Proposition Every inequality a T i x b i for i M that represents a face of Q of dimension less than dim(q) 1 is irrelevant to the description of Q. Thus, later we will emphasize finding facets of the convex hull of S. Mitchell LP Relaxations 23 / 29

39 Minimal representation of polyhedron With these propositions, we can now state two theorems showing how a polyhedron can be minimally represented using linear equalities and inequalities. Theorem (full dimensional case) A full dimensional polyhedron Q has a unique (to within scalar multiplication) minimal representation by a finite set of linear inequalities. In particular, for each facet F i of Q with i = 1,..., t, there is an inequality a T i x b i (unique to within scalar multiplication) representing F i and Q = {x R n : a T i x b i, i = 1,..., t}. Mitchell LP Relaxations 24 / 29

40 Example 0 Full-dimensional example Q := x R 2 : x 1 + x 2 6 M x 2 6 M x 1 x 2 2 M 2x 1 + x 2 10 M x 1 0 M x 2 0 M Mitchell LP Relaxations 25 / 29

41 Example 0 Full-dimensional example Q := x R 2 : x 1 + x 2 6 M x 2 6 M x 1 x 2 2 M 2x 1 + x 2 10 M x 1 0 M x 2 0 M Mitchell LP Relaxations 25 / 29

42 Minimal representation of polyhedron Theorem (not full dimensional) If dim(q) = n k with k > 0 then let t be the number of facets F i of Q, and we have Q = {x R n : a T i x = b i for i = 1,..., k, a T i x b i for i = k + 1,..., k + t}. The set {(a i, b i ) : i = 1,..., k} is a maximal set of linearly independent rows of (A =, b = ). For i = k + 1,..., k + t, (a i, b i ) is any inequality from the equivalence class of inequalities representing the facet F i of Q. Mitchell LP Relaxations 26 / 29

43 Example 1 First example with implicit equality in R 2 Q := x R 2 : x 1 + x 2 3 M = x 1 + x 2 3 M = 3x 1 + x 2 3 M x 1 0 M x 2 0 M Mitchell LP Relaxations 27 / 29

44 Example 1 First example with implicit equality in R 2 Q := x R 2 : x 1 + x 2 = 3 M = x 1 + x 2 3 M = 3x 1 + x 2 3 M x 1 0 M x 2 0 M Mitchell LP Relaxations 27 / 29

45 Example 1 First example with implicit equality in R 2 Q := x R 2 : x 1 + x 2 3 M = x 1 + x 2 = 3 M = 3x 1 + x 2 3 M x 1 0 M x 2 0 M Mitchell LP Relaxations 27 / 29

46 Example 1 First example with implicit equality in R 2 Q := x R 2 : x 1 + x 2 3 M = x 1 + x 2 = 3 M = 3x 1 + x 2 3 M x 1 0 M x 2 0 M Mitchell LP Relaxations 27 / 29

47 Example 2 Second example with implicit equality in R 2 Q := x R 2 : 2x 1 + x 2 4 M = x 1 + x 2 4 M = x 1 0 M = x 2 0 M Mitchell LP Relaxations 28 / 29

48 Example 2 Second example with implicit equality in R 2 Q := x R 2 : 2x 1 + x 2 4 M = x 1 + x 2 = 4 M = x 1 = 0 M = x 2 0 M Mitchell LP Relaxations 28 / 29

49 Example 2 Second example with implicit equality in R 2 Q := x R 2 : 2x 1 + x 2 = 4 M = x 1 + x 2 4 M = x 1 = 0 M = x 2 0 M Mitchell LP Relaxations 28 / 29

50 Example 2 Second example with implicit equality in R 2 Q := x R 2 : 2x 1 + x 2 = 4 M = x 1 + x 2 = 4 M = x 1 0 M = x 2 0 M Mitchell LP Relaxations 28 / 29

51 Summary The aim in a cutting plane approach is to try to develop a minimal description of the polyhedron conv(s), at least in the neighborhood of an optimal point to (IP). This pair of theorems tells us that it is enough to determine all the facets of conv(s), as well as a description of the affine hull of conv(s). Mitchell LP Relaxations 29 / 29