15.081J/6.251J Introduction to Mathematical Programming. Lecture 24: Discrete Optimization
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1 15.081J/6.251J Introduction to Mathematical Programming Lecture 24: Discrete Optimization
2 1 Outline Modeling with integer variables Slide 1 What is a good formulation? Theme: The Power of Formulations 2 Integer Programming 2.1 Mixed IP (MIP) max c x + h y s.t. Ax + By b x Z+(x n 0, x integer) n y (y 0) R + Slide Pure IP (IP) max s.t. Important special case: Binary IP c x Ax b x n Z + Slide 3 (BIP) max c x s.t. Ax b x {0, 1} n 2.3 LP (LP) max s.t. c x By b y R+ n Slide 4 3 Modeling with Binary Variables 3.1 Binary Choice { Slide 5 1, if event occurs x 0, otherwise Example 1: IP formulation of the knapsack problem n : projects, total budget b a j : cost of project j c j : value { of project j Slide 6 1, if project j is selected. x j = 0, otherwise. 1
3 max s.t. n c j x j a j x j b x j {0, 1} 3.2 Modeling relations At most one event occurs Neither or both events occur x j 1 j x 2 x 1 = 0 Slide 7 If one event occurs then, another occurs 0 x 2 x 1 If x = 0, then y = 0; if x = 1, then y is uncontrained 0 y Ux, x {0, 1} 3.3 The assignment problem n people m jobs c ij : cost { of assigning person j to job i. 1 person jis assigned to job i x ij = 0 min c ij x ij n s.t. x ij = 1 each job is assigned m x ij 1 each person can do at most one job. i=1 x ij {0, 1} 3.4 Multiple optimal solutions Generate all optimal solutions to a BOP. Slide 8 Slide 9 max c x s.t. Ax b x {0, 1} n Generate third best? Extensions to MIO? 2
4 3.5 Nonconvex functions How to model min c(x), where c(x) is piecewise linear but not convex? Slide 10 4 What is a good formulation? 4.1 Facility Location Data N = {1... n} potential facility locations I = {1...m} set of clients c j : cost of facility placed at j h ij : cost of satisfying client i from facility j. Decision variables { 1, a facility is placed at location j x ij = 0, otherwise y ij = fraction of demand of client i satisfied by facility j. Consider an alternative formulation. n m n IZ 1 = min c j x j + h ij y ij i=1 n s.t. y ij = 1 y ij x j x j {0, 1}, 0 y ij 1. n m n IZ 2 = min c j x j + h ij y ij i=1 n s.t. y ij = 1 m y ij m x j i=1 x j {0, 1}, 0 y ij 1. Slide 11 Slide 12 Slide 13 Are both valid? Which one is preferable? 4.2 Observations IZ 1 = IZ 2, since the integer points both formulations define are the same. Slide 14 n } 0 x j 1 P 1 = {(x, y) : y ij = 1, y ij x j, 0 y ij 1 3
5 Let n m P 2 = {(x, y) : y ij = 1, y ij m x j, 0 x j 1 0 y ij 1 i=1 Z 1 = min cx + hy, Z 2 = min cx + hy (x, y) P 1 (x, y) P 2 } Slide 15 Z 2 Z 1 IZ 1 = IZ Implications Finding IZ 1 (= IZ 2 ) is difficult. Slide 16 Solving to find Z 1, Z 2 is an LP. Since Z 1 is closer to IZ 1 several methods (branch and bound) would work better (actually much better). Suppose that if we solve mincx + hy, (x, y) P 1 we find an integral solution. Have we solved the facility location problem? Slide 17 Formulation 1 is better than Formulation 2. (Despite the fact that 1 has a larger number of constraints than 2.) What is then the criterion? 4.4 Ideal Formulations Let P be an LP relaxation for a problem Slide 18 Let H = {(x, y) : x {0, 1} n } P Consider Convex Hull (H) i i = {x : x = λ i x, λ i = 1, λ i 0, x H} i i The extreme points of CH(H) have {0, 1} coordinates. Slide 19 So, if we know CH(H) explicitly, then by solving min cx + hy, (x, y) CH(H) we solve the problem. Message: Quality of formulation is judged by closeness to CH(H). CH(H) P 1 P 2 4
6 5 Minimum Spanning Tree (MST) How do telephone companies bill you? It used to be that rate/minute: Boston LA proportional to distance in MST Other applications: Telecommunications, Transportation (good lower bound for TSP) Given a graph G = (V, E) undirected and Costs c e, e E. Find a tree of minimum cost spanning all the nodes. { 1, if edge e is included in the tree Decision variables x e = 0, otherwise The tree should be connected. How can you model this requirement? Let S be a set of vertices. Then S and V \ S should be connected } i S Let δ(s) = {e = (i, j) E : j V \ S Then, x e 1 What is the number of edges in a tree? Then, x e = n 1 e E Slide 20 Slide 21 Slide Formulation IZ MST = min c e x e e E H Is this a good formulation? x e 1 x e = n 1 e E x e {0, 1}. S V, S, V Slide 23 Slide 24 Is P cut the CH(H)? P cut = {x R E : 0 x e, x e = n 1 e E x e 1 S V, S, V } 5
7 5.2 What is CH(H)? Let P sub = {x R E : x e = n 1 e E(S) e E x e S 1 S V, S =, V } Slide 25 { } i S E(S) = e = (i, j) : j S Why is this a valid IP formulation? Slide 26 Theorem: P sub = CH(H). P sub is the best possible formulation. MESSAGE: Good formulations can have an exponential number of con straints. 6 The Traveling Salesman Problem Given G = (V, E) an undirected graph. V = {1,...,n}, costs c e e E. Find a tour that minimizes total length. Slide Formulation I { Slide 28 1, if edge e is included in the tour. x e = 0, otherwise. min c e x e e E s.t. x e 2, S E x e = 2, i V x e {0, 1} 6.2 Formulation II min c e x e s.t. x e S 1, S E e E(S) x e = 2, i V x e {0, 1} Slide 29 Slide 30 6
8 P TSP = {x R E ; cut x e 2, 0 x e 1} P TSP R E sub = {x ; x e = 2 x e S 1 0 x e 1} x e = 2 Slide 31 Theorem: P TSP = P TSP CH(H) cut sub Nobody knows CH(H) for the TSP 7 Minimum Matching Given G = (V, E); c e costs on e E. Find a matching of minimum cost. Slide 32 Formulation: min c e x e s.t. x e = 1, i V x e {0, 1} Let Is the LP ralaxation CH(H)? Theorem: P MAT = CH(H) 8 Observations P MAT = {x R E : x e = 1 x e 1 S = 2k + 1, S = x e 0} For MST, Matching there are efficient algorithms. CH(H) is known. Slide 33 Slide 34 For TSP efficient algorithm. TSP is an NP hard problem. CH(H) is not known. Conjuecture: The convex hull of problems that are polynomially solvable are explicitly known. 7
9 9 Summary 1. An IP formulation is better than another one if the polyhedra of their LP relaxations are closer to the convex hull of the IP. Slide A good formulation can have an exponential number of constraints. 3. Conjecture: Formulations characterize the complexity of problems. If a problem is solvable in polynomial time, then the convex hull of solutions is known. 8
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