Introduction to Integer Programming

Transcription

1 Lecture 3/3/2006 p. /27 Introduction to Integer Programming Leo Liberti LIX, École Polytechnique

2 Lecture 3/3/2006 p. 2/27 Contents IP formulations and examples Total unimodularity The Cutting Planes method Application to Travelling Salesman Problem (TSP) The Branch-and-Bound method

3 Lecture 3/3/2006 p. 3/27 Definitions Mathematical programming formulation: min x s.t. c T x Ax b x 0 x Z n [P] () The constraints x Z n are the integrality constraints The linear (or continuous) relaxation R P of P is obtained by P relaxing (i.e. removing) the integrality constraints Let F(P) be the feasible region of P : we have F(R P ) F(P) Let x be the solution of P and x the solution of R P ; then c T x c T x : R P is a lower bounding problem w.r.t. P

4 Lecture 3/3/2006 p. 4/27 Simple example Consider example: min 2x 3x 2 x + 2x 2 3 6x + 8x 2 5 x,x 2 Z + x 2 x 2 5/8 F(P) 5/8 F(R P ) 3/2 f 2 5/2 3 (2) () x 3/4 3/ f /2 2 5/2 3 (2) () x

5 Lecture 3/3/2006 p. 5/27 Maximum flow problem Given a network on a directed graph G = (V,A) with a source node s, a destination node t, and integer capacities u ij on each arc (i,j). We have to determine the maximum amount of integral material flow that can circulate on the network from s to t. The variables x ij Z, defined for each arc (i,j) in the graph, denote the number of flow units. i V, max x i s i t (i,j) A (i,j) A (i,j) A (s,i) A x si x ij = (j,i) A 0 x ij u ij x ij Z x ji x ji x ij i

6 Lecture 3/3/2006 p. 6/27 Transportation problem Let x ij be the (discrete) number of product units transported from plant i m to customer j n with respective unit transportation cost c ij from plant i to customer j. We model the problem of determining x minimizing the total cost, subject to production limits l i at plant i and demand d j at customer j, as follows: min x i m j n m i= n c ij x ij j= n x ij l i j= m x ij d j i= i,j x ij Z + i l i x ij j d j

7 Lecture 3/3/2006 p. 7/27 Set Covering problem Let x i = if a servicing facility will be built on geographical region i m and 0 otherwise. The cost of building a facility on region i is c i, and a ij = if a facility on region i can serve town j n, and 0 otherwise. We need to determine x {0, } m so that each town is serviced by at least one facility and the total cost is minimum. x i = min x j n m i= c i x i m a ij x i i= i m x i Z + i i x i = 0 a ij = j

8 Lecture 3/3/2006 p. 8/27 Travelling Salesman problem A travelling salesman must decide in which order to visit n cities so that its total travelled distance is minimized. Let c ij be the distance from city i to city j, and x ij = if the travelling salesman goes from city i to city j and 0 otherwise. min x i n j n S {,...,n} i j S c ij x ij i j n x ij = j n x ij = i n x ij S i j n x ij {0, } S i 0 0 i j

9 Lecture 3/3/2006 p. 9/27 Main algorithmic ideas If we can say a priori that x Z n then can solve P by simply solving R P (total unimodularity property) Add constraints to get P such that x Z n (cutting planes algorithm) Solve by smart enumeration of all solutions (Branch-and-Bound algorithm) Most IP methods use LP machinery (simplex algorithm) as their main algorithmic step The Cutting Planes and Branch-and-Bound algorithms progressively add constraints to the problem

10 Lecture 3/3/2006 p. 0/27 Cutting planes: definitions I A constraint C π T x π 0 is valid for P if x F(P) (π T x π 0 ) x 2 f C 2 () (2) x

11 Lecture 3/3/2006 p. /27 Cutting planes: definitions II Let P be problem P with the added valid constraint C. C is a cutting plane for P if F(R P ) F(R P ) x 2 C f () (2) x

12 Lecture 3/3/2006 p. 2/27 Cutting planes: definitions III Let f be the optimal objective function value for R P and f for R P. A cutting plane C is a valid cut for P if f > f x 2 C f f = f = () (2) x

13 Lecture 3/3/2006 p. 3/27 Convex hull Let ˆP be problem P with all possible cutting planes for P as added constraints: F(R ˆP) is the convex hull of F(P) x 2 C f C () (2) Theorem: the convex hull of a discrete set of points in R n is a polytope Computing the convex hull for F(P) is in general harder than solving P x

14 Lecture 3/3/2006 p. 4/27 Total unimodularity I Consider system Bx = b where B = (b ij ) is invertible n n s.t. b ij Z for all i,j Solve for x, get B b From inverse matrix formula, infer B = B C with C integral If B {, } then x = B b = ±Cb Z n A square invertible matrix B s.t. B = ± is unimodular An m n matrix A s.t. every square submatrix has determinant in {, 0, } is totally unimodular (TUM) Theorem: if A is TUM, then for all b R n, every vertex of the polyhedron {x R n Ax b} is integral. Intuititively, every vertex can be written as B b for B square submatrix of A

15 Lecture 3/3/2006 p. 5/27 Total unimodularity II If A is TUM, A T and (A I) are TUM TUM Sufficient conditions. An m n matrix A is TUM if:. for all i m, j n we have a ij {0,, }; 2. each column of A contains at most 2 nonzero coefficients; 3. there is a partition R,R 2 of the set of rows such that for each column j, i R a ij i R 2 a ij = 0. Example: take R = {, 3, 4}, R 2 = {2}

16 Lecture 3/3/2006 p. 6/27 Total unimodularity III Consider digraph G = (V,A) and a nonnegative flow x ij R + on each arc; the flow conservation equations i V x ij x ji = 0 yield a TUM matrix (i,j) A (j,i) A (partition: R = all rows, R 2 =, and work out the details) Maximum flow problem can be solved to integrality by simply solving the continuous relaxation with the simplex algorithm Show that the same can be said for the transportation problem Find counterexamples to show that this technique cannot be applied to the set covering and travelling salesman problem

17 Lecture 3/3/2006 p. 7/27 Cutting Plane Algorithm Overall strategy:. Solve R P, get relaxed solution x 2. If x Z n problem is solved, exit 3. Use solution x of R P to construct a valid cut C for P 4. Add the constraint C to the formulation of P 5. Go back to Crux of the algorithm: point 3 (separation problem) Cutting Plane algorithms may depend on the particular problem structure or be completely general Independent of problem structure: Gomory cutting planes Problem structure: Row generation for the TSP

18 Lecture 3/3/2006 p. 8/27 Row generation for TSP I TSP formulation has an exponential number of constraints (one for each proper subset of {,...,n}) Continuous relaxation solution becomes unmanageable as n grows Try relaxing (i.e., removing) problematic constraints S {,...,n} i j S x ij S Obtain IP with a TUM matrix whose solutions are sets of disjoint cycles Relaxed solution: Optimal solution:

19 Lecture 3/3/2006 p. 9/27 Row generation for TSP II Consider enforcing problematic constraints one by one in a Cutting Plane algorithm: how do we solve the separation problem? In example above, problematic constraint with S = {, 2} yields: x 2 + x 2 The relaxed solution above has x 2 = x 2 = 0 The constraint is valid but it is not a valid cut (i.e. current solution will not change when constraint is added to the continuous relaxation of the problem) Does not solve the separation problem Generate valid cuts by identifying disjoint cycles Usually requires considerably fewer than 2 n added constraints

20 Lecture 3/3/2006 p. 20/27 Row generation for TSP III In relaxed solution, S = {, 3, 5} is a disjoint cycle Enforce constraint for S (at most S = 2 arcs in complete digraph on S): x 3 + x 3 + x 5 + x 5 + x 35 + x 53 2 Get: In above relaxed solution, S = {4, 7} is a disjoint cycle, enforce constraint x 47 + x 74 Get optimal solution

21 Lecture 3/3/2006 p. 2/27 Branch-and-Bound I Applies to IPs in general form (assume WLOG c is an integral vector) Main idea:. Solve R P to get fractionary x 2. Choose j n s.t. x j Z 3. Split in two subproblems A,B 4. A: add constraint x j x j 5. B: add constraint x j x j 6. Recursively solve A, B 5/8 gradf 3/ P 3/ /8 3/ (3) / / A 3/2 2 5/2 3 (2) P (4) gradf /2 2 5/2 3 (2) B () ()

22 Lecture 3/3/2006 p. 22/27 Branch-and-Bound II. Initialize list problem L = {P } 2. Initialize best objective function value f =, x = infeasible 3. If L =, terminate with solution x 4. Select a subproblem Q from L and remove it from L 5. (Bound) Solve R Q to find solution x with objective value f 6. If f f, Q cannot contain optimal solution, back to 3 7. If x is integral and f < f : update x = x,f = f and go back to 3 8. (Branch) Select a fractionary component x j, generate two subproblems from Q with added constraints x j x j and x j x j respectively, add them to L, then back to 3

23 Lecture 3/3/2006 p. 23/27 Branch-and-Bound III How do we choose a subproblem Q from L (step 4)? How do we select a fractionary component x j from x (step 8)? No best answer, depends on problem structure Choice of subproblem: associate LB f to each generated problem, then choose subproblem with minimum LB Choice of fractionary component: choose the component with fractionary value closest to 0.5

24 Lecture 3/3/2006 p. 24/27 BB example I Consider simple example: min 2x 3x 2 x + 2x 2 3 6x + 8x 2 5 x,x 2 Z + Solution of R P is at x = (3/2, 3/4) with f = 2/4 5/8 3/4 3/2 x f F(R P ) 3/2 2 5/2 3 (2) () x

25 Lecture 3/3/2006 p. 25/27 BB example II x i = solution of R Pi, f i = optimal objective value of R Pi, i 5/8 3/2 3/4 3/ x 2 x (3) x 3 (4) f 3/2 2 5/2 3 (2) P 2 P P 3 x 2 = (, ) x = (.5, 0.75) x 3 = (2, 0.375) f 2 = 5 f = 5.25 f3 = 5.25 () L P x x 2 P 2 P 3 Z 2 f2 > f

26 Lecture 3/3/2006 p. 26/27 BB example III Initialization: let L = {P }, f = Choose problem P from L and remove it from L (P P ) Solve R P to get x = (.5, 0.75), f = 5.25 Since x Z 2 and f < f, branch Choose fractionary component x Generate two subproblems P 2,P 3, where P 2 is like P with added constraint x and P 3 is like P with added constraint x 2 Push P 2,P 3 onto L Iterate

27 Lecture 3/3/2006 p. 27/27 Course material C. Papadimitriou, K. Steiglitz, Combinatorial Optimization: Algorithms and Complexity, Dover, New York, 998