Integer Programming (reduced to 180 minutes)

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1 Integer Programming (reduced to 10 minutes) 1 1 Departamento de Estadística e Investigación Operativa Universidad de Murcia doc-course Constructive approximation, Optimization and Mathematical Modeling Sevilla 2010

2 SESSION ONE. MOTIVATION

3 An integer (linear) programming problem is min c x >< s.t. Ax b (A x b ) (A x = b ) x Z n + with c R n, A m n = (a ij ), a ij R i, j, b R m For instance, max 40x x 2 >< s.t. x 1 4 6x x 2 33 x 2 2 x 1, x 2 Z +. n = 2, m = 3 c = ( 40, 55) «1 6 0 A = b 33 2 A

4 By removing the integrity constraints, we get the linear relaxation of this formulation. We know how to solve it in an efficient way: (dual) simplex algorithm. In the example, max 40x x 2 >< s.t. x 1 4 6x x 2 33 x 2 2 x 1, x 2 0 the optimal solution `4, 5 11 is not integer. We should not waste our time rounding it: (4, 1) does not satisfy the second constraint and (4, 0) gives an objective value 160, worse than (3, 1).

5 Some academic examples 1 Decide how to invest c e if there are n projects available. To invest in the j-th project costs c j 0 e and gives a profit b j > c j. Variables x j = ( 1 if project j is chosen 0 otherwise >< s.t. max nx (b j c j )x j j=1 nx c j x j c j=1 x j {0, 1} x j 0 j x j 1 j x j Z j j j = 1,..., n

6 Some academic examples 2 A businessman owns a company with n workers. In some moment, n tasks must be done and each worker must do one task. The profit of assigning worker i to task j is b ij. x ij = ( 1 if task j is assigned to worker i 0 otherwise i, j = 1,..., n max >< s.t. nx nx b ij x ij i=1 j=1 nx x ij = 1 j i=1 nx x ij = 1 i j=1 x ij {0, 1} i, j

7 Some academic examples 3 Given some regions, we want to decide where to install emergency centers. The cost of a center in region j is c j. Not all the regions can be attended from a center installed at a given region and there cannot be unattended regions. Data: Variables: a ij = 1 if region i can be attended from region j x j = 1 if we choose region j to install a center >< min s.t. P j c jx j P n j=1 a ijx j 1 x j {0, 1} j i

8 Some academic examples 4 Find a shortest path from s to t on a directed graph (V, A) with lengths l ij > 0. Variables: x ij = 1 if arc (i, j) belongs to the path. min >< s.t. P i P j l ijx ij P k x sk = 1 P k x ik P k x ki = 0 i s, t P k x kt = 1 x ij {0, 1} i, j

9 Some academic examples 5 Find a shortest tour traversing all nodes on a directed graph (V, A) with lengths l ij > 0. Variables: x ij = 1 if arc (i, j) belongs to the tour min >< s.t. P P i j l ijx ij P j x ij = 1 i P i x ij = 1 j x ij {0, 1} i, j X X x ij S 1 S N, 2 S n 1 i S j S

10 Real life applications Facility location Vehicle routing High schools timetables Pick and delivery Inventory control Sensitive data protection Industrial design Task sequencing Military Supply chain management Recycling Polling districts design Tolls Voting systems design Telecom networks design Clustering Data bases design Insurance Archaeology Psychology Sudokus Many more...

11 Combinatorial auctions How to assign items to bidders when bids are done on bundles of objects?

12 Combinatorial auctions Data: J: bidders I : objects S: bundles (subsets of I ) b js money j J offers by s S M bundles somebody is interested in Variables: y js = ( 1 if bundle s is assigned to bidder j 0 otherwise j J, s M >< max s.t. P j J P j J P s M b jsy js P s M:s i y js 1 i I y js {0, 1} j J, s M X y js 1 j J s M

13 Football! In some league of some country, a double round tournament has been designed to satisfy televisions. The only thing we must decide is if team A receives team B at home either in the first or in the second round, in order to minimize the number of times each team plays two consecutive matches a home or as a visitor. Variables (tricky): y i = 1 if team i plays first day at home v it = 1 if team i goes back home after t-th day s it = 1 if team i goes out after t-th day x it = 1 if team i plays at home day t P P max i t (s it + v it ) s.t. x it + x jt = 1 if i plays against j day t >< s it + v it 1 i, t x it + x i,t+(n 1) = 1 i, t x it = y i + P j<t (v ij s ij ) i, t y i = x i1 i x it, y i, v it, s it {0, 1} i, t

14 Proteins

15 Proteins

16 Proteins comparison Unfolded protein Folded protein Representation by means of a (sorted) graph:

17 Proteins comparison Unfolded protein Folded protein Representation by means of a (sorted) graph:

18 Proteins comparison Joints in the second level are used to compare two proteins G 1 = (V i, E 1) G 2 = (V 2, E 2) V 1 = (1,..., n 1) V 2 = (1,..., n 2) looking for i 1 <... < i k j 1 <... < j k in such a way that the number of times an edge (i a, i b ) in graph 1 meets an edge (j a, j b ) in graph 2 be maximized.

19 Proteins comparison

20 Proteins comparison Data: b i1 j 1 i 2 j 2 = 1/2 if (i 1, i 2) E 1 and (j 1, j 2) E 2. Variables: x ij = 1 if node i in first graph matches node j in the second graph Nonlinear! max >< s.t. Pi 1 Pi 2 Pj 1 P j 2 b i1 j 1 i 2 j 2 x i1 j 1 x i2 j P 2 j x ij 1 i P i x ij 1 j x i1 j 1 + x i2 j 2 1 if edges cross each other x ij {0, 1} i, j

21 Formulations Definition A polyhedron P R n is P = {x R n ; Ax b} where (A, b) has order m (n + 1). Definition A polyhedron P R n is a formulation of a set X Z n iff X = P Z n. Same feasible set, different formulations

22 Formulations Definition Given a finite set X, the convex hull of X, denoted conv(x ), is: conv(x ) = {x : x = tx λ i x i, i=1 tx λ i = 1, λ i 0 i i=1 over all subsets {x 1,..., x t } of X } conv(x ) is a polyhedron All the extreme points of conv(x ) yield in X Consequently we could replace by which is a linear programming problem. max{cx : x X } max{cx : x conv(x )},

23 Formulations A formulation P 1 is better than other formulation P 2 if P 1 P 2 X = {(0, 0, 0, 0), (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (0, 1, 0, 1), (0, 0, 1, 1)}. P 1 = {0 x 1, 3x x x x 4 100} P 2 = {0 x 1, 4x 1 + 3x 2 + 2x 3 + x 4 4} 9 4x 1 + 3x 2 + 2x 3 + x 4 4 >< >= x 1 + x 2 + x 3 1 P 3 = x 1 + x 4 1 >; 0 x 1 conv(x ) = P 3 P 2 P 1

24 SESSION TWO. RESOLUTION

25 Step 1. REFORMULATION Example: p-median problem.

26 Step 1. REFORMULATION Example: p-median problem.

27 Step 1. REFORMULATION Data: n number of points p number of medians distance between i and j d ij Variables: j 1 if j is median y j = 0 otherwise j 1 if point i is allocated to median j x ij = 0 otherwise min s.t. >< P P i j d ijx ij P j x ij = 1 i P i x ij n y j j P j y j = p x ij {0, 1} i, j y j {0, 1} j min s.t. >< P P i j d ijx ij P j x ij = 1 i x ij y j i, j P j y j = p x ij {0, 1} i, j y j {0, 1} j

28 Step 1. REFORMULATION

29 Step 2. CUTTING THE POLYHEDRON Definition An inequality πx π 0, with π R n, π 0 R, is called valid inequality for P R n if it is satisfied by all the points in P.

30 Generating valid inequalities (cutting planes) Start with (P) >< min s.t. cx Ax = b x j Z + Let x be the optimal solution for the linear relaxation of (P). Choose x k such that x k is not integer. Then, an equality x k + X j / B y kj x j = x k can be extracted from the dual simplex final tableau. j

31 Generating valid inequalities (cutting planes) Splitting integer parts and fractional parts x k + X j / B ( y kj + f kj )x j = x k + f k, and then x k + X z } { y kj x k = X f kj x j +f k f k. j / B j / B {z } {z } 0 Z LHS is integer in the feasible region and f k < 1, thus 0 Therefore valid inequality X f kj x j f k j / B 0 can be added to the problem.

32 Example >< max 3x 1 + 4x 2 >< max 3x 1 + 4x 2 s.t. 3x 1 + x 2 + x 3 = 6 s.t. 2x 1 + 3x 2 + x 4 = 9 x x3 1 7 x4 = 9 7 x x x4 = 15 7 From the first equality 3 7 x3 6 7 x From the second equality 5 7 x3 3 7 x4 1 7

33 Generating valid inequalities (cutting planes) 2 min >< s.t. (P) cx Ax b x 0 x j Z j I. Let x be the optimal solution for the linear relaxation of (P). Choose x k, k I, such that x k is not integer. Then, an equality x k + X j / B y kj x j = x k can be extracted from the dual simplex final tableau.

34 Generating valid inequalities (cutting planes) 2 Re-writing: x k x k = f k X j / B y kj x j. We distinguish two cases: 1 If x k x k then x k x k 0 and therefore X y kj x j f k. (1) j / B 2 If x k x k + 1, it follows f k X j / B y kj x j 1 and then X y kj x j f k 1. (2) j / B

35 We define From (1) and From (2) and then B + := {j / B : y kj > 0} B := {j / B : y kj < 0} 0 X z X } { y kj x j y kj x j f k {z} {z} j B + X j B < 0 j B + y kj x j f k. 0 X z X } { y kj x j + y kj x j f k 1 {z} {z} j B j B + > X y kj x j f k 1 X {z } j B < 0 0 j B 0 0 f k 1 f k y kj x j f k.

36 Inequality X j B II : I : X y kj x j f k j B + X f k y kj x j f k 1 f k j B f k y kj x j X 1 f k j B + y kj x j f k is valid since in case I we sum negative terms and in case II we sum also negative terms. Moreover, RHS is negative and then the fractional optimal solution is cut off.

37 Example max 4x 1 + 5x 2 + 3x 3 max 4x 1 + 5x 2 + 3x 3 s.t. 3x 1 + 4x 2 + x 4 = 10 5 >< 2x 1 + x 2 + x 3 + x 5 = 7 >< s.t. 3 x1 + x x5 1 3 x6 = x 1 + 4x 2 + x 3 + x 6 = 12 x1 + x x5 4 3 x6 = x 2, x 4, x 5, x 6 0 x1 + x2 1 3 x x6 = 5 3 x 1, x 3 Z + x i 0 i. From the first equation 5 3 x1 4 3 x5 1 1 {z } 3 2 x6 1 {z } 3. B + B

38 Step 3. CUTTING MORE DEEPLY Definition The dimension of a polyhedron P R n is k if the maximum number of affinely independent points in P is k + 1. P R n has complete dimension if dim(p) = n. Definition If πx π 0 is a valid inequality of P, F = {x P; πx = π 0} is a face of P. Definition A face F of P is a facet of P if dim(f ) = dim(p) 1.

39 Definition Inequality πx π 0 dominates inequality γx γ 0 if a real number µ > 0 exists such that µγ π and π 0 µγ 0. In such a way {x R n +; πx π 0} {x R n +; γx γ 0} Definition A valid inequality of P is maximal if it is not (strictly) dominated by any other inequality of P. Every facet is maximal valid inequality.

40 Example: the set packing problem (SPP) where A is a 0-1 matrix. >< max cx s.t. Ax 1 x {0, 1} n (EX) max >< cx s.t. x 1 + x 2 1 x 1 + x 5 1 x 1 + x 6 1 x 2 + x 3 1 x 2 + x 6 1 x 3 + x 4 1 x 3 + x 6 1 x 4 + x 5 1 x {0, 1} n

41 Why set packing? Auctions problem is set packing: P max >< P s.t. j J j J P s M b jsy js P s M:s i y js 1 i I P s M y js 1 j J y js {0, 1} j J, s M Proteins problem is (almost) set packing: P max Pi 1 Pi 2 Pj 1 j 2 b i1 j 1 i 2 j 2 x i1 j 1 x i2 j P 2 >< s.t. j x ij 1 i P i x ij 1 j x i1 j 1 + x i2 j 2 1 if edges cross each other x ij {0, 1} i, j

42 Why set packing? Consider the partitioning problem with formulation P P >< max i j w ijx ij s.t. x ij x ik + x jk i, j, k x ij {0, 1} i, j where x ij = 1 if nodes i and j are in different clusters. Where are set packing constraints?

43 Why set packing? Defining new variables >< 1 if i / j, k y ijk = x ij x jk = 1 if k / i, j 0 otherwise and adding new constraints like y y (because 1 / 2, 3 implies 1 and 2 are in different clusters 4 / 1, 2 implies 1 and 2 are in the same cluster) we enforce the formulation with set packing constraints.

44 STILL IN Step 3. CUTTING MORE DEEPLY The set packing problem polyhedron has complete dimension: the n + 1 vectors 0 = (0, 0, 0,..., 0) >< e 1 = (1, 0, 0,..., 0) e 2 = (0, 1, 0,..., 0). e n = (0, 0, 0,..., 1) belong to the polyhedron and are affinely independent. Therefore, facets will have dimension n 1.

45 Some facets for SPP (EX) max >< cx s.t. x 1 + x 2 1 x 1 + x 5 1 x 1 + x 6 1 x 2 + x 3 1 x 2 + x 6 1 x 3 + x 4 1 x 3 + x 6 1 x 4 + x 5 1 x {0, 1} n Looking for maximal subsets of variables pairwise incompatible, clique facets are obtained: x 1 + x 2 + x 6 1 x 2 + x 3 + x 6 1 x 1 + x 5 1 x 3 + x 4 1 x 4 + x 5 1

46 Proof for one of them (EX) max >< cx s.t. x 1 + x 2 1 x 1 + x 5 1 x 1 + x 6 1 x 2 + x 3 1 x 2 + x 6 1 x 3 + x 4 1 x 3 + x 6 1 x 4 + x 5 1 x {0, 1} n x 1 + x 2 + x 6 1 Independent points:

47 Example of lifted odd hole facet (EX) max >< cx s.t. x 1 + x 2 1 x 1 + x 5 1 x 1 + x 6 1 x 2 + x 3 1 x 2 + x 6 1 x 3 + x 4 1 x 3 + x 6 1 x 4 + x 5 1 x {0, 1} n x 1 + x 2 + x 3 + x 4 + x 5 + x 6 2 Independent points:

48 Step 4. BRANCH AND BOUND >< min (P) s.t. cx Ax b x j Z + j Solve the linear relaxation of (P). If x a / Z, add x a x a to obtain subproblem P 1 y x a x a to obtain subproblem P 2. Divide and conquer (branching): (P 1) (P) + {x a 4} (P 2) (P) + {x a 5} (P 21) (P 2) + {x b 5} (P 22) (P 2) + {x b 6} Bounding: If the optimal value for the linear relaxation of some subproblem is greater than an upper bound on the optimal value of (P), the optimal solution for (P) is not inside the feasible region of the subproblem.

49 BETTER LOWER BOUNDS: Lagrangian relaxation Start with (PP) >< min s.t. with A m n, b R m, X an arbitrary set. The Lagrangian relaxed problem is: with u R m. (PR(u)) θ(u) = cx Ax b x X, ( min cx + u(ax b) s.t. x X, The Lagrangian dual problem is (PD) ( max θ(u) s.t. u 0.

50 Example min >< s.t. P m P n i=1 j=1 c ijx ij P n j=1 a ijx ij b i i P m i=1 x ij = 1 j x {0, 1} m n Relaxing the first family >< min θ(u) = s.t. P i,j c ijx ij + P i u i( P j a ijx ij b i ) P i x ij = 1 j x ij {0, 1} i, j

51 This problem can be split into n independent subproblems: >< min s.t. j P i (c ij + u i a ij )x ij P i x ij = 1 x ij {0, 1} i with optimal solution Then θ(u) = X j min i {c ij + u i a ij }. min{c ij + u i a ij } X i i b i u i and (PD) max u 0 {X j min{c ij + u i a ij } X i i b i u i }

52 BETTER LOWER BOUNDS: Lagrangian relaxation Denote F ( ) the feasible region of a problem and v( ) its optimal value. Let PP be the linear relaxation of (PP). Theorem (Weak duality) If x F (PP), u F (PD), then cx θ(u). Theorem θ(u) cx + u(ax b) cx. v(pp) v(pd) v(pp) = min {cx : x F (PP)} θ(u) Consequently v(pp) v(pd). v(pp) v(pd) is called duality gap. u F (PD).

53 Corollary If then x is optimal of (PP). x F (P) u F (PD) cx = θ(u)

54 Theorem (Strong duality) If there exist u 0, x such that i) x is optimal for PR(u), ii) x is feasible for (PP), iii) u(ax b) = 0, then x is optimal for (PP). In x the minimum of is reached. Then {cx + u(ax b)} x X θ(u) = cx + u(ax b) = cx.

55 It can be really better bound min cx >< s.t. Ax b, (PP) Qx q, Rx = r, x 0, x j Z j J, min cx >< s.t. Ax b, (PP) Qx q, Rx = r, x 0, with X = {x 0 : Qx q, Rx = r, x j Z j J}. v(pd) = max {min {cx + u(ax b) : Qx q, Rx = r, x 0, x j Z j J}} u 0 max { ub+ min {(c + ua) x : Qx q, Rx = r, x 0}} = u 0 max { ub+ max { qα + rβ : αq + βr c + ua, α 0}} = u 0 max {qα + rβ ub : αq + βr ua c, α 0, u 0} = min {cx : Qx q, Rx = r, Ax b, x 0} = v(pp).

56 Subgradient algorithm Proposition If X is a compact set, then θ is a concave function on R m. Theorem If x is an optimal solution of (PR(u)), then Ax b is a subgradient of θ at u. Subgradient algorithm: u 1 x 1 Ax 1 b u 2... Here u j+1 = u j + t j (Ax j b) with t j a step size.

57 Bibliography Combinatorial auctions S. De Vries and R.V. Vohra, Combinatorial auctions: A survey, INFORMS Journal on Computing 15(3), (2003). L. Escudero, M. Landete, A. Marin. A branch-and-cut algorithm for the Winner Determination Problem. Decision Support Systems 46, , T. Sandholm, S. Suri, A. Gilpin and D. Levine, CABOB: A fast optimal algorithm for winner determination in combinatorial auctions, Management Science 51, (2005). Proteins comparison Caprara, A. y Lancia, G. Structural alignment of large-size proteins via Lagrangian relaxation, Proceedings of the sixth annual international conference on Computational biology, Lancia, G., Carr. R., Walenz, B. e Istrail, S. 101 optimal PDB structure alignments: a branch-and-cut algorithm for the maximum contact map overlap problem Proc. 5th RECOMB, , Lancia, G. e Istrail, S. Protein structure comparison: Algorithms and applications en Protein structure: analysis and design, Springer, Meneses, C., Oliveira, C. y Pardalos, P. Mathematical Programmming formulations for problems in Genomics and Proteomics in: Data Mining in Biomedicine , Springer, 200. Strickland, D., Barnes E. y Sokol, J. Optimal protein structure alignment using maximum cliques, Operations Research 53, , Xie, W. y Sahinidis, N. A reduction-based exact algorithm for the contact map overlap problem Journal of Computational Biology 14, , 2007.

58 Facets for set packing problems Balas, E. y Zemel, E., Critical cutsets of graphs and canonical facets of set-packing polytopes, Maths. of Operations Research 2 (1977) L. Canovas, M. Landete, A. Marin. Facet obtaining procedures for set packing problems. SIAM Journal on Discrete Mathematics 16, , L. Canovas, M. Landete, A. Marin. New facets for the set packing polytope. Operations Research Letters 27, , Cho, D.C., Johnson, E.L., Padberg, M. y Rao, M.R., On the uncapacitated plant location problem. I: Valid inequalities and facets, Mathematics of Operations Research (193) Cho, D.C., Padberg, M. y Rao, M.R., On the uncapacitated plant location problem. II: facets and lifting theorems, Mathematics of Operations Research (193) Chvtal, V., On certain polytopes associated with graphs, Journal of Combinatorial Theory (B), 1 (1975) Nemhauser, G.L. y Trotter, L.E., Properties of vertex packing and independence set polyhedra, Mathematical Programming 6 (1974) Padberg, M., On the facial structure of set packing polyhedra, Mathematical Programming 5 (1973)

59 Lagrangian relaxation Fisher, M.L., The Lagrangian relaxation method for solving integer programming problems, Management Science 27 (191) 1-1. Gavish, B. y Pirkul, H., Algorithms for the multi-resource generalized assignment problem, Management Science 37 (1991) Geoffrion, A.M., Lagrangean relaxation for integer programming, Mathematical Programming Study 2 (1974) Guignard, M. Lagrangean relaxation, Top 11 (2003) Guignard, M. y Kim, S., Lagrangean decomposition for integer programming: theory and applications, R.A.I.R.O. 21 (197) Guignard, M. y Kim, S., Lagrangean decomposition: a model yielding stronger Lagrangean bounds, Mathematical Programming 39 (197)

60 Optimization in sports Ribeiro, OR on the ball: Applications of Combinatorial Optimization in Sports Scheduling and Management, ALIO/EURO, Pars, Schreuder, Combinatorial aspects of construction of competition Dutch Professional Football Leagues, DAM 1992, Trick, A schedule-then-break approach to sports timetabling. Proceedings of the Third PATAT Conference, Integer programming Wolsey. Integer Programming. Wiley, 199. Nemhauser, Wolsey. Integer and Combinatorial Optimization. Wiley, 19. Clique partitioning Borndörfer, Weismantel, Set packing relaxations of some integer programs, Mathematical Programming, , 2000.

61 Exercise 1. Easy FORMULATE: A) Given n points in the plane choose p 1 (centers) in such a way the maximum distance from the n points to their closest center be minimized. B) Given n points in the plane choose p 2 (centers) in such a way the maximum distance from the n points to their second closest center be minimized.

62 Exercise 2. Not so easy FORMULATE: A) Given a n n chess board with benefits associated to the squares (cells), put queens in some cells in such a way that they do not threaten each other and the sum of the benefits of cells with queens be maximized. B) Do the same with horses.

63 Exercise 3. Not so easy Given x ijk = 1 if number k is written in cell (i, j), i, j, k = 1,..., 9 formulate the constraints needed to solve the above sudoku.

64 Exercise 4. Difficult Look for clique facets for the polyhedron given by the constraints of the protein comparison problem. >< P j x ij 1 i P i x ij 1 j x i1 j 1 + x i2 j 2 1 if edges cross each other x ij {0, 1} i, j

65 Exercise 5. More than difficult Build a formulation in order to check if the above sudoku has more than one solution.

66 Exercise 6. Extremely difficult Define variables and formulate the constraints needed to solve the above kenken.

Integer Programming ISE 418. Lecture 12. Dr. Ted Ralphs

Integer Programming ISE 418. Lecture 12. Dr. Ted Ralphs Integer Programming ISE 418 Lecture 12 Dr. Ted Ralphs ISE 418 Lecture 12 1 Reading for This Lecture Nemhauser and Wolsey Sections II.2.1 Wolsey Chapter 9 ISE 418 Lecture 12 2 Generating Stronger Valid