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Menkes van den Briel Member of Research Staff NICTA and ANU menkes@nicta.com.au Combining Linear Programming Based Decomposition Techniques with Constraint Programming

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CP-based column generation ApplicationReference Urban transit crew management T.H. Yunes., A.V. Moura, C.C. de Souza. Solving very large crew scheduling problems to optimality. Proceedings of ACM symposium on Applied Computing, pages 446-451, 2000. T.H. Yunes., A.V. Moura, C.C. de Souza. Hybrid column generation approaches for urban transit crew management problems. Transportation Science 39(2):273-288, 2005. Travelling tournament K. Easton, G.L. Nemhauser, and M.A. Trick. Solving the travelling tournament problem: A combined integer programming and constraint programming approach. Proceedings of Practice and Theory of Automated Timetabling, volume 2740 of Lecture Notes in Computer Science, pages 100-112. Springer, 2002. Two-dimensional bin packing D. Pisinger, M. Sigurd. Using decomposition techniques and constraint programming for solving the two-dimensional bin-packing problem. Journal on Computing 19(1):36-51, 2007. Graph coloringS. Gualandi. Enhancing constraint programming-based column generation for integer programs. PhD thesis, Politechnico di Milano, 2008. Constrained cutting stock T. Fahle, M. Sellmann. Cost based filtering for the constrained knapsack problem. Annals of Operations Research 115(1):73-93, 2002. Employee timetabling S. Demassey, G. Pesant, L.M. Rousseau. A cost-regular based hybrid column generation approach. Constraints 11(4):315-333, 2006. Wireless mesh networks A. Capone, G. Carello, I. Filippini, S. Gualandi, F. Malucelli. Solving a resource allocation problem in wireless mess networks: A comparison between a CP-based and a classical column generation. Networks 55(3):221-233, 2010. Multi-machine scheduling R. Sadykov, L.A. Wolsey. Integer programming and constraint programming in solving a multimachine assignment scheduling problem with deadlines and release dates. Journal on Computing 18(2):209-217, 2006.

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CP-based column generation ApplicationReference Airline crew assignment U. Junker, S.E. Karisch, N. Kohl, B. Vaaben, T. Fahle, M. Sellmann. A framework for constraint programming based column generation. Proceedings of Principles and Practice of Constraint Programming, volume 1713 of Lecture Notes in Computer Science, pages 261-274, 1999. T. Fahle, U. Junker, S.E. Karisch, N. Kohl, M. Sellmann, B. Vaaben. Constraint programming based column generation for crew assignment. Journal of Hueristics 8(1):59-81, 2002. M. Sellmann, K. Zervoudakis, P. Stamatopoulos, T. Fahle. Crew assignment via constraint programming: integrating column generation and heuristic tree search. Annals of Operations Research 115(1):207-225, 2002. Vehicle routing with time windows L.M. Rousseau. Stabilization issues for constraint programming based column generation. Proceedings of Integration of AI and OR techniques in CP for Combinatorial Optimization, volume 3011 of Lecture notes in Computer Science, pages 402-408. Springer, 2004. L.M. Rousseau, M. Gendreau, G. Pesant, F. Focacci. Solving VRPTWs with constraint programming based column generation. Annals of Operations Research 130(1):199-216, 2004.

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CP-based Benders decomposition ApplicationReference Parallel machine scheduling V. Jain, I.E. Grossmann. Algorithms for hybrid MILP/CP models for a class of optimization problems. INFORMS Journal on Computing 13(4):258-276, 2001. Polypropylene batch scheduling C. Timpe. Solving planning and scheduling problems with combined integer and constraint programming. OR Spectrum 24(4):431-448, 2002. Call center scheduling T. Benoist, E. Gaudin, B. Rottembourg. Constraint programming contribution to Benders decomposition: A case study. Principles and Practice of Constraint Programming, volume 2470 of Lecture Notes in Computer Science, pages 603-617. Springer, 2002. Multi-machine scheduling J.N. Hooker. A hybrid method for planning and scheduling. Principles and Practice of Constraint Programming, volume 3258 of Lecture Notes in Computer Science, pages 305-316. Springer, 2004. J.N. Hooker. Planning and scheduling to minimize tardiness. Principles and Practice of Constraint Programming, volume 3709 of Lecture Notes in Computer Science, pages 314-327. Springer, 2005.

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CP versus IP CPIP VariablesFinite domainContinuous, Binary, Integer ConstraintsSymbolic: alldifferent cumulative Linear, algebraic: (+, –, *, =, ≤, ≥) InferenceConstraint propagation LP relaxation Global Optimal Local Feasible

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CP versus IP “MILP is very efficient when the relaxation is tight and models have a structure that can be effectively exploited” “CP works better for highly constrained discrete optimization problems where expressiveness of MILP is a major limitation” “From the work that has been performed, it is not clear whether a general integration strategy will always perform better than either CP or an MILP approach by itself. This is especially true for the cases where one of these methods is a very good tool to solve the problem at hand. However, it is usually possible to enhance the performance of one approach by borrowing some ideas from the other” –Source: Jain and Grossmann, 2001

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Outline Background Introduction Dantzig Wolfe decomposition Benders decomposition Conclusions

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What is your background? Have implemented Benders and/or Dantzig Wolfe decomposition Have heard about Benders and/or Dantzig Wolfe decomposition Have seen Bender and/or Dances with Wolves

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Things to take away A better understanding of how to combine linear programming based decomposition techniques with constraint programming A better understanding of column generation, Dantzig Wolfe decomposition and Benders decomposition A whole lot of Python code with example implementations

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Helpful installations Python 2.6.x or 2.7.x –“Python is a programming language that lets you work more quickly and integrate your systems more effectively” –http://www.python.org/getit/http://www.python.org/getit/ Gurobi (Python interface) –“The state-of-the-art solver for linear programming (LP), quadratic and quadratically constrained programming (QP and QCP), and mixed-integer programming (MILP, MIQP, and MIQCP)” –http://www.gurobi.com/products/gurobi-optimizer/try-for-yourselfhttp://www.gurobi.com/products/gurobi-optimizer/try-for-yourself NetworkX –“NetworkX is a Python language software package for the creation, manipulation, and study of the structure, dynamics, and functions of complex networks” –http://networkx.lanl.gov/download.htmlhttp://networkx.lanl.gov/download.html

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Abbreviations Artificial Intelligence (AI) Constraint Programming (CP) Constraint Satisfaction Problem (CSP) Integer Programming (IP) Linear Programming (LP) Mixed Integer Programming (MIP) Mixed Integer Linear Programming (MILP) Mathematical Programming (MP) Operations Research (OR)

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Outline Background Introduction Dantzig Wolfe decomposition Benders decomposition Conclusions

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What is decomposition? “Decomposition in computer science, also known as factoring, refers to the process by which a complex problem or system is broken down into parts that are easier to conceive, understand, program, and maintain” –Source: http://en.wikipedia.org/wiki/Decomposition_(computer_science)http://en.wikipedia.org/wiki/Decomposition_(computer_science) Decomposition in linear programming is a technique for solving linear programming problems where the constraints (or variables) of the problem can be divided into two groups, one group of “easy” constraints and another of “hard” constraints

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“easy” versus “hard” constraints Referring to the constraints as “easy” and “hard” may be a bit misleading –The “hard” constraints need not be very difficult in themselves, but they can complicate the linear program making the overall problem more difficult to solve –When the “hard” constraints are removed from the problem, then more efficient techniques could be applied to solve the resulting linear program

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Example Shortest path problem (P) Min (i,j) A c ij x ij s.t. 1 for i = s Source j:(i,j) A x ij – j:(j,i) A x ji = 0 for i N – {s, t} Flow -1 for i = t Sink x ij {0, 1} Resource constrained shortest path problem (NP-complete) Min (i,j) A c ij x ij s.t. 1 for i = s Source j:(i,j) A x ij – j:(j,i) A x ji = 0 for i N – {s, t} Flow -1 for i = t Sink (i,j) A d ij x ij ≤ C Capacity x ij {0, 1} G = (N, A), source s, sink t

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Example Assignment problem (P) Max i=1,…, m, j=1,…,n c ij x ij s.t. j=1,…,n x ij = 1for 1 ≤ i ≤ m Job i=1,…,m x ij = 1for 1 ≤ j ≤ n Machine x ij {0, 1} Generalized assignment problem (NP-complete) Max i=1,…, m, j=1,…,n c ij x ij s.t. j=1,…,n x ij = 1for 1 ≤ i ≤ m Job i=1,…,m d ij x ij ≤ C j for 1 ≤ j ≤ n Capacity x ij {0, 1} m jobs, n machines

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Example Consider developing a strategic corporate plan for several production facilities. Each facility has its own capacity and production constraints, but decisions are linked together at the corporate level by budgetary considerations Common constraints Facility 1 Facility 2 Facility n Independent constraints

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“easy” versus “hard” variables Referring to the variables as “easy” and “hard” may be a bit misleading –The “hard” variables need not be very difficult in themselves, but they can complicate the linear program making the overall problem more difficult to solve –When the “hard” variables are removed from the problem, then more efficient techniques could be applied to solve the resulting linear program

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Example Capacitated facility location problem (NP-complete) Min i=1,…,n,j=1,…,m c ij x ij + j=1,…,m f j y j s.t. i=1,…,m x ij ≥ 1 for j = 1,…, n Demand j=1,…,n d i x ij ≤ C i y i for i = 1,…, m Roll x ij ≤ y i for i = 1,…, m j = 1,…, n Flow impl. x ij ≥ 0 y i {0, 1} m facilities, n customers

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Example Consider solving a multi period scheduling problem. Each period has its own set of variables but is linked together through resource consumption variables Independent variables Common variables Period 1 Period 2 Period n

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Outline Background Introduction Dantzig Wolfe decomposition Benders decomposition Conclusions

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Primal Min cx s.t.Ax ≥ b[y] x ≥ 0 Dual Max y T b s.t.y T A ≤ c[x] y ≥ 0 Background

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Primal Min cx s.t.Ax ≥ b[y] x ≥ 0 Dual Max b T y s.t.A T y ≤ c T [x] y ≥ 0 Background cxc A x Axb bTybTybTbT ATAT cTcT ATyATy y

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Travelling salesman G = (N, A), cost c ij 0 1 2 3 4 5 6 7 8 9 xy 02019 111 21715 3146 412 5 3 698 71520 81911 975

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Travelling salesman G = (N, A), cost c ij xy 02019 111 21715 3146 412 5 3 698 71520 81911 975 0 1 2 3 4 5 6 7 8 9 Cost 60.78

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Travelling salesman Variables x ij is 1 if arc (i, j) is on the shortest tour, 0 otherwise Formulation Min (i,j) A c ij x ij s.t. i:(i,j) A x ij = 1for j N Inflow j:(i,j) A x ij = 1for i N Outflow i,j S:(i,j) A x ij ≤ |S| – 1for S N Subtour x ij {0, 1}

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Travelling salesman Variables x ij is 1 if arc (i, j) is on the shortest tour, 0 otherwise Formulation Min (i,j) A c ij x ij s.t. i:(i,j) A x ij = 1for j N Inflow j:(i,j) A x ij = 1for i N Outflow x ij {0, 1}

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Example code

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Travelling salesman G = (N, A), cost c ij xy 02019 111 21715 3146 412 5 3 698 71520 81911 975 1 2 3 4 5 6 7 9 0 8 Subtour 0, 2, 7

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Travelling salesman G = (N, A), cost c ij xy 02019 111 21715 3146 412 5 3 698 71520 81911 975 1 2 3 4 5 6 7 9 0 8 Subtour 0, 8, 1, 9

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Travelling salesman G = (N, A), cost c ij xy 02019 111 21715 3146 412 5 3 698 71520 81911 975 1 2 3 4 5 6 7 9 0 8 Subtour 0, 8, 2, 7

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Travelling salesman G = (N, A), cost c ij xy 02019 111 21715 3146 412 5 3 698 71520 81911 975 0 1 2 3 4 5 6 7 8 9 Cost 79.98

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Travelling salesman G = (N, A), cost c ij xy 02019 111 21715 3146 412 5 3 698 71520 81911 975 0 1 2 3 4 5 6 7 8 9 Cost 60.78

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LPs with many constraints The number of constraints that are tight (or active) is at most equal to the number of variables, so even with many constraints (possibly exponential many) only a small subset will be tight in the optimal solution Active Non-active A

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AAxb Row generation in the primal… cxc x

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y bTbT ATAT … is column generation in the dual bTybTy cTcT ATyATy

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…and vice versa x c A cx bAxAx ATAT ATyATycTcT bTybTybTbT y Column generation in the primal Row generation in the dual =

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Resource constrained shortest path G = (N, A), source s, sink t, for each (i, j) A, cost c ij, resource demand d ij, and resource capacity C 1 2 3 4 5 6 1,10 10,3 1,7 2,2 1,210,1 1,1 12,3 2,3 5,7 ij c ij, d ij Capacity = 14 Source: Desrosiers and Lübbecke, 2005

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Resource constrained shortest path G = (N, A), source s, sink t, for each (i, j) A, cost c ij, resource demand d ij, and resource capacity C 1 6 1,10 10,3 1,7 2,2 1,210,1 1,1 12,3 2,3 5,7 ij c ij, d ij Cost 13 Demand 13 Capacity = 14 2 3 4 5

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Resource constrained shortest path Variables x ij is 1 if arc (i, j) is on the shortest path, 0 otherwise Formulation Min (i,j) A c ij x ij s.t. 1 for i = s Source j:(i,j) A x ij – j:(j,i) A x ji = 0 for i N – {s, t} Flow -1 for i = t Sink (i,j) A d ij x ij ≤ C Capacity x ij {0, 1}

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Example code

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Resource constrained shortest path Variables k is 1 if path k is the shortest path, 0 otherwise Formulation Min k K c k k s.t. k K k = 1 Convex k K d k k ≤ C Capacity k ≥ 0

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Arc variablesPath variables Arc versus path 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6

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Example code

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Revised Simplex method Min cx s.t.Ax ≥ b x ≥ 0 Min z = cx s.t.Ax = b x ≥ 0 Let x be a basic feasible solution, such that x = (x B, x N ) where x B is the vector of basic variables and x N is the vector of non-basic variables Add slack variables

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Revised Simplex method Min z = cx s.t.Ax = b x ≥ 0 Min z = c B x B + c N x N s.t.Bx B + A N x N = b x B, x N ≥ 0 Min z = c B x B + c N x N s.t.x B = B -1 b – B -1 A N x N x B, x N ≥ 0 Rearrange x = (x B, x N ), c = (c B, c N ), A = (B, A N )

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Revised Simplex method Min z = c B x B + c N x N s.t.x B = B -1 b – B -1 A N x N x B, x N ≥ 0 Min z = c B B -1 b + (c N – c B B -1 A N )x N s.t.x B = B -1 b – B -1 A N x N x B, x N ≥ 0 Substitute

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Revised Simplex method Min z = c B B -1 b + (c N – c B B -1 A N )x N s.t.x B = B -1 b – B -1 A N x N x B, x N ≥ 0 At the end of each iteration we have –Current value of non-basic variables x N = 0 –Current objective function value z = c B B -1 b –Current value of basic variables x B = B -1 b –Objective coefficients of basic variables 0 –Objective coefficients of non-basic variables (c N – c B B -1 A N ) are the so-called reduced costs –With a minimization objective we want non-basic variables with negative reduced costs

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Revised Simplex method Simplex algorithm 1.Select new basic variable ( x N to enter the basis) 2.Select new non-basic variable ( x B to exit the basis) 3.Update data structures

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Revised Simplex method Simplex algorithm x S = b (slack variables equal rhs) x \S = 0 (non-slack variables equal 0) while min j {(c j – c B B -1 A j )} < 0 1.Select new basic variable j : (c j – c B B -1 A j ) < 0 2.Select new non-basic variable j’ by increasing x j as much as possible 3.Update data structures by swapping columns between matrix B and matrix A N

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Example Min z = – x 1 – 2x 2 s.t.– 2x 1 + x 2 ≥ 2 – x 1 + 2x 2 ≥ 7 x 1 ≥ 7 x 1, x 2 ≥ 0 Min z = – x 1 – 2x 2 s.t.– 2x 1 + x 2 + x 3 = 2 – x 1 + 2x 2 + x 4 = 7 x 1 + x 5 = 7 x 1, x 2, x 3, x 4, x 5 ≥ 0 Add slack variables

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Simplex methodRevised Simplex method Example bscx1x2x3x4x5rhs -z-1-20000 x3-211002 x420107 x5100013 bscx1x2x3x4x5rhs -z-502004 x2-211002 x430-2103 x5100013 bscx3x4x5rhs -z0000 x31002 x40107 x50013 bscx3x4x5rhs -z2004 x31002 x4-2103 x50013 x2 -2 1 2 0 x1 -5 -2 3 1

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Simplex methodRevised Simplex method Example bscx1x2x3x4x5rhs -z00-3/45/309 x201-1/32/304 x110-2/31/301 x5002/3-1/312 bscx1x2x3x4x5rhs -z0001213 x20101/2 5 x1100013 x3001-1/23/23 bscx3x4x5rhs -z2009 x2-1/32/304 x1-2/31/301 x52/3-1/312 bscx3x4x5rhs -z00013 x201/2 5 x10013 x31-1/23/23 x3 -3/4 -1/3 -2/3 2/3

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Column generation Simplex algorithm x S = b (slack variables equal rhs) x \S = 0 (non-slack variables equal 0) while min j {(c j – c B B -1 A j )} < 0 1.Select new basic variable j : (c j – c B B -1 A j ) < 0 2.Select new non-basic variable j’ by increasing x j as much as possible 3.Update data structures by swapping columns between matrix B and matrix A N In column generation, rather than checking the reduced cost for each variable, a subproblem is solved to find a variable with negative reduced cost

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LPs with many variables The number of basic (non-zero) variables is at most equal to the number of constraints, so even with many variables (possibly exponential many) only a small subset will be in the optimal solution A xBxB xNxN

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(c N – c B B -1 A N ) < 0 (c N – y T A N ) < 0 Column generation Substitute

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(c N – y T A N ) < 0 Primal Min cx s.t.Ax ≥ b x ≥ 0 Dual Max y T b s.t.y T A ≤ c y ≥ 0 Column generation x c A cx bAxAx ATAT ATyATycTcT bTybTybTbT y Column with negative reduced cost Row with violated rhs

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Resource constrained shortest path Variables k is 1 if path k is the shortest path, 0 otherwise Formulation Min k K c k k s.t. k K k = 1 Convex k K d k k ≤ C Capacity k {0, 1}

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Primal Min k K c k k s.t. k K k = 1 [ ] k K d k k ≤ C [ ] k ≥ 0 Dual Max + C s.t. + d k ≤ c k [ k ] = free ≤ 0 Resource constrained shortest path Need to find a path for which c k – – d k < 0 Implicitly search all paths by optimizing Min (i,j) A (c ij – d ij ) s.t. Source, Flow, Sink

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Resource constrained shortest path G = (N, A), source s, sink t, for each (i, j) A, cost c ij, resource demand d ij, and resource capacity C 1 2 3 4 5 6 1 10 1 2 1 1 12 2 5 ij (c ij – d ij ) Capacity = 14

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Resource constrained shortest path Master Min k K c k k s.t. k K k = 1 Convex k K d k k ≤ C Capacity k ≥ 0 Subproblem Min (i,j) A (c ij – d ij )x ij s.t. 1 for i = s Source j:(i,j) A x ij – j:(j,i) A x ji = 0 for i N – {s, t} Flow -1 for i = t Sink Add variable to master if (i,j) A (c ij – d ij )x ij – < 0

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Example code

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Roll width W, m orders of d i rolls of length l i, i = 1,…, m Cutting stock 12 31 36 45 11 x 4 x 2 x lili didi 100

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Roll width W, m orders of d i rolls of length l i, i = 1,…, m Cutting stock 12 31 45 12 36 12 31 45 31 36 12 36 12 31 36 45 11 x 4 x 2 x lili didi Rolls 5 100 98 96

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Cutting stock Variables x ik is the number of times order i is cut from roll k y k is 1 if roll k is used, 0 otherwise Formulation Min k=1,…,K y k s.t. k=1,…,K x ik ≥ d i for i = 1,…, n Demand i=1,…,n l i x ik ≤ Wy k for k = 1,…, K Roll x ik ≥ 0 and integer y k {0, 1}

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Example code

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Cutting stock Variables k is the number of times cutting pattern k is used Formulation Min k K k s.t. k K a ik k ≥ d i for i = 1,…, m Demand k ≥ 0 and integer

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Cutting stock Cutting pattern variables 12 36 k a ik [2, 0, 2, 0] 12 31 36 45 11 x 4 x 2 x 12 31 45 k a ik [2, 1, 0, 1]

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Primal Min k K k s.t. k K a ik k ≥ d i [ i ] k ≥ 0 Dual Max i=1,…,n d i i s.t. i=1,…,n a ik i ≤ 1 [ k ] i ≥ 0 Cutting stock Need to find a cutting pattern for which 1 – i=1,…,n a ik i < 0 Implicitly search all cutting patterns by optimizing Max i=1,…,n a i i s.t. i=1,…,n l i a i ≤ W a i ≥ 0 and integer

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m items with value i and weight l i, i = 1,…, m, maximum allowed weight W Cutting stock $0.50, 45lbs $0.50, 36lbs $0.33, 31lbs $0.125, 12lbs 12 31 36 45 lili ii 0.125 0.33 0.50 100lbs 12 36

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Cutting stock Master Min k K k s.t. k K a ik k ≥ d i for i = 1,…, m Demand k ≥ 0 Subproblem Max i=1,…,m a i i s.t. i=1,…,m l i a i ≤ W a i ≥ 0 and integer Add variable to master if 1 – a i i < 0

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Example code

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Generalized assignment n jobs, m machines, cost c ij, demand d ij, capacity C i 1 2 3 4 5 1 2 j i 36 34 CjCj c ij, d ij Job12 117, 823, 15 221, 1516, 7 322, 1421, 23 418, 2316, 22 524, 817, 11

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Generalized assignment n jobs, m machines, cost c ij, demand d ij, capacity C i Cost 95 30 29 1 2 3 4 5 1 2 j i 36 34 CjCj c ij, d ij Job12 117, 823, 15 221, 1516, 7 322, 1421, 23 418, 2316, 22 524, 817, 11

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Generalized assignment Variables x ij is 1 if job j is assigned to machine i, 0 otherwise Formulation Max i=1,…,m,j=1,…,n c ij x ij s.t. i=1,…,m x ij = 1for 1 ≤ j ≤ n Job j=1,…,n d ij x ij ≤ C i for 1 ≤ i ≤ m Capacity x ij {0, 1}

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Example code

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Generalized assignment Variables ik is 1 if machine i has job assignment k, 0 otherwise Formulation Max i=1,…,m,k=1,…,Ki c ik ik s.t. i=1,…,m,k=1,…,Ki a ijk ik = 1for 1 ≤ j ≤ n Job k=1,…,Ki ik = 1 for 1 ≤ i ≤ m Convexity ik {0, 1}

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Generalized assignment Job assignment variables ik a ijk [1, 0, 1, 0, 1] ik a ijk [0, 1, 0, 1, 0] 1 2 3 4 5 1 2 3 4 5 1 2 1 2

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Generalized assignment Formulation Max i=1,…,m,k=1,…,Ki c ik ik s.t. i=1,…,m,k=1,…,Ki a ijk ik = 1for 1 ≤ j ≤ n Job k=1,…,Ki ik = 1 for 1 ≤ i ≤ m Convexity ik {0, 1} Common constraints Machine 1 Machine 2 Machine n Independent constraints

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Primal Max i=1,…,m,k=1,…,Ki c ik ik s.t. i=1,…,m,k=1,…,Ki a ijk ik = 1 k=1,…,Ki ik = 1 ik ≥ 0 Dual Min j=1,…,n j + i=1,…,m i s.t. j=1,…,n a ijk j + i ≥ c ik j = free i = free Generalized assignment Need to find a cutting pattern for which j=1,…,n (c ik – a ijk j ) – i > 0for i = 1,…,m Implicitly search all cutting patterns by optimizing Max j=1,…,n (c ij – a ij j ) s.t. j=1,…,n d ij a ij ≤ C i a ij ≥ 0 and integer

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$55.00, 8lbs n items with value j and weight d ij, j = 1,…, n, maximum allowed weight W Generalized assignment $52.00, 23lbs $51.00, 14lbs $55.00, 15lbs $44.00, 8lbs 36lbs Job1 144, 8 255, 15 351, 14 452, 23 555, 8 Job2 140, 15 237, 7 343, 23 434, 22 541, 11 36 1 2 3 4 5 1 2 34 1 2 3 4 5 1 2

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Generalized assignment Master Max i=1,…,m,k=1,…,Ki c ik ik s.t. i=1,…,m,k=1,…,Ki a ijk ik = 1for 1 ≤ j ≤ n Job k=1,…,Ki ik = 1 for 1 ≤ i ≤ m Convexity ik {0, 1} Subproblem (for each machine i ) Max j=1,…,n (c ij – a ij j ) s.t. j=1,…,n d ij a ij ≤ C i a ij ≥ 0 and integer Add variable to master if j=1,…,n (c ij – a ij j ) – i > 0

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Example code

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History of column generation 1961: A linear programming approach to the cutting-stock problem P.C. Gilmore and R.E. Gomory 1963: A linear programming approach to the cutting-stock problem–Part II P.C. Gilmore and R.E. Gomory 1960: Decomposition principle for linear programs G.B. Dantzig and P. Wolfe “Credit is due to Ford and Fulkerson for their proposal for solving multicommodity network problems as it served to inspire the present development.” 1958: A suggested computation for maximal multicommodity network flows L.R. Ford and D.R. Fulkerson 1969: A column generation algorithm for a ship scheduling problem L.E. Appelgren

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Solving integer programs by column generation 2000: On Dantzig-Wolfe decomposition in integer programming and ways to perform branching in a branch-and-price algorithm F. Vanderbeck 2005: A primer in column generation J. Desrosiers and M.E. Lubbecke 1998: Branch-and-price: column generation for solving huge integer programs C. Barnhart, E.L. Johnson, G.L. Nemhauser, M.W.P. Savelsbergh and P.H. Vance 1984: Routing with time windows by column generation Y. Dumas, F. Soumis and M. Desrochers 2011: Branching in branch-and-price: a generic scheme F. Vanderbeck

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CP-based column generation 2000: Solving very large crew scheduling problems to optimality T.H. Yunes, A.V. Moura and C.C. de Souza 1999: A framework for constraint programming based column generation U. Junker, S.E. Karisch, N. Kohl, B. Vaaben, T. Fahle and M. Sellmann

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CP-based column generation ApplicationReferenceCP used to solve subproblem CP used within Branch-and-Price Urban transit crew management T.H. Yunes., A.V. Moura, C.C. de Souza. 2000. YY T.H. Yunes., A.V. Moura, C.C. de Souza. 2005. YY Travelling tournament K. Easton, G.L. Nemhauser, and M.A. Trick. 2002. YY Two-dimensional bin packing D. Pisinger, M. Sigurd. 2007.YY Graph coloringS. Gualandi. 2008.YY Constrained cutting stock T. Fahle, M. Sellmann. 2002.YN Employee timetabling S. Demassey, G. Pesant, L.M. Rousseau. 2006. YY Wireless mesh networks A. Capone, G. Carello, I. Filippini, S. Gualandi, F. Malucelli. 2010. YN Multi-machine scheduling R. Sadykov, L.A. Wolsey. 2006.YN Source: Gualandi and Malucelli, 2009

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CP-based column generation ApplicationReferenceCP used to solve subproblem CP used within Branch-and-Price Airline crew assignment U. Junker, S.E. Karisch, N. Kohl, B. Vaaben, T. Fahle, M. Sellmann. 1999. YN T. Fahle, U. Junker, S.E. Karisch, N. Kohl, M. Sellmann, B. Vaaben. 2002. YN M. Sellmann, K. Zervoudakis, P. Stamatopoulos, T. Fahle. 2002. YN Vehicle routing with time windows L.M. Rousseau. 2004. YN L.M. Rousseau, M. Gendreau, G. Pesant, F. Focacci. 2004. YY Source: Gualandi and Malucelli, 2009

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CP-based column generation Typical implementation MasterSubproblem Linear programming Constraint programming Dual information New columns

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Outline Background Introduction Dantzig Wolfe decomposition Benders decomposition Conclusions

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Two-stage optimization Stage 1Stage 2 Solution values

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Benders decomposition Stage 1Stage 2 Solution values Benders cuts

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Benders decomposition “Learn from ones mistakes” Distinguish primary variables from secondary variables Search over primary variables (master problem) For each trial value of primary variables, solve problem over secondary variables (subproblem) If solution is suboptimal/infeasible, find out why and design a constraint that rules out not only this solution but a large class of solutions that are suboptimal/infeasible for the same reason (Benders cut) Add Benders cut to the master problem and resolve MasterSubproblem Solution values Benders cuts

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Capacitated facility location m facilities, n customers, cost c ij, demand d j, capacity C i, fixed cost f i 1 2 3 4 5 1 3 2 i j C i, f i c ij djdj 10, 3 10, 4 6 7 4 8 5 Cust123 1245 2334 3412 4521 5763

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Capacitated facility location m facilities, n customers, cost c ij, demand d j, capacity C i, fixed cost f i 1 2 3 4 5 1 3 2 i j C i, f i c ij djdj 10, 3 10, 4 6 7 4 8 5 Cust123 1245 2334 3412 4521 5763 Cost 21.29

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Capacitated facility location Variables x ij fraction of demand supplied by facility i to cusomter j y i is 1 if facility i is open, 0 otherwise Formulation Min i=1,…,n,j=1,…,m c ij x ij + j=1,…,m f j y j s.t. i=1,…,m x ij ≥ 1 for j = 1,…, n Demand j=1,…,n d i x ij ≤ C i y i for i = 1,…, m Roll x ij ≤ y i for i = 1,…, m j = 1,…, n Flow x ij ≥ 0 y i {0, 1}

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Example code

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MasterSubproblem Solution values Benders cuts Min cx + dy s.t.Ax ≥ b Px + Qy ≥ r x ≥ 0 and integer y ≥ 0 Min cx + s.t.Ax ≥ b x ≥ 0 and integer ≥ 0 Min dy s.t.Qy ≥ r – Px y ≥ 0 Benders decomposition What if the subproblem is infeasible?

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Benders decomposition Primal, dual possibilities OptimalUnboundedInfeasible OptimalYesNo UnboundedNo Yes InfeasibleNoYes Dual Primal

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MasterSubproblem Solution values Benders cuts Min cx + dy s.t.Ax ≥ b Px + Qy ≥ r x ≥ 0 and integer y ≥ 0 Min cx + s.t.Ax ≥ b optimality cuts feasibility cuts x ≥ 0 and integer ≥ 0 Min dy s.t.Qy ≥ r – Px y ≥ 0 Benders decomposition

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Min dy s.t.Qy ≥ r – Px[u] y ≥ 0 Optimal Infeasible Max u T (r – Px) s.t.u T Q ≤ d[y] u ≥ 0 Optimality cut ≥ u k T (r – Px) Infeasibility cut v k T (r – Px) ≤ 0 Benders decomposition

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MasterSubproblem Solution values Benders cuts Min cx + dy s.t.Ax ≥ b Px + Qy ≥ r x ≥ 0 and integer y ≥ 0 Min cx + s.t.Ax ≥ b ≥ u k T (r – Px) v k T (r – Px) ≤ 0 x ≥ 0 and integer ≥ 0 Max dy s.t.Qy ≤ r – Px y ≥ 0 Benders decomposition

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Solve master problem Is optimal? START Solve sub problem Terminate? END Add optimality cut Add feasibility cut yesno yes no

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Capacitated facility location Variables x ij fraction of demand supplied by facility i to cusomter j y i is 1 if facility i is open, 0 otherwise Formulation Min i=1,…,n,j=1,…,m c ij x ij + j=1,…,m f j y j s.t. i=1,…,m x ij ≥ 1 for j = 1,…, n Demand j=1,…,n d i x ij ≤ C i y i for i = 1,…, m Roll x ij ≤ y i for i = 1,…, m j = 1,…, n Flow x ij ≥ 0 y i {0, 1}

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Capacitated facility location Master Min j=1,…,m f j y j + s.t.optimality cuts feasibility cuts y i {0, 1} ≥ 0 Subproblem Min i=1,…,n,j=1,…,m c ij x ij s.t. i=1,…,m x ij ≥ 1 for j = 1,…, n Demand j=1,…,n d i x ij ≤ C i y i for i = 1,…, m Roll x ij ≤ y i for i = 1,…, m j = 1,…, n Flow x ij ≥ 0

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Capacitated facility location Subproblem primal Min i=1,…,n,j=1,…,m c ij x ij s.t. i=1,…,m x ij ≥ 1 [ j ] j=1,…,n d i x ij ≤ C i y i [ i ] x ij ≤ y i [ ij ] x ij ≥ 0 Subproblem dual Max j=1,…,m j + i=1,…,n C i y i i + i=1,…,n,j=1,…,m y i ij s.t. j + d i i + ij ≥ 1 [x ij ] j ≥ 0 i ≤ 0 ij ≤ 0

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Capacitated facility location Master Min j=1,…,m f j y j + s.t. ≥ j=1,…,m j + i=1,…,n C i i y i + i=1,…,n,j=1,…,m ij y i j=1,…,m j + i=1,…,n C i i y i + i=1,…,n,j=1,…,m ij y i ≤ 0 y i {0, 1} ≥ 0

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Example code

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Benders decomposition for stochastic prog. MasterScenario 2 Scenario 1 Scenario 3

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Capacitated facility location m facilities, n customers, cost c ij, demand d j, capacity C i, fixed cost f i 1 2 3 4 5 1 3 2 i j C i, f i c ij djdj 10, 3 10, 4 6 7 4 8 5 Cust123 1245 2334 3412 4521 5763 5 6 3 7 4 4 5 2 6 3

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Example code

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CP-based Benders decomposition Typical implementation(?) MasterSubproblem Constraint programming Linear programming Solution values Benders cuts

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CP-based Benders decomposition Recent developments MasterSubproblem Integer programming Constraint programming Solution values Benders cuts

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CP-based Benders decomposition ApplicationReferenceMaster problemSubproblem Parallel machine scheduling V. Jain, I.E. Grossmann. 2001. MILPCP Polypropylene batch scheduling C. Timpe. 2002. MILPCP Call center schedulingT. Benoist, E. Gaudin, B. Rottembourg. 2002. CPLP Multi-machine scheduling J.N. Hooker. 2004. MILPCP J.N. Hooker. 2005. MILPCP Source: Hooker, 2006

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Nested Benders decomposition –When the subproblem is decomposed into a master and subproblem MasterSub MasterSub MasterSub MasterSub Forward pass Solve master problems Backward pass Solve subproblems and add Benders cuts

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Outline Introduction Background Dantzig Wolfe decomposition Benders decomposition Conclusions

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Why use decomposition? Many real-world systems contain loosely connected components, and as a result, the corresponding mathematical models present a certain structure that can be exploited It may be your only choice when solving the model without decomposition is impossible, because it is too large (memory error or timeout)

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When is decomposition likely most effective? When you have either complicating constraints or complicating variables Dantzig Wolfe decomposition Benders decomposition

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Further reading Column Generation –Guy Desaulniers, Jacques Desrosiers, Marius M. Solomon Decomposition Techniques in Mathematical Programming –Antonio J. Conejo, Enrique Castillo, Roberto Minguez and Raquel Garcia-Bertrand Linear Programming and Network Flows –Mokhtar S. Bazaraa, John J. Jarvis, Hanif D. Sherali

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From imagination to impact

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