Combining Optimization and Constraint Programming

Transcription

1 Combining Optimization and Constraint Programming John Hooker Carnegie Mellon University GE Research Center 7 September 2007 GE 7 Sep 07 Slide 1

2 Optimization and Constraint Programming Optimization: focus on mathematical programming. 50+ years old Constraint programming 20 years old Developed in computer science/ai community Better known in Europe/Asia than USA GE 7 Sep 07 Slide 2

3 Math programming & Constraint programming Mathematical programming methods rely heavily on numerical calculation. Linear programming (LP) Mixed integer/linear programming (MILP) Nonlinear programming (NLP) Constraint programming relies heavily on constraint propagation A form of logical inference GE 7 Sep 07 Slide 3

4 CP: Early commercial successes Circuit design (Siemens) Container port scheduling (Hong Kong and Singapore) Real-time control (Siemens, Xerox) GE 7 Sep 07 Slide 4

5 CP: Applications Employee scheduling Shift planning Assembly line smoothing and balancing Cellular frequency assignment Maintenance planning Airline crew rostering and scheduling Airport gate allocation and stand planning GE 7 Sep 07 Slide 5

6 CP: Applications Production scheduling chemicals aviation oil refining steel lumber photographic plates tires Transport scheduling (food, nuclear fuel) Warehouse management Course timetabling GE 7 Sep 07 Slide 6

7 Math programming & Constraint programming In mathematical programming, equations (constraints) describe the problem but don t tell how to solve it. In constraint programming, each constraint invokes a procedure that screens out unacceptable solutions. GE 7 Sep 07 Slide 7

8 Math programming & Constraint programming Mathematical programming Linear, nonlinear and mixed integer programming, global optimization CPLEX, OSL, Xpress-MP, Excel solver, MINTO, BARON Constraint programming Constraint propagation, domain reduction, interval arithmetic ILOG Scheduler, OPL Studio, CHIP, Mozart, ECLiPSe, Newton GE 7 Sep 07 Slide 8

9 Why unify math programming and constraint programming? One-stop shopping. One solver does it all. GE 7 Sep 07 Slide 9

10 Why unify math programming and constraint programming? One-stop shopping. One solver does it all. Richer modeling framework. Natural models, less debugging & development time. GE 7 Sep 07 Slide 10

11 Why unify math programming and constraint programming? One-stop shopping. One solver does it all. Richer modeling framework. Natural models, less debugging & development time. Computational speedup. A selection of results GE 7 Sep 07 Slide 11

12 Computational Advantage of Integrating MP and CP Using CP + relaxation from MILP Problem Speedup Focacci, Lodi, Milano (1999) Lesson timetabling 2 to 50 times faster than CP Refalo (1999) Piecewise linear costs 2 to 200 times faster than MILP Hooker & Osorio (1999) Flow shop scheduling, etc. 4 to 150 times faster than MILP. Thorsteinsson & Ottosson (2001) Product configuration 30 to 40 times faster than CP, MILP GE 7 Sep 07 Slide 12

13 Computational Advantage of Integrating MP and CP Using CP + relaxation from MILP Problem Speedup Sellmann & Fahle (2001) Automatic recording 1 to 10 times faster than CP, MILP Van Hoeve (2001) Stable set problem Better than CP in less time Bollapragada, Ghattas & Hooker (2001) Beck & Refalo (2003) Structural design (nonlinear) Scheduling with earliness & tardiness costs Up to 600 times faster than MILP. 2 problems: <6 min vs >20 hrs for MILP Solved 67 of 90, CP solved only 12 GE 7 Sep 07 Slide 13

14 Computational Advantage of Integrating MP and CP Using CP-based Branch and Price Problem Speedup Yunes, Moura & de Souza (1999) Easton, Nemhauser & Trick (2002) Urban transit crew scheduling Traveling tournament scheduling Optimal schedule for 210 trips, vs. 120 for traditional branch and price First to solve 8-team instance GE 7 Sep 07 Slide 14

15 Computational Advantage of Integrating MP and CP Using CP/MILP Benders methods Problem Speedup Jain & Grossmann (2001) Thorsteinsson (2001) Min-cost planning & scheduing Min-cost planning & scheduling Timpe (2002) Polypropylene batch scheduling at BASF 20 to 1000 times faster than CP, MILP 10 times faster than Jain & Grossmann Solved previously insoluble problem in 10 min GE 7 Sep 07 Slide 15

16 Computational Advantage of Integrating MP and CP Using CP/MILP Benders methods Problem Speedup Benoist, Gaudin, Rottembourg (2002) Call center scheduling Solved twice as many instances as traditional Benders Hooker (2004) Min-cost, min-makespan planning & cumulative scheduling Hooker (2005) Min tardiness planning & cumulative scheduling times faster than CP, MILP times faster than CP, MILP GE 7 Sep 07 Slide 16

17 Modeling is key CP models explain by revealing problem structure. Through global constraints. This can be extended to a unified framework. Global constraints become metaconstraints. Each metaconstraint knows how to combine MP and CP to exploit its structure. GE 7 Sep 07 Slide 17

18 The basic algorithm Search: Enumerate problem restrictions Tree search (branching) Constraint-based (nogood-based) search GE 7 Sep 07 Slide 18

19 The basic algorithm Search: Enumerate problem restrictions Tree search (branching) Constraint-based (nogood-based) search Infer: Deduce constraints from current restriction GE 7 Sep 07 Slide 19

20 The basic algorithm Search: Enumerate problem restrictions Tree search (branching) Constraint-based (nogood-based) search Infer: Deduce constraints from current restriction Relax: Solve relaxation of current restriction GE 7 Sep 07 Slide 20

21 Unifying framework Existing methods are special cases of this framework. Integrated methods are also special cases. Select an overall search scheme. Select inference methods as needed from CP, OR. Select relaxation methods as needed. GE 7 Sep 07 Slide 21

22 Some existing methods Branching Constraint solvers (CP) Search: Branching on domains Inference: Constraint propagation, filtering Relaxation: Domain store Mixed integer programming (OR) Search: Branch and bound Inference: Cutting planes Relaxation: Linear programming GE 7 Sep 07 Slide 22

23 Some existing methods Constraint-based search SAT solvers (CP) Search: Branching on variables Inference: Unit clause rule, clause learning (nogoods) Relaxation: Conflict clauses Benders decomposition (OR) Search: Enumeration of subproblems Inference: Benders cuts (nogoods) Relaxation: Master problem GE 7 Sep 07 Slide 23

24 Software for integrated methods Cooperating solvers: ILOG s OPL Studio ECLiPSe Integration with low-level modeling: Dash Optimization s Mosel. Integration with high-level modeling: SIMPL (CMU). GE 7 Sep 07 Slide 24

25 Outline Example problems to illustrate integrated approach Simpler modeling Lot sizing and scheduling Branching search Freight transfer Product configuration Continuous global optimization Airline crew scheduling Constraint-based search Machine scheduling Success stories from BASF, Barbot, Peugeot-Citroën GE 7 Sep 07 Slide 25

26 Example: Lot sizing and scheduling Simplified modeling GE 7 Sep 07 Slide 26

27 Lot sizing and scheduling Day: A B A Product At most one product manufactured on each day. Demands for each product on each day. Minimize setup + holding cost. GE 7 Sep 07 Slide 27

28 GE 7 Sep 07 Slide 28 0, {0,1},, all 1,, all,, all,, all,, all 1,, all, 1, all,, all,, all, s.t. min 1, 1, 1, 1, 1,, = + + = + + it it ijt it it i it it it jt ijt t i ijt jt t i ijt t i it it it t i it it it it it t i i t i j ijt ij it it s x z y t y t i Cy x t i y t i y t i y y t i y z t i y z t i y y z t i s d x s q s h δ δ δ δ δ Integer programming model (Wolsey) Many variables

29 Integrated model Minimize holding and setup costs min s.t. s q y + t i i 0, t x x d C, s h s + s, all i, t all i, t ( y i ) ( x = 0 ), all i, t t 1 + it y t 1 t it = it it it i it it 0, Inventory balance Production capacity GE 7 Sep 07 Slide 29

30 Example: Freight Transfer Branch-and-bound search with interval propagation and cutting planes GE 7 Sep 07 Slide 30

31 This example illustrates: Branching on variables, with pruning based on bounds. Propagation based on: Interval propagation. Cutting planes (knapsack cuts). Relaxation based on linear programming. GE 7 Sep 07 Slide 31

32 Freight Transfer Transport 42 tons of freight using 8 trucks, which come in 4 sizes GE 7 Sep 07 Slide 32 Truck size Number available Capacity (tons) Cost per truck

33 Number of trucks of type 1 min 90 x + 60 x + 50 x + 40 x 7 x + 5 x + 4 x + 3 x 42 x + x + x + x x i {0,1,2,3} 8 GE 7 Sep 07 Slide 33 Truck type Number available Capacity (tons) Cost per truck

34 Number of trucks of type 1 Knapsack metaconstraint knows which inference and relaxation techniques to use. GE 7 Sep 07 Slide 34 min 90 x + 60 x + 50 x + 40 x 7 x + 5 x + 4 x + 3 x 42 x + x + x + x x Truck type i {0,1,2,3} Number available Capacity (tons) Cost per truck

35 Number of trucks of type 1 Domain metaconstraint knows how to branch min 90 x + 60 x + 50 x + 40 x 7 x + 5 x + 4 x + 3 x 42 x + x + x + x x i {0,1,2,3} 8 GE 7 Sep 07 Slide 35 Truck type Number available Capacity (tons) Cost per truck

36 Bounds propagation min 90 x + 60 x + 50 x + 40 x 7 x + 5 x + 4 x + 3 x 42 x + x + x + x x i {0,1,2,3} 8 x = 7 1 GE 7 Sep 07 Slide 36

37 Bounds propagation Reduced domain min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x {1,2,3}, x, x, x {0,1,2,3} x = 7 1 GE 7 Sep 07 Slide 37

38 Continuous relaxation min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 Replace domains with bounds This is a linear programming problem, which is easy to solve. Its optimal value provides a lower bound on optimal value of original problem. GE 7 Sep 07 Slide 38

39 Cutting planes (valid inequalities) min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 We can create a tighter relaxation (larger minimum value) with the addition of cutting planes. GE 7 Sep 07 Slide 39

40 Cutting planes (valid inequalities) min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 Cutting plane All feasible solutions of the original problem satisfy a cutting plane (i.e., it is valid). But a cutting plane may exclude ( cut off ) solutions of the continuous relaxation. Feasible solutions GE 7 Sep 07 Slide 40 Continuous relaxation

41 Cutting planes (valid inequalities) min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 {1,2} is a packing because 7x 1 + 5x 2 alone cannot satisfy the inequality, even with x 1 = x 2 = 3. GE 7 Sep 07 Slide 41

42 Cutting planes (valid inequalities) min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 {1,2} is a packing So, 4 x x 4 42 ( ) GE 7 Sep 07 Slide 42

43 Cutting planes (valid inequalities) min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 {1,2} is a packing So, 4 x x 4 42 ( ) which implies 42 ( ) x 3 + x 4 = max { 4,3 } GE 7 Sep 07 Slide 43 Knapsack cut 2

44 Cutting planes (valid inequalities) min 90 x + 60 x + 50 x + 40 x x + 5 x + 4 x + 3 x x + x + x + x x 3, x 1 i 1 8 Maximal Packings Knapsack cuts {1,2} x 3 + x 4 2 {1,3} x 2 + x 4 2 {1,4} x 2 + x 3 3 GE 7 Sep 07 Slide 44 Knapsack cuts corresponding to nonmaximal packings can be nonredundant.

45 Continuous relaxation with cuts min 90 x + 60 x + 50 x + 40 x 7 x + 5 x + 4 x + 3 x 42 x + x + x + x 1 0 x 3, x 1 x + x x + x x + x i Knapsack cuts Optimal value of is a lower bound on optimal value of original problem. GE 7 Sep 07 Slide 45

46 Branch-inferand-relax tree Propagate bounds and solve relaxation of original problem. GE 7 Sep 07 Slide 46 x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓

47 Branch-inferand-relax tree Branch on a variable with nonintegral value in the relaxation. GE 7 Sep 07 Slide 47 x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3

48 Branch-inferand-relax tree Propagate bounds and solve relaxation. Since relaxation is infeasible, backtrack. GE 7 Sep 07 Slide 48 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3

49 Branch-inferand-relax tree Propagate bounds and solve relaxation. Branch on nonintegral variable. GE 7 Sep 07 Slide 49 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3 x 1 { 3} x 2 {0123} x 3 {0123} x 4 {0123} x = (3,2.6,2,0) value = 526 x 2 = 3 x 2 {0,1,2}

50 Branch-inferand-relax tree Branch again. GE 7 Sep 07 Slide 50 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3 x 1 { 3} x 2 {012 } x 3 { 123} x 4 {0123} x = (3,2,2¾,0) value = 527½ x 3 {1,2} x 3 = 3 x 1 { 3} x 2 {0123} x 3 {0123} x 4 {0123} x = (3,2.6,2,0) value = 526 x 2 = 3 x 2 {0,1,2}

51 Branch-inferand-relax tree Solution of relaxation is integral and therefore feasible in the original problem. This becomes the incumbent solution. GE 7 Sep 07 Slide 51 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3 x 1 { 3} x 2 {012 } x 3 { 123} x 4 {0123} x = (3,2,2¾,0) value = 527½ x 1 { 3} x 2 { 12 } x 3 { 12 } x 4 { 123} x = (3,2,2,1) value = 530 feasible solution x 3 {1,2} x 3 = 3 x 1 { 3} x 2 {0123} x 3 {0123} x 4 {0123} x = (3,2.6,2,0) value = 526 x 2 = 3 x 2 {0,1,2}

52 Branch-inferand-relax tree Solution is nonintegral, but we can backtrack because value of relaxation is no better than incumbent solution. GE 7 Sep 07 Slide 52 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3 x 1 { 3} x 2 {012 } x 3 { 123} x 4 {0123} x = (3,2,2¾,0) value = 527½ x 1 { 3} x 2 { 12 } x 3 { 12 } x 4 { 123} x = (3,2,2,1) value = 530 feasible solution x 3 {1,2} x 3 = 3 x 1 { 3} x 2 {0123} x 3 {0123} x 4 {0123} x = (3,2.6,2,0) value = 526 x 2 = 3 x 2 {0,1,2} x 1 { 3} x 2 {012 } x 3 { 3} x 4 {012 } x = (3,1½,3,½) value = 530 backtrack due to bound

53 Branch-inferand-relax tree Another feasible solution found. No better than incumbent solution, which is optimal because search has finished. GE 7 Sep 07 Slide 53 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3 x 1 { 3} x 2 {012 } x 3 { 123} x 4 {0123} x = (3,2,2¾,0) value = 527½ x 1 { 3} x 2 { 12 } x 3 { 12 } x 4 { 123} x = (3,2,2,1) value = 530 feasible solution x 3 {1,2} x 3 = 3 x 1 { 3} x 2 {0123} x 3 {0123} x 4 {0123} x = (3,2.6,2,0) value = 526 x 2 = 3 x 2 {0,1,2} x 1 { 3} x 2 {012 } x 3 { 3} x 4 {012 } x = (3,1½,3,½) value = 530 backtrack due to bound x 1 { 3} x 2 { 3} x 3 {012 } x 4 {012 } x = (3,3,0,2) value = 530 feasible solution

54 Branch-inferand-relax tree Two optimal solutions found. In general, not all optimal solutions are found, GE 7 Sep 07 Slide 54 x 1 { 12 } x 2 { 23} x 3 { 123} x 4 { 123} infeasible relaxation x 1 { 123} x 2 {0123} x 3 {0123} x 4 {0123} x = (2⅓,3,2⅔,0) value = 523⅓ x 1 {1,2} x 1 = 3 x 1 { 3} x 2 {012 } x 3 { 123} x 4 {0123} x = (3,2,2¾,0) value = 527½ x 1 { 3} x 2 { 12 } x 3 { 12 } x 4 { 123} x = (3,2,2,1) value = 530 optimal solution x 3 {1,2} x 3 = 3 x 1 { 3} x 2 {0123} x 3 {0123} x 4 {0123} x = (3,2.6,2,0) value = 526 x 2 = 3 x 2 {0,1,2} x 1 { 3} x 2 { 3} x 3 {012 } x 4 {012 } x = (3,3,0,2) value = 530 optimal solution x 1 { 3} x 2 {012 } x 3 { 3} x 4 {012 } x = (3,1½,3,½) value = 530 backtrack due to bound

55 Other types of cutting planes Lifted 0-1 knapsack inequalities Clique inequalities Gomory cuts Mixed integer rounding cuts Disjunctive cuts Specialized cuts Flow cuts (fixed charge network flow problem) Comb inequalities (traveling salesman problem) Many, many others GE 7 Sep 07 Slide 55

56 Example: Product Configuration Branch-and-bound search with propagation and relaxation of variable indices. From: Thorsteinsson and Ottosson (2001) GE 7 Sep 07 Slide 56

57 This example illustrates: Propagation of variable indices. Variable index is converted to a specially structured element constraint. Specially structured filtering for element. Valid knapsack cuts are derived and propagated. GE 7 Sep 07 Slide 57

58 This example illustrates: Propagation of variable indices. Variable index is converted to a specially structured element constraint. Specially structured filtering for element. Valid knapsack cuts are derived and propagated. Relaxation of variable indices. Element is interpreted as a disjunction of linear systems. Convex hull relaxation for disjunction is used. GE 7 Sep 07 Slide 58

59 The problem Choose what type of each component, and how many Personal computer Memory Memory Memory Memory Memory Disk drive Disk drive Memory Power supply Power supply Power supply Power supply GE 7 Sep 07 Slide 59 Disk drive Disk drive Disk drive

60 Problem data GE 7 Sep 07 Slide 60

61 Model of the problem Unit cost of producing attribute j Amount of attribute j produced (< 0 if consumed): memory, heat, power, weight, etc. min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i Amount of attribute j produced by type t i of component i t i is a variable index Quantity of component i installed GE 7 Sep 07 Slide 61

62 Model of the problem Linear inequality metaconstraint min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i GE 7 Sep 07 Slide 62

63 Model of the problem Indexed linear metaconstraint min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i GE 7 Sep 07 Slide 63

64 To solve it: Branch on domains of t i and q i. Propagate element constraints and bounds on v j. Derive and propagate knapsack cuts. Relax element. Convex hull relaxation for disjunction. GE 7 Sep 07 Slide 64

65 Propagation min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i GE 7 Sep 07 Slide 65 This is propagated in the usual way

66 Propagation min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i GE 7 Sep 07 Slide 66 = v z, all j j i i ( ) element t,( q, A,, q A ), z, all i, j i i ij 1 i ijn i This is rewritten as This is propagated in the usual way

67 Propagation = v z, all j j i i ( ) element t,( q, A,, q A ), z, all i, j i i ij 1 i ijn i This can be propagated by (a) using specialized filters for element constraints of this form GE 7 Sep 07 Slide 67

68 Propagation = v z, all j j i i ( ) element t,( q, A,, q A ), z, all i, j i i ij 1 i ijn i This is propagated by (a) using specialized filters for element constraints of this form, (b) adding knapsack cuts for the valid inequalities: i i { } max A q v, all j k D t i { } min A q v, all j k D t i ijk i ijk i j and (c) propagating the knapsack cuts. j [ v j, v j ] is current domain of v j GE 7 Sep 07 Slide 68

69 Relaxation min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i GE 7 Sep 07 Slide 69 This is relaxed as v j v j v j

70 Relaxation min j c v j j v = q A, all j j i ijt ik L v U, all j j j j i GE 7 Sep 07 Slide 70 = v z, all j j i i ( ) element t,( q, A,, q A ), z, all i, j i i ij 1 i ijn i This is relaxed by relaxing this and adding the knapsack cuts. This is relaxed as v j v j v j

71 Relaxation = v z, all j j i i ( ) element t,( q, A,, q A ), z, all i, j i i ij 1 i ijn i This is relaxed by writing each element constraint as a disjunction of linear systems and writing a convex hull relaxation of the disjunction: z = A q, q = q i ijk ik i ik k D k D t i t i GE 7 Sep 07 Slide 71

72 Relaxation So the following LP relaxation is solved at each node of the search tree to obtain a lower bound: min j v = A q, all j q = q, all i v v v, all j q q q, all i { } knapsack cuts for max A q v, all j { } knapsack cuts for min A q v, all j q 0, all i, k t i j j j ijk ik i k D t i j ik k D j j j i i i ik c v i i k D k D t i t i ijk i j ijk i j GE 7 Sep 07 Slide 72

73 Solution of the example After propagation, the solution of the relaxation is feasible at the root node. No branching needed. min j v = A q, all j q = q, all i v v v, all j q q q, all i { } knapsack cuts for max A q v, all j { } knapsack cuts for min A q v, all j q 0, all i, k t i j j j ijk ik i k D t i j ik k D j j j i i i ik c v i i k D k D t i t i q 1, q 1C = 1 t 1 = C q 2, q 2A = 2 t 2 = A q 3, q 3B = 3 t 3 = B ijk i j ijk i j GE 7 Sep 07 Slide 73

74 Solution of the example After propagation, the solution of the relaxation is feasible at the root node. No branching needed. GE 7 Sep 07 Slide 74 min j v = A q, all j q = q, all i v v v, all j q q q, all i { } knapsack cuts for max A q v, all j { } knapsack cuts for min A q v, all j q 0, all i, k t i j j j ijk ik i k D t i j ik k D j j j i i i ik c v i i k D k D t i t i q 1, q 1C = 1 t 1 = C q 2, q 2A = 2 t 2 = A q 3, q 3B = 3 t 3 = B ijk i j ijk i j q 1 is integral, only one q 1k is positive

75 Solution of the example After propagation, the solution of the relaxation is feasible at the root node. No branching needed. Personal computer Memory B Memory B Memory B Disk drive A Disk drive A q 1, q 1C = 1 t 1 = C q 2, q 2A = 2 t 2 = A q 3, q 3B = 3 t 3 = B Power supply C GE 7 Sep 07 Slide 75

76 Computational Results (Ottosson & Thorsteinsson) x10 16x20 20x24 20x30 CPLEX CLP Hybrid Seconds Problem GE 7 Sep 07 Slide 76

77 Example: Continuous Global Optimization Branching on continuous variables with Lagrangean-based bounds propagation and linear or convex nonlinear relaxation GE 7 Sep 07 Slide 77

78 This example illustrates: Branching by splitting boxes. Propagation based on: Interval arithmetic. Lagrange multipliers. Linear relaxation using McCormick factorization. The best continuous global solvers (e.g., BARON) combine these tecnniques from CP and OR. GE 7 Sep 07 Slide 78

79 Propagation using Lagrange multipliers Supposing min f ( x ) g ( x ) 0 x S has optimal value v*, and each constraint i has Lagrange multiplier λ i *, it can be shown that if constraint i is tight, g ( x ) i U v * λ i * Where U is an upper bound on the optimal value. GE 7 Sep 07 Slide 79

80 Propagation using Lagrange multipliers Supposing min f ( x ) g ( x ) 0 x S has optimal value v*, and each constraint i has Lagrange multiplier λ i *, it can be shown that if constraint i is tight, g ( x ) i U v * λ i * This inequality can be propagated. GE 7 Sep 07 Slide 80

81 Propagation using Lagrange multipliers Supposing min f ( x ) g ( x ) 0 x S has optimal value v*, and each constraint i has Lagrange multiplier λ i *, it can be shown that if constraint i is tight, g ( x ) i U v * λ i * This inequality can be propagated. When applied to bounds on variables, this technique yields bounds propagation. When the relaxation is linear, this becomes reduced cost variable fixing (widely used in integer programming) GE 7 Sep 07 Slide 81

82 The problem max x + x 4 x x = x + x 2 x [0,1], x [0,2] 1 2 Feasible set GE 7 Sep 07 Slide 82 x 2 Global optimum Local optimum x 1

83 The Problem max 4 x x = x + x x + x 2 x [0,1], x [0,2] 1 2 GE 7 Sep 07 Slide 83 Linear inequality metaconstraint

84 The Problem max 4 x x = x + x x + x 2 x [0,1], x [0,2] 1 2 GE 7 Sep 07 Slide 84 Bilinear (in)equality metaconstraint

85 To solve it: Search: split interval domains of x 1, x 2. Each node of search tree is a problem restriction. Propagation: Interval propagation, domain filtering. Use Lagrange multipliers to infer valid inequality for propagation. Relaxation: Use function factorization to obtain linear continuous relaxation. GE 7 Sep 07 Slide 85

86 Interval propagation Propagate intervals [0,1], [0,2] through constraints to obtain [1/8,7/8], [1/4,7/4] GE 7 Sep 07 Slide 86 x 2 x 1

87 Relaxation (function factorization) Factor complex functions into elementary functions that have known linear relaxations. Write 4x 1 x 2 = 1 as 4y = 1 where y = x 1 x 2. This factors 4x 1 x 2 into linear function 4y and bilinear function x 1 x 2. Linear function 4y is its own linear relaxation. GE 7 Sep 07 Slide 87

88 Relaxation (function factorization) Factor complex functions into elementary functions that have known linear relaxations. Write 4x 1 x 2 = 1 as 4y = 1 where y = x 1 x 2. This factors 4x 1 x 2 into linear function 4y and bilinear function x 1 x 2. Linear function 4y is its own linear relaxation. Bilinear function y = x 1 x 2 has relaxation: x x + x x x x y x x + x x x x x x + x x x x y x x + x x x x where domain of x j is [ x j, x j ] GE 7 Sep 07 Slide 88

89 Relaxation (function factorization) The linear relaxation becomes: min 4 y = x + x x + x 2 x x + x x x x y x x + x x x x x x + x x x x y x x + x x x x x x x, j = 1,2 j j j GE 7 Sep 07 Slide 89

90 Relaxation (function factorization) x 2 Solve linear relaxation. GE 7 Sep 07 Slide 90 x 1

91 Relaxation (function factorization) x 2 2 [1,1.75] x Solve linear relaxation. Since solution is infeasible, split an interval and branch. 2 [0.25,1] x GE 7 Sep 07 Slide 91 x 1

92 x 2 [1,1.75] x 2 [0.25,1] x 2 x 2 x 1 GE 7 Sep 07 Slide 92 x 1

93 x 2 [1,1.75] x 2 [0.25,1] x 2 Solution of relaxation is feasible, value = 1.25 x 2 This becomes incumbent solution x 1 GE 7 Sep 07 Slide 93 x 1

94 x 2 [1,1.75] x 2 [0.25,1] x 2 Solution of relaxation is feasible, value = 1.25 This becomes incumbent solution x 2 Solution of relaxation is not quite feasible, value = Also use Lagrange multipliers for domain filtering x 1 x 1 GE 7 Sep 07 Slide 94

95 Relaxation (function factorization) min 4 y = x + x x + x 2 x x + x x x x y x x + x x x x x x + x x x x y x x + x x x x x x x, j = 1,2 j j j Associated Lagrange multiplier in solution of relaxation is λ 2 = 1.1 GE 7 Sep 07 Slide 95

96 Relaxation (function factorization) min 4 y = x + x x + x 2 Associated Lagrange multiplier in solution of relaxation is λ 2 = 1.1 x x + x x x x y x x + x x x x x x + x x x x y x x + x x x x x x x, j = 1,2 j j j GE 7 Sep 07 Slide 96 This yields a valid inequality for propagation: x 1 + x 2 2 = Value of relaxation Lagrange multiplier Value of incumbent solution

97 Example: Airline Crew Scheduling Branch and price in which a linear relaxation of an MILP is solved by CP-based column generation From: Fahle et al. (2002) GE 7 Sep 07 Slide 97

98 This example illustrates: Overall mixed integer programming framework. Linear relaxation solved by CP-based column generation. GE 7 Sep 07 Slide 98

99 Solving an LP by column generation Suppose the LP relaxation of an integer programming problem has a huge number of variables: min 0 cx Ax = b x GE 7 Sep 07 Slide 99

100 Solving an LP by column generation Suppose the LP relaxation of an integer programming problem has a huge number of variables: We will solve a restricted master problem, which has a small subset of the variables: Column j of A GE 7 Sep 07 Slide 100 min 0 cx Ax = b x min j J x j 0 j J j j c x j j A x = b ( λ )

101 Solving an LP by column generation Suppose the LP relaxation of an integer programming problem has a huge number of variables: min 0 cx Ax = b x We will solve a restricted master problem, which has a small subset of the variables: Column j of A min j J x j 0 j J j j c x j j A x = b Adding x k to the problem would improve the solution if x k has a negative reduced cost: r k = c k λ A k < 0 GE 7 Sep 07 Slide 101 Row vector of dual (Lagrange) multipliers ( λ )

102 Column generation Adding x k to the problem would improve the solution if x k has a negative reduced cost: r k = c k λ A k < 0 Computing the reduced cost of x k is known as pricing x k. Cost of column y So we solve the pricing problem: min c y λ y y is a column of A GE 7 Sep 07 Slide 102

103 Column generation Adding x k to the problem would improve the solution if x k has a negative reduced cost: r k = c k λ A k < 0 Computing the reduced cost of x k is known as pricing x k. Cost of column y So we solve the pricing problem: min c y λ y y is a column of A This can often be solved by CP. We hope to find an optimal solution before generating too many columns. GE 7 Sep 07 Slide 103

104 Airline Crew Scheduling We want to assign crew members to flights to minimize cost while covering the flights and observing complex work rules. Flight data A roster is the sequence of flights assigned to a single crew member. The gap between two consecutive flights in a roster must be from 2 to 3 hours. Total flight time for a roster must be between 6 and 10 hours. Start time Finish time GE 7 Sep 07 Slide 104 For example, flight 1 cannot immediately precede 6 flight 4 cannot immediately precede 5. The possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6)

105 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. GE 7 Sep 07 Slide 105

106 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. Rosters that cover flight 1. GE 7 Sep 07 Slide 106

107 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. Rosters that cover flight 2. GE 7 Sep 07 Slide 107

108 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. Rosters that cover flight 3. GE 7 Sep 07 Slide 108

109 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. Rosters that cover flight 4. GE 7 Sep 07 Slide 109

110 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. Rosters that cover flight 5. GE 7 Sep 07 Slide 110

111 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. Rosters that cover flight 6. GE 7 Sep 07 Slide 111

112 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (1,3,5), (1,4,6), (2,3,5), (2,4,6) The LP relaxation of the problem is: Cost c 12 of assigning crew member 1 to roster 2 = 1 if we assign crew member 1 to roster 2, = 0 otherwise. Each crew member is assigned to exactly 1 roster. Each flight is assigned at least 1 crew member. GE 7 Sep 07 Slide 112 In a real problem, there can be millions of rosters.

113 Airline Crew Scheduling We start by solving the problem with a subset of the columns: Optimal dual solution u 1 u 2 v 1 v 2 v 3 v 4 v 5 v 6 GE 7 Sep 07 Slide 113

114 Airline Crew Scheduling We start by solving the problem with a subset of the columns: Dual variables u 1 u 2 v 1 v 2 v 3 v 4 v 5 v 6 GE 7 Sep 07 Slide 114

115 Airline Crew Scheduling We start by solving the problem with a subset of the columns: Dual variables u 1 u 2 v 1 v 2 v 3 v 4 v 5 v 6 GE 7 Sep 07 Slide 115 The reduced cost of an excluded roster k for crew member i is c ik u i v j j in roster k We will formulate the pricing problem as a shortest path problem.

116 Pricing problem Crew member 1 Crew member 2 GE 7 Sep 07 Slide 116 2

117 Pricing problem Crew member 1 Crew member 2 GE 7 Sep 07 Slide 117 Each s-t path corresponds to a roster, provided the flight time is within bounds. 2

118 Pricing problem Crew member 1 Crew member 2 GE 7 Sep 07 Slide 118 Cost of flight 3 if it immediately follows flight 1, offset by dual multiplier for flight 1 2

119 Pricing problem Crew member 1 Dual multiplier omitted to break symmetry Crew member 2 GE 7 Sep 07 Slide 119 Cost of transferring from home to flight 1, offset by dual multiplier for crew member 1 2

120 Pricing problem Crew member 1 Crew member 2 GE 7 Sep 07 Slide 120 Length of a path is reduced cost of the corresponding roster. 2

121 Pricing problem 10 Crew member 1 0 Crew member 2 GE 7 Sep 07 Slide Arc lengths using dual solution of LP relaxation

122 Pricing problem 10 Crew member 1 Reduced cost = 1 Add x 12 to problem. 0 Crew member 2 Reduced cost = 2 Add x 23 to problem. GE 7 Sep 07 Slide Solution of shortest path problems After x 12 and x 23 are added to the problem, no remaining variable has negative reduced cost.

123 Pricing problem The shortest path problem cannot be solved by traditional shortest path algorithms, due to the bounds on total path length. It can be solved by CP: Set of flights assigned to crew member i Path length Graph Path global constraint Setsum global constraint Path( X, z, G ), all flights i i i ( j j ) T f s T min max j X i { } X flights, z < 0, all i i i Duration of flight j GE 7 Sep 07 Slide 123

124 Example: Machine Scheduling Constraint-directed search using logic-based Benders decomposition GE 7 Sep 07 Slide 124

125 This example illustrates: Constraint-based search as Benders decomposition Nogoods are Benders cuts. Solution of the master problem by MILP. Allocate jobs to machines. Solution of the subproblem by CP. Schedule jobs on each machine GE 7 Sep 07 Slide 125

126 Benders decomposition Constraint-directed search in which the master problem contains a fixed set of variables x. Applied to problems of the form min f ( x, y ) S ( x, y ) x D y D x, y When x is fixed to some value, the resulting subproblem is much easier: min f ( x, y ) S ( x, y ) y D y perhaps because it decouples into smaller problems. GE 7 Sep 07 Slide 126

127 Benders decomposition Constraint-directed search in which the master problem contains a fixed set of variables x. Applied to problems of the form min f ( x, y ) S ( x, y ) x D y D x, y When x is fixed to some value, the resulting subproblem is much easier: min f ( x, y ) S ( x, y ) y D y perhaps because it decouples into smaller problems. Nogoods are Benders cuts and exclude solutions no better than x. The Benders cut is obtained by solving the inference dual of the subproblem (classically, the linear programming dual). GE 7 Sep 07 Slide 127

128 Machine Scheduling Assign 5 jobs to 2 machines (A and B), and schedule the machines assigned to each machine within time windows. The objective is to minimize makespan. Time lapse between start of first job and end of last job. Assign the jobs in the master problem, to be solved by MILP. Schedule the jobs in the subproblem, to be solved by CP. GE 7 Sep 07 Slide 128

129 Machine Scheduling Job Data Once jobs are assigned, we can minimize overall makespan by minimizing makespan on each machine individually. So the subproblem decouples. Machine A Machine B GE 7 Sep 07 Slide 129

130 Machine Scheduling Job Data Once jobs are assigned, we can minimize overall makespan by minimizing makespan on each machine individually. So the subproblem decouples. Minimum makespan schedule for jobs 1, 2, 3, 5 on machine A GE 7 Sep 07 Slide 130

131 Machine Scheduling The problem is min M M s + p, all j j x j r s d p, all j j j j x j Start time of job j ( j j = ij j = ) disjunctive ( s x i ),( p x i ), all i Time windows Jobs cannot overlap GE 7 Sep 07 Slide 131 j j

132 Machine Scheduling The problem is min M M s + p, all j j x j r s d p, all j j j j x j ( j j = ij j = ) Indexed linear metaconstraint disjunctive ( s x i ),( p x i ), all i GE 7 Sep 07 Slide 132 j j

133 Machine Scheduling The problem is min M M s + p, all j j x j r s d p, all j j j j x j ( j j = ij j = ) disjunctive ( s x i ),( p x i ), all i Specially-structured indexed linear metaconstraint Disjunctive scheduling metaconstraint GE 7 Sep 07 Slide 133 j j

134 Machine Scheduling The problem is min M M s + p, all j j x j r s d p, all j j j j x j Start time of job j ( j j = ij j = ) disjunctive ( s x i ),( p x i ), all i Time windows Jobs cannot overlap For a fixed assignment x the subproblem on each machine i is GE 7 Sep 07 Slide 134 min M M s + p, all j with x = i j x j j r s d p, all j with x = i j j j x j j ( s j x j = i p ij x j = i ) disjunctive ( ),( ) j j j j

135 Benders cuts Suppose we assign jobs 1,2,3,5 to machine A in iteration k. We can prove that 10 is the optimal makespan by proving that the schedule is infeasible with makespan 9. Edge finding derives infeasibility by reasoning only with jobs 2,3,5. So these jobs alone create a minimum makespan of 10. So we have a Benders cut v B k + 1 ( x ) = 10 if x = x = x = A otherwise GE 7 Sep 07 Slide 135

136 Benders cuts We want the master problem to be an MILP, which is good for assignment problems. So we write the Benders cut v B x k + 1 ( ) = 10 if x = x = x = A otherwise Using 0-1 variables: 10 ( ) A A A v x x x v = 1 if job 5 is assigned to machine A GE 7 Sep 07 Slide 136

137 Master problem The master problem is an MILP: min v j j 5 = 5 1 = 1 p x Aj Aj p x Bj Bj Constraints derived from time windows 10, etc. Constraints derived from release times 10, etc. 5 5 v p x, v 2 + p x, etc., i = A, B ij ij ij ij j = 1 j = 3 v 10( x + x + x 2) v 8 x x ij B 4 { 0,1 } A 2 A 3 A 5 Benders cut from machine A Benders cut from machine B GE 7 Sep 07 Slide 137

138 Computational Results (Jain & Grossmann) Seconds Problem sizes (jobs, machines) 1 - (3,2) 2 - (7,3) 3 - (12,3) 4 - (15,5) 5 - (20,5) Each data point represents an average of 2 instances MILP and CP ran out of memory on 1 of the largest instances Problem size GE 7 Sep 07 Slide 138 MILP CP OPL Hybrid

139 Enhancement Using Branch and Check (Thorsteinsson) Computation times in seconds. Problems have 30 jobs, 7 machines Hybrid Branch & check Seconds Problem GE 7 Sep 07 Slide 139

140 Stronger Benders cuts If all release times are the same, we can strengthen the Benders cuts. We are now using the cut v M ik x ij J ik + 1 j J ik Min makespan on machine i in iteration k Set of jobs assigned to machine i in iteration k A stronger cut provides a useful bound even if only some of the jobs in J ik are assigned to machine i: v M (1 x ) p ik ij ij j J ik These results can be generalized to cumulative scheduling. GE 7 Sep 07 Slide 140

141 Three success stories These are chosen because: They illustrate how scheduling interacts with other aspects of supply chain. And thus how CP can interact with other methods. Since they were part of a government (EU) supported project (LISCOS), a fair amount of detail was released to public. All were solved with help of Dash s Mosel system. GE 7 Sep 07 Slide 141

142 Process scheduling and lot sizing at BASF Manufacture of polypropylenes in 3 stages polymerization intermediate storage extrusion GE 7 Sep 07 Slide 142

143 Process scheduling and lot sizing at BASF Manual planning (old method) Required 3 days Limited flexibility and quality control 24/7 continuous production Variable batch size. Sequence-dependent changeover times. GE 7 Sep 07 Slide 143

144 Process scheduling and lot sizing at BASF Intermediate storage Limited capacity One product per silo Extrusion Production rate depends on product and machine GE 7 Sep 07 Slide 144

145 Process scheduling and lot sizing at BASF Three problems in one Lot sizing based on customer demand forecasts Assignment put each batch on a particular machine Sequencing decide the order in which each machine processes batches assigned to it GE 7 Sep 07 Slide 145

146 Process scheduling and lot sizing at BASF The problems are interdependent Lot sizing depends on assignment, since machines run at different speeds Assignment depends on sequencing, due to restrictions on changeovers Sequencing depends on lot sizing, due to limited intermediate storage GE 7 Sep 07 Slide 146

147 Process scheduling and lot sizing at BASF Solve the problems simultaneously Lot sizing: solve with MIP (using XPRESS-MP) Assignment: solve with MIP Sequencing: solve with CP (using CHIP) The MIP and CP are linked mathematically. Use logic-based Benders decomposition, developed only in the last few years. GE 7 Sep 07 Slide 147

148 Sample schedule, illustrated with Visual Scheduler (AviS/3) Source: BASF GE 7 Sep 07 Slide 148

149 Process scheduling and lot sizing at BASF Benefits Optimal solution obtained in 10 mins. Entire planning process (data gathering, etc.) requires a few hours. More flexibility Faster response to customers Better quality control GE 7 Sep 07 Slide 149

150 Paint production at Barbot Two problems to solve simultaneously Lot sizing Machine scheduling Focus on solvent-based paints, for which there are fewer stages. Barbot is a Portuguese paint manufacturer. Several machines of each type GE 7 Sep 07 Slide 150

151 Paint production at Barbot Solution method similar to BASF case (MIP + CP). Benefits Optimal solution obtained in a few minutes for 20 machines and 80 products. Product shortages eliminated. 10% increase in output. Fewer cleanup materials. Customer lead time reduced. GE 7 Sep 07 Slide 151

152 Production line sequencing at Peugeot-Citroën The Peugeot 206 can be manufactured with 12,000 option combinations. Planning horizon is 5 days GE 7 Sep 07 Slide 152

153 Production line sequencing at Peugeot-Citroën Each car passes through 3 shops. Objectives Group similar cars (e.g. in paint shop). Reduce setups. GE 7 Sep 07 Slide 153 Balance work station loads.

154 Production line sequencing at Peugeot-Citroën Special constraints Cars with a sun roof should be grouped together in assembly. Air-conditioned cars should not be assembled consecutively. Etc. GE 7 Sep 07 Slide 154

155 Production line sequencing at Peugeot-Citroën Problem has two parts Determine number of cars of each type assigned to each line on each day. Determine sequencing for each line on each day. Problems are solved simultaneously. Again by MIP + CP. GE 7 Sep 07 Slide 155

156 Source: Peugeot/Citroën GE 7 Sep 07 Slide 156 Sample schedule

157 Production line sequencing at Peugeot-Citroën Benefits Greater ability to balance such incompatible benefits as fewer setups and faster customer service. Better schedules. GE 7 Sep 07 Slide 157

158 Line balancing at Peugeot-Citroën A classic production sequencing problem Source: Peugeot/Citroën GE 7 Sep 07 Slide 158

159 Line balancing at Peugeot-Citroën Objective Equalize load at work stations. Keep each worker on one side of the car Constraints Precedence constraints between some operations. Ergonomic requirements. Right equipment at stations (e.g. air socket) GE 7 Sep 07 Slide 159

160 Line balancing at Peugeot-Citroën Solution again obtained by an integrated method. MIP: obtain solution without regard to precedence constraints. CP: Reschedule to enforce precedence constraints. The two methods interact. GE 7 Sep 07 Slide 160

161 Source: Peugeot/Citroën GE 7 Sep 07 Slide 161

162 Line balancing at Peugeot-Citroën Benefits Better equalization of load. Some stations could be closed, reducing labor. Improvements needed Reduce trackside clutter. Equalize space requirements. Keep workers on one side of car. GE 7 Sep 07 Slide 162

163 Background Reading Much of this talk is based on: J. N. Hooker, Integrated Methods for Optimization, Springer (2007). J. N. Hooker, Operations research methods in constraint programming, in F. Rossi, P. van Beek and T. Walsh, eds., Handbook of Constraint Programming, Elsevier (2006) Cited references: T. Fahle, U. Junker, S. E. Karish, N. Kohn, M. Sellmann, B. Vaaben. Constraint programming based column generation for crew assignment, Journal of Heuristics 8 (2002) E. Thorsteinsson and G. Ottosson, Linear relaxation and reducedcost based propagation of continuous variable subscripts, Annals of Operations Research 115 (2001) GE 7 Sep 07 Slide 163

164 GE 7 Sep 07 Slide 164