A Framework for Integrating Optimization and Constraint Programming

Transcription

1 A Framework for Integrating Optimization and Constraint Programming John Hooker Carnegie Mellon University SARA 2007 SARA 07 Slide

2 Underlying theme Model vs. solution method. How can we match the model with the solution method? Modeling for integrated methods forces the issue. We can t formulate the problem for a particular solution method. SARA 07 Slide 2

3 An Eercise in Synthesis Analysis takes things apart. Synthesis looks for commonality. Characteristic of 9 th century. I will provide an overview of several eamples. Look for common patterns, not details. SARA 07 Slide 3

4 Why unify CP and Operations Research methods? One-stop shopping. One solver does it all. SARA 07 Slide 4

5 Why unify CP and OR methods? One-stop shopping. One solver does it all. Richer modeling framework. Natural models, less debugging & development time. SARA 07 Slide 5

6 Why join CP and OR methods? One-stop shopping. One solver does it all. Richer modeling framework. Natural models, less debugging & development time. Computational speedup. A selection of results SARA 07 Slide 6

7 Computational Advantage of Integrating CP and OR Using CP + relaation from MILP Focacci, Lodi, Milano (999) Refalo (999) Hooker & Osorio (999) Thorsteinsson & Ottosson (200) Problem Lesson timetabling Piecewise linear costs Flow shop scheduling, etc. Product configuration Speedup 2 to 50 times faster than CP 2 to 200 times faster than MILP 4 to 50 times faster than MILP. 30 to 40 times faster than CP, MILP SARA 07 Slide 7

8 Computational Advantage of Integrating CP and MILP Using CP + relaation from MILP Sellmann & Fahle (200) Van Hoeve (200) Bollapragada, Ghattas & Hooker (200) Beck & Refalo (2003) SARA 07 Slide 8 Problem Automatic recording Stable set problem Structural design (nonlinear) Scheduling with earliness & tardiness costs Speedup to 0 times faster than CP, MILP Better than CP in less time Up to 600 times faster than MILP. 2 problems: <6 min vs >20 hrs for MILP Solved 67 of 90, CP solved only 2

9 Computational Advantage of Integrating CP and MILP Using CP-based Branch and Price Yunes, Moura & de Souza (999) Easton, Nemhauser & Trick (2002) Problem Urban transit crew scheduling Traveling tournament scheduling Speedup Optimal schedule for 20 trips, vs. 20 for traditional branch and price First to solve 8-team instance SARA 07 Slide 9

10 Computational Advantage of Integrating CP and MILP Using CP/MILP Benders methods Jain & Grossmann (200) Thorsteinsson (200) Timpe (2002) Problem Min-cost planning & scheduing Min-cost planning & scheduling Polypropylene batch scheduling at BASF Speedup 20 to 000 times faster than CP, MILP 0 times faster than Jain & Grossmann Solved previously insoluble problem in 0 min SARA 07 Slide 0

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

12 Modeling is the key Models should be eplanatory. As in the natural sciences. CP/OR models eplain by revealing problem structure. Through metaconstraints. Based on CP s global constraints. The model structure drives the solution process. Each metaconstraint knows how to combine CP and OR to eploit its structure. SARA 07 Slide 2

13 The basic algorithm Search: Enumerate problem restrictions Tree search (branching) Constraint-based (nogood-based) search SARA 07 Slide 3

14 The basic algorithm Search: Enumerate problem restrictions Tree search (branching) Constraint-based (nogood-based) search Infer: Deduce constraints from current restriction SARA 07 Slide 4

15 The basic algorithm Search: Enumerate problem restrictions Tree search (branching) Constraint-based (nogood-based) search Infer: Deduce constraints from current restriction Rela: Solve relaation of current restriction SARA 07 Slide 5

16 Unifying framework Eisting 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 relaation methods as needed. SARA 07 Slide 6

17 Color code Techniques from CP Techniques from OR SARA 07 Slide 7

18 Some eisting methods Branching Constraint solvers (CP) Search: Branching on domains Inference: Constraint propagation, filtering Relaation: Domain store Mied integer programming (OR) Search: Branch and bound Inference: Cutting planes Relaation: Linear programming SARA 07 Slide 8

19 Some eisting methods Constraint-based search SAT solvers (CP) Search: Branching on variables Inference: Unit clause rule, clause learning (nogoods) Relaation: Conflict clauses Benders decomposition (OR) Search: Enumeration of subproblems Inference: Benders cuts (nogoods) Relaation: Master problem SARA 07 Slide 9

20 Outline Eample problems to illustrate integrated approach Branching search Freight transfer Product configuration Continuous global optimization Airline crew scheduling Constraint-based search Propositional satisfiability Machine scheduling SARA 07 Slide 20

21 Eample: Freight Transfer Branch-and-bound search with interval propagation and cutting planes SARA 07 Slide 2

22 This eample illustrates: Branching on variables, with pruning based on bounds. Propagation based on: Interval propagation. Cutting planes (knapsack cuts). Relaation based on linear programming. SARA 07 Slide 22

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

24 Number of trucks of type min {0,,2,3} i Truck type Number available Capacity (tons) Cost per truck SARA 07 Slide

25 Number of trucks of type Knapsack metaconstraint knows which inference and relaation techniques to use. SARA 07 Slide 25 min {0,,2,3} Truck type i Number available Capacity (tons) Cost per truck

26 Number of trucks of type Domain metaconstraint knows how to branch min {0,,2,3} i Truck type Number available Capacity (tons) Cost per truck SARA 07 Slide

27 Bounds propagation min {0,,2,3} i = 7 SARA 07 Slide 27

28 Bounds propagation Reduced domain min {,2,3},,, {0,,2,3} = 7 SARA 07 Slide 28

29 Continuous relaation min , i 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. SARA 07 Slide 29

30 Cutting planes (valid inequalities) min , i We can create a tighter relaation (larger minimum value) with the addition of cutting planes. SARA 07 Slide 30

31 Cutting planes (valid inequalities) min , i Cutting plane Continuous relaation All feasible solutions of the original problem satisfy a cutting plane (i.e., it is valid). SARA 07 Slide 3 But a cutting plane may eclude ( cut off ) solutions of the continuous relaation. Feasible solutions

32 Cutting planes (valid inequalities) min , {,2} is a packing i because alone cannot satisfy the inequality, even with = 2 = 3. SARA 07 Slide 32

33 Cutting planes (valid inequalities) min , {,2} is a packing i So, ( ) 3 4 SARA 07 Slide 33

34 Cutting planes (valid inequalities) min , {,2} is a packing i So, ( ) 3 4 Knapsack cut which implies 42 ( ) + = ma{ 4,3} SARA 07 Slide 34

35 Cutting planes (valid inequalities) min , i Maimal Packings {,2} {,3} {,4} Knapsack cuts SARA 07 Slide 35 Knapsack cuts corresponding to nonmaimal packings can be nonredundant.

36 Continuous relaation with cuts min , i Knapsack cuts Optimal value of is a lower bound on optimal value of original problem. SARA 07 Slide 36

37 Branch-inferand-rela tree Propagate bounds and solve relaation of original problem. { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ SARA 07 Slide 37

38 Branch-inferand-rela tree Branch on a variable with nonintegral value in the relaation. { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ {,2} = 3 SARA 07 Slide 38

39 Branch-inferand-rela tree Propagate bounds and solve relaation. Since relaation is infeasible, backtrack. { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ {,2} = 3 SARA 07 Slide 39

40 Branch-inferand-rela tree { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ Propagate bounds and solve relaation. Branch on nonintegral variable. { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation {,2} = 3 { 3} 2 {023} 3 {023} 4 {023} = (3,2.6,2,0) value = {0,,2} 2 = 3 SARA 07 Slide 40

41 Branch-inferand-rela tree { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ Branch again. { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation {,2} = 3 { 3} 2 {023} 3 {023} 4 {023} = (3,2.6,2,0) value = 526 { 3} 2 {02 } 3 { 23} 4 {023} = (3,2,2¾,0) value = 527½ 2 {0,,2} 2 = 3 3 {,2} 3 = 3 SARA 07 Slide 4

42 Branch-inferand-rela tree Solution of relaation is integral and therefore feasible in the original problem. This becomes the incumbent solution. SARA 07 Slide 42 { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ {,2} { 3} 2 {02 } 3 { 23} 4 {023} = (3,2,2¾,0) value = 527½ { 3} 3 {,2} 2 { 2 } 3 { 2 } 4 { 23} = (3,2,2,) value = 530 feasible solution = 3 3 = 3 { 3} 2 {023} 3 {023} 4 {023} = (3,2.6,2,0) value = {0,,2} 2 = 3

43 Branch-inferand-rela tree { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ Solution is nonintegral, but we can backtrack because value of relaation is no better than incumbent solution. { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation {,2} = 3 { 3} 2 {02 } 3 { 23} 4 {023} = (3,2,2¾,0) value = 527½ { 3} 2 {023} 3 {023} 4 {023} = (3,2.6,2,0) value = {0,,2} 2 = 3 SARA 07 Slide 43 { 3} 3 {,2} 2 { 2 } 3 { 2 } 4 { 23} = (3,2,2,) value = 530 feasible solution 3 = 3 { 3} 2 {02 } 3 { 3} 4 {02 } = (3,½,3,½) value = 530 backtrack due to bound

44 Branch-inferand-rela tree { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ Another feasible solution found. No better than incumbent solution, which is optimal because search has finished. SARA 07 Slide 44 { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation {,2} { 3} 2 {02 } 3 { 23} 4 {023} = (3,2,2¾,0) value = 527½ { 3} 3 {,2} 2 { 2 } 3 { 2 } 4 { 23} = (3,2,2,) value = 530 feasible solution = 3 3 = 3 { 3} 2 {023} 3 {023} 4 {023} = (3,2.6,2,0) value = {0,,2} 2 = 3 { 3} 2 {02 } 3 { 3} 4 {02 } = (3,½,3,½) value = 530 backtrack due to bound { 3} 2 { 3} 3 {02 } 4 {02 } = (3,3,0,2) value = 530 feasible solution

45 Branch-inferand-rela tree Two optimal solutions found. In general, not all optimal solutions are found, SARA 07 Slide 45 { 2 } 2 { 23} 3 { 23} 4 { 23} infeasible relaation { 23} 2 {023} 3 {023} 4 {023} = (2⅓,3,2⅔,0) value = 523⅓ {,2} { 3} 2 {02 } 3 { 23} 4 {023} = (3,2,2¾,0) value = 527½ { 3} 3 {,2} 2 { 2 } 3 { 2 } 4 { 23} = (3,2,2,) value = 530 optimal solution = 3 3 = 3 { 3} 2 {023} 3 {023} 4 {023} = (3,2.6,2,0) value = {0,,2} 2 = 3 { 3} 2 {02 } 3 { 3} 4 {02 } = (3,½,3,½) value = 530 backtrack due to bound { 3} 2 { 3} 3 {02 } 4 {02 } = (3,3,0,2) value = 530 optimal solution

46 Other types of cutting planes Lifted 0- knapsack inequalities Clique inequalities Gomory cuts Mied integer rounding cuts Disjunctive cuts Specialized cuts Flow cuts (fied charge network flow problem) Comb inequalities (traveling salesman problem) Many, many others SARA 07 Slide 46

47 Eample: Product Configuration Branch-and-bound search with propagation and relaation of variable indices. From: Thorsteinsson and Ottosson (200) SARA 07 Slide 47

48 This eample illustrates: Propagation of variable indices. Variable inde is converted to a specially structured element constraint. Specially structured filtering for element. Valid knapsack cuts are derived and propagated. SARA 07 Slide 48

49 This eample illustrates: Propagation of variable indices. Variable inde is converted to a specially structured element constraint. Specially structured filtering for element. Valid knapsack cuts are derived and propagated. Relaation of variable indices. Element is interpreted as a disjunction of linear systems. Conve hull relaation for disjunction is used. SARA 07 Slide 49

50 The problem Choose what type of each component, and how many Memory Memory Memory Memory Memory Memory Personal computer Power supply Disk drive Disk drive Disk drive Disk drive Disk drive Power supply Power supply Power supply SARA 07 Slide 50

51 Problem data SARA 07 Slide 5

52 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 v = q A, all j j i ijt ik L v U, all j j j j j j i Amount of attribute j produced by type t i of component i t i is a variable inde Quantity of component i installed SARA 07 Slide 52

53 Model of the problem Linear inequality metaconstraint min j c v v = q A, all j j i ijt ik L v U, all j j j j j j i SARA 07 Slide 53

54 Model of the problem Indeed linear metaconstraint min j c v v = q A, all j j i ijt ik L v U, all j j j j j j i SARA 07 Slide 54

55 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. Rela element. Conve hull relaation for disjunction. SARA 07 Slide 55

56 Propagation min j c v v = q A, all j j i ijt ik L v U, all j j j j j j i This is propagated in the usual way SARA 07 Slide 56

57 Propagation v z, all j j = i i ( ) element t,( q, A,, q A ), z, all i, j i i ij i ijn i min j c v v = q A, all j j i ijt ik L v U, all j j j j j j i This is rewritten as This is propagated in the usual way SARA 07 Slide 57

58 Propagation v z, all j j = i i ( ) element t,( q, A,, q A ), z, all i, j i i ij i ijn i This can be propagated by (a) using specialized filters for element constraints of this form SARA 07 Slide 58

59 Propagation v z, all j j = i i ( ) element t,( q, A,, q A ), z, all i, j i i ij 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: SARA 07 Slide 59 i i ma k D t i min k D t i { } A q v, all j ijk { } i A q v, all j ijk i j and (c) propagating the knapsack cuts. j [ v, v ] j j is current domain of v j

60 Relaation min j c v v = q A, all j j i ijt ik L v U, all j j j j j j i This is relaed as v j v j v j SARA 07 Slide 60

61 Relaation v z, all j j = i i ( ) element t,( q, A,, q A ), z, all i, j i i ij i ijn i min j c v v = q A, all j j i ijt ik L v U, all j j j j j j i This is relaed by relaing this and adding the knapsack cuts. This is relaed as v j v j v j SARA 07 Slide 6

62 Relaation v z, all j j = i i ( ) element t,( q, A,, q A ), z, all i, j i i ij i ijn i This is relaed by writing each element constraint as a disjunction of linear systems and writing a conve hull relaation of the disjunction: z = A q, q = q i ijk ik i ik k D k D ti ti SARA 07 Slide 62

63 Relaation So the following LP relaation is solved at each node of the search tree to obtain a lower bound: min j v = A q, all j j ijk ik i k D q = q, all i j k D t i v v v, all j q q q, all i { } knapsack cuts for ma A q v, all j { } knapsack cuts for min A q v, all j q 0, all i, k j t i ik j j j i i i ik c v j i i k D k D t i t i ijk i j ijk i j SARA 07 Slide 63

64 Solution of the eample After propagation, the solution of the relaation is feasible at the root node. No branching needed. min j v = A q, all j j ijk ik i k D q = q, all i j k D t i v v v, all j q q q, all i { } knapsack cuts for ma A q v, all j { } knapsack cuts for min A q v, all j q 0, all i, k j t i ik j j j i i i ik c v j i i k D k D t i t i q, q C = t = C q 2, q 2A = 2 t 2 = A q 3, q 3B = 3 t 3 = B ijk i j ijk i j SARA 07 Slide 64

65 After propagation, the solution of the relaation is feasible at the root node. No branching needed. SARA 07 Slide 65 Solution of the eample min j v = A q, all j j ijk ik i k D q = q, all i j k D t i v v v, all j q q q, all i { } knapsack cuts for ma A q v, all j { } knapsack cuts for min A q v, all j q 0, all i, k j t i ik j j j i i i ik c v j i i k D k D t i t i q, q C = t = C q 2, q 2A = 2 t 2 = A q 3, q 3B = 3 t 3 = B ijk i j ijk i j q is integral, only one q k is positive

66 Solution of the eample After propagation, the solution of the relaation is feasible at the root node. No branching needed. Personal computer Memory B Memory B Memory B Disk drive A Disk drive A q, q C = t = C q 2, q 2A = 2 t 2 = A q 3, q 3B = 3 t 3 = B Power supply C SARA 07 Slide 66

67 Eample: Continuous Global Optimization Branching on continuous variables with Lagrangean-based bounds propagation and linear or conve nonlinear relaation SARA 07 Slide 67

68 This eample illustrates: Branching by splitting boes. Propagation based on: Interval arithmetic. Lagrange multipliers. Linear relaation using McCormick factorization. The best continuous global solvers (e.g., BARON) combine these tecnniques from CP and OR. SARA 07 Slide 68

69 Propagation using Lagrange multipliers Supposing min f ( ) g( ) 0 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 ( ) i U v λ Where U is an upper bound on the optimal value. * i * SARA 07 Slide 69

70 Propagation using Lagrange multipliers Supposing min f ( ) g( ) 0 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 ( ) i U v λ This inequality can be propagated. * i * SARA 07 Slide 70

71 Propagation using Lagrange multipliers Supposing min f ( ) g( ) 0 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 ( ) i U v λ This inequality can be propagated. * i * When applied to bounds on variables, this technique yields bounds propagation. When the relaation is linear, this becomes reduced cost variable fiing (widely used in integer programming) SARA 07 Slide 7

72 The problem ma 2 4 = [0,], [0,2] 2 2 Global optimum Local optimum Feasible set SARA 07 Slide 72

73 The Problem ma 4 = [0,], [0,2] 2 Linear inequality metaconstraint SARA 07 Slide 73

74 The Problem ma 4 = [0,], [0,2] 2 Bilinear (in)equality metaconstraint SARA 07 Slide 74

75 To solve it: Search: split interval domains of, 2. Each node of search tree is a problem restriction. Propagation: Interval propagation, domain filtering. Use Lagrange multipliers to infer valid inequality for propagation. Relaation: Use function factorization to obtain linear continuous relaation. SARA 07 Slide 75

76 Interval propagation 2 Propagate intervals [0,], [0,2] through constraints to obtain [/8,7/8], [/4,7/4] SARA 07 Slide 76

77 Relaation (function factorization) Factor comple functions into elementary functions that have known linear relaations. Write 4 2 = as 4y = where y = 2. This factors 4 2 into linear function 4y and bilinear function 2. Linear function 4y is its own linear relaation. SARA 07 Slide 77

78 Relaation (function factorization) Factor comple functions into elementary functions that have known linear relaations. Write 4 2 = as 4y = where y = 2. This factors 4 2 into linear function 4y and bilinear function 2. Linear function 4y is its own linear relaation. Bilinear function y = 2 has relaation: + y y where domain of j is [, ] j j SARA 07 Slide 78

79 Relaation (function factorization) The linear relaation becomes: min 4y = y y , j =,2 j j j SARA 07 Slide 79

80 Relaation (function factorization) 2 Solve linear relaation. SARA 07 Slide 80

81 Relaation (function factorization) 2 Solve linear relaation. 2 [,.75] Since solution is infeasible, split an interval and branch. 2 [0.25,] SARA 07 Slide 8

82 2 [,.75] 2 [0.25,] 2 2 SARA 07 Slide 82

83 2 [,.75] 2 [0.25,] 2 Solution of relaation is feasible, value =.25 This becomes incumbent solution 2 SARA 07 Slide 83

84 2 [,.75] 2 [0.25,] 2 Solution of relaation is feasible, value =.25 This becomes incumbent solution 2 Solution of relaation is not quite feasible, value =.854 Also use Lagrange multipliers for domain filtering SARA 07 Slide 84

85 Relaation (function factorization) min 4y = y y , j =,2 j j j Associated Lagrange multiplier in solution of relaation is λ 2 =. SARA 07 Slide 85

86 Relaation (function factorization) min 4y = Associated Lagrange multiplier in solution of relaation is λ 2 =. + y y , j =,2 j j j SARA 07 Slide 86 This yields a valid inequality for propagation: =.45. Value of relaation Lagrange multiplier Value of incumbent solution

87 Eample: Airline Crew Scheduling Branch and price in which a linear relaation of an MILP is solved by CP-based column generation From: Fahle et al. (2002) SARA 07 Slide 87

88 This eample illustrates: Overall mied integer programming framework. Linear relaation solved by CP-based column generation. SARA 07 Slide 88

89 Solving an LP by column generation Suppose the LP relaation of an integer programming problem has a huge number of variables: min c A = b 0 SARA 07 Slide 89

90 Solving an LP by column generation Suppose the LP relaation 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 min c A = b 0 min j J j j 0 j J A j c j = b j ( λ) SARA 07 Slide 90

91 Solving an LP by column generation Suppose the LP relaation 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 min c A = b 0 min j J j j 0 j J A j c j = b j ( λ) Adding k to the problem would improve the solution if k has a negative reduced cost: r = c λa < k k k 0 SARA 07 Slide 9 Row vector of dual (Lagrange) multipliers

92 Column generation Adding k to the problem would improve the solution if k has a negative reduced cost: r = c λa < k k k 0 Computing the reduced cost of k is known as pricing k. Cost of column y So we solve the pricing problem: min y cy λy is a column of A SARA 07 Slide 92

93 Column generation Adding k to the problem would improve the solution if k has a negative reduced cost: r = c λa < k k k 0 Computing the reduced cost of k is known as pricing k. Cost of column y So we solve the pricing problem: This can often be solved by CP. min y cy λy is a column of A We hope to find an optimal solution before generating too many columns. SARA 07 Slide 93

94 Airline Crew Scheduling We want to assign crew members to flights to minimize cost while covering the flights and observing comple work rules. Flight data Start time Finish time SARA 07 Slide 94 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 0 hours. For eample, flight cannot immediately precede 6 flight 4 cannot immediately precede 5. The possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6)

95 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. SARA 07 Slide 95

96 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. Rosters that cover flight. SARA 07 Slide 96

97 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. Rosters that cover flight 2. SARA 07 Slide 97

98 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. Rosters that cover flight 3. SARA 07 Slide 98

99 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. Rosters that cover flight 4. SARA 07 Slide 99

100 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. Rosters that cover flight 5. SARA 07 Slide 00

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

102 Airline Crew Scheduling There are 2 crew members, and the possible rosters are: (,3,5), (,4,6), (2,3,5), (2,4,6) The LP relaation of the problem is: Cost c 2 of assigning crew member to roster 2 = if we assign crew member to roster 2, = 0 otherwise. Each crew member is assigned to eactly roster. Each flight is assigned at least crew member. SARA 07 Slide 02 In a real problem, there can be millions of rosters.

103 Airline Crew Scheduling We start by solving the problem with a subset of the columns: Optimal dual solution u u 2 v v 2 v 3 v 4 v 5 v 6 SARA 07 Slide 03

104 Airline Crew Scheduling We start by solving the problem with a subset of the columns: Dual variables u u 2 v v 2 v 3 v 4 v 5 v 6 SARA 07 Slide 04

105 Airline Crew Scheduling We start by solving the problem with a subset of the columns: Dual variables u u 2 v v 2 v 3 v 4 v 5 v 6 The reduced cost of an ecluded roster k for crew member i is c u v ik i j j in roster k We will formulate the pricing problem as a shortest path problem. SARA 07 Slide 05

106 Pricing problem 2 Crew member Crew member 2 SARA 07 Slide 06

107 Pricing problem Each s-t path corresponds to a roster, provided the flight time is within bounds. 2 Crew member Crew member 2 SARA 07 Slide 07

108 Pricing problem Cost of flight 3 if it immediately follows flight, offset by dual multiplier for flight 2 Crew member Crew member 2 SARA 07 Slide 08

109 Pricing problem Cost of transferring from home to flight, offset by dual multiplier for crew member 2 Crew member Dual multiplier omitted to break symmetry Crew member 2 SARA 07 Slide 09

110 Pricing problem Length of a path is reduced cost of the corresponding roster. 2 Crew member Crew member 2 SARA 07 Slide 0

111 Pricing problem Arc lengths using dual solution of LP relaation Crew member Crew member SARA 07 Slide

112 Pricing problem Solution of shortest path problems Crew member Reduced cost = Add 2 to problem Crew member 2 Reduced cost = 2 Add 23 to problem. SARA 07 Slide After 2 and 23 are added to the problem, no remaining variable has negative reduced cost.

113 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 min i j X i { } i T f s T ma X flights, z < 0, all i i ( j j ) i Duration of flight j SARA 07 Slide 3

114 Eample: Propositional Satisfiability Constraint-directed search using conflict clauses as nogoods SARA 07 Slide 4

115 This eample illustrates: Constraint-based search Branching in Davis-Putnam-Loveland (DPL) method is reinterpreted as constraint-based search. Propagation based on unit clause rule, inference of nogoods. Nogoods are conflict clauses. Relaation that consists of current nogood set. SARA 07 Slide 5

116 Constraint-directed search Search over partial assignments to variables. Net partial assignment must satisfy nogoods so far accumulated (the master problem). Generate nogood for current partial assignment. Try to solve for the values of the remaining variables (i.e., solve the subproblem). If there is no solution, the nogood should eclude this partial assignment and perhaps others. Terminate when the subproblem has a solution. Or in an optimization problem, when the subproblem has the same value as the master problem. SARA 07 Slide 6

117 SARA 07 Slide Find a satisfying solution. The problem

118 SARA 07 Slide Find a satisfying solution. The problem CNF metaconstraint

119 To solve it by branching (DPL): Search: branch on j. Each node of search tree is a problem restriction. SARA 07 Slide 9

120 To solve it by branching (DPL): Search: branch on j. Each node of search tree is a problem restriction. Infer: clause learning, unit clause rule. SARA 07 Slide 20

121 To solve it by branching (DPL): Search: branch on j. Each node of search tree is a problem restriction. Infer: clause learning, unit clause rule. Rela: not used. SARA 07 Slide 2

122 Branching = 0 2 = 0 3 = 0 4 = 0 5 = 0 Branch to here. Unit clause rule proves infeasibility. (, 5 ) = (0,0) creates the conflict. SARA 07 Slide 22

123 Branching = 0 2 = 0 3 = 0 4 = 0 5 = 0 5 Branch to here. Unit clause rule proves infeasibility. (, 5 ) = (0,0) creates the conflict. Add conflict clause to constraint set. SARA 07 Slide 23

124 Branching = 0 2 = 0 3 = 0 4 = 0 = 0 = Backtrack and branch to here. (, 5 ) = (0,) creates the conflict. Generate conflict clause SARA 07 Slide 24

125 Branching = 0 2 = 0 4 = 0 3 = 0 = 0 = Backtrack to here and note that (, 2 ) = (0,0) is enough to create conflict. Generate new conflict clause Backtrack and branch to here. (, 5 ) = (0,) creates the conflict. Generate conflict clause SARA 07 Slide 25

126 Branching = 0 = 0 = = 0 3 = 0 2 Backtrack and branch to here and generate conflict clause = 0 = SARA 07 Slide 26

127 Branching = 0 = 0 = 2 2 Backtrack to here and generate new conflict clause 4 = 0 3 = 0 2 Backtrack and branch to here and generate conflict clause = 0 = SARA 07 Slide 27

128 Branching 2 = 0 = 0 = Backtrack and branch to here and generate final conflict clause 4 = 0 3 = 0 2 = 0 = SARA 07 Slide 28

129 To solve it by constraint-directed search: Search: generate partial assignments. Each leaf node of search tree corresponds to a partial assignment. SARA 07 Slide 29

130 To solve it by constraint-directed search: Search: generate partial assignments. Each leaf node of search tree is a partial assignment. Infer: generate nogoods (conflict clauses). Process nogoods in current relaation with parallel resolution. SARA 07 Slide 30

131 To solve it by constraint-directed search: Search: generate partial assignments. Each leaf node of search tree is a problem restriction. Infer: generate nogoods (conflict clauses). Process nogoods in current relaation with parallel resolution. Rela: Relaation R k consists of current processed nogoods. SARA 07 Slide 3

132 To solve it by constraint-directed search: Search: generate partial assignments. Each leaf node of search tree is a problem restriction. Infer: generate nogoods (conflict clauses). Process nogoods in current relaation with parallel resolution. Rela: Relaation R k consists of current processed nogoods. Select solution of relaation as in chronological backtracking. SARA 07 Slide 32

133 Constraint-Directed Search = 0 3 = 0 2 = 0 k 0 Relaation 5 R k Solution of R (0,0,0,0,0, ) k Nogoods 5 4 = 0 5 = 0 Conflict clause appears as nogood induced by solution of R k. 5 SARA 07 Slide 33

134 Constraint-Directed Search = 0 Consists of processed nogoods 3 = 0 2 = 0 k 0 Relaation 5 R k Solution of R (0,0,0,0,0, ) (0,0,0,0,, ) k Nogoods = 0 = 0 = SARA 07 Slide 34

135 Constraint-Directed Search = 0 Consists of processed nogoods 4 = 0 3 = 0 2 = 0 = 0 = k Relaation R k Solution of R (0,0,0,0,0, ) (0,0,0,0,, ) k Nogoods parallel-resolve to yield 2 parallel-absorbs SARA 07 Slide 35

136 Constraint-Directed Search = 0 = 0 = = 0 3 = 0 = 0 = SARA 07 Slide k Relaation R k Solution of (0,0,0,0,0, ) (0,0,0,0,, ) (0,,,,, ) R parallel-resolve to yield k Nogoods

137 Constraint-Directed Search = 0 = 2 = 0 3 = = 0 = 0 = k Relaation 5 2 R k Solution of R (0,0,0,0,0, ) (0,0,0,0,, ) (0,,,,, ) (,,,,, ) k Nogoods SARA 07 Slide 37 Search terminates

138 Partial-order dynamic backtracking Slight variation of previous algorithm Select solution of relaation that conforms to current nogood set. More freedom than in branching. SARA 07 Slide 38

139 Partial Order Dynamic Backtracking k 0 Relaation 5 R k Solution of R (0,0,0,0,0, ) k Nogoods 5 Arbitrarily choose one variable to be last SARA 07 Slide 39

140 Partial Order Dynamic Backtracking k 0 Relaation 5 R k Solution of R (0,0,0,0,0, ) k Nogoods 5 Other variables are penultimate Arbitrarily choose one variable to be last SARA 07 Slide 40

141 Partial Order Dynamic Backtracking k 0 2 Relaation 5 R k Solution of R (0,0,0,0,0, ) (,,,,0, ) k Nogoods 5 Since 5 is penultimate in at least one nogood, it must conform to nogoods. It must take value opposite its sign in the nogoods. 5 will have the same sign in all nogoods where it is penultimate. SARA 07 Slide 4 This allows more freedom than chronological backtracking.

142 Partial Order Dynamic Backtracking k 0 2 Relaation 5 R k Solution of R (0,0,0,0,0, ) (,,,,0, ) k Nogoods 5 5 Choice of last variable is arbitrary but must be consistent with partial order implied by previous choices. Since 5 is penultimate in at least one nogood, it must conform to nogoods. It must take value opposite its sign in the nogoods. 5 will have the same sign in all nogoods where it is penultimate. SARA 07 Slide 42 This allows more freedom than chronological backtracking.

143 Partial Order Dynamic Backtracking k 0 2 Relaation 5 5 R k Solution of R (0,0,0,0,0, ) (,,,,0, ) k Nogoods 5 5 Choice of last variable is arbitrary but must be consistent with partial order implied by previous choices. 5 5 Parallel-resolve to yield 5 SARA 07 Slide 43 Since 5 is penultimate in at least one nogood, it must conform to nogoods. It must take value opposite its sign in the nogoods. 5 will have the same sign in all nogoods where it is penultimate. This allows more freedom than chronological backtracking.

144 SARA 07 Slide ),,,,0, ( 2 ),0,,, (, ) (0,0,0,0,0, 0 Nogoods Solution of Relaation R R k k k does not parallel-resolve with Partial Order Dynamic Backtracking because 5 is not last in both clauses

145 SARA 07 Slide 45 4 ),,,,, ( 3 ),,,,0, ( 2 ),0,,, (, ) (0,0,0,0,0, 0 Nogoods Solution of Relaation R R k k k Partial Order Dynamic Backtracking Must conform Search terminates

146 SARA 07 Slide 46

147 Eample: Machine Scheduling Constraint-directed search using logic-based Benders decomposition SARA 07 Slide 47

148 This eample 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 SARA 07 Slide 48

149 Benders decomposition Constraint-directed search in which the master problem contains a fied set of variables. Applied to problems of the form min f (, y) S(, y) D, y D y When is fied to some value, the resulting subproblem is much easier: min f (, y ) S(, y) y D y perhaps because it decouples into smaller problems. SARA 07 Slide 49

150 Benders decomposition Constraint-directed search in which the master problem contains a fied set of variables. Applied to problems of the form min f (, y) S(, y) D, y D SARA 07 Slide 50 y When is fied to some value, the resulting subproblem is much easier: min f (, y ) S(, y) y D y perhaps because it decouples into smaller problems. Nogoods are Benders cuts and eclude solutions no better than. The Benders cut is obtained by solving the inference dual of the subproblem (classically, the linear programming dual).

151 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. SARA 07 Slide 5

152 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 SARA 07 Slide 52

153 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, 2, 3, 5 on machine A SARA 07 Slide 53

154 Machine Scheduling The problem is min M M s + p, all j j j r s d p, all j j j j j j Start time of job j j ( j j = ij j = ) disjunctive ( s i ),( p i ), all i Time windows Jobs cannot overlap SARA 07 Slide 54

155 Machine Scheduling The problem is min M M s + p, all j j j r s d p, all j j j j j j j ( j j = ij j = ) Indeed linear metaconstraint disjunctive ( s i ),( p i ), all i SARA 07 Slide 55

156 Machine Scheduling The problem is min M M s + p, all j j j r s d p, all j j j j j j j ( j j = ij j = ) disjunctive ( s i ),( p i ), all i Specially-structured indeed linear metaconstraint Disjunctive scheduling metaconstraint SARA 07 Slide 56

157 Machine Scheduling The problem is min M M s + p, all j j j r s d p, all j j j j j j Start time of job j j ( j j = ij j = ) disjunctive ( s i ),( p i ), all i Time windows Jobs cannot overlap For a fied assignment the subproblem on each machine i is SARA 07 Slide 57 min M M s + p, all j with = i j j j j r s d p, all j with = i j j j j j j ( s j j = i pij j = i ) disjunctive ( ),( )

158 Benders cuts Suppose we assign jobs,2,3,5 to machine A in iteration k. We can prove that 0 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 0. So we have a Benders cut SARA 07 Slide 58 0 if 2 = 3 = 4 = A v Bk + ( ) = 0 otherwise

159 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 ( ) k + 0 if 2 = 3 = 4 = A = 0 otherwise Using 0- variables: v 0( + + 2) v 0 A2 A3 A5 = if job 5 is assigned to machine A SARA 07 Slide 59

160 Master problem The master problem is an MILP: min v 5 j = 5 j = Aj Bj 5 5 ij ij ij ij j = j = 3 B4 { 0,} Constraints derived from time windows 0, etc. Constraints derived from release times 0, etc. v p, v 2 + p, etc., i = A, B v 0( + + 2) v ij p p 8 Aj Bj A2 A3 A5 Benders cut from machine A Benders cut from machine B SARA 07 Slide 60

161 Stronger Benders cuts If all release times are the same, we can strengthen the Benders cuts. We are now using the cut SARA 07 Slide 6 Min makespan on machine i in iteration k v Mik ij Jik + j J ik 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 ( ) p ik ij ij j J These results can be generalized to cumulative scheduling. ik

162 Background Reading Most 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) SARA 07 Slide 62 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) 59-8 E. Thorsteinsson and G. Ottosson, Linear relaation and reducedcost based propagation of continuous variable subscripts, Annals of Operations Research 5 (200) 5-29.

163 SARA 07 Slide 63