How to Relax. CP 2008 Slide 1. John Hooker Carnegie Mellon University September 2008

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1 How to Relax Slide 1 John Hooker Carnegie Mellon University September 2008

2 Two ways to relax Relax your mind and body. Relax your problem formulations. Slide 2

3 Relaxing a problem Feasible set of original problem Feasible set of relaxed problem Slide 3 Drop constraints to make the problem easier to solve.

4 Popular Relaxations Continuous relaxations Linear relaxations are most popular. Highly developed solution technology. Discrete relaxations Already used in CP/AI, perhaps under other names. For example, domain store. Slide 4

5 Why relax a problem? Solution of relaxed problem may solve original problem. Slide 5

6 Why relax a problem? Solution of relaxed problem may solve original problem. It may guide the search in a promising direction. Slide 6

7 Why relax a problem? Solution of relaxed problem may solve original problem. It may guide the search in a promising direction. It may prune the search tree by providing a bound on the optimal value. Slide 7

8 Why relax a problem? Solution of relaxed problem may solve original problem. It may guide the search in a promising direction. It may prune the search tree by providing a bound on the optimal value. It may help filter domains (e.g., with Lagrange multipliers). Slide 8

9 Why relax a problem? Solution of relaxed problem may solve original problem. It may guide the search in a promising direction. It may prune the search tree by providing a bound on the optimal value. It may help filter domains (e.g., with Lagrange multipliers). It can provide a more global view of the problem, because a single relaxation can pool relaxations of several constraints. Slide 9

10 When is Relaxation Useful? Relaxation already plays a central role in constraint programming. The domain store is a relaxation. Relaxations used in many other contexts. It may pay to think about how to strengthen the relaxation. Slide 10

11 When is Relaxation Useful? Relaxation is the workhorse of optimization. Without it, problems would be intractable. Particularly in mathematical programming. Slide 11

12 When is Relaxation Useful? Relaxation is the workhorse of optimization. Without it, problems would be intractable. Particularly in mathematical programming. But it is not the mere existence of an objective function that makes relaxation important. Slide 12

13 When is Relaxation Useful? Relaxation is the workhorse of optimization. Without it, problems would be intractable. Particularly in mathematical programming. But it is not the mere existence of an objective function that makes relaxation important. Often, it is the existence of constraints with many variables. These constraints do not propagate well. Such as cost constraints. Relaxation is particularly valuable for such constraints. Slide 13

14 Outline Continuous relaxations Based on mixed integer modeling. Based on analysis of global constraints. Relaxation duality To strengthen continuous or discrete relaxations. Example: Lagrangean duality Discrete relaxations. Based on the dependency graph. Based on multivalued decision diagrams. Slide 14

15 Outline Continuous relaxations Based on mixed integer modeling. Based on analysis of global constraints. Relaxation duality To strengthen continuous or discrete relaxations. Example: Lagrangean duality Discrete relaxations. Based on the dependency graph. Based on multivalued decision diagrams. Along the way Slide 15 Destinations where you can relax mind and body

16 Relax This is a survey. There is no need to follow everything in detail. Slide 16

17 Slide 17 Great Barrier Reef, Australia

18 Slide 18 Cairns, Australia

19 Continuous Relaxation: Mixed Integer Modeling Mixed Integer Models Example: Facility Location Example: Cumulative Scheduling Cutting Planes Slide 19

20 Mixed Integer Models A mixed integer/linear model has the form min cx + dy Ax + by b x, y 0 y integer First write a problem (or part of it) in this form. Then drop the integrality constraint to get a continuous relaxation. Slide 20

21 Mixed Integer Models A mixed integer/linear model has the form min cx + dy Ax + by b x, y 0 y integer First write a problem (or part of it) in this form. Then drop the integrality constraint to get a continuous relaxation. Strengthen the relaxation with cutting planes. Slide 21

22 Mixed Integer Models A mixed integer/linear model has the form min cx + dy Ax + by b x, y 0 y integer First write a problem (or part of it) in this form. Then drop the integrality constraint to get a continuous relaxation. Strengthen the relaxation with cutting planes. Historical emphasis on inequality constraints is due to success of linear programming. Slide 22

23 Mixed integer representability Theorem. A problem can be given a mixed integer model if its feasible set is the union of finitely many polyhedra having the same recession cone. Recession cone of polyhedron Union of polyhedra with the same recession cone (in this case, the origin) Polyhedron Slide 23

24 Modeling a union of polyhedra Start with a disjunction of linear systems to represent the union of polyhedra. The kth polyhedron is { x A k x b k } min k cx ( k k A x b ) Introduce a 0-1 variable y k that is 1 when x is in polyhedron k. Disaggregate x to create an x k for each k. min cx k k k A x b y, all k k k y x = y k k = 1 x k { 0,1} k Slide 24

25 Convex hull relaxation Good news: continuous relaxation of this model is a convex hull relaxation (tightest linear relaxation). min cx k k k A x b y, all k k y x = k k = 1 k 0 y 1 k x k Slide 25 Convex hull

26 General representability theorem Theorem. A problem can be given a mixed integer model if and only if it is the union of finitely many mixed integer polyhedra having the same recession cone. Recession cone of Q = recession cone of P Mixed integer polyhedron P Union of mixed integer polyhedra with the same recession cone (in this case, the origin) Polyhedron Q Slide 26 JNH, A principled approach to mixed integer problem formulation, 2008

27 Convex hull relaxation More good news: continuous relaxation of this model is a convex hull relaxation, if the polyhedra have convex hull models. k k k A x b y, all k k y k = 1 k x = x k n x R Z 0 y 1 k p k Union of mixed integer polyhedra with convex hull descriptions Slide 27 Convex hull relaxation

28 Example: Facility location m possible factory locations n markets Locate factories to serve markets so as to minimize total factory cost and transport cost. C j D j Fixed cost incurred for each vehicle used. f i Fixed cost Slide 28 i Capacity c ij K ij j Transport cost per vehicle

29 f i Fixed cost Facility location m possible factory locations C j i Capacity Slide 29 c ij K ij n markets j Transport cost per vehicle Disjunctive model: min i i ij ij ij xij Ci j 0 x, all ij 0, all ij Kijw x j ij j =, all i z zi 0 i = f = wij, all j Z x = D, all j i ij z + c w Flow on route (i,j) j Factory at location i No factory at location i

30 f i Fixed cost Facility location m possible factory locations C j i Capacity Slide 30 c ij K ij n markets j Transport cost per vehicle Disjunctive model: min i i ij ij ij xij Ci j 0 x, all ij 0, all ij Kijw x j ij j =, all i z zi 0 i = f = wij, all j Z x = D, all j i ij z + c w Flow on route (i,j) j Factory at location i Number of vehicles from factory i to market j No factory at location i

31 Facility location Problem: Disjuncts don t have same recession cone Recession directions: min i i ij ij ij xij Ci j 0 x, all ij 0, all ij Kijw x j ij j =, all i z zi 0 i = f = wij, all j Z x = D, all j i ij z + c w j x ij, z i bounded x ij, z i bounded w ij + w ij ± Slide 31

32 Facility location Solution: Add innocuous bounds Recession directions: min i i ij ij ij xij Ci j xij = 0, all j 0 xij Kijwij, all j zi = 0, all i zi = f wij 0 wij, all j Z x = D, all j i ij z + c w j x ij, z i bounded x ij, z i bounded w ij + w ij + Slide 32

33 Facility location Slide 33 Disjunctive model: Mixed integer formulation: min xij Ci j xij = 0, all j 0 xij Kijwij, all j zi = 0, all i zi = f wij 0 wij, all j Z x = D, all j min j i i i i ij ij ij x C y, all i ij i i x = D, all j ij j 0 x K w, all i, j ij ij ij y {0,1}, w Z, all i, j i i ij z + c w i i ij ij ij j f y + c w ij

34 Example: Cumulative scheduling Cumulative scheduling constraint: ( s s s p p p c c c C) cumulative (,, ),(,, ),(,, ), C p 3 job 2 A feasible solution: job 1 job 3 c 3 t Slide 34 s 3

35 Cumulative scheduling Disjunctive formulation: t i t: t T Job i must start at some time t si = t x = c, t T x C, all t it it i it it, all i C x it = c i Set of times job i is running if it starts at t job i t { t t T it } Slide 35

36 Cumulative scheduling Disjunctive formulation: t i t: t T si = t x = c, t T x C, all t it it i it it, all i C Slide 36 job i x it = c i t { t t T it } Mixed integer model: t s = ty, all i, t i t x = c y, t T, all i, t i t: t T it it it i it t { } x C, all t it t s = s, all i x i t t it it t t y it = y it i x = 1, all i 0,1, all i, t

37 Cumulative scheduling Disjunctive formulation: t i t: t T si = t x = c, t T x C, all t it it i it it, all i s = ty, all i i i t: t T t y it y Slide 37 t it it c y C, all t i it { } it = 1, all i 0,1, all i, t simplifies Mixed integer model: t s = ty, all i, t i t x = c y, t T, all i, t i t: t T it it it i it t { } x C, all t it t s = s, all i x i t t it it t t y it = y it i x = 1, all i 0,1, all i, t

38 Cutting Planes A cutting plane (cut, valid inequality) for a mixed integer model: is valid - It is satisfied by all feasible solutions of the model. cuts off solutions of the continuous relaxation. - This makes the relaxation tighter. Cutting plane Continuous relaxation Feasible solutions Slide 38

39 Some popular cutting planes Knapsack cuts - Generated for individual inequality constraints. - Sequential lifting. - Sequence-independent lifting. Rounding cuts - Generated for the entire model, they are widely used. - Gomory cuts for integer variables only. - Mixed integer rounding cuts for mixed integer models. Special-purpose cuts - For many problems. Slide 39

40 Slide 40 Les Cevénnes, France

41 Slide 41 Le fromage roquefort, France

42 Continuous Relaxation for Global Constraints Element Alldiff Circuit Cumulative scheduling Slide 42

43 Element The element constraint implements a variable indexed by a variable: x y Example: Channeling constraints for employee scheduling x y y k x i = = k i Slide 43

44 Element The element constraint implements a variable indexed by a variable: x y Example: Channeling constraints for employee scheduling x y y k x i = = k i Shift assigned to employee k Slide 44

45 Element The element constraint implements a variable indexed by a variable: x y Example: Channeling constraints for employee scheduling Employee assigned to shift y k x y y k x i = = k i Shift assigned to employee k Slide 45

46 Element The element constraint implements a variable indexed by a variable: x y Example: Channeling constraints for employee scheduling x y y k x i = = k i Employee assigned to shift i Slide 46

47 Element The element constraint implements a variable indexed by a variable: x y Example: Channeling constraints for employee scheduling Shift assigned to employee x i x y y k x i = = k i Employee assigned to shift i Slide 47

48 Element The element constraint implements a variable indexed by a variable: x y To implement x y : Replace x y with z and add the constraint ( y x x z) element,(,, ), 1 n Assume domain of y is {1,, n}. Slide 48

49 Element The element constraint implements a variable indexed by a variable: To relax View it as a disjunction x y ( y x x z) element,(,, ), 1 i ( x = z) i n Slide 49

50 Element The element constraint implements a variable indexed by a variable: To relax View it as a disjunction x y ( y x x z) element,(,, ), and write the convex hull relaxation z = Assume bounds L x U 1 i n ( x = z) i i i i i Ly x Uy, all i i k x = x, all i i i k i y = 1, y 0, all i i x i i Slide 50 JNH, Logic-based Methods for Optimization, 2000

51 Element The element constraint implements a variable indexed by a variable: x y If each x i has the same bounds L 0 x i U 0 the convex hull relaxation simplifies to i x ( n 1) U z x ( n 1) L i L x U, all i 0 i 0 0 i 0 i Slide 51 JNH, Logic-based Methods for Optimization, 2000

52 Element A specially structured element constraint implements ay x Example: Assignment cost ay x i i i Amount of product i to be manufactured Slide 52

53 Element A specially structured element constraint implements ay x Example: Assignment cost ay x i i i Amount of product i to be manufactured Machine on which product i is manufactured Slide 53

54 Element A specially structured element constraint implements ay x Example: Assignment cost ay x i i i Unit cost of manufacturing product i on machine y i Amount of product i to be manufactured Machine on which product i is manufactured Slide 54

55 Element A specially structured element constraint implements ay x To implement a y x: Replace a y x with z and add the constraint ( y a x a x z) element,(,, ), 1 n Slide 55

56 Element A specially structured element constraint implements ay x To relax element ( y,( a x,, a x), z) 1 n Slide 56

57 Element A specially structured element constraint implements ay x To relax View it as a disjunction ( y a x a x z) element,(,, ), 1 i n ( a x = z) i Slide 57

58 Element A specially structured element constraint implements ay x To relax View it as a disjunction ( y a x a x z) element,(,, ), and write the convex hull relaxation min max 1 { i } min min{ i } i { } max{ } L = min a L U = a U i L = max a L U = a U i i n Umin Lmin ULmin LUmin U L UL LU x + z x + U L U L U L U L where L x U, i ( a x = z) i max max max max max i i Slide 58 JNH, Integrated Methods for Optimization, 2007

59 Alldiff The alldiff polytope (permutation polytope) is the convex hull of feasible solutions of alldiff. The polytope is completely known. The facet-defining inequalities provide a convex hull relaxation Slide 59

60 Alldiff The alldiff polytope (permutation polytope) is the convex hull of feasible solutions of alldiff. The polytope is completely known. The facet-defining inequalities provide a convex hull relaxation For example, the convex hull of x { 1,3,6 } with domains is completely described by i x + x + x = x + x 4, x + x 4, x + x x, x, x ( x x x ) alldiff,, Slide 60 JNH, Logic-based Methods for Optimization, 2000 Williams and Yan, Representations of the all-different predicate, 2001

61 Alldiff Permutation polytope for alldiff(x 1,x 2,x 3 ) with domain {1,3,6} (1,3,6) (3,1,6) (6,1,3) (1,6,3) (3,6,1) (6,3,1) Slide 61 Dimension = 2

62 Alldiff Permutation polytope for alldiff(x 1,x 2,x 3 ) with domain {1,3,6} (1,3,6) (3,1,6) (6,1,3) Affine hull x 1 + x 2 + x 3 = 10 (1,6,3) (3,6,1) (6,3,1) Slide 62 Dimension = 2

63 Alldiff Unfortunately, convex hull relaxation of alldiff is weak. (1,3,6) (3,1,6) (6,1,3) (1,6,3) (3,6,1) (6,3,1) Slide 63 Dimension = 2

64 Circuit The circuit constraint has a tighter convex hull relaxation. The polytope can be almost completely characterized. Slide 64

65 Circuit The circuit constraint has a tighter convex hull relaxation. The polytope can be almost completely characterized. ( ) For example, circuit x, x, x with domains {1,3,6} has solutions Vertex after vertex 1 (x 1,x 2,x 3 ) = (3,6,1) 1 (x 1,x 2,x 3 ) = (6,1,3) Slide 65

66 Circuit The circuit constraint has a tighter convex hull relaxation. The polytope can be almost completely characterized. Other permutations are not circuits, for example (x 1,x 2,x 3 ) = (1,3,6) 1 (x 1,x 2,x 3 ) = (6,1,3) Slide 66

67 Circuit The circuit constraint has a tighter convex hull relaxation. The polytope can be almost completely characterized. Traveling salesman problem using alldiff min i c x x i + 1 ( x x x ) alldiff,, i using circuit min i c ix i ( x x x ) circuit,, nd city in tour City after City 3 in tour Slide 67

68 Circuit The circuit constraint has a tighter convex hull relaxation. The polytope can be almost completely characterized. Traveling salesman problem using alldiff min i c x x i + 1 ( x x x ) alldiff,, i using circuit min i c ix i ( x x x ) circuit,, Distance from ith to (i+1)th city Distance from City i to next city Slide 68

69 Circuit The circuit constraint has a tighter convex hull relaxation. The polytope can be almost completely characterized. Traveling salesman problem Slide 69 using alldiff min i c x x i + 1 ( x x x ) alldiff,, i Filtering is easy but relaxation is weak using circuit min i c ix i ( x x x ) circuit,, Filtering is hard but relaxation is tighter L. Genc-Kaya and JNH, The circuit polytope, 2008

70 Circuit (1,3,6) (3,1,6) (6,1,3) (1,6,3) (3,6,1) (6,3,1) Polytope of circuit(x 1,x 3,x 6 ) Dimension = 1 Slide 70

71 Permutation polytope for domain {1,2,3,4} Projected onto 3-space Dimension = 3 Slide 71

72 Solutions of circuit(x 1,x 2,x 3,x 4 ) Dimension = 3 Slide 72

73 Circuit To identify facet-defining inequalities containing variables x, x j1 j k Solve the combinatorial problem of identifying undominated solution values of x, x j1 j k Solve the numerical problem of checking which hyperplanes defined by undominated solutions correspond to valid inequalities. Slide 73

74 Circuit To identify facet-defining inequalities containing variables x, x j1 j k Solve the combinatorial problem of identifying undominated solution values of x, x j1 j k Solve the numerical problem of checking which hyperplanes defined by undominated solutions correspond to valid inequalities. To simplify presentation, assume we seek facet-defining inequalities with positive coefficients. Any sign pattern can be accommodated. Slide 74

75 Circuit Generate facets of containing variables circuit( x, x ) v v n x, x, x 0 1 v v v Slide 75

76 Circuit There are exactly 4 undominated solutions. v0 v1 v2 v3 v 4 ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v v0 v1 v2 v3 v 4 ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v v0 v1 v2 v3 v 4 ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v v0 v1 v2 v3 v 4 ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v Slide 76

77 Circuit Each set of 3 solutions determines a hyperplane and potential facet. ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v ( x, x, x, x, x ) = ( v, v, v, v, v ) v 0 v1 v2 v3 v The valid inequalities that result are facet-defining. Check each inequality to see if it is violated by the other solution. Slide 77

78 Circuit Solutions (x 0,x 2,x 3 ) for domain {0,2,3, } x 3 (2,0,3) (3,0,1) x 1 x 2 (3,2,0) (1,3,0) Slide 78

79 Circuit Solutions (x 0,x 2,x 3 ) for domain {0,1,2, } x 3 (2,0,3) (3,0,1) x 1 x 2 (1,3,0) (3,2,0) Not a facet. Violated by (1,3,0) Slide 79

80 Circuit Solutions (x 0,x 2,x 3 ) for domain {0,1,2, } x 3 (2,0,3) (3,0,1) x 1 x 2 (3,2,0) Slide 80 (1,3,0) Not a facet. Violated by (3,0,1)

81 Circuit Solutions (x 0,x 2,x 3 ) for domain {0,1,2, } x 3 (2,0,3) (3,0,1) x 1 x 2 (3,2,0) Slide 81 (1,3,0) These are facets.

82 Circuit Facets with positive coefficients and variables x, x, x v v v ( v v )( v v ) x + ( v + v + v v v v v v v v ) x + ( v v )( v v ) x v v v v ( v v ) v v + v + v v v v v v For domain {0,1,2, }: 6x0 + 5x1 + 3x ( v v )( v v ) x + ( v v )( v v ) x + ( v v )( v v ) x v v v v ( v v ) v v + v ( v v ) + v v v For domain {0,1,2, }: x0 + 2x1 + 4x2 7 Slide 82

83 Circuit There are also fast separation heuristics. Slide 83

84 Cumulative scheduling We can derive valid inequalities for ( s s p p c c C) cumulative (,, ),(,, ),(,, ), 1 n 1 n 1 n by energetic reasoning (but not the same reasoning used for domain reduction) Slide 84

85 Cumulative scheduling We can derive valid inequalities for ( s s p p c c C) cumulative (,, ),(,, ),(,, ), 1 n 1 n 1 n by energetic reasoning (but not the same reasoning used for domain reduction) Start times Resource consumption rates Processing times Slide 85

86 Cumulative scheduling We can derive valid inequalities for ( s s p p c c C) cumulative (,, ),(,, ),(,, ), 1 n 1 n 1 n by energetic reasoning (but not the same reasoning used for domain reduction) Start times Resource consumption rates Processing times Energy of job j = p j c j Slide 86

87 Cumulative scheduling Take any subset of jobs J = { j 1,, j k }. Index jobs so that p c p c j j j j 1 1 k k Theorem. The following inequalities (for example) are valid: k k k 1 1 kr + ( k i + 1) p c p s kd ( k i + 1) p c J j j j j J j j C i = 1 i = 1 j J C i = 1 i i i i i Earliest release time in J Latest deadline in J Slide 87 JNH, Integrated Methods for Optimization, 2007

88 Cumulative scheduling Example C job 2 job 3 r 1 job 1 r 2 d 2 d 1 r 3 d 3 t Slide Valid inequalities: s + s + s s + s

89 Slide 89 Cappadocia, Türkiye

90 Slide 90 Subterranean city, Derinkuyu, Türkiye

91 Relaxation Duality Relaxation Dual Lagrangean Dual Example: Fast LP Example: TSP with Time Windows Slide 91

92 Relaxation dual We often search over problem restrictions to find a better solutions. Branching methods Local search Can we search over relaxations to find a better bound? Slide 92

93 Relaxation dual We often search over problem restrictions to find a better solutions. Branching methods Local search Can we search over relaxations to find a better bound? Yes, if we parameterize the relaxations. Then search over parameter values. Slide 93

94 Relaxation dual A relaxation dual seeks a relaxation that provides the tightest bound. The relaxation parameters are dual variables. Solve the dual by searching over values of the dual variables Slide 94

95 Relaxation dual A relaxation dual seeks a relaxation that provides the tightest bound. The relaxation parameters are dual variables. Solve the dual by searching over values of the dual variables Some relaxation duals: Linear programming dual Lagrangean dual Surrogate dual Superadditive dual Roof dual, etc. etc. Slide 95

96 Relaxation dual Given the optimization problem x D { f x C } min ( ) Slide 96

97 Relaxation dual Given the optimization problem x D { f x C } min ( ) Easy constraints Hard constraints Slide 97

98 Relaxation dual Given the optimization problem x D { f x C } min ( ) Easy constraints Hard constraints A parameterized relaxation is { f x λ C λ } min (, ) ( ) x D Lower bound on f(x) Relaxation of C (easier constraints) Slide 98

99 Relaxation dual Given the optimization problem x D { f x C } min ( ) Easy constraints Hard constraints A parameterized relaxation is { f x λ C λ } min (, ) ( ) x D Slide 99 Lower bound on f(x) Relaxation of C vector of dual variables

100 Relaxation dual Given the optimization problem x D { f x C } min ( ) Easy constraints Hard constraints The relaxation dual is { { f x λ C λ }} max min (, ) ( ) λ x D Find relaxation that provides the largest lower bound. Slide 100

101 Relaxation dual Weak duality always holds: Max value of dual problem Min value of original problem Difference = duality gap Slide 101

102 Lagrangean dual Used when the hard constraints are inequality constraints. { f x g x } min ( ) ( ) 0 x D Slide 102

103 Lagrangean dual Used when the hard constraints are inequality constraints. { f x g x } min ( ) ( ) 0 x D The relaxation is parameterized by a vector of Langrange multipliers u. { λ T } min f ( x ) g ( x ) x D Lower bound on f(x) Hard constraints disappear completely Slide 103

104 Lagrangean dual Used when the hard constraints are inequality constraints. { f x g x } min ( ) ( ) 0 x D The relaxation is parameterized by a vector of Langrange multipliers u. { λ T } min f ( x ) g ( x ) x D Lower bound on f(x) Hard constraints disappear completely Dualized constraints Slide 104

105 Lagrangean dual Used when the hard constraints are inequality constraints. { f x g x } min ( ) ( ) 0 x D The Lagrangean dual is u 0 { { }} T f x λ g x max min ( ) ( ) x D Slide 105

106 Lagrangean dual Can use subgradient optimization to solve the dual. Exploit the fact that { f x g x } θ( λ) = min ( ) ( ) 0 x D Is always concave (but not differentiable). Slide 106

107 Lagrangean dual Can use subgradient optimization to solve the dual. Exploit the fact that { f x g x } θ( λ) = min ( ) ( ) 0 x D Is always concave (but not differentiable). A subgradient (direction of steepest ascent) of θ (λ) is g(x*), where x* solves min f ( x) g( x) 0 x D { } Slide 107

108 Lagrangean dual Can use subgradient optimization to solve the dual. Exploit the fact that { f x g x } θ( λ) = min ( ) ( ) 0 x D Is always concave (but not differentiable). A subgradient (direction of steepest ascent) of θ (λ) is g(x*), where x* solves min f ( x) g( x) 0 x D { } Step k of search is ( *) λ k + 1 λ k = + α g x k Slide 108

109 Lagrangean dual Can use subgradient optimization to solve the dual. Exploit the fact that { f x g x } θ( λ) = min ( ) ( ) 0 x D is always concave (but not differentiable). A subgradient (direction of steepest ascent) of θ (λ) is g(x*), where x* solves min f ( x) g( x) 0 x D { } Step k of search is ( *) λ k + 1 λ k = + α g x k Slide 109 stepsize, perhaps α k = 1/k

110 Example min 4x + 5x + 6x subject to 3x + 4x + 5x x + 2x + 3x 24 x x x x integer } easy } hard Slide 110

111 Example min 4x + 5x + 6x subject to 3x + 4x + 5x x + 2x + 3x 24 x x x x integer } easy } hard Slide 111 4x1 + 5x2 + 6x3 θ( λ) = min + λ1(47 3x1 4x2 5 x3 ) x1 2 x2 3 + λ2(24 x1 2x2 3 x3 ) x = min x1 2 + (6 5λ1 3 λ2 ) x3 + 47λ1 + 24λ2 x 3 x (4 3 λ λ ) x + (5 4λ 2 λ ) x Solve by inspection for fixed λ

112 Subgradient search in λ-space: λ Start λ 1 Slide 112 Value of Lagrangean dual = 57.6 < 58 = optimal value

113 Lagrangean dual The Lagrangean dual of a linear programming problem is the classical linear programming dual. Slide 113

114 Lagrangean dual The Lagrangean dual of a linear programming problem is the classical linear programming dual. Optimization duals can also be conceived as inference duals. - As in Benders decomposition, sensitivity analysis, etc. Slide 114

115 Example: Fast Linear Programming In CP, it is best to process each node of the search tree very rapidly. Lagrangean relaxation allows fast calculation of a lower bound on the LP value at each node. Slide 115

116 Example: Fast Linear Programming In CP, it is best to process each node of the search tree very rapidly. Lagrangean relaxation allows fast calculation of a lower bound on the LP value at each node. Solve the Lagrangean dual at the root node (which is an LP) and use the same Lagrange multipliers to get an LP bound at other nodes. Slide 116

117 Example: Fast LP At root node, solve min cx Dualize Special structure, e.g. variable bounds Ax Dx b d x 0 ( λ) The (partial) LP dual solution λ* solves the Lagrangean dual in which x 0 { cx λ Ax b } θ( λ) = min ( ) Dx d Slide 117

118 Example: Fast LP At root node, solve Dualize Special structure, e.g. variable bounds min cx Ax b ( u) Dx d x 0 Slide 118 The (partial) LP dual solution u* solves the Lagrangean dual in which x 0 { T } θ( u) = min cx u ( Ax b) Dx d At another node, the LP is Lower bound θ (λ*) is quickly calculated by solving a specially structured LP. min Ax b ( u) Dx Hx x 0 cx d h Branching constraints, etc.

119 Example: TSP with Time Windows A salesman must make several stops and return home, subject to time windows at each stop. Objective: Minimize travel time. Use Lagrange multipliers for cost-based filtering. Travel time c ij Stop i Stop j Slide 119

120 TSP with time windows Assignment relaxation min ij ij ij j j x ij x = x = 1, all i { } c x ij ji 0,1, all i, j = 1 if stop i immediately precedes stop j Stop i is preceded and followed by exactly one stop. Slide 120

121 TSP with time windows Assignment relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij Because this problem is totally unimodular, it can be solved as an LP. Slide 121

122 TSP with time windows Assignment relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij Because this problem is totally unimodular, it can be solved as an LP. The relaxation provides a very weak lower bound on the optimal value. Slide 122

123 TSP with time windows Assignment relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij Because this problem is totally unimodular, it can be solved as an LP. The relaxation provides a very weak lower bound on the optimal value. But cost-based filtering can be effective. Slide 123

124 TSP with time windows Assignment relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij We know x ij U v λ ij * Slide 124

125 TSP with time windows Assignment relaxation Known upper bound on shortest distance Value of LP relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij We know x ij U v λ ij * Slide 125

126 TSP with time windows Assignment relaxation Known upper bound on shortest distance Value of LP relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij We know x ij U v λ Lagrange multiplier ij * Slide 126

127 TSP with time windows Assignment relaxation Known upper bound on shortest distance Value of LP relaxation min ij ij ij j j ij x = x = 1, all i 0 x 1, all i, j ( λ ) ij c x ji ij We know x ij U v λ Lagrange multiplier ij * If this upper bound is < 1, we can fix x ij to 0. Known in OR as reduced-cost variable fixing Slide 127 Focacci, Lodi & Milano, Solving TSP with time windows with constraints, 1999

128 Slide 128 Éméi Shān, Sichuan Province, China

129 Slide 129 Tài jí quán

130 Discrete Relaxation: Dependency Graph Dependency Graph and Induced Width Example: Relaxing the Constraint Dual Slide 130

131 Dependency Graph and Induced Width Complexity of solving a problem is at worst exponential in the induced width of the dependency graph Slide 131

132 Dependency Graph and Induced Width Complexity of solving a problem is at worst exponential in the induced width of the dependency graph The idea has surfaced independently in several contexts. Nonserial dynamic programming (1972) Belief logics (1986) k-trees (1986) Bayesian networks (1988) Pseudo-boolean optimization (1990) Location analysis (1994) Bucket elimination (1996) Slide 132

133 Dependency graph and induced width Let s relax the problem by thinning out the dependency graph. Reduce the induced width while maintaining a valid relaxation. Use relaxation duality to find the best relaxation. Slide 133

134 Example: Relaxing the Constraint Dual min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} Solution: (x 1,,x 4 ) = (0,1,0,0) Optimal value is 2. Slide 134

135 Example: Relaxing the constraint dual Dependency graph has induced width 3: x 1 x 2 min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} x 4 x 3 Slide 135

136 Example: Relaxing the constraint dual Relax the problem by removing edges from dependency graph. and form mini-buckets min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} Slide 136 min f ( x, x, x ) + f ( x, x, x ) + f ( x, x, x ) f ( x, x, x ) f ( x, x, x ) f ( x, x, x ) x + 2x + 3 x if x + x + x 1 otherwise 0x + 0x + 4 x if x + x + x 1 otherwise 0x + 0x + 0 x if x + x + x 1 otherwise = = = Dechter & Rish, Mini-buckets: A general scheme for approximating inference, 1998

137 Example: Relaxing the constraint dual The problem separates into 3 problems with induced width 2: x x 1 2 x 1 x2 x 2 x 3 x x4 x 4 3 Slide 137 min f ( x, x, x ) + f ( x, x, x ) + f ( x, x, x ) f ( x, x, x ) f ( x, x, x ) f ( x, x, x ) x + 2x + 3 x if x + x + x 1 otherwise 0x + 0x + 4 x if x + x + x 1 otherwise 0x + 0x + 0 x if x + x + x 1 otherwise = = =

138 Example: Relaxing the constraint dual But the resulting bound is weak x x 1 2 x 1 x2 x 2 x 3 x x4 x 4 3 Slide 138 min f ( x, x, x ) + f ( x, x, x ) + f ( x, x, x ) = = 1< 2 x1 + 2x2 + 3 x3 if x1 + x2 + x3 1 f1( x1, x2, x3 ) = otherwise 0x1 + 0x2 + 4 x4 if x1 + x2 + x4 1 f2( x1, x2, x4 ) = otherwise 0x2 + 0x3 + 0 x4 if x2 + x3 + x4 1 f3( x2, x3, x4 ) = otherwise

139 Example: Relaxing the constraint dual Let s apply relaxation duality to the constraint dual: x x = 2 = = x = x x x x x x = ( x, x, x ) D ( x, x, x ) D ( x, x, x ) D min x + 2x + 3x + 4x x + x + x x + x + x 1 x j x + x + x 1 {0,1} Standardize apart variables in different constraints and equate them in binary constraints. Slide 139 where D = 3 {0,1} {(0,0,0)}

140 Example: Relaxing the constraint dual Dependency graph for constraint dual: 1 x 1 1 x 2 1 x 3 min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} x 1 2 x 2 2 x 4 Induced width of dependency graph is 3. 3 x 2 3 x 3 3 x 4 Any deletion of vertical edges yields a valid relaxation. Slide 140

141 Example: Relaxing the constraint dual So let s delete vertical edges, which equate variables. 1 x 1 1 x 2 1 x 3 min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} x 1 2 x 2 2 x 4 The previous mini-bucket scheme deletes all vertical edges. 3 x 2 3 x 3 3 x 4 Slide 141

142 Example: Relaxing the constraint dual Let s search other deletion patterns to find a tighter relaxation (i.e., solve relaxation dual) 1 x 1 1 x 2 1 x 3 min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} x 1 2 x 2 2 x 4 By deleting only one edge we can reduce induced width to 2 3 x 2 3 x 3 3 x 4 Slide 142

143 Example: Relaxing the constraint dual Let s search other deletion patterns to find a tighter relaxation (i.e., solve relaxation dual). 1 x 1 1 x 2 1 x 3 min x + 2x + 3x + 4x x1 + x2 + x3 1 x1 + x2 + x4 1 x + x + x 1 x j {0,1} x 1 2 x 2 2 x 4 By deleting only one edge we can reduce induced width to 2. 3 x 2 3 x 3 3 x 4 Bound is now tight. Slide 143 JNH, Duality in optimization and constraint satisfaction, 2006

144 Slide 144 Great Smoky Mountains, USA

145 Slide million years old

146 Discrete Relaxation: MDDs Domain Store Relaxation Multivalued Decision Diagrams MDD Relaxation Propagation in Relaxed MDDs Slide 146

147 Domain Store Relaxation A domain store can be viewed as a (weak) relaxation. We propagate constraints in the domain store by using them to refine it. Slide 147

148 Domain Store Relaxation A domain store can be viewed as a (weak) relaxation. We propagate constraints in the domain store by using them to refine it. Question: Can we propagate in a tighter relaxation? Slide 148

149 Domain Store Relaxation A domain store can be viewed as a (weak) relaxation. We propagate constraints in the domain store by using them to refine it. Question: Can we propagate in a tighter relaxation? Proposal: Use a relaxed multivalued decision diagram. This can reduce branching by propagating more information. Slide 149

150 Domain Store Relaxation A domain store can be viewed as a (weak) relaxation. We propagate constraints in the domain store by using them to refine it. Question: Can we propagate in a tighter relaxation? Proposal: Use a relaxed multivalued decision diagram. This can reduce branching by propagating more information. Can also justify more thorough processing of each constraint. Slide 150

151 Multivalued Decision Diagrams A reduced ordered MDD is the result of superimposing all isomorphic subtrees in an enumeration tree and removing redundant edges. Example: What is the reduced ordered MDD for 1 3 x1 x2 x = 3 3 x1 + x2 5 x j x + x = { 1,2,3 } 4 (,, ) (1,1,2) Slide 151

152 x + x = x1 x2 x = 3 3 x1 + x2 5 Multivalued Decision Diagrams (,, ) (1,1,2) x 1 {1} u 1 {2} {3} Edge domain x 2 u 2 u 3 {1} {2,3} {1,2} x 3 u 4 u 5 u 6 {2} {3} {1} Slide 152 1

153 Multivalued Decision Diagrams Each path represents a cartesian product of solutions x 1 {1} u 1 {2} {3} {1} {2,3} {3} x 2 u 2 u 3 {1} {2,3} {1,2} x 3 u 4 u 5 u 6 {2} {3} {1} Slide 153 1

154 Multivalued Decision Diagrams Each path represents a cartesian product of solutions x 1 {1} u 1 {2} {3} {2} {1,2,3} {2} x 2 u 2 u 3 {2,3} {1} {1,2} x 3 u 4 u 5 u 6 {2} {3} {1} Slide 154 1

155 MDD Relaxation Let s use relaxed MDD with max width of 2 Full MDD: 8 solutions Relaxed MDD: 14 solutions x 1 x 2 {1} u 1 {2} {3} u 2 u 3 {2,3} {1} {1,2} x 3 u 4 u 5 {3} {1,2} Slide 155 1

156 MDD Relaxation MDD relaxation with width 1 is just the domain store. Full MDD: 8 solutions Relaxed MDD: 14 solutions Domain store: = 27 solutions x 1 x 2 x 3 u 1 u 2 u 5 {1,2,3} {1,2,3} {1,2,3} Slide 156 1

157 Branching search using relaxed MDD x 2 {1,2,3} x 1 {1,2,3} {1} x 1 x 2 {1} u 1 {2} {3} u 2 u 3 {2,3} {1} {1,2} x 3 u 4 u 5 {3} {1,2} Slide 157 1

158 Branching search using relaxed MDD x 2 {1,2,3} {1} x 3 {1,2} x 1 {1,2,3} {1} x 1 x 2 {1} u 1 {2} {3} u 2 u 3 {2,3} {1} {1,2} x 3 u 4 u 5 {3} {1,2} Slide 158 1

159 Branching search using relaxed MDD x 2 {1,2,3} x 1 {1,2,3} {1} {1} {2,3} x 3 {1,2} x 3 {3} x 1 u 1 u 2 u 3 {2,3} {1} {1,2} x 2 {1} {2} {3} x 3 u 4 u 5 {3} {1,2} Slide 159 1

160 Branching search using relaxed MDD x 1 {1,2,3} {1} {2,3} x 2 {1,2,3} x 2 {1,2,3} {1} {2,3} x 3 {1,2} x 3 {3} x 1 x 2 {1} u 1 {2} {3} u 2 u 3 {2,3} {1} {1,2} Slide 160 And so forth. Less branching than with domain store. x 3 u 4 u 5 {1,2} {3} 1

161 Propagation in MDDs We can propagate a constraint in an MDD relaxation by edge domain filtering and refinement (node splitting). We take care not to exceed max width. Example: We will propagate alldiff in an MDD relaxation of width 3. Slide 161 Andersen, Hadzic, Hooker & Tiedemann, A constraint store based on multivalued decision diagrams, 2007

162 Propagation in MDDs Current MDD relaxation {1} {2} u 1 {3} u 2 u 3 {1,2} {1} u 4 {1,2,3} u 5 {1,2} {3} Slide 162 u 6 u 6

163 Propagation in MDDs First filter edge domains using alldiff {1} {2} u 1 {3} u 2 u 3 {1,2} {1} u 4 {1,2,3} u 5 {1,2} {3} Slide 163 u 6 u 6

164 Propagation in MDDs First filter edge domains using alldiff {1} {2} u 1 {3} u 2 u 3 {2} {1} u 4 {1,2} u 5 {1,2} {3} Slide 164 u 6 u 6

165 Propagation in MDDs To split u 5 : Identify equivalence classes of incoming edges {1} {2} u 1 {3} u 2 u 3 {2} {1} u 4 {1,2} u 5 {1,2} {3} Slide 165 u 6 u 6

166 Propagation in MDDs To split u 5 : Identify equivalence classes of incoming edges {1} {2} u 1 {3} These are equivalent for alldiff because they lead to the same set of feasible paths. u 2 u 3 {1} {2} {1} u 5 {1,2} {3} u 4 {2} Slide 166 u 6 u 6

167 Propagation in MDDs To split u 5 : Identify equivalence classes of incoming edges. {1} {2} u 1 {3} Split u 5 to receive 3 equivalence classes. u 2 u 3 {1} {2} {1} u 4 {2} u 5 u 5 u 5 {1,2} {3} Slide 167 u 6 u 6

168 Propagation in MDDs Duplicate outgoing edges. {1} {2} u 1 {3} u 2 u 3 {1} {2} {1} u 4 {2} u 5 u 5 u 5 {1,2} {3} Slide 168 u 6 u 6

169 Propagation in MDDs Duplicate outgoing edges. {1} {2} u 1 {3} u 2 u 3 {1} {2} {1} u 4 {2} Slide 169 u 5 u 5 u 5 {1,2} {3} {1,2} {3} {3} u 6 {1,2} u 6

170 Propagation in MDDs Filter domains. {1} {2} u 1 {3} u 2 u 3 {1} {2} {1} u 4 {2} Slide 170 u 5 u 5 u 5 {1,2} {3} {1,2} {3} {3} u 6 {1,2} u 6

171 Propagation in MDDs Filter domains. {1} {2} u 1 {3} u 2 u 3 {1} {2} {1} u 4 {2} u 5 u 5 u 5 {2} {3} Slide 171 u 6 {1} u 6

172 Propagation in MDDs Alldiff has now been propagated. {1} {2} u 1 {3} u 2 u 3 {1} {2} {1} u 4 {2} Hadzic, Hooker, O Sullivan & Tiedemann, Approximate compilation of constraints into multivalued decision diagrams, 2008 Slide 172 u 5 u 5 u 5 {2} {3} u 6 {1} u 6

173 Slide 173 That s it. Have a relaxing day!