Global Constraints for the Mean Aboslute Deviation and the Variance: Application to the Vertical Line Balancing

Transcription

1 1/51 Global Constraints for the Mean Aboslute Deviation and the Variance: Application to the Vertical Line Balancing Pierre Schaus and Yves Deville, Jean-Charles Régin also collaborated Belgian Constraints UCLouvain January 16, 2008

2 2/51 Overview 1 Line Balancing Problems 2 Introduction to CP 3 Global Constraints for Variance and Mean Absolute Deviation 4 Conclusion

3 3/51 Assembly Line of the Ford-T

4 An Assembly Line is... A set of stations or working stations (WS) achieving part of assembly operations of a product. linked such that the product transit is is possible, (partial product of complete product). Taken from slides Lignes d assemblages, Conception, Organisation et gestion (Thomas L Eglise and Benoit Raucent). 4/51

5 5/51 Precedence graph (PG): DAG Taken from slides Lignes d assemblages, Conception, Organisation et gestion (Thomas L Eglise and Benoit Raucent).

6 6/51 Precedence Graph cont. Each node is a task and each task comes with a duration to achieve it.

7 7/51 Assembly Line at work Each time cycle, one product leaves the line. T c is the time between the ending of two product at the end of the line.

8 8/51 Our objective: Assign each task to a working station such that all precedences are satisfied

9 9/51 SALBP-2: Maximize The Cadence of the line The Cyle Time is fixed by the slowest WS: max(x i ) where X i is the sum of processing time of tasks in WS i. Optimization problem: Find a schedule minimizing max(x i ).

10 10/51 Vertical Line Balacing: Equalize the duration of each working stations. s = n i=1 X i is the sum of the durations of all the tasks. n is the number of working stations. Minimization of: n i=1 X i s/n n i=1 (X i s/n) 2 max{x 1...X n } max{x 1...X n } min{x 1...X n } max{ X i s/n } {X i X i s/n}...

11 11/51 Vertical Line Balacing: Equalize the duration of each working stations. In this presentation we study the 3 first: n i=1 X i s/n n i=1 (X i s/n) 2 max{x 1...X n } A 90-year-old debate:the advantages of the mean deviation by STEPHEN GORARD, University of York British Journal of Educational Studies, December 2005

12 12/51 Study of the Three Criteria on Hahn instance from Scholl Benchmark 53 Tasks, min duration of 40, max duration of 1775.

13 13/51 No criterion subsumes the others... Minimization of: Measure of: max i {X i } i (X i s/n) 2 /n i X i s/n /n max i {X i } 2336 (+3%)2400 (+4%)2418 i(x i s/n) 2 /n (+53%) (+9%)738 i X i s/n /n (+34%)298 (+0%) but i (X i s/n) 2 /n approximates very well the other two.

14 14/51 Consider now that each tasks can have a stochastic time following a gaussian distribution N d i, p d i t i d i

15 15/51 The height of each bin also follows a gaussian distribution N d tot, p d tot t 3 d 3 d tot t 2 d 2 t 1 d 1

16 The height of each bin also follows a gaussian distribution X 1 N 1 X 2 N 2 X 3 N 3 X 4 N 4 X 5 N 5 X 6 N 6 X 7 N 7 The cycle time depends on the distribution of Y = max{x 1, X 2,...} 16/51

17 17/51 Finding the distribution of Y = max{x 1, X 2,...}. Cumulative distribution: F Y (y) = P(Y y) = P(X 1 y) P(X 2 y)... P(X n y) The probability density function f Y (y) = d dy F Y (y)

18 Comparison of the 3 solutions with i : µ = d i, σ i = 0.3 d i f Y(y) max{x i } i 2e 04 4e 04 6e 04 8e 04 n (Xi s n) 2 n 1 n Xi s n n Expectation of the maximum cycle time is smaller for the solution minimizing the variance. 18/51

19 19/51 In Conclusion for the 3 criteria n i=1 (X i s/n) 2 approximates the best the other two criteria. The minimization of n i=1 (X i s/n) 2 or n i=1 X i s/n gives more robust solutions when the distribution of the tasks is stochastic (often the case when achieved by humans).

20 20/51 Introduction to Constraint Programming 1 Line Balancing Problems 2 Introduction to CP 3 Global Constraints for Variance and Mean Absolute Deviation 4 Conclusion

21 21/51 Some definitions of Constraint Programming A Constraint Satisfaction Problem (CSP) consist of: a set of variables X = {X 1,..., X n }, for each variable X i, a finite set Dom(X i ) of possible values (its domain), and a set of constraints restricting the values that the variables can simultaneously take. A solution to a CSP is an assignment of a value from its domain to every variable, in such a way that every constraint is satisfied. We may want to find: just one solution, with no preference as to which one, all solutions, an optimal, or at least a good solution, given some objective function defined in terms of some or all of the variables.

22 22/51 Constraint Each constraint is responsible to remove as much inconsistent values as possible of the domains. Example X 1 {1, 2, 5, 6}, X 2 {0, 2} a constraint C := X 1 X 2. C can remove {1, 2, 5/, 6/} {0/, 2, 4, 7} We say that a constraint filters the domains of the variables. The algorithm responsible to filter the domains for a constraint is called a propagator.

23 23/51 Searching for a solution: branching and propagation Dom(X1)={1,3} Dom(X2)={2,3,5} Dom(X3)={2,3,4} X1>X2 X2 X3 X2 X3 X1 X2 X1 X3

24 24/51 Searching for a solution: branching and propagation Dom(X1)={1,3} Dom(X2)={2,3,5} Dom(X3)={2,3,4} X1>X2 X2 X3 X2 X3 X1 X2 X1 X3

25 25/51 Searching for a solution: branching and propagation Dom(X1)=3 Dom(X2)={2,3,5} Dom(X3)={2,3,4} X1>X2 X2 X3 X2 X3 X1 X2 X1 X3

26 26/51 Searching for a solution: branching and propagation X2=2 Dom(X1)=3 Dom(X2)={2,5} Dom(X3)={2,4} X1>X2 X2 X3 X2 X3 X1 X2 X1 X3 Dom(X1)=3 Dom(X2)=2 Dom(X3)={2,4}

27 27/51 Searching for a solution: branching and propagation X2=2 Dom(X1)=3 Dom(X2)={2,5} Dom(X3)={2,4} X1>X2 X2 X3 X2 X3 X1 X2 X1 X3 Dom(X1)=3 Dom(X2)=2 Dom(X3)={2,4}

28 28/51 Searching for a solution: branching and propagation X2=2 Dom(X1)=3 Dom(X2)={2,5} Dom(X3)={2,4} X2 2 X1>X2 X2 X3 X2 X3 X1 X2 X1 X3

29 29/51 Searching for a solution: branching and propagation X2=2 Dom(X1)=3 Dom(X2)={2,5} Dom(X3)={2,4} X2 2 Dom(X1)=3 Dom(X2)=5 Dom(X3)={2,4} X1>X2 X2 X3 X2 X3 X1 X2 X1 X3

30 30/51 Different levels of consistency reached by a propagator C Dom(Y) C C Dom(X)

31 Different levels of consistency reached by a propagator: Arc-Consistency and Bound-Consistency C C Dom(Y) C C C C Dom(X) Dom(X) For some constraints, arc consistency is too costly e.g:x X n = S For others it is possible very efficiently e.g.:alldifferent(x 1,..., X n ) 31/51

32 32/51 Global Constraints and decomposition Some constraints can be decomposed into simpler constraints. e.g. AllDiff (X 1, X 2, X 3 ) is equivalent to X 1 X 2, X 1 X 3, X 2 X 3. But it is often better to consider the constraint globally to filter more values. Example Dom(X 1 ) = {1, 2}, Dom(X 2 ) = {1, 2}, Dom(X 3 ) = {1, 2, 3}. X 1 X 2, X 1 X 3, X 2 X 3 are separately arc consistent (nothing to filter). AllDiff (X 1, X 2, X 3 ) is able to filter Dom(X 3 ) = {1/, 2/, 3}

33 33/51 Modeling Line Balancing in CP Variables: ws j is the workstation of task j. b ij {0, 1}, b ij = 1 if task j scheduled at WS i. b ij = 0 otherwise. l i is the load of WS i Constraints: Reification: b ij = 1 ws j = i l i = i b ij d j where d j is the duration of task j. j b ij = 1 (each task is scheduled only once). ws j ws k if (j, k) is an edge of PG. Our three possible Objectives: min max i {l i }. min (l i ( l i )/n) 2 ). min l i ( l i )/n.

34 34/51 Improved model An easy but important redundant constraint: l i = d j i j Redundant global constraints for the bin packing: IloPack A constraint for bin packing, Paul Shaw (CP2004). IloPack+Precedences

35 35/51 Two global constraints: spread and deviation to realize our objective functions Given X = (X 1,..., X n ), one integer value s, one finite domain variable (1), (2) deviation(x, s, (1) ) holds if and only if n X i = s and (1) i=1 n n X i s. i=1 spread(x, s, (2) ) holds if and only if n n X i = s and (2) n ( Xi 2 ) s 2. i=1 i=1

36 36/51 Two global constraints: spread and deviation deviation(x, s, (1) ) holds if and only if n X i = s and (1) i=1 n n X i s. i=1 spread(x, s, (2) ) holds if and only if n n X i = s and (2) n ( Xi 2 ) s 2. i=1 It means that spread and deviation can be expressed with two simple sum constraints (decomposition). i=1 Is it really necessary to have a global constraint? Will the filtering be stronger than with the decomposition?

37 37/51 Implementation of deviation by decomposition Dom(X 1 ) = Dom(X 2 ) = [0, 10] Dom( (1) ) = [0, 8] deviation ({X 1, X 2 }, s = 10, (1) ) 10 = X 1 + X 2 and 2 X X 2 10 (1)

38 38/51 Bound-consistency on 10 = X 1 + X 2 The constraint is already bound-consistent. Nothing to prune. 10 Dom X 2 5 X 1 X 2 =10 5 Dom X 1 10

39 39/51 BC on 2 X X 2 10 (1) Dom( (1) ) = [0, 8] 10 9 Dom X 2 5 X 1 5 X Dom X 1 10

40 40/51 Now, deviation is bound-consistent 10 9 X 1 5 X X 1 X 2 =10 7 Dom X Dom X 1

41 41/51 Efficient algorithms to achieve bound-consistency deviation(x, s, (1) ) Bound consistency achieved in O(n). spread(x, s, (2) ) Bound consistency achieved in O(n log n) Time constraint, we only present the filtering of (2) for spread.

42 42/51 Filtering of Dom( (2) ) in spread(x, s, (2) ) n Q (2) = min{n (x[i] s/n) 2 x i=1 s.t. and n x[i] = s i=1 Xi min x[i] X max i i} Filtering is: Dom( (2) ) Dom( (2) ) [ Q (2), + ) Question: How to compute efficiently Q (2)?

43 43/51 Important theorem characterizing optimal solution n Q (2) = min{n (x[i] s/n) 2 x i=1 s.t. and n x[i] = s i=1 Xi min x[i] X max i Optimal solution is a ν-centered assignment [Régin et al. ] i}

44 44/51 Important theorem characterizing optimal solution Yes but the constraint n ν-centered assignment. i=1 x[i] = s must also be satisfied by the =? x[i]/ n s/n So we have to find ν such that the red line equals the green line.

45 45/51 Finding the ν-centered assignment of sum s 1 Consider a splitting of at most n 1 intervals based on the bounds of the domains. 2 For each interval, determine if there exists a ν-centered assignment of sum s with ν inside the interval. I 11 I 10 I 9 I 8 I 7 I 6 5 I 4 I 3 I 2 I 1

46 46/51 Is there a ν-centered assignment of sum s with ν I 7? ES I 7 =. I 11 I 10 I 9 I 8 I 7 I 6 5 I 4 I 3 I 2 I 1 ν must satisfy: 3 ν + ES(I 7 ) = s Answer yes if s ES(I 7) 3 I 7. Otherwise ν / I 7, try another interval.

47 47/51 Complexity to filter (2) in spread(x, s, (2) ) O(n) once intervals I are known and for each ES(I ) + the number of overlapping domains. Intervals I, ES(I ) and number of overlapping variables for each I can be computed in O(n) once the bounds of variables are sorted (not so trivial). O(n log n) to sort the bound. Hence overal complexity is dominated by O(n log n) to sort the bounds. Filtering of all X i X in O(n log n) (not explained here).

48 48/51 Complexity to filter (1) in deviation(x, s, (1) ) All domains can be filtered in O(n) (not explained here).

49 49/51 Implementation and results We have implemented both propagators in Ilog Solver and we give some results. Backtracks Time(s) #Tasks #stations Max Spread Devi Max Spread Devi Buxey Hahn Wee-Mag Lutz Mulkherje

50 50/51 Conclusion We have shown that the common criterion used in line balancing that is minimization of max{x 1...X n } was not the best to equalize the loads of stations. We have show that when the tasks durations are stochastic, minimizing n i=1 (X i s/n) 2 obtain solutions with an even smaller expected cycle time than solutions minimizing max{x 1...X n }. We have introduced two global constraints and their propagators for n i=1 (X i s/n) 2 and n i=1 X i s/n respectively spread and deviation. We have implemented the propagators in Ilog and tested it on real line balancing problems.

51 51/51 Thank you for your attention!