1. Jean-Charles REGIN Michel RUEHER regin@ilog.fr rueher@essi.fr ILOG Sophia Antipolis Université de Nice – Sophia Antipolis A global constraint combining a sum constraint and binary inequalities 2 August, 2000 1

2. Global constraint IS Combination of: A sum constraint: y=x i and binary Inequalities: x i  x j + c ij Example: x 1 + x 2 = y & x 1  x 2 + 1 2 August, 2000 2

3. Motivations to improve “back” propagation when solving Optimization Problems in the CP Framework 2 August, 2000 3

4. Optimization Problems Objective function : a sum z=x i CP Framework  solving a sequence of decision problems (B&B) where z < z* (each solution must be better than the previous one) Local consistency algorithms on each constraint poorly propagate (not very effective) Global constraint IS : to improve the back propagation when the bounds of z are modified 2 August, 2000 4

5. Minimizing mean flow time F=1/n  j=1…n (C j -r j ) Minimizing tardiness D=1/n  j=1…n D j Where D j =max(C j -d j,0) C j, r j, d j : completion time, ready time, and duration of task T j D ifference constraints : precedence/distance constraints between the tasks 2 August, 2000 5 Applications in deterministic scheduling problems

6. D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[1,13] C 1 : x 1 + x 2 =y C 2 : x 1  x 2 - 1 This system is arc consistent : each value belongs to a solution of every constraint If the lower bound of y is set to 6 this system remains arc consistent  no pruning can be achieved 2 August, 2000 6 Example 1

7. D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[6,13] C 1 : x 1 + x 2 =y C 2 : x 1  x 2 - 1 However, when x 2 belongs to [1,3] C 1 cannot be satisfied Global constraint IS will delete [1,3] from the domain D (x 2 ) 2 August, 2000 7 Example 1 (continued)

8. Constraint S um : y=  j=1…n x j Inequalities I neq : { x j - x j  c ji (i,j  [1,n])} Domain constraints D om : { l j  x j  u j (i  [1,n])} Global constraint IS = {S um }  D om  I neq Algorithm (scheme): If a bound of x is modified  Filtering D om  I neq by interval consistency If a bound of some y is modified  Filtering S um by interval consistency Updating the bounds of every x j with respect to IS 2 August, 2000 8 Summary of our framework

9.  Filtering S um by interval consistency : 0(n)  Filtering I neq by interval consistency : 0(mn)  Filtering IS by interval consistency : O(n(m + nlogn)) 2 August, 2000 9 Summary of our framework Contribution

10. Sum constraint y=x i is interval consistent iff: (1)min(y)   j=1…n min(x j ) (2)max(y)   j=1…n max(x j ) (3) x i : min(x i )  min(y) -  j i max(x j ) (4) x i : max(x i )  max(y) -  j i min(x j ) 2 August, 2000 10

11. Sum constraint (continued) Checking interval consistency of y=x i is in 0(n) (3)  x i : min(x i )  min(y) -  j i max(x j )  j i max(x j )=  j=1…n max(x j )- max(x i ) 2 August, 2000 11

12. Binary inequalities Filtering by AC  complexity depends on the size of the domains Filtering by interval consistency can be achieved in O(mn)  simple temporal CSP (Dechter et al) 2 August, 2000 12

13. Simple Temporal CSP Shortest path :  (s,x j )   (s,x i ) + c (x i, x j ) Distance Graph G=(N,E) N: source node s with D (s) ={0} + one node for each variable E:  x i  x j + c ji  arc (x j, x i ) with cost c ji  D (x) =[min x, max x ] s  x – min(x)  arc (x,s) with cost -min(x) x  s + max(x)  arc (s,x) with cost max(x) 2 August, 2000 13

14. Example D (x i ) =[1,6] D (x j ) =[2,5] x j  x i - 3 s i j 6 5 -2 -3 2 August, 2000 14 Distance graph (continued)

15. x i is interval consistant if D (x i ) =[-  (x i, s),  (s,x i ) ] Example D (x i ) =[1 5,6] D (x j ) =[2,5 3 ] x j  x i - 3 s i j -1 -5 6 5 3 -2 -3 2 August, 2000 15 Distance graph (continued) Computing shortest path : 0(nm) but after ONE computation we can use reduced costs  0(m+n log( n ))

16. IS constraint ( Sum & binary Inequalities ) (3) x i : min(x i )  min(y) -  ji max (x j ) is too weak to enforce interval consistency on IS Example: D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[6,13] C 1 : x 1 + x 2 =y C 2 : x 1  x 2 - 1 min(x 2 )  min(y) - max (x 1 ) = 0 although IS cannot be satisfied when x 1  3 2 August, 2000 16

17. IS constraint ( Sum & binary Inequalities ) Interval consistent iff: min(y)   i=1…n min(x i ) max(y)   i=1…n max(x i ) (3b) x i : min(x i )  min(y) -  ji max x i  min(x i ) (x j ) (4b) x i : max(x i )  max(y) -  ji min x i  max(x i ) (x j ) max x i  min(x i ) (x j ) : maximum value of D (xj) which satisfies I neq when x i is set to min(x i ) 2 August, 2000 17

18. IS constraint ( Sum & binary Inequalities ) (3b) x i : min(x i )  min(y) -  ji max x i  min(x i ) (x j ) Example: D (x 1 ) =[0,6] D (x 2 ) =[1,7] D(y)=[6,13] C 1 : x 1 + x 2 =y C 2 : x 1  x 2 - 1 min(x 2 )  min(y) - max x 2  min(x 2 ) (x 1 ) = 4 2 August, 2000 18

19. How to compute min(x i ) ? (3b) x i : min(x i )  min(y) -  ji max x i  min(x i ) (x j )  x i = min(y) -  j i max x i  x i (x j ) 2 August, 2000 19

20. Computing x i max (x j )=min(  (s,x j ), x i +  ’ (x i,x j ) ) (1) xi  xi where  ’ (x i,x j ) ) is the shortest path from x i to x j in G-{s} and thus x i = min(y) -  j i min( (s,x j ), x i +  ’ (x i,x j ) ) (2) 2 August, 2000 20 max (x j ) depends only on the upper-bounds of the variables x k that belong to a shortest path from s to x j

21. Computing min(x i ) (continued) x i = min(y) -  j i min( (s,x j ), x i +  ’ (x i,x j ) ) (2) 2 August, 2000 21 algorithm (1) Computing  ’ (x i,x j ) ) for all j i and be  (s,x j ) -  (x i,x j ) =  j (2) L  sorted sorted list of  j S  { x j :  j  0} S is the set of the x j ’s for which min( (s,x j ), x i +  ’ (x i,x j ) ) =  (s,x j ) Loop x i = min(y) -  j i min( (s,x j ), x i +  ’ (x i,x j ) ) S  S  { x j :  j  x i } until of S does no more change

22. Computing min(x i ) (continued) 2 August, 2000 22   (x i,x j ) can be computed on the graph of reduced costs of G-{s} in O(nlogn)  Identifying the shortest paths from x i to x j which go through s  iteration step : O(nlogn)  Filtering of IS by interval consistency : O(n(m + nlogn)) (no propagation step is required when min(x i ) is increased)

23. Extensions Works also for  i x i =y (3b) x i  1 (min(y) -  ji  j  (x i,x j ) ) |X| Works also if all the variables do not occur in the sum constraint 2 August, 2000 23

24. Conclusion  An original combination and an efficient algorithm for a new global constraint  to improve propagation in optimization problems  Further work : implementation & experimentation 2 August, 2000 24

25. Computing min(x i ) (continued) x i  1 (min(y) -  ji  (x i,x j ) ) |X| 2 August, 2000 25   (x i,x j ) can be computed on the graph of reduced costs of G-{s}  Identifying the shortest paths from x i to x j which go through s  iteration step : O(nlogn)  Filtering of IS by interval consistency : O(n(m + nlogn)) (no propagation step is required when min(x i ) is increased)

26. Computing x i max (x j )=min(  (s,x j ), x i +  (x i,x j ) ) (1) xi  xi and thus max (x j )  x i +  (x i,x j ) (2) xi  xi and x i  1 (min(y) -  ji  (x i,x j ) ) (3) |X| 2 August, 2000 26 max (x j ) depends only on the upper-bounds of the variables x k that belong to a shortest path from s to x j

27. Computing min(x i ) (continued) max (x j )= x i +  (x i,x j ) (2) xi  min(xi) 2 August, 2000 27 Proof (scheme) Two cases : 1. x i belongs to a shortest path from s to x j when x i is set to the value of x i we are searching for  max xi  min(xi) (x j )=x i +  (x i,x j ) 2. x i does not belong to a shortest from s to x j (whatever value is assigned to x i ) x i =min(x i ) =-  (x i,s) max xi  min(xi) (x j )=  (s,x j ) = x i +  (x i,s) +  (s,x j ) and s belongs to a shortest path from s to x j

28. Computing min(x i ) (improvement) 3b) requires x i  = 1 (min(y) -  j i  (x i,x j ) ) |X| 2 August, 2000 28  (x i,x j ) can be approximated by its lower bound max(x j )-max(x i ) Computation can be stopped earlier