Collection Depots Facility Location Problems in Trees R. Benkoczi, B. Bhattacharya, A. Tamir

1. Collection Depots Facility LocationProblems in TreesR. Benkoczi, B. Bhattacharya, A. Tamir 陳冠伶‧王湘叡‧李佳霖‧張經略 Jun 12, 2007

2. Outline

3. By 陳冠伶 INTRODUCTION

4. Client (demand service) Settings Facility (service center) Collection Depots

5. Cost of Service Trip F P2 D P1 Service Cost 2(P1+P2)‧w(c) C

6. Application (1) Express Transportation

7. Application (2) Garbage collection

8. Problem • IN: given a tree and • points of clients • points of collection depots • an integer k • OUT • Optimal placements of kfacilities • that minimizes some global function of the service cost for all clients.

9. Objective – Minimax • Minimize the service cost of the most expensive client C C D D D C D F D C C

10. Minimax – center problems 1-center Minimize the maximum distance to the facility

11. Minimax – center problems k-center Minimize the maximum distance to the closest facility

12. Objective – Minisum • Minimize the total service cost C C D D D C D F D C C

13. Minisum – median problems 1-median Minimize the average distance to the facility

14. Minisum – median problems k-median Minimize the average distance to the closest facility

15. Classifications

16. Summary of Results • Unrestricted 1-center problem • O(n) • Unrestricted median problems • 1-median: O(nlogn) • k-median: O(kn3) • Restricted k-median problem • NP-complete • Facility setup costs are not identical

17. BY 王湘叡 1-CENTER PROBLEM

18. Prune and Search • Every iteration, eliminate a fraction of impossible instances. • Binary Search • T(n)=T(n/2)+1 • T(n)=O(lg n) • How about

19. Observation • c(f)=max min r(f, vi) • Service cost is non-decreasing when the facility goes away from the client.

20. Where could the facility be? • A linear time algorithm could determine! T1 T2 Ti Tk

21. Initial tree depot client

22. Divide T(i) into S1 and S2 • Find the centroid and partition the tree into two parts centroid • S1 > 1/3 |T(i)| • S2 > 1/3 |T(i)|

23. Find the Xmax • Find the client Xmax with the largest service cost from the centroid. Xmax f S2 S1 foptmust be in S1

24. Xmax Special case • Centroid is the optimal Should be optimal X’max

25. Partition the clients • Compute all depot distance • Find the median δmed • Separate all clients into two sets, K+ (red) and K- (blue) δmed • S2

26. Consider f’ in S1, that depot distance δ(f’)< δmed δ(f’)< δmed f’ • S1

27. Partition S1 by δmed • Find all f’, they form trees T1, T2, …,Tn • There are two cases, fopt is in ∪Tior not f T1 T2 T3

28. fopt is in ∪Ti • If fopt in red, consider K+, δ(fopt)<δmed<δ(K+) • For a facility F’ in S1 and a client in S2, δ(fopt, u) is in S1 δ(f’, u) δ(f’, u) fopt f’

29. fopt is in not ∪Ti • If fopt is not in red, consider K-, δ(K-)<δmed <δ(fopt) • For a facility F’ in Sand a client in S2, δ(fopt, u) is in S2 • Similar to previous case • Only fopt in ∪Ti is considered. δ(fopt, u) f’ fopt

30. Details on fopt is in ∪Ti • Arbitrarily paired clients in K+ • For each pair (u, v), Compute tuv s.t. w(v)(tuv+d(c,v))=w(u).(tuv+d(c,u)) • Compare tmed and d(fopt, c)+d((fopt,c),p(fopt, c)) δ(f’, u) fopt fopt

31. d(fopt, c)+d((fopt,c),p(fopt, c)) < tmed • consider tmed<tuv • d(fopt, c)+d((fopt,c),p(fopt, c))<tmed<tuv δ(f’, u) fopt fopt

32. d(fopt, c)+d((fopt,c),p(fopt, c)) > tmed • consider tmed>tuv • d(fopt, c)+d((fopt,c),p(fopt, c))>tmed>tuv • ¼ K+ can be removed δ(f’, u) fopt fopt

33. BY 李佳霖 1-MEDIAN PROBLEM

34. The 1-median Problem • Find a placement for facility to minimize the cost of all tours. • i.e. minimize the sum of weighted distances of the facility to client, then to optimal depot, and return to facility. • For the path of a facility to a client, the closest depot can be found efficiently. • Brute Force: Ο(n2) • Using Spine decomposition and pre-sorting: Ο(nlogn)

35. The Spine Decomposition r0 5 3 3 3 2

36. Construct Search Tree r0

37. Search Tree of SD r0

38. Super-path of Search Tree r0 f

39. Cost of Subtree d d2 v dnew d4 f c2 d1 cj d3 c1 c3 c4

40. Complexity • Construction for the SD has time complexity Ο(n) and space complexity Ο(n) • Costs of the subtrees can be evaluated in constant timeonce j is determined. • If we use binary search with dnew, we spend Ο(logn) time for every subtree. So Ο(log2n). • Use the sequential search in sorted order. So Ο(logn). • The 1-median collection depots problem in tree can be sloved in Ο(nlogn) time and Ο(n) space.

41. BY 張經略 UNRESTRICTEDK-MEDIAN PROBLEM

42. The objective • To minimize the sum of facility opening costs plus service costs for servicing the clients.

43. The “自給自足” property (1/4) • We fixed an arbitrary optimal solution and explore its structure.

44. The “自給自足” property (2/4) • Consider an arbitrary vertex v. • xv: minimize the trip cost of serving v • yv:be a closest facility to v. yv Assumed (for contradiction) servicing facility for client C xv v Tleft Tright client C

45. The “自給自足” property (3/4) Assumed (for contradiction) servicing facility for client C yv xv v client C Tright Tleft

46. The “自給自足” property (4/4) • The blue part of the following graph is proven by symmetry. yv xv v Tleft Tright

47. The intuition… (1/2) • The total cost can be partitioned into four categories: the red, yellow, blue cost and v. yv xv v Tleft Tright

48. The intuition… (2/2) • The optimal solution has to be a combination of optimal substructures • You have to be “optimal” in the red (to minimize the red cost) and the yellow (to minimize the yellow cost). • This almost leads to Dynamic Programming already!

49. The technical things • Due to some complications, the final Dynamic Programming is much more complicated… • But the proof requires no special technique beyond the “自給自足” property. • The challenge is to devise the “right” recurrences to carry out the aforementioned intuitive approach.

50. Simple intuition, complicated recurrences… take a look