THE BI-CRITERIA DOUBLY WEIGHTED CENTER-MEDIAN PATH PROBLEM ON A TREE

1. Presenter: 陳思佑 鄭仲鈞吳佳欣 高新綸 THE BI-CRITERIA DOUBLY WEIGHTED CENTER-MEDIAN PATH PROBLEM ON A TREE 2007.06.13 J. Puerto, A.M. Rodriguez-Chia, A. Tamir, & D. Perez-Brito

2. Outline • Abstract • Critical value for the weighted path center problem • Obtaining the weighted center-median non-dominated set • The non-dominated set with length constraint

3. Abstract Presenter: 陳思佑

4. Problem • Given • T: a tree network with n node • P*: a given subset of simple path in T • PL: the subset of all discrete paths (length < L) • Consider a path P, P in PL • P: a path-shaped facility • Tree node: the customers • How to calculate the total transportation cost from customers to facility

5. From title… 5 • THE BI-CRITERIA DOUBLY WEIGHTED CENTER-MEDIAN PATH PROBLEM ON A TREE • Doubly weighted • Node: center-weight u, median-weight w (nonnegative) • Bi-criteria • Weighted path center criterion • maximum center-weighted distance • MAX(P) = max ubd(vb, p), vb in V • Weighted path median criterion • sum of the median-weighted distance • SUM(P) = Σwbd(vb, p), b = 1 to n

6. Path: (MAX(P), SUM(P)) • View as a outcome space • Path P -> (MAX(P), SUM(P)) • Finding all the non-dominated pointForm a set, NDO(P*) • P is non-dominate, if no path P’ • MAX(P’) <= MAX(P), SUM(P’) <= SUM(P) • MAX(P’) + SUM(P’) < MAX(P) + SUM(P) SUM(P) MAX(P)

7. In this paper 7 • # of efficient paths: Ω(n2) • At most 2n non-dominated outcome • Generate in O(nlogn) optimal time SUM(P) MAX(P)

8. Solve other model • Cent-dian model • Objective: convex combination of weighted center and weighted median function • Restricted model • Objective: minimize one of MAX or SUM, subject to an upper bound on the other one • After the set of non-dominated outcome has been obtained, all these problems are solve in linear time • Overall, O(nlogn)

9. Summary of results

10. Critical value for the weighted path center problem Presenter: 陳思佑

11. vk Path P longest vb Critical value • A path P, vk in P • For P, if the maximum center-weighted distance to P is attained at vk, vk is critical • Max ubd(vb, P) = max ubd(vb, vk) • Compute the set of all the critical value=> compute all the different MAX(P) • Will prove that the cardinality is at most 2n=> |NDO(P*)| <= 2n

12. In this problem v1 • T = (V, E) • V = {v1, …, vn} • E = {e2, …, en} • An edge ej is identified as an interval of length lj,We can refer to its interior points, continuous • Discrete path • The endpoints of path are nodes of T, not arbitrary point e2 v2

13. Continuous weighted 1-center problem v1 = cu • Min max ubd(x, vb) = max ubd(cu, vb) • cu can be obtained in O(n), center • Add cu, and assume T is rooted at cu • Identifies bottleneck nodes, vs, vt • v1 in Pst, usd(vs, v1) = utd(vt, v1) = max uid(vi, v1) • S ~> t, diameter v2 vt vs

14. v1 = cu • Consider any pair of subtrees, Ts and Tt • vs in Ts, vt in Tt • Ts∩Tt = {v1}, Ts∪Tt = T • 3 sets • Ps = {P in P*, P in Ts} • Pt = {P in P*, P in Tt} • P’ = {Pij in P*, vi in Ts, vj in Tt} v2 vt vs Tt Ts

15. Theorem • A denote the set of distinct MAX(P) for all path P • |A| <= 2n

16. Prove (1) • |P*| <= n(n+1)/2 • Pijin P’ can be represented as the union of Pi1, Pj1 • βsi = max ukd(vk, Pi1), vk in Ts • βtj = max ukd(vk, Pj1), vk in Tt • MAX(Pij) = max(βsi, βtj) v1 = cu v2 vt vs Tt Ts

17. Prove (2) • P in Ps • v1 is a continuous weighted center • Let vk be the closet node to v1 in P • MAX(P) = max uad(va, vk), vk is critical • Define δk = max uad(va, vk) • MAX(P) is an element in {δk: vk in Ts} • P in Pt, same v1 = cu vk v2 vt vs Tt Ts

18. Prove (3) • P’: MAX(Pij) = max(βsi, βtj) • Ps: MAX(P) is an element in {δk: vk in Ts} • Pt: MAX(P) is an element in {δk: vk in Tt} • βsi = max ukd(vk, Pi1), vk in Ts • βtj = max ukd(vk, Pj1), vk in Tt • δk = max uad(va, vk) • Define A* = {βsi: vi in Ts} ∪{βtj: vj in Tt} ∪{δk: vk in V} • A <= A* n n

19. Conclusion • The set A* is a superset of A • We next refine the definition of a superset contaning A

20. Critical value for the weighted path center problem (2) Presenter: 鄭仲鈞

21. Notation Definition • Optimalweighted path center : Ppq • Optimal means: • 假設這樣的path不只一條，取其交集 這樣的交集一樣是optimal

22. Optimal path

23. Optimal path

24. Notation Definition • Va(k) : vk這個節點，延path往下走的Children所構成的點集合。vk屬於Ppq上去除v1的點集合。

25. Notation Definition • For v1: 分為Ts方向和Tt方向，分別是 • For vp and vq: 因為已經走到path的末端 => Va(p)就直接相當於vp的Children的點集合 同理對vq也是一樣

26. Notation Definition • γk : vk延path走下去，離vk最遠點和vk的距 離。即: • γ1s , γ1t : 因為v1往path兩邊走都可，因此 有兩個γ值。即: 但因為v1是continuous weighted center point => γ1s =γ1t => 令γ1=γ1s =γ1t

27. Graph of γk , γ1s , γ1t

28. Notation Definition • εs, εt : Ts和Tt上，離optimal path Ppq 最遠的點，到的距離Ppq。即: 根據定義，明顯的: MAX(Ppq) = max{εs, εt }

29. Graph of εs, εt

30. Properties of Ppq • 此一性質即是說明，對任一點vk來說， 往path處走去的最遠Children的距離(γk)， 必定比不往path處走去之最遠Children還遠: 並且γk必定比εs, εt還要大: Why?

31. Properties of Ppq 假若此性質不成立，代表 optimal path不再是optimal。 =>path應該往較長的那個方 向走，才可以有效的減短 MAX(Ppq) = max{εs, εt } 也就是說，這是optimal path自然產生出的性質。 ”往path走的最遠子節 點，才有可能是其最 遠的子節點。”

32. Properties of Ppq _ • Lemma:任何一條屬於P(端點跨過Ts-v1-Tt 的路徑之集合)的路徑，其Critical node 必定在Ppq之上。

33. Critical node

34. Proof • 假設我們一行一行的看證明，那真的是 有點繁瑣: 所以，來個簡單的說明: 首先假設Critical node不在Optimal path上

35. The graph of proof 明顯的vm到vk比 vr到vk長，而根據 Ppq的特性， 也就是vg到vk應該要比 vr到vm長 =>vk才是真正的critical node. 和vr是critical node的假設 相矛盾

36. Conclusion of Ppq • 既然所有屬於P的路徑，他們的Critical node 都在Ppq之上，那就代表Ppq之上的節點vk ， 它們所對應到的γk，以及εs, εt ，涵蓋了所 有可能的MAX(P)。若再加上之前Ps以及Pt ， 計算出來的MAX(P)之值域，即可形成A** :

37. Conclusion of Ppq 由Critical point看去 往γk走去的才可能是MAX(P) (往另外一個方向走較短) 否則MAX(P) 就必定是εs或 εt (在P包含vp的狀況之下)

38. Generating A** • 有演算法可以在O(n logn)的時間內產生 A**這個值域內的所有值。 • 假如是identical weighted的樹: =>可以在O(n)的時間內完成

39. Presenter: 吳佳欣 Obtaining the weighted center-median non-dominated set 39

40. To generate NDO(P*) we compute a planar set W = {(x,y)},satisfying |W|= O(n) and • The set of first coordinate x of W = A** • We’ll minimize SUM(P) under constraint MAX(P)=x

41. Example 8 (4,2) (3,7) 6 5 4 (2,7) 6 3 2 3 (1,4) 8 (3,3) 7 4 (5,6) 2 4 5 7 3 (3,2) 9 10 0 1 2 (2,9) (1,5) (2,2) (1,9)

42. Weighted center V1 2.5 5.5 (4,2) (3,7) 6 5 4 (2,7) 6 3 2 3 (1,4) 8 (3,3) 7 4 (5,6) 2 4 5 7 3 (3,2) 9 10 0 1 2 (2,9) (1,5) (2,2) (1,9)

43. Weighted Path Center V1 2.5 5.5 (4,2) (3,7) 6 5 4 (2,7) 6 3 2 3 (1,4) 8 (3,3) 7 4 (5,6) 2 4 5 7 3 (3,2) 9 10 0 1 2 (2,9) (1,5) (2,2) (1,9)

44. Critical Value for weighted path center problem γv1=37.5 V1 γ6=22 2.5 5.5 (4,2) (3,7) 6 5 γ5=30 4 (2,7) 6 3 2 γ8=10 3 (1,4) γ4=10 8 (3,3) 7 4 (5,6) 2 4 5 7 3 (3,2) 9 10 0 1 2 (2,9) (1,5) γ0=8 (2,2) (1,9)

45. Case 1: MAX(P)=δk • If MAX(P)=δk, or • The critical node vk of P (i.e. maxi ukd(vk,vi)=MAX(P)) isthe closest point to weighted 1-center point. • efficient path with critical node vk應該是掛在vk上的一條帶子，往兩個最能減少SUM(P)的方向垂下。

46. Sum vi vi vj (vi, vj相鄰)

47. Sum vi vj 取min vb 從vi往vj走可以減少的SUM(P)值

48. SUM(P) • Let SUM(P) be the minimum of all SUM(P’) of P’ with critical node vk and MAX(P’)=δk • SUMt(), t=0,1,2,3. Can be compute in O(n) • Time: O(Σv|Neighbor(v)|)=O(2n-2)=O(n) • add (δk, SUM(P)) into W 前兩大 SUM3(vk,vb) 其實就是

49. Result V1 403 371 2.5 5.5 (4,2) (3,7) 6 5 4 (2,7) 6 3 2 595 3 (1,4) 8 (3,3) 547 7 676 4 (5,6) 2 4 5 7 3 622 740 (3,2) 9 10 0 1 2 (2,9) (1,5) (2,2) (1,9) 100 853 740 735

50. Case 2: MAX(P)=γk or {єs,єt} • 如果MAX(P)= γk, 則P包含v1, critical node是P與path center交集的端點之一。 • 所以我們要查的是掛在Path(vk,vb)上, 從vk與vb垂下的路徑(vk, vb都在path center上, 且一個在Ts, 一個在Tt). 因為vk才是critical node, 所以γk>= γb. • 設path center為Path(vp,vq) • 把P從v1切開，只看Ts這半的話是Path(vi, v1), 從vk開始離開 Path(vp,v1), 以 表示 • 這類的Path中, 在Ts的最小weight sum(另一半另外考慮)