1. 1 Fair Cost-Sharing Method for the Minimum Spanning Tree Game 邱冠凱 胡啟政 劉宗灝 文國煒

2. 2 Outline Coalition Game Dissatisfaction MWD MAD

3. 3 Coalition Game 邱冠凱

4. 4 Example Players = {1,2,3} All nonempty subset (named as coalition) –{1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} A cost function c related to all coalitions. –c({1}) = v1, c({2}) = v2,..., c({1,2,3}) = v7 c(S) is the amount that the players in the coalition S have to pay collectively in order to have access to a service.

5. 5 Core The problem is to find the core of this coalition game. Core is a cost distribution of the grand coalition such that no other coalition can obtain an outcome better for all its members than the current assignment. There may not exist any core. –Emptynness of the core. There may exist many cores. –Some players would unhappy with the cost allocation.

6. 6 Example We want to find the cost allocation {x1, x2, x3} such that –x1+x2+x3 = c({1,2,3}) –x1 ≦ c({1}) –x2 ≦ c({2}) –x3 ≦ c({3}) –x1+x2 ≦ c({1, 2}) –x1+x3 ≦ c({1, 3}) –x2+x3 ≦ c({2, 3})

7. 7 Cooperative Game Given a solution in the core, there is no incentive for a player to leave the grand coalition. Coalition game is also named as cooperative game. The minimum spanning tree game is a classical cooperative game problem.

8. 8 Minimum Spanning Tree Game r a b c d e f 3 3 3 3 3 3 21 21 G: connected network r: service provider V = {a, b, c, d, e, f}: clients c({a, b, d, e}) = 10 c({b, d, e, f}) = 9

9. 9 The core of MST game Bird proposed a cost allocation rule known as Bird’s rule. Bird’s rule ensures that no coalition has incentive to be formed. –We don’t prove it here. r a b c d e f 3 3 3 3 3 3 21 21 x a = 2, x b = 3, x c = 1, x d = 2, x e = 3, x f = 1, x a + x b + x c + x d + x e + x f = 12 = c({a, b, c, d, e, f}).

10. 10 Outline Coalition Game Dissatisfaction MWD MAD

11. 11 Dissatisfaction 胡啟政

12. 12 Dissatisfaction Definition: For the cost a player have to pay in one solution, dissatisfaction is the ratio of the current cost to the best cost (the smallest one) of all solutions. Dissatisfaction of an agent i in a solution is:

13. 13 Cost Sharing with Fairness Two optimization problems: – MWD: minimize the worst dissatisfaction i.e. Find an allocation which minimizes – MAD: minimize the average dissatisfaction i.e. Find an allocation which minimizes

14. 14 With Bird’s Rule… The dissatisfaction of a vertex v to a mcst T is: Two optimization problems become –SPANNING TREE-MWD: Find a mcst that minimizes the worst dissatisfaction –SPANNING TREE-MAD: Find a mcst that minimizes the average dissatisfaction → Polynomial time algorithms for them.

15. 15 Work With Trees… The fee of a vertex depends on the path between the root and it. β(T1, a) = 41 β(T1, b) = 11 β(T1, c) = 12 β(T1, d) = 39 β(T1, e) = 10 T1 T2 β(T2, a) = 11 β(T2, b) = 12 β(T2, c) = 39 β(T2, d) = 10 β(T2, e) = 41

16. 16 FEE To determine the dissatisfaction of a vertex v ∈ V, it’s necessary to know the set of fees it can have to pay, which is F(v). From Bird’s rule, β(B, v) can only take values in L(v) = { | [x, v] ∈ E } β(B, v)

17. 17 FEE Example: G: L(x) = {1, 2, 3} L(y) = {1, 4} L(z) = {2, 4} F(x) = {3} F(y) = {1} F(z) = {2}

18. 18 To compute F(v)… Find all mcsts and then trace them?

19. 19 Arborescences Assume arborescences are oriented from the root to the leaves. For convenience to deal with the problem, we can build from G=(,E,c) a weighted digraph H=(,A,c). For each edge [x,y] ∈ E, there are two arcs (x,y) and (y,x) in A with cost

20. 20 Arborescences Example: G: H:

21. 21 Arborescences Thus, given a mcst T of an undirected connected graph G, there exists in H a corresponding mcsa B, and C(T) = C(B).

22. 22 FEE Input: A digraph H = (, A, c), a vertex v ∈ V Step 1: Compute any mcsa of H and let be its total cost Step 2: L := { | (x, v) ∈ A } Step 3: F(v) :=  Step 4: For each l of L do A’ := A \ {(x,v) | ≠ l } H’ := (, A, c ) Compute a mcsa B’ in H’ If B’ exists and its total cost is Then F(v) := F(v) U { l } End For Output: F(v)

23. 23 FEE Example: To compute F(x)… C = 6, L(x) = { 2, 3 } r x yz 3 3 2 2 1 1 l = 2 3 F(x) = { }2 l = 3, 3

24. 24 FEE Computing a mcsa : O(mn) ( m = |A|, n = | | ) Total time : O(mn ) After F(v) is computed, F (v) and F (v) can be determined. Also, dissatisfactions are determined. And we can start to solve optimization problems now!!

25. 25 Outline Coalition Game Dissatisfaction MWD MAD

26. 26 The Minimum Worst Dissatisfaction Problem 劉宗灝

27. 27 MWD – Minimum Worst Dissatisfaction Problem The set of allocations from Bird’s rule v 在 B 上的 cost v 在所有 mcsa r 中最小的 cost

28. 28 MWD – Minimum Worst Dissatisfaction Problem The maximum F max (v) / F min (v) among all vertices is an upper bound on the worst dissatisfaction. The idea of the algorithm is decrease the upper bound until it reached the optimal.

29. 29 How to decrease the upper bound? Removing some arcs each time. Which arc? –Arcs that cause worst dissatisfaction.

30. 30 Algorithm 1.Input: A digraph H = (V r, A, c) 2. 用 FEE 算每個點的 F(v) 3. 算出一棵 H 的 mcsa r ，其 cost 為 C 1 4. 在 H 上挑一個最大不滿意度 (F max (v) / F min (v)) 最高 的點 v 5. 把造成 F max (v) 的邊從 H 上移除 6. 在 H 上重算一棵 mcsa r 1. 若不存在，或其 cost > C 1 ，則回傳 C 1 對應之 mcsa r 。 2. 否則 goto step 4

31. 31 Example r x y z 3 3 3 2 1

32. 32 Digraph r x y z 3 3 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2,3} F(z) = {1,3} mcsa r s of this digraph have cost of 6

33. 33 Pick node z r x y z 3 3 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2,3} F(z) = {1,3}

34. 34 (r,z) removed. r x y z 3 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2,3} F(z) = {1}

35. 35 mcsa r still exists r x y z 3 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2,3} F(z) = {1}

36. 36 Pick node y r x y z 3 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2,3} F(z) = {1}

37. 37 (r,y) removed. r x y z 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2} F(z) = {1}

38. 38 mcsa r still exists r x y z 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2} F(z) = {1}

39. 39 Pick node y r x y z 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2} F(z) = {1}

40. 40 (x,y) removed. r x y z 3 2 1 1 F(x) = {2,3} F(y) = {1} F(z) = {1}

41. 41 Any more mcsa r ? r x y z 3 2 1 1 F(x) = {2,3} F(y) = {1} F(z) = {1} NO

42. 42 Recover (x,y), find mcsa r. r x y z 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2} F(z) = {1}

43. 43 Solution r x y z 3 2 1 1 2 F(x) = {2,3} F(y) = {1,2} F(z) = {1}

44. 44 Worst dissatisfaction = 2. r x y z 3 2 1 F(x) = {2,3} F(y) = {1,2} F(z) = {1}

45. 45 Outline Coalition Game Dissatisfaction MWD MAD

46. 46 The Minimum Average Dissatisfaction Problem 文國煒

47. 47 MAD Problem Give a graph, among all spanning arborescences of minimum cost, find one that minimizes the average dissatisfaction. Minimize the average dissatisfaction also minimize the worst dissatisfaction. However, this is not always true.

48. 48 Example 1 r ae bcd 41 11 10 1239 TT` F max (a)/F min (a)=41/11 F max (b)/F min (b)=1 F max (c)/F min (c)=1 F max (d)/F min (d)=39/10 F max (e)/F min (e)=1 The Worst:3.9 Average:2.12 F max (a)/F min (a)=1 F max (b)/F min (b)=12/11 F max (c)/F min (c)=39/12 F max (d)/F min (d)=1 F max (e)/F min (e)=41/10 The Worst:4.1 Average:2.08

49. 49 MAD Problem Minimize the sum or the average dissatisfaction are equivalent. The SPANNING TREE-MAD problem can be described as follows:

50. 50 Solving Procedure First –The initial graph G=(V r, E, c) is turned into a diagraph H=(V r, A, c). –For each edge[x, y] belongs to E, A has corresponding arcs (x, y) and (y, x) –c (x, y) =c (y, x) =c [x, y]

51. 51 Solving Procedure Second –Remove each arc that c(x, y)<F min (y) Third –For each arc (x, y) belongs to A, has a weight w (x, y) = c (x, y) /F min (y). –The weight w (x, y) captures the dissatisfaction of y if the arc (x, y) is chosen.

52. 52 Solving Procedure Forth –Computing a spanning arborescence rooted in r which minimizes the total weight can provide the optimal solution.

53. 53 Example 2 r h ab c d e f g 4 4 321 321 3 TT`T``

54. 54 Example 2 r h abcd e f g 4,4/3 3,3/2 3,1 3,3/2 2,2 2,1 2,2 1,1 3,3 TAverage dissatisfaction: (4/3+3/2+2+1+3+1+1+1)/8≈1.47 T`Average dissatisfaction: (4/3+3/2+2+1+3+1+1+1)/8≈1.47 Average dissatisfaction: (4/3+3/2+2+1+1+2+3/2+4/3)/8≈1.45 T`` T`` is not a minimum cost spanning tree!! cost, weight

55. 55 Solving Procedure Forth –For each arc (x, y), combines c (x, y) and w (x, y) in to a composite cost and generates the new H *. –Computing a spanning arborescence rooted in r with which minimizes the total cost can provide the optimal solution.

56. 56 Algorithm

57. 57 Example 3 h ab c d e f g 4 4 321 321 3 G 4,4/3 3,3/2 3,1 2,2 2,1 2,2 2,1 1,1 3,3 1,1 H A mcsa r B is determined and its sum of dissatisfactions is D=4/3+3/2+1+3+1+1+1=71/6 λ= 71/77 4λ+(1- λ)4/3 3λ+(1- λ)3/2 3λ+(1- λ) 3λ+(1- λ)3/2 3λ+(1- λ) 2λ+(1- λ) 2 2 1 1 1 1 3 3 H*H* B*B* Optimal solution

58. 58 Thank you!