1. On the Price of Stability for Designing Undirected Networks with Fair Cost Allocations M.Sc. Thesis Defense Svetlana Olonetsky

2. Definition of the problem Graph G=(V,E) n players Player i wants to connect vertices s i, t i S i – some path that connects s i to t i (S i is called the strategy of player i) State S=(S 1,S 2,…,S n )

3. Cost definition c(e) – cost of edge e x s (e) – number of users that use edge e in state S cost to the player: total cost: w C(v) = 8 $2 $6 $5 C(w)= 5 r u v C(v) = ? C(w)= ?

4. Nash Equilibrium State S is a Nash equilibrium if for every state S ’ =(S 1,…,S i-1, S ’ i, S i+1,…,S n )

5. Price of Stability Price of Stability = C(best NE) C(OPT) (Min cost Steiner forest)

6. Price of Stability For this game on directed graphs: Price of stability Θ(log n) “The Price of Stability for Network Design with Fair Cost Allocation “ [E. Anshelevich, A. Dasgupta, J. Kleinberg,E. Tardos, T. Roughgarden ]

7. Example: High Price of Stability 1 1 n 1 2 1 3 123n t 0000 1+ ... n-1 0 1

8. Example: High Price of Stability 1 1 n 1 2 1 3 123n t 0000 1+ ... n-1 0 1 C(OPT) = 1+ε

9. Example: High Price of Stability 1 1 n 1 2 1 3 123n t 0000 1+ ... n-1 0 1 C(OPT) = 1+ε …but not a NE: player n pays (1+ε)/n, could pay 1/n

10. Example: High Price of Stability 1 1 n 1 2 1 3 123n t 0000 1+ ... n-1 0 1 so player n would deviate

11. Example: High Price of Stability 1 1 n 1 2 1 3 123n t 0000 1+ ... n-1 0 1 now player n-1 pays (1+ε)/(n-1), could pay 1/(n-1)

12. Example: High Price of Stability 1 1 n 1 2 1 3 123n t 0000 1+ ... n-1 0 1 so player n-1 deviates too

13. Example: High Price of Stability 1 1 n 1 2 1 3 123 n t 0000 1+ ... n-1 0 1 Continuing this process, all players defect. This is a NE! (the only Nash) cost = 1 + + … + Price of Stability is H n = Θ(log n) ! 1 2 n

14. Potential function This game is a special case of congestion games, therefore has a potential function: If user i changes its strategy from S i to S ’ i :

15. Upper bound on the Price of Stability

16. Summary Results of Anshelevich et. al: Price of stability on directed graphs  (log n) Open problem: Price of stability on undirected graphs

17. Our results We consider a restricted version of a game: –undirected graph –all players want to connect to the same vertex r –every vertex v has at least one player associated with it Theorem: The Price of Stability for this game is O(loglog n).

18. Overview of the proof Start with OPT tree (OPT is MST) Schedule sequence of improvement moves When no moves are possible => NE Bound cost of NE

19. Improvement moves r

20. r Edges in OPT Edges in graph

21. Improvement moves r Edges in OPT Edges in graph

22. EE move r v Edges in OPT Edges in graph v - change of strategy

23. EE move r v Edges in OPT Edges in graph no new edges were added by v v - change of strategy

24. OPT move r v Edges in OPT Edges in graph v - change of strategy

25. OPT move r v Edges in OPT Edges in graph v - change of strategy new OPT edge was added

26. w r w Edges in OPT Edges in graph - change of strategy - move

27. r w Edges in OPT Edges in graph - change of strategy - move w new edge, not in OPT, was added

28. Given a state S, and user u, improvement moves for u can be classified as follows: EE – All edges in the path u->r are in S. OPT – All edges in the path u->r are in S  OPT. – The first edge e=(u,v) of S’u is neither in S nor in OPT, all other edges are in S. Other – All other improvement moves Improvement moves We never schedule Other improvement moves

29. EE moves do not increase the total cost OPT moves increase the Price of Stability by at most a factor of 2 moves can increase the total cost Every move adds one new edge to S EE, OPT, and moves

30. Lemma : If no EE moves possible  S is a tree Lemma : If no EE, OPT, or moves possible  state S is in Nash equilibrium

31. Scheduling algorithm The scheduler works in phases In the beginning of a phase no OPT or EE moves are possible.

32. Scheduling phase r OPT edges graph edges dashed edges unused in S

33. Scheduling phase r u OPT edges graph edges

34. Scheduling phase r u OPT edges graph edges u performs move x

35. Scheduling phase 1 r u OPT edges graph edges x loop on dist OPT (u,w)

36. Scheduling phase 1 2 r u OPT edges graph edges x loop on dist OPT (u,w)

37. Scheduling phase 1 2 r u OPT edges graph edges x 6 3 4 5 unused edge unused edge loop on dist OPT (u,w)

38. Scheduling phase r OPT edges graph edges x x/8 1 2 u 6 3 4 5

39. Scheduling phase 1 2 r u OPT edges graph edges x 6 3 4 5 x/8

40. Scheduling phase 1.Player u performs move 2.Loop over players w in increasing order dist OPT (u,w): –If strategy S ’ w that consist of Path OPT (w,u) followed by current strategy of u is an improvement move perform it 3.While possible, schedule OPT and EE moves

41. Scheduling algorithm properties (1) Let e=(u,v), e  OPT, that was added to S. It must have been added by an move that started a phase. Lemma: During the remainder of the phase –All users w within dist OPT (u,w) ≤ c(e)/8 use the strategy u  …  r as the tail of their strategy. –When each of these users modify their strategy to include u  …  r, the potential drops by a constant fraction of c(e)

42. Proof: r u v w SwSw SvSv SuSu u performs move Used OPT edges Unused OPT edges

43. Proof: r u v w SwSw SvSv SuSu dist OPT (u,w) <x/8 x S’uS’u u performs move S’’ w Used OPT edges Unused OPT edges

44. Proof: r u v w SwSw SvSv SuSu x S’uS’u player w with dist OPT (u,w)≤x/8 will decrease potential by at least x/4 u performs move S’’ w C S (u) < C S (w) + dist OPT (u,w) C S (w) > C S (u) – x/8 C S’’ (w) < x/8 + C S’ (u) – x/2 C S’’ (w) < C S’ (w) - x/4 dist OPT (u,w) <x/8

45. Scheduling algorithm properties (2) Let e 1 =(u 1,v 1 ), e 2 =(u 2,v 2 ) be two edges that belong to Nash, e 1  OPT and e 1  OPT. Lemma:

46. Proof : w r v OPT edges graph edges dashed edges unused in S e1e1 e2e2 c(e 1 )≤c(e 2 ) Suppose dist OPT (v,w)≤c(e 1 )/8. c(e 1 )/8 dist OPT (v,w) c(e 2 )/8

47. Definition Let u  v  …  r be the strategy for u in the final state (Nash equilibrium). Classify edge e = (u,v)  OPT with c(e) = x, as either –a light edge – if there are ≤ log n vertices within dist OPT ≤x/8 of u, or –a crowded edge - otherwise

48. Lemma: The total cost of all crowded edges is  (OPT) Proof: –In the phase such an edge was added to S, the potential drops by at least  (c(e)log n). –Thus, the total drop in potential during phases with crowded first edges

49. Lemma: The total cost for all light edges is  (OPT loglog n) Proof: Let u  v  …  r be the strategy for u in the final state and let e=(u,v) be a light edge, define the cost of u to be the cost of e=(u,v)/16. Call u light vertex.

50. Light edge amortization r OPT tree

51. Light edge amortization r OPT tree light vertices remove all vertices that are not light and don’t have light descendants

52. Light edge amortization r OPT tree light vertices remove all vertices that are not light and don’t have light descendants

53. Light edge amortization r OPT tree light vertices remove all vertices that are not light and don’t have light descendants

54. Light edge amortization r v OPT tree light vertices take furthest vertex from r

55. Light edge amortization r v OPT tree light vertices mark v's cost in r-direction

56. Light edge amortization r v OPT tree light vertices mark subtree mark v's cost in r-direction

57. Light edge amortization r v OPT tree light vertices amortize the cost of subtree and remove it from tree

58. Light edge amortization r OPT tree light vertices continue with the rest of the tree mark its cost in r-direction take furthest vertex from r amortize the cost of subtree remove from tree

59. Light edge amortization r OPT tree light vertices continue with the rest of the tree

60. Subtree amortization Lemma: The total cost of light edges starting from vertices in a subtree is at most loglog n times the cost of the subtree

61. Subtree amortization v Direct a path from u towards v, of length equal to the cost of vertex u Paths starting at vertices of the same cost don’t intersect The total cost of vertices of equal cost with a subtree is no more than the cost of the subtree.

62. Subtree amortization Proof: –suppose all costs are powers of 2 –at most C(subtree)/ 2 i vertices with cost 2 i –at most logn vertices So the cost: loglog n C(subtree) v Theorem finished!!!

63. Open problems We believe that the price of stability for this version is constant. Can our result be applied to a single source setting where there may not be an agent in every node? Generalization to the case where agents want to connect to different sources?