1. The Price Of Stability for Network Design with Fair Cost Allocation Elliot Anshelevich, Anirban Dasgupta, Jon Kleinberg, Eva Tardos, Tom Wexler, Tim Roughgarden Presented by : Kobi Yablonka (most of the slides are taken from Elliot Anshelevich ’ s presentation “ The price of Stability for Network Design ” – www.cs.princeton.edu/~eanshele)

2. The context A network Design Game Number of independent agents Each agent try to minimize the cost

3. The Model Directed garph G=(V,E), with each edge e having a nonegative cost Each player i has a set of terminal nodes that he wants to connect Strategy for player i is s.t connects all node in All players using an edge split up the cost of the edge equally

4. The Model cont. vector of the players strategies - the number of players using edge e The cost to player i is The total edge cost of the network is The cost to a player is affected from the strategies of other players

5. Nash Equilibrium A Nash Equilibrium (NE) is a set of payments for players such that no player wants to deviate. When considering deviations, player i assumes that other player payments are fixed. Given a solution consisting of a vector of strategies S there is no strategy for player i s.t

6. Price of stability The best Nash equilibrium relative to the global optimum Stands in contrasts to the “ price of anarchy ” which is the ratio of the worst Nash equilibrium to the optimum

7. Congestion Games This is a congestion game! Usual congestion games have latency/delay/load: cost per player increases as the number of players sharing an edge increases. Fair Connection Game has edge costs: cost per player decreases as the number of players sharing an edge increases.

8. Related Work Shapley value cost sharing [Feigenbaum, Papadimitriou, Shenker; Herzog, Shenker, Estrin] Price of anarchy in routing and congestion games [Roughgarden, Tardos] Potential games [Monderer, Shapley]

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

10. Example: High Price of Stability 1 1 k 1 2 1 3 123k t 0000 1+ ... k-1 0 1 cost(OPT) = 1+ε

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

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

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

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

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

16. To Show The H k Price of Stability is worst case possible. Proof uses the idea of a Potential Game [Monderer and Shapley]. Extend results to many natural generalizations of the Fair Connection Game.

17. Potential Games A game is a potential game if there exists a function Ф(S) mapping the current game state S to a real value s.t. If player i moves, i ’ s improvement = change in Ф(S). Such games have pure NE: just do Best Response! The Fair Connection Game is a potential game! We extend analysis to bound Price of Stability.

18. A Potential Function Define Ф e (S) = c e [1+ 1/2 + 1/3 + … 1/k e ] where k e is # players using e in S. H k Let Ф(S) = Σ Ф e (S) Consider some solution S (set of edges for each player). Suppose player i is unhappy and decides to deviate. What happens to Ф(S)? e є S

19. Tracking Player Happiness Ф e (S) = c e [1+ 1/2 + 1/3 + … 1/k e ] Suppose player i ’ s new path includes e. i pays c e /(k e +1) to use e. Ф e (S) increases by the same amount. Likewise, if player i leaves an edge e ’, Ф e ’ (S) exactly reflects the change in i ’ s payment. e e’e’ c e [1+ 1/2 +… +1/k e ] c e’ [1+ 1/2 +… +1/k e’ ] i

20. Tracking Player Happiness e e’e’ c e [1+ 1/2 +… +1/k e ]+c e /(k e +1) c e’ [1+ 1/2 +… +1/k e’ ] -c e ’ /k e ’ i Ф e (S) = c e [1+ 1/2 + 1/3 + … 1/k e ] Suppose player i ’ s new path includes e. i pays c e /(k e +1) to use e. Ф e (S) increases by the same amount. Likewise, if player i leaves an edge e ’, Ф e ’ (S) exactly reflects the change in i ’ s payment.

21. Bounding Price of Stability Consider starting from OPT (central optimum). From OPT, players will settle on some Nash NE.

22. Bounding Price of Stability Consider starting from OPT (central optimum). From OPT, players will settle on some Nash NE. Ф(NE) < Ф(OPT) _

23. Bounding Price of Stability Consider starting from OPT (central optimum). From OPT, players will settle on some Nash NE. Ф(NE) < Ф(OPT) for any S, cost(S) < Ф(S) < H k cost(S). So cost(NE) < Ф(NE) < Ф(OPT) < H k cost(OPT). _ __ ___

24. Extensions Take a fair connection game with each edge having a nondecreasing concave cost function c e (x),where x is the number of players using edge e. Then the price of stability is at most H k The proof is analogous to the previous proof.

25. Extensions All results hold if edges have capacities. Incorporate distance: cost to player i = c i (P i ) + length(P i ) Utility function of player i can depend on both cost and the set S i picked by i –cost to player i = Σ c e (k e )/k e + f i (S i ) –PoS is still within log(k) if c e is concave e є S i

26. More Questions Cost and Latency Only Latency –Nash exist (same potential argument) –Best NE costs at most OPT w/ twice as many players. Best Response Dynamics –Can construct games with k players so that a certain ordering of moves takes 2 O(k) time. Weighted Game

27. Adding Latency What if we want to model congestion? … marginal cost increases, so not buy-at-bulk. Every edge has increasing delay function d e (k e ). Cost of edge e for player i is c e (k e )/k e +d e (k e ). Total cost of edge is c e (k e ) + k e  d e (k e ).

28. Cost + Latency From earlier proof, we know that if for all S, cost(S) < A  Ф(S) < AB  cost(S), then the price of stability is < AB. if c e is concave, d e is nondecreasing for all e, and for all e and x e then the price of stability is at most AH k (separate cost and latency) E.g. if c e is concave, d e is polynomial with degree m, then Price of Stability is < (m+1)  log(k). - --

29. Latency In this case Nash Equilibria can be computed. d(x) d(1) d(2) d(3) … All edges capacity 1 Claim: A min cost flow corresponds to a NE. Idea: Since d is increasing, flow will use d(1), then d(2), etc, mirroring a potential function. [Fabrikant, Papadimitriou, Talwar] Convert all edges

30. Latency Results (with single source) Nash exist (same potential argument) Best NE costs at most OPT w/ twice as many players.

31. Best Response Dynamics How long before players settle on a NE? In games with 2 players, O(n) time, since shared segment grows monotonically. Can construct games with k players so that a certain ordering of moves takes 2 O(k) time. Can 3-player games run for exponential time? Can k-player games be scheduled to be polytime?

32. Weighted Game If some player has more traffic, should pay more … In a weighted game, player i has weight w(i). Players pay for edges proportionally to their weight. No potential function exists. Do NE always exist? Best Response converges for single commodity. Games with at most 2 players per edge have NE. If NE do exist, Price of Stability will be >> log(k)

33. Games with at most 2 players per edge have NE For edge with one player F (s) = w i c i if I uses e 0 otherwise When player i moves the change in F (S) is equal to the change in player's i payment scaled up by w i For edge e used by players i and j

34. Best Response converges for single commodity All players have the same source and sink For every s-t path P the marginal weight is Order the paths of the players in tuple in according their marginal weight(lexicographic order) Show that in every step the lexicographic size of the tuple decrees If the move is P1 -> P2 P* = group of paths that have edges in either P1 or P2 C ’ (P2) – the cost of P2 after the move show that :

35. If NE do exist, Price of Stability will be >> log(k) w i = 2 i-1 1/2 1 2 1 2 1 2 123k t 0000 1... k-1 0 1 2

36. Thank you.