Presentation on theme: "Best Response Dynamics in Multicast Cost Sharing"— Presentation transcript:
1Best Response Dynamics in Multicast Cost Sharing Seffi NaorMicrosoft Research and TechnionBased on papers with:C. Chekuri, J. Chuzhoy, L. Lewin-Eytan, and A. Orda [EC 2006]M. Charikar, C. Mattieu, H. Karloff, M. Saks TexPoint fonts used in EMF.Read the TexPoint manual before you delete this box.: AAAAAAAA
2MotivationTraditional networks – single entity, single control objective.Modern networking – many entities, different parties.Users act selfishly, maximizing their objective function.Decisions of each user are based on the state of the network, which depends on the behavior of the other users. non-cooperative network games.
3Our Framework A network shared by a finite number of users. Each edge has a fixed cost.Cost sharing method defines the rules of the game: determines the mutual influence between players.Performance of a user is total payment =sum of payments for all the edges it uses.Two fundamental models:The congestion model.The cost sharing model.
4Common in unicast routing. Edge cost: Congestion ModelCost Sharing ModelCommon in unicast routing.Edge cost:Modeled by a load dependent function.Non-decreasing in the total flow of the edge.Each user has a negative effect on the performance of other users.Common in multicast routing.Fixed costCost sharing mechanism determines how the cost is shared by the users.Each user has a favorable effect on the performance of other users (cross monotonicty).In both models:Each user routes its traffic over a minimum-cost path.Splittable routing model vs. unsplittable routing model.
5The Multicast GameA special root node r, and a set of n receivers (players).A player’s strategy is a routing decision – the choice of a single path to r.Egalitarian (Shapley( cost sharing mechanism: the cost of each edge is evenly split among the players using it. Each player on edge e with ne players pays: ce / neThe goal of the players is to connect to the root by making a routing decision minimizing their payment.
6The Multicast Game Two different models: The integral model: each player connects to the root through a single path.The fractional model: each player is allowed to split its connection to the root into several paths (fractions add up to 1).
7Nash Equilibrium Players are rational. Each player knows the rules of the underlying game.Nash Equilibrium: no player can unilaterally improve its cost by changing its path to the root. Cost of a path takes into account cost sharing.Nash equilibrium solutions are stable operating points.
8The Price of AnarchyNash equilibrium outcomes do not necessarily optimize the overall network performance.Price of Anarchy: The ratio between the cost of the worst Nash equilibrium and the (social) optimum.Quantifies the “penalty” incurred by lack of cooperation.
9The Integral Multicast Game Potential function Φ of a solution T [Rosenthal `73]:Exact potential: change in cost of a connection of player i to the root is equal to the change in the potential Φ.If edge e is deleted from T: Φ = Φ - ce / ne(T)If edge e is added to T: Φ = Φ + ce / (ne(T)+1)
10The Integral Multicast Game Finite strategy space Φ has an optimal value.Φ Nash equilibrium existence.Global / Local optima of Φ correspond to a NE.A Nash equilibrium solution is a tree rooted at r spanning the players.Special case of a congestion game.
11Price of Anarchy vs. Price of Stability Price of anarchy can be as bad as (n).Price of stability – ratio between the cost of best Nash solution to the cost of OPT.Outcome of scenarios in the ‘middle ground’ between centrally enforced solutions and non-cooperative games.E.g.: central entity can enforce the initial operating point.[Anshelevich et al., FOCS 2004]Directed graphs - price of stability is θ(log n).Undirected graphs – upper bound on the price of stability is O(log n). PoS can be reached from a 2-approximate Steiner tree configuration:C(TNash) Φ(TNash) Φ(T2-Seiner) log n ∙C (T2-Steiner)rt
12Best Response Dynamics Each player, in its turn, selects a path minimizing its cost (best response).Eventually, an equilibrium point is reached.PoA strongly depends on the choice of the initial configuration.Starting from a near-optimal solution may be hard to enforce: requires relying on a central trusted authority.Question: What happens if we start from an empty configuration? [Chekuri, Chuzhoy, Lewin-Eytan, Naor, and Orda, EC 2006]Round 1: Players arrive one by one, each player plays best response.Round 2: Best response dynamics continue in arbitrary order till NE.Question: Can a good equilibrium always be achieved as a consequence of best-response dynamics in this model?
13… Can a good equilibrium be achieved as a consequence Cost of user 1: c (r, x, 1) = 1+εc (r , 1) = 1rrrrrrrrrrGreedy cost of 3, … ,n = 1Cost of user 2: c (r , x, 2) = 1+εc (r, 1, x, 2 ) = 1+2εc (r, 2) = 1113/41111xxxxxxxxxxxxxxx¼ + ε¼ + ε¼ + ε¼ + ε…11111111111111111222222222223333333nn-2n-1Price of anarchy = 4Can a good equilibrium be achieved as a consequenceof best-response dynamics?
14Results The integral multicast game for undirected graphs: Upper bound of O(log3n) on the PoA of best-response dynamics in the two-round game starting from an “empty” configuration. (Improving over the bound of [CCLNO-EC06] of )Upper bound of O(log2+ n) on the cost of the solution at the end of the first round.Lower bound of (log n) on the PoA of this game.Computing a Nash equilibrium minimizing Rosenthal’s potential function is NP-hard.
15Price of Anarchy: Lower Bound Theorem: Price of anarchy of our game ¸ (logn).Proof: Adaptation of lower bound proof for the online Steiner problem [Imaze&Waxman]Online Steiner problem: used edges have cost 0Take hard instance [IW] and replace each terminal by a star of n2 + 1 terminals at zero distanceThe n2 + 1 terminals choose the same path to rootcost of used edge becomes negligible
16PoA in Undirected Graphs: Upper Bound Analysis is performed in two steps:Round 1: players connect one by one to the root via best response.Solution T is reached after a sequence of arrivals t1 , t2 ,…, tn . We show: O(log3n )∙ c(TOPT )c(T) · O(log2+ n )∙ c(TOPT )Round 2: players play in arbitrary order till NE is reached c(TNash) Φ(TNash) Φ(T)
17The First RoundChoose a threshold 2 (0,1).Terminal t is -good if cost of next terminal using the same path as t · (1-) ¢ cost(t).Otherwise, terminal t is -bad.Idea: bound separately the contribution to Φ of the -good terminals and the -bad terminals.
18Charging -good Terminals Suppose there is a tree T’ spanning the -good terminals.Upon arrival t pays: cost(t)Upon arrival, t’ pays at most:cost(t’) d + (1-) ¢ cost(t)cost(t)tt’dt and t’ are -good terminalst arrives first, then t’ arrives
19Charging -good Terminals (contd.) Charges decay at an exponential rate along a root – leaf path in T’Upon arrival of ti it pays at most:cost(ti) · di + (1-)¢cost(ti-1)cost(tk) · dk + dk-1(1-)+ dk-2(1-)2 + … + d1(1-)k-1rd1t1d2t2d3t3d4t4dktkThe charge to each edge e 2 T’:· d(e) ¢ i (1-)i ¢ ne(i)t1,…, tk are -good terminalsarrival order: t1,…, tk
20Auxiliary TreeTOPT is transformed to an auxiliary tree T’ defined on the set of -good terminals:The descendants of terminal t in T’ are terminals which have arrived after t.c(T’) · O(1/ ¢ logn) ¢ c(TOPT )The depth of T’ · O(1/ ¢ logn)
21Contribution of -good Terminals Theorem: The contribution of the -good terminals to Φ in the first phase is bounded as follows:Proof: Follows from the properties of T’ together with the exponential decay of the charges to the edges of T’ of the -good terminals.
22Contribution of -bad Terminals Theorem: The contribution of the -bad terminals to Φ in the first phase is bounded as follows:Intuition: The cost of the edges “opened” for the first time by -bad terminals constitutes only a small part of the sum of the costs of the -bad paths.Setting = O(1/logn) O(log4n ) upper bound on PoASetting cleverly O(log3n ) upper bound on PoA
23The Fractional Multicast Game Players split their connection to the source.A splittable multicast model.rFlow on (r, y):¼ unit of flow is shared by users 1 & 2.½ unit of flow is used only by user 2.Flow on (r, x):¼ unit of flow is shared by users 1 & 2.½ unit of flow is used only by user 1.3/43/41/41/4xy12
24The Fractional Multicast Game The cost of each flow fraction is split evenly between its users. ce·fe,n_e is the total cost of edge e.fe,1 = 1/4fe,2 = 3/4fe,3 = 7/81/43/47/8Total cost of user 3: ce· (1/12 + 1/4 + 1/8).
25The Fractional Multicast Game The potential function Φ of the fractional model:Φ is an exact potential.A fractional flow configuration defining a local minimum of Φ corresponds to a NE.
26Results: Fractional Multicast Game Nash equilibrium existence.NE - minimizing the potential function:Can be computed in polynomial time (using LP).It is NP-hard in the case of an integral Nash.PoA of the computed NE is O(log n).
27Computing a Minimum Potential NE Create a new graph G’ = (V, E’) by replacing each edge e by n copies e1, e2, …, en.Copy ej : “should” be used if j players use edge e.The cost of a unit flow on copy j of edge e is ce / j.The undirected graph is replaced by a directed flow network.
28Computing a Minimum Potential NE The linear program:Objective function = potential function:The capacity of edge ej is 0 xe_j 1.Variables of the LP:Flows of the users on the edges ejE’.Capacities of the edges in E’.Constraints:Non-aggregating flow constraint: (flow of user i on ej ) xe_j .Aggregating flow constraint: (total flow on ej ) = j ∙ xe_j .
29The Linear ProgramTheorem 5: There exists an optimal solution to the linear program that corresponds to a fractional multicast flow.Heavily depends on the non-increasing property of the cost function.LP can be used:For any cost sharing mechanism that is cross-monotonic.In case players are not restricted to have a common source.PoA of a minimum potential fractional NE is O(log n).
30Integral vs. Fractional Potential Minimization There exists an instance where the gap between the integral and fractional minimum potential solutions is a small constant.Finding an integral Nash equilibrium that minimizes the potential function is NP-hard.Building block: a variation of the Lund-Yannakakis hardness proof of approximating the set cover problem.
31The Weighted Multicast Game Each player i has a positive weight wi .The cost share of each player is proportional to its weight.Integral: cost share of player i = ce· (wi / We)(We – weight of players currently using e)Fractional: weighted sharing on each fraction.Theorem: A NE always exists for the weighted fractional model.Note: NE does not necessarily exists for the weighted integral model [Chen-Roughgarden, SPAA 06].