The price of anarchy of finite congestion games Kapelushnik Lior Based on the articles: “ The price of anarchy of finite congestion games ” by Christodoulou.

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Presentation transcript:

The price of anarchy of finite congestion games Kapelushnik Lior Based on the articles: “ The price of anarchy of finite congestion games ” by Christodoulou & Koutsoupias “ the price of routing unsplittable flow ” by Awerbuch, Azar & Epstein

Agenda Congestion games description Price of anarchy definitions Linear latency functions PoA upper and lower bounds Polynomial latency functions PoA upper and lower bounds

Network congestion game A directed graph G=(V,E) For each edge exists a latency function n users, user j have request Request j assigned to path which is the strategy of player j Users are none cooperative

Network game example s1t1 t3t2 s2 s3

Network game example agent paths (strategies) s1t1 t3t2 s2 s3 Agent 2 path 1 Agent 2 path 2

Network congestion game Consider a strategy profile Denote Ne(A) as the load on e in A Mark as the possible paths for user i Player j cost is the total cost of it ’ s strategy path Strategy A is a NE if no player has reason to deviate from his strategy

Congestion game More generalized than a network congestion game N players, a set of facilities E Each player i has a strategy chosen from several sets of facilities Facility j have cost (latency) function

Congestion game definitions Symmetric game (single-commodity) – all players choose from the same strategies Asymmetric game (multi-commodity) – different players may have different strategy options Mixed strategy for player i – a probability distribution over

Congestion game example f1f2f3f4F5F6

Congestion game example f1f2f3f4F5F6

Congestion game example f1f2f3f4F5F6

Congestion game example f1f2f3f4F5F6

Congestion games Every network congestion game is a congestion game Each edge will be represented by a facility Each strategy path of a player will be replaced by the set of edges in the path

Network to congestion game s t e1e2 e3 fe1fe2fe3 Instead of possible path strategies (e1->e2) and (e3) The congestion game strategies are ({fe1,fe2}, {fe3}}

Social cost Two possible definitions Definition 1: Definition 2: or for weighted requests When considering PoA the social cost definition of sum is equivalent to average (just divide by n)

Price of anarchy The worst-case ratio between the social cost of a NE and the optimal social cost Definition 1: Definition 2:

Linear latency function If an equivalent problem can be described with function Duplicate an edge times

Avg. social cost PoA Asymmetric case Unweighted requests pure strategy Mixed strategy Weighted requests Pure + mixed strategies Symmetric case Unweighed pure strategy

Upper PoA bounds Sketch of proof 1. Compare agent ’ s delay to the delay that would be encountered at the optimal path 2. Combine the bounds and transform to a relation between a total NE delay and the total optimal delay

Upper bound unweighted requests, fe(x)=x Lemma: for a pair of nonnegative integers a,b

Upper bound unweighted requests, fe(x)=x In a NE A and an optimal P allocation The inequality holds since moving from a NE does not decrease the cost Summing for all players we get

Upper bound unweighted requests, fe(x)=x cont ’ Using the lemma we get And thus And the upper bound is proven

Upper bound weighted requests Notations: J(e) – set of agents using e P – NE strategies profile Qj – request j path in P X* - value X in optimal state l – load vector of a system

Upper bound weighted requests Lemma 1: (follows from Cauchy-Schwartz inequality) Lemma 2: for any

Upper bound weighted requests since deviation from NE doesn ’ t decrease cost Multiplying by Wj we get

Upper bound weighted requests Summing for all agents we get Changing summation order we get Using lemma 2 we get

Upper bound weighted requests Using lemma 1 in previous expression

Lower bounds, unweighted requests, congestion game Assume N≥3 agents, 2N facilities fe(x)=x Facilities Agent i strategies Optimal allocation: each agent i chooses Worst NE agent i choose The cost for each agent is 2 in the optimal allocation and 5 in the NE, PoA is 5/2

Lower bounds, unweighted requests, network game S1 S2 S3 t1 t2 t3

Lower bounds, weighted requests, network game U W V fe(x)=0 fe(x)=x

Lower bounds, weighted requests, network game U W V fe(x)=0 fe(x)=x Agent 1 Optimal cost, player 1 use UV, player 2 use UW, Player 3 use VW, player 4 use WV Total cost: Agent 2 Agent 4 Agent 3

Lower bounds, weighted requests, network game U W V fe(x)=0 fe(x)=x Agent 1 NE cost, player 1 use UWV, player 2 use UVW, Player 3 use VUW, player 4 use WUV Total cost: Agent 2 Agent 4 Agent 3

Linear congestion symmetric games lower bound of PoA The upper bound for asymmetric games with avg. social cost also holds for symmetric games The lower bound both max and avg. social cost is (5N-2)/(2N+1) Next is a game description which achieves this PoA for N players

Lower bound game construction for symmetric games The facilities will be in N sets of the same size P1,P2, …,Pn Each Pi is a pure strategy and in optimal allocation each player i plays Pi Each Pi contains facilities At NE player i plays alone facilites of each Pj At NE each pair of players play together facilities of each Pj

Lower bound game construction for symmetric games cont ’ At NE A, We want that at NE no players will switch to Pj For NE we need Which proves the PoA of (5N-2)/(2N+1)

Max social cost PoA Unweighted pure strategy cases only Symmetric case Lower bound already shown Asymmetric case

Asymmetric case upper bound Let A be a NE, P optimal allocation, w.l.o.g Max(A)=c1(A), the NE imply Denote the players in A that use facilities of P1 The avg. social cost lower bound showed

Asymmetric case upper bound Combining the last 2 inequalities substitute in the first inequality

Asymmetric case lower bound V1V2V3VkVk+1 k k k k One player wants to go from V1 to Vk+1 For each i k-1 players wants to go from Vi to Vi+1 Between Vi and Vi+1 one path with length 1, k-1 paths with length k Opt – all players use different paths, cost is k NE – all players use the 1 length paths cost is PoA is

Symmetric case upper bound Let A be a NE, P optimal allocation, w.l.o.g Max(A)=c1(A), the NE imply Summing for all possible j in P and using the lemma Using the avg. social cost lower bound

Polynomial latency function The latency functions are polynomials of bounded degree p The proofs for PoA of linear latency functions are quite similar to those of polynomial latencies

Polynomial latencies cost PoA For polynomials of degree p, nonnegative coefficients Avg. social cost weighted requests, unweighted requests, symmetric games, asymmetric games, pure strategies, mixed strategies Max social cost Pure strateties symmetric games Pure strateties asymmetric games

Upper bound unweighted requests polynomial latencies Instead of the lemma for linear functions for a pair of nonnegative integers a,b A new lemma is used, if f(x) polynomial in x with nonnegative coefficients, of degree p, for nonegative x and y Where

Upper bound unweighted requests polynomial latencies cont ’ In a NE A and an optimal P allocation The inequality holds since moving from a NE does not decrease the cost Summing for all players we get

Upper bound unweighted requests polynomial latencies cont ’ Using the lemma we get And thus And the upper bound is proven

Lower bound game construction for symmetric games The facilities will be in N sets of the same size P1,P2, …,Pn Each Pi is a pure strategy and in optimal allocation each player i plays Pi Each Pj contains N facilities At NE player i plays

Lower bound game construction for symmetric games cont ’ At NE A, We want that at NE no players will switch to Pj For NE we need to select N such that For opt The PoA is Which proves the PoA of when choosing N that satisfies the equation

Asymmetric case lower bound almost like in the linear case V1V2V3V(k^p) V(k^p+1) k k k One player wants to go from V1 to For each i k-1 players wants to go from Vi to Vi+1 Between Vi and Vi+1 a path with length 1, k-1 paths with length Opt – all players use different paths, cost is NE – all players use the 1 length paths cost is