The price of anarchy of finite congestion games

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## The price of anarchy of finite congestion games

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**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**s1 t1 t3 t2 s2 s3**Network game exampleagent paths (strategies)**Agent 2 path 2 s1 t1 t3 t2 s2 Agent 2 path 1 s3**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**f1 f2 f3 f4 F5 F6**Congestion game example**f1 f2 f3 f4 F5 F6**Congestion game example**f1 f2 f3 f4 F5 F6**Congestion game example**f1 f2 f3 f4 F5 F6**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**e1 e2 fe1 fe2 fe3 s t e3 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 • Compare agent’s delay to the delay that would be encountered at the optimal path • 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 t1 S2 t2 S3 t3**Lower bounds, weighted requests, network game**V fe(x)=0 U fe(x)=x W**Lower bounds, weighted requests, network game**V fe(x)=0 Agent 1 U Agent 4 fe(x)=x Agent 3 Agent 2 W Optimal cost, player 1 use UV, player 2 use UW, Player 3 use VW, player 4 use WV Total cost:**Lower bounds, weighted requests, network game**Agent 2 V fe(x)=0 Agent 3 U fe(x)=x Agent 1 Agent 4 W NE cost, player 1 use UWV, player 2 use UVW, Player 3 use VUW, player 4 use WUV Total cost:**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**k k k V1 V2 V3 Vk Vk+1 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 boundalmost like in the linear case**k k k V1 V2 V3 V(k^p) V(k^p+1) • 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