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Inefficiency of equilibria, and potential games Computational game theory Spring 2008 Michal Feldman TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.:

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Inefficiency of equilibria Outcome of rational behavior might be inefficient How to measure inefficiency? – E.g., prisoners dilemma Define an objective function – Social welfare (= sum of players payoffs): utilitarian – Maximize min i u i (egalitarian) – … 0,53,3 1,15,0

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Inefficiency of equilibria To measure inefficiency we need to specify: – Objective function – Definition of approximately optimal – Definition of an equilibrium – If multiple equilibria exist, which one do we consider?

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Common measures Price of anarchy (poa)=cost of worst NE / cost of OPT Price of stability (pos)=cost of best NE / cost of OPT – Note: poa, pos 1 (by definition) Approximation ratio: Measures price of limited computational resources Competitive ratio: Measures price of not knowing future Price of anarchy: Measures price of lack of coordination

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Price of anarchy Example: in prisoners dilemma, poa = pos = 3 – But can be as large as desired Wish to find games in which pos or poa are bounded – NE approximates OPT – Might explains Internet efficiency. Suppose we define poa and pos w.r.t. NE in pure strategies – we first need to prove existence of pure NE 0,53,3 1,15,0 Prisoners dilemma

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Max-cut game Given undirected graph G = (V,E) players are nodes v in V An edge (u,v) means u hates v (and vice versa) Strategy of node i: s i {Black,White} Utility of node i: # neighbors of different color Lemma: for every graph G, corresponding game has a pure NE

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Proof 1 Claim: OPT of max-cut defines a NE Proof: – Define strategies of players by cut (i.e., one side is Black, other side is White) – Suppose a player i wishes to switch strategies: is benefit from switching = improvement in value of the cut – Contradicting optimality of cut u i =1 u i =2

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Proof 2 Algorithm greedy-find-cut (GFC): – Start with arbitrary partition of nodes into two sets – If exists node with more neighbors in other side, move it to other side (repeat until no such node exists) Claim 1: GFC provides 2-approx. to max-cut, and runs in polynomial time Proof: – Poly time: GFC terminates within at most |E| steps (since every step improves the value of the solution in at least 1, and |E| is a trivial upper bound to solution) – 2-approx.: Each node ends up with more neighbors in other side than in own side, so at least |E|/2 edges are in cut (since #edges in cut > #edges not in cut)

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Proof 2 (contd) Claim 2: cut obtained by GFC defines a NE Proof: obvious, as each player stops only if his strategy is the best response to the other players strategies Conclusion: max-cut game admits a NE in pure strategies

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Extensions What would happen if the edges were weighted? Say, +5 – hate a lot, -5 love a lot? What would happen if love/hate were not symmetric? Home work assignment – I dont know: - Find NE? Complexity?

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Potential games Definition: a game is an ordinal potential game if there exists :S1×…×Sn R, s.t. i,s i,s -i,s i, c i (s i,s -i ) > c i (s i,s -i ) IFF (s i,s -i ) > (s i,s -i ) Note: G is an exact potential game if c i (s i,s -i ) - c i (s i,s -i ) = (s i,s -i ) - (s i,s -i ) Example: max-cut is an exact potential game, where is the cut size – Unfortunately, is not always so natural

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Potential games Lemma: a game is a potential game IFF local improvements always terminate proof: – Define a directed graph with a node for each possible pure strategy profile – Directed edge (u,v) means v (which differs from u only in the strategy of a single player, i) is a (strictly) better action for i, given the strategies of the other players – A potential function exists IFF graph does not contains cycles If cycle exists, no potential function; e.g., (a,b,c,a) means f(a)

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Examples direction of local improvement 1,-1-1,1 1,-1 0,53,3 1,15,0 0,02,2 3,30,0 Matching pennies Prisoners dilemma Coordination game C D DC col row Which are potential games? Exact potential games? Are the potential functions unique? 0,02,1 1,20,0 Battle of the sexes

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Properties of potential games Admit a pure strategy Nash equilibrium Best-response dynamics converge to NE Price of stability is bounded

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Existence of a pure NE Theorem: every potential game admits a pure NE Proof: we show that the profile minimizing is a NE – Let s be pure profile minimizing – Suppose it is not a NE, so i can improve by deviating to a new profile s – (s) - (s) = c i (s) – c i (s) < 0 – Thus (s) < (s), contradicting s minimizes More generally, the set of pure-strategy Nash equilibria is exactly the set of local minima of the potential function –Local minimum = no player can improve the potential function by herself

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Best-response dynamics converge to a NE Best-response dynamics: –Start with any strategy profile –If a player is not best-responding, switch that players strategy to a better response (must decrease potential) –Terminate when no player can improve (thus a NE) –Alas, no guarantee on the convergence rate

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17 Multicast (and non-multicast) Routing Multicast routing: Given a directed graph G = (V, E) with edge costs c e 0, a source node s, and k agents located at terminal nodes t 1, …, t k. Agent j must construct a path P j from node s to its terminal t j. Routing: Given a directed graph G = (V, E) with edge costs c e 0, and k agents seeking to connect s j,t j pairs, Agent j must construct a path P j from node sj to its terminal t j. Fair share: If x agents use edge e, they each pay c e / x. Slides on cost sharing based on slides by Kevin Wayne Pearson-Addison Wesley. All rights reserved.

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18 Multicast Routing : Shapley price sharing (fair cost sharing) outer 2 middle 4 1 pays /2 + 1 middle4 1 outer middle outer 8 2 pays 8 5/ s t1t1 v t2t

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19 Nash Equilibrium Example: – Two agents start with outer paths. – Agent 1 has no incentive to switch paths (since ). – Once this happens, agent 1 prefers middle path (since 4 > 5/2 + 1). – Both agents using middle path is a Nash equilibrium. s t1t1 v t2t

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Recall price of anarchy and stability Price of anarchy (poa)=cost of worst NE / cost of OPT Price of stability (pos)=cost of best NE / cost of OPT

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Socially Optimum Social optimum: Minimizes total costs of all agents. Observation: In general, there can be many Nash equilibria. Even when it is unique, it does not necessarily equal the social optimum. s t1t1 v t2t Social optimum = 7 Unique Nash equilibrium = 8 s t k 1 + Social optimum = 1 + Nash equilibrium A = 1 + Nash equilibrium B = k k agents pos=1, poa=kpos=poa=8/7

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Price of anarchy Claim: poa k Proof: – Let N be the worst NE – Suppose by contradiction c(N) > k OPT – Then, there exists a player i s.t. c i (N) > OPT – But i can deviate to OPT (by paying OPT alone), contradicting that N is a NE Note: bound is tight (lower bound in prev. slide)

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23 Price of Stability What is price of stability in multicast routing? Lower bound of log k: s t2t2 t3t3 tktk t1t /2 1/31/k /2 + … + 1/k Social optimum: Everyone Takes bottom paths. Unique Nash equilibrium: Everyone takes top paths. Price of stability: H(k) / (1 + ). upper bound will follow..

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24 Finding a potential function Attempt 1: Let (s) = j=1 k cost(t i ) be the potential function. A problem: The potential might increase when some agent improve. Example: When all 3 agents use the right path, each pays 4/3 and the potential (total cost) is 4. After one agent moves to the left path the potential increases to 5. s t agents

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25 Finding a potential function Consider a set of paths P 1, …, P k. – Let x e denote the number of paths that use edge e. – Let (P 1, …, P k ) = e E c e · H(x e ) be a potential function. – Consider agent j switching from path P j to path P j '. – Change in agent js cost: H(0) = 0,

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26 Potential function – increases by – decreases by – Thus, net change in is identical to net change in player js cost

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27 Bounding the Price of Stability Claim: Let C(P 1, …, P k ) denote the total cost of selecting paths P 1, …, P k. For any set of paths P 1, …, P k, we have Proof: Let x e denote the number of paths containing edge e. – Let E + denote set of edges that belong to at least one of the paths.

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28 Bounding the Price of Stability Theorem: There is a Nash equilibrium for which the total cost to all agents exceeds that of the social optimum by at most a factor of H(k) (i.e., price of stability H(k)). Proof: – Let (P 1 *, …, P k * ) denote set of socially optimal paths. – Run best-response dyn algorithm starting from P *. – Since is monotone decreasing (P 1, …, P k ) (P 1 *, …, P k * ). previous claim applied to P previous claim applied to P*

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Local search and PLS (polynomial local search) Local optimization problem: find a local optimum (i.e., no improvement in neighborhood Local optimization problem is in PLS if exists an oracle that for every instance and solution s decides if s is a local optimum; if not returns a better solution s in neighborhood of s Finding NE in potential games is in PLS – Define neighborhood of a profile s to be profiles obtained by deviation of a single player – s is local optimum for c(s) = (s) iff s is a NE

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Congestion games [Rosenthal 73] There is a set of resources R Agent is set of actions (pure strategies) A i is a subset of 2 R, representing which subsets of resources would meet her needs –Note: different agents may need different resources There exist cost functions c r : {1, 2, 3, …} such that agent is cost for a = (a i, a -i ) is Σ r a i c r (n r (a)) –n r (a) is the number of agents that chose r as one of their resources in the profile a

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Example: multicast routing Resources = edges Each resource r has a cost c r Player 1s action set: {{A}, {C,D}} Player 2s action set: {{B}, {C,E}} For all resources r, c r (n r (a)) = c r / n r (a) s t1t1 v t2t2 E A 4 C D B

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Every congestion game is an exact potential game Use potential (a) = Σ r Σ 1 i nr(a) c r (i) –One interpretation: the sum of the costs that the agents would have received if each agent were unaffected by all later agents Why is this a correct potential function? Suppose an agent changes action: stop using some resources (R-), start using others (R+) increase in the agents cost equals Σ r R+ c r (n r (a) + 1) - Σ r R- c r (n r (a)) This is exactly the change in the potential function above –Conclusion: congestion games are exact potential games TexPoint Display

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Computational Game Theory: Network Creation Game Arbitrary Payments (Not a congestion game) Credit to Slides To Eva Tardos Modified/Corrupted/Added to by Michal Feldman and Amos Fiat

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Network Creation Game – Arbitrary Cost partition G = (V,E) is an undirected graph with edge costs c(e). There are k players. Each player i has a source s i and a sink t i he wants to have connected. s1s1 t3t3 t1t1 t2t2 s2s2 s3s3

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Model (cont) Player i picks payment p i (e) for each edge e. e is bought if total payments c(e). Note: any player can use bought edges s1s1 t3t3 t1t1 t2t2 s2s2 s3s3

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The Game Each player i has only 2 concerns: 1) Must be a bought path from s i to t i s1s1 t3t3 t1t1 t2t2 s2s2 s3s3 bought edges

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The Game Each player i has only 2 concerns: 1) Must be a bought path from s i to t i 2) Given this requirement, i wants to pays as little as possible. s1s1 t3t3 t1t1 t2t2 s2s2 s3s3

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Nash Equilibrium A Nash Equilibium (NE) is set of payments for players such that no player wants to deviate. Note: player i doesnt care whether other players connect. s1s1 t3t3 t1t1 t2t2 s2s2 s3s3

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An Example One NE: Each player pays 1/k to top edge. Another NE: Each player pays 1 to bottom edge. Note: No notion of fairness; many NE that pay unevenly for the cheap edge. s 1 …s k t 1 …t k c(e) = 1 c(e) = k

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Three Observations 1) The bought edges in a NE form a forest. 2) Players only contribute to edges on their s i -t i path in this forest. 3) The total payment for any edge e is either c(e) or 0.

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Example 2: No Nash s1s1 t1t1 t2t2 s2s2 all edges cost 1 a b c d

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Example 2: No Nash s1s1 t1t1 t2t2 s2s2 We know that any NE must be a tree: WLOG assume the tree is a,b,c. all edges cost 1 a b c d

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Example 2: No Pure Nash s1s1 t1t1 t2t2 s2s2 We know that any NE must be a tree: WLOG assume the tree is a,b,c. Only player 1 can contribute to a. all edges cost 1 a b c d

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Example 2: No Pure Nash s1s1 t1t1 t2t2 s2s2 We know that any NE must be a tree: WLOG assume the tree is a,b,c. Only player 1 can contribute to a. Only player 2 can contribute to c. all edges cost 1 a b c d

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Example 2: No Pure Nash s1s1 t1t1 t2t2 s2s2 We know that any NE must be a tree: WLOG assume the tree is a,b,c. Only player 1 can contribute to a. Only player 2 can contribute to c. Neither player can contribute to b, since d is tempting deviation. all edges cost 1 a b c d

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