CSCI 3210: Computational Game Theory Inefficiency of Equilibria - - PDF document

csci 3210 computational game theory inefficiency of
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CSCI 3210: Computational Game Theory Inefficiency of Equilibria - - PDF document

5/4/17 CSCI 3210: Computational Game Theory Inefficiency of Equilibria & Routing Games Ref: Ch 17, 18 [AGT] Mohammad T . Irfan Email: mirfan@bowdoin.edu Web: www.bowdoin.edu/~mirfan Split or steal game u NE outcome vs. socially


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CSCI 3210: Computational Game Theory

Mohammad T . Irfan Email: mirfan@bowdoin.edu Web: www.bowdoin.edu/~mirfan

Inefficiency of Equilibria & Routing Games

Ref: Ch 17, 18 [AGT]

Split or steal game

u NE outcome vs. socially best/optimal outcome Payoff matrix

Split Steal Split

$33K, $33K $0+fr., $66K

Steal

$66K, $0+fr. $0, $0

Lucy Tony

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Prisoner's "dilemma" game

u Again: NE outcome vs socially optimal outcome Payoff matrix

Not Confess Confess Not Confess

1, 1 10, 0

Confess

0, 10 5, 5

Suspect 1 Suspect 2 Costs (negative of payoffs)

Measuring the inefficiency of NE

u What is the objective function to compare

different outcomes?

u Utilitarian u Egalitarian

u How to deal with multiplicity of NE?

u Inefficiency of which NE? u Price of anarchy vs. price of stability

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Price of Anarchy (PoA)

u PoA =

Worst objective function value of a NE Objective function value of optimal outcome

Price of Stability (PoS)

u PoS =

Best objective function value of a NE Objective function value of optimal outcome

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Example

u Calculate PoA and PoS Payoff matrix

Not Confess Confess Not Confess

1, 1 10, 0

Confess

0, 10 5, 5

Suspect 1 Suspect 2 Costs (negative of payoffs)

Example

u Calculate PoA and PoS Payoff matrix

L R U

21, -1 10, 0

D

100, 10 7, 8

Row player Column player Costs (negative of payoffs)

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PoA vs. PoS

u Consider costs u PoA and PoS will be >= 1 u PoA = PoS when all NE have the same cost

(e.g., unique NE)

u In general, PoA >= PoS u PoA: worst case guarantee in a system of

independent agents

u PoS: measures benefit of a protocol or

proposed outcome

Pigou's example

u Assume one unit of traffic from s to t u Objective function: average travel time u NE: all traffic in lower edge. Avg cost = 1. u Optimal: ½ and ½ . Avg cost = ½ * 1 + ½ * ½. u PoA, PoS = ?

x = amount of traffic on edge

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Routing Games

Model: nonatomic selfish routing

u Multicommodity flow network

u Directed network with multiple (source, sink) pairs u Each (source, sink) pair is called a commodity u ri amount of traffic for each commodity i

u Each edge e has a delay or cost function ce

u Every car going through an edge gets same delay

u Cost of a path = sum of edge costs

u Note: cost doesn't depend on identity of players u Congestion games

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Equilibrium flow

u Let f be a feasible flow (combining all

commodities)

u f is equilibrium flow if

u A commodity uses a path P in its flow within f

=> All detours have higher (or equal) delay

equilibrium flow

More complex graphs

u What is the traffic flow on an (s,t) path?

u Amount of traffic using that path fully (not

partially) to go from s à t u What's the cost (or delay) of a path?

u Total delay on that path– traffic using part of that

path also contribute to delay

s1 s2 t1 t2

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Cost of an (s,t) path

u Sum of edge costs (or delays) on that path

flow (all the flows, not just on P) cost function

  • f edge e

flow on edge e path

Cost of a flow

u Cost (or delay) of a flow

= sum of (cost of each path * flow on that path) = sum of (cost of each edge * flow on that edge)

flow cost function

  • f edge e

flow on edge e

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Example: nonlinear Pigou

u Consider large number p; 1 unit of traffic u Equilibrium: All down. cost = 1. u Optimal: ε up, (1-ε) down

u Cost = ε * 1 + (1-ε) * (1-ε)p à 0

equilibrium flow PoA à +∞

Example: Braess' paradox

u 1 unit of traffic

equilibrium flow ½ ½

Equilibrium cost = ½ * ( ½ + 1) + ½ * (1 + ½) =3/2

PoA = 1

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Braess' paradox

u New super highway between v and w

PoA = 4/3

  • Equilibrium cost = 1 + 1 = 2
  • Optimal: ½ and ½ (ignore superhighway)

Another model: atomic selfish routing

u Restricted version of multicommodity flow

u ri amount of traffic for each commodity i u Each s à t flow must go through a single path

(difference with nonatomic selfish routing) u Each player/commodity chooses the

minimum delay path w.r.t. others' choices

u Equilibrium flow: No better path for any

player

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Drawback of the atomic model

u There may not exist any equilibrium flow u 2 players: both s à t u r1 = 1, r2 = 2 u Proof: If player 2 goes s à t, then player 1's

BR is sàvàwàt, to which player 2's BR is sàwàt. (and so on)

Nonatomic model: equilibrium existence

u Prove that equilibrium flow exists

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Price of anarchy: nonatomic

u Consider linear cost functions only u Pigou's example gives a lower bound of 4/3 u NE: all traffic in lower edge. Avg cost = 1. u Optimal: ½ and ½ .

Avg cost = ½ * 1 + ½ * ½ = 3/4.

u PoA = 1 / (3/4) = 4/3

x = amount of traffic on edge

Surprise!

u Price of anarchy for any nonatomic routing

game with linear costs <= 4/3