Guest Lecture: Prof. Allan Borodin Game Theory (Cost sharing & - - PowerPoint PPT Presentation

CSC304 Lecture 4 Guest Lecture: Prof. Allan Borodin Game Theory

(Cost sharing & congestion games, Potential function, Braess’ paradox)

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Recap

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• Finding pure and mixed Nash equilibria

➢ Best response diagrams ➢ Indifference principle

• Price of Anarchy (PoA) and Price of Stability (PoS)

➢ How does the Nash equilibrium compare to the social

• ptimum, in the worst case and in the best case?

Cost Sharing Game

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• 𝑜 players on directed weighted graph 𝐻
• Player 𝑗

➢ Wants to go from 𝑡𝑗 to 𝑢𝑗 ➢ Strategy set 𝑇𝑗 = {directed 𝑡𝑗 → 𝑢𝑗 paths} ➢ Denote his chosen path by 𝑄𝑗 ∈ 𝑇𝑗

• Each edge 𝑓 has cost 𝑑𝑓 (weight)

➢ Cost is split among all players taking edge 𝑓 ➢ That is, among all players 𝑗 with 𝑓 ∈ 𝑄𝑗

1 1 1 1 𝑡1 𝑢1 10 𝑡2 𝑢2 10 10

Cost Sharing Game

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• Given strategy profile 𝑄, cost 𝑑𝑗 𝑄 to player 𝑗

is sum of his costs for edges 𝑓 ∈ 𝑄𝑗

• Social cost 𝐷 𝑄 = σ𝑗 𝑑𝑗 𝑄

➢ Note that 𝐷 𝑄 = σ𝑓∈𝐹 𝑄 𝑑𝑓, where

𝐹(𝑄)={edges taken in 𝑄 by at least one player}

• In the example on the right:

➢ What if both players take the direct paths? ➢ What if both take the middle paths? ➢ What if only one player takes the middle path while

the other takes the direct path?

1 1 1 1 𝑡1 𝑢1 10 𝑡2 𝑢2 10 10

Cost Sharing: Simple Example

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• Example on the right: 𝑜 players
• Two pure NE

➢ All taking the n-edge: social cost = 𝑜 ➢ All taking the 1-edge: social cost = 1

• Also the social optimum
• In this game, price of anarchy ≥ 𝑜
• We can show that for all cost sharing

games, price of anarchy ≤ 𝑜

s t 𝑜 1

Cost Sharing: PoA

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• Theorem: The price of anarchy of a cost sharing

game is at most 𝑜.

• Proof:

➢ Suppose the social optimum is (𝑄

1 ∗, 𝑄2 ∗, … , 𝑄 𝑜 ∗), in which

the cost to player 𝑗 is 𝑑𝑗

∗.

➢ Take any NE with cost 𝑑𝑗 to player 𝑗. ➢ Let 𝑑𝑗

′ be his cost if he switches to 𝑄𝑗 ∗.

➢ NE ⇒ 𝑑𝑗

′ ≥ 𝑑𝑗

(Why?)

➢ But : 𝑑𝑗

′ ≤ 𝑜 ⋅ 𝑑𝑗 ∗ (Why?)

➢ 𝑑𝑗 ≤ 𝑜 ⋅ 𝑑𝑗

∗ for each 𝑗 ⇒ no worse than 𝑜 × optimum

∎

Cost Sharing

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• Price of anarchy

➢ All cost-sharing games: PoA ≤ 𝑜 ➢ Example game where PoA = 𝑜

• Price of stability? Later…
• Both examples we saw had

pure Nash equilibria

➢ What about more complex

games, like the one on the right? 10 players: 𝐹 → 𝐷 27 players: 𝐶 → 𝐸 19 players: 𝐷 → 𝐸 E D A

7

B C

60 12 32 10 20

Good News

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• Theorem: All cost sharing games have a pure Nash

eq.

• Proof:

➢ Via “potential function” argument

Step 1: Define Potential Fn

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• Potential function: Φ ∶ ς𝑗 𝑇𝑗 → ℝ+

➢ For all pure strategy profiles 𝑄 = 𝑄

1, … , 𝑄 𝑜 ∈ ς𝑗 𝑇𝑗, …

➢ all players 𝑗, and … ➢ all alternative strategies 𝑄𝑗

′ ∈ 𝑇𝑗 for player 𝑗…

𝑑𝑗 𝑄𝑗

′, 𝑄−𝑗 − 𝑑𝑗 𝑄 = Φ 𝑄𝑗 ′, 𝑄−𝑗 − Φ 𝑄

• When a single player changes his strategy, the

change in his cost is equal to the change in the potential function

➢ Do not care about the changes in the costs to others

Step 2: Potential Fn → pure Nash Eq

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• All games that admit a potential function have a

pure Nash equilibrium. Why?

➢ Think about 𝑄 that minimizes the potential function. ➢ What happens when a player deviates?

• If his cost decreases, the potential function value must also

decrease.

• 𝑄 already minimizes the potential function value.
• Pure strategy profile minimizing potential function

is a pure Nash equilibrium.

Step 3: Potential Fn for Cost-Sharing

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• Recall: 𝐹(𝑄) = {edges taken in 𝑄 by at least one player}
• Let 𝑜𝑓(𝑄) be the number of players taking 𝑓 in 𝑄

Φ 𝑄 = ෍

𝑓∈𝐹(𝑄)

෍

𝑙=1 𝑜𝑓(𝑄) 𝑑𝑓

𝑙

• Note: The cost of edge 𝑓 to each player taking 𝑓 is

𝑑𝑓/𝑜𝑓(𝑄). But the potential function includes all fractions: 𝑑𝑓/1, 𝑑𝑓/2, …, 𝑑𝑓/𝑜𝑓 𝑄 .

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Φ 𝑄 = ෍

𝑓∈𝐹(𝑄)

෍

𝑙=1 𝑜𝑓(𝑄) 𝑑𝑓

𝑙

• Why is this a potential function?

➢ If a player changes path, he pays

𝑑𝑓 𝑜𝑓 𝑄 +1 for each new

edge 𝑓, gets back

𝑑𝑔 𝑜𝑔 𝑄 for each old edge 𝑔.

➢ This is precisely the change in the potential function too. ➢ So Δ𝑑𝑗 = ΔΦ.

∎

Step 3: Potential Fn for Cost-Sharing

Potential Minimizing Eq.

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• There could be multiple pure and multiple mixed

Nash equilibria

➢ Pure Nash equilibria are “local minima” of the potential

function.

➢ A single player deviating should not decrease the

function value.

• Minimizing the potential function just gives one of

the pure Nash equilibria

➢ Is this equilibrium special? Yes!

Potential Minimizing Eq.

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෍

𝑓∈𝐹(𝑄)

𝑑𝑓 ≤ Φ 𝑄 = ෍

𝑓∈𝐹(𝑄)

෍

𝑙=1 𝑜𝑓(𝑄) 𝑑𝑓

𝑙 ≤ ෍

𝑓∈𝐹(𝑄)

𝑑𝑓 ∗ ෍

𝑙=1 𝑜 1

𝑙

Social cost

∀𝑄, 𝐷 𝑄 ≤ Φ 𝑄 ≤ 𝐷 𝑄 ∗ 𝐼 𝑜 𝐷 𝑄∗ ≤ Φ 𝑄∗ ≤ Φ 𝑃𝑄𝑈 ≤ 𝐷 𝑃𝑄𝑈 ∗ 𝐼(𝑜)

Harmonic function 𝐼(𝑜) = σ𝑙=1

𝑜

1/𝑜 = 𝑃(log 𝑜) Potential minimizing eq. Social optimum

Potential Minimizing Eq.

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• Potential minimizing equilibrium gives 𝑃(log 𝑜)

approximation to the social optimum

➢ Price of stability is 𝑃(log 𝑜) ➢ Compare to the price of anarchy, which can be 𝑜

Congestion Games

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• Generalize cost sharing games
• 𝑜 players, 𝑛 resources (e.g., edges)
• Each player 𝑗 chooses a set of resources 𝑄𝑗 (e.g.,

𝑡𝑗 → 𝑢𝑗 paths)

• When 𝑜𝑘 player use resource 𝑘, each of them get a

cost 𝑔

𝑘(𝑜𝑘)

• Cost to player is the sum of costs of resources used

Congestion Games

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• Theorem [Rosenthal 1973]: Every congestion game

is a potential game.

• Potential function:

Φ 𝑄 = ෍

𝑘∈𝐹(𝑄)

෍

𝑙=1 𝑜𝑘 𝑄

𝑔

𝑘 𝑙

• Theorem [Monderer and Shapley 1996]: Every

potential game is equivalent to a congestion game.

Potential Functions

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• Potential functions are useful for deriving various

results

➢ E.g., used for analyzing amortized complexity of

algorithms

• Bad news: Finding a potential function that works

may be hard.

The Braess’ Paradox

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• In cost sharing, 𝑔

𝑘 is decreasing

➢ The more people use a resource, the less the cost to each.

• 𝑔

𝑘 can also be increasing

➢ Road network, each player going from home to work ➢ Uses a sequence of roads ➢ The more people on a road, the greater the congestion,

the greater the delay (cost)

• Can lead to unintuitive phenomena

The Braess’ Paradox

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• Parkes-Seuken Example:

➢ 2000 players want to go from 1 to 4 ➢ 1 → 2 and 3 → 4 are “congestible” roads ➢ 1 → 3 and 2 → 4 are “constant delay” roads

1 4 2 3

The Braess’ Paradox

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• Pure Nash equilibrium?

➢ 1000 take 1 → 2 → 4, 1000 take 1 → 3 → 4 ➢ Each player has cost 10 + 25 = 35 ➢ Anyone switching to the other creates a greater

congestion on it, and faces a higher cost 1 4 2 3

The Braess’ Paradox

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• What if we add a zero-cost connection 2 → 3?

➢ Intuitively, adding more roads should only be helpful ➢ In reality, it leads to a greater delay for everyone in the

unique equilibrium! 1 4 2 3

𝑑23 𝑜23 = 0

The Braess’ Paradox

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• Nobody chooses 1 → 3 as 1 → 2 → 3 is better

irrespective of how many other players take it

• Similarly, nobody chooses 2 → 4
• Everyone takes 1 → 2 → 3 → 4, faces delay = 40!

1 4 2 3

𝑑23 𝑜23 = 0

The Braess’ Paradox

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• In fact, what we showed is:

➢ In the new game, 1 → 2 → 3 → 4 is a strictly dominant

strategy for each firm! 1 4 2 3

𝑑23 𝑜23 = 0