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Congestion Games Example: Network Routing x s t 10 n = 10 players want to travel from s to t Each edge e is labeled with its (flow-dependent) delay function d e ( x ) Question: What are the pure Nash equilibria here? Georgios

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Congestion Games

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Example: Network Routing

s t x 10

  • n = 10 players want to travel from s to t
  • Each edge e is labeled with its (flow-dependent)

delay function de(x) Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 2

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SLIDE 3

Example: Network Routing

s t x 10

  • n = 10 players want to travel from s to t
  • Each edge e is labeled with its (flow-dependent)

delay function de(x) Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 2

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SLIDE 4

Example: Network Routing

s t x 10

  • n = 10 players want to travel from s to t
  • Each edge e is labeled with its (flow-dependent)

delay function de(x) Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 2

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SLIDE 5

Example: Network Routing

s t x 10

  • n = 10 players want to travel from s to t
  • Each edge e is labeled with its (flow-dependent)

delay function de(x) Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 2

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SLIDE 6

Example: Network Routing

s t x 10 10

  • n = 10 players want to travel from s to t
  • Each edge e is labeled with its (flow-dependent)

delay function de(x) Question: What are the pure Nash equilibria here? (10, 0)

Georgios Amanatidis Social Networks & Online Markets 2020 2

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SLIDE 7

Example: Network Routing

s t x 10 9 1

  • n = 10 players want to travel from s to t
  • Each edge e is labeled with its (flow-dependent)

delay function de(x) Question: What are the pure Nash equilibria here? (10, 0), (9, 1)

Georgios Amanatidis Social Networks & Online Markets 2020 2

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SLIDE 8

Example: The El Farol Bar Problem

100 people consider visiting the El Farol Bar on a Thursday night. They all have identical preferences:

  • If 60 or more people show up, it’s nicer to be at home.
  • If fewer than 60 people show up, it’s nicer to be at the bar.

Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 3

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SLIDE 9

Example: The El Farol Bar Problem

100 people consider visiting the El Farol Bar on a Thursday night. They all have identical preferences:

  • If 60 or more people show up, it’s nicer to be at home.
  • If fewer than 60 people show up, it’s nicer to be at the bar.

Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 3

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SLIDE 10

Example: The El Farol Bar Problem

100 people consider visiting the El Farol Bar on a Thursday night. They all have identical preferences:

  • If 60 or more people show up, it’s nicer to be at home.
  • If fewer than 60 people show up, it’s nicer to be at the bar.

Question: What are the pure Nash equilibria here?

Georgios Amanatidis Social Networks & Online Markets 2020 3

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Example: The El Farol Bar Problem

100 people consider visiting the El Farol Bar on a Thursday night. They all have identical preferences:

  • If 60 or more people show up, it’s nicer to be at home.
  • If fewer than 60 people show up, it’s nicer to be at the bar.

Question: What are the pure Nash equilibria here? 59 people at the bar.

Georgios Amanatidis Social Networks & Online Markets 2020 3

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Example: Congestion Game

A congestion game is a tuple N, R, A, d , where

  • N = {1, . . . , n} is a finite set of players
  • R = {1, . . . , m} is a finite set of resources
  • A = A1 × · · · × An is a finite set of action profiles a = (a1, . . . , an),

with Ai ⊆ 2R \ {∅} being the set of actions available to player i

  • d = (d1, . . . , dm) is a vector of delay functions dr : N → R.

Goal: player i ∈ N chooses a subset of resources ai ∈ Ai Given an action profile a = (a1, . . . , an), the cost of player i is ci(a) =

  • r∈ai

dr(nr(a)) where nr(a) = |{i ∈ N : r ∈ ai}|. Note: ui(a) = −ci(a) for every i here.

Georgios Amanatidis Social Networks & Online Markets 2020 4

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Example: Congestion Game

A congestion game is a tuple N, R, A, d , where

  • N = {1, . . . , n} is a finite set of players
  • R = {1, . . . , m} is a finite set of resources
  • A = A1 × · · · × An is a finite set of action profiles a = (a1, . . . , an),

with Ai ⊆ 2R \ {∅} being the set of actions available to player i

  • d = (d1, . . . , dm) is a vector of delay functions dr : N → R.

Goal: player i ∈ N chooses a subset of resources ai ∈ Ai Given an action profile a = (a1, . . . , an), the cost of player i is ci(a) =

  • r∈ai

dr(nr(a)) where nr(a) = |{i ∈ N : r ∈ ai}|. Note: ui(a) = −ci(a) for every i here.

Georgios Amanatidis Social Networks & Online Markets 2020 4

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Example: Congestion Game

A congestion game is a tuple N, R, A, d , where

  • N = {1, . . . , n} is a finite set of players
  • R = {1, . . . , m} is a finite set of resources
  • A = A1 × · · · × An is a finite set of action profiles a = (a1, . . . , an),

with Ai ⊆ 2R \ {∅} being the set of actions available to player i

  • d = (d1, . . . , dm) is a vector of delay functions dr : N → R.

Goal: player i ∈ N chooses a subset of resources ai ∈ Ai Given an action profile a = (a1, . . . , an), the cost of player i is ci(a) =

  • r∈ai

dr(nr(a)) where nr(a) = |{i ∈ N : r ∈ ai}|. Note: ui(a) = −ci(a) for every i here.

Georgios Amanatidis Social Networks & Online Markets 2020 4

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Modelling the Examples

Congestion Game:

  • players N = {1, 2, . . . , 10}
  • resources R = {↑, ↓}
  • action spaces Ai = {{↑}, {↓}} representing the two routes
  • delay functions d↑ : x → x and d↓ : x → 10

El Farol Bar Problem:

  • players N = {1, 2, . . . , 100}
  • resources R = {, ↸1, ↸2, . . . , ↸100}
  • action spaces Ai = {{}, {↸i}}
  • delay functions d : x → 1x60 and d↸i : x → 1

2

Remark: neither example uses the full generality of congestion games (actions correspond to singleton resource sets only)

Georgios Amanatidis Social Networks & Online Markets 2020 5

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Modelling the Examples

Congestion Game:

  • players N = {1, 2, . . . , 10}
  • resources R = {↑, ↓}
  • action spaces Ai = {{↑}, {↓}} representing the two routes
  • delay functions d↑ : x → x and d↓ : x → 10

El Farol Bar Problem:

  • players N = {1, 2, . . . , 100}
  • resources R = {, ↸1, ↸2, . . . , ↸100}
  • action spaces Ai = {{}, {↸i}}
  • delay functions d : x → 1x60 and d↸i : x → 1

2

Remark: neither example uses the full generality of congestion games (actions correspond to singleton resource sets only)

Georgios Amanatidis Social Networks & Online Markets 2020 5

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Modelling the Examples

Congestion Game:

  • players N = {1, 2, . . . , 10}
  • resources R = {↑, ↓}
  • action spaces Ai = {{↑}, {↓}} representing the two routes
  • delay functions d↑ : x → x and d↓ : x → 10

El Farol Bar Problem:

  • players N = {1, 2, . . . , 100}
  • resources R = {, ↸1, ↸2, . . . , ↸100}
  • action spaces Ai = {{}, {↸i}}
  • delay functions d : x → 1x60 and d↸i : x → 1

2

Remark: neither example uses the full generality of congestion games (actions correspond to singleton resource sets only)

Georgios Amanatidis Social Networks & Online Markets 2020 5

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Existence of Pure Nash Equilibria

Good news:

Theorem (Rosenthal, 1973)

Every congestion game has at least one pure Nash equilibrium.

R.W. Rosenthal. A Class of Games Possessing Pure-Strategy Nash Equilibria. Inter- national Journal of Game Theory, 2(1):65–67, 1973.

Georgios Amanatidis Social Networks & Online Markets 2020 6

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Inefficiency of Equilibria

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Inefficiency of Equilibria

Prisoner’s Dilemma: Nash equilibria might be suboptimal! Question: Can we quantify how “bad” Nash equilibria are? Social cost: define the social cost of strategy profile a as SC(a) =

  • i∈N

ci(a) → let a∗ be a strategy profile minimizing SC(·) (social optimum) → a∗ is best possible outcome if one could coordinate the players Note: consider social welfare SW =

i ui for utility maximizing players

Idea: measure worst case loss in social cost due to lack of coordination POA = max

a∈PNE

SC(a) SC(a∗) → termed the price of anarchy by Koutsoupias & Papadimitriou (1999)

Georgios Amanatidis Social Networks & Online Markets 2020 14

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Inefficiency of Equilibria

Prisoner’s Dilemma: Nash equilibria might be suboptimal! Question: Can we quantify how “bad” Nash equilibria are? Social cost: define the social cost of strategy profile a as SC(a) =

  • i∈N

ci(a) → let a∗ be a strategy profile minimizing SC(·) (social optimum) → a∗ is best possible outcome if one could coordinate the players Note: consider social welfare SW =

i ui for utility maximizing players

Idea: measure worst case loss in social cost due to lack of coordination POA = max

a∈PNE

SC(a) SC(a∗) → termed the price of anarchy by Koutsoupias & Papadimitriou (1999)

Georgios Amanatidis Social Networks & Online Markets 2020 14

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Inefficiency of Equilibria

Prisoner’s Dilemma: Nash equilibria might be suboptimal! Question: Can we quantify how “bad” Nash equilibria are? Social cost: define the social cost of strategy profile a as SC(a) =

  • i∈N

ci(a) → let a∗ be a strategy profile minimizing SC(·) (social optimum) → a∗ is best possible outcome if one could coordinate the players Note: consider social welfare SW =

i ui for utility maximizing players

Idea: measure worst case loss in social cost due to lack of coordination POA = max

a∈PNE

SC(a) SC(a∗) → termed the price of anarchy by Koutsoupias & Papadimitriou (1999)

Georgios Amanatidis Social Networks & Online Markets 2020 14

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SLIDE 23

Inefficiency of Equilibria

Prisoner’s Dilemma: Nash equilibria might be suboptimal! Question: Can we quantify how “bad” Nash equilibria are? Social cost: define the social cost of strategy profile a as SC(a) =

  • i∈N

ci(a) → let a∗ be a strategy profile minimizing SC(·) (social optimum) → a∗ is best possible outcome if one could coordinate the players Note: consider social welfare SW =

i ui for utility maximizing players

Idea: measure worst case loss in social cost due to lack of coordination POA = max

a∈PNE

SC(a) SC(a∗) → termed the price of anarchy by Koutsoupias & Papadimitriou (1999)

Georgios Amanatidis Social Networks & Online Markets 2020 14

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SLIDE 24

Inefficiency of Equilibria

Prisoner’s Dilemma: Nash equilibria might be suboptimal! Question: Can we quantify how “bad” Nash equilibria are? Social cost: define the social cost of strategy profile a as SC(a) =

  • i∈N

ci(a) → let a∗ be a strategy profile minimizing SC(·) (social optimum) → a∗ is best possible outcome if one could coordinate the players Note: consider social welfare SW =

i ui for utility maximizing players

Idea: measure worst case loss in social cost due to lack of coordination POA = max

a∈PNE

SC(a) SC(a∗) → termed the price of anarchy by Koutsoupias & Papadimitriou (1999)

Georgios Amanatidis Social Networks & Online Markets 2020 14

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SLIDE 25

Inefficiency of Equilibria

Prisoner’s Dilemma: Nash equilibria might be suboptimal! Question: Can we quantify how “bad” Nash equilibria are? Social cost: define the social cost of strategy profile a as SC(a) =

  • i∈N

ci(a) → let a∗ be a strategy profile minimizing SC(·) (social optimum) → a∗ is best possible outcome if one could coordinate the players Note: consider social welfare SW =

i ui for utility maximizing players

Idea: measure worst case loss in social cost due to lack of coordination POA = max

a∈PNE

SC(a) SC(a∗) → termed the price of anarchy by Koutsoupias & Papadimitriou (1999)

Georgios Amanatidis Social Networks & Online Markets 2020 14

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Network Routing Games n = 10 Si = {↑, ↓} s t d↑(x) = x d↓(x) = 10

Georgios Amanatidis Social Networks & Online Markets 2020 15

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Network Routing Games n = 10 Si = {↑, ↓} s t d↑(x) = x d↓(x) = 10 10 Nash equilibrium: SC(s) = 10 · 10 = 100

Georgios Amanatidis Social Networks & Online Markets 2020 15

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Network Routing Games n = 10 Si = {↑, ↓} s t d↑(x) = x d↓(x) = 10 5 5 social optimum: SC(s∗) = 5 · 5 + 5 · 10 = 75

Georgios Amanatidis Social Networks & Online Markets 2020 15

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Network Routing Games n = 10 Si = {↑, ↓} s t d↑(x) = x d↓(x) = 10 5 5 price of anarchy:

SC(s) SC(s∗) = 100 75 = 4 3

Georgios Amanatidis Social Networks & Online Markets 2020 15

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Price of Anarchy for Congestion Games

Good news:

  • POA is independent of the network structure
  • POA depends on the class of delay functions

→ POA = 5

2 for affine functions (proof on next slide)

→ POA = O(1) for quadratic, cubic, . . . functions Bad news:

  • POA increases with the “steepness” of delay functions
  • in general: unbounded!
  • even worse: unbounded for practically relevant delay functions

Georgios Amanatidis Social Networks & Online Markets 2020 16

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Price of Anarchy for Congestion Games

Good news:

  • POA is independent of the network structure
  • POA depends on the class of delay functions

→ POA = 5

2 for affine functions (proof on next slide)

→ POA = O(1) for quadratic, cubic, . . . functions Bad news:

  • POA increases with the “steepness” of delay functions
  • in general: unbounded!
  • even worse: unbounded for practically relevant delay functions

Georgios Amanatidis Social Networks & Online Markets 2020 16

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POA for Congestion Games

Theorem

The price of anarchy of congestion games with affine delay functions is 5

2.

Proof: All delay functions are of the form dr(x) = px + q with p, q ∈ Z≥0. Can assume wlog that dr(x) = x ∀r (think about it!). Let a be a PNE and let a∗ be a social optimum. We have SC(a) =

  • i

ci(ai, a−i) ≤

  • i

ci(a∗

i , a−i) =

  • i
  • r∈a∗

i

dr(nr(a∗

i , a−i))

=

  • i
  • r∈a∗

i

nr(a∗

i , a−i) ≤

  • i
  • r∈a∗

i

(nr(a) + 1) =

  • r∈R

(nr(a) + 1)

  • i:r∈a∗

i

1 =

  • r∈R

nr(a∗)(nr(a) + 1) ≤ 5 3

  • r∈R

(nr(a∗))2 + 1 3

  • r∈R

(nr(a))2 = 5 3SC(a∗) + 1 3SC(a)

Georgios Amanatidis Social Networks & Online Markets 2020 17

See proof at the end!

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Tight Example

Instance:

  • N = [3]
  • R = R1 ∪ R2, where R1 = {h1, h2, h3} and R2 = {g1, g2, g3}
  • delay function dr(x) = x for every r ∈ R
  • each player i has two strategies: {hi, gi} and {hi−1, hi+1, gi+1}

(modulo 3). Social optimum: every player selects his first strategy: SC(a∗) = 6 Nash equilibrium: every player chooses his second strategy: SC(a) =

i ci(a) = 3 · 5 = 15

Georgios Amanatidis Social Networks & Online Markets 2020 18

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Tight Example

Instance:

  • N = [3]
  • R = R1 ∪ R2, where R1 = {h1, h2, h3} and R2 = {g1, g2, g3}
  • delay function dr(x) = x for every r ∈ R
  • each player i has two strategies: {hi, gi} and {hi−1, hi+1, gi+1}

(modulo 3). Social optimum: every player selects his first strategy: SC(a∗) = 6 Nash equilibrium: every player chooses his second strategy: SC(a) =

i ci(a) = 3 · 5 = 15

Georgios Amanatidis Social Networks & Online Markets 2020 18

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Tight Example

Instance:

  • N = [3]
  • R = R1 ∪ R2, where R1 = {h1, h2, h3} and R2 = {g1, g2, g3}
  • delay function dr(x) = x for every r ∈ R
  • each player i has two strategies: {hi, gi} and {hi−1, hi+1, gi+1}

(modulo 3). Social optimum: every player selects his first strategy: SC(a∗) = 6 Nash equilibrium: every player chooses his second strategy: SC(a) =

i ci(a) = 3 · 5 = 15

Georgios Amanatidis Social Networks & Online Markets 2020 18

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Unbounded POA

In general, POA for congestion games is unbounded. Question: Can we improve the POA? → natural idea: infrastructure improvement (add new edges)

Georgios Amanatidis Social Networks & Online Markets 2020 19

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Unbounded POA

In general, POA for congestion games is unbounded. Question: Can we improve the POA? → natural idea: infrastructure improvement (add new edges)

Georgios Amanatidis Social Networks & Online Markets 2020 19

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SLIDE 38

Unbounded POA

In general, POA for congestion games is unbounded. Question: Can we improve the POA? → natural idea: infrastructure improvement (add new edges)

Georgios Amanatidis Social Networks & Online Markets 2020 19

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x 5 5

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x 5 5 player’s delay: 15

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x 10

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Braess Paradox

n = 10 s u v t de(x) = x 10 10 x player’s delay: increases(!) to 20 10

Georgios Amanatidis Social Networks & Online Markets 2020 20

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Price of Anarchy

So: In a congestion game, the natural better-response dynamics will always lead us to a pure NE. Nice. But: How good is that equilibrium? Recall our traffic congestion example: s t

x

10

10 people overall top delay = # on route bottom delay = 10 minutes

If x 10 players use top route, social welfare (sum of utilities) is: sw(x) = −[x · x + (10 − x) · 10] = −[x2 − 10x + 100] This function is maximal for x = 5 and minimal for x = 0 and x = 10. In equilibrium, 9 or 10 people will use the bottom route (10 is worse). The so-called price of anarchy of this game is:

sw(10) sw(5) = −100 −75 = 4 3.

Thus: not perfect, but not too bad either (for this example).

Georgios Amanatidis Social Networks & Online Markets 2020 22

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Braess’ Paradox

Something to think about. 10 people have to get from s to t: s u v t x

10 10

x If the delay-free link from u to v is not present:

  • In equilibrium, 5 people will use the top route s–u–t and 5 people the

bottom route s–v–t. Everyone will take 15 minutes. Now, if we add the delay-free link from u to v, this happens:

  • In the worst equilibrium, everyone will take the route s–u–v–t and take

20 minutes! (Other equilibria are only slightly better.)

Georgios Amanatidis Social Networks & Online Markets 2020 23

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Price of Anarchy for Linear/Affine Congestion Games

Based on slides by A. Voudouris

1

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  • Recall that a state 𝒕 = (𝑡1, … , 𝑡𝑜) is an equilibrium if for each player 𝑗

the strategy 𝑡𝑗 minimizes her personal cost, given the strategies of the other players

  • 𝒕−𝑗 = (𝑡1, … , 𝑡𝑗−1, 𝑡𝑗+1, … , 𝑡𝑜)
  • 𝒕 is an equilibrium if for each player i, the strategy 𝑡𝑗 is such that

cost𝑗(𝑧, 𝒕−𝑗) is minimized for 𝑧 = 𝑡𝑗

  • Alternatively, for every possible strategy 𝑧 of player 𝑗:
  • We have one such inequality for every player

A general technique for PoA bounds

cost𝑗 𝑡𝑗, 𝒕−𝑗 ≤ cost𝑗(𝑧, 𝒕−𝑗)

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SLIDE 56
  • By adding these inequalities, we get
  • We can get an upper bound of 𝜇 on the price of anarchy if there exists

a strategy 𝑧𝑗 for every player 𝑗 such that

  • The goal is to pinpoint the strategy 𝑧𝑗 for each player 𝑗, which will

allow us to prove an inequality like this

A general technique for PoA bounds

SC 𝒕 = ෍

𝑗∈𝑂

cost𝑗 𝑡𝑗, 𝒕−𝑗 ≤ ෍

𝑗∈𝑂

cost𝑗 𝑧, 𝒕−𝑗 ෍

𝑗∈𝑂

cost𝑗 𝑧𝑗, 𝒕−𝑗 ≤ 𝜇 ⋅ SC(𝒕𝑃𝑄𝑈)

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SLIDE 57
  • 𝒕 = (𝑡1, … 𝑡𝑜) is an equilibrium state
  • 𝒛 = (𝑧1, … , 𝑧𝑜) is an arbitrary state

Linear congestion games: PoA

Theorem The price of anarchy of linear congestion games is at most 5/2 SC 𝒕 = ෍

𝑗∈𝑂

cost𝑗 𝑡𝑗, 𝒕−𝑗 ≤ ෍

𝑗∈𝑂

cost𝑗 𝑧𝑗, 𝒕−𝑗 = ෍

𝑓∈𝐹

෍

𝑗∈𝑂:𝑓∈𝑧𝑗

(𝑏𝑓 ⋅ 𝑜𝑓 𝑧𝑗, 𝒕−𝑗 + 𝑐𝑓)

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  • (𝑧𝑗, 𝒕−𝑗) differs from 𝒕 = (𝑡𝑗, 𝒕−𝑗) only in the strategy of player 𝑗

⇨ 𝑜𝑓 𝑧𝑗, 𝒕−𝑗 ≤ 𝑜𝑓(𝒕) + 1 for every resource 𝑓 ∈ 𝐹

Linear congestion games: PoA

SC 𝒕 ≤ ෍

𝑓∈𝐹

෍

𝑗∈𝑂:𝑓∈𝑧𝑗

(𝑏𝑓 ⋅ 𝑜𝑓 𝑧𝑗, 𝒕−𝑗 + 𝑐𝑓) ≤ ෍

𝑓∈𝐹

෍

𝑗∈𝑂:𝑓∈𝑧𝑗

(𝑏𝑓 ⋅ (𝑜𝑓 𝒕 + 1) + 𝑐𝑓) = ෍

𝑓∈𝐹

𝑜𝑓 𝒛 (𝑏𝑓 ⋅ (𝑜𝑓 𝒕 + 1) + 𝑐𝑓) = ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 𝑜𝑓 𝒛 ⋅ 𝑜𝑓 𝒕 + 1 + 𝑐𝑓𝑜𝑓 𝒛

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  • (𝑧𝑗, 𝒕−𝑗) differs from 𝒕 = (𝑡𝑗, 𝒕−𝑗) only in the strategy of player 𝑗

⇨ 𝑜𝑓 𝑧𝑗, 𝒕−𝑗 ≤ 𝑜𝑓(𝒕) + 1 for every resource 𝑓 ∈ 𝐹

Linear congestion games: PoA

SC 𝒕 ≤ ෍

𝑓∈𝐹

෍

𝑗∈𝑂:𝑓∈𝑧𝑗

(𝑏𝑓 ⋅ 𝑜𝑓 𝑧𝑗, 𝒕−𝑗 + 𝑐𝑓) ≤ ෍

𝑓∈𝐹

෍

𝑗∈𝑂:𝑓∈𝑧𝑗

(𝑏𝑓 ⋅ (𝑜𝑓 𝒕 + 1) + 𝑐𝑓) = ෍

𝑓∈𝐹

𝑜𝑓 𝒛 (𝑏𝑓 ⋅ (𝑜𝑓 𝒕 + 1) + 𝑐𝑓) = ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 𝑜𝑓 𝒛 ⋅ 𝑜𝑓 𝒕 + 1 + 𝑐𝑓𝑜𝑓 𝒛

slide-60
SLIDE 60

SC 𝒕 ≤ ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 𝑜𝑓 𝒛 𝑜𝑓 𝒕 + 1 + 𝑐𝑓𝑜𝑓 𝒛 ≤ ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 1 3 5𝑜𝑓 𝒛 2 + 𝑜𝑓 𝒕 2 + 𝑐𝑓𝑜𝑓 𝒛 = ෍

𝑓∈𝐹

5 3 𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛 + 1 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒕 2 ≤ 5 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛 + 1 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒕 2 + 𝑐𝑓𝑜𝑓 𝒕

  • For every pair of integers 𝛿, 𝜀 ≥ 0: 𝛿 𝜀 + 1 ≤ 1

3 (5𝛿2 + 𝜀2)

  • Set 𝛿 = 𝑜𝑓 𝒛 and 𝜀 = 𝑜𝑓(𝒕)

Linear congestion games: PoA

slide-61
SLIDE 61
  • For every pair of integers 𝛿, 𝜀 ≥ 0: 𝛿 𝜀 + 1 ≤ 1

3 (5𝛿2 + 𝜀2)

  • Set 𝛿 = 𝑜𝑓 𝒛 and 𝜀 = 𝑜𝑓(𝒕)

Linear congestion games: PoA

SC 𝒕 ≤ ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 𝑜𝑓 𝒛 𝑜𝑓 𝒕 + 1 + 𝑐𝑓𝑜𝑓 𝒛 ≤ ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 1 3 5𝑜𝑓 𝒛 2 + 𝑜𝑓 𝒕 2 + 𝑐𝑓𝑜𝑓 𝒛 = ෍

𝑓∈𝐹

5 3 𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛 + 1 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒕 2 ≤ 5 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛 + 1 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒕 2 + 𝑐𝑓𝑜𝑓 𝒕

slide-62
SLIDE 62
  • For every pair of integers 𝛿, 𝜀 ≥ 0: 𝛿 𝜀 + 1 ≤ 1

3 (5𝛿2 + 𝜀2)

  • Set 𝛿 = 𝑜𝑓 𝒛 and 𝜀 = 𝑜𝑓(𝒕)

Linear congestion games: PoA

SC 𝒕 ≤ ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 𝑜𝑓 𝒛 𝑜𝑓 𝒕 + 1 + 𝑐𝑓𝑜𝑓 𝒛 ≤ ෍

𝑓∈𝐹

𝑏𝑓 ⋅ 1 3 5𝑜𝑓 𝒛 2 + 𝑜𝑓 𝒕 2 + 𝑐𝑓𝑜𝑓 𝒛 = ෍

𝑓∈𝐹

5 3 𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛 + 1 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒕 2 ≤ 5 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛 + 1 3 ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒕 2 + 𝑐𝑓𝑜𝑓 𝒕

slide-63
SLIDE 63
  • Since

we obtain

  • Since this holds for any 𝒛, it also holds for 𝒕𝑃𝑄𝑈

▢

Linear congestion games: PoA

SC 𝒕 ≤ 5 3 SC 𝒛 + 1 3 SC 𝒕 ⇒ SC(𝒕) SC(𝒛) ≤ 5 2 SC 𝒛 = ෍

𝑓∈𝐹

𝑏𝑓𝑜𝑓 𝒛 2 + 𝑐𝑓𝑜𝑓 𝒛

slide-64
SLIDE 64
  • To show a lower bound, it suffices to construct a specific instance and

prove that the social cost of the equilibrium is 5/2 times the optimal social cost

Can we do any better?

Theorem The price of anarchy of linear congestion games is at least 5/2

slide-65
SLIDE 65
  • Equilibrium: each player 𝑗 uses two edges to connect 𝑨𝑗 to 𝑢𝑗
  • Players 1 and 2 (red, blue) have cost 3, while players 3 and 4 (green,
  • range) have cost 2
  • By changing to the direct edge, all players would still have the same

cost, so there is no reason for them to deviate

Can we do any better?

𝑨1, 𝑨2 𝑨4, 𝑢2, 𝑢3 𝑨3, 𝑢1, 𝑢4

𝑦 𝑦 𝑦 𝑦

slide-66
SLIDE 66
  • Optimal: each player 𝑗 uses the direct edge between 𝑨𝑗 and 𝑢𝑗
  • All players have cost 1
  • SC(equilibrium) = 10 vs. SC(optimal) = 4 ⇨ PoA = 5/2

▢

Can we do any better?

𝑨1, 𝑨2 𝑨4, 𝑢2, 𝑢3 𝑨3, 𝑢1, 𝑢4

𝑦 𝑦 𝑦 𝑦