Efficiency of equilibria Non-atomic routing games Non-atomic - - PowerPoint PPT Presentation

Algorithmic game theory

Ruben Hoeksma October 29, 2018

Efficiency of equilibria

Non-atomic routing games

Non-atomic routing games

Definition: Non-atomic routing games

◮ Directed graph G = (V , E). Edge e ∈ E has a non-negative,

non-decreasing cost function ce : R+ → R+.

◮ Set N of n commodities. Commodity i: amount of flow ri; origin

& destination (oi, di).

◮ Each player controls an infinitesimal amount of flow. ◮ Each player in commodity i chooses (oi, di)-path Pi. ◮ feasible flow: routes all flow ri from oi to di for all i ∈ N. I.e.,

ri =

Pi∈i fPi, with fPi the amount of flow on Pi. ◮ Flow on edge e is fe = e∈E: e∈Pi fPi. ◮ Cost of path Pi is cPi(f ) = e∈Pi ce(fe).

This is not a simultaneous move game

Non-atomic routing games

Atomic routing game potential function

Φ(f ) =

• e∈E

fe

• j=1

ce(j) .

Non-atomic routing game potential function (not exact)

Φ(f ) =

• e∈E

fe

j=0

ce(j) dj . Potential function is closed and convex, if some δ > 0 amount of users can profitably change paths, then Φ(f ′) − Φ(f ) < 0. ⇒ NE always exists (at argmin Φ(f )). Also called Wardrop Equilibrium.

Example

• d

r = 1 x

x 2 1 2

1 What does a NE look like?

◮ If the upper path has flow, the lower path is at least as expensive. ◮ If the lower path has flow, the upper path is at least as expensive. ◮ If both paths have flow, both have equal cost.

Example

• d

r = 1 x

x 2 1 2

1 What does a NE look like? Let x be the flow on the upper path, then 1 − x is the flow on the bottom path. Then, x +

1 2 = 1 − x 2

+ 1 ⇔

3x 2

= 1

Example

• d

r = 1 x

x 2 1 2

1 What does a NE look like?

Definition Nash/Wardrop equilbrium in non-atomic routing games

A flow f is a Nash/Wardrop equilbrium if for all commodities i ∈ N, any path P ∈ Σi with fP > 0, and any path Q ∈ Σi cP(f ) ≤ cQ(f ) .

Efficiency of equilibria

Braess’s paradox

Braess’s paradox

• d

r = 1 x 1 1 x One commodity with r = 1; origin o; destination d What is the NE flow? ctop(f ) = cbottom(f ) ⇔ ftop = fbottom =

1 2 .

Average cost/travel time: 1.5 (same for all users).

Braess’s paradox

• d

r = 1 x 1 1 x Is same flow (f ) still a NE? No, since czig-zag(f ) = 1

2 + 1 2 = 1 < 3 2.

All flow on the zig-zag path is the new NE. What is the average cost/travel time? 2. Total cost increased!

What if they closed 42nd street and nobody noticed?

[New York Times Dec. 25, 1990]

◮ On earth day 1990 NYC closed down

traffic heavy 42nd str.

◮ Everyone expected chaos, but. . . ◮ . . . congestion improved ◮ Real example of Braess’s paradox ◮ Of course, when 42nd was reopened

the next they, it became worse again

◮ There are many documented

examples of traffic congestion improving by closing down streets, or worsening when adding new ones

Physics: springs and strings

1kg https://youtu.be/ekd2MeDBV8s?t=6

Efficiency of equilibria

Price of anarchy

Braess’s paradox

• d

r = 1 x 1 1 x One commodity with r = 1; origin o; destination d Define: Social cost is a cost function C : Σ → R. E.g. total cost C(f ) =

i∈N

• Pi∈Σi fPicPi(f ) =

e∈E ce(fe).

Optimal flow? min

f

• i∈N
• Pi∈Σi

fPicPi(f )

Braess’s paradox

• d

r = 1 x 1 1 x Let x be flow on top path, y flow on bottom path, then 1 − x − y is flow on zig-zag. C(x, y) = x(1 − y + 1) + y(1 − x + 1) + (1 − x − y)(1 − y + 1 − x) = 2 + y2 + x2 − x − y ∂C(x, y) ∂x = 2x − 1; ∂C(x, y) ∂y = 2y − 1 ⇒ x = y = 1

2

Braess’s paradox

• d

r = 1 x 1 1 x Optimal total cost: C(f OPT) =

3 2.

NE routes all flow over the zig-zag path. Total NE cost: C(f NE) = 2. Is this “bad”?

Price of anarchy (POA)

Definition: Price of anarchy of a game

Let G be a cost minimization game. The Price of anarchy of G is given by POA(G) = sup

s∈NE(G)

C(s) C(OPT) .

Definition: Price of anarchy of a class of games

Let Γ be a class of cost minimization games. The Price of anarchy of Γ is given by POA(Γ) = sup

G∈Γ

sup

s∈NE(G)

C(s) C(OPT) . In the above C(OPT) = inf

s∈Σ C(s), the minimum value of the social

cost that can possibly be reached.

Price of anarchy (POA)

Definition: Price of anarchy of a game

Let G be a welfare maximization game. The Price of anarchy of G is given by POA(G) = sup

s∈NE(G)

W (OPT) W (s) .

Definition: Price of anarchy of a class of games

Let Γ be a class of welfare maximization games. The Price of anarchy

• f Γ is given by

POA(Γ) = sup

G∈Γ

sup

s∈NE(G)

W (OPT) W (s) . In the above W (·) is the welfare function and W (OPT) = sup

s∈Σ

W (s), the maximum value of the welfare that can possibly be reached.

Price of anarchy for non-atomic routing games

POA(Braess’s paradox) = 2 / 3

2 = 4 3

How about (all) other routing games?

• d

r = 1 x 1 Pigou’s network Q: What is a Nash equilibrium? ctop(f ) ≤ cbottom(f ) Q: What is an optimal solution? min0≤x≤1 x2 + (1 − x)1 ⇒ x =

1 2

POA = C(NE) / C(OPT) = 1 / ( 1

2 · 1 2 + 1 2 · 1 ) = 1 / 3 4 = 4 3

Generalized Pigou-network

• d

r = 1 x2 1 Pigou’s network Q: What is a Nash equilibrium? ctop(f ) ≤ cbottom(f ) Q: What is an optimal solution? min0≤x≤1 x3 + (1 − x)1 ⇒ x =

1 √ 3

POA = C(NE) / C(OPT) = 1 /

• 1

√ 3 · 1 3 +

• 1 − 1

√ 3

• · 1
• =

3 √ 3 3 √ 3 − 2 ≈ 1.6

Generalized Pigou-network

• d

r = 1 xp 1 Pigou’s network NE: ctop(f ) ≤ cbottom(f ) OPT: min0≤x≤1 xp+1 + (1 − x)1 ⇒ x =

1

p

√p+1

POA = C(NE)/C(OPT) = 1/

• 1

p

√p + 1

p+1

+

• 1 −

1

p

√p + 1

• · 1
• =

(p + 1) p √p + 1 (p + 1) p √p + 1 − p → ∞ as p → ∞.

Price of anarchy for non-atomic routing games

Lemma

POA for non-atomic routing games is unbounded. We can make a finer subdivision of non-atomic routing games.