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 c e : R + → R + . ◮ Set N of n commodities. Commodity i : amount of flow r i ; origin & destination ( o i , d i ). ◮ Each player controls an infinitesimal amount of flow. ◮ Each player in commodity i chooses ( o i , d i )-path P i . ◮ feasible flow: routes all flow r i from o i to d i for all i ∈ N . I.e., r i = � P i ∈ i f P i , with f P i the amount of flow on P i . ◮ Flow on edge e is f e = � e ∈ E : e ∈ P i f P i . ◮ Cost of path P i is c P i ( f ) = � e ∈ P i c e ( f e ). This is not a simultaneous move game
Non-atomic routing games Atomic routing game potential function f e � � Φ( f ) = c e ( j ) . j =1 e ∈ E Non-atomic routing game potential function (not exact) � f e � Φ( f ) = c e ( j ) dj . j =0 e ∈ E 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 1 x 2 r = 1 o d x 1 2 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 1 x 2 r = 1 o d x 1 2 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, 1 1 − x 3 x x + 2 = + 1 = 1 ⇔ 2 2
Example 1 x 2 r = 1 o d x 1 2 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 ∈ Σ i with f P > 0, and any path Q ∈ Σ i c P ( f ) ≤ c Q ( f ) .
Efficiency of equilibria Braess’s paradox
Braess’s paradox 1 x r = 1 o d x 1 One commodity with r = 1; origin o ; destination d What is the NE flow? 1 c top ( f ) = c bottom ( f ) f top = f bottom = 2 . ⇔ Average cost/travel time: 1 . 5 (same for all users).
Braess’s paradox x 1 0 r = 1 o d x 1 Is same flow ( f ) still a NE? No, since c zig-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 x 1 0 r = 1 o d x 1 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 ) = � � P i ∈ Σ i f P i c P i ( f ) = � e ∈ E c e ( f e ). i ∈ N Optimal flow? � � min f P i c P i ( f ) f i ∈ N P i ∈ Σ i
Braess’s paradox 1 x 0 r = 1 o d x 1 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 + y 2 + x 2 − x − y ∂ C ( x , y ) ∂ C ( x , y ) ⇒ x = y = 1 = 2 x − 1; = 2 y − 1 2 ∂ x ∂ y
Braess’s paradox x 1 0 r = 1 o d x 1 3 Optimal total cost: C ( f OPT ) = 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 C ( s ) POA( G ) = sup C (OPT) . s ∈ NE( G ) 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 C ( s ) POA(Γ) = sup sup C (OPT) . G ∈ Γ s ∈ NE( G ) 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 W (OPT) POA( G ) = sup . W ( s ) s ∈ NE( G ) Definition: Price of anarchy of a class of games Let Γ be a class of welfare maximization games. The Price of anarchy of Γ is given by W (OPT) POA(Γ) = sup sup . W ( s ) G ∈ Γ s ∈ NE( G ) In the above W ( · ) is the welfare function and W (OPT) = sup W ( s ), s ∈ Σ 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 4 2 = 3 How about (all) other routing games? x r = 1 o d 1 Pigou’s network Q: What is a Nash equilibrium? c top ( f ) ≤ c bottom ( f ) Q: What is an optimal solution? min 0 ≤ x ≤ 1 x 2 + (1 − x )1 ⇒ x = 1 2 POA = C (NE) / C (OPT) = 1 / ( 1 2 · 1 1 2 · 1 ) = 1 / 3 4 2 + 4 = 3
Generalized Pigou-network x 2 r = 1 o d 1 Pigou’s network Q: What is a Nash equilibrium? c top ( f ) ≤ c bottom ( f ) Q: What is an optimal solution? min 0 ≤ x ≤ 1 x 3 + (1 − x )1 ⇒ x = 1 √ 3 1 3 · 1 1 − 1 � � � � POA = C (NE) / C (OPT) = 1 / 3 + · 1 √ √ 3 √ 3 3 = √ 3 − 2 ≈ 1 . 6 3
Generalized Pigou-network x p r = 1 o d 1 Pigou’s network NE: c top ( f ) ≤ c bottom ( f ) OPT: min 0 ≤ x ≤ 1 x p +1 + (1 − x )1 ⇒ x = 1 √ p +1 p �� � p +1 � 1 1 � � POA = C (NE) / C (OPT) = 1 / + 1 − · 1 √ p + 1 √ p + 1 p p √ p + 1 ( p + 1) p = √ p + 1 − p → ∞ as p → ∞ . ( p + 1) 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.
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