Balázs Sziklai Selfish Routing in Non-cooperative Networks

Historical Outline J.G. Wardrop (1952): Some theoretical aspects of road traffic research, Proc. of the Institute of Civil Engineers D. Braess (1968): Über ein Paradox der Verkehrsplanung, Unternehmensforschung E. Koutsoupias and C. Papadimitriou (1999): Worst Case Equilibria, Proc. Of the 16th International Symposium on Theoretical Aspects of Computer Science Feldmann et al. (2003): Selfish Routing in Non-cooperative Networks: A Survey, Lecture Notes in Computer Science

The Wardrop Model A problem instance consists of a triple a directed graph a set of routing tasks and a set of edge latency functions For a fixed flow f the latency of an edge e is defined as the product of the latency function on e when routing traffic f times the traffic f itself. The social cost of a flow is defined to be the sum of the edge latencies. st x + 1

Pigou’s example We would like to route a total of one unit of flow from s to t. To obtain the optimal solution we have to compute the minimum of the following quadratic polinomial: Simple calculation shows that the social optimum is to route ½ of the traffic on the upper edge and ½ of the traffic on the lower edge. However the latency cost is lower on the upper edge for any user, which means it cant be a Nash equilibrium. The unique NE point is to route all the traffic on the upper edge. st x 1

Braess Paradox We would like to route 6 unit of traffic on the following graph: 10x x + 50 s t

Braess Paradox The social cost of the flow at the Nash Equilibrium: 2 ∙[3 ∙(10∙3+(3+50))] = 6 ∙ 83 = s t

Braess Paradox Now we connect the two middle points with an edge. 10x x + 50 x +10 s t

Braess Paradox Using the new edge some of the users can be better off by changing their path from s to t = (3 ∙10) + (3+50) > (3 ∙10) + (1+10) + (4 ∙10) = 81 s t

Braess Paradox Using the new edge some of the users can be better off by changing their path from s to t = (3+50) + (4 ∙10) > (4 ∙10) + (2+10) + (4 ∙10) = 92 s t

Braess Paradox Again we arrived to a NE point. No user can improve its private latency by unilaterally changing its route The new social cost is 4 ∙(10∙4+(2+50))+ 2 ∙(10∙4+(2+10)+ 10∙4) = 6 ∙ 92 = 558 > 496 s t

How to resolve the Braess Paradox To find the ‘bad’ edges in a network is NP-hard (Roughgarden 2002). Capacity should be added across the network rather than on a local scale (Korilis et al. 1995). Stackelberg approach: controlling just a small portion of the traffic the system can be driven close into the network optimum (Korilis et al. 1995, Roughgarden 2001).

A physical representation of the Braess Paradox 1 kg

Price of Anarchy Introduced by Papadimitriou in a conference paper (2001). By definitions it is the ratio of the (worst case) Nash equilibrium social cost and the optimal social cost. In the Wardrop model the coordination ratio cannot be bounded from above by a constant, when arbitrary edge latency functions are allowed. However when we restrict ourselves to linear edge latency functions the coordination ratio is bounded from above by 4/3. This bound is tight (see Pigou’s example).

KP-model Named after Koutsoupias and Papadimitriou (1999). Simple network with m paralell links and n users. Each user i has a weight w i and these traffics are unsplittable. A pure strategy of a user is a specific link, and a mixed strategy is a probability distribution over the set of links. st

KP-model The social cost is the expected maximum latency on a link, where the expectation is taken over all random choices of the users. Basic model – identical links. General model – related links (different links have different capacities).

FMNE conjecture Consider the instance of three identical links and three users with traffic weight of 2. st A B C Prob. dist. Link ALink BLink C User 14/91/94/9 User 22/31/30 User 301/32/3 Social Cost = 3,53 Prob. dist. Link ALink BLink C User 11/3 User 21/3 User 31/3 Social Cost = 3,77 Prob. dist. Link ALink BLink C User 1100 User 2010 User 3001 Social Cost = 2

FMNE conjecture Fully Mixed Nash Equilibrium is a NE where every user routes along every possible edge. Consider the model of arbitrary traffics and related links. Then any traffic vector w such that the fully mixed Nash Equilibrium F exists, and for any Nash equilibrium P, SC(w, P) ≤ SC(w, F).

Differences between the two models Wardrop model defines the social cost as the sum of the edge latencies while the KP-model as the expected maximum edge latency on a link. While in the Wardrop-model traffic can be splitted into infinitely tiny pieces, traffic rates are unsplittable in the KP-model. All Nash Equilibria are pure and have the same social cost in the Wardrop-model.

Unsolved problems Give upper and lower bounds for the Price of Anarchy. How to design a network free from Braess paradox? Is the FMNE conjecture true? Find better algorithms to compute NE points.

Thank you for your attention!