1. How Bad is Selfish Routing? Tim Roughgarden & Eva Tardos Presented by Wonhong Nam wnam@cis.upenn.edu

2. 2003-4-24CIS 6202 Introduction The routing traffic problem Given the rate of traffic between each pair of nodes in a network, find an assignment of traffic to paths so that the total latency is minimized. It may be expensive or impossible to impose optimal or near- optimal routing strategies on the traffic in a network. How much does network performance suffer from this lack of regulation? Assumptions Users act in a purely selfish manner. We can view network users as independent agents participating in a noncooperative game. We can expect the routes chosen by users to form a Nash equilibrium.

3. 2003-4-24CIS 6203 Model We call the triple (G, r, l) an instance. A direct graph G = (V, E) k source-destination pairs {s 1, t 1 }, …,{s k, t k }. A finite and positive rate r i with each pair {s i, t i } For each edge e, a latency function l e.

4. 2003-4-24CIS 6204 Example Traffic rate: r=1, one source-sink

5. 2003-4-24CIS 6205 Flow P i is the set of simple s i -t i paths. P = U i P i. A flow is a function f:P  R. f p = amount routed on s i -t i path p. f e = amount routed on an edge e.

6. 2003-4-24CIS 6206 Cost of a flow The latency of a path P w.r.t. a flow f: l P (f) is the sum of the latencies of the edges in the path. The cost of a flow f(total latency): C(f) = ∑ p f p * l p (f)

7. 2003-4-24CIS 6207 Nash Equilibrium A flow is at Nash equilibrium(or is a Nash flow) if no agent can improve its path latency by changing its path. Def: A flow f for instance (G, r, l) is at Nash equilibrium if for all i ∈ {1, …,k}, P 1, P 2 ∈ P i, and δ ∈ [0, f P1 ], we have l P1 (f) <= l P2 (f ’ ), where f P ’ = f P – δif P = P 1 f P ’ = f P + δif P = P 2 f P ’ = f P if P is not P 1 nor P 2.

8. 2003-4-24CIS 6208 Nash flows & Social welfare Lemma: A flow f is a Nash flow if and only if all flow travels along minimum- latency paths(w.r.t f). Central question: What is the cost of the lack of coordination in a Nash flow?

9. 2003-4-24CIS 6209 The Braess paradox Cost of Nash flow = 1.5 Cost of Nash flow = 2 The intuitively helpful action of adding a new zero-latency link may negatively impact all of the agents!

10. 2003-4-24CIS 62010 The result for linear latency Thm.1: The latency function of each edge e is linear in the edge congestion; that is, l e (x) = a e x + b e, the cost of the optimal latency flow is at least 3/4 times that of a Nash flow.

11. 2003-4-24CIS 62011 General latency Bad Example: (r=1, l(x)=x i ) The cost of Nash flow is 1. The cost of optimal flow is 1-i*(i+1) -(i+1)/i It assigns (i+1) -1/i units to the upper link. As i  ∞, the total cost tends to 0. The ratio cannot be bounded.

12. 2003-4-24CIS 62012 Result for general latency Thm.2: In any network (G, r, l) with latency functions which are continuous, non-decreasing, the cost of a Nash flow with rates r i for i = 1, …, k is at most the cost of a optimal cost flow for the network (G, 2r, l) with rates 2r i.

13. 2003-4-24CIS 62013 Proof sketch of Thm.1 (1/3) Thm.1: The optimal cost in networks with linear latency is at least ¾ of a Nash flow cost. We prove: The optimal at rate r/2 is f/2  C(f/2) >= ¼ C(f). The cost of augmenting from rate r/2 to r >= ½ C(f).

14. 2003-4-24CIS 62014 Proof sketch of Thm.1(2/3) Lemma: Suppose (G, r, l) has linear latency functions and f is a Nash flow. Then, the flow f/2 is optimal for (G, r/2, l). Lemma: Let (G, r, l) be an instance with edge latency functions l e (x) = a e x + b e for each e ∈ E. Then, A flow f is a Nash flow in G if and only if for each source-sink pair i and P, P ’ ∈ Pi with fp > 0, ∑ (e ∈ P) a e f e + b e <= ∑ (e ∈ P ’ ) a e f e + b e A flow f * is optimal in G if and only if for each source- sink pair i and P, P ’ ∈ Pi with fp > 0, ∑ (e ∈ P) 2a e f e * + b e <= ∑ (e ∈ P ’ ) 2a e f e * + b e

15. 2003-4-24CIS 62015 Proof sketch of Thm.1(3/3) Consider the Nash flow f, let L be the s-t latency in f, then cost of flow f is C(f) = r*L. Optimal flow at rate r/2 is flow f/2, and gradient along flow paths in flow is L. The marginal cost of increasing flow from flow f/2 is L. Cost of increasing flow amount by r/2 is at least (r/2)*L = ½ *C(f).

16. 2003-4-24CIS 62016 Proof sketch of Thm.2 Thm.2: The cost of a Nash flow with rates r i for i = 1, …, k is at most the cost of the optimal cost flow for the network (G, 2r, l) with rates 2r i. Augmenting Nash to optimal? Idea: Gradient is at least the latency. Marginal cost to increase Nash? But, Nash can be improved. Idea: Separate effects of increased and decreased flow.

17. 2003-4-24CIS 62017 Extensions Agent can often only evaluate path latency approximately, rather than exactly. Thm.3: If f is at  -approximate Nash equilibrium with  < 1 for (G,r,l) and f * is for (G,2r,l), then C(f) <= (1+  /1-  )C(f * ). We expect to encounter a finite number of agents, each controlling a strictly positive amount of flow. Thm.4: If f is at Nash equilibrium for the finite splittable instance (G,r,l) with x*l e (x) convex for each e, and f * is for the finite splittable instance (G,2r,l), then C(f) <= C(f * ).

18. 2003-4-24CIS 62018 Conclusion We quantify the degradation in network performance due to unregulated traffic. If the latency of each edge is a linear function of its congestion, then the total latency of the routes chosen by self network users is at most 4/3 times the minimum possible total latency. It is no more than the total latency incurred by optimally routing twice as much traffic.