Braess’s Paradox, Fibonacci Numbers, and Exponential Inapproximability

1. Braess’s Paradox, Fibonacci Numbers, and Exponential Inapproximability Henry Lin* Tim Roughgarden** Éva Tardos† Asher Walkover†† *UC Berkeley **Stanford University †Cornell University ††Google

2. Overview • Selfish routing model and Braess’s Paradox • New lower and upper bounds on Braess’s Paradox in multicommodity networks • Connections to the price of anarchy with respect to the maximum latency objective • Open questions

3. Routing in congested networks • a directed graph: G = (V,E) • for each edge e, a latency function: ℓe(•) • nonnegative, nondecreasing, and continuous • one or more commodities: (s1, t1, r1) … (sk, tk, rk) • for i=1 to k, a rate ri of traffic to route from si to ti Single Commodity Example (k=1): r1=1 v ℓ(x)=1 ℓ(x)=x Flow = ½ s1 t1 ℓ(x)=x Flow = ½ ℓ(x)=1 u

4. Selfish Routing and Nash Flows How do we model selfish behavior in networks? Def: A flow is at Nash equilibrium (or is a Nash flow) if all flow is routed on min-latency paths [at current edge congestion] • Note: at Nash Eq., all flow must have same s to t latency • Always exist & are unique [Wardrop, Beckmann et al 50s] An example Nash flow: v k=1, r1=1 ℓ(x)=1 ℓ(x)=x Flow = ½ s1 t1 ℓ(x)=x Flow = ½ ℓ(x)=1 u

5. Braess’s Paradox • Common latency is 1.5 • Adding edge increased latency to 2! • Replacing x with xd yields more severe example where latency increases from 1 to 2 v ½ 1 ½ 1 x 0 s t ½ 1 ½ x 1 u

6. Previous results on Braess’s Paradox In single-commodity networks: • Thm: [R 01]Adding 1 edge to a graph can increase common latency by at least a factor of 2 • Thm: [LRT 04]Adding 1 edge to a graph can increase common latency by at most a factor of 2 What about multicommodity networks?

7. New results for BPin multicommodity networks In a network with k ≥ 2 commodities, n nodes, m edges: • Thm: Adding 1 edge to a graph can increase common latency by at least a factor of 2Ω(n) or 2Ω(m), even if k = 2 • Thm: Adding 1 edge to a graph can increase common latency at most a factor of 2O(m·logn)or 2O(kn),whichever is smaller

8. Braess’s Paradox in MC networks t2 r1 = r2 = 1 • All unlabelled edges have 0 latency (at current flow) • Only edge leaving s1 has latency 1 • Latency between s1 and t1 is 1 • Latency between s2 and t2 is 0 1 s1 t1 s2

9. Braess’s Paradox in MC networks t2 r1 = r2 = 1 • All unlabelled edges have 0 latency (at current flow) 1 s1 t1 1 -½ flow +½ flow s2

10. Braess’s Paradox in MC networks t2 r1 = r2 = 1 • All unlabelled edges have 0 latency (at current flow) 1 s1 t1 1 1 -¼ flow +¼ flow s2

11. Braess’s Paradox in MC networks t2 r1 = r2 = 1 • All unlabelled edges have 0 latency (at current flow) 1 1 s1 t1 1 1 -⅛ flow +⅛ flow s2

12. Braess’s Paradox in MC networks t2 r1 = r2 = 1 • All unlabelled edges have 0 latency (at current flow) 1 1 s1 t1 2 1 1 -1/16 flow +1/16 flow s2

13. Braess’s Paradox in MC networks t2 r1 = r2 = 1 • All unlabelled edges have 0 latency (at current flow) 3 -1/32 flow +1/32 flow 1 1 s1 t1 2 1 1 s2

14. Braess’s Paradox in MC networks t2 -1/64 flow +1/64 flow 3 1 1 s1 t1 2 5 1 1 s2 • All unlabelled edges have 0 latency (at current flow)

15. Braess’s Paradox in MC networks t2 8 -1/128 flow +1/128 flow 3 1 1 s1 t1 2 5 1 1 s2 • All unlabelled edges have 0 latency (at current flow)

16. Braess’s Paradox in MC networks t2 • Latency between s1 and t1 increased from 1 to 9 • Latency between s2 and t2 increased from 0 to 13 8 3 1 1 s1 t1 2 5 1 1 s2 • All unlabelled edges have 0 latency (at current flow)

17. t2 8 3 1 1 s1 t1 2 5 1 1 s2 Braess’s Paradox in MC networks • In a general network with O(p) nodes: • Latency between s1 and t1 can increase from 1 to Fp-1+1 • Latency between s2 and t2 can increased from 0 to Fp • (where Fp is the pth fibonacci number) • In fact, adding 1 edge is enough to cause this bad example

18. Proving Upper Bounds To prove 2O(m·logn)bound, let: f be the flow before edges were added g be the flow after edges were added Main Lemma: For any edge e: ℓe(ge) ≤ 2O(m·logn)·maxe’єE(ℓe’(fe’))

19. Proving Main Lemma Main Lemma: For any edge e: ℓe(ge) ≤ 2O(m·logn)·maxe’єE(ℓe’(fe’)) Proof (sketch):Let f, g, and ℓe(fe) be fixed. Resulting latencies ℓe(ge) must be: • nonnegative • nondecreasing • at Nash equilibrium Requirements can be formulated as a set of linear constraints on ℓe(ge)

20. Proving Main Lemma Main Lemma: For any edge e: ℓe(ge) ≤ 2O(m·logn)·maxe’єE(ℓe’(fe’)) Proof (sketch):Let f, g, and ℓe(fe) be fixed. In fact, finding maximum ℓe(ge) can be formulated as a linear program • can show maximum occurs at extreme point • can bound extreme point solution with Cramer’s rule and a bound on the determinant

21. Price of Anarchy with respect to Maximum Latency Objective In the Braess’s Paradox example: • The maximum si-ti latency at Nash Eq. is 2Ω(n) • An optimal flow avoiding the extra edges can have maximum si-ti latency equal to 1 New Thm:The price of anarchy wrt to the maximum latency is at least 2Ω(n). Disproves conjecture that PoA for multicommodity networks is no worse than for single-commodity networks

22. Price of Anarchy with respect to Maximum Latency Objective • Linear programming technique not specific to Braess’s Paradox • Provides same bound for price of anarchy wrt maximum latency New Thm:The price of anarchy wrt to the maximum latency is at most 2O(m·logn) or 2O(kn), whichever is smaller

23. Open Questions • Can the upper bounds be improved to 2O(n)or 2O(m)? • Can the lower bounds be improved to 2Ω(m·logn) or 2Ω(kn)? • What are upper and lower bounds on Braess’s Paradox and price of anarchy for atomic splittable instances?