A Stronger Bound on Braess’s Paradox Henry Lin * Tim Roughgarden * Éva Tardos † *UC Berkeley † Cornell University

Overview Selfish Routing – users attempt to minimize personal latency traveling in a traffic network Braess’s Paradox – adding an edge can increase latency for all users How bad can Braess’s Paradox be if k edges are added?

Routing in congested networks a directed graph G = (V,E) a rate r of traffic to route from source s to destination t for each edge e, a latency function ℓ e () –assumed continuous, nondecreasing st ℓ(x)=x Flow = ½ ℓ(x)=1 Example: r=1

Nash flows Some assumptions: agents represent a small unit of flow want to minimize personal latency Def: A flow is at Nash equilibrium (or is a Nash flow) if all flow is routed on min-latency paths [given current edge congestion] –At Nash Eq., all flow will have same s to t latency –have existence, uniqueness [Wardrop, Beckmann et al 50s] x st 1 Flow =.5 st 1 Flow = 0 Flow = 1 x Example: This flow is envious! r=1

Braess’s Paradox Common latency is 1.5 Adding edge increased latency to 2! Replacing x with x d yields more severe example where latency increases from 1 to 2 s t x1 ½ x 1 ½ ½ ½1 1 0 v u

A bigger Braess’s Paradox s t ⅓ flow along each path Common latency is 1 Adding 2 edges increased latency to 3! ½ flow along each path

The severity of Braess’s Paradox increases as more edges are added Thm: [Roughgarden 01] Adding k edges to a graph can increase common latency by a k+1 factor But, can adding a single edge cause a large increase in delay? When adding a single edge, are there more severe examples than Braess’s original example?

Bounding Braess’s Paradox New result: Adding k edges to a network increases common latency by at most a k+1 factor Note: This bound is tight

Key idea in proof Look at how the flow changed after adding edges Remember: Flow travels along shortest s to t path [given current congestion] s t x1 light x 1 heavy 0 v u added

Outline of proof We can show that there must be a path consisting of only light forward edges and heavy backward edges, called an alternating path, from s to t [use a cut argument] Show that the latency to nodes in an alternating path can be bounded

A simple example with light edges Latency along light edges has decreased Latency traveling from s to v 1, v 2, and t has improved Common latency must improve since shortest s to t path got shorter s v1v1 v2v2 t light added

What about heavy edges? Flow chose s-u-v route, despite better s-v path s-u latency must have improved, since u-v has increased latency u-t has decreased latency, so s-t latency has improved s t light heavy v u “light”

More formally d’(v) represents s-v latency before edge added d(v) represents s-v latency after edge added Along a path of light forward edges and heavy backward edges with no added edges, we can prove by induction that d(v) ≤ d’(v) If alternating path reaches t without crossing an added edge, then overall latency decreases

Crossing an added edge Can no longer conclude d(u) ≤ d’(u) Can only conclude d(u) ≤ d’(u) + d’(t) In the worst case, s-u latency increases from 0 to d’(t) Latency to node after added edge may increase by d’(t) But can still bound d(t) ≤ d’(t) + d’(t) so d(t) ≤ 2*d’(t) s t light added v u

Bound when k edges are added sv1v1 v2v2 v3v3 v4v4 v5v5 d(s) ≤ d’(s) d(v 1 ) ≤ d’(v 1 ) d(v 2 ) ≤ d’(v 2 ) d(v 3 ) ≤ d’(v 3 ) d(v 4 ) ≤ d’(v 4 ) + d’(t) d(v 5 ) ≤ d’(v 5 ) + d’(t) d(v 6 ) ≤ d’(v 6 ) + 2  d’(t) v6v6 v7v7 v8v8 t d(t) ≤ d’(t) + k  d’(t) d(t) ≤ (k+1)  d’(t) blue = light red = heavy green = added etc…

Corollaries Decreasing the flow rate will decrease common latency [Hall 78] Adding edges incident to s or t will not increase latency The increase in latency is bounded by the size of the maximal node disjoint set of edges added that avoids s and t This gives alternate proof of the n/2 bound in [Roughgarden 01]