Braess’s Paradox, Fibonacci Numbers, and

Exponential Inapproximability

Henry Lin*

Tim Roughgarden**

Éva Tardos†

Asher Walkover††

*UC Berkeley **Stanford University †Cornell University ††Google

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

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

ℓ(x)=x

Flow = ½

Flow = ½

ℓ(x)=1

Single Commodity Example (k=1):r1=1

s1 t1

v

uℓ(x)=1

ℓ(x)=x

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]

ℓ(x)=x

Flow = ½

Flow = ½

ℓ(x)=1

An example Nash flow:k=1, r1=1

s1 t1

v

uℓ(x)=1

ℓ(x)=x

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

s t

x 1

½

x1

½

½

½1

1

0

v

u

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?

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

Braess’s Paradox in MC networks

s1 t1

s2

t2

r1 = r2 = 1

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

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

-½ flow+½ flow

r1 = r2 = 1

- All unlabelled edges have 0 latency (at current flow)

1

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

r1 = r2 = 1

- All unlabelled edges have 0 latency (at current flow)

1

-¼ flow+¼ flow

1

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

r1 = r2 = 1

- All unlabelled edges have 0 latency (at current flow)

1 1

-⅛ flow+⅛ flow

1

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

r1 = r2 = 1

- All unlabelled edges have 0 latency (at current flow)

1 1

-1/16 flow+1/16 flow

1

2

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

r1 = r2 = 1

- All unlabelled edges have 0 latency (at current flow)

1 1

-1/32 flow+1/32 flow

1

2

3

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

- All unlabelled edges have 0 latency (at current flow)

1 1

-1/64 flow+1/64 flow

1

2

3

5

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

1 1

-1/128 flow+1/128 flow

1

2

3

5

8

- All unlabelled edges have 0 latency (at current flow)

Braess’s Paradox in MC networks

s1 t1

s2

t2

1

- Latency between s1 and t1 increased from 1 to 9- Latency between s2 and t2 increased from 0 to 13

1 1

1

2

3

5

8

- All unlabelled edges have 0 latency (at current flow)

Braess’s Paradox in MC networksIn 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)

s1 t1

s2

t2

1

1 1

1

2

3

5

8

- In fact, adding 1 edge is enough to cause this bad example

Proving Upper Bounds

To prove 2O(m·logn) bound, let:f be the flow before edges were addedg be the flow after edges were added

Main Lemma: For any edge e:ℓe(ge) ≤ 2O(m·logn)·maxe’єE(ℓe’(fe’))

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)

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

Price of Anarchy with respect to Maximum Latency ObjectiveIn 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

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

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?