Braessâs Paradox, Fibonacci Numbers, and Exponential Inapproximability

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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

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### Braess’s Paradox, Fibonacci Numbers, and Exponential Inapproximability

Henry Lin*

Tim Roughgarden**

Éva Tardos†

Asher Walkover††

*UC Berkeley **Stanford University

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

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

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

• 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

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

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

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

t2

r1 = r2 = 1

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

1

s1

t1

1

-½ flow

+½ flow

s2

t2

r1 = r2 = 1

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

1

s1

t1

1

1

-¼ flow

+¼ flow

s2

t2

r1 = r2 = 1

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

1

1

s1

t1

1

1

-⅛ flow

+⅛ flow

s2

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

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

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)

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)

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)

t2

8

3

1

1

s1

t1

2

5

1

1

s2

• 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
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’))

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 Objective

• 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?