Braess’s Paradox, Fibonacci Numbers, and Exponential Inapproximability

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Braess’s Paradox, Fibonacci Numbers, and Exponential Inapproximability

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Braess’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

- 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

- 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

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
What about 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

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’))

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)

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

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

- 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

- 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?