Braesss Paradox, Fibonacci Numbers, and Exponential Inapproximability

1
Braesss 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 Braesss Paradox
• New lower and upper bounds on Braesss 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 i1 to k, a rate ri of traffic to route from
  si to ti

Single Commodity Example (k1)
r11
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
k1, r11
l(x)1
l(x)x
Flow ½
s1
t1
l(x)x
Flow ½
l(x)1
u
5
Braesss 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
½
1
1
x
0
s
t
½
½
1
x
1
u
6
Previous results on Braesss 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 2O(n) or 2O(m),
  even if k 2
• Thm Adding 1 edge to a graph can increase common
  latency at most a factor of 2O(mlogn) or 2O(kn),
  whichever is smaller

8
Braesss 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
Braesss 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
Braesss 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
Braesss Paradox in MC networks
t2
r1 r2 1

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

1
1
s1
t1
1
1
-? flow ? flow
s2
12
Braesss 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
Braesss Paradox in MC networks
t2
r1 r2 1

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

-1/32 flow 1/32 flow
3
1
1
s1
t1
2
1
1
s2
14
Braesss 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
Braesss 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
Braesss 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
Braesss Paradox in MC networks

• In a general network with O(p) nodes
• Latency between s1 and t1 can increase from 1 to
  Fp-11
• 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(mlogn) 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(mlogn)maxe?E(le(fe))

19
Proving Main Lemma

• Main Lemma For any edge e
• le(ge) 2O(mlogn)maxe?E(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(mlogn)maxe?E(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 Cramers
  rule and a bound on the determinant

21
Price of Anarchy with respect to Maximum Latency
Objective

• In the Braesss Paradox example
• The maximum si-ti latency at Nash Eq. is 2O(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 2O(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
  Braesss 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(mlogn) 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 2O(mlogn) or
  2O(kn)?
• What are upper and lower bounds on Braesss
  Paradox and price of anarchy for atomic
  splittable instances?