The Price of Anarchy is Independent of the Network Topology

Tim Roughgarden

(presented by Aleksandr Yampolskiy)

Outline

Motivation The Model Results Conclusion

Q: Which route would you take?

suburb

train station

wide, circuitous road:

1 hour delay

narrow, straight road:

20 minute delay

A: Most drivers would take the narrow road

suburb

train station

resulting in traffic

congestion

What if…there were a dictator who told the cars (or Internet

packets) which route to take?

suburb

train station

red sedan must take the longer

route

Outline

Motivation Model Results Conclusion

The Model

A directed graph G = (V, E) k source-destination pairs {s1, t1}, …, {sk, tk}

A rate ri ≥ 0 of traffic from si to ti

Each edge e has a latency function le(▪) (continuous, non-decreasing)

Pi = a set of simple si-ti paths and

(G, r, l)

s1 t1

v

x

x1

1

0

r1 = 1

w

Simple Paths

s1→v→t1

s1→v→w→t1

s1→w→t1

Flows Flow f: P → R+ = routes of many non-cooperative

agents fp = flow per path

fe = flow per edge =

s1 t1

v

x

x1

1

0

r1 = 1

w

fp2 = ½

fp1 = ½

edge e = (w, t1):fe = ½ + ½ = 1le(fe) = 1

The Cost of a Flow Latency of a path P = sum of latencies of

edges in P

Cost of a flow f = total latency incurred by f :

s1 t1

v

x

x1

1

0

r1 = 1

w

fp2 = ½

fp1 = ½

lp1(f) = ½ + 1 = 1.5

lp2(f) = ½ + 0 + 1 = 1.5

C(f) = ½ * 1.5 + ½ * 1.5 = 1.5

Some Assumptions Finding an optimal routing is difficult Flows behave “selfishly” and “greedily” Each network user controls a negligible

fraction of the overall traffic

In other words: Network routing is a non-

cooperative game and the routes form a

Nash Equilibrium.

Nash FlowsDef: A flow f is a Nash flow iff all traffic is routed on

minimum-latency paths. Formally,

8 i 2 {1, ... , k} and P1, P2 2 Pi with fP1 > 0,

lP1(f) · lP2

(f)

Lemma: [BMW ’56] An acyclic Nash flow exists and is essentially unique.

Lemma: [RT ’00] In a Nash flow f, all si-ti flow paths have equal latency Li(f).

More on Nash Flows

Fact: Nash flow does not optimize the total latency (cf. Prisoner’s Dilemma).

s1 t1

1

x

r1 = 1 flow = ½

flow = ½

s1 t1

1

x

r1 = 1 flow = 0

flow = 1

C(f*) = ½ * 1 + ½ * ½ = ¾ C(f) = 0 * 1 + 1 * 1 = 1

optimal flow Nash flow[Pigou 1920]

Optimal Flows

Def: Marginal latency

Lemma: [BMW ’56] An optimal flow is a Nash flow for the marginal latencies.

s1 t1

1

x

r1 = 1

s1 t1

1

2x

r1 = 1

latency functions marginal cost functions

flow = ½

flow = ½

flow = ½

flow = ½

optimal flow Nash flow

The Price of Anarchy

Def: [KP ’99] Price of anarchy is the worst-case ratio between the cost of a Nash and of an optimal flow.

ρ(G, r, l) =

Outline

Motivation Model Results Conclusion

Linear Latency Functions

Thm: [RT ’00] For linear latency functions, ρ(G, r, l) = C(f) / C(f*) = 4/3.

s1 t1

1

2x

r = 1 flow = ½

flow = ½

s1 t1

1

x

r = 1 flow = 0

flow = 1

C(f*) = ½ * 1 + ½ * ½ = ¾C(f) = 0 * 1 + 1 * 1 = 1

latency functions marginal cost functions

Nash flow: Optimal flow:

Linear Latency Functions (cont.) Linear latencies have the form le(x) = aex + be for

some ae, be ¸ 0. Marginal cost function is le*(x) = 2aex + be.

le*(x/2) = 2ae(x/2) + be = le(x) for each edge e. Thus, lP*(f/2) = lP(f) for each path P.

Corollary: [RT ’00] The Nash flow f/2 is optimal for rate r/2.

Proof Idea Goal: Lower bound C(f*) in terms of C(f) Idea: We create an optimal flow f* for (G, r, l)

via a two-step process:1. Send a scaled down Nash flow f/2 through G. By

corollary, it will be optimal for (G, r/2, l).

2. Augment f/2 to a flow optimal for (G, r, l).

s1 t1

1

x

r = 1

flow = ½

flow = 0

flow = ½

flow =0

r = ½

Proof Idea

cost of f* at rate r

= cost of f/2 at rate r/2

+ cost of augmenting flow to rate r

≥ ¼ C(f) (easy) ≥ ½ C(f) (hard)

≥ ¾ C(f)

Goal: Lower bound C(f*) in terms of C(f). Idea: We create an optimal flow f* for (G, r, l)

via a two-step process:1. Send a scaled down Nash flow f/2 through G. By

corollary, it will be optimal for (G, r/2, l).

2. Augment f/2 to a flow optimal for (G, r, l).

General Latency Functions Can we find a similar bound on ρ(G, r, l) for general

latency functions? Unfortunately, for a general l(·), the price of anarchy may

be much larger than 4/3 even in a simple network:

s1 t1

1

xp

r = 1

s1 t1

1 r = 1

latency functions marginal cost functions

(p +1)xp

flow = 1

flow = 0

flow = (p + 1)-1/p

flow = 1 – (p + 1)-1/p

Nash flow: Optimal flow:

C(f) = 1 C(f*) = 1 – p¢(p +1)-(p+1)/p = ( )

Main Theorem

Def: anarchy value (L) 2 [1, 1) = how “nice” latency functions in class L are

Thm: Price of anarchy is independent of network topology: ρ(G, r, l) = (L) for a standard class of functions L.

Upper Bound: ρ(G, r, l) ≤ (L)

Idea: Mimic the proof for linear latencies. Scale down Nash flow f/c and then augment it to an optimal flow.

Problem: For non-linear latency functions, there is no constant c for which f/c is optimal for reduced rate r/c.

Upper Bound: ρ(G, r, l) ≤ (L) (cont.)

Idea: Scale down Nash flow f by different factors on different edges.

Problem: This violates conservation constraints!

Upper Bound: ρ(G, r, l) ≤ (L) (cont.)

Combining two previous ideas, we create an optimal

flow f* for (G, r, l) via a two-step process:

1. Send a pseudoflow {efe}e2 E such that le*(efe) = le(fe).

2. Augment the pseudoflow to an optimal flow f*

s1 t1

1

x2

r = 1

flow = 1/√3

flow = 1 - 1/√3

l(x) = x2, l*(x) = 3x2.

Want 3( x)2 = x2 ) = √1/3

Upper Bound: ρ(G, r, l) ≤ (L) (cont.)

Lemma: C(f*) ¸ e le(efe)efe + e (fe* - efe)le(fe)

Lower Bound: ρ(G, r, l) ≥ (L)

Computing the Price of Anarchy

So, the price of anarchy, ρ(G, r, l) for l 2 L is easy to compute

It is simply (L ) = the worst-possible ratio of in a two-node network:

s1 t1

constant

·l

r

Computing the Price of Anarchy (cont.)

Conclusion