Transcript How Bad is Selfish Routing? Tim Roughgarden & Eva Tardos Presented by Wonhong Nam [email protected].

How Bad is Selfish Routing?
Tim Roughgarden & Eva Tardos
Presented by Wonhong Nam
[email protected]
Introduction

The routing traffic problem






Given the rate of traffic between each pair of nodes in a
network,
find an assignment of traffic to paths
so that the total latency is minimized.
It may be expensive or impossible to impose optimal or
near-optimal routing strategies on the traffic in a network.
How much does network performance suffer from this lack
of regulation?
Assumptions



Users act in a purely selfish manner.
We can view network users as independent agents participating
in a noncooperative game.
We can expect the routes chosen by users to form a Nash
equilibrium.
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Model

We call the triple (G, r, l) an instance.

A direct graph G = (V, E)

k source-destination pairs {s1, t1},…,{sk, tk}.
A finite and positive rate ri with each pair {si,
ti }
 For each edge e, a latency function le.

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Example

Traffic rate: r=1, one source-sink
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Flow
Pi is the set of simple si-ti paths.
 P = Ui Pi.
 A flow is a function f:PR.

 fp
= amount routed on si-ti path p.
 fe = amount routed on an edge e.
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Cost of a flow

The latency of a path P w.r.t. a flow f:
 lP(f)
is the sum of the latencies of the edges
in the path.
 The cost of a flow f(total latency):

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C(f) = ∑p fp * lp(f)
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Nash Equilibrium

A flow is at Nash equilibrium(or is a Nash flow) if
no agent can improve its path latency by changing
its path.

Def: A flow f for instance (G, r, l) is at Nash
equilibrium if for all i ∈ {1,…,k}, P1, P2 ∈ Pi, and
δ ∈[0, fP1], we have lP1(f) <= lP2(f’), where



fP’ = fP – δ
fP’ = fP + δ
fP’ = fP
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if P = P1
if P = P2
if P is not P1 nor P2.
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Nash flows & Social welfare
Lemma: A flow f is a Nash flow if and
only if all flow travels along
minimum-latency paths(w.r.t f).
 Central question:


What is the cost of the lack of coordination
in a Nash flow?
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The Braess paradox
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Cost of Nash flow = 1.5

Cost of Nash flow = 2

The intuitively helpful action of adding a new zero-latency
link may negatively impact all of the agents!
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The result for linear latency

Thm.1: The latency function of each
edge e is linear in the edge
congestion; that is, le(x) = aex + be,

the cost of the optimal latency flow is at
least 3/4 times that of a Nash flow.
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General latency
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Bad Example: (r=1, l(x)=xi)
The cost of Nash flow is 1.
 The cost of optimal flow is 1-i*(i+1)-(i+1)/i




It assigns (i+1)-1/i units to the upper link.
As i ∞, the total cost tends to 0.
The ratio cannot be bounded.
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Result for general latency

Thm.2: In any network (G, r, l) with
latency functions which are
continuous,
 non-decreasing,


the cost of a Nash flow with rates ri
for i = 1,…, k is at most the cost of a
optimal cost flow for the network (G,
2r, l) with rates 2ri.
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Proof sketch of Thm.1 (1/3)

Thm.1: The optimal cost in networks with
linear latency is at least ¾ of a Nash flow
cost.

We prove:


The optimal at rate r/2 is f/2  C(f/2) >= ¼ C(f).
The cost of augmenting from rate r/2 to r >= ½ C(f).
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Proof sketch of Thm.1(2/3)

Lemma: Suppose (G, r, l) has linear
latency functions and f is a Nash flow.


Then, the flow f/2 is optimal for (G, r/2, l).
Lemma: Let (G, r, l) be an instance with
edge latency functions le(x) = aex + be for
each e ∈ E. Then,

A flow f is a Nash flow in G if and only if for each
source-sink pair i and P, P’ ∈ Pi with fp > 0,


∑(e ∈ P) aefe + be <= ∑(e ∈ P’) aefe + be
A flow f* is optimal in G if and only if for each
source-sink pair i and P, P’ ∈ Pi with fp > 0,

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∑(e ∈ P) 2aefe* + be <= ∑(e ∈ P’) 2aefe* + be
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Proof sketch of Thm.1(3/3)

Consider the Nash flow f,


let L be the s-t latency in f, then cost of flow f is
C(f) = r*L.
Optimal flow at rate r/2 is flow f/2, and
gradient along flow paths in flow is L.


The marginal cost of increasing flow from flow f/2
is L.
Cost of increasing flow amount by r/2 is at least
(r/2)*L = ½*C(f).
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Proof sketch of Thm.2

Thm.2: The cost of a Nash flow with rates ri for i =
1,…, k is at most the cost of the optimal cost flow
for the network (G, 2r, l) with rates 2ri.

Augmenting Nash to optimal?


Idea: Gradient is at least the latency.
Marginal cost to increase Nash?


But, Nash can be improved.
Idea: Separate effects of increased and decreased flow.
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Extensions

Agent can often only evaluate path latency
approximately, rather than exactly.


Thm.3: If f is at -approximate Nash equilibrium
with  < 1 for (G,r,l) and f* is for (G,2r,l), then C(f)
<= (1+ /1- )C(f*).
We expect to encounter a finite number of
agents, each controlling a strictly positive
amount of flow.

Thm.4: If f is at Nash equilibrium for the finite
splittable instance (G,r,l) with x*le(x) convex for
each e, and f* is for the finite splittable instance
(G,2r,l), then C(f) <= C(f*).
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Conclusion

We quantify the degradation in
network performance due to
unregulated traffic.
If the latency of each edge is a linear
function of its congestion, then the total
latency of the routes chosen by self
network users is at most 4/3 times the
minimum possible total latency.
 It is no more than the total latency incurred
by optimally routing twice as much traffic.

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