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2003-4-24 CIS 620 2
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 equilibriu
m.
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ModelWe 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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ExampleTraffic 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:PR.
fp = amount routed on si-ti path p. fe = amount routed on an edge e.
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Cost of a flowThe 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):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 a
gent 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 if P = P1
fP’ = fP + δ if P = P2
fP’ = fP if P is not P1 nor P2.
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Nash flows & Social welfareLemma: 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 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 latencyThm.1: The latency function of each edg
e 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 latencyBad 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 latencyThm.2: In any network (G, r, l) with laten
cy 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 line
ar 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 fu
nctions 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,
∑(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 ap
proximately, 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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ConclusionWe 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.
