How Bad is Selfish Routing Tim Roughgarden Eva Tardos presented by - - PowerPoint PPT Presentation

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How Bad is Selfish Routing Tim Roughgarden Eva Tardos presented by - - PowerPoint PPT Presentation

How Bad is Selfish Routing Tim Roughgarden Eva Tardos presented by Yajun Wang (yalding@cs.ust.hk) for COMP670O Spring 2006, HKUST 1-1 Problem Formulation: Traffic Model Given the rate of traffic between each pair of nodes in a network,


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How Bad is Selfish Routing

Tim Roughgarden Eva Tardos

presented by Yajun Wang (yalding@cs.ust.hk)

for COMP670O Spring 2006, HKUST

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Problem Formulation: Traffic Model

  • Given the rate of traffic between each pair of

nodes in a network, find an assignment of traffic to minimize the total latency.

  • On each edge, the latency is load dependent

v s t w

l(x) = x l(x) = x l(x) = 1 l(x) = 1

v s t w

l(x) = x l(x) = x l(x) = 1 l(x) = 1 l(x) = 0

Braess’s Paradox

  • Each player controls a negligible fraction of the
  • verall traffic.
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Formal Model

  • Graph G = (V, E) and k source-destination pairs {si, ti}
  • P = ∪iPi
  • Pi denotes the set of (simple) si − ti paths, and
  • A flow is a function:

f : P → R+

  • A flow is feasible if :
  • P ∈Pi fP = ri
  • Each edge has a nonnegative, differentiable, nondecreasing

latency function le(·)

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Cost for Flows

  • Let (G, r, l) be an instance , and f is a flow.
  • Latency of a path P
  • Cost of a flow f:
  • Players are small flows behave ”greedily” and ”selfishly”

lP (f) =

e∈P le(fe)

C(f) =

P ∈P lP (f)fP = e∈E le(fe)fe

There are infinite number of players, each carry a negligible amout of flow.

fe =

P :e∈P fP

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Flows at Nash Equilibrium

  • Definition (Nash Equilibrium):

A flow f is feasible 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 + δ if P = P2 fP if P / ∈ {P1, P2}

  • Lemma: A flow f feasible for instance (G, r, l) is at Nash

Equilibrium if and only if for all i ∈ {1, . . . , k}, P1, P2 ∈ Pi with fP1 > 0, lP1(f) ≤ lP2(f).

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Optimal Flows via Convex Programming

  • NonLinear Programming Formulation
  • P ∈Pi

fP = ri ∀i ∈ {1, . . . , k} fe =

  • P ∈P:e∈P

fP ∀e ∈ E fP ≥ 0 ∀P ∈ P Min

  • e∈E

ce(fe) subject to:

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Characteristic of Optimal Flows

  • Lemma: A flow f is optimal for a convex program of the

previous form if and only if for every i ∈ {1, . . . , k} and P1, P2 ∈ Pi with fP1 > 0, c′

P1(f) ≤ c′ P2(f).

c′

P (f) = e∈P c′ e(fe)

Let c′

e be the derivative d dxce(x)

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Characteristic of Optimal Flows

  • Lemma: A flow f is optimal for a convex program of the

previous form if and only if for every i ∈ {1, . . . , k} and P1, P2 ∈ Pi with fP1 > 0, c′

P1(f) ≤ c′ P2(f).

c′

P (f) = e∈P c′ e(fe)

Let c′

e be the derivative d dxce(x)

  • Lemma: A flow f feasible for instance (G, r, l) is at Nash

Equilibrium if and only if for all i ∈ {1, . . . , k}, P1, P2 ∈ Pi with fP1 > 0, lP1(f) ≤ lP2(f).

C(f) =

k

  • i=1

Li(f)ri

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Nash Equilibrium and Optimal Flow

  • Corollary: Let (G, r, l) be an instance in which x · le(x)

is a convex function for each edge e, with marginal cost functions l∗

  • e. Then a flow f feasible for (G, r, l) is optimal

if and only if it is at Nash equilibrium for the instance (G, r, l∗)

l∗

e(fe) = (le(fe)fe)′ = le(fe) + l′ e(fe)fe

Marginal cost function:

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Nash Equilibrium and Optimal Flow (cont’)

  • Lemma: An instance (G, r, l) with continuous, nonde-

creasing latency functions admits a feasible flow at Nash

  • equilibrium. Moreover, if f, ˜

f are flows at Nash equilib- rium, then C(f) = C( ˜ f).

Proof:

  • P ∈Pi

fP = ri ∀i ∈ {1, . . . , k} fe =

  • P ∈P:e∈P

fP ∀e ∈ E fP ≥ 0 ∀P ∈ P

Min

  • e∈E

he(fe)

Set he(x) = x

0 le(t)dt

Note, h′

e(x) = le(x)

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”Unique” Nash Equilibrium

  • Lemma: An instance (G, r, l) with continuous, nonde-

creasing latency functions admits a feasible flow at Nash

  • equilibrium. Moreover, if f, ˜

f are flows at Nash equilib- rium, then C(f) = C( ˜ f).

Proof (cont’) :

Min

  • e∈E

he(fe)

Set he(x) = x

0 le(t)dt

If fe = ˜ fe, the function he(x) must be linear and le is a constant function This implies le(fe) = le( ˜ fe). C(f) = k

i=1 Li(f)ri = C( ˜

f).

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Nontrivial Upper Bound for Price of Anarchy ρ = ρ(G, r, l) =

C(f) C(f ∗)

ρ(G, r, l) ≤ α For instance (G, r, l), let f ∗ be an optimal flow and f be a flow at Nash equilibrium.

Corollary: Suppose the instance (G, r, l) and the constant α ≥ 1 satisfy:

x · le(x) ≤ α · x

0 le(t)dt

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Nontrivial Upper Bound for Price of Anarchy (cont’) ρ(G, r, l) ≤ α

Corollary: Suppose the instance (G, r, l) and the constant α ≥ 1 satisfy:

x · le(x) ≤ α · x

0 le(t)dt

Proof:

C(f) =

  • e∈E

le(fe)fe ≤ α

  • e∈E

fe le(t)dt ≤ α

  • e∈E

f ∗

e

le(t)dt ≤ α

  • e∈E

le(f ∗

e )f ∗ e

= α · C(f ∗)

N.E optimizes this ob- jective function.

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Upper Bound for Polynomial Latency Function ρ(G, r, l) ≤ p + 1

Corollary: Suppose the instance (G, r, l) has the latency func- tions:

le(x) = p

i=0 ae,ixi

Remarks: It is not tight. ae,i ≥ 0 le(x) = aex + be for ae, be ≥ 0 ρ ≤ 2 For higher degree polynomial latency functions: ρ = O( p

ln p)

Tight Bound: ρ ≤ 4/3

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A Bicriteria Result for General Latency Functions

Negative Result: :

If l(x) = xp: 1 − p(p + 1)−(p+1)/p → 0 s t

l(x) = 1

l(x) = x

ρ = 4/3 Optimal flows assgins (p + 1)−1/p on the lower link, which has a total latency: ρ → ∞

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Augment Analysis for General Latency Function Let

  • Theorem: If f is a flow at Nash equilibrium for (G, r, l)

and f ∗ is feasible for (G, 2r, l), then C(f) ≤ C(f ∗)

¯ le(x) =

  • le(fe)

if x ≤ fe le(x) if x ≥ fe

  • e

¯ le(f ∗

e )f ∗ e − C(f ∗)

=

  • e∈E

f ∗

e (¯

le(f ∗

e ) − le(f ∗ e ))

  • e∈E

le(fe)fe = C(f)

  • e

¯ lP (f ∗)f ∗

P

  • i
  • P ∈Pi

Li(f)f ∗

P

=

  • i

2Li(f)ri = 2C(f)

¯ lP (f∗) ≥ ¯ lP (f0) ≥ Li(f)

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Worst-Case Ratio with Linear Latency Fuctions

  • Lemma: If (G, r, l) be an instance with edge latency func-

tions le(x) = aex + be for each edge e ∈ E. Then

le = aex + be with ae, be ≥ 0 l∗

e = 2aex + be

(a) a flow f is at Nash equilibrium in G if and only if for P, P ′ ∈ Pi with fP > 0,

  • e∈P aefe + be ≤

e∈P ′ aefe + be

(b) a flow f ∗ is (globally) Optimal in G if and only if for P, P ′ ∈ Pi with f ∗

P > 0,

  • e∈P 2aef ∗

e + be ≤ e∈P ′ 2aef ∗ e + be

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Worst-Case Ratio with Linear Latency Fuctions (cont’)

  • Lemma: Suppose (G, r, l) has linear latency functions and

f is a flow at Nash equilibrium. Then (a) The flow f/2 is optimal for (G, r/2, l) (b) the marginal cost of increasing the flow on a path P for f/2 equals the latency of P for f

Creating optimal flow in two steps: (f is at Nash equilibrium)

(1) Send a flow optimal for instance (G, r/2, l). C(f)/4 (2) Augment to one optimal for instance (G, r, l). C(f)/2

l∗

P (f/2) = lP (f)

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Augment Cost for Linear Latency Functions

  • Lemma: (G, r, l) has linear latency functions and f ∗ is

an optimal flow. Let L∗

i (f ∗) be the minimum marginal

cost for si − ti paths. For any δ > 0, a feasible flow f for (G, (1 + δ)r, l):

x · le(x) = aex2 + be is convex. C(f) ≥ C(f ∗) + δ k

i=1 L∗ i (f ∗)ri

le(fe)fe ≥ le(f ∗

e )f ∗ + (fe − f ∗)l∗ e(f ∗ e )

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Augment Cost for Linear Latency Functions

  • Proof:

C(f) =

  • e∈E

le(fe)fe ≥

  • e∈E

le(f ∗

e )f ∗ e +

  • e∈E

(fe − f ∗

e )l∗ e(f ∗ e )

= C(f ∗) +

k

  • i=1
  • P ∈Pi

l∗

P (f ∗)(fP − f ∗ P )

≥ C(f ∗) +

k

  • i=1

L∗

i (f ∗)

  • P ∈Pi

(fP − f ∗

P )

= C(f ∗) + δ

k

  • i=1

L∗

i (f ∗)ri

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Worst-Case Ratio with Linear Latency Fuctions (cont’)

  • Lemma:

If (G, r, l) has linear latency functions, then ρ(G, r, l) ≤ 4/3

Let f be a flow at N.E. f/2 is optimal for (G, r/2, l). Moreover, L∗

i (f/2) = Li(f).

Proof:

C(f ∗) ≥ C(f/2) +

k

  • i=1

L∗

i (f/2)ri

2 = C(f/2) + 1 2

k

  • i=1

Li(f)ri = C(f/2) + 1 2C(f) ≥ 3 4C(f) C(f/2) = 1 4aef 2

e + 1

2befe ≥ 1 4

  • e

(aef 2

e + befe)

= 1 4C(f)

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Extensions:

  • Approximate Nash Equilibrium:

If f is at ǫ N.E, and f ∗ is feasible for (G, 2r, l), then C(f) ≤ 1+ǫ

1−ǫC(f ∗).

  • Finite Agents: Splittable Flow

C(f) ≤ C(f ∗).

  • Finite Agents: Unsplittable Flow

If for some α < 2, le(x+ri) ≤ α·le(x), x ∈ [0,

j=i rj]

C(f) ≤

α 2−αC(f ∗).