MS&E 246: Lecture 18 Network routing. Ramesh Johari

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Transcription:

MS&E 246: Lecture 18 Network routing Ramesh Johari

Network routing Last lecture: a model where N is finite Now: assume N is very large Formally: Represent the set of users as a continuous interval, [0, B] B represents the total amount of flow Each user is infinitesimal (also called nonatomic)

Network routing We ll consider networks of parallel links. l 1 (n 1 ) s l 2 (n 2 )... d l J (n J ) J = # of links; n j = flow on link j

The network routing game Strategy space of each user: Links from s to d Cost to each user: Delay experienced on chosen link

The network routing game Formally: Delay on link j = l j (n j ) 0 (assume l j is strictly increasing) Strategy space = { 1,, J } Payoff = l j (n j ) if link j chosen

The network routing game What is a pure NE in this game? Each infinitesimal user must have chosen the best path available, given links chosen by all others. Formally: If n j > 0, then l j (n j ) l k (n k ) for all k j

Wardrop equilibrium A Wardrop equilibrium n is a pure NE of this game: (1) j n j = B (2) If n j > 0, then l j (n j ) l k (n k ) for all k j Does a WE exist? Is it unique?

Wardrop equilibrium Recall last lecture: we found a potential Intuition: replace sum by an integral.

Wardrop equilibrium Theorem: There exists a unique WE n. It is the unique solution to: minimize J j = 1 nj 0 l j (z) dz subject to J j = 1 n j = B

Wardrop equilibrium: proof Define V (n) = J j = 1 nj 0 l j (z) dz Suppose n is optimal, and n j > 0. Consider moving δ > 0 units of flow to link k j. What is the change in V?

Wardrop equilibrium: proof The j th term falls by: l j (n j ) δ The k th term increases by: l k (n k ) δ Since n is optimal, this change cannot reduce the cost. So: l k (n k ) δ - l j (n j ) δ 0

Wardrop equilibrium: proof The j th term falls by: l j (n j ) δ The k th term increases by: l k (n k ) δ Since n is optimal, this change cannot reduce the cost. So: l j (n j ) l k (n k ) for all k j

Wardrop equilibrium: proof Conclude: Any optimal solution is a Wardrop equilibrium. At least one optimal solution exists (the feasible set is closed and bounded). So at least one WE exists.

Wardrop equilibrium: proof To check uniqueness: Since l j is strictly increasing, each term of V is strictly convex in n j (strictly convex for all α [0,1], x, x, f(α x + (1 - α) x ) < αf(x) + (1 - α) f(x ) second derivative is strictly positive)

Wardrop equilibrium: proof Since l j is strictly increasing, each term of V is strictly convex in n j First implication: Any WE is also an optimal solution. (Second order conditions automatically hold.)

Wardrop equilibrium: proof Since l j is strictly increasing, each term of V is strictly convex in n j Second implication: If n n are two solutions, then (n + n )/2 is a feasible solution with strictly lower cost. So optimal solution (and hence WE) is unique.

Total delay We again expect Wardrop equilibria will not be Pareto efficient. One way to find a Pareto efficient point: minimize total delay. Total delay = TD(n) = j = 1J n j l j (n j )

Total delay Let s assume l j is affine for each j, i.e., l j (n j ) = a j + b j n j, a j 0, b j 0 Then: TD(n) = J j = 1 a j n j + b j n 2 j

Total delay Let s assume l j is affine for each j, i.e., l j (n j ) = a j + b j n j, a j 0, b j 0 Then: TD(n) = J j = 1 nj 0 (a j + 2 b j z ) dz

Total delay Let s assume l j is affine for each j, i.e., l j (n j ) = a j + b j n j, a j 0, b j 0 Then: TD(n) = J j = 1 nj 0 (a j + 2 b j z ) dz So a flow allocation that minimizes delay is a WE with respect to different latencies: m j (n j ) = a j + 2 b j n j

Pigovian taxes In general, WE does not minimize total delay. Pigovian taxes charge tolls to provide the right incentives to users. (Named for the economist Pigou.)

Pigovian taxes Suppose on link j, a toll is charged, T j (n j ). Also suppose that users value time and money identically. Then a user of link j has payoff: l j (n j ) + T j (n j )

Pigovian taxes Notice that if we choose: T j (n j ) = b j n j, then l j (n j ) + T j (n j ) = m j (n j ) Moral: A Nash equilibrium with these tolls minimizes total delay.

Pigovian taxes More generally: If l j is strictly convex and strictly increasing, the Pigovian tax on link j is: T j (n j ) = n j l j (n j )

Efficiency: example Example: l 1 (n 1 ) = 1 1 unit of flow s d l 2 (n 2 ) = n 2

Efficiency: example Wardrop equilibrium: n 1 = 0, n 2 = 1 Total delay = 1 Minimizing total delay: n 1 = 1/2, n 2 = 1/2, Total delay = 3/4 So WE cost is 4/3 higher than the optimum.

Efficiency When l j are affine, the 1/3 increase in total delay is the worst possible (over all choices of l j ) (Roughgarden and Tardos, 2002)

Other directions More complex user models: 1. Users may not be infinitesimal 2. Users may have different preferences for money and time 3. Users may care about the variance of their delay All these are much harder than WE.

Other directions Partial optimization in the network: ISPs frequently reroute traffic inside their own networks to improve performance. How does this interact with selfish routing from source to destination?

Other directions An ISP might set prices to maximize profit, not achieve minimum total delay. How efficient is profit maximization?

Other directions A service provider may wish to invest to improve the performance of existing links. What is the equilibrium of the resulting game of investment, pricing, and routing?

Other directions Service providers form contracts with each other to share traffic (e.g., peering). What contracts should they use to: -maximize profit? -maximize network performance?