How Much Can Taxes Help Selfish Routing?

How Much Can Taxes Help Selfish Routing?Richard Cole Yvgeniy Dodis Tim Roughgarden

Presented By: Omry Tuval

Some examples are taken form Tim Roughgarden’s slides

Selfish Routing

a directed graph G=(V,E) Source vertex s, sink vertex t Each edge has a latency function Infinitely large population Traffic is selfish – minimize latency Social goal – total latency

Selfish Routing - Example

Flows the fraction of traffic flowing on path P from s to t. The flow function is the routing of the traffic

Interested in the stable state considering selfish traffic. (Nash Equilibrium)

pf

Nash Flow – Def .

A flow is a “Nash Flow” if all traffics is routed along minimum cost paths given current edge congestion

Existence and Uniqueness [wardrope, beckmann 1950s]

f

Price Of Anarchy – How Bad Is Selfish Routing?

POA is unbounded

POA in linear latency networks is 4/3

Economic Incentives - Taxes

Encouraging desired behavior Each edge has usage tax: Users selfishly try to minimize

personal cost = personal latency + personal tax

Hopefully will result in better global cost

eT

Marginal Cost Taxes [Beckmann et al. 1956]

Each user should pay a tax equal to the additional delay other users experience because of its presence With regard to a flow Formal:

f

)(' eeee flfT

MCT Theorem [beckmann 1959]

Let be the optimal flow of original network. Let be the MCT with regard to The Nash flow in the taxed network is

MCT minimized total latency Social injustice Total cost = total latency + total tax Total cost might be higher then original network

Should study taxation with regard to total cost!

*f

*feT *f )(' **

eeee flfT

Let’s get Mathematical…

G=(V,E) directed graph with source s and sink t P: the set of simple s-t paths in G Flow is a function Given a flow we define

Given a traffic amount r, a flow f is feasible if

RPf :

pePppe ff

|

rfPp

p

Let’s get more Mathematical...

Each edge has a continuous, non negative, non decreasing latency function

Each edge has a non negative tax We denote the cost of a feasible flow

We denote an instance of the game by and a taxed instance by

eleT

Ee

eee TflfTfC e ))((),(

),,( lrG),,( TlrG

Nash Flow properties

Claim 1: let be a Nash flow for then there is a constant such that:

Proof: straight from definition. We will denote that constant as Claim 2: is non decreasing in r Proof: [Hall 1978]

f ),,( TlrG c

crTfC ),( )(0 cTflf pppp

),,( TlrGc ),,( TlrGc

MCT can be bad for you

Result: In all linear latency networks, using MCT can only increase the cost of the Nash flow

Theorem: let be an instance with linear latency functions and be the corresponding marginal cost taxes. Let be the Nash flow for and be the Nash flow for then:

Proof

),,( lrGT

f Tf),,( TlrG

),,( lrG

),()0,( TfCfC T

Taxes – How good can it get?

For linear latency networks, taxes can improve Nash cost (at best) by a factor of 4/3

Proof

For general latency networks, taxes can improve Nash cost at best by a factor of n/2

Proof: extension of [Roughgarden 2001]

Taxes vs. Edge Removal

A different approach to improve selfish routing Studied in [Roughgarden FOCS 2001] Taxes at least as powerful as Edge Removal

Infinitely large tax means practically removing the edge

When are taxes actually better than Edge Removal?

Taxes vs. Edge RemovalLinear networks Theorem: for an instance with linear latency functions,

the optimal tax is a tax Proof sketch

For linear networks, taxes never improve over edge removal.

/0

Taxes vs. Edge RemovalGeneral networks Result: for each n, exists a selfish routing

instance where taxes are better than edge removal by a factor of n/2.

Proof

Taxes can be usefulComputability?

We’ve seen that in general latency networks, taxes can improve the Nash cost, and even improve on edge removal

Can the optimal tax be computed efficiently? The problem is NP-hard What about approximations?

“Forget about it” theorem

If there is NO approximation algorithm for linear latency

networks approximation algorithm for general

latency networks The trivial algorithm (return T=0 as approx.) is:

approximation algorithm for linear latency networks

approximation algorithm for general latency networks

NPP

3

4

)( 1 nO

3

4

2/n

Conclusion

The problemLinearGeneral

Can marginal cost taxes help?

NoYes

Maximum benefit of taxes

4/3n/2

Taxes better than edge removal?

noyes

Approximability of optimal taxes

4/3=<n/2

Not sublinear

MCT can be bad for youproof Let be the optimal min. latency flow for Linear latency: MCT:

Define: For each edge:

Therefore is Nash not only of but also of and with the same cost

*f ),,( lrGeee bxaxl )(

*** )(' eeeeee faflfT

eee bxaxl 2)(*

),,( *lrG),,( TlrG

eeeeeeee Tflbfafl )(2)( ****

),,(),,(

),,(),,(),,(),,(*

**

TlrGclrGc

TlrGNashCostTlrGcrlrGcrlrGNashCost

Tf

MCT can be bad for youproof is the Nash flow of

Path cost with regard to flow in is identical to path cost with regard to flow in

Therefore is the Nash flow in and

QED

f ),,( lrG

2/f ),2/,( *lrG

eee bxaxl )( eee bxaxl 2)(*

),,(),,(

),,(),,(

),,(),,(),2/,(),,( **

TlrGNashCostlrGNashCost

TlrGcrlrGcr

TlrGclrGclrGclrGc

f ),,( lrG

2/f ),2/,( *lrG

Taxes in linear networks

For linear latency networks, taxes can improve Nash cost (at best) by a factor of 4/3

Let be an instance with linear latency with Nash flow and optimal flow .

POA for linear latency is 4/3

),,( lrG*f

f

e e

eeeeee flfflf )(4

3)( **

Taxes in linear networks

Let be the optimal taxes, and the Nash flow forTf*T),,( *TlrG

),,(4

3),,(

),,(4

3

)(4

3)()(

))((),,(

*

**

**

lrGNashCostTlrGNashCost

lrGNashCost

flfflfflf

TflfTlrGNashCost

eeee

eeee

e

Tee

Te

ee

Tee

Te

QED

Taxes vs. Edge RemovalLinear networks Theorem: for an instance with linear latency functions,

the optimal tax is a tax Assume false, look at minimal counterexample Look at counterexample optimal tax that has the smallest

sum (existence if proven by minimality) Understand how Nash flow change under local changes

in the tax (linear equations) Perturbing to a smaller tax will increase cost Opposite perturbation lower costs (contradiction) QED

/0

Taxes vs. Edge RemovalGeneral networks Theorem:

For each integer there is an instance such that for all subgraphs of it holds that but for some tax it holds that

for simplicity n is even, and n=2k+2 Our network will be the Braess graph,

2n ),,( lrG

2),,(n

lrHc H G1),,( TlrGcT

k kB

Kth Braess Graph

Vertices Edges:

Type A: Type B: Type C:

},,....,,,...,,{ 11 twwvvs kk

),( ii wv),(),,(),,( 11 kii wstvwv

),(),,( 1 iki vstw

Latency functions

0)( xlA

)1/(1

2

)1/(11)(

kxn

kxxlB

kxn

kxi

kx

xl

vstw

iC

iki

/112

/11

)1/(110

)(

),(),,(for

,

1

Lemma

If is a subgraph of and a Nash flow saturates an edge in then H kB

2),,(n

lrHc ),,( lrH

s-t paths in the graph

k paths Pi of the form:

k-1 paths Qi of the form:

Q1 is the path:

QK is the path:

twvs ii twvs ii 1

tvs 1

tws k

Taxes are good

),( ii wv),1,( TlkBk Consider

Suppose we tax each edge with 1 unit of tax, and 0 elsewhere

The following flow is Nash flow: 1 unit of flow on Pi 1/(k+1) units of flow on Qi

This shows 1),1,( TlkBc k

Edge Removal is bad

We now must show that for every subgraph of

Suppose Nash Flow

1+1/k units of flow on Pi

Also true if removes only type B edges

H kB2/),1,( nlkHc

kBH

2/),1,( nlkHc H

Type C removal

Suppose removes a type C edge, say How much flow can leave S without saturation? At Most

An edge adjacent to S is saturated!

),( ivsH

)1

1)(1(1

1 k

kk

1)1

1

1(

kkkk

kkk

2),,(n

lrHc

Type A removal

Suppose removes some type A edge, say How much flow can leave S without saturation? At most

An edge in the graph is saturated!

H ),( ii wv

)1

1)(2()1

1(2

kk

k

1)1

1

1(22

kkekk

kkk

2),,(n

lrHc