Post on 13-Jan-2016
description
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
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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:
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)(' 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!
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*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