Transcript Selfish Routing in Non

Selfish Routing in Non-cooperative
Networks
Balázs Sziklai
Historical Outline
 J.G. Wardrop (1952): Some theoretical aspects of road traffic
research, Proc. of the Institute of Civil Engineers
 D. Braess (1968): Über ein Paradox der Verkehrsplanung,
Unternehmensforschung
 E. Koutsoupias and C. Papadimitriou (1999): Worst Case
Equilibria, Proc. Of the 16th International Symposium on Theoretical
Aspects of Computer Science
 Feldmann et al. (2003): Selfish Routing in Non-cooperative
Networks: A Survey, Lecture Notes in Computer Science
The Wardrop Model
 A problem instance consists of a triple
 a directed graph
 a set of routing tasks
 and a set of edge latency functions
 For a fixed flow f the latency of an edge e is defined as the product
of the latency function on e when routing traffic f times the traffic
f itself.
 The social cost of a flow is defined to be the sum of the edge
latencies.
s
x+1
t
Pigou’s example
 We would like to route a total of one unit of flow from s to t.
 To obtain the optimal solution we have to compute the minimum
of the following quadratic polinomial:
 Simple calculation shows that the social optimum is to route ½ of
the traffic on the upper edge and ½ of the traffic on the lower
edge.
 However the latency cost is lower on the upper edge for any user,
which means it cant be a Nash equilibrium.
x
 The unique NE point is to route
all the traffic on the upper edge. s
t
1
Braess Paradox
 We would like to route 6 unit of traffic on the following graph:
s
10x
x + 50
x + 50
10x
t
Braess Paradox
 The social cost of the flow at the Nash Equilibrium:
2 ∙[3 ∙(10∙3+(3+50))] = 6 ∙ 83 = 496
s
3
3
3
3
t
Braess Paradox
 Now we connect the two middle points with an edge.
s
10x
x + 50
x +10
x + 50
10x
t
Braess Paradox
 Using the new edge some of the users can be better off by
changing their path from s to t.
s
3
3
1
2
4
t
83 = (3 ∙10) + (3+50) >
(3 ∙10) + (1+10) + (4 ∙10) = 81
Braess Paradox
 Using the new edge some of the users can be better off by
changing their path from s to t.
s
4
2
2
2
4
t
93 = (3+50) + (4 ∙10) >
(4 ∙10) + (2+10) + (4 ∙10) = 92
Braess Paradox
 Again we arrived to a NE point. No user can improve its private
latency by unilaterally changing its route.
s
4
2
2
2
4
t
The new social cost is
4 ∙(10∙4+(2+50))+
2 ∙(10∙4+(2+10)+ 10∙4) =
6 ∙ 92 = 558 > 496
How to resolve the Braess Paradox
 To find the ‘bad’ edges in a network is NP-hard (Roughgarden 2002).
 Capacity should be added across the network rather than on a local scale
(Korilis et al. 1995).
 Stackelberg approach: controlling just a small portion of the traffic the
system can be driven close into the network optimum (Korilis et al.
1995, Roughgarden 2001).
A physical representation of the
Braess Paradox
1 kg
1 kg
Price of Anarchy
 Introduced by Papadimitriou in a conference paper (2001).
 By definitions it is the ratio of the (worst case) Nash equilibrium
social cost and the optimal social cost.
 In the Wardrop model the coordination ratio cannot be bounded
from above by a constant, when arbitrary edge latency functions
are allowed.
 However when we restrict ourselves to linear edge latency
functions the coordination ratio is bounded from above by 4/3.
 This bound is tight (see Pigou’s example).
KP-model
 Named after Koutsoupias and Papadimitriou (1999).
 Simple network with m paralell links and n users.
 Each user i has a weight wi and these traffics are unsplittable.
 A pure strategy of a user is a specific link, and a mixed strategy is a
probability distribution over the set of links.
s
t
KP-model
 The social cost is the expected maximum latency on a link, where
the expectation is taken over all random choices of the users.
 Basic model – identical links.
 General model – related links (different links have different
capacities).
FMNE conjecture
 Consider the instance of three identical links and three users with
traffic weight of 2.
Prob.
dist.
Link A
Link B
Link C
Prob.
dist.
Link A
Link B
Link C
User 1
1
0
0
User 1
4/9
1/9
4/9
User 2
0
1
0
User 2
2/3
1/3
0
User 3
0
0
1
User 3
0
1/3
2/3
Social Cost = 2
Social Cost = 3,53
A
s
B
C
t
Prob.
dist.
Link A
Link B
Link C
User 1
1/3
1/3
1/3
User 2
1/3
1/3
1/3
User 3
1/3
1/3
1/3
Social Cost = 3,77
FMNE conjecture
 Fully Mixed Nash Equilibrium is a NE where every user routes
along every possible edge.
 Consider the model of arbitrary traffics and related links. Then any
traffic vector w such that the fully mixed Nash Equilibrium F
exists, and for any Nash equilibrium P, SC(w, P) ≤ SC(w, F).
Differences between the two models
 Wardrop model defines the social cost as the sum of the edge
latencies while the KP-model as the expected maximum edge
latency on a link.
 While in the Wardrop-model traffic can be splitted into infinitely
tiny pieces, traffic rates are unsplittable in the KP-model.
 All Nash Equilibria are pure and have the same social cost in the
Wardrop-model.
Unsolved problems
 Give upper and lower bounds for the Price of Anarchy.
 How to design a network free from Braess paradox?
 Is the FMNE conjecture true?
 Find better algorithms to compute NE points.
Thank you for your attention!