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Algorithmic Game Theory September 15, 2008
Lecture 2: Selfish Routing
Lecturer: Sergei Vassilvitskii Scribe:Etienne Vouga
Recall from the first lecture that we are working graphs G : (V, E) with vertices V and directed edges E. We also specify two distinct special vertices, s ∈ V and t ∈ V , termed the source and sink respectively, and associate monotonic edge latency functions !e : [0, 1] → R +^ to each edge e ∈ E. We let P = {p 1 ,... , p (^) n } be the set of paths from s to t, and define a flow f to be an assignment to each path p (^) i a nonnegative real number fpi , with
∑n i=1 fpi^ = 1. One intuitive interpretation of flow is to imagine 1 unit of infinitely splittable supply starting at s, and trying to travel to t. Then an fpi fraction of supply travels on each path p (^) i. We consider only simple paths, that is each path p (^) i has no cycles, and passes through each edge at most once. We write e ∈ p (^) i if p (^) i passes through edge e. Then for a flow f of G, we can compute the flow fe through an edge e by summing the contribution of each path through e: fe =
i|e∈pi
fpi.
The latency contributed by edge e is then !e (fe ), and the cost cpi (fpi ) induced by f on a path p (^) i is cpi (f ) =
e∈pi
fpi !e (fe ).
The total cost C(f ) is then the sum of the costs of all of the paths: C(f ) =
i
cpi (f )
i
e∈pi
fpi !e (fe )
e∈E
pi ∈P
fpi !e (fe )
e∈E
fe !e (fe )
The last equality follows after switching the order of summations, and will prove useful to us later. Given the above structure, there are two canonical problems we can pose: first, we can try to find an optimal routing, a flow f ∗^ which minimizes the total cost C(f ∗^ ). Second, we can look for a Nash equilibrium f , in which, intuitively, every infinitesimal “piece” of supply acts selfishly and takes the cheapest available path to the sink, regardless of the consequences this choice has on the rest of the supply. We give a more precise formulation of Nash equilibria below. Given the two solutions flows f ∗^ and f , we can consider the ratio (^) CC((ff ∗^ ) (^) ) , the price paid by the supply for acting greedily instead of in concert. This is also known as the price of anarchy. The remainder of the lecture proves the following theorem:
Theorem 1 In the case of linear latency functions, l (^) e , the price of anarchy, (^) CC((ff ∗^ ) (^) ) ≤ 43.
1 Nash Equilibrium
Suppose we have a flow f on G, and any two paths p 1 , p 2 with fp 1 > 0. If you’re flowing along p 1 and acting greedily, and flowing along p 2 instead would be cheaper, you would switch; hence, a necessary condition on f being a Nash equilibrium is that ∑
e∈p 1
!e (fe ) ≤
e∈p 2
!e (fe ) ∀p 2 ∈ P, p 1 ∈ P |fp 1 > 0. (1)
It can be shown the above condition is also sufficient.
2 Optimal Solution
Again, consider a flow f on G, p 1 , p 2 paths, with fp 1 > 0. Suppose we were to move some amount of flow δ from p 1 to p 2. On the one hand, the cost along p 1 will decrease since less supply is moving through it, but on the other hand, the const along p 2 will increase. If f is optimal, such a switch cannot improve the total cost, so the benefit cannot outweigh the cost. Writing it down formally, the total cost of the original flow f is:
C(f ) =
e
ce (fe ) =
e∈p 1 e∈p 2
ce (fe ) +
e∈p 1 e#∈p 2
ce (fe ) +
e#∈p 1 e∈p 2
ce (fe ) +
e#∈p 1 e#∈p 2
ce (fe ).
Consider a flow f ′^ where we take a δ fraction of flow from p 1 and route it on p 2 instead. Notice that for edges e contained both in p 1 and p 2 or neither p 1 nor p 2 the flow doesn’t change. Formally, :
f (^) e′ =
fe e ∈ p 1 , e ∈ p (^2) fe − δ e ∈ p 1 , e &∈ p (^2) fe + δ e &∈ p 1 , e ∈ p (^2) fe e &∈ p 1 , e &]inp 2
The total cost of f ′^ is:
C(f ′^ ) =
e
ce (f (^) e′ ) =
e∈p 1 e∈p 2
ce (fe ) +
e∈p 1 e#∈p 2
ce (fe − δ) +
e#∈p 1 e∈p 2
ce (fe + δ) +
e#∈p 1 e#∈p 2
ce (fe ).
If f is the optimal solution, then:
C(f ) ≤ C(f ′^ ) ∑
e∈p 1 e#∈p 2
ce (fe ) +
e#∈p 1 e∈p 2
ce (fe ) ≤
e∈p 1 e#∈p 2
ce (fe − δ) +
e#∈p 1 e∈p 2
ce (fe + δ)
e∈p 1 e#∈p 2
(ce (fe ) − ce (fe − δ)) ≤
e#∈p 1 e∈p 2
(ce (fe + δ) − ce (fe ))
Lemma 1 Let f ∗^ be an optimal partial flow of some amount k < 1 of supply, and f any flow of k + δ supply. Then C(f ) ≤ C(f ∗^ ) + δL∗^ (f ∗^ ),
where L∗^ (f ∗^ ) is the marginal cost of increasing the flow starting with f ∗^ , L∗^ (f ∗^ ) =
i c ′ pi (f^ ∗ pi ).
Proof. Notice that since f ∗^ is optimal, we know that for any two paths p 1 and p 2 , the marginal costs are the same,
e∈p 1 c ′ e (f^ ∗ e ) =^
e∈p 2 c ′ e (f^ ∗ e ) =^ L ∗ (^) (f ∗ (^) ). Consider a flow f ,
with cost
C(f ) =
e∈E
fe !e (fe )
e∈E
f (^) e∗ !e (fe ) +
e∈E
(fe − f (^) e∗ )c′ e (f (^) e∗ )
The inequality follows due tot he convexity of the cost function. It is each to check that fe > f (^) e∗ then ce (fe ) > ce (f (^) e∗ ) + (fe − f (^) e∗ )c′ e (fe ). On the other hand, if fe < f (^) e∗ then ce (fe ) < ce (f (^) e∗ ) + (fe − f (^) e∗ )c′ e (fe ). Continuing, we have:
C(f ) ≥
e∈E
f (^) e∗ !e (fe ) +
e∈E
(fe − f (^) e∗ )c′ e (f (^) e∗ )
= C(f ∗^ ) +
pi ∈P
(fpi − f (^) p∗i )c′ pi (f (^) p∗i )
= C(f ∗^ ) +
pi ∈P
(fpi − f (^) p∗i )L∗^ (f ∗^ )
= C(f ∗^ ) + δL∗^ (f ∗^ )
! We are now ready to prove the theorem. Proof of Theorem 1: Letting f ∗^ be the optimal flow, f the Nash flow, and f 2 half of the Nash flow. We know from the discussion above that for the path p (^) i , the marginal cost
c′ pi
f 2 pi
is equal to the Nash latency !ei (feI ). Thus L∗
f 2
= C(f ), and so, applying the
lemma with δ = 12 ,
C(f ∗^ ) ≥ C
f 2
L∗
f 2
≥ C
f 2
C(f ).
It remains to calculate C
f 2
C
f 2
e
fe 2
l (^) e
fe 2
e
fe 2
ae fe 2
e
ae f (^) e^2 4
b (^) e fe 2
e
ae f (^) e^2 + b (^) e fe
C(f ),
so C(f ∗^ ) ≥ 34 C(f ), and C(f ) C(f ∗^ )
