How bad is selfish routing?

1
How bad is selfish routing?

• ??? ????? ????.
• 027256569
• ??"? ???? ?? ???? ????? ???? ??-??????.

2
The problem

• We are given
• Network.
• Rate of traffic between nodes.
• Latency time.
• We should route the traffic such that the sum of
  all travel time is minimized.
• Difficult aspect when link traverse is
  load-dependent.

3
Assumptions

• Finding optimal routing is difficult or even
  impossible.
• No network regulations.
• Users are purely selfish, but not malicious.
• Users know the links congestions.
• In other words Network routing is no-cooperative
  game and the routes form Nash-Equilibrium.
• Most of the time each user/agent controls
  negligible fraction of the traffic.

4
Brasss Paradox

• Consider the following network When 1 (divided
  to many fractions and agents) unit of flow need
  to be routed.
• In the upper network, Nash equilibrium Which is
  the optimal flow is half of the traffic goes up
  and half down. Total latency time is 10.51.5
• In the lower network, we added 0 edge. The new
  optimal flow does not change, but Nashs flow is
  s-gtv-gtw-gtt. And the total latency is 112

v
x
1
s
t
1
x
w
v
x
1
0
s
t
1
x
w
5
Model

• Directed network Graph G(V,E) V-vertex,
  E-edges.
• K source-destination vertex pairs s1,t1 ..
  sk,tk.
• f i is a set of simple si-ti path. f Ui f i.
• A flow f f -gt
• fe
• ri The rate - the amount of flow from si to
  ti.
• f is feasible if for all i
• Each edge e has latency function le()
• The latency of path p with flow s is
• le is non-negative, continuous and
  non-decreasing.
• The cost C(f) of flow f is
• The Cost ca also be written as

6
Flows and Nash equilibrium, definition

• Flows behave selfishly and greedily
• Definition A flow f in G is at Nash equilibrium
  if for all i ?1, . . . , k, P 1 , P 2 ? P i ,
  and d ?0, fP1 , we have lP1 (f) lt lP2 (ƒ),
  where ƒp is
• If PP1 fp- d
• If PP2 fp d
• Else fp
• And of course we let d -gt0

7
Flows and Nash equilibrium

• Lemma 1 A flow f is at Nash equilibrium if and
  only if for every i ? 1, . . . , k and P1,P2?P
  i with fP1 gt 0, lP1(f) lt lP2 (f).
• Conclusion if f is at Nash equilibrium then the
  latencies of all paths from si to ti (that have
  positive flow), Li(f), are equal.
• Lemma 2 If f is feasible flow at Nash
  equilibrium then
• The definition correspond to pure strategy since
  every agent is negligible fraction from the
  traffic.

8
Optimal flow

• We should find the minimum of
• Subject to (NLP)
• fpgt0
• Where ce(fe)le (fe)fe
• This is a non-linear program (NLP).

9
Local optimal flow

• We expect a flow to be locally optimal if and
  only if moving flow from one path to another
  increase the total flow.
• In other words the gradient on any path from
  si-ti should be no greater then any other paths
  gradient from si-ti.
• By applying Karush-Kuhn-Tucker Theorem we know
  that local optima coincide global optima in
  convex function.
• Let
• Lemma 3 A flow f is optimal for a convex
  program of the form (NLP) if and only if for
  every i ? 1, . . . , k and P1,P2 ? f i with fP1
  gt 0, cP1(f) cP2(f).
• We can see the similarity to Nash equilibrium.
  Optimal flow can be regarded as an Nash
  equilibrium with respect to latency function c.
• Lemma 4 A network G with continuous,
  non-decreasing latency functions admits a
  feasible flow at Nash equilibrium. Moreover, if
  f, f are flows at Nash equilibrium, then C(f)
  C(f).

10
General latency function

• For network G admitting an optimal flow f and a
  flow at Nash equilibrium f, we denote the ratio
  C(f)/C(f) by ? ?(G, r, l)
• Braesss paradox ? 2/1.5 4/3.
• In this new simple example Nash-(0,1), optimal
  (½, ½) -gt ?1/0.754/3.
• If the function is not linear, Xk. optimal flow
  assign (k1)(-1/k) on the lower link. The total
  latency is 1-k(k1)(k1)/k, that tends to 0
  if k-gt 8.
• We will consider what happens if we increase the
  r. when rgt1, 1 unit of flow will be routed in
  the lower link and r-1 in the upper.
• We will show that the optimal flows latency of
  2r is not less then the latency at Nash
  equilibrium with rate r.

1
X
11
General latency function

• Theorem If f is a flow at Nash equilibrium for
  (G, r, l) and f is feasible for (G, 2r, l), then
  C(f) lt C(f).
• Proof Suppose we have f, f. The cost of the
  flow is
• We will define a close latency that can easily
  lower bound the cost
• le(x)
• le(fe) if xltfe
• le(x) if xgtfe

12
General latency function

• We compare f under l to the original l (C(f)).
• For any edge e, le(x) - le(x) is zero for x gt fe
  and bounded above by le(fe) for x lt fe, so
  xle(x) - le(x) lt le(fe)fe.
• On the other hand, if f0 denotes the zero flow in
  G, then by construction lP(f0) gtLi(f) for any
  path p. and therefore lP(f)gtLi(f) for each path
  P.
• The cost of f with respect to l can be bounded
  below
• Which leads to
• That can be generalized if f is Nash eq at
  (G,r,l) and f at (g,(1d),l) then C(f)ltC(f)/
  d.

13
Linear latency

• Linear latency le(fe) aefebe for ae,begt0.
• The total latency is Se aefe2befe.
• Since aegt0 then (NLP) is convex. And le(fe) is
  2aefebe.
• Lemma 5 Let (G, r, l) be an instance with edge
  latency functions leaex be for each e ? E.
  Then
• (a) a flow f is at Nash equilibrium in G if and
  only if for each i and P,Pi ? f i with fP gt
  0,
• (b) a flow f is (globally) optimal in G if and
  only if for each i and P,Pi ? f i with fP gt 0
• Conclusion Let G be a network in which each edge
  latency function e is of the form le(fe) aex.
  Then for any rate vector r, a flow feasible for
  (G, r,l) is optimal if and only if it is at Nash
  equilibrium.

14
Linear latency

• Lemma 6 Suppose (G, r, l) has linear latency
  functions and f is a flow at Nash equilibrium.
  Then,
• (a) the flow f/2 is optimal for (G, r/2, l)
• (b) the marginal cost of increasing the flow on a
  path P with respect to f/2 equals the latency of
  P with respect to f.
• Proof (b) If edge e has latency function le(x)
  aex be then e has marginal cost function le(x)
  2aex be. Thus, le(fe/2) le (fe) for each
  edge e and hence lp (f/2) lp(f) for each path
  P.

15
Linear latency

• Lemma 7 Suppose (G, r, l) is an instance with
  linear latency functions for which f is an
  optimal flow. Let Li (f) be the minimum
  marginal cost of increasing flow on an si-ti path
  with respect to f. Then for any d gt 0, a
  feasible flow for the problem instance (G,
  (1d)r, l) has cost at least
• Proof fix d gt 0, suppose f is feasible for
  (G,(1d)r,l). fe may be smaller or larger then
  fe. x le(x) aex2bex implies that
• le (fe)fe gt le (fe)fe (fe - fe) le
  (fe).
• (fe - fe)le(fe) is a bound to the cost of
  changing the flow value (if fegtfe it
  underestimate the increasing, and if feltfe it
  overestimate the decreasing)
• Since Li (f) is minimal (unless fp0), we
  obtain

16
Linear latency

• Theorem If (G, r, l) has linear latency
  functions, then ?(G, r, l) 4/3.
• Proof
• f-Nash equilibrium flow. Li(f) is latency on
  si-ti so
• f/2 is optimal solution to (G, r/2,l), and
  Li(f/2) Li(f) (lemma 6).
• Now we will lower bound f/2

17
Extensions

• We have some Drawback in practice
• agents can often only evaluate path latency
  approximately, rather than exactly.
• We typically expect to encounter a finite number
  of agents, each controlling a strictly positive
  amount of flow.

18
Extensions(1)

• We cant expect agents to evaluate latency with
  arbitrary precision.
• We estimate that agents can distinguish between
  latencies that differ in more than 1e, we change
  definition
• Definition A flow f in G is e-approximate Nash
  equilibrium if for all i?1, . . . , k, P 1 , P
  2 ? P i , and d ?0, fP1 , we have lP1(f) lt
  (1e) lP2(ƒ), where ƒp is
• If PP1 fp- d
• If PP2 fp d
• Else fp
• Lemma 8 A flow f is at e-approximate Nash
  equilibrium if and only if for every i ? 1, . .
  . , k and P1,P2?P i with fP1 gt 0, lP1(f) lt
  (1e) lP2 (f).
• Conclusion if f is at e-approximate Nash
  equilibrium for (G,r,l) and f is feasible for
  (G,2r,l) the C(f)lt(1e)C(f)

19
Extensions(2)

• In practice we dont have infinite number of
  agents,but finitely many agents, each of whom
  controls a strictly positive amount of flow. we
  allow an agent to split flow along any number of
  paths.
• The changes are
• k agents. We assume that agent i intends to send
  ri units of flow from source si to destination
  ti.
• We call the instance (G,r,l) finite splittable.
• A flow f now consists of k functions, with one
  function f(i) f i -gtR
• For a flow f, we denote by Ci(f).
• A flow f is at Nash equilibrium if and only if
  for each i, f(i) minimizes Ci(f) given f(j) for j
  ! i.
• If f is at Nash equilibrium for the finite
  splittable instance (G, r, l), and f is feasible
  for the finite splittable instance (G, 2r, l),
  then C(f)ltC(f).

20
Extensions(3) unsplittable flow

• A flow f (now consisting only of k paths) is at
  Nash equilibrium if and only if for each i, agent
  i routes its flow on a path minimizing P (f),
  given the paths chosen by the other k - 1 agents.
• Example 2 agents,
• optimal 1-(s,v,t), 2-(s,w,t) total cost less
  then 4, (it also Nash)
• Nash 1-(s,v,w,t) 2- (s,t). Cost defends on e,
  unlimited.

21
Extensions(3) unsplittable flow

• A flow f (now consisting only of k paths) is at
  Nash equilibrium if and only if for each i, agent
  i routes its flow on a path minimizing P (f),
  given the paths chosen by the other k - 1 agents.
• Example 2 agents,
• optimal 1-(s,v,t), 2-(s,w,t) total cost less
  then 4, (it also Nash)
• Nash 1-(s,v,w,t) 2- (s,t). Cost defends on e,
  unlimited.

22
Extensions(3) unsplittable flow

• This example was very bad for routing, because
  routing a strictly positive amount of additional
  flow on an edge may increase the latency of that
  edge by an arbitrarily large amount.
• If the largest possible change in edge latency
  resulting from a single agent rerouting its flow
  is not too large, we can conclude
• Theorem Suppose f is at Nash equilibrium in the
  finite unsplittable instance (G, r, l), and for
  some a lt 2, we have le(xri) lt a le(x) for all
  agents i ? 1, . . . , k, edges e ? E, and x ?
  0,sum(rj j!i) . Then for any flow f
  feasible for (G, 2r, l), C(f) lt a C(f).