Price Of Anarchy: Routing
Lecturer: Yishay Mansour
Ido Trivizki and Mille Gandelsman
Routing – Lecture Overview

Optimize the performance of a congested and
unregulated network:





Network.
Rate of traffic between each pair of nodes.
Latency function.
Selfish behavior does not perform as well as
an optimized regulated network.
Investigating the price of anarchy (PoA) by
exploring the characteristics of Nash
Equilibrium and mininal latency optimal flow.
Lecture overview – cont.

We will prove that:



If the latency of each edge is a linear function, PoA
is at most 4/3.
In atomic routing the PoA is bounded by 2.6.
I the latency function is known to be continuous,
non-decreasing and differentiable, there is no
bounded coordination ratio.
Introduction



Job scheduling, discussed last time, can be
viewed as a private case of routing.
Each player has to choose exactly one line to
pass his traffic through.
Parallel lines routing:
Introduction – cont.

Today – the problem of routing traffic in a
network:


Given a rate of traffic between pairs of nodes in
the network, find an assignment of the traffic to
paths so that the total latency is minimized.
It is often impossible to impose regulation, so
we are interested in those settings where each
user selects his minimum latency path:

Non-cooperative game in which each player plays
best response – expect the routes to form a Nash
Equilibrium.
Two player models


Non-atomic: we can split up traffic to several
paths.
Atomic: each user chooses a single path on
which he transports all of his traffic.
The global target function in both cases is to
minimize the total latency suffered by all users.
Reminder: The Price of Anarchy

The ration between the worst value of an
equilibrium and that of the optimal:
PoA  max xPNE


C ( x)
OPT
Where OPT denotes the minimum
latency among all feasible flows and
C(x) is total cost of flow x.
Our goal is to bound the PoA.
Example 1:
routing on parallel lines




n players with weights w , i  1,..., n (w  0)
m lines with speeds s , i  1,..., m
The players are allowed to split their
flow between different lines.
Nash Equilibrium is achieved when the
load on each line is:
 w W
i
i
i
n
Li (a ) 

i 1
m
j 1
i
sj

S
Example 1 – cont.
Routing on parallel lines


The optimum is achieved by dividing
the flow equally between the lines.
Therefore – we achieve a PoA=1
Example 2:
Pigou’s Network
C(x)=1

T
S
C(x)=x

Nash flow will only
traverse in the
lower path.
OPT will divide the
flow equally
among the two
paths.
Example 2 – cont.


The target function is 1 (1  x)  x  x and it
reaches minimum with value ¾, when
x=1/2, giving a PoA of 4/3.
Combing the example with the tighter
upper bound to be shown, it is a
demonstration of a tight bound of 4/3
for linear latency functions.
Example 3:
Pigou’s Non-Linear Network
C(x)=1

T
S
c( x)  x p
The flow at Nash
will continue to
use only the lower
path.
Example 3 – cont.



Let (1-x) be the flow on the upper path
in the optimum solution, and x - the
flow on the lower path, respectively.
The overall cost is (1  x) 1 x  x  1  x  x .
Proof that lim OPT  0:
p
p 

if we choose
(1  x) 
log p
:
p 1
log p 1
 0
p  p  1
p
lim OPT  lim (1  x  x p 1 )  lim
p 
p 
p 1
Example 3 – summary


In this case lim PoA  
It means that the PoA cannot be
bounded from above in some cases
when nonlinear latency functions are
allowed.
p 
Example 4 – Braess’s paradox

There are exactly two disjoint paths
from s to t, each of them follows exactly
two edges.
Example 4 – cont.

The optimal flow coincides with the Nash
equilibrium:




half of the traffic takes the upper path, and the half
the lower.
The latency perceived by each user is 3/2.
In any other non-equal distribution, there will
be a difference in the total latency.
Users will be motivated to reroute to the less
congested path.
Example 4 – cont.



Consider adding a fifth edge with
latency 0.
The optimal flow stays 3/2.
Nash will only
occur by routing the
entire traffic on the
single svwt
path.
Example 4 – cont.


The latency each user experiences
increases to 2.
Amazingly, adding a new zero latency
link had a negative effect for all agents.
Formal Definition of the Problem


Consider a directed graph
Input:


G  (V , E )
k pairs of source and destination vertices (si , ti )
Demand r ( the amount of required flow between si
and ti ). Assume: r  0 .
Each edge e  E is given a load dependant non
decreasing and differentiable latency function l : R  R
i
i


e

Output:

Flow f - a function that defines for each path
flow f . f induces flow on edge e : f   f
p
e
p:e p
p
p
a

Formal Definition – cont.

We denote the set of simple paths connecting
the pair (si , ti ) by  and let     .
Solution is feasible if i :  f  r .
The latency of the a path is defines as:
l (f) l (f )
i
i
i


pi
p

i
e
Our goal is to find a flow that will minimize
the total social cost of a flow is defined as:
C( f )   l ( f )  f   l ( f )  f
pP p

e p e
p
p
eE e
e
e
The cost of player i : ci ( f )   pP l p ( f )  f p
i
Flows at Nash equilibrium

Lemma:

A feasible flow f for instance (G, r , l ) is Nash
Equilibrium if for every i {1,..., k} and p, p' i :
if f p  0 then l p ( f )  l p ' ( f )

Corollary:

is a flow at Nash Equilibrium for instance
(G , r , l ) if and only if C ( f )   Li ( f )  ri , where
i
f
Li ( f )  min
pi
lp ( f )
Optimal Solution – flow

Our goal is find a feasible flow f that will
minimize the total cost.





Let ce ( x)  le ( x)  x .
Clearly, it follows that C ( f )  eE ce ( f e )
To find the optimal flow c*e ( x) , we will look
at ce ' ( x)  le ( x)  x  le ' ( x) .
We assume that for each edge e  E: ce (x) is
convex and therefore C ( f ) is also convex.
ce (x) is differentiable.
The Optimality Condition


Define:
and c p ' ( x)  e p ce ' ( x)
Let (G, r , l ) be a dividable game. For each edge
e  E the function ce ( x)  le ( x)  x is convex,
continuous and differentiable function.
A flow f is optimal for (G, r , l ) if and only if:
ce ' ( x ) 
d
ce ( x )
dx
p, p' i : f p  0  c p ' ( f )  c p ' ' ( f )
The Optimality Condition – cont.



Notice the resemblance between the
characterization of optimality conditions and
Nash Equilibrium.
An optimal flow can be interpreted as a Nash
Equilibrium with respect to a different edge
latency functions.
We will use this resemblance to reach the
bound on PoA.
The Optimality Condition – cont.

le * (c)  ce ' ( x)  (le ( x)  x)'  le ( x)  x  le ' ( x)
Let: l * ( x)  
Corollary:
p



l * ( x)
e p e
is an optimal flow for (G, r , l ) if and only if it is
Nash Equilibrium for the instance (G , r , l*)
f
Proof:

By the optimality condition: f is optimal for
l if and only if i, p, p'  : f  0  c ' ( f )  c ' ( f ) , if and
only if (by def.) i, p, p'  : f  0  l * ( f )  l * ( f ) , if and
only if f is Nash Equilibrium for l * .
i
p
i
p
p
p'
p
p'
The optimality condition - proof

Definition: a set S is called a convex set if
A, B  S ,0    1, A  (1   ) B  S

Intuitively it means that a
set S is convex if the linear
segment connecting two
points in the set, is entirely
in the set.
The optimality condition - proof

Definition: a function f is called convex
function if
x, y,0    1 : f (x  (1   ) y )  f ( x)  (1   ) f ( y )
The optimality condition - proof


Let F (x) be a convex function, and S a convex
set.
A convex programming is of the form:
min F ( x), s.t. x  S

Lemma:


If F (x ) is strictly convex, then the solution is unique.
Proof:



Assume that x  y are both minimum solutions.
1
1
z

x

y
z S .
Let
2
2 , because S is convex:
1
1
F
(
z
)

F
(
x
)

F ( y) ,
F
(x
)
Since
is strictly convex:
2
2
contradicting F (x) and F ( y ) being minimal .
The optimality condition - proof

Lemma:


Lemma:


If F (x) is convex, then the solution set U is convex.
If F (x) is convex and y is not optimal then y is not
a local minimum. Consequently, any local
minimum is also a global minimum.
Proof:

Assume that y is not optimal, i.e. x : F ( x)  F ( y )
let z  x  (1   ) y , Since F is convex:
F ( z )  F ( x)  (1   ) F ( y )  F ( y )
for every 0    1 .
Existence of flows at Nash
Equilibrium

Theorem: For every splittable game
(G, r , l )


There exists at least one Nash Equilibrium
If f and f ' are Nash equilibria then for
every e, ce ( f )  ce ( f )
Existence of flows at Nash
Equilibrium - Proof


Define: h ( x)  l ( y)dy , so that h 'e ( x)  le ( x)
Further define a potential function:
x
e
0 e
( f )   he ( f e )
e


is non-negative, monotonous,
increasing and differential.  Is a
convex function.
Nash equillibrium flows are global
minimizers of 
he
Existence of flows at Nash
Equilibrium – Proof Cont.


By Weierstrass’s Theorem,  has a
minimum, and therefore Nash
equilibrium exists.
Let f and f ' be Nash equillibria:

Define g   f  (1   ) f  for   [0,1]

 is convex  ( g )  ( f )  (1   )( f )

f and f ' minimize , and we get ( f )  ( f )  ( g )

 is sum of convex functions, and therefore it’s possible only
if all members of the sum are equal, and therefore:
ce ( g )  ce ( f )  ce ( f )
Bounding the Price of Anarchy

PoA 
Nash
OPT
Theorem: If there exists a constant   1
such that x :  he ( x)  ce ( x) then PoA  

Nash  C ( f )   ce ( f e )  ( by assumption)
e
  he ( f e )  ( f is OPT for he )
e
  he ( f e *)  (le is non decreasing  he ( x)  ce ( x))
e
  ce ( f e *)   C ( f *)   OPT
e

Corollary: If the latency function is polynomial
function of degree d , then PoA  d 1
A tight bound for linear
latency functions





A natural example for such a model:
Network with congestion control (e.g.:
TCP)
Using the corollary, we get a bound of 2
We’ll show a bound of 4/3 (which is
tight, as we’ve seen in Example 2)
le  ae x be
When le be , both Nash and OPT (equal) will route
all the flow in the shortest paths.
A tight bound for linear
latency functions - Proof

Lemma (proof is trivial and omitted):
2
y
xy  x 2 
4
is a flow at Nash equilibrium, and f * is
an optimal flow. Given a flow f let
 f
l  ae fe  be and C ( x)   l e xe
f
e
f
f
A tight bound for linear
latency functions – Proof Cont
C f ( x)   (ae f e  be )xe   (ae f e xe  be xe )  (by the lemma)
 (ae xe2  be xe )  ae fe2
1
1
 C ( x)  C ( f )
4
4
As x : C f ( f )  C f ( x) :
1
C f ( f )  C ( f )  C ( f *)  C ( f )
4
3
C ( f )  C ( f *)
4
4
C ( f )  C ( f *)
3
4
PoA 
3
Unsplittable (Atomic) Routing

Example 1:
4 players, all with
demand 1 (r = 1):
(U,V), (U,W),
(V,W), (W,V)
 An optimal and Nash equilibrium flow
would use only edges with l ( x)  x at total
cost of 4

Unsplittable (Atomic) Routing
– Exmaple 1 Cont.

Example 1 (cont.):
But the optimal
solution is not the
only NE.
 Another Nash equilibrium:
Player 1: U->W->V; Player 2: U->V->W;
Player 3: V->U->W; Player 4: W->U->V
With total cost of 10, which gives PoA = 2.5.

Unsplittable (Atomic) Routing
– Exmaple 2
Both players have
s=S and t=T, but,
player 1 has r=1,
While player 2, r=2.
 Possible paths S->T:
p1: S->T; p2: S->V->T; p3: S->W->T;
p4: S->V->W->T

Unsplittable (Atomic) Routing
– Exmaple 2 Cont.
In this example
there is no pure
equilibrium.
 Easy to show the
following facts:





If player 2 chooses p1 or p2, player 1 will choose p4.
If player 1 chooses p4, player 2 will choose p3.
If player 2 chooses p3 or p4, player 1 will choose p1.
If player 1 chooses p1, player 2 will choose p2.
Unsplittable (Atomic) Routing
– Existence of Nash Equilibrium


We’ve shown that NE does not always
exist.
Theorem: If (G, r , l ) is an unsplittable game
with i  N : ri  1 then there exists a Nash
equilibrium.
Proof:
f (e)
Define a potential function a ( f )    le (i)
e i 1
When player I moves from p to p’:

l p ( f )  l p ( f ) 

e p  p
l ( f e  1) 

e p  p
l ( fe )
Unsplittable (Atomic) Routing –
Existence of Nash Equilibrium
Cont.
And:
a ( f ')   a ( f ) 
le ( fe  1) if e  p  p
le ( fe ) if e  p  p '
0 otherwise
So: l p ( f )  l p ( f )   a ( f )   a ( f )
So when the players plays “best response” the
potential decreases, and as it’s non-negative,
s series of “best responses” will converge to a
Nash equilibrium.
Bounding the price of anarchy for
unsplittable linear games

Theorem: let (G, r , l ) be an unsplittable
routing game with linear cost functions, then
3 5
PoA 
3

2.618
Proof:


Let f be a nash equillibirim (we assume it exists)
flow, and f * be an optimal flow.
c pi ( f )   (ae fe  be ) 
e pi
 (a ( f
e p*i
e
e
 ri )  be )
Bounding the price of anarchy for
unsplittable linear games – Cont.

Lemma: C ( f )  C ( f * )  ae fe fe*

Proof (of lemma):
e
C ( f )   ri (  ae ( f e  ri )  be ) 
e pi*
i
*
*
*
r
(
a
(
f

f
)

b
)

(
a
(
f

f
)

b
)
f
i  e e e e  e e e e e
eE
e pi*
i
  (ae ( f e* ) 2  be f e*   ae f e f e*
eE
eE
 C ( f * )   ae f e f e*
eE

Using Cauchy-Schwartz:
*
a
f
f
 eee
e
2
a
f
 ee
e
* 2
*
a
(
f
)

C
(
f
)
C
(
f
)
 e e
e
Bounding the price of anarchy for
unsplittable linear games – Cont.

We get:
C ( f )  C ( f * )  C ( f )C ( f * )
C( f )
C( f )
 1
*
C( f )
C( f *)

C( f )
Solving the equation for C ( f * )
We get :
C( f ) 3  5

*
C( f )
2
2.618