Tolls
Our scope is to set tolls that “transform” the system optimum to an
equilibrium
Tolls are set on the edges: edge e gets a τe
Player usingX
path p gets a delay cost cp (f) =
¿e as tolls.
pay ¿p =
e2p
X
ce (fe ) and has to
e2p
Player i has a sensitivity αi to latency. Her total cost is ai lp (f) + ¿p
Social Cost is not affected
A “magic” LP program
Assume g is a (feasible) congestion that we want to enforce.
Consider the following LP and its Dual:
(feasible) g is minimal if inequality 1 is tight
g is enforceable if there are tolls to enforce it on equilibrium.
The Theorem
Theorem: g minimal iff g enforceable
Proof:
 There is an optimal solution f with a complementary
optimal solution (t,z), for which 1 is tight :
The Theorem
Theorem: g minimal iff g enforceable
Proof:
 Consider eq. flow f and tolls t. f is an equilibrium:
f and (t,z) are complementary (and feasible) and so
they are both optimal.
A minimal and optimal g? Where?
g is called minimally feasible if:
– it is feasible and
– reducing any ge (for any e) results to infeasibility
A minimally feasible g has optimal solutions for which 1 is tight
Let g be the optimal congestion, the one that we want to enforce.
Reduce the ge’ s, stopping whenever feasibility “stops”
g* is minimally feasible + optimal (
)
Braess’ Paradox
s
x
0
1
1
x
t
Braess’ Paradox
s
1
x
0
1
1
1
x
t
s
½
x
1
WE and OPT flows
½ 0
½ 1
½
x
t
Braess’ Paradox
s
1
x
0
1
1
1
x
t
s
½
x
1
½ 0
½ 1
½
x
t
WE and OPT flows
Removing the “middle” edge:
s
x
1
1
x
t
Braess’ Paradox
s
1
x
0
1
1
1
x
t
s
½
x
1
½ 0
½ 1
½
x
t
WE and OPT in the original network
Removing the “middle” edge:
OPT and WE in the
subnetwork
s
½
x
1
1
½
x
t
Braess’ Paradox
s
1
x
0
1
1
1
x
PoA=4/3
t
s
½
x
1
½ 0
½ 1
½
x
t
WE and OPT in the original network
Removing the “middle” edge:
OPT and WE in the
subnetwork
PoA=1 x ½
1
s
1
½
x
t
Braess’ Paradox
s
1
x
0
1
1
1
x
PoA=4/3
t
s
½
x
1
½ 0
½ 1
½
x
t
WE and OPT in the original network
Removing the “middle” edge:
OPT and WE in the
subnetwork
PoA=1 x ½
1
s
1
Is it possible to detect the Braess’ Paradox?
½
x
t
Thank you!
(and Roughgarden)