Given a flow of six (6) units from origin “o” to destination “r”. Flow on each route ab is designated with Qab in the Time Function. Apply Wardrop's Network Equilibrium Principle (Users Equalize Travel Times on all used routes)
A. What is the flow and travel time on each link?
(complete the table below) for Network A
Link Attributes
Link
Link Performance Function
Flow
Time
o-p
p-r
o-q
q-r
B. What is the system optimal assignment?
C. What is the Price of Anarchy?
Solution:
Part A
What is the flow and travel time on each link? Complete the table below for Network A:
Link Attributes
Link
Link Performance Function
Flow
Time
o-p
p-r
o-q
q-r
These four links are really 2 links O-P-R and O-Q-R, because by conservation of flow Qop = Qpr and Qoq = Qqr.
Link Attributes
Link
Link Performance Function
Flow
Time
o-p-r
o-q-r
By Wardrop's Equilibrium Principle, the travel time (cost) on each used route must be equal. Therefore
.
OR
By the conservation of flow principle
By substitution
Check
Check (within rounding error)
Link Attributes
Link
Link Performance Function
Flow
Time
o-p-r
2.84
42.01
o-q-r
3.15
42.01
or expanding back to the original table:
Link Attributes
Link
Link Performance Function
Flow
Time
o-p
2.84
14.2
p-r
2.84
27.84
o-q
3.15
26.3
q-r
3.15
15.75
User Equilibrium: Total Delay = 42.01 * 6 = 252.06
Part B
What is the system optimal assignment?
Conservation of Flow:
Analytic Solution requires minimizing total delay
And we can compute the SO travel times on each path
Note that unlike the UE solution,
Total Delay = 3.04(25+ 6*3.04) + 2.96(20+7*2.96) = 131.45+120.53= 251.98
Note: one could also use software such as a "Solver" algorithm to find this solution.
Part C
What is the Price of Anarchy?
User Equilibrium: Total Delay =252.06
System Optimal: Total Delay = 251.98