# Fundamentals of Transportation/Route Choice/Solution

Problem:

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

o-p ${\displaystyle C_{op}=5*Q_{op}}$
p-r ${\displaystyle C_{pr}=25+Q_{pr}}$
o-q ${\displaystyle C_{oq}=20+2*Q_{oq}}$
q-r ${\displaystyle C_{qr}=5*Q_{qr}}$

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:

o-p ${\displaystyle C_{op}=5*Q_{op}}$
p-r ${\displaystyle C_{pr}=25+Q_{pr}}$
o-q ${\displaystyle C_{oq}=20+2*Q_{oq}}$
q-r ${\displaystyle C_{qr}=5*Q_{qr}}$

These four links are really 2 links O-P-R and O-Q-R, because by conservation of flow Qop = Qpr and Qoq = Qqr.

o-p-r ${\displaystyle C_{opr}=25+6*Q_{opr}}$
o-q-r ${\displaystyle C_{oqr}=20+7*Q_{oqr}}$

By Wardrop's Equilibrium Principle, the travel time (cost) on each used route must be equal. Therefore ${\displaystyle C_{opr}=C_{oqr}}$.

OR

${\displaystyle 25+6*Q_{opr}=20+7*Q_{oqr}}$

${\displaystyle 5+6*Q_{opr}=7*Q_{oqr}}$

${\displaystyle Q_{oqr}=5/7+6*Q_{opr}/7}$

By the conservation of flow principle

${\displaystyle Q_{oqr}+Q_{opr}=6}$

${\displaystyle Q_{opr}=6-Q_{oqr}}$

By substitution

${\displaystyle Q_{oqr}=5/7+6/7(6-Q_{oqr})=41/7-6*Q_{oqr}/7}$

${\displaystyle 13*Qoqr=41}$

${\displaystyle Q_{oqr}=41/13=3.15}$

${\displaystyle Q_{opr}=2.84}$

Check

${\displaystyle 42.01=25+6(2.84)}$

${\displaystyle 42.05=20+7(3.15)}$

Check (within rounding error)

o-p-r ${\displaystyle C_{opr}=25+6*Q_{opr}}$ 2.84 42.01
o-q-r ${\displaystyle C_{oqr}=20+7*Q_{oq}}$ 3.15 42.01

or expanding back to the original table:

o-p ${\displaystyle C_{op}=5*Q_{op}}$ 2.84 14.2
p-r ${\displaystyle C_{pr}=25+Q_{pr}}$ 2.84 27.84
o-q ${\displaystyle C_{oq}=20+2*Q_{oq}}$ 3.15 26.3
q-r ${\displaystyle C_{qr}=5*Q_{qr}}$ 3.15 15.75

User Equilibrium: Total Delay = 42.01 * 6 = 252.06

Part B

What is the system optimal assignment?

Conservation of Flow:

${\displaystyle Q_{opr}+Q_{oqr}=6\,\!}$

${\displaystyle TotalDelay=Q_{opr}(25+6*Q_{opr})+Q_{oqr}(20+7*Q_{oqr})\,\!}$

${\displaystyle 25Q_{opr}+6Q_{opr}^{2}+(6-Q_{opr})(20+7(6-Q_{opr}))\,\!}$

${\displaystyle 25Q_{opr}+6Q_{opr}^{2}+(6-Q_{opr})(62-7Q_{opr}))\,\!}$

${\displaystyle 25Q_{opr}+6Q_{opr}^{2}+372-62Q_{opr}-42Q_{opr}+7Q_{opr}^{2}\,\!}$

${\displaystyle 13Q_{opr}^{2}-79Q_{opr}+372\,\!}$

Analytic Solution requires minimizing total delay

${\displaystyle \delta C/\delta Q=26Q_{opr}-79=0\,\!}$

${\displaystyle Q_{opr}=79/26=3.04\,\!}$

${\displaystyle Q_{oqr}=6-Q_{opr}=2.96\,\!}$

And we can compute the SO travel times on each path

${\displaystyle C_{opr,SO}=25+6*3.04=43.24\,\!}$

${\displaystyle C_{oqr,SO}=20+7*2.96=40.72\,\!}$

Note that unlike the UE solution, ${\displaystyle C_{opr,SO}>C_{oqr,SO}\,\!}$

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

Price of Anarchy = 252.06/251.98 = 1.0003 < 4/3