Last modified on 17 September 2014, at 15:00

Fundamentals of Transportation/Route Choice/Solution

TProblem
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

Link Attributes
Link Link Performance Function Flow Time
o-p C_{op} = 5 *Q_{op}
p-r C_{pr} = 25 + Q_{pr}
o-q C_{oq} = 20 + 2*Q_{oq}
q-r C_{qr} = 5 *Q_{qr}

B. What is the system optimal assignment?

C. What is the Price of Anarchy?

Example
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 C_{op} = 5 *Q_{op}
p-r C_{pr} = 25 + Q_{pr}
o-q C_{oq} = 20 + 2*Q_{oq}
q-r 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.

Link Attributes
Link Link Performance Function Flow Time
o-p-r C_{opr} =25+ 6 *Q_{opr}
o-q-r C_{oqr} = 20 + 7*Q_{oqr}

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

OR

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

5+ 6*Q_{opr}= 7*Q_{oqr}

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


By the conservation of flow principle

Q_{oqr}+Q_{opr} = 6

Q_{opr} = 6 - Q_{oqr}


By substitution

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

13*Qoqr = 41

Q_{oqr} = 41/13 = 3.15

Q_{opr}=2.84


Check

42.01 = 25+6(2.84)

42.05 = 20+7(3.15)


Check (within rounding error)

Link Attributes
Link Link Performance Function Flow Time
o-p-r C_{opr} =25+ 6 *Q_{opr} 2.84 42.01
o-q-r C_{oqr} = 20 + 7*Q_{oq} 3.15 42.01

or expanding back to the original table:

Link Attributes
Link Link Performance Function Flow Time
o-p C_{op} = 5 *Q_{op} 2.84 14.2
p-r C_{pr} = 25 + Q_{pr} 2.84 27.84
o-q C_{oq} = 20 + 2*Q_{oq} 3.15 26.3
q-r 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:

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

Total Delay = Q_{opr}(25+ 6*Q_{opr}) + Q_{oqr}(20+7*Q_{oqr}) \,\!

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

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

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

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

Analytic Solution requires minimizing total delay

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

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

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

And we can compute the SO travel times on each path

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

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

Note that unlike the UE solution,  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