# Fundamentals of Transportation/Route Choice/Solution2

The Marcytown - Rivertown corridor was served by 3 bridges, according to the attached map. The bridge over the River on the route directly connecting Marcytown and Citytown collapsed, leaving two alternatives, via Donkeytown and a direct. Assume the travel time functions Cij in minutes, Qij in vehicles/hour, on the five links routes are as given.
Marcytown - Rivertown Cmr = 5 + Qmr/1000
Marcytown - Citytown (prior to collapse) Cmc = 5 + Qmc/1000
Marcytown - Citytown (after collapse) Cmr = ∞
Citytown - Rivertown Ccr = 1 + Qcr/500
Marcytown - Donkeytown Cmd = 7 + Qmd/500
Donkeytown - Rivertown Cdr = 9 + Qdr/1000
Also assume there are 10000 vehicles per hour that want to make the trip. If travelers behave according to Wardrops user equilibrium principle.
A) Prior to the collapse, how many vehicles used each route?
B) After the collapse, how many vehicles used each route?
C) After the collapse, public officials want to reduce inefficiencies in the system, how many vehicles would have to be shifted between routes? What is the “price of anarchy” in this case?
A) Prior to the collapse, how many vehicles used each route?
Route A (Marcytown-Rivertown) = Ca = 5 + Qa/1000
Route B (Marcytown-Citytown-Rivertown) = Cb = 5 + Qb/1000 + 1 + Qb/500 = 6 + 3Qb/1000
Route C (Marcytown-Donkeytown-Rivertown)= Cc = 7 + Qc/500 + 9 + Qc/1000 = 16 + 3Qc/1000
At equilibrium the travel time on all three used routes will be the same: Ca = Cb = Cc
We also know that Qa + Qb + Qc = 10000
Solving the above set of equations will provide the following results:
Qa = 8467;Qb = 2267;Qc = −867
We know that flow cannot be negative. By looking at the travel time equations we can see a pattern.
Even with a flow of 0 vehicles the travel time on route C(16 minutes) is higher than A or B. This indicates that vehicles will choose route A or B and we can ignore Route C.
Solving the following equations:
Route A (Marcytown-Rivertown) = Ca = 5 + Qa /1000
Route B (Marcytown-Citytown-Rivertown) = Cb = 6 + 3Qb /1000
Ca = Cb
Qa + Qb = 10000
We can the following values:
Qa = 7750; Qb = 2250; Qc = 0
B) After the collapse, how many vehicles used each route?
We now have only two routes, route A and C since Route B is no longer possible. We could solve the following equations:
Route A (Marcytown-Rivertown) = Ca = 5 + Qa /1000
Route C (Marcytown- Donkeytown-Rivertown) = Cc = 16 + 3Qc /1000
Ca = Cc
Qa+ Qc= 10000
But we know from above table that Route C is going to be more expensive in terms of travel time even with zero vehicles using that route. We can therefore assume that Route A is the only option and allocate all the 10,000 vehicles to Route A.
If we actually solve the problem using the above set of equations, you will get the following results:
Qa = 10250; Qc = -250
which again indicates that route C is not an option since flow cannot be negative.
C) After the collapse, public officials want to reduce inefficiencies in the system, how many vehicles would have to be shifted between routes? What is the “price of anarchy” in this case?
User Equilibrium
TotalDelayUE =(15)(10,000)=150,000
System Optimal
TotalDelaySO =(Qa)(5+Qa/1000)+(Qc)(16+3Qc/1000)
Using Qa + Qc = 10,000
TotalDelaySO =(Qa2)/250−71Qa+460000
Minimize total delay ∂((Qa2)/250 − 71Qa + 460000)/∂Qa = 0
Qa/125−7 → Qa = 8875 Qc = 1125 Ca = 13,875 Cc = 19,375
TotalDelaySO =144938
Price of Anarchy = 150,000/144,938 = 1.035