# Fundamentals of Transportation/Route Choice/Solution2

< Fundamentals of Transportation | Route Choice**Problem:**

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?

**Solution:**

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