16-Civ-A6 Highway Design, Construction, and Maintenance · December 2014

Question 6 of 7: User-Equilibrium Assignment and the Braess Paradox

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination — 98-Civ-A6 Transportation Planning & Engineering, December 2014. Closed book (one two-sided aid sheet), 3 hours. Seven questions of equal value (20 marks); any five constitute a complete paper. All seven are solved here as a study resource.

Reference texts (subject):

Papacostas, C.S. & Prevedouros, P.D., Transportation Engineering and Planning, 3rd ed. (Prentice Hall).
Roess, R.P., Prassas, E.S. & McShane, W.R., Traffic Engineering, 4th ed. (Pearson).
Mannering, F.L. & Washburn, S.S., Principles of Highway Engineering and Traffic Analysis (Wiley).
Ortúzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. (Wiley).
Sheffi, Y., Urban Transportation Networks (Prentice Hall).
Transportation Association of Canada (TAC), Geometric Design Guide for Canadian Roads (Canadian context).

Question 6: User-Equilibrium Assignment and the Braess Paradox (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Link performance functions $t_1=10+V_1/200$, $t_2=20+V_2/100$, $t_3=15+V_3/150$ (min); total demand $Q=3200$ veh/h.

Find. UE volumes and travel times for the 2-route network (a) and the 3-route network (b); and whether adding a route always helps (c).

Parallel-route network between the commercial (origin) and residential (destination) zones; Route 3 is the proposed addition.

Approach. Wardrop's first principle: at UE every used route carries equal (and minimal) travel time. Set the used-route times equal, impose the flow-conservation constraint, and solve.

(a) Equate used-route times. With both routes used, $t_1=t_2$ and $V_1+V_2=3200$: $$10+\frac{V_1}{200}=20+\frac{3200-V_1}{100}.$$
(a) Solve. Multiplying by 200: $V_1-2(3200-V_1)=2000\Rightarrow 3V_1=8400$, so $$\boxed{V_1=2800,\ V_2=400\ \text{veh/h};\quad t_1=t_2=24\ \text{min}.}$$ Both volumes are positive, confirming both routes are used.
(b) Three routes at a common time $t$. Invert each function: $V_1=200(t-10)$, $V_2=100(t-20)$, $V_3=150(t-15)$. Conservation $V_1+V_2+V_3=3200$ gives $$450\,t-6250=3200\;\Rightarrow\; t=21\ \text{min}.$$
(b) Back-substitute. $$\boxed{V_1=2200,\ V_2=100,\ V_3=900\ \text{veh/h};\quad t_1=t_2=t_3=21\ \text{min}.}$$ All three volumes are positive, so all routes are used; the common travel time falls from 24 to 21 minutes.
(c) Does adding a route always help? No. Here the extra capacity lowered the equilibrium time, but adding a link can increase everyone's travel time at UE — the Braess paradox. Because drivers choose routes selfishly (minimizing their own time, not total system time), the UE flow pattern is generally not the system optimum; introducing a new link can shift flows to a worse equilibrium. Whether a new route helps must therefore be checked by re-solving the equilibrium, not assumed.

Question 6 — user equilibrium
Case	$V_1$	$V_2$	$V_3$	Travel time
(a) Two routes	2800	400	—	24 min
(b) Three routes	2200	100	900	21 min

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