16-Civ-A6 Highway Design, Construction, and Maintenance · May 2015

Question 6 of 7: User-Equilibrium Traffic Assignment

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

Notes on this paper

National Examination — 98-Civ-A6 Transportation Planning & Engineering, May 2015. 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, 6th ed. (Wiley).
Ortúzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. (Wiley).
Sheffi, Y., Urban Transportation Networks (Prentice Hall).

Question 6: User-Equilibrium Traffic Assignment (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.

These values yield clean equilibria with whole-minute route times.

Given. Two parallel routes between A and B with linear (BPR-type) travel-time functions and a fixed O–D demand.

Given data
Route	Travel-time function (min)
Bypass	$t_b = 15 + 0.005\,V_b$
Town centre	$t_{tc} = 10 + 0.02\,V_{tc}$
Alternative (part b)	$t_{t2} = 12 + 0.02\,V_{t2}$
Total demand	$Q = 4000$ veh/h

Find. The UE volumes and common travel time with two routes (a) and three routes (b), and a discussion of whether new capacity always helps (c).

Route network from A to B: two parallel routes (a) plus the added alternative (b).

Approach. At UE every used route has equal and minimal travel time (Wardrop's first principle); write that equality with the flow-conservation constraint and solve.

(a) Impose equal travel times. With $V_b+V_{tc}=4000$, set $t_b=t_{tc}$: $$15+0.005V_b = 10+0.02V_{tc},\qquad V_b=4000-V_{tc}.$$ Substituting: $15+0.005(4000-V_{tc}) = 10+0.02V_{tc}\Rightarrow 35-0.005V_{tc}=10+0.02V_{tc}\Rightarrow 25=0.025V_{tc}.$
(a) Solve. $$V_{tc}=1000\ \text{veh/h},\qquad V_b=3000\ \text{veh/h}.$$ $$\boxed{V_b=3000,\ V_{tc}=1000,\ t=15+0.005(3000)=30\ \text{min (both routes).}}$$
(b) Three used routes at a common time $t$. Invert each cost function for volume at travel time $t$: $$V_b=\frac{t-15}{0.005}=200t-3000,\quad V_{tc}=\frac{t-10}{0.02}=50t-500,\quad V_{t2}=\frac{t-12}{0.02}=50t-600.$$ Sum to demand: $(200t-3000)+(50t-500)+(50t-600)=4000\Rightarrow 300t=8100.$
(b) Solve. $t=27$ min, so $$V_b=2400,\quad V_{tc}=850,\quad V_{t2}=750\ \text{veh/h}\ (\text{sum }4000).$$ $$\boxed{V_b=2400,\ V_{tc}=850,\ V_{t2}=750,\ t=27\ \text{min.}}$$ All three route times equal 27 min ($15+0.005(2400)=27$, $10+0.02(850)=27$, $12+0.02(750)=27$), so all three are used. Adding the route cut travel time from 30 to 27 min.

Question 6 — UE volumes and travel times
Route	(a) two routes	(b) three routes
Bypass $V_b$	3000 veh/h	2400 veh/h
Town centre $V_{tc}$	1000 veh/h	850 veh/h
Alternative $V_{t2}$	—	750 veh/h
Common travel time	30 min	27 min

(c) Does a new route always reduce travel time?

No. Here the extra route helped (30 → 27 min), but that is not guaranteed. In a congestible network the Braess paradox shows that adding a link can raise everyone's equilibrium travel time. Because UE is a selfish (Nash) equilibrium, drivers reroute to minimise their own time without regard to the congestion they impose on others; a new link can attract flow onto shared bottleneck segments and push the whole network to a worse equilibrium than before. The outcome depends on the specific cost functions and demand: new capacity helps only when it relieves, rather than feeds, the binding congestion. This is why network additions must be evaluated by re-solving the UE assignment, not assumed beneficial.

Topic: Deterministic user-equilibrium (Wardrop) traffic assignment on parallel routes.

Key relations: Wardrop UE — all used routes share the minimum travel time; unused routes cost at least as much. Solve $t_r(V_r)$ equal across used routes subject to $\sum V_r = Q$.

Why this works: If any used route were slower, drivers on it would switch to a faster one until times equalise — the equilibrium condition. Inverting the linear cost functions to $V_r(t)$ and summing turns the problem into one equation in the common time $t$.

Common pitfall: Assuming all routes are used (a route with a high free-flow time may carry zero flow — check $V_r\ge 0$), and assuming more links always help (Braess paradox).

Source: Sheffi, Urban Transportation Networks, Ch. 1–3; Ortúzar & Willumsen, Modelling Transport, Ch. 10 (assignment).

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