User‑equilibrium traffic assignment: 16-Civ-A6 Highway Design, Construction, and Maintenance, May 2013, Q6

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.

Given. Link performance (volume–delay) functions and a fixed total demand.

Given data
Route	Travel‑time function (min)	Free‑flow time (min)
1	$t_1=22+2V_1/225$	22
2	$t_2=12+V_2/100$	12
3 (part b)	$t_3=14+4V_3/225$	14
Total demand	$\sum V_i=7200$ veh/h	—

Find. UE volumes and travel times for two routes (a) and for three routes (b); and whether adding a route always helps (c).

Two zones (O–D) joined by parallel, non‑overlapping routes; 7200 veh/h is split so that every used route carries equal travel time (Wardrop first principle).

Approach. Apply Wardrop user‑equilibrium principle: all used routes have equal (and minimum) travel time. Set the route times equal and impose flow conservation, then solve the linear system.

UE conditions, two routes (a). Equal times and conservation:
$$22+\frac{2V_1}{225}=12+\frac{V_2}{100},\qquad V_1+V_2=7200.$$
Substituting $V_2=7200-V_1$ and solving,
$$0.018889\,V_1=62\ \Rightarrow\ V_1=3282,\quad V_2=3918.$$
Equilibrium time (a).
$$t=22+\frac{2(3282)}{225}=12+\frac{3918}{100}=\boxed{51.18\ \text{min};\quad V_1\approx3282,\ V_2\approx3918\ \text{veh/h}.}$$
UE with three routes (b). Write each volume as a function of the common time $t$:
$$V_1=112.5(t-22),\quad V_2=100(t-12),\quad V_3=56.25(t-14).$$
Summing to the demand,
$$268.75\,t-4462.5=7200\ \Rightarrow\ t=43.40\ \text{min}.$$
Three‑route volumes (b). Back‑substitute the equilibrium time:
$$\boxed{V_1\approx2407,\ V_2\approx3140,\ V_3\approx1653\ \text{veh/h};\quad t=43.40\ \text{min}.}$$
All three routes are used (each volume positive), so the equal‑time solution is valid; the new route cuts the equilibrium time from 51.18 to 43.40 min.

Question 6 — user‑equilibrium results
Case	$V_1$	$V_2$	$V_3$	Travel time
(a) Two routes	3282	3918	—	51.18 min
(b) Three routes	2407	3140	1653	43.40 min

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

No. Here the new, independent route lowered the equilibrium time, but this is not guaranteed. In a network where routes share links, adding a link or route can actually raise everyone equilibrium travel time — the celebrated Braess paradox. The reason is that user equilibrium is a selfish (Nash) equilibrium: each driver minimizes their own time with no regard for the congestion externality they impose on others. A new route can attract flow onto shared bottleneck links in a way that worsens the system outcome. Adding capacity reliably helps only when the new route is genuinely parallel and non‑overlapping (as in this problem); in a connected network the outcome must be checked, not assumed.

Topic: Traffic assignment — Wardrop user equilibrium on parallel routes.

Key relations: Wardrop I — all used routes have equal, minimal time $t_i(V_i)=t^{*}$; flow conservation $\sum V_i=Q$; linear BPR‑type link functions.

Why this works: At equilibrium no driver can reduce their travel time by unilaterally switching routes, which forces the used routes to equal times. With linear time–flow functions this becomes a small linear system: invert each function to $V_i(t)$, sum to the demand, and solve for the single unknown equilibrium time. Any route whose free‑flow time exceeds $t^{*}$ would carry zero flow (check the signs).

Common pitfall: Assuming an equal volume split, or assuming more routes always help. Equilibrium equalizes time, not volume, and Braess paradox shows extra capacity can hurt.

Source: Sheffi, Urban Transportation Networks, ch. 1–3 (user equilibrium); Ortuzar & Willumsen, ch. 11.

Question 6 of 7: User‑equilibrium traffic assignment

Question 6: User‑equilibrium traffic assignment (20 marks)

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