NivaarExam PrepOfficial exam papers ↗

16-Civ-A6 Highway Design, Construction, and Maintenance · December 2013

Question 6 of 7: User-equilibrium traffic assignment and the Braess paradox

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

Notes on this paper

Paper format. National Examination, 98‑Civ‑A6 Transportation Planning & Engineering (December 2013). Closed book, one two‑sided aid sheet, 3 hours. Seven questions; any five constitute a complete examination and each is of equal value (20 marks). All seven are solved below as a study resource.

Reference texts (subject).



Question 6: User-equilibrium traffic 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. A single origin–destination pair served by two (then three) non‑overlapping routes with linear link‑performance functions and a fixed demand of 1,800 veh/h.

Given data
RoutePerformance function (min)
1$t_1 = 11 + V_1/225$
2$t_2 = 6 + V_2/200$
3 (added in b)$t_3 = 7 + V_3/112.5$
Total demand1,800 veh/h

Find. The UE volumes and travel times for the two‑route and three‑route networks, and an explanation of the Braess paradox.

ODRoute 1: t1 = 11 + V1/225Route 2: t2 = 6 + V2/200Route 3 (added): t3 = 7 + V3/112.5total demand = 1800 veh/h, non-overlapping routes
Residential origin O to CBD destination D served by two parallel non‑overlapping routes, with a third route added in part (b).

Approach. By Wardrop’s first principle, all used routes carry equal travel time at UE. Invert each linear function to $V_i(t)$, impose that the volumes sum to the demand, solve for the common time $t$, then back‑substitute.

  1. Equal‑time condition (a). Set $t_1=t_2=t$ and use $V_1+V_2=1800$. Writing $V_1=225(t-11)$ and $V_2=200(t-6)$ and summing: $$225(t-11)+200(t-6)=1800\ \Rightarrow\ 425\,t=5475\ \Rightarrow\ t=\boxed{12.88\ \text{min}}.$$
  2. Two‑route volumes. $V_1=225(12.88-11)=\boxed{424\ \text{veh/h}}$ and $V_2=1800-424=\boxed{1376\ \text{veh/h}}$. Check: $t_1=11+424/225=12.88=t_2=6+1376/200$. ✓
  3. Add Route 3 (b). With $V_3=112.5(t-7)$, all three used at common time $t$: $$225(t-11)+200(t-6)+112.5(t-7)=1800\ \Rightarrow\ 537.5\,t=6262.5\ \Rightarrow\ t=\boxed{11.65\ \text{min}}.$$
  4. Three‑route volumes. $V_1=225(11.65-11)=\boxed{147}$, $V_2=200(11.65-6)=\boxed{1130}$, $V_3=112.5(11.65-7)=\boxed{523}$ veh/h (sum = 1800). All three carry the same 11.65 min, so every route’s travel time is reduced from 12.88 to 11.65 min — here the new route helps.
Final results — Question 6
Case$V_1$$V_2$$V_3$Common time
(a) two routes4241376—12.88 min
(b) three routes147113052311.65 min

(c) When a new route makes everyone worse off (Braess paradox)

Adding capacity does not guarantee lower travel times at user equilibrium. In the Braess paradox a new link changes the set of available paths so that self‑interested (UE) route choice drives traffic onto a combination of links whose congested times are higher for everyone than before the link existed. The mechanism is that UE minimises each traveller’s own time, not the system total; because each driver ignores the congestion externality they impose on others, the equilibrium can settle at a worse collective outcome than the system optimum. When the added route is a genuinely parallel alternative (as in part b), it relieves congestion and helps; but when it creates a new through‑connection that tempts drivers to overload shared bottleneck links, the paradox appears. It illustrates the general gap between user‑equilibrium and system‑optimal assignment, and is why network additions must be evaluated with an assignment model rather than assumed beneficial.