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

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

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Notes on this paper

Paper format. National Examination, 98-Civ-A6 Transportation Planning & Engineering, May 2014 — 3 hours, closed book (one two-sided aid sheet). Seven questions; any five constitute a complete examination and all are of equal value (20 marks). All seven are solved below as a study resource.

Reference texts (subject).

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. Two parallel routes with $t_1=14+V_1/600$, $t_2=15+V_2/1400$ (minutes); total demand $Q=5000$ veh/h. Part (b) adds Route 3 with $t_3=16+V_3/2800$.

Find. UE volumes and travel times with two routes (a) and three routes (b); and whether a new route always lowers travel time (c).

O D Route 1: t₁=14+V₁/600 Route 2: t₂=15+V₂/1400 Route 3 (part b): t₃=16+V₃/2800
Non‑overlapping parallel routes between origin O and destination D. Route 3 (green, dashed) is added in part (b).

Approach. Apply Wardrop's first principle: at user equilibrium every used route has equal (and minimal) travel time. Invert each linear time function to $V_i(t)$, impose $\sum V_i=Q$, and solve for the common time $t$.

  1. UE condition, two routes (a). Set $t_1=t_2$ with $V_1+V_2=5000$: $14+\dfrac{V_1}{600}=15+\dfrac{5000-V_1}{1400}$. Solving, $V_1=\boxed{1920}$ and $V_2=3080\ \text{veh/h}$.
  2. Equilibrium time (a). $t_1=14+1920/600=17.2$ min and $t_2=15+3080/1400=17.2$ min — equal, confirming UE. The common travel time is $\boxed{17.2\ \text{min}}$.
  3. UE with three routes (b). Write each volume as a function of the common time: $V_1=600(t-14)$, $V_2=1400(t-15)$, $V_3=2800(t-16)$. Summing to 5000: $600(t-14)+1400(t-15)+2800(t-16)=5000$, i.e. $4800\,t-74\,200=5000$, giving $t=\boxed{16.5\ \text{min}}$.
  4. Volumes (b). Back‑substitute: $V_1=600(2.5)=1500$; $V_2=1400(1.5)=2100$; $V_3=2800(0.5)=1400$ veh/h (all positive, so all three routes are used and sum to 5000). Every route now takes $16.5$ min — a $0.7$ min improvement over part (a).
  5. Does a new route always help? (c). No. Here Route 3 lowered travel time, but adding capacity does not guarantee improvement at UE — this is the Braess paradox. Because equilibrium is reached by individually optimising travellers (not a system optimum), a new link can attract flow in a way that raises congestion on shared segments and leaves everyone worse off. It only helps when, as here, the added route provides genuinely independent capacity between the same origin and destination.
Question 6 — user-equilibrium results
CaseV₁V₂V₃Travel time
(a) Two routes19203080—17.2 min
(b) Three routes15002100140016.5 min