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

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

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

Notes on this paper

Paper format: 98-Civ-A6 Transportation Planning & Engineering, National Examination, December 2015. Seven questions, each of equal value (20 marks); any five constitute a complete examination. Closed book, one two-sided aid sheet permitted. Three hours. All seven questions are solved below as a study resource.

Reference texts. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis (Wiley); Papacostas & Prevedouros, Transportation Engineering and Planning (Prentice Hall); Roess, Prassas & McShane, Traffic Engineering (Pearson); Ortúzar & Willumsen, Modelling Transport (Wiley) for the demand-model chapters (trip generation, distribution, mode choice, assignment).

Question 6: User-Equilibrium Traffic Assignment and 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. Parallel routes with linear volume-delay functions $t_i=10+20(V_i/C_i)$ and a fixed total demand.

Given data
Highway	Capacity $C$ (veh/h)	Travel-time function
1	2,200	$t_1=10+20\,V_1/2200$
2	3,000	$t_2=10+20\,V_2/3000$
3 (part b)	2,800	$t_3=10+20\,V_3/2800$
Total demand $Q$	8,000 veh/h

Find. UE volumes and travel times with two routes (a) and three routes (b); an explanation of Braess’ paradox (c).

Approach. At Wardrop user equilibrium every used parallel route carries equal travel time. Because each free-flow term is the same (10 min), equal travel time forces equal $V_i/C_i$; letting that common ratio be $k$, the flow-conservation equation $\sum V_i=Q$ solves directly.

(a) Equal-time condition. Setting $t_1=t_2$ cancels the common 10 and 20, leaving $V_1/C_1=V_2/C_2\equiv k$. Then $V_1+V_2=Q$ gives $$k(C_1+C_2)=Q\;\Rightarrow\;k=\frac{8000}{2200+3000}=\frac{8000}{5200}=1.538.$$
(a) Volumes and time. $$V_1=2200k=3384.6\ \text{veh/h},\qquad V_2=3000k=4615.4\ \text{veh/h},$$ $$\boxed{t=10+20(1.538)=40.8\ \text{min on both routes}}$$ Note $V_1$ and $V_2$ exceed the nominal capacities: with this BPR-type function that simply drives $V/C>1$ (oversaturation), not an infeasibility.
(b) Add Highway 3. The same argument gives $V_1/C_1=V_2/C_2=V_3/C_3=k$, so $$k=\frac{Q}{C_1+C_2+C_3}=\frac{8000}{2200+3000+2800}=\frac{8000}{8000}=1.000.$$
(b) New volumes and time. $$V_1=2200,\quad V_2=3000,\quad V_3=2800\ \text{veh/h},$$ $$\boxed{t=10+20(1.000)=30.0\ \text{min on all three routes}}$$ The travel time falls from 40.8 min to 30.0 min — every route improves.
(c) Whether the reduction always occurs. Here it does, because the new link adds pure parallel capacity. In a network where routes share links, adding a link can instead raise everyone's travel time — Braess’ paradox — as explained below.

(c) Braess’ paradox. User equilibrium is selfish: each driver minimises only their own travel time, ignoring the congestion externality they impose on others. When a new link is added to a general network, it can offer individual drivers a tempting shortcut whose use loads a shared, congestion-sensitive segment more heavily. At the new equilibrium every driver is still individually best-responding, yet the extra loading on the shared link can make the common travel time higher than before the link existed. The system optimum and the user equilibrium do not coincide, so adding capacity does not guarantee improvement. In this particular problem the three highways are fully independent parallel routes with no shared bottleneck, so the paradox cannot arise and the added capacity strictly helps.

Question 6 — UE assignment
Case	$V_1$	$V_2$	$V_3$	Travel time
(a) Two routes	3384.6	4615.4	—	40.8 min
(b) Three routes	2200	3000	2800	30.0 min

Topic: Wardrop’s first principle (user equilibrium) traffic assignment and Braess’ paradox.

Key relations: UE $\Rightarrow$ equal travel time on all used routes; with equal free-flow terms, $V_i/C_i=k$ and $k=Q/\sum C_i$; link performance $t_i=t_{0}+a(V_i/C_i)$.

Why this works: If one used route were faster, drivers would switch to it until the times equalise — that balancing is the equilibrium. The identical intercepts let the equal-time condition collapse to equal degree of saturation, which conservation then pins down in one line. Adding a purely parallel route raises $\sum C_i$, lowering $k$ and hence every travel time.

Common pitfall: Assuming $V/C$ cannot exceed 1 — the BPR-type delay function is defined for oversaturation and simply returns a large delay. A second trap is concluding that new capacity always helps: that is true for independent parallel routes but false in shared networks (Braess).

Source: Roess, Prassas & McShane, Traffic Engineering, Ch. on traffic assignment; Ortúzar & Willumsen, Modelling Transport, Ch. 10 (assignment; Braess’ paradox).

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