16-Civ-A6 Highway Design, Construction, and Maintenance · December 2016

Question 6 of 7: User-Equilibrium Traffic Assignment

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

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. Two, later three, non-overlapping routes between one origin–destination pair, each with a linear volume–delay function, and a fixed peak-hour demand.

Find. The equilibrium volumes and the common travel time with two routes, then with three; and whether adding the third route helps.

Approach. Impose Wardrop's first principle — every used route carries the same travel time and no unused route is faster — by inverting each linear time function to $V_i(t)$ and summing to the known demand, which reduces the problem to one equation in the common time $t$.

1. State the equilibrium conditions. Wardrop's user-equilibrium requires $$t_1=t_2=t \quad\text{(both used)},\qquad V_1+V_2=Q=3600,\qquad V_i\ge0$$ At UE no driver can reduce their own travel time by switching, which is exactly the condition that all used routes share one travel time.
2. Invert each volume–delay function. Solving each linear function for volume as a function of the common time, $$V_1=\frac{225}{2}\,(t-22)=112.5\,(t-22),\qquad V_2=100\,(t-12)$$ Each is valid only while non-negative, i.e. while $t$ exceeds that route's free-flow time.
3. Solve for the equilibrium time with two routes. Substituting into the demand constraint, $$112.5\,(t-22)+100\,(t-12)=3600$$ $$212.5\,t=3600+2475+1200=7275$$ $$\boxed{t=34.24\ \text{min}}$$ Both routes are used, since $34.24$ exceeds both free-flow times (22 and 12 min), so the two-route solution is admissible and no route needs to be dropped.
4. Back-substitute for the volumes. $$V_1=112.5\,(34.24-22)=1376.5\ \text{veh/h},\qquad V_2=100\,(34.24-12)=2223.5\ \text{veh/h}$$ $$\boxed{V_1=1{,}376.5\ \text{veh/h},\quad V_2=2{,}223.5\ \text{veh/h},\quad t_1=t_2=34.24\ \text{min}}$$ The volumes sum to 3,600 veh/h as required. Route 2 carries the larger share despite its steeper congestion coefficient, because its 10-minute free-flow advantage dominates.
5. Repeat with Route 3 in the network. Adding $V_3=\frac{225}{4}(t-14)=56.25\,(t-14)$ to the demand constraint, $$112.5\,(t-22)+100\,(t-12)+56.25\,(t-14)=3600$$ $$268.75\,t=3600+2475+1200+787.5=8062.5$$ $$\boxed{t=30.00\ \text{min exactly}}$$ $$V_1=112.5(8)=900,\qquad V_2=100(18)=1800,\qquad V_3=56.25(16)=900\ \text{veh/h}$$ All three volumes are positive, so all three routes are used, and each returns $t_i=30.00$ min — confirming the equilibrium. (Had any volume come out negative, that route would be unused and the problem would have to be re-solved on the remaining routes, with Wardrop's second condition then verified.)
6. Assess the effect of the new route. Comparing the two equilibria, $$\Delta t = 34.24-30.00=4.24\ \text{min saved per traveller, a }12.4\%\ \text{reduction}$$ Total system travel time falls from $123{,}247$ to $108{,}000$ veh-min/h, a saving of $15{,}247$ veh-min/h — about 254 vehicle-hours every peak hour. Route 1's volume drops from 1,376.5 to 900 veh/h and Route 2's from 2,223.5 to 1,800 veh/h, so the new route relieves both existing routes as intended. $$\boxed{\text{Yes — the new route reduces travel time, from }34.24\text{ to }30.00\text{ min.}}$$ This outcome is not automatic. Adding capacity to a congested network can raise equilibrium travel time (Braess's paradox), because user equilibrium is not system-optimal; here the added route is a genuine parallel alternative rather than a shortcut that induces a worse routing pattern, so no paradox arises.

(c) Limitation of the shortest-path assumption, and how to overcome it

Deterministic user-equilibrium assignment rests on three assumptions that real drivers violate. First, perfect information and perfect perception: every driver is assumed to know the travel time on every route exactly. In reality drivers estimate travel times from limited experience, so two drivers facing identical conditions may rank the same pair of routes differently, and the observed flow pattern is spread across routes that the deterministic model would leave empty. Second, a homogeneous population: all drivers are assumed to value time identically and to minimise the same objective. Real drivers differ in value of time (a courier and a commuter respond to a toll quite differently) and weight attributes the model ignores — travel-time reliability, number of signals and stops, tolls, road-surface quality, truck presence, scenery, ease of navigation and simple familiarity or habit. Third, travel time as the sole disutility, when route choice is really made on a generalised cost. A model built on these assumptions will over-concentrate flow on the nominally fastest routes, under-predict flow on parallel arterials, and mis-forecast the response to any policy whose effect is on reliability or cost rather than mean time.

The remedies form a natural progression. Replacing travel time with a generalised cost $c_i = t_i + \theta\,(\text{toll}) + \phi\,(\text{reliability measure}) + \dots$ retains the equilibrium framework while admitting the attributes that matter. Introducing perception error gives stochastic user equilibrium (SUE), in which route choice follows a logit or probit model of perceived cost, so every route receives a positive share that falls smoothly as its cost rises; this is the direct fix for the "all-or-nothing" behaviour of the deterministic model and it is what commercial planning packages implement. Because a logit route-choice model inherits the independence-of-irrelevant-alternatives problem, overlapping paths must be handled explicitly by path-size logit, C-logit or a probit formulation — the present question sidesteps this by stipulating that the routes do not overlap, which is precisely the condition that makes plain logit defensible. Allowing the value of time to vary gives multi-class assignment, with a separate user class (and its own generalised cost) for commuters, freight and tolled traffic. Finally, where the concern is day-to-day variability or within-peak dynamics, a day-to-day learning or dynamic traffic-assignment model reproduces the adjustment process itself rather than assuming the network is always at equilibrium. In each case the analyst should also calibrate against observed link counts and, where possible, probe data, since the choice among these formulations is ultimately empirical.