User-Equilibrium Traffic Assignment: 16-Civ-B7 Transportation Planning and Engineering, May 2017, Q7

Question 7: 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, and later three, parallel routes between the same origin and destination, each with a linear link-performance function, carrying a fixed total demand.

Given data — route performance functions
Route	Travel-time function (minutes)	Free-flow time	Congestion coefficient
Existing bypass	$t_b = 12 + 0.004 V_b$	12 min	0.004 min/veh
Town-centre route	$t_t = 8 + 0.016 V_t$	8 min	0.016 min/veh
New bypass (part b)	$t_{b'} = 16 + 0.002 V_{b'}$	16 min	0.002 min/veh
Total demand	$V = 2500$ vehicles

Find. The volume and travel time on each route at user equilibrium, before and after the second bypass is opened, and a discussion of the perfect-information assumption that underlies the equilibrium.

Figure 7.1 — Travel time on each route against the volume assigned to the bypass. User equilibrium is the intersection, where both routes take 19.2 minutes.

Approach. Apply Wardrop's first principle — at user equilibrium every used route between an origin and destination has the same travel time, and no unused route offers a shorter one — then solve the resulting equal-time condition together with conservation of the total demand.

Part (a) — state the equilibrium condition for two routes. If both routes carry traffic, Wardrop's first principle requires $t_b = t_t$, while conservation requires $V_b + V_t = 2500$. Substituting $V_t = 2500 - V_b$ into the equal-time condition, $$12 + 0.004 V_b = 8 + 0.016(2500 - V_b)$$
Solve for the split. Expanding the right-hand side gives $12 + 0.004V_b = 48 - 0.016V_b$, so $$0.020\,V_b = 36 \quad\Longrightarrow\quad \boxed{V_b = 1800 \ \text{veh}, \qquad V_t = 700 \ \text{veh}}$$
Evaluate the common travel time. Substituting back into either function, $$t_b = 12 + 0.004(1800) = 19.2 \ \text{min}, \qquad t_t = 8 + 0.016(700) = 19.2 \ \text{min}$$ The two agree, which confirms the equilibrium: $$\boxed{t_b = t_t = 19.2 \ \text{minutes}}$$ The bypass carries 72 per cent of the demand although its free-flow time is 50 per cent longer, because its congestion coefficient is only a quarter of the town-centre route's — that is what "high capacity" means numerically. The total travel time in the system is $2500 \times 19.2 = 48\,000$ vehicle-minutes.
Part (b) — generalise the equilibrium to three routes. With three parallel routes the same principle applies: at equilibrium all used routes share a common travel time $t$, so each volume can be written in terms of it by inverting its own performance function, $$V_b = \frac{t - 12}{0.004}, \qquad V_t = \frac{t - 8}{0.016}, \qquad V_{b'} = \frac{t - 16}{0.002}$$ This inversion is the efficient route to the answer: it turns a three-unknown problem into one equation in $t$.
Impose conservation and solve for the common time. Requiring the three volumes to sum to 2500, $$\frac{t-12}{0.004} + \frac{t-8}{0.016} + \frac{t-16}{0.002} = 2500$$ $$250(t-12) + 62.5(t-8) + 500(t-16) = 2500$$ $$812.5\,t - 11\,500 = 2500 \quad\Longrightarrow\quad \boxed{t = 17.23 \ \text{minutes}}$$ Before using this value, check that all three routes are in fact used: $t = 17.23$ minutes exceeds the largest free-flow time of 16 minutes, so the new bypass does attract traffic and the assumption that all three are used is self-consistent. Had $t$ come out below 16 minutes, the new bypass would have to be dropped and the two-route solution retained.
Recover the three volumes. Substituting $t = 17.2308$ minutes into each inverted function, $$V_b = \frac{5.2308}{0.004} = 1307.7, \qquad V_t = \frac{9.2308}{0.016} = 576.9, \qquad V_{b'} = \frac{1.2308}{0.002} = 615.4$$ $$\boxed{V_b = 1308 \ \text{veh}, \quad V_t = 577 \ \text{veh}, \quad V_{b'} = 615 \ \text{veh}}$$ The three sum to 2500 vehicles exactly, which is the conservation check.
Quantify the benefit. Every traveller now experiences 17.23 minutes instead of 19.2, so the total travel time falls from 48 000 to $2500 \times 17.2308 = 43\,077$ vehicle-minutes, a saving of 4923 vehicle-minutes or 10.3 per cent. The new bypass draws 492 vehicles from the old bypass and 123 from the town centre, so although it was built to relieve the town centre it takes four times as much traffic from the parallel bypass — a result that is obvious from the algebra but frequently surprises the sponsors of such projects.

Final results — Question 7
Quantity	(a) Two routes	(b) Three routes
Existing bypass volume $V_b$	1800 veh	1307.7 veh
Town-centre volume $V_t$	700 veh	576.9 veh
New bypass volume $V_{b'}$	—	615.4 veh
Equilibrium travel time on every used route	19.2 min	17.23 min
Total system travel time	48 000 veh-min	43 077 veh-min
Saving from the new bypass	4923 veh-min, or 10.3 per cent

(c) The perfect-information assumption

Deterministic user equilibrium assumes that every driver knows the travel time on every route exactly and chooses the fastest one, which is why the equilibrium is characterised by exactly equal travel times. Real drivers do not satisfy this. They perceive travel times with error, and the error differs from driver to driver according to familiarity with the network, trip purpose, departure time and the information available to them. They also value attributes the model omits entirely — reliability, signal density, road type, scenery, tolls, the presence of schools or truck traffic — so two routes of equal travel time are not equally attractive. Habit and inertia keep many drivers on a familiar route long after conditions have changed, and travel times fluctuate day to day so that no single deterministic value is ever "the" travel time. The observable consequence is that in reality some traffic uses routes that are demonstrably slower than the best available, which a deterministic assignment can never reproduce: it loads nothing at all onto a route whose travel time exceeds the equilibrium value, as it would have done here had the new bypass's free-flow time exceeded 17.23 minutes.

The standard remedy is stochastic user equilibrium, which replaces the perceived travel time with a random variable, $C_r = c_r + \varepsilon_r$, and assigns each route a choice probability rather than a share of zero or all. With Gumbel-distributed errors this yields a logit route-choice model of exactly the form used in Question 6, with a dispersion parameter that controls how sharply traffic concentrates on the best route; a probit formulation is preferred where overlapping routes make the errors correlated, since route overlap is the network analogue of the IIA problem. The equilibrium condition then becomes that no driver believes a better route is available, which is a weaker and more defensible statement than the deterministic one, and the assignment is solved by the method of successive averages rather than by Frank-Wolfe. Two refinements are usually needed alongside it. Route overlap must be corrected explicitly, through the C-logit or path-size logit commuted-cost term, so that two nearly identical paths are not double-counted as independent alternatives. And where the interest is in how travellers respond to changing conditions rather than in a long-run average, a day-to-day learning or dynamic-assignment model represents the process by which drivers update their beliefs from experience and from the traveller-information systems discussed in Question 1(c) — which, by supplying real-time travel times, actually push the network toward the deterministic equilibrium the model assumes.

Topic. Deterministic user-equilibrium traffic assignment on parallel routes, and the case for a stochastic formulation.

Key relations. Wardrop's first principle: all used routes between an origin and destination have equal and minimum travel time. With linear performance functions $t_r = a_r + b_r V_r$, inverting to $V_r = (t - a_r)/b_r$ and imposing $\sum_r V_r = V$ gives the closed form $t = \left(V + \sum_r a_r/b_r\right)\big/\sum_r (1/b_r)$, valid provided $t > \max_r a_r$; otherwise the routes with $a_r \ge t$ are unused and the calculation is repeated without them.

Why this works. User equilibrium is the fixed point of a behavioural process: any driver on a slower route can improve by switching, so switching continues until no improvement is available, at which point all used routes are equally fast. Because each performance function is increasing in its own volume, the equilibrium exists and is unique, and it can be shown to minimise the Beckmann integral $\sum_r \int_0^{V_r} t_r(x)\,dx$ — which is not the total travel time, and this is why user equilibrium generally differs from the system optimum.

Common pitfall. Forgetting to test whether a route is used at all: mechanically solving the three-route equation without checking $t > 16$ minutes will happily return a negative volume on a route that should simply have been excluded. A second, more conceptual error is to solve for the split that minimises total travel time — that is the system optimum, which requires marginal rather than average costs to be equalised, and it is not what "user equilibrium" asks for.

Source. Ortuzar & Willumsen, Modelling Transport, 4th ed., Ch. 10–11 (assignment, stochastic user equilibrium); Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 12; Papacostas & Prevedouros, Transportation Engineering and Planning, 3rd ed., Ch. 11.

Question 7 of 7: User-Equilibrium Traffic Assignment

Question 7: User-Equilibrium Traffic Assignment (20 marks)

(c) The perfect-information assumption