Question 6 of 7: User-Equilibrium Assignment and the Value of a New Route

Question 6: User-Equilibrium Assignment and the Value of a New Route (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.

Total demand from zone A to zone B, Q = 4,400 veh/h, fixed and independent of the travel time. The routes are separate facilities, so each performance function depends only on its own volume.

Find. The user-equilibrium volume and travel time on each route, first with two routes and then with three; and a critical discussion of the shortest-path assumption.

Approach. Apply Wardrop's first principle — at user equilibrium every used route between an origin–destination pair carries the same travel time, and no unused route offers less — then solve the resulting linear system against the demand constraint, checking that every volume comes out non-negative.

1. Part (a) — check first whether both routes are used. Route 1 has much the lower free-flow time (1.50 min against 4.25 min), so an all-or-nothing assignment would load all 4,400 veh/h onto it, giving t₁ = 1.5 + 0.015(4400) = 67.5 min — far above route 2's free-flow 4.25 min. Route 2 is therefore certainly used, and both routes carry traffic at equilibrium. This one-line pre-check is worth making on every assignment problem, because a route whose back-substituted volume is negative must be dropped before re-solving.
2. Impose Wardrop's first principle and solve. Setting t₁ = t₂ with V₂ = Q − V₁, $$1.5+0.015\,V_1=4.25+0.0125\,(4400-V_1)$$ $$1.5+0.015\,V_1=59.25-0.0125\,V_1 \;\Longrightarrow\; 0.0275\,V_1=57.75$$ $$\boxed{\begin{aligned}V_1=2100\ \text{veh/h} \\ V_2=4400-2100=2300\ \text{veh/h}\end{aligned}}$$
3. Back-substitute for the equilibrium travel time. $$\begin{aligned}t_1&=1.5+0.015(2100)=1.5+31.5=33.0\ \text{min}, \\ t_2&=4.25+0.0125(2300)=4.25+28.75=33.0\ \text{min}\end{aligned}$$ The two agree, as the equilibrium condition requires, so $$\boxed{t_1=t_2=33.0\ \text{minutes}}$$ The total travel time consumed by the corridor is 4,400 × 33.0 = 145,200 veh·min = 2,420 veh·h per hour of operation.
4. Part (b) — extend the equilibrium condition to three routes. With all three used, all three carry the common travel time t, so each performance function can be inverted for its own volume: $$\begin{aligned}V_1&=\frac{t-1.5}{0.015}, \\ V_2&=\frac{t-4.25}{0.0125}, \\ V_3&=\frac{t-0.4}{0.008}\end{aligned}$$ Summing to the fixed demand gives a single equation in t: $$\frac{t-1.5}{0.015}+\frac{t-4.25}{0.0125}+\frac{t-0.4}{0.008}=4400$$ $$(66.667t-100)+(80t-340)+(125t-50)=4400 \;\Longrightarrow\; 271.667\,t=4890$$ $$\boxed{t_1=t_2=t_3=18.0\ \text{minutes}}$$
5. Back-substitute for the three volumes and confirm the demand closes. $$\begin{aligned}V_1&=\frac{18.0-1.5}{0.015}=1100, \\ V_2&=\frac{18.0-4.25}{0.0125}=1100, \\ V_3&=\frac{18.0-0.4}{0.008}=2200\end{aligned}$$ All three are positive, so the assumption that every route is used is validated, and 1100 + 1100 + 2200 = 4,400 veh/h reproduces the demand exactly. Checking each performance function returns 18.0 min in every case.

What the new route buys. The equilibrium travel time falls from 33.0 min to 18.0 min, a saving of 15.0 min or 45.5 per cent for every traveller in the corridor. Total travel time falls from 145,200 to 4,400 × 18.0 = 79,200 veh·min, a saving of 66,000 veh·min = 1,100 vehicle-hours in each hour of operation. The distributional effect is much larger than the average one and is worth a sentence: route 1 loses 1,000 veh/h, 47.6 per cent of its traffic, and route 2 loses 1,200 veh/h, while route 3 takes 2,200 veh/h — half the corridor — because it is simultaneously the fastest at free flow and the least congestible of the three.

Part (c) — the limitation of the shortest-path assumption, and how to overcome it

The deterministic user-equilibrium model just solved rests on three behavioural assumptions that are all demonstrably false in detail. It assumes that every driver has perfect information about the travel time on every route; that every driver perceives that time identically, so a route one second faster captures the traveller entirely; and that route choice is made on travel time alone. Real drivers work from incomplete and out-of-date information, they perceive the same route differently according to familiarity, habit and the day's experience, and they weigh reliability, tolls, fuel, road type, scenery, merging difficulty and the number of turns as well as time. The visible consequence is that a deterministic model puts zero flow on any route even slightly slower than the best, whereas observed traffic spreads across all reasonable routes. In this corridor the result is a cliff: with two routes the model assigns nothing to route 2 until route 1 is loaded past 183 veh/h, which no observation would support.

The standard remedy is stochastic user equilibrium (SUE), which replaces the actual travel time with a perceived travel time equal to the actual time plus a random error term, and defines equilibrium as the state in which no driver believes he can improve his perceived time by switching. Implemented with a logit route-choice kernel, route r receives the share exp(−θt_r) / Σ exp(−θt_s), where the dispersion parameter θ is calibrated from observed route splits; a large θ recovers the deterministic solution and a small θ approaches an even spread. Applied here, SUE would leave a modest but non-zero flow on the slower routes at all demand levels, which matches observation. Probit-based SUE (Burrell or Monte-Carlo assignment) is the alternative when the error terms need to be correlated.

Two refinements complete a professional answer. First, a plain logit kernel suffers the route-overlap problem — the independence-of-irrelevant-alternatives property discussed in Question 7 — so two routes sharing most of their length are treated as independent alternatives and jointly attract too much traffic; path-size logit, C-logit or a link-nested formulation corrects it. In this particular problem the paper states that the routes are separate facilities with independent performance functions, so overlap does not arise, but on a real network it always does. Second, the objective should be generalised cost rather than travel time — monetised time plus toll plus operating cost, with a reliability term such as the 95th-percentile travel time — and the model should be segmented by traveller type, because a commuter with a fixed arrival time and a discretionary traveller value reliability quite differently.

It is also worth answering the question that the pairing of parts (b) and (c) invites. Adding a route does not always help: under Braess's paradox, adding a link to a network can raise the travel time on every route, because user equilibrium equalises average cost while the socially efficient assignment equalises marginal cost, and individually rational route switching can move the network to a worse collective state. That cannot happen here, because the three routes do not overlap, so each performance function depends only on its own volume, the assignment is separable, and a parallel route can only add capacity — which is precisely why part (b) shows every route improving. The price of that self-interest is nevertheless measurable: solving the two-route network for the system optimum, where the marginal costs a_i + 2b_iV_i rather than the average costs are equalised, gives V₁ = 2,050 and V₂ = 2,350 veh/h and a total travel time of 145,131 veh·min against the equilibrium's 145,200 — so drivers acting in their own interest waste about 69 veh·min per hour on this particular network, a small price of anarchy because the two routes are closely matched.