Question 7 of 7: Wardrop User Equilibrium on Parallel Routes

Question 7: Wardrop User Equilibrium on Parallel Routes (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.

Find. The equilibrium volumes and travel times on two routes, then on three, and a discussion of the behavioural limitation of the shortest-path assumption.

Approach. Apply Wardrop's first principle: at user equilibrium no traveller can reduce their own travel time by unilaterally switching routes, so every used route carries the same travel time and no unused route offers less. On a set of independent parallel routes this reduces to equating the travel-time functions subject to the demand constraint — a linear system, since the cost functions here are linear.

1. (a) State the equilibrium conditions for two routes. Both routes are plainly attractive (Route 1's free-flow time of 12 min is only 2 min above Route 2's, and 2800 veh/h on Route 2 alone would take $10 + 0.006(2800) = 26.8$ min), so both will be used and their times must be equal: $$t_1 = t_2\qquad\text{and}\qquad V_1 + V_2 = 2800$$
2. Solve the two-route system. Substituting $V_2 = 2800 - V_1$: $$12 + 0.010\,V_1 = 10 + 0.006\left(2800 - V_1\right) = 10 + 16.8 - 0.006\,V_1$$ $$0.010\,V_1 + 0.006\,V_1 = 26.8 - 12 = 14.8 \;\Longrightarrow\; 0.016\,V_1 = 14.8$$ $$\boxed{V_1 = 925\ \text{veh/h},\qquad V_2 = 2800 - 925 = 1875\ \text{veh/h}}$$ Checking both travel times: $$t_1 = 12 + 0.010(925) = 21.25\ \text{min},\qquad t_2 = 10 + 0.006(1875) = 21.25\ \text{min}\;\checkmark$$ $$\boxed{t_1 = t_2 = 21.25\ \text{min}}$$ The equilibrium is skewed toward Route 2, which takes 67 per cent of the demand, because it is both faster in free flow and less sensitive to volume. Total system travel time is $2800 \times 21.25 = 59\,500$ veh·min.
3. (b) Set up the three-route problem in a form that scales. With linear cost functions $t_i = a_i + b_iV_i$, express each volume in terms of the common equilibrium time $t$: $$V_i = \frac{t - a_i}{b_i}$$ and impose the demand constraint $\sum_i V_i = Q$: $$t\sum_i\frac{1}{b_i} - \sum_i\frac{a_i}{b_i} = Q\;\Longrightarrow\;t = \frac{Q + \sum_i a_i/b_i}{\sum_i 1/b_i}$$ This one-line result is worth remembering: it solves any number of parallel linear routes without simultaneous equations, and it makes the check for unused routes trivial.
4. Evaluate the equilibrium travel time. $$\sum_i\frac{1}{b_i} = \frac{1}{0.010} + \frac{1}{0.006} + \frac{1}{0.006} = 100 + 166.667 + 166.667 = 433.333$$ $$\sum_i\frac{a_i}{b_i} = \frac{12}{0.010} + \frac{10}{0.006} + \frac{10.2}{0.006} = 1200 + 1666.667 + 1700 = 4566.667$$ $$t = \frac{2800 + 4566.667}{433.333} = \frac{7366.667}{433.333}$$ $$\boxed{t_1 = t_2 = t_3 = 17.00\ \text{min}}$$
5. Recover the three volumes. $$V_1 = \frac{17.00 - 12}{0.010} = 500\ \text{veh/h},\qquad V_2 = \frac{17.00 - 10}{0.006} = 1166.7\ \text{veh/h},\qquad V_3 = \frac{17.00 - 10.2}{0.006} = 1133.3\ \text{veh/h}$$ $$\boxed{V_1 = 500,\quad V_2 = 1166.7,\quad V_3 = 1133.3\ \text{veh/h}}$$ All three volumes are strictly positive, which confirms that all three routes are used and that the assumed equilibrium condition (equal times on all three) is the correct one; had any $V_i$ come out negative, that route would have had to be dropped from the set and the calculation repeated. The volumes sum to $500 + 1166.7 + 1133.3 = 2800$ veh/h as required, and substituting back gives $t_1 = 12+0.010(500) = 17.0$, $t_2 = 10+0.006(1166.7) = 17.0$ and $t_3 = 10.2+0.006(1133.3) = 17.0$ min.
6. Quantify the benefit of Route 3. Travel time falls from 21.25 to 17.00 min for every commuter, a reduction of 4.25 min or 20.0 per cent. Aggregated over the demand, $$\Delta = 2800(21.25) - 2800(17.00) = 59\,500 - 47\,600 = 11\,900\ \text{veh}\cdot\text{min} = 198.3\ \text{veh}\cdot\text{h per hour of the peak}$$ Route 1's volume nearly halves, from 925 to 500 veh/h, because it is the steepest and least attractive of the three — the new route draws most heavily from the route that was worst served. It is worth noting explicitly that adding capacity here helped; that is not guaranteed. In a network where the new link is not a simple parallel alternative, adding it can raise equilibrium travel time for everyone, which is the Braess paradox. Because Route 3 is stated not to overlap Routes 1 or 2, the network stays a set of independent parallel paths and the paradox cannot arise.
7. (c) State the limitation of the shortest-path assumption. Deterministic user equilibrium assumes that every driver knows the travel time on every route exactly, evaluates all of them identically, and always chooses the minimum. Four things are wrong with that in practice. Drivers have imperfect information: they know the routes they habitually use, and estimates of alternatives are guesses. Drivers perceive travel times with error and differ from one another in that error, so two drivers facing an identical choice may choose differently. Drivers do not all minimise the same thing — value of time, reliability, tolls, road type, number of turns, signal count, scenery and route familiarity all enter, and a heterogeneous population has a heterogeneous objective function. And drivers exhibit inertia and bounded rationality: a saving of half a minute on a 20-minute trip is below the threshold at which most people will change a habitual route at all. In this problem the deterministic model predicts that Route 1 carries exactly 500 veh/h and not one vehicle more; in reality the split would be dispersed around that figure, and with Routes 2 and 3 differing by only 0.2 min in free-flow time the two would in practice be treated as interchangeable.
8. Set out how to overcome it. The direct remedy is stochastic user equilibrium (SUE), in which each driver minimises perceived travel time, $\tilde{t}_i = t_i + \varepsilon_i$, and the resulting route shares follow a discrete-choice model — a logit route-choice model $P_i = e^{-\theta t_i}\big/\sum_j e^{-\theta t_j}$ if the perception errors are taken as Gumbel, or a probit model if a full correlation structure is wanted. The dispersion parameter $\theta$ controls how tightly flows concentrate on the fastest route, and as $\theta\to\infty$ SUE converges to the deterministic equilibrium computed above, so the deterministic answer is the limiting case rather than a different model. Four supporting measures complete the answer. Use multi-class assignment, with separate user classes for different values of time, so that a tolled or premium route is chosen by the class that values time most rather than by an average driver who does not exist. Where routes overlap physically, use a path-size or C-logit correction to stop the model treating two nearly identical paths as two independent choices — the same over-counting defect as the IIA problem in Question 6, and not needed here only because the question states that Route 3 does not overlap. Model day-to-day dynamics or bounded rationality, with an indifference band inside which drivers do not switch, if the question of interest is how quickly the network re-equilibrates after a change. And recognise that real-time traveller information — the connected-vehicle systems discussed in Question 1(c) — systematically shrinks the perception error, so a network operating with high-quality guidance behaves more like the deterministic model, and one without it behaves less like it.
9. Add the system-optimum contrast, which is the natural close. Wardrop's second principle asks instead for the assignment that minimises total system travel time, which requires equal marginal costs rather than equal average costs. On the two-route network of part (a), setting $d(V_1t_1)/dV_1 = d(V_2t_2)/dV_2$ gives $12+0.020V_1 = 10+0.012V_2$, and with $V_1+V_2=2800$ this yields $V_1 = 987.5$ and $V_2 = 1812.5$ veh/h, with a total system travel time of $59\,437.5$ veh·min against $59\,500$ at user equilibrium. The user-equilibrium inefficiency — the price of anarchy — is therefore only 62.5 veh·min, or 0.105 per cent, on this network. That small figure is itself the useful conclusion: on a simple parallel network with similar routes there is very little to be gained by trying to steer drivers away from their selfish choices, and the 20 per cent saving delivered by building Route 3 dwarfs it by two orders of magnitude. Congestion pricing and system-optimal routing earn their keep on networks with severe bottlenecks and dissimilar routes, not on this one.