Question 6 of 7: Wardrop User Equilibrium on Parallel Routes

Question 6: 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. A fixed peak-hour demand assigned across non-overlapping parallel routes with linear, separable link performance functions.

Find. The equilibrium volumes and travel times on two routes and then on three, whether the new route reduces travel time on every route, and why adding a route can sometimes make everyone worse off.

Approach. Impose Wardrop’s first principle — every used route carries the same travel time, and no unused route is faster — then invert each linear performance function to express volume as a function of that common time and solve the single flow-conservation equation.

1. Check that both routes are used before assuming they are. If all 1,800 veh/h took Route 2 its travel time would be \(6 + 1800/200 = 15\) minutes, while an empty Route 1 offers 11 minutes. Route 1 is therefore attractive and both routes carry flow, so the equal-time condition applies rather than a corner solution.
2. Impose the equal-travel-time condition (part a). Setting \(t_1 = t_2\) with \(V_2 = Q - V_1\), \[11 + \frac{V_1}{225} = 6 + \frac{1800 - V_1}{200} = 15 - \frac{V_1}{200},\] \[V_1\left(\frac{1}{225} + \frac{1}{200}\right) = 4 \quad\Longrightarrow\quad V_1 = \frac{4 \times 225 \times 200}{225 + 200} = \frac{180{,}000}{425} = 423.53\ \text{veh/h}.\]
3. Recover the second volume and the common travel time. \[V_2 = 1800 - 423.53 = 1376.47\ \text{veh/h},\] \[t_1 = 11 + \frac{423.53}{225} = 12.88\ \text{min},\qquad t_2 = 6 + \frac{1376.47}{200} = 12.88\ \text{min}.\] The two times agree, which confirms the equilibrium. \[\boxed{V_1 = 424\ \text{veh/h},\quad V_2 = 1{,}376\ \text{veh/h},\quad t_1 = t_2 = 12.88\ \text{min}}\] Total system travel time is \(1800 \times 12.88 = 23{,}188\) veh·min.
4. Set up the three-route equilibrium (part b). Rewriting Route 3 as \(t_3 = 7 + V_3/112.5\) and inverting all three functions about the common equilibrium time \(t\), \[V_1 = 225(t - 11),\qquad V_2 = 200(t - 6),\qquad V_3 = 112.5(t - 7).\]
5. Solve the single conservation equation for the common time. Substituting into \(V_1 + V_2 + V_3 = 1800\), \[(225 + 200 + 112.5)\,t - \big(225(11) + 200(6) + 112.5(7)\big) = 1800,\] \[537.5\,t = 1800 + 4462.5 = 6262.5 \quad\Longrightarrow\quad \boxed{t = 11.65\ \text{minutes}}\]
6. Back-substitute for the three volumes and confirm all are positive. \[V_1 = 225(11.651 - 11) = 146.51,\qquad V_2 = 200(11.651 - 6) = 1130.23,\qquad V_3 = 112.5(11.651 - 7) = 523.26\ \text{veh/h}.\] All three are positive, so the assumption that every route is used is self-consistent, and they sum to 1,800 veh/h as required. Had any come out negative, that route would have been dropped and the remaining equations re-solved.
7. Answer the question actually asked: is travel time reduced on each route? Yes. At user equilibrium every used route carries the same travel time by construction, so there is a single number to compare: it falls from 12.88 to 11.65 minutes, a saving of 1.23 minutes or 9.6 per cent for every traveller on the corridor. Total system travel time falls from 23,188 to 20,972 veh·min, a saving of 2,216 veh·min or about 36.9 vehicle-hours in the peak hour. The distributional effect is large: Route 1, the slowest free-flow option, loses 65.4 per cent of its volume as travellers reassign to the new route.

(c) Why adding a route can increase travel times — the Braess paradox

User equilibrium is the outcome of every traveller choosing selfishly, and a selfish equilibrium is not in general the outcome that minimises total travel time. Each additional vehicle on a route imposes a delay on every other vehicle already there, and the driver choosing that route pays only their own travel time, not that external cost. The equilibrium therefore sits where average costs are equal, whereas the system optimum sits where marginal costs are equal. Even on the simple two-route network of part (a) the gap is real: minimising \(V_1 t_1(V_1) + V_2 t_2(V_2)\) puts 688 veh/h on Route 1 and 1,112 veh/h on Route 2 for a total of 22,527 veh·min, which is 662 veh·min better than what selfish routeing delivers.

Adding a link enlarges the set of choices available to those selfish travellers. If the new link is on a shared part of the network, the reassignment it triggers can load a link that other travellers also depend on, and the new equilibrium can be worse for everybody — the Braess paradox. The classic construction has two long-and-insensitive routes joined by a short new cross-link: adding the cross-link makes a hybrid path attractive, everyone switches to it, both congestible segments become over-loaded, and the equilibrium travel time rises above what it was before the link existed. Nothing is wrong with the arithmetic; the paradox is a property of selfish equilibrium, and the same construction has been observed in reverse when closing a street in a dense downtown network improved travel times.

The reason no paradox occurs in part (b) is structural and worth stating explicitly: the question specifies that the three routes do not overlap. Each performance function therefore depends only on its own volume, the assignment problem is separable, and adding a route can only add capacity in parallel. On a network of strictly parallel routes with increasing, separable cost functions, more routes never hurt. The paradox needs shared links, so that the flow one traveller adds to a new path also degrades a path someone else is using. The practical lesson for Canadian network planning is that the effect of a new link must always be tested by re-running the assignment on the full network rather than reasoned about link by link, and that the remedies — congestion pricing set at the marginal external cost, ramp metering, or simply not building the link — work by closing the gap between average and marginal cost.