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

Question 6 of 7: User-Equilibrium Assignment and the Braess Paradox

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

Question 6: User-Equilibrium Assignment and the Braess Paradox (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. Three non-overlapping parallel routes between a common origin and a common destination, each with a linear volume-delay function, carrying a fixed peak-hour demand.

Find. The equilibrium volumes and common travel time with two routes, then with three routes, and an explanation of why adding capacity can make everyone worse off.

Approach. Apply Wardrop's first principle: at user equilibrium every used route between the same origin–destination pair carries the same travel time, and no unused route offers a shorter one. Because each performance function is linear, invert each one to express volume as a function of the common time and sum to the demand.

1. State the equilibrium conditions for two routes. With both routes used, $$ t_1 = t_2 = t^\ast, \qquad V_1 + V_2 = 1\,800 $$ Substituting the performance functions, $$ 15 + \frac{V_1}{200} = 5 + \frac{1\,800-V_1}{150} $$
2. Solve for the volume on Route 1. Multiplying through by 600 to clear the denominators, $$ 9\,000 + 3V_1 = 3\,000 + 4(1\,800-V_1) = 10\,200 - 4V_1 $$ so \(7V_1 = 1\,200\) and $$ \boxed{V_1 = 171.4\ \text{veh/h}}, \qquad \boxed{V_2 = 1\,800 - 171.4 = 1\,628.6\ \text{veh/h}} $$
3. Evaluate the common travel time and check it — part (a). $$ t_1 = 15 + \frac{171.4}{200} = 15.86\ \text{min}, \qquad t_2 = 5 + \frac{1\,628.6}{150} = 15.86\ \text{min} $$ The two agree, confirming the equilibrium: $$ \boxed{t^\ast = 15.86\ \text{minutes on both routes}} $$ The lopsided split — Route 2 carries more than nine times the volume of Route 1 — is the direct consequence of Route 2's ten-minute advantage in free-flow time, which it can spend on congestion before drivers become indifferent.
4. Test whether all three routes are used in part (b). Inverting each function, \(V_1 = 200(t-15)\), \(V_2 = 150(t-5)\) and \(V_3 = 75(t-10)\). If all three were used, $$ 200(t-15)+150(t-5)+75(t-10) = 1\,800 \;\Longrightarrow\; 425t = 6\,300 \;\Longrightarrow\; t = 14.82\ \text{min} $$ which would give \(V_1 = 200(14.82-15) = -35\) veh/h. A negative volume is inadmissible, so Route 1 must be unused at the new equilibrium. This is Wardrop's second condition in action: Route 1's free-flow time of 15 min already exceeds the equilibrium time available on the other two routes.
5. Re-solve with Routes 2 and 3 only. $$ 150(t-5)+75(t-10) = 1\,800 \;\Longrightarrow\; 225t = 3\,300 \;\Longrightarrow\; t = 14.667\ \text{min} $$ Back-substituting, $$ V_2 = 150(14.667-5) = 1\,450\ \text{veh/h}, \qquad V_3 = 75(14.667-10) = 350\ \text{veh/h} $$ and the two sum to 1 800 as required.
6. Verify that leaving Route 1 empty is consistent. At \(V_1 = 0\) the notional travel time on Route 1 is its free-flow time of 15.00 min, which is greater than the 14.67 min available on Routes 2 and 3. No driver can improve by switching to Route 1, so the solution satisfies both Wardrop conditions: $$ \boxed{V_1 = 0,\quad V_2 = 1\,450,\quad V_3 = 350\ \text{veh/h},\qquad t^\ast = 14.67\ \text{minutes}} $$
7. Answer the question asked: is travel time reduced? Yes. Every traveller now experiences 14.67 min instead of 15.86 min, a saving of 1.19 min or 7.5 per cent per trip. Aggregated over the 1 800 vehicles this is $$ 1\,800 \times 1.19 = 2\,143\ \text{vehicle-minutes} \approx 35.7\ \text{vehicle-hours saved per peak hour} $$ Note that Route 1 does not have a "reduced" travel time in the ordinary sense — it is abandoned entirely, and its nominal time falls from 15.86 min to its free-flow 15.00 min because it carries no traffic. Route 2 sheds 178.6 veh/h to the new route and its time falls accordingly. This particular network is therefore not an instance of the paradox described in part (c).

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

The result in part (b) is the intuitive one, but it is not guaranteed. In 1968 Dietrich Braess showed that adding a link to a congested network can raise the equilibrium travel time on every route, and the effect has been observed in practice — the closure of 42nd Street in New York and of the Cheonggyecheon expressway in Seoul both improved traffic conditions.

The mechanism has two ingredients. The first is that user equilibrium is not a system optimum. Each driver chooses the route that minimises their own travel time and ignores the marginal delay that their presence imposes on everyone else already on that link. The equilibrium therefore settles where average costs are equalised, whereas the system optimum requires marginal costs to be equalised. The gap between them — the price of anarchy — means the equilibrium flow pattern is generally inefficient, and there is no reason why making the network larger should shrink that inefficiency.

The second ingredient is that a new link changes the set of paths, not merely the set of routes. In the network of this question the three routes are explicitly stated not to overlap, so each carries its own independent flow and Wardrop's conditions can only redistribute demand onto more capacity — the equilibrium time can never rise. The paradox needs shared links. When a new connector allows drivers to combine the uncongested first half of one route with the uncongested second half of another, that combined path is individually attractive, so traffic floods onto it. But it now loads both of the shared links simultaneously, and because volume-delay functions are convex and steeply rising near capacity, the added congestion on those two links can exceed the saving from the shortcut. Every driver, acting rationally, ends up worse off than before the link existed, and no driver can improve by switching back — that is precisely what makes it an equilibrium.

The practical lessons for the transportation engineer are three. Adding capacity to a congested network must be tested by re-running the assignment on the whole network, never by reasoning locally about the new link. Demand should be treated as elastic, since a faster network induces additional trips that erode the benefit. And congestion pricing set at the marginal external cost realigns the user equilibrium with the system optimum, which both eliminates the paradox and generally delivers more benefit per dollar than the corresponding capacity expansion.