Question 6 of 7: Wardrop User Equilibrium Assignment

Question 6: Wardrop User Equilibrium 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, then three, parallel and non-overlapping routes between a single origin and a single destination, each with a linear performance function sharing the same 10-minute free-flow time, and a fixed demand.

Find. The equilibrium volumes and travel times with two routes and then with three, whether every route improves, and the reason a new route can sometimes make everyone worse off.

Approach. Apply Wardrop's first principle — all used routes carry equal travel time, no unused route is cheaper — by equating the performance functions, solving jointly with the demand conservation equation, and then repeating with the third route added.

1. Part (a) — check whether both routes will be used before solving. The all-or-nothing pre-check is worth doing on every problem of this type: load the whole demand onto the cheaper route and see whether it stays cheaper. Putting all 8,000 veh/h on Highway 2 would give $t_2 = 10+8000/150 = 63.3$ min against a free-flow $t_1$ of 10 min, so Highway 1 is certainly used, and by symmetry so is Highway 2. Both routes are therefore in the equilibrium solution.
2. Impose Wardrop's first principle. At user equilibrium every used route has the same travel time, so $$t_1 = t_2 \;\Longrightarrow\; 10+\frac{V_1}{110}=10+\frac{V_2}{150}\;\Longrightarrow\;\frac{V_1}{110}=\frac{V_2}{150}$$ Because both routes share the same free-flow time, the equal-time condition reduces to a simple statement that the congestion term is common to both. Writing that common term as $x$ gives $V_1 = 110x$ and $V_2 = 150x$.
3. Close the system with demand conservation and solve. All demand must be assigned, so $$V_1+V_2 = 110x+150x = 260x = 8{,}000 \;\Longrightarrow\; x = \frac{8000}{260}=30.769$$ $$\boxed{V_1 = 3{,}384.6 \text{ veh/h}, \qquad V_2 = 4{,}615.4 \text{ veh/h}}$$ The wider route carries the larger share, in the exact ratio of the capacity parameters, 150 to 110.
4. Evaluate the equilibrium travel time and confirm it is common. Substituting back, $$\begin{aligned}t_1 &= 10+\frac{3384.6}{110}=10+30.77\\ t_2 &= 10+\frac{4615.4}{150}=10+30.77\end{aligned}$$ $$\boxed{t_1 = t_2 = 40.77 \text{ min}}$$ The equality is the check that the assignment is a genuine equilibrium, and the total travel time in the corridor is $8{,}000\times 40.77 = 326{,}154$ vehicle-minutes per hour.
5. Part (b) — add the third route and re-apply the principle. All three free-flow times are 10 min, so the third route will certainly attract traffic and the equal-time condition again reduces to a common congestion term: $$\frac{V_1}{110}=\frac{V_2}{150}=\frac{V_3}{140}=x \;\Longrightarrow\; V_1=110x,\; V_2=150x,\; V_3=140x$$
6. Solve for the new equilibrium. Conservation of demand now gives $$110x+150x+140x=400x=8{,}000 \;\Longrightarrow\; x=20$$ $$\boxed{V_1 = 2{,}200 \text{ veh/h},\quad V_2 = 3{,}000 \text{ veh/h},\quad V_3 = 2{,}800 \text{ veh/h}}$$ The three volumes sum to 8,000 veh/h, and each is again in proportion to its own capacity parameter.
7. Evaluate the new travel times and answer the question asked. Since the congestion term is common, $$t_1=t_2=t_3=10+20$$ $$\boxed{t_1=t_2=t_3=30.0 \text{ min}}$$ Yes — the travel time on every highway is reduced, from 40.77 min to 30.00 min, a saving of 10.77 min for every commuter. Total corridor travel time falls from 326,154 to 240,000 vehicle-minutes per hour, a saving of 86,154 vehicle-minutes or about 1,436 vehicle-hours in each peak hour.
8. Note the distributional effect, which is larger than the average effect. Highway 1 loses 1,184.6 veh/h, i.e. 35 per cent of the traffic it carried before, and Highway 2 loses 1,615.4 veh/h; put the other way, Highway 1 previously carried 53.8 per cent more traffic than it does now. Every user gains the same 10.77 min, but the relief is concentrated on the routes that were most heavily loaded. This is the ordinary and expected outcome when a parallel route is added to a network of independent links.
9. Part (c) — explain why a new route can make things worse, and why it cannot here. The phenomenon is Braess's paradox, and it arises because user equilibrium is not the same as the system optimum. Each driver chooses the route that minimises his or her own travel time and ignores the marginal delay imposed on everyone else already on that route; the equilibrium therefore equalises average costs when efficiency would require equalising marginal costs. When the network contains shared links — so that a new connection allows drivers to combine two links that were previously on different paths — the individually rational reroute can load a common link so heavily that the resulting equilibrium is worse for everybody, even though no driver could improve by changing route unilaterally. The paradox needs interaction between routes: it cannot occur when the performance function of each route depends only on that route's own volume.
10. Apply that test to this network. Here the three highways are parallel and separate, and each performance function contains only its own volume, so the assignment problem is separable and adding a route can only add capacity. That is why part (b) shows an unambiguous improvement while part (c) is nonetheless a real phenomenon on a general network. Two further points complete the answer. First, on this particular network the system optimum coincides with the user equilibrium: setting the marginal costs $10 + 2V_i/b_i$ equal across routes yields the same volume ratio as setting the average costs equal, because all three routes share the same free-flow intercept — so the price of anarchy is zero here, though it is generally positive. Second, the practical remedies where the paradox does bite are marginal-cost (congestion) pricing, which makes drivers face the delay they impose on others, and network design that anticipates the equilibrium response rather than assuming that traffic patterns will stay fixed after the new link opens.