Question 6 of 7: User Equilibrium versus System Optimal Assignment

Question 6: User Equilibrium versus System Optimal 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 non-overlapping parallel routes between one origin–destination pair, each with a linear volume–delay function.

Find. The UE route volumes, route travel times and total vehicle travel time; the same three quantities at the system optimum; and a discussion of the gap and of the demand-management measures that close it.

Approach. Wardrop's first principle equalises the travel time on all used routes, which with two linear functions gives one linear equation; Wardrop's second principle equalises the marginal travel time, obtained by differentiating \(V_it_i(V_i)\), and gives a second linear equation of the same form.

1. Confirm both routes are used. Before solving, load the whole demand onto the cheaper route: at \(V_2=3600\) the time on Route 2 is \(12+36=48\) min, which exceeds Route 1's free-flow time of 22 min. Route 1 is therefore attractive and both routes carry traffic at equilibrium; had the all-or-nothing time stayed below the other route's free-flow time, that route would be unused and would have to be dropped before solving.
2. (a) Apply Wardrop's user-equilibrium condition. At UE no traveller can reduce their own travel time by switching, so both used routes have equal time: $$t_1=t_2 \;\Longrightarrow\; 22+\frac{2V_1}{225}=12+\frac{3600-V_1}{100}$$ $$22+0.0088889V_1=48-0.0100000V_1 \;\Longrightarrow\; 0.0188889V_1=26$$ $$\boxed{V_1=1376.47 \text{ veh},\qquad V_2=3600-1376.47=2223.53 \text{ veh}}$$
3. Evaluate the equilibrium travel times. Substituting back into each function, $$t_1=22+\frac{2(1376.47)}{225}=22+12.235=34.24 \text{ min}$$ $$t_2=12+\frac{2223.53}{100}=12+22.235=34.24 \text{ min}$$ The two agree, confirming the equilibrium. Note that the shorter free-flow route ends up carrying the larger volume — capacity, not free-flow time alone, decides the split.
4. Compute the UE total vehicle travel time. Because both routes have the same travel time, the sum collapses to the demand times that common value: $$T_{UE}=\frac{V_1t_1+V_2t_2}{60}=\frac{3600 \times 34.2353}{60}$$ $$\boxed{T_{UE}=2054.12 \text{ vehicle-hours}}$$
5. (b) Set up the system-optimal condition. The system optimum minimises \(Z=V_1t_1(V_1)+V_2t_2(V_2)\) subject to \(V_1+V_2=Q\). Differentiating each route's total travel time gives its marginal cost, $$\frac{\partial}{\partial V_i}\bigl[V_it_i(V_i)\bigr]=t_i+V_i\frac{dt_i}{dV_i}=a_i+2b_iV_i$$ and the first-order condition of the constrained minimisation is that these marginal costs be equal across routes: $$22+\frac{4V_1}{225}=12+\frac{2(3600-V_1)}{100}$$
6. Solve for the system-optimal split. $$22+0.0177778V_1=84-0.0200000V_1 \;\Longrightarrow\; 0.0377778V_1=62$$ $$\boxed{V_1=1641.18 \text{ veh},\qquad V_2=1958.82 \text{ veh}}$$ The travel times are now unequal: $$t_1=22+\frac{2(1641.18)}{225}=36.59 \text{ min},\qquad t_2=12+\frac{1958.82}{100}=31.59 \text{ min}$$ and the marginal costs are equal at \(22+4(1641.18)/225 = 51.18\) min and \(12+2(1958.82)/100 = 51.18\) min, which is the check that the optimum has been found.
7. Compute the SO total vehicle travel time. $$T_{SO}=\frac{1641.18(36.5882)+1958.82(31.5882)}{60}=\frac{60\,048+61\,876}{60}$$ $$\boxed{T_{SO}=2032.06 \text{ vehicle-hours}}$$
8. (c) Quantify the gap. The system optimum saves $$\Delta T=T_{UE}-T_{SO}=2054.12-2032.06=22.06 \text{ vehicle-hours per peak period}$$ which is 1.07 per cent of the total — the price of anarchy on this network. At an all-purpose value of time of about CAD 20 per hour that is roughly CAD 440 per peak period, or on the order of CAD 110 000 per year over 250 working days.

(c) Why the two conditions differ, and how to move from UE to SO

Why they differ. A traveller choosing a route considers only their own travel time — the average cost of the route. Adding one vehicle to Route 2, however, also slows every other vehicle already on it, imposing an external cost of \(V_2\,dt_2/dV_2\) that the entrant does not pay. The route's marginal social cost therefore exceeds its average cost, and because the two routes have different congestion slopes (\(0.00889\) versus \(0.01000\) min/veh) the size of that unpriced externality differs between them. Equalising average costs (UE) is not the same as equalising marginal costs (SO), so the selfish equilibrium over-loads the route with the steeper delay function. Here UE puts 2223.53 vehicles on Route 2 where the optimum wants 1958.82 — 264.7 too many — and the correction transfers exactly that number to Route 1.

How to close the gap. The theoretically exact instrument is marginal-cost (congestion) pricing: charge each route a toll equal to the externality its users impose, \(\tau_1=V_1\,dt_1/dV_1=(2/225)(1641.18)=14.59\) min-equivalent and \(\tau_2=V_2\,dt_2/dV_2=(1958.82)/100=19.59\) min-equivalent. Under those tolls the generalised costs are \(36.59+14.59=51.18\) and \(31.59+19.59=51.18\) — equal, so the tolled user equilibrium coincides exactly with the system optimum. Only the difference matters in practice, so a differential toll of 5.0 min-equivalent on Route 2 (about CAD 1.67 at a value of time of CAD 20/h) achieves the same split. Practical alternatives and complements are: HOV or HOT lanes on the over-loaded route, which price the externality indirectly; advanced traveller information systems and dynamic route guidance, which nudge drivers toward the under-used route; ramp metering or signal-timing changes that ration entry to Route 2; parking pricing and employer TDM programs that remove vehicle-trips altogether; and staggered work hours that spread the 3600-vehicle demand over a longer peak.

Two caveats complete the discussion. First, the SO assignment is not Pareto-improving: Route 1 users are made worse off (34.24 → 36.59 min) while Route 2 users gain (34.24 → 31.59 min), which is precisely why the split will not occur voluntarily and why revenue recycling is politically necessary. Second, because the question states the two routes do not overlap, each travel-time function depends only on its own volume, the assignment is separable, and Braess's paradox — where adding a link can increase everyone's travel time — cannot arise on this network; it requires shared links whose flows interact.