Question 2 of 7: Deterministic Queueing at a Freeway Incident

Question 2: Deterministic Queueing at a Freeway Incident (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 deterministic D/D/1 queue with a constant arrival rate and a service rate that changes twice as the incident is cleared, with the clock started at the instant of the accident.

Find. The cumulative arrival and departure curves, the instant at which the queue clears, the maximum queue length and maximum waiting time, and the total and average delay attributable to the closure.

Approach. Build the cumulative arrival function $A(t)$ and the piecewise cumulative departure function $D(t)$, find the clearance instant by equating them on the branch where the queue is discharging, and read the queue length as the vertical gap, the waiting time as the horizontal gap and the total delay as the area between the two curves.

1. Part (a) — write the cumulative arrival function. Arrivals are undisturbed by the incident, because the queue forms upstream of the blockage and vehicles keep joining it at the prevailing demand rate: $$A(t)=\lambda t = 12t \qquad (t \text{ in min})$$ This is the straight line of slope 12 veh/min in the figure above.
2. Build the cumulative departure function branch by branch. Nothing gets past the incident while both lanes are blocked, one lane discharges at 6 veh/min for the next 25 minutes, and the full section capacity of 18 veh/min applies once the incident is cleared: $$D(t)=\begin{cases}0, & 0\le t\le 10\\[2pt] 6\,(t-10), & 10\le t\le 35\\[2pt] 150+18\,(t-35), & t\ge 35\end{cases}$$ Evaluating the middle branch at its upper limit gives the number of vehicles released during the single-lane period, $D(35)=6\times 25=\boxed{150 \text{ veh}}$, which is the value carried forward into the third branch.
3. Test each interval before solving for the clearance instant. The queue can only clear where the departure rate exceeds the arrival rate, and that happens only on the third branch: during $0\le t\le 10$ the queue grows at 12 veh/min, during $10\le t\le 35$ it still grows, at $12-6=6$ veh/min, and only after $t=35$ does it shrink, at $18-12=6$ veh/min. Setting $A(t)=D(t)$ on the third branch, $$12t = 150 + 18\,(t-35) \;\Longrightarrow\; 12t = 18t - 480 \;\Longrightarrow\; 6t = 480$$ $$\boxed{t_c = 80 \text{ min after the accident}}$$ The root satisfies $t_c \ge 35$, so it lies on the branch it was derived from — the interval test that a candidate must perform, because a root obtained from the wrong branch is meaningless. As a closure check, cumulative departures at that instant are $150+18(80-35)=960$ veh and cumulative arrivals are $12\times 80=960$ veh, so the two curves meet exactly.
4. Part (b) — read the maximum queue length as the largest vertical gap. The queue is $Q(t)=A(t)-D(t)$, and because the arrival rate exceeds the service rate throughout the first 35 minutes and falls below it afterwards, the vertical gap is largest at the moment the second lane reopens: $$Q_{\max}=A(35)-D(35)=12\times 35-150=420-150$$ $$\boxed{Q_{\max}=270 \text{ vehicles}}$$ For reference the queue is already $12\times10=120$ veh when the first lane reopens, so more than half of the growth occurs during the single-lane period even though the road is only half closed.
5. Read the maximum waiting time as the largest horizontal gap. Under first-in first-out discipline, the delay of a vehicle is the horizontal distance between the curves at its own cumulative number. That gap grows while the vehicle is served at 6 veh/min and shrinks once service reaches 18 veh/min, so the worst-delayed vehicle is the one that departs exactly at the capacity change, namely vehicle number 150. It arrived at $$t_{arr}=\frac{150}{12}=12.5 \text{ min}$$ and departed at $t=35$ min, so its delay is $$\boxed{w_{\max}=35-12.5=22.5 \text{ min}}$$
6. Part (c) — total delay is the area enclosed between the curves. Splitting the region at the two rate changes gives one triangle and one trapezoid of growth plus one triangle of recovery: $$\text{0 to 10 min:}\quad \tfrac12 \times 120 \times 10 = 600 \text{ veh}\cdot\text{min}$$ $$\text{10 to 35 min:}\quad \tfrac12\,(120+270)\times 25 = 4{,}875 \text{ veh}\cdot\text{min}$$ $$\text{35 to 80 min:}\quad \tfrac12 \times 270 \times 45 = 6{,}075 \text{ veh}\cdot\text{min}$$ Adding the three contributions, $$\boxed{D_{tot}=11{,}550 \text{ veh}\cdot\text{min}=192.5 \text{ veh}\cdot\text{h}}$$
7. Divide by every vehicle that was affected, not only those that arrived while over-saturated. Every vehicle arriving between the accident and the clearance instant is delayed, because the queue has not dissipated until $t_c$: $$N = \lambda\,t_c = 12 \times 80 = 960 \text{ vehicles}$$ $$\bar{d}=\frac{11{,}550}{960}$$ $$\boxed{\bar{d}=12.03 \text{ min per vehicle}}$$ Dividing instead by the 420 vehicles that arrived during the closure itself would overstate the average delay by more than a factor of two. As a Canadian planning-level interpretation, 192.5 vehicle-hours of delay at a conventional value of travel time of roughly 20 CAD per vehicle-hour represents on the order of 3,900 CAD of user cost from a single 35-minute lane blockage — the standard justification for rapid incident-response programmes on urban freeways.