Question 3 of 7: Deterministic Queueing at a Bridge Lane Closure

Question 3: Deterministic Queueing at a Bridge Lane Closure (20 marks — 4 each)

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 lane closure that reduces one direction of a four-lane bridge to a single 1350 vph lane, with the demand profile tabulated above. Parts (a) to (e) all concern the morning peak.

Find. The maximum queue, the longest delay any vehicle suffers, the clock time at which the queue dissipates, the total and average delay, and whether a specific 08:30 departure reaches work by 09:00.

Approach. Build the cumulative arrival and departure curves with time measured from 07:00, read the maximum vertical gap for the queue, the maximum horizontal gap for the worst individual delay, the intersection of the curves for the dissipation time, and the enclosed area for total delay.

1. Set up the cumulative curves. With $t$ in hours from 07:00, arrivals accumulate as $$A(t) = \begin{cases} 1500\,t, & 0 \le t \le 2 \\ 3000 + 800(t-2), & t \gt 2 \end{cases}$$ and departures accumulate at the lane capacity for as long as a queue exists, $D(t) = 1350\,t$. The queue at any instant is the vertical gap $Q(t) = A(t) - D(t)$.
2. Part (a) — maximum queue. During the peak the queue grows at the difference between demand and capacity, $1500 - 1350 = 150$ veh/h, and it grows for the whole two hours because demand exceeds capacity throughout. After 09:00 demand falls below capacity, so the queue is longest exactly at 09:00: $$Q_{max} = (\lambda_1 - \mu)\,T = (1500 - 1350)(2) = \boxed{300 \text{ veh}}$$ At an average of 7 m per queued vehicle that is a little over 2 km of standing traffic, which is the number that drives the traffic-management plan for the closure.
3. Part (b) — longest time any vehicle spends in the queue. The delay of a vehicle arriving at time $t$ is the horizontal gap, $w(t) = A(t)/\mu - t$. On the rising limb this is $w(t) = t(\lambda_1/\mu - 1) = t/9$, which increases with $t$; after 09:00 it becomes $w(t) = 1.0370 - 0.4074\,t$, which decreases. The worst-off vehicle is therefore the one arriving exactly at 09:00: $$w_{max} = T\left(\frac{\lambda_1}{\mu} - 1\right) = 2\left(\frac{1500}{1350} - 1\right) = 0.2222 \text{ h} = \boxed{13.3 \text{ min}}$$ that is 13 min 20 s. Scanning $w(t)$ over every arrival instant from 07:00 to dissipation confirms no vehicle waits longer, which is worth doing rather than reasoning about which vehicle ought to be worst.
4. Part (c) — when the queue dissipates. After 09:00 the queue shrinks at $\mu - \lambda_2 = 1350 - 800 = 550$ veh/h. Setting $A(t) = D(t)$, $$t_{clear} = \frac{(\lambda_1 - \lambda_2)T}{\mu - \lambda_2} = \frac{(1500-800)(2)}{550} = 2.5455 \text{ h}$$ which is 2 h 32.7 min after 07:00, so the queue clears at $\boxed{09{:}32{:}44}$, call it 09:33. Note that the recovery takes only 33 minutes although the queue took two hours to build: the recovery rate is governed by the surplus capacity, 550 veh/h, not by the size of the disruption.
5. Part (d) — total delay. The queue is triangular in time — zero at 07:00, 300 veh at 09:00, zero at 09:32.7 — so the area between the curves is $$D_{tot} = \tfrac{1}{2}\,Q_{max}\,t_{clear} = \tfrac{1}{2}(300)(2.5455) = \boxed{381.8 \text{ veh}\cdot\text{h}}$$ which is 22\,909 veh·min. The number of vehicles delayed is every one discharged up to dissipation, $\mu\,t_{clear} = 1350(2.5455) = 3436$ veh, so the average delay is $381.8/3436 = 0.1111$ h $= 6.67$ min per vehicle — exactly half the worst-case 13.3 min, as it must be for a triangular queue.
6. Part (e) — the 08:30 departure. Leaving home at 08:30 and driving five minutes to the bridge entrance means joining the queue at 08:35, that is $t = 1.5833$ h after 07:00. The queueing delay for that arrival is $$w(1.5833) = \frac{1.5833}{9} = 0.1759 \text{ h} = 10.56 \text{ min}$$ The journey is then 5 min to the bridge, 10.56 min of queueing and 10 min beyond the exit, a total of 25.56 min, giving an arrival at $\boxed{08{:}55{:}33}$. Work starts at 09:00, so yes — the trip succeeds, with about 4.4 minutes to spare. The margin is thin enough to be worth saying out loud: a departure just 4.0 minutes later, at 08:34:00, arrives exactly on the hour, and anyone leaving after that is late.