Question 2 of 7: Deterministic Queueing at a Single-Lane Bottleneck

Question 2: Deterministic Queueing at a Single-Lane Bottleneck (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.

Find. The cumulative arrival and departure diagram to the instant the queue clears; the maximum queue length and the maximum time any vehicle waits; and the total and average delay accumulated over the whole congested period.

Approach. Treat the bottleneck as a deterministic D/D/1 queue: build the cumulative arrival curve A(t) and the cumulative departure curve D(t), then read the queue as the vertical gap, the individual delay as the horizontal gap, and the total delay as the enclosed area.

1. Part (a) — establish that a queue forms and when it ends. Measure time t in hours after 9:00 a.m. During the first period the arrival rate exceeds the capacity, so the bottleneck is oversaturated and a queue grows at λ₁ − μ = 1500 − 1250 = 250 veh/h. After 11:00 a.m. the arrival rate falls below capacity, so the queue discharges at μ − λ₂ = 1250 − 750 = 500 veh/h. A queue therefore builds for exactly two hours and then drains.
2. Write the two cumulative curves. The arrival curve is piecewise linear and the departure curve runs at capacity for as long as a queue exists: $$A(t)=\begin{cases}1500\,t, & 0 \le t \le 2\\ 3000+750\,(t-2), & t > 2\end{cases}$$ while the departure curve runs at capacity for as long as a queue is present, $$D(t)=1250\,t$$ Both are measured in vehicles, with t in hours after 9:00 a.m.
3. Locate the instant the queue clears. The queue vanishes where the two cumulative curves meet, A(t) = D(t): $$3000+750\,(t-2)=1250\,t \;\Longrightarrow\; 1500=500\,t \;\Longrightarrow\; \boxed{t_c = 3.0\ \text{h, i.e. 12:00 noon}}$$ The root lies in the interval t > 2 over which the second branch of A(t) is valid, so it is admissible — a check worth making, because equating the curves on the wrong branch is the standard error on this question.
4. Part (b) — maximum queue length. The vertical gap Q(t) = A(t) − D(t) grows at 250 veh/h until 11:00 and shrinks thereafter, so it peaks exactly at the break point t = 2 h: $$Q_{max}=A(2)-D(2)=3000-2500=\boxed{500\ \text{vehicles}}$$ Note that the peak is at the rate crossover, not at the end of the analysis period.
5. Maximum waiting time. Under first-in-first-out the delay of an individual vehicle is the horizontal gap between the curves at that vehicle's cumulative position. The worst-off vehicle is the last one to arrive during the build-up, number n = A(2) = 3,000, which arrives at t = 2.0 h and is served when the departure curve reaches 3,000 vehicles: $$t_{depart}=\frac{3000}{1250}=2.40\ \text{h} \;\Longrightarrow\; w_{max}=2.40-2.00=0.40\ \text{h}=\boxed{24\ \text{minutes}}$$ Every earlier vehicle waits less because it joined a shorter queue, and every later one waits less because the queue is by then draining, so 24 min is the maximum.
6. Part (c)(1) — total vehicle delay. The total delay is the area between the two cumulative curves. That area is a triangle on each side of the peak, with a common height equal to the maximum queue: $$D_{total}=\tfrac12\,(2.0)(500)+\tfrac12\,(1.0)(500)=500+250=\boxed{750\ \text{veh}\cdot\text{h}}$$ Equivalently, one may integrate A(t) − D(t) from 0 to 3 h; the numerical integral returns the same 750 veh·h, which is a useful independent check.
7. Part (c)(2) — average delay per vehicle. The delay is shared among every vehicle that arrives during the congested period. That count is 1500(2.0) + 750(1.0) = 3,750 vehicles, which agrees with the departures μt_c = 1250(3.0) = 3,750 — the curves must close at the clearance instant, so this is a free arithmetic check. $$\bar d=\frac{D_{total}}{N}=\frac{750}{3750}=0.20\ \text{h}=\boxed{12\ \text{minutes per vehicle}}$$

For interpretation: a mean delay of 12 min (720 s) is more than eight times the 80 s/veh control-delay threshold that defines level of service F at a signalised intersection, so this bottleneck is severely oversaturated for three hours and the 24-minute worst-case wait is what drivers will actually report. Note also the two spare capacities are unequal — the lane is 250 veh/h short during the peak but has 500 veh/h of surplus afterwards — which is why the recovery takes only half as long as the build-up.