Question 3 of 7: Deterministic Queueing at a Lane Closure

Question 3: Deterministic Queueing at a 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.

Find. The maximum queue, the longest individual delay, the clock time at which the queue clears, the total delay accumulated from 16:00, and the average delay per affected vehicle.

Approach. Demand exceeds capacity by a known constant rate over a known interval, so this is a deterministic (D/D/1) queue. Draw the cumulative arrival and departure curves, and read every answer off their geometry rather than from a queueing formula.

1. Part (a) — Maximum queue. While demand exceeds capacity the queue grows at the difference of the two rates, $$\frac{dQ}{dt} = \lambda_{1} - \mu = 2800 - 2580 = 220\ \text{veh/h}$$ The over-saturated period lasts from 16:00 to 18:00, so the queue is longest at 18:00, the instant demand falls: $$Q_{\max} = (\lambda_{1} - \mu)\,T = 220 \times 2 = \boxed{440\ \text{vehicles}}$$ At an average spacing of about 7.5 m per vehicle in a stopped queue that is roughly 3.3 km of standing traffic in the single open lane, which is worth stating because it determines how far upstream advance warning signs and any diversion must be placed.
2. Part (b) — Longest delay experienced by any vehicle. Under first-in-first-out operation, the worst-off driver is the one who joins exactly when the queue is longest, at 18:00; that driver must wait for the 440 vehicles ahead to be discharged at the bottleneck's capacity rate. On the cumulative diagram this is the widest horizontal gap between the two curves: $$w_{\max} = \frac{Q_{\max}}{\mu} = \frac{440}{2580} = 0.1705\ \text{h} = \boxed{10.2\ \text{min}}$$ Note that the divisor is the capacity, not the arrival rate: once a queue exists the bottleneck discharges at $\mu$ regardless of what is arriving behind.
3. Part (c) — When the queue dissipates. After 18:00 arrivals fall below capacity, so the queue is served off at $$\mu - \lambda_{2} = 2580 - 1350 = 1230\ \text{veh/h}$$ The time needed to clear the 440-vehicle queue is $$t_{d} = \frac{Q_{\max}}{\mu - \lambda_{2}} = \frac{440}{1230} = 0.3577\ \text{h} = 21.5\ \text{min}$$ so the queue clears at $18{:}00 + 21.5\ \text{min} \approx \boxed{18{:}21.5 \ (\text{about } 6{:}21\ \text{PM})}$. Confirming this on the cumulative curves: total arrivals by that instant are $5600 + 1350(0.3577) = 6083$ vehicles and total departures are $2580(2.3577) = 6083$ vehicles — the two curves meet, which is the definition of the queue having dissipated.
4. Part (d) — Total delay. Total delay is the area enclosed between the arrival and departure curves, and that area is two triangles sharing the ordinate $Q_{\max}$. Over the growth phase, $$D_{1} = \tfrac{1}{2}\,T\,Q_{\max} = \tfrac{1}{2}(2)(440) = 440\ \text{veh}\cdot\text{h}$$ and over the decay phase, $$D_{2} = \tfrac{1}{2}\,t_{d}\,Q_{\max} = \tfrac{1}{2}(0.3577)(440) = 78.7\ \text{veh}\cdot\text{h}$$ Adding them, $$D = D_{1} + D_{2} = \boxed{518.7\ \text{veh}\cdot\text{h}}$$ Priced at a typical Ontario value of travel time this single Friday-evening closure costs the travelling public in the order of ten thousand dollars of delay, which is the usual justification for night-time or off-peak staging.
5. Part (e) — Average delay per vehicle. Every vehicle arriving between 16:00 and the instant the queue clears is delayed, so the affected population is $$N = \lambda_{1}T + \lambda_{2}t_{d} = 2800(2) + 1350(0.3577) = 5600 + 483 = 6083\ \text{vehicles}$$ which agrees with the departure count computed in part (c). Dividing the total delay, $$\overline{d} = \frac{D}{N} = \frac{518.7}{6083} = 0.08527\ \text{h} = \boxed{5.12\ \text{min}}$$ The contrast with part (b) is the useful observation: the worst driver loses 10.2 min but the average driver loses only 5.1 min, because the queue spends most of its life well short of its maximum.