16-Civ-A6 Highway Design, Construction, and Maintenance · May 2014

Question 2 of 7: Deterministic Queueing at a Lane Closure

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, 98-Civ-A6 Transportation Planning & Engineering, May 2014 — 3 hours, closed book (one two-sided aid sheet). Seven questions; any five constitute a complete examination and all are of equal value (20 marks). All seven are solved below as a study resource.

Reference texts (subject).

Garber, N.J. & Hoel, L.A., Traffic and Highway Engineering, 5th ed. — traffic flow, deterministic queueing, shock waves.
Papacostas, C.S. & Prevedouros, P.D., Transportation Engineering and Planning, 3rd ed. — the four‑step urban travel‑demand model.
Ortuzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. — trip generation, trip distribution, discrete mode choice, traffic assignment.
Sheffi, Y., Urban Transportation Networks — user‑equilibrium assignment and Wardrop’s principles.

Question 2: Deterministic Queueing at a Lane Closure (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 one‑direction, two‑lane freeway section carrying an arrival flow of 1400 veh/h; per‑lane capacity 1120 veh/h; one lane closed for the first 30 min then re‑opened.

Given data (deterministic D/D/1 bottleneck)
Quantity	Symbol	Value
Arrival flow	λ	1400 veh/h
Service rate, one lane open	μ_closed	1120 veh/h
Service rate, both lanes open	μ_open	2240 veh/h
Closure duration	t_c	0.5 h (30 min)

Find. The time the queue clears; the maximum queue length; and the total and average vehicle delay caused by the closure.

Cumulative arrival A(t) and departure D(t) curves. During the closure D rises at 1120 veh/h; after re‑opening it rises at 2240 veh/h until it overtakes A. The vertical gap is the queue; the horizontal gap is an individual vehicle's delay.

Approach. Model the bottleneck as a deterministic D/D/1 queue: build cumulative arrival and departure curves, read the queue as the vertical gap and total delay as the area between the curves.

Queue growth during the closure. With one lane open, arrivals exceed service, so the queue builds at $\lambda-\mu_{\text{closed}}=1400-1120=280\ \text{veh/h}$. Over the 30‑minute closure the queue reaches $Q_{\max}=280\times0.5=\boxed{140\ \text{veh}}$, its maximum, at $t=30$ min.
Queue discharge after re‑opening. With both lanes open the service rate is $2240\ \text{veh/h}$ while arrivals continue at $1400\ \text{veh/h}$, so the queue shrinks at $2240-1400=840\ \text{veh/h}$.
Time to clear. Clearing the 140‑vehicle queue takes $140/840=0.1667\ \text{h}=10\ \text{min}$ after re‑opening, i.e. at $t=30+10=\boxed{40\ \text{min}}$ measured from the start of the closure. Equivalently, setting cumulative arrivals equal to cumulative departures, $1400\,t=560+2240\,(t-0.5)$ gives $t=0.667\ \text{h}$.
Total delay (area between the curves). The queue‑versus‑time trace is a triangle that rises to 140 veh at 30 min and returns to zero at 40 min. Its area is $D_{\text{tot}}=\tfrac12\,(t_{\text{clear}})\,(Q_{\max})=\tfrac12\,(0.667\ \text{h})(140\ \text{veh})=\boxed{46.7\ \text{veh‑h}}$, or $46.7\times3600\approx1.68\times10^{5}\ \text{veh‑s}$.
Average delay per affected vehicle. The vehicles that experience delay are those arriving before the queue clears: $N=1400\times0.667=933\ \text{veh}$. Hence $\bar d = D_{\text{tot}}/N = 46.7/933 = 0.050\ \text{h}=\boxed{3.0\ \text{min}}$ per vehicle.

Question 2 — results
Quantity	Value
Queue clears at (from closure start)	40 min
Maximum queue length	140 veh
Total vehicle delay	46.7 veh‑h (≈ 168 000 veh‑s)
Vehicles delayed	933 veh
Average delay per vehicle	3.0 min

Topic. Deterministic (D/D/1) queueing analysis of a temporary bottleneck.

Key relations. Queue $Q(t)=[\lambda-\mu(t)]\,\Delta t$; total delay $=\int Q(t)\,dt=$ area between cumulative curves; average delay $=$ total delay $\div$ number delayed.

Why this works. When arrival and service rates are treated as deterministic, cumulative counts are straight line segments. The vertical distance between the arrival and departure curves is the instantaneous number in queue, and the horizontal distance is the time a given vehicle waits; the enclosed area is the sum of all waiting time. Because service is FIFO, area over count gives the mean wait directly.

Common pitfall. Placing the maximum queue at re‑opening but computing delay only over the discharge phase, or dividing total delay by all vehicles rather than only those that actually queued (before $t_{\text{clear}}$).

Source. Garber & Hoel, Traffic and Highway Engineering, ch. 5 (queueing analysis).

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