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

Question 2 of 7: Deterministic Queueing at a Parking Lot

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

Notes on this paper

National Examination — 98-Civ-A6 Transportation Planning & Engineering, December 2014. Closed book (one two-sided aid sheet), 3 hours. Seven questions of equal value (20 marks); any five constitute a complete paper. All seven are solved here as a study resource.

Reference texts (subject):

Papacostas, C.S. & Prevedouros, P.D., Transportation Engineering and Planning, 3rd ed. (Prentice Hall).
Roess, R.P., Prassas, E.S. & McShane, W.R., Traffic Engineering, 4th ed. (Pearson).
Mannering, F.L. & Washburn, S.S., Principles of Highway Engineering and Traffic Analysis (Wiley).
Ortúzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. (Wiley).
Sheffi, Y., Urban Transportation Networks (Prentice Hall).
Transportation Association of Canada (TAC), Geometric Design Guide for Canadian Roads (Canadian context).

Question 2: Deterministic Queueing at a Parking Lot (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 time $t$ measured in minutes after 7:00 am. Arrival rate $\lambda(t)=8$ veh/min for $0\le t\le 20$, $\lambda=0$ for $20\lt t\le 30$, and $\lambda=2$ veh/min for $t\gt 30$; service (departure) rate $\mu=4$ veh/min whenever a queue exists.

Given data
Interval (min after 7:00)	Arrival rate $\lambda$	Cumulative arrivals $A(t)$
0 – 20	8 veh/min	$8t$
20 – 30	0 (access blocked)	$160$
> 30	2 veh/min	$160+2(t-30)$
service, all $t$	$\mu=4$ veh/min	$D(t)=4t$

Find. The time the queue clears; the maximum queue length and maximum wait; and the total and average delay.

Cumulative arrival $A(t)$ and departure $D(t)$ curves. The vertical gap is the queue length; the horizontal gap is the waiting time. Both peak at the vehicle arriving at $t=20$ min (7:20 am).

Approach. Build $A(t)$ and $D(t)$; the queue clears where they meet; the maximum vertical gap gives the maximum queue, the maximum horizontal gap the maximum wait, and the enclosed area the total delay.

Establish the cumulative curves. Arrivals accumulate as $A(t)=8t$ up to $t=20$ (reaching 160), stay flat at 160 through the 10-minute blockage, then rise at 2 veh/min. Departures accumulate at the attendant's rate $D(t)=4t$ so long as anyone is waiting.
Time the queue clears. The queue vanishes when $A(t)=D(t)$ on the last segment: $$160+2(t-30)=4t \;\Rightarrow\; 100-2t=0 \;\Rightarrow\; t=50\ \text{min}.$$ The queue clears $\boxed{50\ \text{min after 7:00 am, i.e. 7:50 am}}$ (the server stays busy the whole interval, so $D=4t$ is valid throughout).
Maximum queue length. The queue $Q(t)=A(t)-D(t)$ builds at $8-4=4$ veh/min until arrivals stop, so it peaks at $t=20$: $$Q_{\max}=A(20)-D(20)=160-80=\boxed{80\ \text{vehicles}}.$$ After $t=20$ the queue shrinks (at 4 veh/min during the blockage, then 2 veh/min).
Maximum waiting time. Under FIFO the wait is the horizontal gap. The vehicle at the queue peak is arrival number 160 (arriving at $t=20$); it is served when $D=160$, i.e. $t=40$: $$w_{\max}=40-20=\boxed{20\ \text{minutes}}.$$
Total delay. Integrate the queue (area between the curves) over the three segments: $$\int_0^{20}\!4t\,dt=800,\quad \int_{20}^{30}\!(160-4t)\,dt=600,\quad \int_{30}^{50}\!(100-2t)\,dt=400.$$ Summing, total delay $=800+600+400=\boxed{1800\ \text{veh-min}}$ (equivalently 30 veh-h).
Average delay per vehicle. All $D(50)=4(50)=200$ vehicles that queued are cleared, so $$\bar w=\frac{1800}{200}=\boxed{9\ \text{min per vehicle}}.$$

Question 2 — results
Quantity	Value
Queue clears	$t=50$ min → 7:50 am
Maximum queue length	80 vehicles (at 7:20 am)
Maximum waiting time	20 minutes
Total vehicle delay	1800 veh-min (30 veh-h)
Average delay per vehicle	9 min/veh

Topic: Deterministic (D/D/1) queueing and cumulative-arrival diagrams.

Key relations: $Q(t)=A(t)-D(t)$; queue clears when $A=D$; total delay $=\int[A(t)-D(t)]\,dt$ = area between curves; average delay = total delay / number served.

Why this works: When arrival and service rates are known deterministic functions of time, the cumulative curves fully describe the queue: their vertical separation is the instantaneous number waiting and their horizontal separation is the time each vehicle spends waiting (FIFO). Because the system is a bottleneck (service 4 < peak arrival 8 veh/min), a queue forms the instant arrivals begin; the access-road accident simply reduces later arrivals and so helps the queue drain.

Common pitfall: Assuming the maximum queue and maximum wait occur at different vehicles, or measuring delay from 7:30 instead of integrating the whole area; and forgetting that the server keeps discharging at 4 veh/min during the blockage.

Source: Mannering & Washburn, Principles of Highway Engineering and Traffic Analysis, Ch. 5 (queueing); Roess, Prassas & McShane, Traffic Engineering, Ch. 5.

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