16-Civ-A6 Highway Design, Construction, and Maintenance · May 2016

Question 2 of 7: Deterministic Queueing at a Signalised Approach

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

Notes on this paper

Paper format: 98-Civ-A6 Transportation Planning & Engineering, National Examination May 2016. Seven questions, all of equal value (20 marks); any five constitute a complete examination. Closed book, one two-sided aid sheet permitted. Three hours. All seven questions are solved below as a study resource.

Reference texts. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis (Wiley) — queueing, shock waves and traffic-stream models; Papacostas & Prevedouros, Transportation Engineering and Planning (Prentice Hall) — the four-step demand model; Ortuzar & Willumsen, Modelling Transport (Wiley) — trip generation, distribution, mode choice and assignment; Roess, Prassas & McShane, Traffic Engineering (Pearson) — signalised-intersection delay.

Question 2: Deterministic Queueing at a Signalised Approach (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 single signalised approach observed over two consecutive 80 s cycles, each cycle consisting of 40 s of red followed by 40 s of green, with the queue discharging at saturation flow from the instant green begins.

Given data, converted to vehicles per second
Quantity	Symbol	Value	In veh/s
Cycle length	$C$	80 s	—
Effective red / effective green	$r\ /\ g$	40 s / 40 s	—
Arrival rate, cycle 1	$\lambda_1$	1 170 veh/h	0.325
Arrival rate, cycle 2	$\lambda_2$	630 veh/h	0.175
Saturation flow rate	$s$	1 800 veh/h	0.500

Find. The cumulative arrival and departure curves over the 160 s observation period, the maximum queue length in vehicles, and the total and average vehicle delay.

Deterministic queueing diagram for the signalised approach: cumulative arrivals A(t) and cumulative departures D(t) over two 80 s cycles. The vertical gap is the queue length; the enclosed area is total delay.

Approach. Treat the approach as a deterministic D/D/1 queue: build the cumulative arrival curve $A(t)$ and the cumulative departure curve $D(t)$, read the queue as the vertical gap between them, and obtain total delay as the area enclosed between the two curves.

The first thing to establish is whether the first cycle is undersaturated, because that decides the entire shape of the diagram. The green interval can discharge at most

$$ s\,g = 0.500 \times 40 = 20\ \text{veh per cycle} $$

whereas cycle 1 delivers $\lambda_1 C = 0.325 \times 80 = 26$ vehicles. The approach is therefore oversaturated in cycle 1 and a residual queue of six vehicles is carried across the cycle boundary. Cycle 2 delivers only $\lambda_2 C = 0.175 \times 80 = 14$ vehicles against the same capacity of 20, so the residual is worked off during cycle 2.

Build the cumulative arrival curve. Arrivals are constant within each cycle, so $A(t)$ is a two-segment polyline through the origin: $$ A(t)=\begin{cases}0.325\,t, & 0 \le t \le 80\\[2pt] 26 + 0.175\,(t-80), & 80 \le t \le 160\end{cases} $$ giving $A(80)=26$ veh and $A(160)=26+14=40$ veh over the two cycles.
Build the cumulative departure curve. No vehicle leaves on red, and on green the queue discharges at the saturation rate for as long as a queue exists. Over green 1 the queue never empties (shown in step 4), so departures run at $s$ for the whole 40 s: $$ D(40)=0,\qquad D(80)=0+0.500\times 40 = 20\ \text{veh} $$ Red 2 holds $D$ at 20 veh until $t=120\ \text{s}$, after which the queue again discharges at saturation: $$ D(160)=20+0.500\times 40 = 40\ \text{veh} $$ The two curves meet exactly at $t=160\ \text{s}$, so the approach clears precisely at the end of the second green.
Locate the queue maxima. The queue at any instant is the vertical separation $Q(t)=A(t)-D(t)$, which can only reach a local maximum at the end of a red interval. At the end of red 1, $$ Q(40) = 0.325\times 40 - 0 = 13\ \text{veh} $$ and at the end of red 2, using $A(120)=26+0.175\times 40 = 33$ veh against $D(120)=20$ veh, $$ Q(120) = 33 - 20 = 13\ \text{veh} $$
Confirm the queue behaviour through each green. During green 1 the queue drains at the net rate $s-\lambda_1 = 0.500-0.325 = 0.175\ \text{veh/s}$; over 40 s that removes only 7 vehicles, leaving $$ Q(80) = 13 - 7 = 6\ \text{veh} $$ as the residual carried into cycle 2, which confirms the oversaturation found above. During green 2 the net drain rate is $s-\lambda_2 = 0.500-0.175 = 0.325\ \text{veh/s}$, so the 13-vehicle queue needs $13/0.325 = 40\ \text{s}$ — exactly the available green — and the queue reaches zero at $t=160\ \text{s}$.
State the maximum queue. Both red intervals end with the same queue, so $$ \boxed{Q_{\max} = 13\ \text{vehicles}} $$ occurring at $t = 40\ \text{s}$ and again at $t = 120\ \text{s}$.
Integrate the area between the curves for total delay. The queue trace is piecewise linear, so the enclosed area is the sum of four trapezoids, one per signal interval: $$ \begin{aligned} \text{red 1}:&\ \tfrac{1}{2}(0+13)(40) = 260\\ \text{green 1}:&\ \tfrac{1}{2}(13+6)(40) = 380\\ \text{red 2}:&\ \tfrac{1}{2}(6+13)(40) = 380\\ \text{green 2}:&\ \tfrac{1}{2}(13+0)(40) = 260 \end{aligned} $$ Summing the four contributions gives $$ \boxed{D_{\text{total}} = 1\,280\ \text{veh}\cdot\text{s} = 21.3\ \text{veh}\cdot\text{min}} $$
Divide by the number of vehicles served. Over the two cycles $40$ vehicles arrive and all 40 are discharged, so $$ \bar d = \frac{D_{\text{total}}}{N} = \frac{1\,280}{40} \quad\Longrightarrow\quad \boxed{\bar d = 32.0\ \text{s per vehicle}} $$

The average delay of 32 s per vehicle is close to two-thirds of the 40 s red interval, which is the signature of an approach running at or slightly beyond capacity in the first cycle: the uniform-delay component alone for an undersaturated approach would be about 20 s, and the extra twelve seconds are the overflow delay contributed by the six-vehicle residual queue.

Final results — Question 2
Quantity	Value
Capacity per cycle vs. arrivals, cycle 1	20 veh vs. 26 veh → oversaturated
Residual queue at end of cycle 1	6 veh
Maximum queue length	13 vehicles (at $t=40$ s and $t=120$ s)
Time the approach clears	$t=160$ s (end of the second green)
Total vehicle delay	1 280 veh·s = 21.3 veh·min
Vehicles served in two cycles	40
Average delay per vehicle	32.0 s

Topic. Deterministic (D/D/1) queueing analysis of a signalised approach using cumulative arrival and departure curves.

Key relations. Queue length $Q(t)=A(t)-D(t)$; capacity per cycle $= s\,g$; undersaturation requires $\lambda C \le s\,g$; total delay $=\int [A(t)-D(t)]\,dt$ = area between the curves; average delay $\bar d = D_{\text{total}}/N$; the horizontal gap between the curves at a given cumulative count is the delay experienced by that individual vehicle.

Why this works. With deterministic arrivals and departures both curves are piecewise linear, so every quantity of interest is a triangle or trapezoid and no probabilistic queueing theory is needed. The vertical gap is a count of vehicles present and the horizontal gap is a waiting time, so the enclosed area — vehicles multiplied by seconds — is exactly total delay. Because $D(t)$ can never exceed $A(t)$, the departure curve is the lower envelope: on red it is flat, on green it rises at the saturation rate until it meets $A(t)$, after which it simply tracks $A(t)$.

Common pitfall. Assuming the queue clears within every green and so restarting each cycle from zero. Here cycle 1 arrives 26 vehicles against a 20-vehicle capacity, and missing the six-vehicle residual understates total delay by roughly forty per cent. A second frequent error is drawing the departure curve rising during red, or letting it rise at the saturation rate after the queue has emptied — once $D=A$, departures are limited by arrivals, not by the signal.

Source. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis, Ch. 5 (queueing theory and traffic analysis) and Ch. 7 (signalised intersections); Roess, Prassas & McShane, Traffic Engineering, Ch. 20.

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Quantity	Symbol	Value	In veh/s
Cycle length	\(C\)	80 s	—
Effective red / effective green	\(r\ /\ g\)	40 s / 40 s	—
Arrival rate, cycle 1	\(\lambda_1\)	1 170 veh/h	0.325
Arrival rate, cycle 2	\(\lambda_2\)	630 veh/h	0.175
Saturation flow rate	\(s\)	1 800 veh/h	0.500