Deterministic Queueing at a Signalised Approach: 16-Civ-B7 Transportation Planning and Engineering, May 2018, Q2

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 operated over two consecutive cycles, with a constant arrival rate within each cycle and discharge at saturation flow from the start of green.

Signal and demand data
Quantity	Symbol	Value
Cycle length	C	60 s (30 s red + 30 s green; yellow ignored)
Arrival rate, cycle 1	λ₁	1,080 veh/h = 0.30 veh/s
Arrival rate, cycle 2	λ₂	720 veh/h = 0.20 veh/s
Saturation flow rate	s	1,800 veh/h = 0.50 veh/s
Analysis period	—	two cycles, 0 to 120 s

Find. The cumulative arrival and departure diagram over the two cycles, the maximum queue length, and the total and average vehicle delay.

Cumulative arrival curve A(t) and departure curve D(t) for the two cycles. The vertical gap between the curves is the queue length at that instant; the shaded area between them is the total delay. The queue reaches 9 vehicles at the end of each red interval and clears exactly at the end of the second green.

(a) The queueing diagram. The figure above is the sketch part (a) asks for. It carries everything the rest of the question needs: cumulative arrivals rise at \(\lambda_1\) then at \(\lambda_2\); cumulative departures are flat through each red interval and rise at the saturation flow \(s\) through each green; the vertical gap between the two curves is the queue at that instant, the horizontal gap is the delay to the vehicle at that point in the cumulative count, and the enclosed area is the total delay. The two red intervals are shaded, and the queue is annotated at the end of each.

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 \(A(t) - D(t)\) and the total delay as the area enclosed between them.

Convert every rate to vehicles per second so the diagram can be built on one time axis. Dividing each hourly rate by 3600 s/h gives \[\lambda_1 = \frac{1080}{3600} = 0.30\ \text{veh/s},\qquad \lambda_2 = \frac{720}{3600} = 0.20\ \text{veh/s},\qquad s = \frac{1800}{3600} = 0.50\ \text{veh/s}.\] The cycle is taken as red first (0 to 30 s) then green (30 to 60 s), which is the convention the question implies when it speaks of the queue “formed on red” passing through the “subsequent green”.
Build the queue through the first red interval. Nothing departs while the signal is red, so the queue grows at the arrival rate: \[Q(30) = \lambda_1 R = 0.30 \times 30 = 9\ \text{veh}.\] Cumulative arrivals at that instant are also 9 vehicles, and cumulative departures are zero.
Discharge the queue through the first green interval. Vehicles leave at the saturation flow rate while arrivals continue, so the queue shrinks at the net rate \(s - \lambda_1 = 0.50 - 0.30 = 0.20\) veh/s. Clearing 9 vehicles would need \[t_{\text{clear}} = \frac{9}{0.20} = 45\ \text{s},\] which is longer than the 30 s of green available. The first cycle therefore does not clear, and the residual queue carried into cycle 2 is \[Q(60) = 9 - 0.20 \times 30 = 3\ \text{veh}.\] Cumulative departures at the end of green 1 are \(0.50 \times 30 = 15\) veh against 18 cumulative arrivals, which is the same 3-vehicle gap read off the diagram.
Carry the residual through the second red interval. Arrivals have fallen to 0.20 veh/s, and the residual queue simply grows on top of itself: \[Q(90) = 3 + 0.20 \times 30 = 9\ \text{veh}.\] The second red therefore ends at exactly the same queue as the first, which is the feature the diagram is built to show.
Discharge through the second green interval and confirm the queue clears. The net discharge rate is now \(s - \lambda_2 = 0.50 - 0.20 = 0.30\) veh/s, so \[t_{\text{clear}} = \frac{9}{0.30} = 30\ \text{s},\] exactly the green available. The queue reaches zero at \(t = 120\) s, where cumulative arrivals and cumulative departures both equal 30 vehicles. This coincidence is the arithmetic check that the diagram is right.
Read the maximum queue off the diagram (part b). The queue is largest at the end of a red interval, and the two red intervals both end at 9 vehicles: \[\boxed{Q_{\max} = 9\ \text{vehicles}}\] occurring at \(t = 30\) s and again at \(t = 90\) s. Note that this answer is independent of whether the first cycle is taken as red-first or green-first, so it is safe under either reading of the question.
Integrate the area between the curves for the total delay (part c1). The queue profile is piecewise linear, so the area is four trapezoids, one per interval: \[D = \tfrac{1}{2}(0+9)(30) + \tfrac{1}{2}(9+3)(30) + \tfrac{1}{2}(3+9)(30) + \tfrac{1}{2}(9+0)(30)\] \[D = 135 + 180 + 180 + 135 = \boxed{630\ \text{veh}\cdot\text{s} = 10.5\ \text{veh}\cdot\text{min}}\]
Divide by the vehicles served for the average delay (part c2). Over the two cycles \(0.30 \times 60 + 0.20 \times 60 = 30\) vehicles arrive, and all 30 have departed by \(t = 120\) s, so \[d = \frac{D}{N} = \frac{630}{30} = \boxed{21\ \text{s per vehicle}}\]

The average delay is worth interpreting rather than just reporting. The green split gives the approach a capacity of \(s \times g/C = 1800 \times 30/60 = 900\) veh/h, so the first cycle is over-saturated at a degree of saturation of \(1080/900 = 1.20\) while the second runs at 0.80. That is exactly why the first cycle leaves a residual: the approach is being asked to pass more than it can. A control delay of 21 s per vehicle corresponds to level of service C for a signalised intersection under the HCM, which is acceptable here only because the over-saturated period lasts a single cycle.

Final results — Question 2
Quantity	Result
(a) Queueing diagram	see the cumulative A(t) / D(t) figure above
Queue at the end of red 1 (t = 30 s)	9 veh
Residual queue at the end of green 1 (t = 60 s)	3 veh
Queue at the end of red 2 (t = 90 s)	9 veh
(b) Maximum queue length	9 vehicles
(c1) Total vehicle delay	630 veh·s (10.5 veh·min)
(c2) Average delay per vehicle	21 s/veh
Vehicles served in the two cycles	30 veh (queue clears exactly at t = 120 s)
Approach capacity and degree of saturation	900 veh/h; 1.20 (cycle 1), 0.80 (cycle 2)

Check: the question states that “all vehicles in the queue formed on red pass through the intersection during the subsequent green interval”. At 1,080 veh/h this is arithmetically impossible — 45 s of green would be needed and only 30 s exist — so that sentence is read as a description of the discharge behaviour (departures begin immediately at saturation flow, with no start-up lost time) rather than as a guarantee that the queue clears. The residual of 3 vehicles is carried forward, which is the standard D/D/1 treatment. The red-first convention is also assumed; taking green first leaves the maximum queue unchanged at 9 vehicles but leaves 6 vehicles unserved at t = 120 s, which the “during the two cycles” wording of part (c) does not support.

Topic. Deterministic (D/D/1) queueing analysis of a signalised approach.

Key relations. Queue length \(Q(t) = A(t) - D(t)\); total delay \(D = \int [A(t) - D(t)]\,dt\) = area between the curves; approach capacity \(c = s\,g/C\); degree of saturation \(X = v/c\); average delay \(d = D/N\).

Why this works. With deterministic arrivals and departures the cumulative curves are piecewise linear, so every quantity the examiner can ask for is a length, a vertical gap or a polygon area on one diagram. The vertical distance between the curves at any instant is the number of vehicles that have arrived but not yet departed — the queue — and the horizontal distance is the waiting time of the vehicle at that point in the cumulative count. Integrating the vertical gap therefore gives total vehicle-seconds of delay directly, with no need for a queueing formula. The method is exact for the idealised signal and is the analytical core of the HCM uniform-delay term.

Common pitfall. Solving \(A(t) = D(t)\) algebraically and accepting a root that lies outside the green interval it was derived for — here the naive root falls beyond the first green and must be rejected. Always test the root against the interval whose equations produced it. A second pitfall is dividing the total delay by the vehicles that departed in a cycle rather than by all vehicles that arrived in the analysis period.

Source. Garber and Hoel, Traffic and Highway Engineering, 5th ed., §5.4 and §8.4 (queueing and signalised-intersection delay); TRB, Highway Capacity Manual, 6th ed., Ch. 19 (signalised intersections, uniform delay and LOS thresholds).

Question 2 of 7: Deterministic Queueing at a Signalised Approach

Question 2: Deterministic Queueing at a Signalised Approach (20 marks)