Total Delay at a Signal Approach: 16-Civ-B10 Traffic Engineering, May 2017, Q5

Question 5: Total Delay at a Signal 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.

Question 5 — given data
Quantity	Symbol	Value
Approach saturation flow rate	$s$	2770 vph = 0.7694 veh/s
Arrival rate	$q$	500 vph = 0.1389 veh/s
Queue already present at the start of effective red	$Q_0$	4 veh
Queue size that calls the green	$M$	10 veh
Delay between that call and the start of effective green	$N$	10 s
Effective green	$g$	15 s

Given. One approach of a signal, with a standing queue of four vehicles at the start of the effective red and the data tabulated above. The effective red is not stated directly; it is defined implicitly by the queue-size trigger.

Find. The total delay, in vehicle-seconds, incurred by the vehicles on this approach.

Cumulative arrival and departure diagram. The first effective green is 3.06 s too short to discharge the standing queue, so 1.93 veh carry over and the delay polygon closes only during the second green.

Approach. Recover the effective red from the queue trigger, build the D/D/1 cumulative diagram, test whether the green is long enough to discharge the queue, and integrate the queue over time until it returns to zero.

Recover the effective red length. The queue starts at four vehicles and grows at the arrival rate, so it reaches ten vehicles after $$t_{10} = \frac{M - Q_0}{q} = \frac{10 - 4}{500/3600} = 43.2 \text{ s}$$ The effective green begins ten seconds later, so the effective red is $$r = t_{10} + N = 43.2 + 10 = \boxed{53.2 \text{ s}}$$ and the cycle length is $C = r + g = 53.2 + 15 = 68.2$ s. This is the step the question hides: “ten seconds after the queue reaches ten vehicles” defines the RED, not an offset within the green.
Queue at the end of the red, and the maximum queue. Vehicles accumulate throughout the red on top of the four already waiting: $$Q_r = Q_0 + q\,r = 4 + 0.1389(53.2) = 11.39 \text{ veh}$$ which is also the maximum queue, since discharge begins the instant the green does.
Test whether the green can clear it. During the green the queue shrinks at the difference between the saturation flow and the continuing arrivals, $s - q = 0.7694 - 0.1389 = 0.6306$ veh/s, so clearing the standing queue needs $$t_c = \frac{Q_r}{s - q} = \frac{11.39}{0.6306} = 18.06 \text{ s}$$ against only 15 s of green. The queue does not clear: this cycle is oversaturated by 3.06 s of green, and $$Q_{end} = Q_r + q\,g - s\,g = 11.39 + 2.08 - 11.54 = 1.93 \text{ veh}$$ carry over into the next cycle. Note that the signal itself is not over capacity — $c = s\,g/C = 2770(15/68.2) = 609$ vph against 500 vph of demand — so the overflow is caused by the four-vehicle standing queue, and it will be worked off.
Delay accumulated inside the stated cycle. The delay is the area between the cumulative curves. On the red limb the queue rises linearly from 4 to 11.39 veh, and on the green limb it falls linearly from 11.39 to 1.93 veh, so both limbs are trapezoids: $$d_{red} = \tfrac{1}{2}(4 + 11.39)(53.2) = 409.3 \text{ veh}\cdot\text{s} \qquad d_{green} = \tfrac{1}{2}(11.39 + 1.93)(15) = 99.9 \text{ veh}\cdot\text{s}$$ giving $509.2$ veh·s, or 8.49 veh·min, within the cycle the question describes.
Carry the diagram forward to closure. Because the polygon has not closed, the total delay is not yet complete. Taking the signal as fixed-time, the next cycle repeats $r = 53.2$ s and $g = 15$ s. The residual grows through the second red to $Q = 1.93 + 0.1389(53.2) = 9.32$ veh, and the second green clears it in $9.32/0.6306 = 14.78$ s — just inside the 15 s available. The two further areas are $$d_{red,2} = \tfrac{1}{2}(1.93 + 9.32)(53.2) = 299.3 \qquad d_{green,2} = \tfrac{1}{2}(9.32)(14.78) = 68.9$$ both in vehicle-seconds.
Total delay. Summing the four areas, the delay from the start of the effective red until the queue is fully discharged at $t = 136.2$ s is $$D_{tot} = 409.3 + 99.9 + 299.3 + 68.9 = \boxed{877 \text{ veh}\cdot\text{s}}$$ that is 14.6 veh·min. Over that period $Q_0 + q\,t = 4 + 0.1389(136.2) = 22.9$ vehicles are served — which equals $s(g + 14.78)$, confirming the diagram closes — so the average delay is $877/22.9 = 38.3$ s per vehicle. Scanning every position in the queue, the worst-served vehicle is the twelfth, held over to the second green, which waits 64.4 s.

Check: the boxed 877 veh·s is the total delay to full queue dissipation, and it assumes the signal repeats the same 53.2 s red and 15 s green in the following cycle, which is what “an effective red of one cycle” implies for a fixed-time controller. If the question intends only the cycle it describes, the answer is the 509 veh·s (8.49 veh·min) accumulated between the start of that effective red and the end of its green; both figures are reported above. One further reading is worth flagging: if the controller is queue-actuated, so that each red again ends ten seconds after the queue reaches ten vehicles, the cycle stretches to 83.1 s, capacity falls to exactly the 500 vph demand, and the 1.93 veh residual never clears — the total delay is then unbounded, which is itself the argument for reading the signal as fixed-time.

Question 5 — final results
Quantity	Result
Effective red recovered from the queue trigger	53.2 s
Cycle length	68.2 s
Maximum queue (end of red)	11.39 veh
Green needed to clear it / green available	18.06 s / 15 s
Residual queue carried into the next cycle	1.93 veh
Delay within the stated cycle	509 veh·s (8.49 veh·min)
Total delay to full dissipation	877 veh·s (14.6 veh·min)
Vehicles served / average delay	22.9 veh / 38.3 s per vehicle
Longest individual delay	64.4 s (the 12th vehicle)

Concept & Theory

Topic. Uniform (D/D/1) delay at a signal approach with an initial queue.

Key relations. $r = (M - Q_0)/q + N$; $Q_r = Q_0 + qr$; $t_c = Q_r/(s-q)$, which must be compared with $g$; capacity $c = s\,g/C$; delay $= \int Q\,dt$, a trapezoid on the red limb plus a trapezoid or triangle on the green limb.

Why this works. The initial queue is what makes this question more than a textbook triangle. With Q0 = 0 the delay polygon is a simple triangle and one cycle always closes when capacity exceeds demand; with Q0 = 4 the red limb becomes a trapezoid and the clearance test can fail even though the signal has spare capacity over the full cycle. Recognising that distinction — a cycle can be oversaturated while the signal is not — is the examinable point, and it is why the clearance test t_c against g must be done rather than assumed.

Common pitfall. Reading the ten seconds as an offset inside the green, or forgetting that arrivals continue during the green so that the discharge rate against the queue is s minus q and not s. A third slip is quoting the delay per cycle as if the polygon had closed: if t_c exceeds g, the area is not the answer and the diagram must be carried into the next cycle.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §8.5 (signal delay, the D/D/1 approximation) and §5.2.

Question 5 of 7: Total Delay at a Signal Approach

Question 5: Total Delay at a Signal Approach (20 marks)