Total Delay on a Signal Approach: 16-Civ-B10 Traffic Engineering, December 2015, Q5

Question 5: Total Delay on 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.

Given. One approach of a signalised intersection, analysed over a single cycle.

Quantity	Symbol	Value
Saturation flow rate	\(s\)	3250 veh/h = 0.9028 veh/s
Arrival flow rate	\(q\)	800 veh/h = 0.2222 veh/s
Queue present when effective red begins	\(Q_0\)	5 vehicles
Effective green	\(g\)	17 s
Green start	—	7 s after the queue reaches 10 vehicles

Find. The total delay accumulated on this approach during the cycle, in vehicle-seconds.

Queue accumulation polygon for the signal approach. The five vehicles present at the start of effective red lift the red limb into a trapezoid; the enclosed area is the total delay of 342 veh·s.

Approach. The length of the effective red is not given directly — it must be deduced from the condition that green begins 7 s after the queue reaches ten vehicles. Once the red is known, the queue is a simple polygon: it starts at 5, rises at \(q\) throughout the red, then falls at \(s-q\) during the green. Total delay is the area of that polygon.

Find when the queue reaches ten vehicles. Five vehicles are already waiting when the effective red begins, so five more must arrive. At \(q = 800/3600 = 0.2222\) veh/s, \[ t_{10} = \frac{10 - 5}{q} = \frac{5}{0.2222} = 22.5\ \text{s} \] after the start of the effective red.
Deduce the length of the effective red. Green begins 7 s later, and the effective red ends at exactly that moment, so \[ r = t_{10} + 7 = 22.5 + 7 = \boxed{29.5\ \text{s}} \] The cycle segment being analysed therefore runs 29.5 s of effective red followed by 17 s of effective green.
Size the queue at the start of green. The queue grows from its initial value at the arrival rate for the whole of the red: \[ Q_r = Q_0 + q\,r = 5 + 0.2222 \times 29.5 = 5 + 6.56 = 11.56\ \text{vehicles} \] This is the peak queue; it is not an integer because the arrival rate is a continuous average, which is the usual convention in this idealised model.
Check that the queue clears within the green. During green, vehicles depart at \(s\) while continuing to arrive at \(q\), so the queue shrinks at the net rate \(s - q = 0.9028 - 0.2222 = 0.6806\) veh/s. The time required is \[ t_c = \frac{Q_r}{s-q} = \frac{11.56}{0.6806} = 16.98\ \text{s} \] which is just inside the 17 s of effective green available. The approach is therefore not oversaturated: no residual queue is carried into the next cycle, and the whole delay of the cycle is contained in one polygon. The degree of saturation is \(16.98/17 = 0.999\) — the approach is operating right at the edge of its capacity.
Compute the delay on the red limb. Over the red the queue is a trapezoid, rising linearly from 5.00 to 11.56 vehicles over 29.5 s: \[ D_r = \frac{Q_0 + Q_r}{2}\,r = \frac{5.00 + 11.56}{2}(29.5) = 244.2\ \text{veh}\cdot\text{s} \]
Compute the delay on the green limb and total. Over the green the queue falls linearly from 11.56 to zero in 16.98 s, a triangle: \[ D_g = \tfrac{1}{2}\,Q_r\,t_c = \tfrac{1}{2}(11.56)(16.98) = 98.1\ \text{veh}\cdot\text{s} \] so the total delay for the signal in this cycle is \[ D = D_r + D_g = 244.2 + 98.1 = \boxed{342\ \text{veh}\cdot\text{s}} \] that is 5.70 vehicle-minutes, or 0.095 vehicle-hours.
Express the result per vehicle as a sanity check. The number of vehicles served in the cycle is \(s\,t_c = 0.9028 \times 16.98 = 15.33\), which must equal the five initially queued plus the arrivals over the 46.5 s analysed, \(5 + 0.2222 \times 46.48 = 15.33\) — the two agree, so no vehicle has been lost or double-counted. The average delay is therefore \[ \bar{d} = \frac{342.3}{15.33} = 22.3\ \text{s/veh} \] a plausible figure for a short-cycle approach running at its capacity.

Quantity	Result
Effective red deduced from the queue condition	29.5 s
Queue at the start of green	11.56 vehicles
Time for the queue to clear (against 17 s available)	16.98 s — clears
Delay during effective red	244.2 veh·s
Delay during effective green	98.1 veh·s
Total delay for the signal	342 veh·s (5.70 veh·min)
Vehicles served / average delay	15.33 veh / 22.3 s per vehicle

Topic. Uniform (deterministic) delay at an isolated signal approach with an initial queue.

Key relations. \(Q(t) = Q_0 + qt\) during red; \(Q(t) = Q_r - (s-q)t\) during green; clearance time \(t_c = Q_r/(s-q)\); delay \(= \tfrac12 (Q_0+Q_r) r + \tfrac12 Q_r t_c\); average delay \(= D/(s\,t_c)\).

Why this works. The queue accumulation polygon is the vertical gap between a cumulative arrival line of slope \(q\) and a cumulative departure line that is flat during red and rises at slope \(s\) during green. Because both rates are constant, the gap is piecewise linear and its area — the total delay — is the sum of two elementary shapes. The initial five vehicles simply lift the whole polygon, converting the first triangle into a trapezoid; they represent an overflow queue left by the previous cycle or a mid-block source. The condition that the queue must clear within the green (\(t_c \le g\)) is precisely the statement that the degree of saturation \(x = q C/(s\,g)\) does not exceed unity, which is what makes a single-cycle analysis legitimate.

Common pitfall. Taking the effective red as 22.5 s and forgetting the extra 7 s, or taking it as 7 s alone; the wording defines the red as running until the green starts, which is 7 s after the queue reaches ten. A second error is to compute the green-limb delay over the full 17 s of green rather than the 16.98 s that the queue actually lasts — delay stops accruing the moment the queue empties, even though the green continues.

Source. Garber & Hoel, Traffic and Highway Engineering, Ch. 8, “Delay at signalised intersections — the uniform delay term”; HCM, signalised-intersection control-delay model, uniform component \(d_1\).

Question 5 of 7: Total Delay on a Signal Approach

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