Deterministic (D/D/1) queueing at a signalised approach: 16-Civ-B10 Traffic Engineering, December 2017, Q6

Question 6 — Deterministic (D/D/1) queueing at a signalised approach (a) to (h), 2.5 marks each — 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.

Quantity	Symbol	Value	In veh/s
Saturation (departure) flow	$s$	2500 veh/h	0.6944
Approach (arrival) flow	$q$	500 veh/h	0.1389
Cycle time	$C$	80 s	—
Effective green	$g$	25 s	—
Effective red	$r = C - g$	55 s	—

Find. The eight quantities (a)–(h): the capacity check, queue clearance time, the proportion of the cycle with a queue, the proportion of vehicles stopped, the longest queue, the total delay per cycle, the average delay per vehicle and the greatest delay suffered by any single vehicle.

Figure 6.1 — Deterministic queueing diagram for one 80 s cycle. The vertical gap between $A(t)$ and $D(t)$ is the instantaneous queue; the horizontal gap is an individual vehicle's delay; the shaded area is the total delay.

Approach. Draw the cumulative arrival and departure curves for one cycle. Arrivals accumulate at the constant rate $q$; departures are zero through the effective red and then discharge at the saturation rate $s$ until the accumulated queue is exhausted. Every requested quantity is then a geometric feature of that diagram — a vertical gap, a horizontal gap, an intersection point or an area.

Convert the flows to a per-second basis and identify the red time. $q = \frac{500}{3600} = 0.1389\ \text{veh}\,\text{s}^{-1}, \qquad s = \frac{2500}{3600} = 0.6944\ \text{veh}\,\text{s}^{-1}$ $r = C - g = 80 - 25 = 55\ \text{s}$
(a) Verify that capacity exceeds the arrival rate. The approach capacity is the saturation flow prorated by the green ratio: $c = s\,\frac{g}{C} = 2500 \times \frac{25}{80} = \boxed{781\ \text{veh}\,\text{h}^{-1} \gt 500\ \text{veh}\,\text{h}^{-1}}$ The degree of saturation is $x = q/c = 500/781.25 = 0.64 \lt 1$, so the approach is undersaturated and the queue clears within every cycle. The equivalent test on the ratios confirms it: the flow ratio $q/s = 0.200$ is less than the green ratio $g/C = 0.3125$.
(b) Time to queue clearance after the start of the effective green. At the start of green the queue is $q\,r$ vehicles, and it is worked off at the net rate $s - q$ (departures minus continuing arrivals): $t_0 = \frac{q\,r}{s - q} = \frac{500 \times 55}{2500 - 500} = \frac{27\,500}{2000} = \boxed{13.75\ \text{s}}$ The queue therefore clears 13.75 s into the 25 s green, at $t = 68.75$ s from the start of red — the point marked in Figure 6.1 where $D(t)$ meets $A(t)$.
(c) Proportion of the cycle with a queue. A queue exists throughout the red and for $t_0$ into the green: $P_q = \frac{r + t_0}{C} = \frac{55 + 13.75}{80} = \frac{68.75}{80} = \boxed{0.859 \;(85.9\%)}$
(d) Proportion of vehicles stopped. Every vehicle arriving while a queue is present must join it. The number discharged during the clearance period is $s\,t_0$, and the number arriving in a cycle is $q\,C$, so $P_s = \frac{s\,t_0}{q\,C} = \frac{0.6944 \times 13.75}{0.1389 \times 80} = \frac{9.549}{11.111} = \boxed{0.859 \;(85.9\%)}$ This equals the answer to (c), and not by coincidence: the clearance condition $s\,t_0 = q(r + t_0)$ makes $P_s$ and $P_q$ algebraically identical in a D/D/1 system. About 9.5 of the 11.1 vehicles arriving each cycle are stopped; only the 1.6 arriving after $t = 68.75$ s pass through without stopping.
(e) Maximum number of vehicles in the queue. The vertical gap between the two curves is largest at the end of the red, immediately before discharge begins: $Q_{max} = q\,r = 0.1389 \times 55 = \boxed{7.64\ \text{veh}\;(\text{say } 8\ \text{vehicles})}$ For storage design the figure is rounded up: at about 6.5 m per queued passenger car this is roughly 52 m of storage required on the approach.
(f) Total vehicle delay per cycle. The total delay is the area between $A(t)$ and $D(t)$, the shaded region of Figure 6.1. That region is a triangle of base $(r + t_0)$ and height $q\,r$: $D_{tot} = \tfrac{1}{2}\,(q\,r)\,(r + t_0) = \tfrac{1}{2}(7.639)(68.75) = \boxed{262.5\ \text{veh}\cdot\text{s per cycle}}$ Equivalently 4.4 vehicle-minutes per cycle, or with 45 cycles per hour about 3.3 vehicle-hours of delay per hour on this approach.
(g) Average delay per vehicle. Dividing by the number of vehicles arriving in a cycle, $q\,C = 0.1389 \times 80 = 11.11$ veh: $\bar d = \frac{D_{tot}}{q\,C} = \frac{262.5}{11.11} = \boxed{23.6\ \text{s per vehicle}}$ By the HCM criteria for a signalised approach this corresponds to level of service C — acceptable operation, consistent with the degree of saturation of 0.64 found in part (a).
(h) Maximum delay of any vehicle. An individual vehicle's delay is the horizontal gap between the curves. A vehicle arriving at time $t$ during the red is the $q\,t$-th in the queue, so it departs at $r + q\,t/s$ and its delay is $d(t) = r + \frac{q\,t}{s} - t = r - t\left(1 - \frac{q}{s}\right)$ which decreases monotonically in $t$. The greatest delay therefore falls on the first vehicle to arrive after the onset of red, at $t = 0$: $d_{max} = r = \boxed{55\ \text{s}}$ Note the contrast with part (g): the worst-off driver waits 55 s while the average is 23.6 s, so quoting only the average conceals a factor of more than two in the individual experience.

Part	Quantity	Result
(a)	Capacity $c = s\,g/C$ versus arrival rate	781 veh/h > 500 veh/h; $x = 0.64$ — undersaturated
(b)	Queue clearance time after start of green, $t_0$	13.75 s (clears at $t = 68.75$ s)
(c)	Proportion of the cycle with a queue	0.859 (85.9 %)
(d)	Proportion of vehicles stopped	0.859 (85.9 %)
(e)	Maximum queue, $Q_{max} = q\,r$	7.64 veh (8 vehicles, about 52 m of storage)
(f)	Total delay per cycle	262.5 veh·s (4.4 veh·min)
(g)	Average delay per vehicle	23.6 s/veh (LOS C)
(h)	Maximum delay of any vehicle	55.0 s (the first arrival after the onset of red)

Check — parts (c) and (d) give the same number, and that is correct

The proportion of the cycle with a queue and the proportion of vehicles stopped are computed from different formulas — a time ratio and a vehicle ratio — yet both return 0.859. This is an identity of the deterministic model, not an error: the queue-clearance condition $s\,t_0 = q(r+t_0)$ makes $s\,t_0/(qC)$ equal to $(r+t_0)/C$ exactly. In a stochastic (M/D/1 or M/M/1) treatment the two would differ, because arrivals during the clearance period would be random and some cycles would carry over a residual queue. Both derivations are shown above so the distinct reasoning behind each part is visible.

Concept & Theory

Topic. Deterministic (D/D/1) queueing at a signalised approach — uniform arrivals, uniform service, one channel — and the cumulative-curve construction that makes every delay statistic a piece of geometry.

Key relations. $c = s\,g/C$; $x = q/c$; $t_0 = qr/(s-q)$; $Q_{max} = q\,r$; $P_q = (r+t_0)/C$; $P_s = s\,t_0/(qC)$; $D_{tot} = \tfrac{1}{2}q\,r\,(r+t_0)$; $\bar d = D_{tot}/(qC)$; $d_{max} = r$.

Why this works. On a cumulative diagram, arrivals $A(t)$ and departures $D(t)$ are both non-decreasing, so at any instant the vertical separation is the number of vehicles in the queue and the horizontal separation is the time an individual vehicle spends in it. The area between the curves is the integral of the queue over time, which is by definition the total delay in vehicle-seconds. Because both curves are straight lines here, every one of those measures reduces to elementary geometry — and the same construction extends unchanged to freeway incidents, toll plazas and ramp meters, which is why it is worth drawing the diagram before writing any formula.

Common pitfall. Three recurring errors. First, using the cycle length instead of the effective red in $t_0 = qr/(s-q)$, which inflates the clearance time by a factor of $C/r$. Second, taking the maximum queue at the moment the queue clears rather than at the end of red — the queue is shrinking throughout the clearance period. Third, confusing the average delay with the maximum delay: the maximum falls on the first vehicle to arrive after the onset of red and equals the full red time, whereas the average is roughly $r/2$ scaled by the fraction stopped. It is also worth confirming $q/s \lt g/C$ before starting: if that inequality fails the queue never clears, the single-cycle triangle does not close, and an oversaturated (multi-cycle) analysis is required instead.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §6.5 (deterministic queueing) and §8.5 (signal delay and level of service); TRB Highway Capacity Manual, uniform-delay term of the signalized-intersection delay model.

Question 6 of 7: Deterministic (D/D/1) queueing at a signalised approach

Question 6 — Deterministic (D/D/1) queueing at a signalised approach (a) to (h), 2.5 marks each — 20 marks