Question 4 of 7: D/D/1 queueing at a signalised approach

Question 4 — 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.

Find. (a) capacity against demand; (b) queue clearance time $t_0$; (c) proportion of the cycle with a queue; (d) proportion of vehicles stopped; (e) maximum queue; (f) total delay per cycle; (g) average delay per vehicle; (h) maximum individual delay.

Approach. Build the cumulative arrival and departure curves for one cycle. Arrivals accumulate at the uniform rate $q$ throughout; departures are zero during the effective red and then run at the saturation flow $s$ until the queue is exhausted, after which they track the arrivals. Every one of the eight answers is a geometric property of the triangle enclosed between the two curves.

1. Part (a) — verify the approach has enough capacity. The capacity of a signalised approach is its saturation flow prorated by the green ratio:
   $$c=s\,\frac{g}{C}=2750\times\frac{30}{85}=970.6\ \text{veh/h}$$
   Since $c=970.6>q=550$ veh/h, the approach is undersaturated with a degree of saturation of $x=q/c=550/970.6=0.567$. The equivalent form of the same test, which is the one used in the steps that follow, compares the flow ratio with the green ratio:
   $$\frac{q}{s}=\frac{550}{2750}=0.200\ <\ \frac{g}{C}=\frac{30}{85}=0.353$$
   Both statements confirm that a queue formed during red will clear within the green, so the single-cycle triangle closes and no residual queue carries over. This is the precondition for every subsequent part, which is why the paper asks for it first.
2. Part (b) — time to clear the queue after the start of the green. During the effective red, $q\,r$ vehicles arrive and none depart. Once the green begins, departures run at $s$ while arrivals continue at $q$, so the queue drains at the net rate $s-q$. Letting $t_0$ be the clearance time measured from the start of the green, the number discharged equals the number that has arrived:
   $$s\,t_0=q\left(r+t_0\right)\quad\Longrightarrow\quad t_0=\frac{q\,r}{s-q}$$
   $$t_0=\frac{550\times55}{2750-550}=\frac{30250}{2200}=\boxed{13.75\ \text{s}}$$
   This is comfortably inside the 30 s green, consistent with part (a).
3. Part (c) — proportion of the cycle with a queue present. A queue exists from the onset of the effective red until clearance, a span of $r+t_0$:
   $$P_q=\frac{r+t_0}{C}=\frac{55+13.75}{85}=\frac{68.75}{85}=\boxed{0.809}$$
   So for 80.9 per cent of the cycle — 68.75 s of every 85 s — at least one vehicle is stopped or slowing on the approach.
4. Part (d) — proportion of vehicles stopped. Every vehicle that arrives before clearance joins the queue, and those are exactly the vehicles discharged during $t_0$ at the saturation rate. As a fraction of the $qC$ vehicles arriving per cycle,
   $$P_s=\frac{s\,t_0}{q\,C}=\frac{2750\times13.75}{550\times85}=\frac{37812.5}{46750}=\boxed{0.809}$$
5. Note the identity between parts (c) and (d), which is a property of the model rather than a coincidence. The clearance condition $s\,t_0=q(r+t_0)$ can be substituted directly into the expression for $P_s$:
   $$P_s=\frac{s\,t_0}{q\,C}=\frac{q\left(r+t_0\right)}{q\,C}=\frac{r+t_0}{C}=P_q$$
   so the two parts are algebraically the same number reached by two different routes — one a ratio of times, the other a ratio of vehicles. Both derivations are shown because the paper awards 2.5 marks to each, but no arithmetic discrepancy should be hunted for. The equality holds only because arrivals are deterministic and uniform; under M/D/1 or M/M/1 a residual queue survives in some cycles, vehicles are stopped that arrive after nominal clearance, and the two proportions separate.
6. Part (e) — maximum number of vehicles in the queue. The queue grows for the whole of the effective red and shrinks thereafter, so it peaks precisely at the end of the red — not at clearance:
   $$Q_{max}=q\,r=\frac{550}{3600}\times55=0.15278\times55=\boxed{8.40\ \text{veh}}$$
   In practice this is 8 to 9 vehicles, requiring roughly 8.4 × 6.5 = 55 m of storage on the approach, which is the number a designer would take to the lane-length check.
7. Part (f) — total vehicle delay per cycle. The aggregate delay is the area enclosed between the arrival and departure curves, a triangle of base $r+t_0$ and height $Q_{max}$. Evaluating it in the standard closed form,
   $$D=\frac{q\,r^{2}}{2\left(1-q/s\right)}=\frac{0.15278\times55^{2}}{2\left(1-0.200\right)}=\frac{462.15}{1.600}$$
   $$D=\boxed{288.9\ \text{veh}\cdot\text{s}}$$
   Equivalently, $\tfrac12(r+t_0)Q_{max}=\tfrac12(68.75)(8.403)=288.8$ veh·s, which is the same area computed geometrically — a useful cross-check because the two forms share no intermediate quantity. Expressed in the units a delay study would report, 288.9 veh·s is 0.080 veh·h per cycle, or 3.40 veh·h per hour of operation over the 42.4 cycles in an hour.
8. Part (g) — average delay per vehicle. Dividing the cycle's total delay by the number of vehicles arriving in the cycle,
   $$n=q\,C=0.15278\times85=12.99\ \text{veh}$$
   $$d_{avg}=\frac{D}{n}=\frac{288.9}{12.99}=\boxed{22.2\ \text{s/veh}}$$
   By the HCM control-delay criteria, 22.2 s/veh corresponds to level of service C for a signalised approach.
9. Part (h) — maximum delay of any vehicle. A vehicle arriving $t$ seconds after the onset of the effective red waits until its turn to discharge, which gives the delay function
   $$d(t)=r-t\left(1-\frac{q}{s}\right)$$
   This decreases monotonically in $t$, so the worst-off vehicle is the first one to arrive after the onset of the red, not the last vehicle to arrive during it. Setting $t=0$,
   $$d_{max}=r=\boxed{55\ \text{s}}$$
   which is simply the whole effective red: the first vehicle stops at the stop line and moves off at the start of the green with no queue ahead of it. The last vehicle to join the queue, arriving at $t=r$, waits only $55-55(0.8)=11.0$ s, and the average of 22.2 s/veh sits between the two as it must.

Results.