D/D/1 Queueing at a Signalised Approach: 16-Civ-B10 Traffic Engineering, Undated paper, Q7

Question 7: D/D/1 Queueing at a Signalised Approach (8 × 2.5 = 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. Saturation flow $s = 2500$ veh/h, approach flow $q = 500$ veh/h, cycle $C = 80$ s, effective green $g = 25$ s, hence effective red $r = C - g = 55$ s. Deterministic arrivals and deterministic departures, single channel (D/D/1).

Find. Capacity check, queue clearance time, proportion of cycle with a queue, proportion of vehicles stopped, maximum queue, total and average delay, and the longest delay suffered by any individual vehicle.

Cumulative arrival–departure diagram for one cycle. Arrivals rise uniformly at q = 500 vph; departures are held at zero through the 55 s effective red and then discharge at the 2500 vph saturation rate until the curves meet at 68.75 s. The shaded triangle is the 262.6 veh·s of total delay; its vertical extent at t = 55 s is the maximum queue and its horizontal extent at the origin is the maximum individual delay.

Approach. Build the cumulative arrival–departure diagram for one cycle. Arrivals accumulate at a uniform $q$ throughout; departures are zero during the effective red and then run at the saturation rate $s$ until the queue is exhausted, after which departures track arrivals. Every requested quantity is a length, an intercept or an area on that diagram.

Part (a) — Verify the capacity exceeds the arrival rate. The approach is served for a fraction $g/C$ of every cycle at the saturation rate, so $$c=s\,\frac{g}{C}=2500\times\frac{25}{80}=\boxed{781.25\ \text{veh/h}}\;>\;q=500\ \text{veh/h}$$ Equivalently $q/s = 0.200 < g/C = 0.3125$, and the degree of saturation is $X = q/c = 500/781.25 = 0.64$. The single-cycle triangle therefore closes and no residual queue carries over — this test must be passed before any of the remaining parts is meaningful.
Part (b) — Time to queue clearance after the start of effective green. At the start of green the queue holds the vehicles that arrived during the red. The queue shrinks at the rate $(s-q)$ once discharge begins, so it clears at $$t_0=\frac{q\,r}{s-q}=\frac{500(55)}{2500-500}=\frac{27\,500}{2000}=\boxed{13.75\ \text{s}}$$ That is, the queue is exhausted 13.75 s into a 25 s green, leaving 11.25 s of green during which arrivals pass without stopping.
Part (c) — Proportion of the cycle with a queue. A queue exists from the start of the effective red until clearance, a span of $r + t_0$: $$P_q=\frac{r+t_0}{C}=\frac{55+13.75}{80}=\frac{68.75}{80}=\boxed{0.859}$$
Part (d) — Proportion of vehicles stopped. The vehicles that stop are exactly those discharged at the saturation rate during the clearance interval, $s\,t_0$, out of the $qC$ that arrive in a cycle: $$P_s=\frac{s\,t_0}{q\,C}=\frac{(2500/3600)(13.75)}{(500/3600)(80)}=\frac{9.549}{11.111}=\boxed{0.859}$$ This is the same number as part (c), and that is not a coincidence: the clearance condition $s\,t_0 = q(r+t_0)$ makes $P_s = s t_0/(qC)$ identically equal to $P_q = (r+t_0)/C$. One value, reached by two different routes — one a ratio of times, the other a ratio of vehicles.
Part (e) — Maximum number of vehicles in the queue. The queue is longest at the end of the effective red, immediately before discharge begins, not at clearance: $$Q_{\max}=q\,r=\frac{500}{3600}(55)=\boxed{7.64\ \text{veh}}$$ For storage design this is rounded up to 8 vehicles, requiring roughly $8 \times 6$ m $= 48$ m of queue storage on the approach.
Part (f) — Total vehicle delay per cycle. Total delay is the area between the cumulative arrival and departure curves — a triangle of base $(r+t_0)$ and height $Q_{\max}$: $$D=\tfrac{1}{2}(r+t_0)\,Q_{\max}=\tfrac{1}{2}(68.75)(7.6389)=\boxed{262.6\ \text{veh}\cdot\text{s}}$$ The closed form confirms it: $D = q r^{2}/[2(1-q/s)] = (0.13889)(3025)/(2\times 0.8) = 262.6$ veh·s.
Part (g) — Average delay per vehicle. The number of vehicles arriving in one cycle is $n = qC = (500/3600)(80) = 11.11$ veh, so $$\bar d=\frac{D}{n}=\frac{262.59}{11.111}=\boxed{23.63\ \text{s/veh}}$$ The closed form agrees exactly: $\bar d = r^{2}/[2C(1-q/s)] = 3025/(2\times 80\times 0.8) = 23.63$ s. Note that this average is taken over all arrivals, including the 14% that are not stopped at all; the average delay of a stopped vehicle is $23.63/0.859 = 27.5$ s.
Part (h) — Maximum delay of any vehicle. Individual delay is $d(t) = r - t\,(1 - q/s)$ for a vehicle arriving $t$ seconds after the onset of red, which decreases monotonically in $t$. The worst-delayed vehicle is therefore the first arrival after the start of the effective red, which waits the entire red: $$d_{\max}=r=\boxed{55\ \text{s}}$$ It is the first vehicle in the queue, not the last, that suffers most — a result that is obvious on the diagram and routinely got backwards without one.

Check — the model and where it stops being true. D/D/1 assumes perfectly uniform arrivals and a perfectly uniform discharge, which is why parts (c) and (d) collapse to one number and why the queue clears in every cycle. Under a stochastic arrival process (M/D/1 or M/M/1) some cycles receive more than $q C$ arrivals, a residual queue survives into the next cycle, the two proportions separate, and an overflow-delay term must be added to Webster's uniform delay. At $X = 0.64$ that overflow term is small (a few seconds per vehicle), so the deterministic answers above are a sound design estimate here; at $X > 0.85$ they would understate delay substantially. Effective green and effective red are used throughout as given, so no separate treatment of amber or start-up lost time is required.

Part	Quantity	Symbol	Result
(a)	Approach capacity	$c = sg/C$	781.25 veh/h > 500 veh/h ✓ $(X = 0.64)$
(b)	Time to queue clearance after start of green	$t_0$	13.75 s
(c)	Proportion of the cycle with a queue	$P_q$	0.859
(d)	Proportion of vehicles stopped	$P_s$	0.859 (identically equal to $P_q$)
(e)	Maximum queue (at end of red)	$Q_{\max}$	7.64 veh → 8 veh of storage
(f)	Total delay per cycle	$D$	262.6 veh·s
(g)	Average delay per vehicle	$\bar d$	23.63 s/veh (27.5 s per stopped vehicle)
(h)	Maximum delay of any vehicle	$d_{\max}$	55 s (the first arrival after the onset of red)

Topic. Deterministic (D/D/1) queueing analysis of an under-saturated signalised approach, using the cumulative arrival–departure diagram.

Key relations. $c = sg/C$; clearance $t_0 = qr/(s-q)$; $P_q = (r+t_0)/C \equiv P_s = st_0/(qC)$; $Q_{\max} = qr$; total delay $D = qr^{2}/[2(1-q/s)]$; average delay $\bar d = r^{2}/[2C(1-q/s)]$ (Webster's uniform-delay term); $d_{\max} = r$.

Why this works. Every one of these results is a feature of one triangle. Cumulative arrivals form a straight line of slope $q$; cumulative departures are flat for $r$ seconds and then rise at slope $s$ until they intercept the arrival line. The horizontal gap between the curves at any ordinate is the delay of that particular vehicle, the vertical gap at any abscissa is the queue at that instant, and the enclosed area is total delay in vehicle-seconds. That single picture answers all eight parts, which is why drawing it first is faster than memorising eight formulas — and why the identity between parts (c) and (d) is obvious rather than suspicious.

Common pitfall. Reporting the maximum queue at clearance rather than at the end of red, and reporting the maximum individual delay as belonging to the last vehicle in the queue rather than the first. Both follow directly from the diagram and both are routinely inverted. A third pitfall is failing the $q/s < g/C$ test first: if it fails, the single-cycle triangle never closes and every formula above is invalid.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §8.3 (deterministic queueing at signalised approaches) and §8.6 (Webster's delay model).

Part	Quantity	Symbol	Result
(a)	Approach capacity	\(c = sg/C\)	781.25 veh/h > 500 veh/h ✓ \((X = 0.64)\)
(b)	Time to queue clearance after start of green	\(t_0\)	13.75 s
(c)	Proportion of the cycle with a queue	\(P_q\)	0.859
(d)	Proportion of vehicles stopped	\(P_s\)	0.859 (identically equal to \(P_q\))
(e)	Maximum queue (at end of red)	\(Q_{\max}\)	7.64 veh → 8 veh of storage
(f)	Total delay per cycle	\(D\)	262.6 veh·s
(g)	Average delay per vehicle	\(\bar d\)	23.63 s/veh (27.5 s per stopped vehicle)
(h)	Maximum delay of any vehicle	\(d_{\max}\)	55 s (the first arrival after the onset of red)

Question 7 of 7: D/D/1 Queueing at a Signalised Approach

Question 7: D/D/1 Queueing at a Signalised Approach (8 × 2.5 = 20 marks)