Deterministic (D/D/1) Queueing at a Signalized Approach: 16-Civ-B10 Traffic Engineering, December 2016, Q7

Question 7: Deterministic (D/D/1) Queueing at a Signalized Approach (20 marks — 2.5 each)

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. A signalized approach with the following timing and flow parameters.

Approach parameters
Quantity	Symbol	Value
Cycle length	$C$	90 s
Effective green	$g$	27 s
Effective red	$r = C - g$	63 s
Saturation flow	$s$	2,800 veh/h = 0.7778 veh/s
Approach arrival flow	$\lambda$	600 veh/h = 0.1667 veh/s

Find. The eight quantities (a) to (h) that describe the deterministic queue cycle: the capacity check, the clearance time, the proportion of the cycle and of the vehicles affected, the maximum queue, and the total, average and maximum delays.

Cumulative arrival and departure curves for one cycle. Arrivals accumulate at the uniform rate λ throughout; departures are zero during the red and then occur at the saturation flow until the queue clears at t₀. The shaded area between the curves is the total delay per cycle.

Approach. Build the deterministic cumulative arrival and departure diagram for one cycle: arrivals accumulate uniformly throughout, departures are zero during the red and occur at the saturation rate until the queue clears, and every quantity asked for is a distance, an intersection point or an area on that diagram.

Part (a) — verify the capacity. An approach can only discharge during its effective green, so its hourly capacity is the saturation flow prorated by the green ratio: $$c = s\,\frac{g}{C} = 2{,}800 \times \frac{27}{90} = \boxed{840\ \text{veh/h}}$$ Since $840 > 600$, capacity exceeds demand, the degree of saturation is $X = 600/840 = 0.714$, and the approach is undersaturated — every cycle’s queue clears within its own green, which is the precondition for the whole D/D/1 analysis that follows.
Part (e) — maximum queue length (taken first, since it anchors the diagram). Nothing departs during the red, so the queue grows for the whole of $r$ at the arrival rate: $$Q_{max} = \lambda\,r = 0.1667 \times 63 = \boxed{10.5\ \text{vehicles}}$$ This is the queue standing at the stop line the instant the green begins, and it is the vertical distance between the two curves at $t = r$ on the figure.
Part (b) — time to clear the queue after the start of green. Once the green begins, departures run at $s$ while arrivals continue at $\lambda$, so the queue shrinks at the net rate $(s - \lambda)$. Clearing $\lambda r$ vehicles at that net rate takes $$t_0 = \frac{\lambda\,r}{s - \lambda} = \frac{0.1667 \times 63}{0.7778 - 0.1667} = \frac{10.5}{0.6111} = \boxed{17.18\ \text{s}}$$ The queue therefore clears 17.18 s into a 27 s green, leaving 9.8 s of green during which arriving vehicles pass without stopping.
Part (c) — proportion of the cycle with a queue. A queue exists from the start of the red until the moment of clearance, a span of $r + t_0$: $$P_q = \frac{r + t_0}{C} = \frac{63 + 17.18}{90} = \frac{80.18}{90} = \boxed{0.891 = 89.1\%}$$
Part (d) — proportion of vehicles stopped. Every vehicle that arrives while a queue is present must join it, and because arrivals are uniform the count of those vehicles is $\lambda(r + t_0)$ out of the $\lambda C$ that arrive in the cycle: $$P_s = \frac{\lambda(r + t_0)}{\lambda C} = \frac{0.1667(80.18)}{0.1667(90)} = \frac{13.36}{15.0} = \boxed{0.891 = 89.1\%}$$ With uniform arrivals the two proportions are numerically identical — 13.4 of the 15 vehicles arriving each cycle are stopped — but they answer different questions and would differ if arrivals were not uniform.
Part (f) — total vehicle delay per cycle. Total delay is the area enclosed between the cumulative arrival and departure curves, a triangle of base $(r + t_0)$ and height $Q_{max}$: $$D_{tot} = \tfrac{1}{2}(r + t_0)\,\lambda r = \tfrac{1}{2}(80.18)(10.5) = \boxed{421.0\ \text{veh}\cdot\text{s}}$$ The closed form $D_{tot} = \lambda r^2 / [2(1 - \lambda/s)] = 0.1667(63)^2/[2(0.7857)] = 421.0$ veh·s confirms it.
Part (g) — average delay per vehicle. Spreading the cycle’s total delay over every vehicle that arrives in the cycle, stopped or not, $$d_{avg} = \frac{D_{tot}}{\lambda C} = \frac{421.0}{0.1667 \times 90} = \frac{421.0}{15.0} = \boxed{28.1\ \text{s/veh}}$$ On the HCM control-delay scale this corresponds to level of service C for a signalized intersection approach.
Part (h) — maximum delay of any vehicle. Under a deterministic first-in-first-out queue the worst-off vehicle is the one arriving an instant after the green ends; it waits out the entire red and is then the first to be released, so $$d_{max} = r = C - g = \boxed{63\ \text{s}}$$ Note that this is not the vehicle that waits longest in the queue in space — that is the last vehicle to join before clearance — but it is the one whose individual delay is greatest, and it is the horizontal distance between the two curves at the origin.

Taken together the eight answers describe an approach that is working but not comfortably: capacity exceeds demand by 40 per cent, yet a queue is present for 89 per cent of every cycle and nearly nine vehicles in ten are stopped. That combination is characteristic of a short green ratio — only 30 per cent here — and points to the split, not the cycle length, as the variable worth revisiting if this approach is the one generating complaints.

Question 7 — final results
Part	Quantity	Value
(a)	Capacity $c = s\,g/C$	840 veh/h > 600 veh/h ($X = 0.714$) — verified
(b)	Time to queue clearance $t_0$	17.18 s after start of green
(c)	Proportion of cycle with a queue	0.891 = 89.1%
(d)	Proportion of vehicles stopped	0.891 = 89.1% (13.4 of 15 veh)
(e)	Maximum queue $Q_{max}$	10.5 vehicles
(f)	Total delay per cycle	421.0 veh·s
(g)	Average delay per vehicle	28.1 s/veh
(h)	Maximum delay of any vehicle	63 s

Concept & Theory

Topic. Deterministic D/D/1 queueing analysis of an undersaturated signal approach.

Key relations. $c = s\,g/C$; $X = \lambda/c$; $Q_{max} = \lambda r$; $t_0 = \lambda r/(s - \lambda)$; $P_q = P_s = (r + t_0)/C$ for uniform arrivals; $D_{tot} = \lambda r^2/[2(1 - \lambda/s)]$; $d_{avg} = D_{tot}/(\lambda C)$; $d_{max} = r$.

Why this works. The whole analysis is one picture: plot cumulative arrivals against time as a straight line of slope $\lambda$, and cumulative departures as a line that is flat through the red and then rises at slope $s$ until it meets the arrival line. The vertical gap between the curves at any instant is the queue, the horizontal gap is the delay of the vehicle at that ordinate, and the enclosed area is the total delay. Every one of the eight quantities is a reading off that figure, which is why sketching it first is faster and far safer than memorising eight formulas.

Common pitfall. Using $C$ instead of $r$ as the queue-building interval, which inflates $Q_{max}$ and every quantity downstream. Also common is dividing the total delay by the number of stopped vehicles rather than by all arrivals in the cycle — average delay per vehicle is conventionally taken over every vehicle on the approach — and reporting the maximum delay as $r + t_0$; the vehicle arriving at the start of red waits exactly $r$, since it is first out when the green begins.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §5.5 and §8.5 — “D/D/1 queueing at signalized intersections: delay, queue length and stops”.

Question 7 of 7: Deterministic (D/D/1) Queueing at a Signalized Approach

Question 7: Deterministic (D/D/1) Queueing at a Signalized Approach (20 marks — 2.5 each)