16-Civ-B10 Traffic Engineering · December 2019

Question 7 of 7: D/D/1 deterministic queueing at a signal approach

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2019, 16-Civ-B10 Traffic Engineering. Three-hour duration, OPEN BOOK (any non-communicating calculator permitted). Seven questions on four pages, all of equal value at 20 marks each; the paper requires five solutions and marks only the first five as they appear in the answer book. Because the set is a study resource, all seven questions are solved here. The paper’s own Note 1 invites a clear statement of any assumptions made and Note 2 permits any required-but-not-given data to be assumed; every assumption used below is stated explicitly where it is introduced. Page 2 reproduces the AASHTO 2001 metric stopping-sight-distance table, whose footnote fixes the brake-reaction time at 2.5 s and the deceleration at 3.4 m/s²; Question 1 is built around it.

Reference texts.

Garber, N.J. and Hoel, L.A., Traffic and Highway Engineering, 5th ed., Cengage — Ch. 4 (traffic-engineering studies, moving-vehicle method), Ch. 6 (fundamental principles of traffic flow and deterministic queueing), Ch. 8 (intersection control and signal timing), Ch. 15 (geometric design of highways, crest vertical curves).
Transportation Research Board, Highway Capacity Manual — signalised-intersection capacity, saturation flow, lane-group definition and the pedestrian-interval (minimum green) method.
AASHTO, A Policy on Geometric Design of Highways and Streets, 2001 metric edition — stopping sight distance and crest vertical curves; the table printed on page 2 of this paper is AASHTO 2001, Table 3-1.
Transportation Association of Canada, Geometric Design Guide for Canadian Roads — the Canadian design-controls document, and the governing reference for Canadian practice.
Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) — signal displays, actuated control and pedestrian intervals.
Webster, F.V. and Cobbe, B.M., Traffic Signals, Road Research Technical Paper No. 56, HMSO — the optimum-cycle and average-delay relations used in Questions 2 and 5.
Hillier, F.S. and Lieberman, G.J., Introduction to Operations Research — the M/M/1 birth-and-death results used in Question 4.

Canadian context. These are Engineers Canada national examinations, so the answers are framed for Canadian practice: the TAC Geometric Design Guide for Canadian Roads and the MUTCDC are the governing documents and metric design controls are used throughout. Question 1 supplies the AASHTO 2001 metric table directly, so that table is used as printed — the TAC guide adopts the same 2.5 s / 3.4 m/s² stopping model, so the two agree here. Lane widths, walking speed and crosswalk geometry follow TAC and MUTCDC practice.

Question 7: D/D/1 deterministic queueing at a signal 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. A single signalised approach with uniform (deterministic) arrivals and uniform departures at the saturation rate during green.

Quantity	Symbol	Value
Saturation flow	s	2500 veh/h
Approach flow	q	500 veh/h
Cycle length	C	80 s
Effective green	g	25 s
Effective red	r = C − g	55 s

Find. Eight quantities describing the cyclic queue: the capacity check, the clearance time, two proportions, the maximum queue, the total and average delay, and the longest individual delay.

Question 7: the cumulative arrival–departure diagram. Arrivals accumulate at a constant 500 veh/h through the 55 s red; departures leave at 2500 veh/h once the green starts, and the two lines meet 13.75 s into the green. The shaded area is the total delay.

Approach. Build the cumulative arrival–departure diagram: arrivals are a straight line of slope q throughout, departures are zero during red and a straight line of slope s during green. Every one of the eight answers is a coordinate, a length, an area or a ratio taken from that single triangle.

Part (a) — capacity check. The approach can discharge at the saturation rate only while the green is displayed, so its capacity is $$c = s\,\frac{g}{C} = 2500\left(\frac{25}{80}\right) = \boxed{781.25\ \text{veh/h}} \gt q = 500\ \text{veh/h}$$ Equivalently, the single-cycle clearance condition $q/s = 0.200 \lt g/C = 0.3125$ holds, so the queue formed in each red is fully served within the following green and no residual carries over. The degree of saturation is $x = q/c = 500/781.25 = 0.640$.
Part (b) — time to queue clearance. During the red, $q$ accumulates for $r$ seconds; during the green the queue shrinks at the net rate $s - q$. Clearance therefore occurs at $$t_{0} = \frac{q\,r}{s-q} = \frac{500(55)}{2500-500} = \boxed{13.75\ \text{s}}$$ after the start of the effective green. Since $t_0 = 13.75 \lt g = 25$ s, the triangle closes inside the green, confirming part (a) geometrically.
Part (c) — proportion of the cycle with a queue. A queue exists from the onset of red until clearance, a span of $r + t_0$: $$P_{q} = \frac{r + t_{0}}{C} = \frac{55 + 13.75}{80} = \boxed{0.8594}$$ so a queue is present for 85.94 % of the cycle, i.e. 68.75 s out of 80 s.
Part (d) — proportion of vehicles stopped. Every vehicle that joins the queue is one that departs during the clearance period, so the stopped vehicles number $s\,t_0/3600$ while the cycle serves $q\,C/3600$ in total: $$P_{s} = \frac{s\,t_{0}}{q\,C} = \frac{2500(13.75)}{500(80)} = \boxed{0.8594}$$ This is identically the same number as part (c), and that is not a coincidence to be hunted down as an error: the clearance condition $s\,t_0 = q(r + t_0)$ makes the two expressions algebraically equal. The paper awards 2.5 marks to each because they are different derivations — one a ratio of times, the other a ratio of vehicle counts — and both should be shown.
Part (e) — maximum queue. The queue is longest at the end of the red, immediately before discharge begins: $$Q_{max} = q\,r = \frac{500}{3600}(55) = \boxed{7.64\ \text{veh}}$$ At an average 6.5 m of storage per vehicle that is about 50 m of queue, which comfortably fits a normal approach and does not reach back to an upstream intersection.
Part (f) — total delay per cycle. The total delay is the area between the cumulative arrival and departure curves, a triangle of base $r + t_0$ and height $Q_{max}$: $$D = \tfrac{1}{2}(r+t_{0})\,Q_{max} = \frac{q\,r^{2}}{2\left(1-\frac{q}{s}\right)} = \frac{(500/3600)(55)^{2}}{2(1-0.200)} = \boxed{262.6\ \text{veh}\cdot\text{s}}$$ Both forms give 262.59, which is a useful internal check that the triangle has been set up correctly. Over an hour of 45 cycles that is 3.28 vehicle-hours of delay on this approach alone.
Part (g) — average delay per vehicle. The cycle serves $n = qC/3600 = 500(80)/3600 = 11.11$ vehicles, so $$d_{avg} = \frac{D}{n} = \frac{262.59}{11.11} = \boxed{23.63\ \text{s/veh}}$$ Averaged over all arrivals, including the 14 % that arrive on green and are not delayed at all. On the HCM level-of-service scale for signalised intersections this is level of service C — entirely acceptable for an approach running at $x = 0.64$.
Part (h) — maximum delay of any vehicle. The delay of a vehicle arriving $t$ seconds after the onset of red is $d(t) = r - t\left(1 - q/s\right)$, which decreases in t. The maximum is therefore at $t = 0$: $$d_{max} = r = \boxed{55.0\ \text{s}}$$ the full effective red, suffered by the first vehicle to arrive after the signal turns red. Note that this is not the last vehicle to join the queue: that vehicle arrives at $t = r$ and waits only $r(q/s) = 11.0$ s. The distinction is worth 2.5 marks and is the part of this question most often got backwards.

Question 7 — final results
Part	Quantity	Value
(a)	Capacity c = sg/C	781.25 veh/h > 500 veh/h; x = 0.640
(b)	Time to queue clearance t₀	13.75 s
(c)	Proportion of the cycle with a queue	0.8594 (68.75 s of 80 s)
(d)	Proportion of vehicles stopped	0.8594 (identically equal to (c))
(e)	Maximum queue Q_max	7.64 veh (≈ 50 m)
(f)	Total delay per cycle	262.6 veh·s (3.28 veh·h per hour)
(g)	Average delay per vehicle	23.63 s/veh (LOS C)
(h)	Maximum individual delay	55.0 s, the first arrival after the onset of red

Topic. Deterministic (D/D/1) queueing at an under-saturated signal approach.

Key relations. $c = sg/C$; clearance condition $q/s \lt g/C$; $t_0 = qr/(s-q)$; $P_q = (r+t_0)/C$; $P_s = st_0/(qC)$; $Q_{max} = qr$; $D = qr^{2}/[2(1-q/s)]$; $d_{avg} = D/(qC)$; $d_{max} = r$.

Why this works. Assuming both arrivals and departures are uniform turns the whole cycle into one triangle on a cumulative-vehicle diagram, and every question about the queue becomes a question about that triangle. Vertical distance between the curves is queue length, horizontal distance is an individual vehicle’s delay, and enclosed area is total delay — so the maximum queue is at the point of maximum vertical separation (the end of red) and the maximum individual delay is at the point of maximum horizontal separation (the start of red). The identity between parts (c) and (d) falls straight out of the geometry: the clearance point is where the departure line catches the arrival line, which forces the count discharged to equal the count that arrived.

Common pitfall. Three recur. Reporting the maximum individual delay as the wait of the last vehicle in the queue reverses the geometry and gives 11.0 s instead of 55.0 s. Placing the maximum queue at clearance rather than at the end of red gives zero. And treating the equality of (c) and (d) as an arithmetic mistake wastes time hunting a bug that is not there — though it is worth saying in the answer that the equality is a property of the deterministic model and would break under M/D/1 or M/M/1, where a residual queue survives some cycles.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 6 §6.5 (deterministic queueing) and Ch. 8 (signal delay); TRB, Highway Capacity Manual, uniform-delay term of the signalised-intersection delay model.

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