16-Civ-B10 Traffic Engineering · December 2018

Question 5 of 7: D/D/1 queueing at a signal approach (8 × 2.5 marks)

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

Notes on this paper

Paper format. National Examinations, December 2018, 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, which Question 2 is built around.

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 queueing), Ch. 8 (intersection control and signal timing), Ch. 15 (geometric design of highways, 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 SSD 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, pedestrian intervals and the Canadian signal-warrant procedure.
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 3 and 7.

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(e) names the US MUTCD explicitly, so its eight warrants are answered as printed; the Canadian equivalent procedure is noted alongside, which is polish rather than a correction.

Question 5: D/D/1 queueing at a signal approach (8 × 2.5 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.

Question 5 data
Quantity	Symbol	Value
Saturation flow	$s$	2600 veh/h
Approach flow	$q$	650 veh/h
Cycle length	$C$	80 s
Effective green	$g$	25 s
Effective red (derived)	$r = C - g$	55 s

Find. The eight quantities (a) to (h), all from the deterministic D/D/1 model.

Question 5. Cumulative arrival and departure curves for one cycle. The shaded triangle is the total delay; its base is the interval during which a queue exists, and the vertical gap at the end of red is the maximum queue.

Approach. Draw the cumulative diagram once. Arrivals accumulate at a constant $q$ throughout; departures are zero during the effective red, then run at the saturation flow $s$ until the queue clears, then track the arrivals. Every one of the eight answers is a feature of that one triangle.

Part (a) — verify capacity exceeds demand. The approach can only discharge during the effective green, so $$c = s\,\frac{g}{C} = 2600\left(\frac{25}{80}\right) = \boxed{812.5\ \text{veh/h}} \;>\; q = 650\ \text{veh/h}$$ Equivalently and more usefully, $q/s = 650/2600 = 0.250$ against $g/C = 25/80 = 0.3125$: the flow ratio is below the green ratio, so the queue that builds during red must clear within the green. The degree of saturation is $x = q/c = 650/812.5 = 0.800$, so the approach runs at 80 per cent of capacity with 20 per cent reserve.
Part (b) — time to queue clearance after the start of green. During the clearance interval $t_0$ the queue is served at $s$ while new vehicles keep arriving at $q$, so the departures must catch up with everything that has arrived since the onset of red: $s\,t_0 = q(r + t_0)$, hence $$t_0 = \frac{q\,r}{s - q} = \frac{650(55)}{2600 - 650} = \frac{35\,750}{1950} = \boxed{18.33\ \text{s}}$$ and $t_0 = 18.33$ s is comfortably inside the 25 s green, confirming part (a) geometrically.
Part (c) — proportion of the cycle with a queue. A queue exists from the onset of the effective red until clearance, a span of $r + t_0$: $$P_q = \frac{r + t_0}{C} = \frac{55 + 18.33}{80} = \frac{73.33}{80} = \boxed{0.917}$$ so a vehicle arriving at a random instant finds a queue 91.7 per cent of the time.
Part (d) — proportion of vehicles stopped. Every vehicle that joins the queue is discharged during the clearance interval, so the number stopped per cycle is $s\,t_0/3600$ while the number arriving per cycle is $q\,C/3600$: $$P_s = \frac{s\,t_0}{q\,C} = \frac{2600(18.33)}{650(80)} = \frac{47\,666.7}{52\,000} = \boxed{0.917}$$ This is the same number as part (c), and not by accident. Substituting $s\,t_0 = q(r + t_0)$ into the expression for $P_s$ gives $P_s = q(r+t_0)/(qC) = (r+t_0)/C = P_q$ identically, so the paper awards 2.5 marks each to a time ratio and a vehicle ratio that a deterministic model forces to coincide. Both derivations are shown because the question asks for both.
Part (e) — maximum number of vehicles in the queue. The queue grows for the whole of the effective red and shrinks thereafter, so its maximum is at the end of red, not at clearance: $$Q_{max} = \frac{q}{3600}\,r = \frac{650}{3600}(55) = \boxed{9.93\ \text{veh}}$$ At about 6.5 m per queued vehicle that is 64.6 m of storage, which the approach must provide upstream of the stop line if the queue is not to block the adjacent intersection.
Part (f) — total vehicle delay per cycle. The total delay is the area of the triangle between the arrival and departure curves. Taking it as $\tfrac12 \times \text{base} \times \text{height}$ with base $r + t_0$ and height $Q_{max}$, or in closed form, $$D = \frac{q\,r^{2}}{2\left(1 - q/s\right)} = \frac{(650/3600)(55)^{2}}{2(1 - 0.250)} = \frac{546.18}{1.500} = \boxed{364.1\ \text{veh}\cdot\text{s}}$$ The geometric route agrees exactly: $\tfrac12(73.33)(9.93) = 364.1$. Over the 45 cycles in an hour that is 4.55 vehicle-hours of delay on this approach alone.
Part (g) — average delay per vehicle. Divide the total delay by the number of vehicles arriving in a cycle, $n = qC/3600 = 650(80)/3600 = 14.44$ veh: $$\bar d = \frac{D}{n} = \frac{364.1}{14.44} = \boxed{25.2\ \text{s/veh}}$$ Note this averages over all arrivals, including the 8.3 per cent that arrive after clearance and are not delayed at all.
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 worst-off vehicle is therefore the first arrival after the onset of red, and it waits the whole effective red: $$d_{max} = r = \boxed{55\ \text{s}}$$ A numerical scan over every arrival instant in the cycle confirms the maximum is 55.0 s at $t = 0$. For contrast, the last vehicle to join the queue (the one arriving at the end of red) waits only $r - r(1 - q/s) = r\,q/s = 55(0.250) = 13.75$ s.

Question 5 — final results
Quantity	Symbol	Result
(a) Approach capacity	$c = s\,g/C$	812.5 veh/h > 650 veh/h; equivalently q/s = 0.250 < g/C = 0.3125
(a) Degree of saturation	$x$	0.800
(b) Time to queue clearance	$t_0$	18.33 s after the start of green
(c) Proportion of cycle with a queue	$P_q$	0.917
(d) Proportion of vehicles stopped	$P_s$	0.917 — algebraically identical to (c)
(e) Maximum queue	$Q_{max}$	9.93 veh (about 64.6 m of storage)
(f) Total delay per cycle	$D$	364.1 veh·s (4.55 veh·h per hour)
(g) Average delay per vehicle	$\bar d$	25.2 s
(h) Maximum delay of any vehicle	$d_{max}$	55 s — the first arrival after the onset of red

Topic. Deterministic (D/D/1) queueing at an undersaturated signal approach.

Key relations. $c = s\,g/C$; clearance $t_0 = qr/(s-q)$ from $s t_0 = q(r+t_0)$; $P_q = (r+t_0)/C \equiv P_s = s t_0/(qC)$; $Q_{max} = qr$ at the end of red; $D = qr^{2}/[2(1-q/s)]$; $\bar d = D/(qC)$; $d(t) = r - t(1-q/s)$ so $d_{max} = r$.

Why this works. With deterministic arrivals and deterministic service the cumulative curves are straight lines, so the whole cycle reduces to one triangle and every question about it is a question about that triangle’s base, height or area. The model is exact whenever $q/s < g/C$, because then the queue always clears within the green and no residual carries into the next cycle.

Common pitfall. Reporting the queue at clearance rather than at the end of red as the maximum, and looking for the maximum individual delay among the last arrivals. Because $d(t)$ falls with $t$, the first vehicle after the onset of red waits longest — 55 s here against only 13.75 s for the last vehicle to join the queue. Also: check $q/s < g/C$ before starting, since a failed test means the single-cycle triangle never closes and an overflow term is needed.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 6 (fundamental principles of traffic flow: deterministic queueing) and Ch. 8 (signal delay); TRB Highway Capacity Manual, uniform-delay term.

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