16-Civ-B10 Traffic Engineering · December 2019

Question 5 of 7: The same intersection with 10 % lower saturation flows and a 20 % higher PHF

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 5: The same intersection with 10 % lower saturation flows and a 20 % higher PHF (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.

Check: the question number in the stem. The paper prints "Repeat question 1", but Question 1 is the crest-vertical-curve problem, which contains neither saturation flow rates nor a peak-hour factor. It is therefore read as a typographical error for "question 2", and this is declared under the paper’s Note 1. Everything below repeats the Question 2 design with the two stated changes.

Given. Exactly the Question 2 data, with two changes. The peak-hour factor rises 20 % from 0.75 to 0.90, and every saturation flow falls 10 %.

Quantity	Question 2	Question 5
Peak-hour factor	0.75	0.90
Left-through saturation flow	2015 vphpl	1813.5 vphpl
Through-right saturation flow	2250 vphpl	2025.0 vphpl
Lost time per phase	3.5 s	3.5 s
All-red per phase	1.5 s	1.5 s
Approach volumes, pedestrians, widths	unchanged

Find. The redesigned cycle and phase lengths, and an explanation of how the two changes act on the cycle length.

Approach. Rather than re-running the whole design blind, exploit the structure: both changes act on every flow ratio by the same constant factor, so the lane balance, the critical approaches and the green split are provably unchanged and only Y moves. Then decompose the answer into four rows — base, each change alone, and both together — because the question asks how the changes affect the cycle, not merely what the new cycle is.

Both changes scale every flow ratio by the same factor. Each flow ratio is $y = q/s = (V/\text{PHF})/s$. Multiplying the PHF by 1.20 and every saturation flow by 0.90 gives $$y' = \frac{V/(1.20\,\text{PHF})}{0.90\,s} = \frac{1}{1.20 \times 0.90}\,y = 0.925926\,y$$ Because the factor is common to all four approaches, the lane-balance equation is unchanged (the constant cancels on both sides), the same approaches stay critical, and the green split $y_i/Y$ is provably identical to Question 2. A redesign that produces a different split has an arithmetic error in it, not a result.
New flow ratios and the new sum. Scaling the Question 2 values: $y_N = 0.3242$, $y_S = 0.3141$, $y_E = 0.2663$, $y_W = 0.2446$, so north and east remain critical and $$Y' = 0.925926 \times 0.6378 = \boxed{0.5905}$$
New vehicle optimum. The lost time is unchanged at 10.0 s, so $$C_{o}' = \frac{1.5(10.0)+5}{1-0.5905} = 48.84\ \text{s}$$ against 55.21 s before — the vehicle optimum falls by 6.4 s.
The pedestrian bound does not move at all. The pedestrian inequality depends on the crosswalk lengths, the walking speed, the pedestrian volumes, the lost time and the green split — and every one of those is unchanged. The bound is therefore still 39.96 s for the north–south phase and 49.67 s for the east–west phase, so $C_{ped} = 49.67$ s exactly as before. This is the fact that decides the whole question.
The governing control flips. In Question 2 the vehicle optimum (55.21 s) exceeded the pedestrian bound (49.67 s). Now 48.84 s falls below 49.67 s, so $$C' = 5\left\lceil \frac{\max(48.84,\ 49.67)}{5} \right\rceil = \boxed{50\ \text{s}}$$ and the design has become pedestrian-controlled. The cycle drops from 60 s to 50 s, a reduction of 16.7 %.
New phase lengths. With $C' = 50$ s the green available is $50 - 10 = 40.0$ s, split in the unchanged proportion 0.5490 / 0.4510: $$g_{NS} = 21.96\ \text{s}, \qquad g_{EW} = 18.04\ \text{s}$$ giving displayed greens of 25.46 s and 21.54 s, each followed by 1.5 s of all-red. The pedestrian intervals now required are 16.64 s and 17.90 s, against 21.96 s and 18.04 s of effective green: both still pass, but the east–west margin has shrunk to 0.14 s, which is the signature of a design sitting exactly on its pedestrian constraint. The critical degree of saturation is $x_{crit} = 0.5905(50)/40 = 0.7381$, slightly better than the 0.7653 of Question 2, and Webster delay falls to 15.8 s/veh on the critical north lane and 18.8 s/veh on the critical east lane.
Decompose the effect — this is the answer to "how do these changes affect the cycle length". The two changes pull in opposite directions, so a single new number would hide the mechanism. Running each alone:

Question 5 — four-row decomposition of the cycle
Case	Y	Vehicle C_o	Pedestrian bound	Adopted C	Governing control
Base (Question 2)	0.6378	55.21 s	49.67 s	60 s	vehicle
Saturation flows −10 % only	0.7086	68.64 s	49.67 s	70 s	vehicle
PHF +20 % only	0.5315	42.69 s	49.67 s	50 s	pedestrian
Both changes (Question 5)	0.5905	48.84 s	49.67 s	50 s	pedestrian

Question 5: which control binds in each case, and the cycle it demands. The saturation-flow loss alone pushes the cycle up to 70 s; the peak-hour-factor rise alone pulls it down to 50 s and hands control to the pedestrians.

The decomposition tells the whole story. Losing 10 % of saturation flow is a pure capacity loss: Y rises to 0.7086 and, because $C_o = (1.5L+5)/(1-Y)$ is hyperbolic in Y, the optimum leaps from 55.21 s to 68.64 s and the cycle would have to be lengthened to 70 s. A 20 % higher peak-hour factor is the opposite — a flatter peak means the same hourly volume arrives more evenly, so the flow rate the signal must serve drops by a sixth, Y falls to 0.5315 and the vehicle optimum collapses to 42.69 s. Acting alone, the second change would hand control to the pedestrians and stop the cycle falling below 49.67 s. Acting together, the peak-hour improvement is the larger of the two effects, so the net movement is downward: 60 s becomes 50 s, and the binding constraint changes from the vehicles to the pedestrians.

Two consequences deserve stating explicitly. First, once the design is pedestrian-controlled, any further gain in vehicle capacity buys nothing at all — it only lowers an already non-binding optimum. If the cycle needed to fall below 50 s the engineering answer would be a wider crosswalk or a median refuge that shortens $L_c$, not more green. Second, the intersection is comfortably operating well within capacity at $x_{crit} = 0.7381$, so the shorter cycle is a genuine improvement: average delay falls on both critical approaches, and shorter cycles reduce the maximum wait a pedestrian or a side-street driver experiences.

Question 5: the redesigned 50 s cycle. The east–west phase now clears its pedestrians with only 0.14 s to spare — the visual signature of a pedestrian-controlled design.

Question 5 — final results
Quantity	Question 2	Question 5
Flow-ratio scale factor	—	0.925926
Sum of critical flow ratios Y	0.6378	0.5905
Webster optimum cycle	55.21 s	48.84 s
Pedestrian minimum cycle	49.67 s	49.67 s (unchanged)
Governing control	vehicle	pedestrian
Adopted cycle	60 s	50 s (−16.7 %)
Effective green, N–S / E–W	27.45 / 22.55 s	21.96 / 18.04 s
Displayed green, N–S / E–W	30.95 / 26.05 s	25.46 / 21.54 s
Green split (unchanged by construction)	0.5490 / 0.4510	0.5490 / 0.4510
Critical degree of saturation	0.7653	0.7381
Webster delay, critical N / E lane	17.6 / 21.1 s/veh	15.8 / 18.8 s/veh

Topic. Sensitivity of a Webster design to saturation flow and peak-hour factor, and the complementarity of the vehicle and pedestrian constraints.

Key relations. $y = (V/\text{PHF})/s$, so a uniform change scales every flow ratio identically; $C_o = (1.5L+5)/(1-Y)$ is hyperbolic in Y; the pedestrian bound depends on $L_c, S_p, v_{ped}, L$ and the green split only; adopted $C = \max(C_o, C_{ped})$.

Why this works. The adopted cycle is a maximum of two independent bounds, so a change pays only when it relieves the bound that is currently binding. Scaling every saturation flow by a common factor cancels out of the lane-balance equation, which is why the split survives untouched and only Y and $C_o$ move; the pedestrian bound does not involve saturation flow or PHF at all and therefore cannot move. That structure is what makes the four-row decomposition the right presentation: it exposes both the direction of each effect and the point at which control changes hands, neither of which is visible in the single new number.

Common pitfall. The commonest error is recomputing the green split and getting a different answer — it cannot change here, and a different split means the lane balance was mis-solved. The second is checking only the combined case: a candidate who reports "the cycle falls to 50 s" without decomposition misses that the saturation loss alone would have raised it to 70 s, which is the more interesting half of the answer. The third is assuming a higher PHF means more traffic; a higher PHF means a flatter peak and therefore a lower design flow rate.

Source. Webster & Cobbe, Traffic Signals, RRL Technical Paper 56; Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 8 §8.4–8.5; TRB, Highway Capacity Manual, pedestrian minimum green.

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