16-Civ-B10 Traffic Engineering · December 2018

Question 7 of 7: Repeat Question 3 with three changes (20 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 7: Repeat Question 3 with three changes (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. Every input of Question 3, modified by three factors:

The three stated changes, resolved into their effect on each input
Input	Question 3	Question 7	Factor
Left-through saturation flow	1775 vphpl	1952.5 vphpl	×1.10
Through-right saturation flow	1895 vphpl	2084.5 vphpl	×1.10
Governing E–W pedestrian flow	1200 ped/h	1164 ped/h	×0.97
Governing N–S pedestrian flow	905 ped/h	878 ped/h	×0.97
Crosswalk length, N–S phase	18 m	10 m	stated
Crosswalk length, E–W phase	15 m	10 m	stated
Turning volumes, PHF, lost time, all-red	unchanged	unchanged	—

Find. The redesigned cycle and phase lengths, and an explanation of how the three changes move the cycle length.

Approach. Do not simply recompute and report a number. A uniform change in the saturation flows and a change in the pedestrian demand act on different constraints, so the honest answer is a four-row decomposition — base, each change alone, and both together — naming which control binds in each row. That is what identifies the mechanism.

A uniform saturation-flow factor leaves the lane split untouched. Substituting $s_1 \to f s_1$ and $s_2 \to f s_2$ into the lane-balance solution, $$x = \frac{f s_1 (v_T + v_R) - f s_2 v_L}{f s_1 + f s_2} = \frac{s_1 (v_T + v_R) - s_2 v_L}{s_1 + s_2}$$ — the factor cancels exactly. So every through split is unchanged (South still 294.52, East still 433.42 veh/h), the same approaches remain critical, and every flow ratio simply scales by $1/f$. A changed green split in this question would be an arithmetic error, not a result.
New flow ratios and Y. Dividing by 1.10, $y_S = 0.3969/1.10 = 0.36083$ and $y_E = 0.3781/1.10 = 0.34376$, so $$Y_7 = 0.36083 + 0.34376 = \boxed{0.7046} = \frac{0.7750}{1.10}$$ and the split $y_S/Y = 0.51211$ is provably identical to Question 3.
New Webster optimum. $$C_{o,7} = \frac{1.5(10.0) + 5}{1 - 0.7046} = \frac{20.0}{0.2954} = \boxed{67.70\ \text{s}}$$ a fall of 21.2 s, or 23.9 per cent, from 88.91 s. The 10 per cent capacity gain buys a disproportionately large reduction because $C_o$ is hyperbolic in $Y$.
New pedestrian bound. Two of the three changes act here, and both act downward: the crosswalks shorten from 18 m and 15 m to 10 m, cutting $a = 3.2 + L_c/S_p$ to $3.2 + 10/1.2 = 11.533$ s on both phases, and the pedestrian volumes fall 3 per cent. Because the green split is unchanged, $k$ is unchanged too, so for the N–S phase $b = 2.7(1164)/[3600(4.0)] = 0.21825$ and $$C \ge \frac{11.533 + 0.51211(10.0)}{0.51211 - 0.21825} = \frac{16.654}{0.29386} = 56.67\ \text{s}$$ against 50.77 s for the E–W phase. The governing pedestrian bound is $\boxed{56.67\ \text{s}}$, down 30.2 per cent from 81.23 s.
Adopt the new cycle. The vehicle bound of 67.70 s again exceeds the pedestrian bound of 56.67 s, so control remains vehicular and $$C_7 = \boxed{70\ \text{s}}$$ down from 90 s — a reduction of 20 s, or 22.2 per cent.
New phase lengths. With $g_{tot} = 70 - 10 = 60.0$ s and the unchanged split, $$g_{NS} = 0.51211(60.0) = \boxed{30.73\ \text{s}},\qquad g_{EW} = 0.48789(60.0) = \boxed{29.27\ \text{s}}$$ so the displayed greens are $30.73 + 3.5 = \boxed{34.23}$ s and $29.27 + 3.5 = \boxed{32.77}$ s, and $34.23 + 1.5 + 32.77 + 1.5 = 70.0$ s closes the cycle.
Pedestrian and capacity checks at C = 70 s. The intervals needed are $G_p = 3.2 + 8.333 + 2.7(1164 \times 70/3600)/4.0 = 26.81$ s against 30.73 s available, and $23.06$ s against 29.27 s — both pass. The new critical-lane capacities are $c_{NS} = 1952.5(30.73)/70 = 857.05$ and $c_{EW} = 1952.5(29.27)/70 = 816.52$ veh/h, and both critical lanes again land on the same $x_{crit} = 0.7046(70)/60 = \boxed{0.8220}$, down from 0.8719. The Webster delays fall to 23.4 s and 24.6 s per vehicle, from 32.5 s and 34.1 s — a 28 per cent improvement, and a move from LOS C–D to solid LOS C.
Now decompose the effect, which is what the question actually asks. Run each change on its own and name the binding control.

The four-row decomposition: which constraint binds, and the cycle it demands
Case	Vehicle optimum C_o (s)	Pedestrian bound (s)	Governing control	Adopted cycle (s)
Base case (Question 3)	88.91	81.23	vehicle	90
Saturation flows +10 per cent only	67.70	81.23	pedestrian	85
Pedestrians −3 per cent and 10 m walks only	88.91	56.67	vehicle	90
All three changes together (Question 7)	67.70	56.67	vehicle	70

The decomposition answers the question properly, and the answer is more interesting than the single number. Neither change on its own delivers the benefit. Raising the saturation flows by 10 per cent drops the vehicle optimum from 88.91 s to 67.70 s, but at that point the pedestrians take over as the binding constraint at 81.23 s, so the cycle falls only from 90 s to 85 s — five seconds out of a possible twenty. Conversely, shortening the crosswalks to 10 m and cutting the pedestrian volumes by 3 per cent slashes the pedestrian bound from 81.23 s to 56.67 s, but the vehicle optimum was already binding at 88.91 s, so the cycle does not move at all: it stays at 90 s. Only together do the two changes work, because each one removes the constraint that would otherwise block the other. Taken jointly the cycle falls the full 20 s to 70 s, the degree of saturation falls from 0.872 to 0.822, and the average delay falls by about 28 per cent.

Of the two pedestrian changes, the geometry does nearly all the work. Shortening the crosswalks to 10 m alone would bring the pedestrian bound to 58.01 s, whereas the 3 per cent volume cut alone would bring it only to 79.36 s. That is the expected ranking: the crosswalk length enters the pedestrian green as $L_c/S_p$, a fixed 6.7 s saving here, whereas the volume enters through $2.7N_{ped}/W_E$, and 3 per cent of it is small.

Question 7. The four cases side by side. The red tick on each row is the adopted cycle. Row 2 is the instructive one: the capacity gain alone is throttled because the pedestrian bound takes over as the binding constraint.

Question 7. Adopted timing for the 70 s cycle. The green split is provably identical to Question 3, because a uniform saturation-flow factor cancels out of the lane balance.

Check: the 10 m approach width. The question reduces every approach to 10 m, which will not hold four 3.5 m lanes ($10/4 = 2.5$ m each). Two readings are possible and the arithmetic settles it. If 10 m meant one lane per approach, the shared left-through-right saturation flow of 2366 vphpl, even after the 10 per cent uplift, gives $Y = 1.093 > 1$ — infeasible at any cycle length. The four-phase plan is likewise still infeasible at $Y = 1.348$. So the two-lane-per-approach layout of Question 3 must be retained, and the 10 m figure enters the calculation only where a width belongs: as the crosswalk length $L_c$. That is also the natural reading of the question, which changes no volume, no lane type and no saturation-flow classification. It is noted for completeness that 2.5 m lanes are substandard against the TAC Geometric Design Guide (3.0–3.7 m for an urban arterial) and would only be acceptable as a temporary or severely constrained cross-section.

Engineering recommendation. A 70 s cycle with $x_{crit} = 0.822$ is a comfortable, well-proportioned design, and it sits well inside the roughly 120 s practical maximum used in Canadian urban practice. The lesson for the client is the row-2 result: buying capacity without also shortening the pedestrian crossing wastes three-quarters of it. If the 10 m cross-section is not achievable, the same effect can be bought by widening the crosswalks instead — because the HCM term is $N_{ped}/W_E$, a given percentage increase in crosswalk width cancels exactly the same percentage increase in pedestrian volume.

Question 7 — final results
Quantity	Symbol	Result
Uplifted saturation flows	$s_{LT},\ s_{TR}$	1952.5 and 2084.5 vphpl
Flow ratios (scaled by 1/1.10)	$y_S,\ y_E$	0.36083 and 0.34376
Sum of critical flow ratios	$Y_7$	0.7046 (exactly Y/1.10)
Green split	$y_S/Y$	0.51211 / 0.48789 — provably unchanged
Webster optimum cycle	$C_{o,7}$	67.70 s (was 88.91 s, −23.9 per cent)
Pedestrian minimum cycle	$C_{ped,7}$	56.67 s (was 81.23 s, −30.2 per cent)
Governing control	—	vehicle (67.70 s > 56.67 s)
Adopted cycle length	$C_7$	70 s, down 20 s (−22.2 per cent) from 90 s
Effective greens	$g_{NS},\ g_{EW}$	30.73 s and 29.27 s
Displayed greens	$G_{NS},\ G_{EW}$	34.23 s and 32.77 s, plus 1.5 s all-red each
Pedestrian intervals delivered	$G_p$	26.81 s of 30.73 s; 23.06 s of 29.27 s — both pass
Critical-lane capacities	$c$	857.05 and 816.52 veh/h
Critical degree of saturation	$x_{crit,7}$	0.8220 (was 0.8719)
Webster average delay	$d$	23.4 s and 24.6 s (was 32.5 s and 34.1 s)
How the changes act	—	saturation alone: 90 → 85 s only (pedestrians take over); pedestrians and width alone: no change at all; together: 90 → 70 s

Topic. Sensitivity of a Webster design to saturation flow, pedestrian demand and crosswalk geometry.

Key relations. A uniform saturation factor $f$ cancels out of the lane balance, so $y_i \to y_i/f$, $Y \to Y/f$ and the split $y_i/Y$ is invariant; $C_o = (1.5L+5)/(1-Y)$ is hyperbolic in $Y$; the pedestrian bound $C \ge (a + kL)/(k - b)$ has $a = 3.2 + L_c/S_p$ linear in crosswalk length and $b = 2.7v_{ped}/(3600W_E)$ linear in pedestrian volume; and the adopted cycle is the larger of the two bounds.

Why this works. The two bounds respond to different inputs and by different laws, which is why a change can be entirely wasted. A capacity gain moves only $C_o$, and once $C_o$ has fallen below the pedestrian bound, further capacity buys nothing. A pedestrian or geometry improvement moves only the pedestrian bound, and while the vehicle bound is the larger, it buys nothing either. Because the adopted cycle is a maximum of the two, benefit accrues only when the currently binding bound is the one being relieved — which is exactly what the four-row decomposition exposes.

Common pitfall. Recomputing $C_o$, reporting the new cycle and stopping. That answers “what is the new cycle” but not “how do these changes affect it”, and it misses the fact that either change alone delivers little or nothing. The second pitfall is recomputing the green split and getting a different answer: a uniform saturation factor provably cannot change it.

Source. Webster & Cobbe, Traffic Signals, RRL Technical Paper No. 56; Garber & Hoel, 5th ed., Ch. 8; TRB Highway Capacity Manual, pedestrian minimum green and critical-movement analysis; TAC Geometric Design Guide for Canadian Roads, Part 2 (urban lane widths).

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