16-Civ-B10 Traffic Engineering · May 2014

Question 2 of 6: Cycle length, green splits, pedestrian check, capacity and delay

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

Notes on this paper

Paper format. National Examinations, May 2014 — 98-Civ-B10 Traffic Engineering. Three-hour, open-book examination; any non-communicating calculator is permitted. Six questions are printed and five complete solutions are required, all questions being of equal value (20 marks each). The printed grading scheme is Q1 (a) to (d) 5 marks each; Q2 (a) to (e) 4 marks each; Q3 (a) to (e) 4 marks each; Q4 (a) and (b) 10 marks each; Q5 (a) to (e) 4 marks each; Q6 (10 + 5 + 5) marks. The paper states that if doubt exists as to the interpretation of a question the candidate should submit with the answer paper a clear statement of any assumptions made, and that any data required but not given can be assumed. All six questions are worked below.

Reference texts. Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering, 5th ed. — Ch. 4 (traffic engineering studies: volume studies, peak-hour factor), Ch. 6 (fundamental principles of traffic flow: Poisson arrivals, deterministic and stochastic queueing, M/M/1 and M/M/N channels), Ch. 8 (intersection control: cycle length, phasing, change and clearance intervals, progression and time–space diagrams) and Ch. 10 (capacity and level of service at signalised intersections). This is the principal reference for the subject. Webster, F. V. and Cobbe, B. M., Traffic Signals, Road Research Laboratory Technical Paper No. 56 — the optimum-cycle and three-term delay formulae used throughout Questions 1 to 3. Transportation Research Board, Highway Capacity Manual — saturation-flow adjustment factors and the pedestrian-green requirement. Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) and Geometric Design Guide for Canadian Roads — Canadian signal-timing, clearance-interval and crosswalk practice. Institute of Transportation Engineers, Traffic Engineering Handbook — controller types and coordinated timing plans.

Check: the four values assumed under the paper's own NOTE 2 (“any data required, but not given, can be assumed”). They are declared here once and used consistently in every question.

Bus passenger-car equivalent $E_B = 2.0$ pcu (HCM value for buses on level terrain). The paper gives bus occupancies but no pcu equivalent; note that the pcu-based and vehicle-based flow ratios in Question 1 agree exactly for any choice of $E_B$, so this assumption does not affect Questions 2 or 3.
Pedestrian walking speed $S_p = 1.2$ m/s (HCM 2000 / MUTCDC design value; the more recent 1.1 m/s would lengthen the pedestrian interval by about 1 s).
Lost time convention: the whole intergreen (amber + all-red) is taken as lost, so effective green equals displayed green. This is the conservative reading and needs no assumed start-up lost time. Webster's alternative, $L = n\ell + R$ with $\ell \approx 2$ s, would give $L = 10$ s instead of 12 s and lengthen each green by about 1 s.
Delay model: Webster's three-term formula, with the first two terms reported separately in Question 3(b) as the uniform and overflow components.

Question 2: Cycle length, green splits, pedestrian check, capacity and delay (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. Carried forward from Question 1:

Quantity	Value
Critical flow ratios $y_{NS}$ / $y_{EW}$	0.2306 (NB) / 0.1944 (WB)
Intersection flow ratio $Y$	0.425
Intergreen $I$ / lost time $L$	6 s per phase / 12 s per cycle
Pedestrian crossing distance $L_c$	$4 \times 3.75 = 15.0$ m, no refuge
Crosswalk width $W_E$ / pedestrian demand	3.0 m / 140 per hour per crosswalk
Walking speed $S_p$	1.2 m/s (assumed, HCM design value)
Lane demands (pcu/h/lane)	NB 415, SB 362, EB 250, WB 350
Chosen cycle for parts (b) to (e)	$C = 75$ s

Find. The shortest cycle that still lets a pedestrian cross without a refuge; the two green intervals at $C = 75$ s; confirmation that those greens also serve the pedestrians; the per-lane capacity and degree of saturation on each approach; and the flow-weighted average delay per vehicle.

Approach. Size the pedestrian interval first, because it sets a floor on the cycle that the vehicular calculation cannot see; then split the available effective green in proportion to the critical flow ratios; then convert green time into capacity and capacity into delay with Webster's formula.

Part (a) — write the pedestrian green requirement. With no refuge the whole 15 m must be crossed in one movement. The HCM / MUTCDC requirement for a crosswalk not wider than 3.0 m is $$G_p = 3.2 + \frac{L_c}{S_p} + 0.27\,N_{\text{ped}},$$ where 3.2 s is the pedestrian start-up (perception and reaction) time, $L_c/S_p$ is the crossing time and the last term is the extra time a platoon needs to discharge from the kerb. Here $L_c/S_p = 15.0/1.2 = 12.5$ s and the platoon size depends on the cycle itself, $N_{\text{ped}} = 140\,C/3600$, so $$G_p = 3.2 + 12.5 + 0.27\left(\frac{140\,C}{3600}\right) = 15.7 + 0.0105\,C .$$
Part (a) — close the loop on the cycle length. Both crossings are 15 m long, so each of the two phases must carry one pedestrian interval, and each phase also carries its intergreen. Requiring the pedestrian interval to fit inside the displayed green gives $$C = 2\left(G_p + I\right) = 2\left(15.7 + 0.0105\,C\right) + 12 \;\Longrightarrow\; C\,(1 - 0.021) = 43.4,$$ $$\boxed{\;C_{\min,\text{ped}} = \frac{43.4}{0.979} = 44.3\ \text{s} \;\rightarrow\; 45\ \text{s}\;}$$ For comparison, Webster's optimum cycle for the vehicular demand is $C_o = (1.5L + 5)/(1 - Y) = (18 + 5)/0.575 = 40$ s, so on this intersection the pedestrians, not the vehicles, set the shortest workable cycle. Signal timings are normally rounded up to a multiple of 5 s, hence 45 s.
Part (b) — distribute the effective green in proportion to $y$. At $C = 75$ s the green available to traffic is what the lost time leaves: $$\sum g = C - L = 75 - 12 = 63\ \text{s}.$$ Webster's split gives each phase a share equal to its share of $Y$: $$g_i = \frac{y_i}{Y}\sum g \;\Rightarrow\; g_{NS} = \frac{0.2306}{0.425}(63) = 34.2\ \text{s},\qquad g_{EW} = \frac{0.1944}{0.425}(63) = 28.8\ \text{s}.$$ Rounding to whole seconds and checking that the cycle still closes, $$\boxed{\;G_{NS} = 34\ \text{s},\quad G_{EW} = 29\ \text{s},\quad 34 + 29 + 2(3 + 3) = 75\ \text{s}\;}$$

Question 2(b) — the resulting two-phase timing plan. Each phase carries a 3 s amber and a 3 s all-red, so 12 s of the 75 s cycle is lost time.

Part (c) — check the greens against the pedestrian requirement. Now that $C$ is fixed, the platoon term is definite: $N_{\text{ped}} = 140(75)/3600 = 2.92$ pedestrians per cycle, so $$G_p = 3.2 + 12.5 + 0.27(2.92) = 16.5\ \text{s}.$$ Pedestrians crossing the north and south legs walk on the E–W green (29 s) and those crossing the east and west legs walk on the N–S green (34 s). Both exceed 16.5 s, so $$\boxed{\;G_{EW} = 29\ \text{s} > 16.5\ \text{s and } G_{NS} = 34\ \text{s} > 16.5\ \text{s} \;\Rightarrow\; \text{satisfactory}\;}$$ The controller would be set up with a 7 s WALK and a 12.5 s flashing DON'T WALK on each phase, the clearance portion being allowed to run into the intergreen.
Part (d) — convert green time into lane capacity. A lane discharges at saturation only while it is green, so $$c = s_0\,\frac{g}{C}, \qquad x = \frac{q}{c}.$$ On the N–S phase $c = 1800(34/75) = 816$ pcu/h/lane and on the E–W phase $c = 1800(29/75) = 696$ pcu/h/lane. Dividing each lane demand by its capacity, $$\boxed{\;x_{NB} = \tfrac{415}{816} = 0.509,\quad x_{SB} = \tfrac{362}{816} = 0.444,\quad x_{EB} = \tfrac{250}{696} = 0.359,\quad x_{WB} = \tfrac{350}{696} = 0.503\;}$$ Every approach is around or below half capacity, so the intersection is comfortably within its useful range and the delay formula of part (e) is well inside its valid domain.
Part (e) — evaluate Webster's average delay. Webster's expression for the average delay per vehicle on an approach is $$d = \frac{C\,(1 - \lambda)^2}{2\,(1 - \lambda x)} + \frac{x^2}{2\,q\,(1 - x)} - 0.65\left(\frac{C}{q^{2}}\right)^{1/3} x^{\,2 + 5\lambda}, \qquad \lambda = \frac{g}{C},$$ with $q$ in vehicles per second. Taking the north-bound lane as the worked case, $\lambda = 34/75 = 0.453$, $x = 0.509$ and $q = 415/3600 = 0.1153$ veh/s give $14.57 + 2.28 - 0.65 = 16.20$ s/pcu. Repeating for the other three approaches gives 15.39 (SB), 17.54 (EB) and 19.26 (WB) s/pcu. Weighting by the approach demands in pcu/h, $$d_{\text{overall}} = \frac{\sum q_i d_i}{\sum q_i} = \frac{830(16.20) + 724(15.39) + 500(17.54) + 700(19.26)}{2754},$$ $$\boxed{\;d_{\text{overall}} = 17.0\ \text{s/pcu}\;}$$ That is comfortably inside the 20 s/veh boundary between level of service B and C for a signalised intersection, which is the right outcome for a plan whose cycle was set by the pedestrians rather than by congestion.

Final Results

Part	Quantity	Result
(a)	Pedestrian green requirement	$G_p = 15.7 + 0.0105\,C$ s
(a)	Minimum cycle for pedestrians	44.3 s → adopt 45 s (Webster optimum for traffic is only 40 s)
(b)	Effective green available	63 s
(b)	Green times at $C = 75$ s	N–S 34 s, E–W 29 s
(c)	Pedestrian requirement at $C = 75$ s	16.5 s — both greens satisfactory
(d)	Lane capacity, N–S / E–W	816 / 696 pcu/h/lane
(d)	Degree of saturation NB / SB / EB / WB	0.509 / 0.444 / 0.359 / 0.503
(e)	Delay NB / SB / EB / WB	16.20 / 15.39 / 17.54 / 19.26 s/pcu
(e)	Average overall delay	17.0 s/pcu (level of service B)

Topic. Webster cycle-length and green-split design, the pedestrian constraint on cycle length, and the capacity–delay relationship.

Key relations. Pedestrian green $G_p = 3.2 + L_c/S_p + 0.27 N_{\text{ped}}$; Webster optimum cycle $C_o = (1.5L + 5)/(1 - Y)$; green split $g_i = (y_i/Y)(C - L)$; capacity $c = s\,g/C$; degree of saturation $x = q/c$; Webster three-term delay.

Why this works. The green split is chosen so that every phase ends up with the same degree of saturation: if $g_i \propto y_i$ then $x_i = y_i C/g_i$ is the same on each phase, which is the condition that minimises total delay for a given cycle. The cycle length itself is a compromise — a long cycle wastes less time on the fixed 12 s of intergreen, but makes every red longer, and Webster's $C_o$ is the stationary point of that trade-off. The pedestrian requirement acts from the other side: it is a hard floor that has nothing to do with vehicular demand, and on a wide crossing with no refuge it frequently governs, exactly as it does here.

Common pitfall. Splitting the whole cycle rather than the cycle less lost time — that overstates both greens and makes the timing plan fail to close. The other common slip is checking the pedestrian interval against the green of the parallel movement: a pedestrian crossing the north leg is walking across the N–S traffic stream and therefore needs the E–W green, not the N–S one.

Source. Webster & Cobbe, Traffic Signals, RRL Technical Paper No. 56, §§3–5; Garber & Hoel, 5th ed., §8.4 (signal timing by the Webster method) and §10.5 (delay and level of service); HCM, pedestrian-timing requirements.

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