16-Civ-B10 Traffic Engineering · May 2014

Question 6 of 6: Peak-hour factor, controller types and a coordinated timing plan

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 6: Peak-hour factor, controller types and a coordinated timing plan (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. Ten consecutive 15-minute counts on one approach, with passenger-car equivalents of 1.5 for a straight-through truck, 1.5 for a right turn and 2.5 for a left turn:

Interval	Left turns	Right turns	ST trucks	ST cars	Flow (pcu)
4:00–4:15	5	10	6	30	66.5
4:15–4:30	6	15	8	26	75.5
4:30–4:45	4	7	10	35	70.5
4:45–5:00	7	16	8	40	93.5
5:00–5:15	10	13	6	49	102.5
5:15–5:30	9	12	12	55	113.5
5:30–5:45	14	15	8	65	134.5
5:45–6:00	12	12	10	50	113.0
6:00–6:15	10	9	8	39	89.5
6:15–6:30	9	12	4	30	76.5

Find. (a) the peak hour, its volume, the peak-hour factor and the design flow rate; (b) a description of the three controller families; (c) a time–space diagram of a coordinated plan.

Approach. Convert each interval to passenger-car units, slide a four-interval window over the series to find the peak hour, then form the peak-hour factor from the busiest quarter within it. Parts (b) and (c) are descriptive and graphical.

Part (a) — convert each interval to passenger-car units. Each count is weighted by its equivalent: $$q_{15} = 2.5\,N_{LT} + 1.5\,N_{RT} + 1.5\,N_{TR} + 1.0\,N_{\text{car}} .$$ For the first interval, $2.5(5) + 1.5(10) + 1.5(6) + 30 = 12.5 + 15 + 9 + 30 = 66.5$ pcu. The full column is tabulated in the Given above; the largest single interval is 5:30–5:45 at 134.5 pcu.
Part (a) — slide a four-interval window to find the peak hour. The peak hour is the highest sum of four consecutive quarters, not the four highest quarters. The seven candidate windows total 306.0, 342.0, 380.0, 444.0, 463.5, 450.5 and 413.5 pcu, so the peak hour runs from 5:00 to 6:00 pm and $$\boxed{\;V = 102.5 + 113.5 + 134.5 + 113.0 = 463.5\ \text{pcu/h}\;}$$

Question 6(a) — the ten 15-minute counts in passenger-car units. The peak hour is the highest four consecutive intervals (5:00 to 6:00 pm) and the peak 15-minute period within it is 5:30 to 5:45.

Part (a) — form the peak-hour factor and the design flow rate. The peak-hour factor compares the hour's volume with the rate implied by its busiest quarter, $$PHF = \frac{V}{4\,V_{15}} = \frac{463.5}{4\,(134.5)} = \frac{463.5}{538.0} = 0.861,$$ and the design flow rate is the hourly volume inflated back to that short-term rate, $$\boxed{\;q_{\text{design}} = \frac{V}{PHF} = \frac{463.5}{0.861} = 538\ \text{pcu/h} \;\;(= 4 \times 134.5)\;}$$ A $PHF$ of 0.86 is typical of an urban approach with a pronounced commuter peak; a value near 1.0 would mean uniform demand across the hour and a value near 0.25 that all the traffic arrived in a single quarter. Capacity and signal-timing calculations use the 538 pcu/h design rate, not the 463.5 pcu/h average, because the intersection must survive the worst fifteen minutes rather than the mean.

Part (b) — pre-timed, semi-actuated and actuated control. A pre-timed (fixed-time) controller runs a fixed cycle length, a fixed phase sequence and fixed green intervals, repeating identically regardless of who is waiting. It needs no detection, is the cheapest to install and maintain, and is entirely predictable, which is what makes it the only controller that can be coordinated rigidly with its neighbours — the timing plan of Questions 1 to 3 is a pre-timed plan. Its weakness is that it serves the demand it was designed for and nothing else: a green runs its full length with no vehicles present, and demand that grows after the plan was set is served no better than before. Modern installations mitigate this with several time-of-day plans.

A semi-actuated controller places detectors on the minor street (and on the pedestrian pushbuttons) but none on the major street. The major-street green is the resting state and is held indefinitely; it is terminated only when a detector or pushbutton registers a call, after which the minor phase runs for the time actually needed and control returns to the major street. This suits a busy arterial crossed by a lightly used side road, gives the arterial the maximum possible green, and — importantly — keeps a fixed background cycle, so a semi-actuated signal can still be coordinated in a progression.

A fully actuated controller has detection on every approach. Every phase has a minimum green (to clear the vehicles already stored between the detector and the stop line), a unit extension or passage time that lengthens the green while vehicles keep arriving, and a maximum green that caps it; a phase with no call is skipped altogether. The cycle length therefore varies from cycle to cycle in response to real demand, which minimises delay at an isolated intersection with irregular or unbalanced flows. The cost is higher capital and maintenance (loops or video detection to keep working) and the loss of a fixed cycle, which makes a fully actuated signal difficult to hold in a coordinated system unless it is run in a coordinated-actuated mode with a fixed background cycle and a force-off point for each phase.

Part (c) — time–space diagram of a coordinated timing plan. A time–space diagram plots distance along the arterial vertically and time horizontally, showing each signal's green and red as a band on its own horizontal line. A vehicle travelling at constant speed is a straight line of slope equal to that speed; a coordinated plan is one in which a whole family of such lines — the through band — passes every signal on green. The illustration below uses the 75 s cycle and 34 s arterial green designed in Question 2, three signals spaced 400 m apart and a progression speed of 50 km/h (13.9 m/s), which requires each signal to be offset from its upstream neighbour by

$$t_{\text{offset}} = \frac{d}{v} = \frac{400}{13.9} = 28.8\ \text{s}.$$

Question 6(c) — time-space diagram for a simple progressive (flexible progressive) system. Green bands are offset by the 28.8 s travel time between signals, so the through band passes all three signals without stopping.

The band width here equals the arterial green, 34 s, giving an efficiency of $34/75 = 45\ \%$ — that is, 45 % of every cycle is usable by an uninterrupted platoon. Three points are worth making on the diagram in an exam answer. First, the offsets are progressive: each signal opens later than the one upstream by the travel time, so this is a simple progressive system rather than a simultaneous one (all offsets zero, used only when blocks are very short) or an alternate one (offsets of half a cycle, which fixes the progression speed at $2d/C$). Second, the band is limited by the shortest green on the route, so a single signal with a long side-street demand throttles the whole arterial. Third, a two-way street needs the band to work in both directions at once, which is possible only when the block spacing and the cycle length are compatible; where they are not, the designer must favour the peak direction, and the resulting inbound-morning / outbound-evening plan pair is the normal outcome of a coordination study.

Final Results

Part	Quantity	Result
(a)	Peak hour	5:00 to 6:00 pm
(a)	Peak hour volume $V$	463.5 pcu/h
(a)	Peak 15-minute flow $V_{15}$	134.5 pcu (5:30 to 5:45)
(a)	Peak hour factor	0.861
(a)	Design (actual) flow rate	538 pcu/h
(b)	Controller families	Pre-timed (fixed cycle, no detection, coordinatable); semi-actuated (minor-street detection, arterial rests in green); fully actuated (all approaches detected, variable cycle)
(c)	Offset for 400 m at 50 km/h	28.8 s per signal
(c)	Through-band width / efficiency	34 s / 45 %

Topic. Volume studies and the peak-hour factor; traffic-controller families; progression and the time–space diagram.

Key relations. Passenger-car conversion $q = \sum E_i N_i$; peak hour = maximum of four consecutive 15-minute counts; $PHF = V/(4 V_{15})$; design flow $q_{\text{design}} = V/PHF = 4 V_{15}$; offset $t_{\text{offset}} = d/v$; band efficiency = band width / cycle length.

Why this works. Capacity is a rate, but demand is measured as a count over an interval, and the two only agree if the demand is uniform. The peak-hour factor is the correction between them: it is exactly the ratio of the average rate over the hour to the rate in the busiest quarter, so dividing by it recovers the short-term rate that the intersection actually has to survive. The identity $V/PHF = 4V_{15}$ is not a coincidence but a restatement of the definition, and it is the quickest arithmetic check available on part (a). In the coordination problem the geometry is just as direct: the offset is the travel time, and a platoon released at one stop line arrives at the next exactly $d/v$ later, so setting the offsets to $d/v$ is what makes the green bands line up along a straight trajectory.

Common pitfall. Picking the four largest intervals rather than the four consecutive ones — here that would wrongly give 5:00–5:15, 5:15–5:30, 5:30–5:45 and 5:45–6:00 the same answer by luck, but on many data sets it does not. A second pitfall is applying the passenger-car equivalents only to the trucks and forgetting that the question also makes right turns 1.5 and left turns 2.5, which changes the peak-hour volume by nearly 20 %. In part (c), a diagram with all the green bands vertically aligned describes a simultaneous system, not a coordinated progression.

Source. Garber & Hoel, 5th ed., §4.3 (volume studies and the peak-hour factor) and §8.5–8.6 (controller types, progression, time–space diagrams); ITE Traffic Engineering Handbook, signal-system design; TAC MUTCDC, Part B, signal coordination.

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