Question 6 of 6: Peak-Hour Volume, Peak-Hour Factor and Design Flow Rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2014 — 98-Civ-B10 Traffic Engineering, 3-hour duration, OPEN BOOK (any non-communicating calculator permitted). Six questions are set; the paper requires a total of five solutions and states that all questions are of equal value (20 marks each, split as shown in the paper's own grading scheme). Because the paper is a study resource, all six questions are solved below. The paper's NOTE 1 invites a clear statement of any assumptions made, and NOTE 2 states that any data required but not given may be assumed — every assumed factor is therefore declared explicitly where it is first used.
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, spot-speed studies), Ch. 6 (fundamental principles of traffic flow: the two mean speeds, the Greenshields model, Poisson arrivals, negative-exponential headways, single- and multi-channel queueing) and Ch. 8 (intersection control: saturation flow, change and clearance intervals, cycle length). This is the principal reference for the subject. Transportation Research Board, Highway Capacity Manual — saturation-flow adjustment factors and the passenger-car-equivalent convention. Transportation Association of Canada, Geometric Design Guide for Canadian Roads and Manual of Uniform Traffic Control Devices for Canada (MUTCDC) — Canadian lane widths, crosswalk practice and clearance-interval policy. Institute of Transportation Engineers, Traffic Engineering Handbook — queueing applications to parking and drive-up facilities. Webster, F. V. and Cobbe, B. M., Traffic Signals, Road Research Laboratory Technical Paper No. 56 — the optimum-cycle relation quoted as a closing check in Question 5.
Given. Ten consecutive 15-minute counts on one intersection approach, disaggregated into left turns, right turns, straight-through trucks and straight-through passenger cars, with passenger-car equivalents of 2.5 for a left turn and 1.5 for both a right turn and a truck.
Interval
LT
RT
ST trucks
ST cars
Total veh
Equivalent (pcu)
5:00–5:15
6
15
8
26
55
75.5
5:15–5:30
5
10
6
30
51
66.5
5:30–5:45
7
16
8
40
71
93.5
5:45–6:00
4
7
10
35
56
70.5
6:00–6:15
9
12
12
55
88
113.5
6:15–6:30
10
13
6
49
78
102.5
6:30–6:45
12
12
10
50
84
113.0
6:45–7:00
14
15
8
65
102
134.5
7:00–7:15
9
12
4
30
55
76.5
7:15–7:30
10
9
8
39
66
89.5
Find. (a) the peak hour volume, (b) the peak-hour factor and (c) the design flow rate for the approach, all expressed in the passenger-car equivalents the question supplies.
The ten 15-minute equivalent volumes. The dashed window is the peak hour identified by the moving four-interval sum; the red bar is the peak 15-minute count within it, which sets the design flow rate.
Approach. Convert each 15-minute count into passenger-car equivalents, slide a four-interval window across the record to locate the peak hour, form the peak-hour factor from the peak hour and its own busiest quarter, and expand that quarter to an hourly rate to obtain the design flow.
Part (a) — convert each interval to passenger-car equivalents. Applying the stated equivalents,
$$V_{15} = 2.5(\text{LT}) + 1.5(\text{RT}) + 1.5(\text{trucks}) + 1.0(\text{cars})$$
For the 6:45–7:00 interval, for example, $2.5(14) + 1.5(15) + 1.5(8) + 1.0(65) = 35 + 22.5 + 12 + 65 = 134.5$ pcu. The full column is tabulated above.
Slide a four-interval window to find the peak hour. The peak hour is the highest sum of any four consecutive intervals, not the four highest intervals. The seven candidate hours give
$$306.0,\; 344.0,\; 380.0,\; 399.5,\; 463.5,\; 426.5,\; 413.5\ \text{pcu/h}$$
for the hours beginning 5:00, 5:15, 5:30, 5:45, 6:00, 6:15 and 6:30 respectively. The largest is the fifth, so the peak hour is 6:00–7:00 and
$$V = 113.5 + 102.5 + 113.0 + 134.5 = \boxed{463.5\ \text{pcu/h}}$$
Part (b) — identify the busiest quarter inside the peak hour. Of the four intervals in the peak hour, the busiest is 6:45–7:00 at $V_{15,\max} = 134.5$ pcu. The peak-hour factor compares the hourly volume with what the hour would have carried had it sustained that busiest quarter throughout:
$$PHF = \frac{V}{4\,V_{15,\max}} = \frac{463.5}{4(134.5)} = \frac{463.5}{538.0} = \boxed{0.8615}$$
A PHF of 0.86 indicates moderate but real peaking within the hour — typical of an urban intersection approach, where values commonly run between 0.85 and 0.95.
Part (c) — expand the peak quarter to an hourly design rate. The design (actual) flow rate is the rate at which traffic actually arrived during the busiest 15 minutes, expressed as an hourly rate:
$$v = 4\,V_{15,\max} = 4(134.5) = \boxed{538\ \text{pcu/h}}$$
Equivalently and as a check, $v = V/PHF = 463.5/0.8615 = 538$ pcu/h. The design rate exceeds the peak-hour volume by 74.5 pcu/h, or 16 %; designing the approach to the hourly average would leave it 16 % short for a quarter of an hour every evening, which is exactly the queue-building period that governs whether the approach fails.
Cross-check on the raw vehicle counts. Repeating the exercise on unweighted vehicles gives the same peak hour (6:00–7:00) and the same peak quarter (6:45–7:00), with $V = 352$ veh/h, $V_{15,\max} = 102$ veh, $PHF = 352/408 = 0.8627$ and a design rate of 408 veh/h. That the two routes agree on which hour and quarter are critical, and on the PHF to within 0.001, confirms that the turning and truck mix does not shift the peak; the equivalents change the magnitude of the answer, not its timing.
Check: the question supplies passenger-car equivalents immediately before parts (a) to (c), so all three answers are reported in pcu/h — the units in which an approach is actually designed. The unweighted vehicle figures (352 veh/h, PHF 0.8627, 408 veh/h) are given in step 5 for completeness in case the equivalents are read as applying only to part (c).
