16-Civ-B10 Traffic Engineering · May 2015

Question 7 of 7: Peak Hour Factor from Interval Counts, and Spot-Speed Statistics

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

Notes on this paper

Paper format. 98-Civ-B10 Traffic Engineering, National Examinations, May 2015. Three-hour duration, OPEN BOOK, any non-communicating calculator permitted. Seven questions, all of equal value (20 marks each) with the mark split printed in the grading scheme; the paper requires a total of five solutions. All seven are worked below, because this document is a study resource rather than an exam script.

Reference texts. Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering — Ch. 4 (traffic-engineering studies: volume counts, peak-hour factor, spot-speed studies), Ch. 6 (fundamental principles of traffic flow: the Greenshields model, deterministic and stochastic queueing, the moving-observer method) and Ch. 8 (intersection control: saturation flow, change and clearance intervals). Transportation Research Board, Highway Capacity Manual — basic-freeway-segment and two-lane-highway level-of-service criteria and the lane-width, lateral-clearance and heavy-vehicle adjustment factors. Transportation Association of Canada, Geometric Design Guide for Canadian Roads and the Manual of Uniform Traffic Control Devices for Canada — Canadian lane widths, crosswalk practice and clearance-interval policy. Institute of Transportation Engineers, Traffic Engineering Handbook — queueing applied to parking and toll facilities.

Check — declared assumptions, invoked under the paper’s own NOTE 1 and NOTE 2 (“any data required, but not given, can be assumed”, with a clear statement of the assumption). This paper is open-book and deliberately withholds several standard table values. The following are used consistently throughout and each is restated where it is first applied:

Bus (heavy-vehicle) passenger-car equivalent on the level intersection approaches of Question 1: E_B = 2.0.
Two-phase signal operation at the Question 1 intersection (one phase for the north–south pair, one for the east–west pair), with no start-up lost time given, so the whole intergreen is taken as lost time.
Maximum service flow rates and volume-to-capacity ratios for basic freeway segments at a 100 km/h design speed: 700/1000/1450/1800/2000 pc/h/lane is the 110 km/h family, so the 100 km/h family 600/1000/1450/1800/2000 pc/h/lane (v/c = 0.30/0.50/0.72/0.90/1.00) is used, with an ideal capacity of 2000 pc/h/lane.
Two-lane-highway ideal capacity 2800 pc/h both directions, with level-terrain v/c ratios 0.15/0.27/0.43/0.64/1.00 at 0 % no-passing zones, and a 50/50 directional split (f_d = 1.00).
Passenger-car equivalents on specific grades: E_T = 5 for trucks on both the 4 % / 1.0 km grade and the 3 % / 1.5 km grade, E_R = 3 for recreational vehicles on the 4 % grade, E_B = 1.6 for intercity buses on grades of 4 % or flatter, and E_T = 2 for trucks in general level terrain.

Different editions of the capacity manual tabulate these cells slightly differently. Where an alternative reading changes an answer materially, the alternative is computed and stated alongside; the method and the arithmetic are what carry the marks.

Question 7: Peak Hour Factor from Interval Counts, and Spot-Speed Statistics (2 × 10 = 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. Part (a): twelve consecutive 5-minute counts through the peak hour, totalling 4100 vehicles. Part (b): a grouped spot-speed distribution of twelve 5 km/h classes with the frequencies tabulated above.

Find. (a) The peak hour factor on a 15-minute basis. (b) The mean speed, the standard deviation of the individual observations, the standard deviation of the mean, and the 90th-percentile speed read from the cumulative distribution.

Approach. Part (a) aggregates the 5-minute counts into 15-minute groups, identifies the busiest group and forms the ratio of the hourly volume to four times that group. Part (b) uses the grouped-data moment sums to get the mean and variance, divides by the root of the sample size for the standard error, and interpolates the cumulative curve at 90 %.

Part (a) — aggregate the 5-minute counts into 15-minute groups and confirm the total. Three consecutive 5-minute counts make one 15-minute interval: $$\begin{aligned} V_1 &= 400+300+200 = 900\\ V_2 &= 600+500+400 = 1500\\ V_3 &= 300+600+200 = 1100\\ V_4 &= 200+100+300 = 600 \end{aligned}$$ The four groups are therefore 900, 1500, 1100 and 600 vehicles, and they sum to $900+1500+1100+600 = 4100$, matching the stated hourly volume. The busiest 15 minutes is the second group, $$V_{15} = 1500 \text{ vehicles}$$
Form the peak hour factor. The peak hour factor compares the hourly volume with the hourly rate implied by the busiest 15 minutes: $$PHF = \frac{V}{4\,V_{15}} = \frac{4100}{4(1500)} = \frac{4100}{6000}$$ $$\boxed{\;PHF = 0.683\;}$$ Equivalently, the peak rate of flow is $4 \times 1500 = 6000$ veh/h, well above the hourly volume of 4100. A peak hour factor of 0.68 is very low — typical urban freeway values run 0.85 to 0.95 — and it says the demand is sharply spiked rather than sustained: the facility must be sized for 6000 veh/h even though only 4100 vehicles use it in the hour.
Check that the fixed grouping has not missed a busier window. Dividing the hour into four fixed blocks can in principle miss a peak that straddles a boundary, so the ten possible rolling three-interval windows are worth scanning: 900, 1100, 1300, 1500, 1200, 1300, 1100, 1000, 500, 600. The maximum is 1500, the same window already found, so the fixed grouping is safe here and $PHF = 0.683$ stands. Had a rolling window exceeded 1500, the question’s wording (“based on 15-minute interval”) would still point to the fixed grouping, but the discrepancy would be worth reporting.

Part (b) turns from volume to speed, and from a count to a sample whose dispersion must be characterised.

Part (b) — form the grouped-data moment sums. Treating each observation as sitting at its class mid-point $u_i$ with frequency $f_i$, the two sums needed are $$n = \sum f_i = 1+2+4+11+17+20+22+21+11+5+2+1 = 117$$ $$\sum f_i u_i = 40 + 90 + 200 + 605 + 1020 + 1300 + 1540 + 1575 + 880 + 425 + 180 + 95 = 7950$$ $$\sum f_i u_i^2 = 1600 + 4050 + 10\,000 + 33\,275 + 61\,200 + 84\,500 + 107\,800 + 118\,125 + 70\,400 + 36\,125 + 16\,200 + 9025 = 552\,300$$
Compute the mean speed. The arithmetic mean of the grouped sample is $$\bar{u} = \frac{\sum f_i u_i}{n} = \frac{7950}{117} = 67.949 \text{ km/h}$$ $$\boxed{\;\bar{u} = 67.9 \text{ km/h}\;}$$ This is the time-mean speed, since the observations are spot speeds taken at a point.
Compute the standard deviation of the individual observations. Using the computing form of the sample variance, $$s^2 = \frac{\sum f_i u_i^2 - n\bar{u}^2}{n-1} = \frac{552\,300 - 117(67.949)^2}{116} = \frac{552\,300 - 540\,190.4}{116} = \frac{12\,109.6}{116} = 104.39$$ $$s = \sqrt{104.39} = 10.217 \text{ km/h}$$ $$\boxed{\;s = 10.2 \text{ km/h}\;}$$ Recomputing from the deviation form $\sum f_i(u_i-\bar{u})^2/(n-1)$ gives the same 104.39, which guards against the cancellation error that the computing form is prone to when the mean is large relative to the spread.
Compute the standard deviation of the mean. The standard error of the mean is the sample standard deviation reduced by the root of the sample size: $$s_{\bar{u}} = \frac{s}{\sqrt{n}} = \frac{10.217}{\sqrt{117}} = \frac{10.217}{10.817} = 0.945 \text{ km/h}$$ $$\boxed{\;s_{\bar{u}} = 0.94 \text{ km/h}\;}$$ The two dispersions answer different questions and must not be confused: 10.2 km/h describes how much individual drivers differ from one another, while 0.94 km/h describes how precisely this sample of 117 has pinned down the population mean. The mean speed is known to within roughly $\pm 1.9$ km/h at 95 % confidence.
Build the cumulative distribution and read the 90th percentile. Accumulating the frequencies and expressing them as percentages of $n = 117$: $$\begin{array}{lcccccccccccc} u \text{ (km/h)} & 40 & 45 & 50 & 55 & 60 & 65 & 70 & 75 & 80 & 85 & 90 & 95\\ \Sigma f & 1 & 3 & 7 & 18 & 35 & 55 & 77 & 98 & 109 & 114 & 116 & 117\\ \% & 0.9 & 2.6 & 6.0 & 15.4 & 29.9 & 47.0 & 65.8 & 83.8 & 93.2 & 97.4 & 99.1 & 100 \end{array}$$ The 90 % level falls between 75 km/h (83.76 %) and 80 km/h (93.16 %), so interpolating linearly on the plotted curve, $$u_{90} = 75 + 5\left(\frac{90 - 83.76}{93.16 - 83.76}\right) = 75 + 5(0.664) = 78.32 \text{ km/h}$$ $$\boxed{\;u_{90} = 78.3 \text{ km/h}\;}$$

Question 7(b) — cumulative distribution of the 117 spot speeds, with the 90th-percentile speed of 78.3 km/h read off by linear interpolation between the 75 km/h (83.8 %) and 80 km/h (93.2 %) ordinates.

Interpret the percentile in design terms. The 90th-percentile speed of 78.3 km/h is the speed only one driver in ten exceeds, and it is the conventional upper anchor for setting the design speed of geometric elements and for defining the upper limit of a speed-enforcement tolerance. For comparison, the same interpolation at the 85 % level — the value normally used to set a regulatory speed limit — gives $$u_{85} = 75 + 5\left(\frac{85 - 83.76}{93.16-83.76}\right) = 75.7 \text{ km/h}$$ so a posted limit of 75 km/h would fit this stream, while the observed mean of 67.9 km/h sits about 8 km/h below it. The 90th and 85th percentiles are close together because the distribution is compact through its upper classes.

Question 7 — results
Quantity	Result
(a) 15-minute groups	900 / 1500 / 1100 / 600 vehicles (sum 4100)
(a) Peak 15-minute volume	1500 vehicles
(a) Peak rate of flow	6000 veh/h
(a) Peak hour factor	0.683
(b) Sample size, n	117 observations
(b) Mean speed	67.9 km/h
(b) Variance / standard deviation	104.4 (km/h)² / 10.2 km/h
(b) Standard deviation of the mean	0.94 km/h
(b) 90th-percentile speed	78.3 km/h
(b) 85th-percentile speed (for comparison)	75.7 km/h

Topic. Volume studies and the peak hour factor; spot-speed studies and the statistics of a grouped speed sample.

Key relations. $PHF = V/(4\,V_{15})$, with peak rate of flow $= 4V_{15}$. Grouped data: $\bar{u} = \sum f_iu_i/n$; $s^2 = \left(\sum f_iu_i^2 - n\bar{u}^2\right)/(n-1)$; standard error of the mean $s_{\bar{u}} = s/\sqrt{n}$; percentile speeds by linear interpolation on the cumulative distribution.

Why this works. The peak hour factor exists because a facility is loaded by the worst few minutes, not by the hour as a whole: queues form and dissipate on a timescale of minutes, so the peak-15-minute rate of flow is the demand a capacity analysis must actually compare against. Expressing that rate as a fraction of the hourly volume is a compact measure of how spiky the demand is, and it is the bridge between a forecast hourly volume and the flow rate that appears in every level-of-service calculation — which is exactly how it was used in Question 6(a). On the statistical side, the two dispersion measures are related but answer opposite questions, and the $\sqrt{n}$ separating them is the central-limit result: individual variability is a property of the driver population and does not shrink with more data, while the uncertainty in the estimated mean shrinks as the root of the sample size. The percentile speeds close the loop back to design: geometric elements are dimensioned for the fast tail of the distribution, regulatory limits are set near the 85th percentile on the argument that the great majority of drivers choose a reasonable speed, and the gap between the mean and the 85th percentile measures how much a limit set at the mean would criminalise ordinary behaviour.

Common pitfall. In part (a), taking the peak 5-minute count instead of the peak 15-minute count — using 600 gives a peak rate of 7200 veh/h and a peak hour factor of 0.57, a different and wrong answer to a different question. In part (b), dividing the sum of squared deviations by $n$ rather than $n-1$; forgetting to weight by the frequencies (averaging the twelve mid-points gives 67.5 km/h, close enough here to hide the error but wrong in principle); and above all reporting the standard deviation of the observations where the standard deviation of the mean is asked for, or the reverse — a factor of $\sqrt{117} \approx 10.8$ apart. When interpolating for the 90th percentile, note that the cumulative percentage is plotted at the class upper boundary in some conventions; here the mid-points are used consistently, as the question tabulates them.

Source. Garber & Hoel, Traffic and Highway Engineering, Ch. 4 (volume studies and the peak hour factor; spot-speed studies, grouped-data statistics, cumulative distributions and percentile speeds); TAC Geometric Design Guide for Canadian Roads, use of the 85th-percentile speed in design and speed-limit setting.

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