16-Civ-B10 Traffic Engineering · December 2019

Question 6 of 7: Moving-vehicle (Wardrop) volume and travel-time study

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

Notes on this paper

Paper format. National Examinations, December 2019, 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, whose footnote fixes the brake-reaction time at 2.5 s and the deceleration at 3.4 m/s²; Question 1 is built around it.

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 deterministic queueing), Ch. 8 (intersection control and signal timing), Ch. 15 (geometric design of highways, crest 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 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 and pedestrian intervals.
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 2 and 5.
Hillier, F.S. and Lieberman, G.J., Introduction to Operations Research — the M/M/1 birth-and-death results used in Question 4.

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 supplies the AASHTO 2001 metric table directly, so that table is used as printed — the TAC guide adopts the same 2.5 s / 3.4 m/s² stopping model, so the two agree here. Lane widths, walking speed and crosswalk geometry follow TAC and MUTCDC practice.

Question 6: Moving-vehicle (Wardrop) volume and travel-time study (5 × 4 = 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.

Note on the printed sub-part text. The page is printed in two overlapping impressions, and separating them recovers the missing fragments intact — "Westbound tra|ffic volu|me (vehicles/hour)", "Average travel |time of |eastbound traffic (mi|nutes)" and the same for westbound. No text has been reconstructed or guessed: both halves of every phrase are printed on the page, and the wording above is the merge of the two impressions.

Given. Sixteen test-car runs over the same section, eight in each direction. For each run the table records the travel time, the number of vehicles met in the opposing stream, the number that overtook the test car, and the number the test car overtook.

Run	Westbound				Eastbound
	t (min)	Met	Overtook	Overtaken	t (min)	Met	Overtook	Overtaken
1	3.04	108	2	1	2.68	93	2	2
2	2.80	103	3	2	2.85	109	1	2
3	3.15	119	1	2	2.94	108	0	2
4	2.95	105	3	2	2.59	94	3	1
5	3.47	116	0	1	2.79	90	1	0
6	3.51	111	2	2	2.48	105	2	1
7	3.28	125	2	1	2.96	111	1	1
8	3.17	107	3	1	2.77	104	2	2

Find. The four run averages in each direction, the traffic volume in each direction, and the true mean travel time of the traffic stream in each direction.

Question 6: the moving-vehicle method. The test car in one direction counts the stream it meets head-on, which is the stream of interest for the other direction — the direction-crossing that the volume formula encodes.

Approach. Average the four recorded quantities over the eight runs in each direction, then apply Wardrop’s two relations, taking care that the volume of a given direction is built from the opposing count made by the test car travelling against it.

Part (a) — run averages. Averaging each column over the eight runs gives, for the westbound runs, a mean travel time $t_w = 3.1713$ min with $M_w = 111.75$ vehicles met, $O_w = 2.00$ overtaking the test car and $P_w = 1.50$ overtaken by it; for the eastbound runs, $t_e = 2.7575$ min with $M_e = 101.75$, $O_e = 1.50$ and $P_e = 1.375$. The eastbound runs are consistently about 25 seconds quicker over the same section, which already signals the directional imbalance the volumes will confirm.
Set up Wardrop’s volume relation. For a two-way section the volume in one direction is $$q = \frac{M + O - P}{t_{w} + t_{e}}$$ where $M$ is the number met by the test car running in the opposite direction, and $O$ and $P$ are the overtaking and overtaken counts recorded by the test car running with the stream. The denominator is the sum of the two mean travel times, $t_w + t_e = 3.1713 + 2.7575 = 5.9288$ min, and it is the same for both directions.
Part (b) — eastbound volume. The eastbound stream is the one the westbound test car meets, so $M = M_w = 111.75$, while $O$ and $P$ come from the eastbound runs: $$q_{E} = \frac{111.75 + 1.50 - 1.375}{5.9288} = 18.870\ \text{veh/min} = \boxed{1132.2\ \text{veh/h}}$$
Part (c) — westbound volume. Symmetrically, the westbound stream is the one the eastbound test car meets: $$q_{W} = \frac{101.75 + 2.00 - 1.50}{5.9288} = 17.246\ \text{veh/min} = \boxed{1034.8\ \text{veh/h}}$$ The two-way volume is 2167.0 veh/h, with the eastbound direction carrying 52.2 % of it.
Parts (d) and (e) — mean travel time of the traffic stream. The test car’s own mean travel time is not the stream’s: a test car that is overtaken more often than it overtakes is travelling slower than the traffic around it, and vice versa. Wardrop corrects for that with $$\bar{t} = t - \frac{O - P}{q}$$ with $q$ expressed in vehicles per minute. Eastbound, $$\bar{t}_{E} = 2.7575 - \frac{1.50 - 1.375}{18.870} = \boxed{2.751\ \text{min}} \;(165.1\ \text{s})$$ and westbound, $$\bar{t}_{W} = 3.1713 - \frac{2.00 - 1.50}{17.246} = \boxed{3.142\ \text{min}} \;(188.5\ \text{s})$$ Both corrections are small — a few seconds — because the test-car drivers evidently floated close to the prevailing speed, which is exactly what the method asks of them.

Check: the direction-crossing is the whole question. The single easiest way to lose this problem is to build each volume from the "met" count recorded in the same direction. Doing that here would give an eastbound volume of 1031.0 veh/h instead of 1132.2 veh/h — an error of about 9 %, and one that produces an entirely plausible-looking answer. The physical reason is that a test car cannot count the stream it is travelling in; it can only count the stream coming towards it. The overtaking and overtaken counts, by contrast, belong to the direction the test car is actually in.

Question 6 — final results
Quantity	Westbound	Eastbound
(a) Mean travel time of the test car	3.1713 min	2.7575 min
(a) Mean number met, M	111.75	101.75
(a) Mean number overtaking the test car, O	2.00	1.50
(a) Mean number overtaken by the test car, P	1.50	1.375
Sum of mean travel times	5.9288 min
(b), (c) Traffic volume	1034.8 veh/h	1132.2 veh/h
(d), (e) Mean travel time of the stream	3.142 min (188.5 s)	2.751 min (165.1 s)
Two-way volume	2167.0 veh/h (eastbound share 52.2 %)

Topic. The moving-vehicle (Wardrop) method for simultaneous volume and travel-time estimation on a two-way road.

Key relations. $q = (M + O - P)/(t_w + t_e)$ with M taken from the opposing run; $\bar{t} = t - (O-P)/q$; two-way volume $= q_E + q_W$; mean speed $= \text{section length}/\bar{t}$.

Why this works. Consider the test car running west. In the time it takes to traverse the section, it meets every eastbound vehicle that was in the section when it started plus every one that entered while it was travelling — a count that depends on the eastbound flow and on the sum of the two travel times, not on either alone. That is the numerator and denominator of the volume relation. The overtaking correction is a separate idea: net overtakings measure how far the test car’s speed departs from the stream mean, and dividing the net count by the flow converts it back into a time correction. The elegance of the method is that a single pair of runs delivers both the volume and the travel time, with no roadside equipment at all.

Common pitfall. Crossing the directions wrongly — using M from the same-direction run — is the dominant error and it is silent, because the answer remains dimensionally sensible. A second error is applying the overtaking correction with q in vehicles per hour while t is in minutes; the units of the correction term must match the units of t. A third is averaging the ratios rather than ratioing the averages: the method requires the mean counts and mean times to be formed first, then combined.

Source. Wardrop, J.G. and Charlesworth, G., "A method of estimating speed and flow of traffic from a moving vehicle", Proc. ICE; Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 4 §4.3 (travel-time and delay studies, moving-vehicle technique).

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