Question 2 of 7: Moving-Vehicle (Wardrop) Volume and Travel-Time Study

Question 2: Moving-Vehicle (Wardrop) Volume and Travel-Time Study (20 marks — 4 each)

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. Sixteen test-vehicle runs over the same section — eight eastbound and eight westbound — each recording the run time, the number of vehicles met travelling in the opposite direction, the number of vehicles that overtook the test vehicle, and the number the test vehicle overtook.

Find. The run averages for each direction, then the eastbound and westbound flows in vehicles per hour and the mean travel time of each stream in minutes.

Approach. Average the eight runs in each direction, then apply Wardrop’s moving-observer relations, taking the opposite-direction run as the source of the meeting count and the same-direction run as the source of the overtaking counts.

1. Part (a) — average the eight runs in each direction. Each column is a simple arithmetic mean of eight observations, e.g. for the eastbound travel time $$\bar{t}_E = \frac{2.71+2.50+2.81+2.63+3.10+3.13+2.93+2.83}{8} = \frac{22.64}{8} = 2.83\ \text{min}$$ and the same operation on the remaining three columns gives the table below.
2. Tabulate the averages that feed the formulas. Writing $M$ for the mean number of vehicles met in the opposite direction, $O$ for the mean number that overtook the test vehicle and $P$ for the mean number the test vehicle overtook:

   Part (a) — run averages

   Direction of test run	$\bar{t}$ (min)	$M$ (veh)	$O$ (veh)	$P$ (veh)
   Eastbound	2.83	103.5	2.00	1.50
   Westbound	3.03	107.0	1.50	1.25

3. State Wardrop’s relations. For the stream being measured, let the subscript $w$ denote the test run made with that stream and $a$ the run made against it. The flow and the mean travel time of the stream are$$q = \frac{M_a + O_w - P_w}{t_a + t_w}, \qquad \bar{t} = t_w - \frac{O_w - P_w}{q}$$The meeting count must come from the against run, because only on that run do the vehicles of the measured stream travel in the opposite direction to the observer.
4. Part (b) — eastbound flow. The eastbound stream is met during the westbound runs, so $M_a = 107.0$, while the overtaking counts come from the eastbound runs, $O_w = 2.00$ and $P_w = 1.50$, with $t_a = 3.03$ min and $t_w = 2.83$ min:$$q_E = \frac{107.0 + 2.00 - 1.50}{3.03 + 2.83} = \frac{107.5}{5.86} = 18.345\ \text{veh/min}$$Converting to an hourly rate, $q_E = 18.345 \times 60$, so$$\boxed{q_E = 1{,}101\ \text{veh/h}}$$
5. Part (c) — westbound flow. The roles reverse: the westbound stream is met during the eastbound runs, so $M_a = 103.5$, and the overtaking counts are the westbound ones, $O_w = 1.50$ and $P_w = 1.25$, with $t_a = 2.83$ min and $t_w = 3.03$ min:$$q_W = \frac{103.5 + 1.50 - 1.25}{2.83 + 3.03} = \frac{103.75}{5.86} = 17.705\ \text{veh/min}$$and therefore$$\boxed{q_W = 1{,}062\ \text{veh/h}}$$
6. Part (d) — mean eastbound travel time. The test vehicle was net-overtaken eastbound ($O_w - P_w = 0.50 > 0$), which means it ran slightly slower than the stream, so the stream mean must be a little below the test-run mean:$$\bar{t}_E = 2.83 - \frac{2.00 - 1.50}{18.345} = 2.83 - 0.027 = \boxed{2.80\ \text{min}}$$
7. Part (e) — mean westbound travel time. Identically, with $O_w - P_w = 0.25$ and $q_W = 17.705$ veh/min,$$\bar{t}_W = 3.03 - \frac{1.50 - 1.25}{17.705} = 3.03 - 0.014 = \boxed{3.02\ \text{min}}$$In both directions the correction is small, which is the expected result when the test driver has been instructed to float with the stream.

The two directions are close to balanced, with the eastbound stream about 4 per cent heavier and about 7 per cent quicker over the section. If the test section length were known, dividing it by these mean travel times would give the space mean speed of each stream directly, which is the quantity the moving-vehicle method is normally run to obtain.