Question 3 of 6: Poisson Arrivals and Negative-Exponential Headways
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. Observed volume $q = 360$ veh/h at a point, with Poisson arrivals. The gap (headway) between successive vehicles is therefore negative-exponentially distributed.
Find. (a) $P(h < 8\ \text{s})$ and (b) $P(8\ \text{s} \le h \le 10\ \text{s})$.
The negative-exponential headway density for 360 veh/h. The blue area is the probability asked for in part (a); the orange band between 8 s and 10 s is part (b).
Approach. Convert the hourly count to a mean arrival rate per second, use the equivalence between “no arrival in t seconds” under the Poisson counting model and “a headway longer than t”, and evaluate the resulting exponential cumulative distribution at the two limits.
Convert the volume to a mean arrival rate and a mean headway. The rate parameter of the arrival process is
$$\lambda = \frac{q}{3600} = \frac{360}{3600} = 0.100\ \text{veh/s}, \qquad \bar{h} = \frac{1}{\lambda} = \frac{3600}{360} = 10.0\ \text{s}$$
A mean headway of 10 s is the natural sanity check: 360 vehicles distributed over 3600 s must average one every 10 s.
Establish the headway distribution from the counting distribution. If arrivals are Poisson, the probability of exactly zero arrivals in an interval of length t is
$$P(0\ \text{in}\ t) = \frac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t}$$
and “no arrival within t seconds of the last one” is the same event as “the headway exceeds t”. Hence the headway survivor function and cumulative distribution are
$$P(h \ge t) = e^{-\lambda t} = e^{-t/\bar{h}}, \qquad P(h < t) = 1 - e^{-t/\bar{h}}$$
This is why a Poisson counting assumption and a negative-exponential headway assumption are not two separate assumptions but one.
Part (a) — evaluate the cumulative distribution at 8 s. Substituting $t = 8$ s and $\bar{h} = 10$ s:
$$P(h < 8) = 1 - e^{-8/10} = 1 - e^{-0.8} = 1 - 0.4493 = \boxed{0.5507}$$
so about 55.1 % of headways are shorter than 8 s. In an hour carrying 360 vehicles that is roughly $0.5507 \times 360 \approx 198$ headways below 8 s — a useful figure when 8 s is the critical gap for a manoeuvre, because it says how many acceptable gaps a driver will not get.
Part (b) — take the difference of two survivor values. A probability over an interval is the difference of the survivor function at its two ends:
$$P(8 \le h \le 10) = P(h \ge 8) - P(h \ge 10) = e^{-0.8} - e^{-1.0}$$
$$P(8 \le h \le 10) = 0.4493 - 0.3679 = \boxed{0.0815}$$
about 8.15 % of headways, or some 29 of the 360 headways in the hour. Working the difference in the survivor form rather than subtracting two cumulative distributions avoids the sign slip that catches most candidates.
Confirm the answers are mutually consistent. The three regions must exhaust the sample space:
$$P(h<8) + P(8 \le h \le 10) + P(h > 10) = 0.5507 + 0.0815 + 0.3679 = 1.0001 \approx 1$$
the small residue being four-figure rounding only. Note also that $P(h > \bar{h}) = e^{-1} = 0.368$ for any exponential headway distribution: because the distribution is strongly skewed, only about 37 % of headways exceed the mean headway, whatever the volume.
