16-Civ-B10 Traffic Engineering · May 2014

Question 4 of 6: Multi-channel drive-in facility and Poisson arrivals at a toll plaza

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

Notes on this paper

Paper format. National Examinations, May 2014 — 98-Civ-B10 Traffic Engineering. Three-hour, open-book examination; any non-communicating calculator is permitted. Six questions are printed and five complete solutions are required, all questions being of equal value (20 marks each). The printed grading scheme is Q1 (a) to (d) 5 marks each; Q2 (a) to (e) 4 marks each; Q3 (a) to (e) 4 marks each; Q4 (a) and (b) 10 marks each; Q5 (a) to (e) 4 marks each; Q6 (10 + 5 + 5) marks. The paper states that if doubt exists as to the interpretation of a question the candidate should submit with the answer paper a clear statement of any assumptions made, and that any data required but not given can be assumed. All six questions are worked below.

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), Ch. 6 (fundamental principles of traffic flow: Poisson arrivals, deterministic and stochastic queueing, M/M/1 and M/M/N channels), Ch. 8 (intersection control: cycle length, phasing, change and clearance intervals, progression and time–space diagrams) and Ch. 10 (capacity and level of service at signalised intersections). This is the principal reference for the subject. Webster, F. V. and Cobbe, B. M., Traffic Signals, Road Research Laboratory Technical Paper No. 56 — the optimum-cycle and three-term delay formulae used throughout Questions 1 to 3. Transportation Research Board, Highway Capacity Manual — saturation-flow adjustment factors and the pedestrian-green requirement. Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) and Geometric Design Guide for Canadian Roads — Canadian signal-timing, clearance-interval and crosswalk practice. Institute of Transportation Engineers, Traffic Engineering Handbook — controller types and coordinated timing plans.

Check: the four values assumed under the paper's own NOTE 2 (“any data required, but not given, can be assumed”). They are declared here once and used consistently in every question.

Bus passenger-car equivalent $E_B = 2.0$ pcu (HCM value for buses on level terrain). The paper gives bus occupancies but no pcu equivalent; note that the pcu-based and vehicle-based flow ratios in Question 1 agree exactly for any choice of $E_B$, so this assumption does not affect Questions 2 or 3.
Pedestrian walking speed $S_p = 1.2$ m/s (HCM 2000 / MUTCDC design value; the more recent 1.1 m/s would lengthen the pedestrian interval by about 1 s).
Lost time convention: the whole intergreen (amber + all-red) is taken as lost, so effective green equals displayed green. This is the conservative reading and needs no assumed start-up lost time. Webster's alternative, $L = n\ell + R$ with $\ell \approx 2$ s, would give $L = 10$ s instead of 12 s and lengthen each green by about 1 s.
Delay model: Webster's three-term formula, with the first two terms reported separately in Question 3(b) as the uniform and overflow components.

Question 4: Multi-channel drive-in facility and Poisson arrivals at a toll plaza (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.

Quantity	Value
(a) Arrival rate $\lambda$	30 veh/h, Poisson
(a) Mean service time $1/\mu$	3 min, exponential ⇒ $\mu = 20$ veh/h per lane
(a) Criterion	average waiting time in the queue $< 5$ min
(b) Mean arrivals per 10-minute period $m$	9 passenger cars
(b) Required probabilities	$P(0)$, $P(10)$, $P(20)$

Find. (a) the smallest number of parallel service lanes $N$ that holds the mean queueing time below five minutes; (b) three ordinates of the Poisson distribution with mean 9.

Approach. Model the drive-in facility as an M/M/$N$ system — one common queue feeding $N$ identical exponential servers — test the stability condition first, then evaluate $W_q$ for increasing $N$ until the criterion is met. Part (b) is a direct application of the Poisson probability mass function.

Part (a) — test stability before anything else. With a mean service time of 3 min each lane serves $\mu = 60/3 = 20$ veh/h, so the offered traffic intensity is $$a = \frac{\lambda}{\mu} = \frac{30}{20} = 1.5\ \text{erlangs}, \qquad \rho = \frac{a}{N} = \frac{\lambda}{N\mu}.$$ A single lane gives $\rho = 1.5 > 1$: the queue grows without bound and no finite waiting time exists. At least two lanes are therefore necessary on stability grounds alone, and the question becomes whether two are sufficient.
Part (a) — evaluate the M/M/$N$ queue for two lanes. The idle-state probability and the mean queue for $N$ servers are $$P_0 = \left[\sum_{n=0}^{N-1}\frac{a^{n}}{n!} + \frac{a^{N}}{N!\,(1-\rho)}\right]^{-1}, \qquad L_q = \frac{P_0\,a^{N}\rho}{N!\,(1-\rho)^{2}}, \qquad W_q = \frac{L_q}{\lambda}.$$ With $N = 2$, $a = 1.5$ and $\rho = 0.75$: $P_0 = [1 + 1.5 + 2.25/(2 \times 0.25)]^{-1} = 1/7 = 0.1429$, hence $L_q = 0.1429(2.25)(0.75)/[2(0.0625)] = 1.93$ vehicles and $$W_q = \frac{1.93}{30}\ \text{h} = 0.0643\ \text{h} = 3.86\ \text{min}.$$

Question 4(a) — mean queueing time against the number of lanes. One lane is unstable; two lanes already clear the five-minute criterion.

Part (a) — compare against the criterion and check the next size up. Since $3.86 < 5$ min the two-lane facility satisfies the requirement: $$\boxed{\;N = 2\ \text{drive-in lanes}\;\;(W_q = 3.86\ \text{min} < 5\ \text{min})\;}$$ A third lane would reduce $\rho$ to 0.50 and $W_q$ to 0.47 min — an eightfold improvement in waiting for a 50 % increase in capital and staffing, which is the kind of diminishing return that Question 5(e) asks about explicitly. If the client's five-minute criterion were meant as the total time in the system (queueing plus the 3 min service), then $W = W_q + 1/\mu = 6.86$ min at $N = 2$ would fail and three lanes ($W = 3.47$ min) would be required; the interpretation should be stated with the answer, as the paper's NOTE 1 requires.
Part (b) — apply the Poisson mass function. For a counting process with mean $m = 9$ vehicles per 10-minute period, $$P(x) = \frac{e^{-m}\,m^{x}}{x!}, \qquad m = 9 .$$ Evaluating at the three requested counts: $$P(0) = e^{-9} = 1.234 \times 10^{-4},\qquad P(10) = \frac{e^{-9}\,9^{10}}{10!} = 0.1186,\qquad P(20) = \frac{e^{-9}\,9^{20}}{20!} = 6.17 \times 10^{-4}.$$ $$\boxed{\;P(0) = 0.012\ \%,\quad P(10) = 11.9\ \%,\quad P(20) = 0.062\ \%\;}$$ The shape is what the distribution predicts: the mode sits at 8 and 9 vehicles, so a count of 10 is very likely, while an empty period and a period of double the mean are both rare. For toll-plaza design the useful reading is that although the average is nine cars per ten minutes, roughly one period in 1600 will deliver twenty — the booth must have queue storage for the tail, not for the mean.

Final Results

Part	Quantity	Result
(a)	Traffic intensity $a = \lambda/\mu$	1.5 erlangs — a single lane is unstable
(a)	$N = 2$: $\rho$, $L_q$, $W_q$	0.75, 1.93 veh, 3.86 min
(a)	$N = 3$: $\rho$, $L_q$, $W_q$	0.50, 0.24 veh, 0.47 min
(a)	Lanes required	2 (3 if the criterion is total time in the system)
(b)	$P(0)$	$1.234 \times 10^{-4}$
(b)	$P(10)$	0.1186
(b)	$P(20)$	$6.17 \times 10^{-4}$

Topic. Multi-channel exponential queueing (M/M/$N$) for service-facility sizing, and the Poisson counting distribution for arrival forecasting.

Key relations. $a = \lambda/\mu$, $\rho = a/N$ with stability requiring $\rho < 1$; Erlang-C form of $P_0$ and $L_q$ above; Little's law $W_q = L_q/\lambda$; Poisson $P(x) = e^{-m}m^{x}/x!$.

Why this works. Exponential inter-arrival and service times make the number in the system a birth–death Markov chain, whose stationary distribution is the product form used above; the queue only exists when all $N$ servers are busy, which is why $L_q$ carries the factor $a^{N}/N!$. The reason $N$ shared lanes beat $N$ separate single-lane facilities is pooling: with one common queue no server is ever idle while a vehicle waits. Splitting the same 30 veh/h into two independent M/M/1 lanes of 15 veh/h each would give $W_q = \rho/(\mu - \lambda) = 9$ min — more than twice the pooled figure, and a failure of the criterion. State which queue discipline you are modelling.

Common pitfall. Reporting $W$ (time in the system) when $W_q$ (time in the queue) is asked, or the reverse — here the two differ by the full 3 min service time and change the answer from two lanes to three. The other classic error is computing $\rho = \lambda/\mu$ instead of $\lambda/(N\mu)$ for a multi-server system, which makes every queue look unstable.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §6.5 (queueing analysis, single- and multi-channel formulations, worked drive-in examples); HCM, toll-plaza and service-facility queueing.

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