16-Civ-B10 Traffic Engineering · December 2018

Question 6 of 7: M/M/1 single cashier (5 × 4 marks)

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

Notes on this paper

Paper format. National Examinations, December 2018, 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, which Question 2 is built around.

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 queueing), Ch. 8 (intersection control and signal timing), Ch. 15 (geometric design of highways, 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 SSD 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, pedestrian intervals and the Canadian signal-warrant procedure.
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 3 and 7.

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(e) names the US MUTCD explicitly, so its eight warrants are answered as printed; the Canadian equivalent procedure is noted alongside, which is polish rather than a correction.

Question 6: M/M/1 single cashier (5 × 4 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. A single server, Poisson arrivals at 3 customers per 6 minutes and exponential service at 4 customers per 6 minutes. Converting both to the same unit, $\lambda = 3/6 = 0.500$ customers per minute and $\mu = 4/6 = 0.667$ customers per minute.

Find. $P_0$; the mean number waiting; the mean number in the system; the mean waiting time and the mean service time; and the probability that the queue exceeds three customers.

Question 6. The single-channel queue and its geometric state distribution. The shaded states are those in which a second cashier would be opened.

Approach. This is a textbook M/M/1 (single channel, Poisson arrivals, exponential service, FIFO, infinite queue). Get the traffic intensity first; every other result is a function of it alone.

Traffic intensity and stability. $$\rho = \frac{\lambda}{\mu} = \frac{0.500}{0.667} = \frac{3}{4} = 0.750$$ Since $\rho < 1$ the queue is stable and the steady-state formulae apply. The cashier is busy 75 per cent of the time.
Part (a) — probability the cashier is free. The cashier is free exactly when the system is empty, and for M/M/1 the state probabilities are geometric, $P(n) = (1-\rho)\rho^{n}$, so $$P_0 = 1 - \rho = 1 - 0.750 = \boxed{0.250}$$ i.e. the cashier is idle a quarter of the time. Note that idle time and queueing coexist: the cashier is idle 25 per cent of the time and the average customer still waits, because arrivals are random rather than evenly spaced.
Part (b) — average number waiting to be processed. This is the queue length excluding whoever is at the till: $$L_q = \frac{\rho^{2}}{1 - \rho} = \frac{0.750^{2}}{0.250} = \frac{0.5625}{0.250} = \boxed{2.25\ \text{customers}}$$
Part (c) — average number of customers in line. Read against part (b), this part must mean the number in the system — the people waiting plus the one being served — otherwise it would repeat (b) verbatim: $$L = \frac{\rho}{1 - \rho} = \frac{0.750}{0.250} = \boxed{3.00\ \text{customers}}$$ The consistency check is immediate and exact: $L - L_q = 3.00 - 2.25 = 0.75 = \rho$, which is the expected number in service, as it must be for a single server.
Part (d) — average wait, and average service time. By Little's law applied to the waiting line, $$W_q = \frac{L_q}{\lambda} = \frac{2.25}{0.500} = \boxed{4.50\ \text{min}} = 270\ \text{s}$$ and the mean service time is simply the reciprocal of the service rate, $$\frac{1}{\mu} = \frac{1}{0.667} = \boxed{1.50\ \text{min}} = 90\ \text{s}$$ Little's law closes on the total: $W = W_q + 1/\mu = 4.50 + 1.50 = 6.00$ min, and independently $W = L/\lambda = 3.00/0.500 = 6.00$ min. So a customer spends six minutes in the shop, 75 per cent of it queueing — the same 75 per cent as the traffic intensity, which is a property of the single-server model.
Part (e) — probability a second cashier is opened. The trigger is “the line of customers is longer than 3”, i.e. more than three customers present, so the event is $n \ge 4$. The geometric tail of an M/M/1 system sums neatly: $$P(n \ge k) = \sum_{n=k}^{\infty}(1-\rho)\rho^{n} = \rho^{k}$$ so $$P(n \ge 4) = \rho^{4} = 0.750^{4} = \frac{81}{256} = \boxed{0.316}$$ A second cashier would be opened roughly 32 per cent of the time — often enough that the shop should probably schedule two.
Check the tail against the state probabilities. Summing $(1-\rho)\rho^{n}$ from $n = 4$ upward numerically returns 0.3164, matching $\rho^{4}$ to four decimals, which confirms both the state distribution and the tail identity.

Check: reading of parts (b), (c) and (e). Parts (b) and (c) are worded almost identically (“waiting to be processed” and “in line”) yet carry 4 marks each, so they must be asking for different quantities: (b) is taken as $L_q$, the number waiting, and (c) as $L$, the number in the system. Likewise part (e) says “longer than 3 customers”, taken as $n \ge 4$ in the system, giving $\rho^4 = 0.316$. If instead the trigger were more than three waiting (i.e. $n \ge 5$), the answer would be $\rho^5 = 0.237$. The first reading is adopted because it matches the wording of the trigger and the convention of parts (b) and (c).

Question 6 — final results
Quantity	Symbol	Result
Traffic intensity	$\rho$	0.750 (cashier busy 75 per cent of the time)
(a) Probability the cashier is free	$P_0$	0.250
(b) Average number waiting	$L_q$	2.25 customers
(c) Average number in the system	$L$	3.00 customers
(d) Average wait in the queue	$W_q$	4.50 min (270 s)
(d) Average service time	$1/\mu$	1.50 min (90 s)
(d) Total time in the system	$W$	6.00 min, of which 75 per cent is waiting
(e) Probability a second cashier opens	$P(n \ge 4)$	0.316 (= 81/256; 0.237 on the queue-based reading)

Topic. Single-channel (M/M/1) queueing.

Key relations. $\rho = \lambda/\mu$ with $\rho < 1$ for stability; $P(n) = (1-\rho)\rho^{n}$; $P_0 = 1-\rho$; $L_q = \rho^{2}/(1-\rho)$; $L = \rho/(1-\rho) = L_q + \rho$; Little's law $L = \lambda W$ and $L_q = \lambda W_q$, so $W = W_q + 1/\mu$; geometric tail $P(n \ge k) = \rho^{k}$.

Why this works. Poisson arrivals and exponential service make the number in the system a birth-death chain with constant rates, whose balance equations give a purely geometric distribution. Everything else — means, waiting times, tail probabilities — is then a property of a geometric series in $\rho$ alone, which is why one number determines the whole answer and why the tail probability is simply $\rho^{k}$.

Common pitfall. Mixing up $L$ and $L_q$, and $W$ and $W_q$; the pairs differ by $\rho$ and by $1/\mu$ respectively, which here is a whole customer and a whole 1.5 minutes. Also: convert $\lambda$ and $\mu$ to the same time unit before forming $\rho$ — “3 per 6 minutes” and “4 per 6 minutes” are given per six minutes here, and forming a ratio of raw counts only happens to work because the intervals match.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 6 (queueing theory: single-channel, undersaturated infinite queues); Little, J.D.C., A Proof for the Queuing Formula L = λW, Operations Research, 1961.

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