16-Civ-B10 Traffic Engineering · December 2019

Question 4 of 7: M/M/1 queue at a single cashier

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 4: M/M/1 queue at a single cashier (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.

Given. A single service channel with Poisson arrivals and exponential service times, served first-in first-out with an unlimited waiting area — the standard M/M/1 system.

Quantity	Symbol	Value
Arrival rate	λ	1 per 2 min = 0.5 customer/min
Service rate	μ	6 per 10 min = 0.6 customer/min
Number of channels	N	1
Traffic intensity	ρ = λ/μ	0.8333

Find. The idle probability, the mean queue and system contents, the mean waiting and service times, and the probability that the queue trigger for a second cashier is reached.

Question 4: the single-channel queue and its state-probability distribution. The shaded band collects the states from n = 6 upward, whose total probability is the answer to part (e).

Approach. Put both rates in the same units, confirm the system is stable, then read off the standard M/M/1 birth-and-death results in the order the question asks for them, using Little’s law to connect the counts to the times.

Common units and the stability test. One arrival per 2 minutes is $\lambda = 0.5$ customer/min; six services per 10 minutes is $\mu = 0.6$ customer/min. The traffic intensity is $$\rho = \frac{\lambda}{\mu} = \frac{0.5}{0.6} = 0.8333$$ Because $\rho \lt 1$ the single channel can keep up on average and steady-state results exist. Note that 0.8333 is a high utilisation: the queue results below are correspondingly sensitive to the service rate, which is exactly why the store has a trigger for a second cashier.
Part (a) — probability the cashier is free. The cashier is free whenever the system holds no customers, so $$P_{0} = 1 - \rho = 1 - 0.8333 = \boxed{0.1667}$$ i.e. the cashier is idle about 16.7 % of the time and busy 83.3 % of the time.
Part (b) — average number waiting to be processed. This is the mean queue length, excluding the customer currently at the till: $$L_{q} = \frac{\rho^{2}}{1-\rho} = \frac{(0.8333)^{2}}{0.1667} = \boxed{4.167\ \text{customers}}$$
Part (c) — average number of customers in line. Read as the number in the system — the customers waiting plus the one being served — this is $$L = \frac{\rho}{1-\rho} = \frac{0.8333}{0.1667} = \boxed{5.000\ \text{customers}}$$ and the two answers are consistent because $L = L_q + \rho = 4.167 + 0.833 = 5.000$. Reporting both, and saying which is which, is what parts (b) and (c) are testing: the difference between them is exactly the probability that the server is busy.
Part (d) — waiting time and service time. Little’s law converts each count into the corresponding time by dividing by the arrival rate: $$W_{q} = \frac{L_{q}}{\lambda} = \frac{4.167}{0.5} = \boxed{8.333\ \text{min}}, \qquad \frac{1}{\mu} = \frac{1}{0.6} = \boxed{1.667\ \text{min}}$$ so a customer waits about 8 min 20 s in line and spends 1 min 40 s at the till, giving a total time in the system of $W = W_q + 1/\mu = 10.000$ min — which the independent route $W = L/\lambda = 5.000/0.5$ reproduces exactly. Five sixths of a customer’s visit is spent waiting rather than being served, which is the practical message of a system running at $\rho = 0.83$.
Part (e) — probability the second cashier opens. For M/M/1 the state probabilities are geometric, $P(n) = (1-\rho)\rho^{n}$, and the tail sums to a single power: $$P(n \ge k) = \sum_{n=k}^{\infty}(1-\rho)\rho^{n} = \rho^{k}$$ "The line of customers is longer than five" is read as more than five customers present, i.e. $n \ge 6$: $$P(n \ge 6) = \rho^{6} = (0.8333)^{6} = \boxed{0.3349}$$ so the second cashier is opened about a third of the time. Summing the geometric series term by term from $n = 6$ reproduces 0.3349 exactly, which confirms the closed form.

Check: which count the trigger refers to. "Longer than five customers" is ambiguous between five customers in the system (giving $P(n \ge 6) = \rho^{6} = 0.3349$) and five customers waiting in line, which means six in the system and gives $P(n \ge 7) = \rho^{7} = 0.2791$. The first reading is adopted because the question’s own part (c) uses "in line" for the system count, and it is declared here under the paper’s Note 1. The conclusion — that a second cashier is needed for roughly a third of the peak — is unchanged either way.

Question 4 — final results
Quantity	Symbol	Value
Traffic intensity	ρ	0.8333
(a) Probability the cashier is free	P₀	0.1667
(b) Average number waiting	L_q	4.167 customers
(c) Average number in the system	L	5.000 customers
(d) Average wait in line	W_q	8.333 min
(d) Average service time	1/μ	1.667 min
(d) Average total time in the system	W	10.000 min
(e) Probability a second cashier opens	P(n ≥ 6)	0.3349
(e) Alternative reading, five waiting	P(n ≥ 7)	0.2791

Topic. Single-channel M/M/1 queueing applied to a service counter.

Key relations. $\rho = \lambda/\mu$; $P_0 = 1-\rho$; $P(n) = (1-\rho)\rho^{n}$; $L = \rho/(1-\rho)$; $L_q = \rho^{2}/(1-\rho)$; $W = L/\lambda$, $W_q = L_q/\lambda$ (Little’s law); $P(n \ge k) = \rho^{k}$.

Why this works. Poisson arrivals and exponential service make the number in the system a birth-and-death process whose balance equations force the state probabilities to be geometric in $\rho$. Everything else follows mechanically: the mean of a geometric distribution gives L, subtracting the expected number in service gives $L_q$, and Little’s law converts either count into a time. The geometric tail is what makes part (e) a one-line answer — the probability of exceeding any threshold is just $\rho$ raised to that threshold, with no summation needed. Note how sharply the results depend on $\rho$: raising the service rate here from 0.6 to 0.7 per minute would cut the mean queue from 4.17 to 1.25 customers.

Common pitfall. The first is unit mismatch — "1 per 2 minutes" and "6 per 10 minutes" must both be converted to customers per minute before dividing, and a candidate who forms $\rho = 1/6$ or $2/10$ gets a plausible-looking but wrong stable system. The second is confusing $L$ with $L_q$ (and $W$ with $W_q$); the two differ by exactly $\rho$ and by $1/\mu$ respectively, and this question deliberately asks for both.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 6 §6.5 (queueing analysis, single-channel results); Hillier & Lieberman, Introduction to Operations Research, queueing-theory chapter (birth-and-death derivation of the geometric state distribution).

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