Question 5 of 7: Single-Channel (M/M/1) Queueing at a Drive-Through
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2016 — 98-Civ-B10 Traffic Engineering. Three-hour duration; OPEN BOOK, any non-communicating calculator permitted. Seven questions, all of equal value (20 marks each), with the mark split printed in the paper’s own grading scheme: Q1 and Q2 and Q5 are (a) to (e) at 4 marks each, Q3 and Q4 are single 20-mark questions, Q6 is (a) 6 marks with (b) and (c) 7 marks each, and Q7 is (a) to (h) at 2.5 marks each. The paper states that a total of five solutions is required and that only the first five in the answer book will be marked. All seven questions are solved here, because this set is a study resource rather than a sitting. The paper also permits assumptions — “Any data required, but not given, can be assumed” — and every assumption made below is stated explicitly where it is used.
Reference texts. Garber, N. J. & Hoel, L. A., Traffic and Highway Engineering, 5th ed. (Cengage) — the core reference for this exam code; Transportation Association of Canada, Geometric Design Guide for Canadian Roads (TAC GDG); Institute of Transportation Engineers (Canadian District), Canadian Capacity Guide for Signalized Intersections, 3rd ed.; Transportation Research Board, Highway Capacity Manual (HCM) — pedestrian crossing-time model; AASHTO, A Policy on Geometric Design of Highways and Streets (the “Green Book”, 2001 edition — the source of the stopping-sight-distance table printed on page 4 of this paper); Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC). Canadian practice governs wherever the paper does not name a specific standard.
Question 5: Single-Channel (M/M/1) Queueing at a Drive-Through (20 marks — 4 each)
Given. A single service channel with mean service rate $\mu = 46$ customers per hour, and a probability of 0.08 that the server is idle — that is, that the system holds zero customers, $P(0) = 0.08$. Arrivals are Poisson and service times negative-exponential, so the system is M/M/1 with an unlimited first-in-first-out queue.
Find. The arrival rate, the mean queue length, the mean number in the system, the mean waiting and total times, and whether $P(n > 5)$ exceeds 0.55.
The drive-through as a single-channel M/M/1 system. The manager’s idle probability is exactly the probability that the system is empty, which fixes the traffic intensity.
Approach. Recover the traffic intensity from the idle probability, then read every remaining quantity off the standard M/M/1 results, and finish with the geometric tail probability.
Part (a) — recover the traffic intensity and the arrival rate. In an M/M/1 system the probability of an empty system is $P(0) = 1 - \rho$, and an empty system is precisely the condition under which the manager is free to clean. Hence $$\rho = 1 - P(0) = 1 - 0.08 = 0.92$$ and since $\rho = \lambda/\mu$,$$\lambda = \rho\,\mu = 0.92 \times 46 = \boxed{42.32\ \text{customers/h}}$$ The system is stable because $\rho < 1$, but only just — a utilisation of 0.92 is deep into the region where queueing measures rise very steeply.
Part (b) — mean number waiting to order. The mean queue length excludes the customer in service: $$L_q = \frac{\rho^2}{1 - \rho} = \frac{0.92^2}{0.08} = \frac{0.8464}{0.08} = \boxed{10.58\ \text{customers}}$$
Part (c) — mean number in the drive-through line. Including the customer being served, $$L = \frac{\rho}{1 - \rho} = \frac{0.92}{0.08} = \boxed{11.5\ \text{customers}}$$ The difference $L - L_q = 0.92$ is exactly $\rho$, the long-run probability that someone is at the window, which is a useful arithmetic check on both figures.
Part (d) — mean waiting and total times. Little’s law converts the two lengths into times at the same arrival rate. The wait before reaching the window is $$W_q = \frac{L_q}{\lambda} = \frac{10.58}{42.32} = 0.250\ \text{h} = \boxed{15.0\ \text{min}}$$ and the total time in the drive-through is $$W = \frac{L}{\lambda} = \frac{11.5}{42.32} = 0.2717\ \text{h} = \boxed{16.3\ \text{min}}$$ The two differ by $1/\mu = 60/46 = 1.30$ min, the mean service time, exactly as they must.
Part (e) — test the expansion trigger. For M/M/1 the number in the system is geometrically distributed, $P(n) = (1-\rho)\rho^n$, so the tail probability is $$P(n > N) = \rho^{\,N+1}$$ With $N = 5$, $$P(n > 5) = 0.92^6 = 0.6064 = \boxed{60.6\%}$$ Since $60.6\% > 55\%$, the trigger is met and the second drive-through lane is warranted.
Confirm the conclusion is robust to the wording. Read strictly, “more than 5 customers waiting in line” excludes the one at the window and would mean $n > 6$, giving $P(n > 6) = 0.92^7 = 0.5578 = 55.8\%$. That still exceeds 55 per cent, so the recommendation is unchanged under either interpretation — though the margin narrows from six percentage points to under one, which is worth stating in the recommendation.
The result is easy to defend on operational grounds as well as arithmetic ones. At $\rho = 0.92$ the mean total time of 16.3 min is far beyond the two to three minutes a drive-through is designed around, and the mean queue of 10.6 vehicles will overflow the stacking lane at most sites and spill onto the street. Adding a second channel roughly halves the offered load per server and would drop the mean queue by more than an order of magnitude — a second order window is the correct response, and it is cheaper than the site reconstruction that a spillback onto the roadway would eventually force.
