Question 2 of 7: Single-Channel Queueing at a Customs Booth

Question 2: Single-Channel Queueing at a Customs Booth (20 marks — 4 each)

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.

Find. The idle probability, the mean queue length and mean number in the system, the mean waiting and service times, and the probability that more than five vehicles are present (the trigger for opening a second booth).

Approach. Poisson arrivals, negative-exponential service and a single channel make this a standard M/M/1 queue, so every part follows from $\rho = \lambda/\mu$ and the steady-state relations; part (e) uses the geometric distribution of system state.

1. Part (a) — Probability the booth is free. The booth is idle whenever the system holds zero vehicles, and for M/M/1 the state probabilities are geometric, $P(n) = \rho^{n}(1-\rho)$. Setting $n = 0$, $$P(0) = 1 - \rho = 1 - \frac{25}{28} = \boxed{0.1071}$$ so the officer is free about 10.7 per cent of the time, and correspondingly busy 89.3 per cent of the time — the utilisation factor $\rho$ is exactly the fraction of time the server is occupied.
2. Part (b) — Average number waiting to be processed. The mean queue length excludes the vehicle at the booth, $$L_{q} = \frac{\lambda^{2}}{\mu(\mu - \lambda)} = \frac{25^{2}}{28(28-25)} = \frac{625}{84} = \boxed{7.44\ \text{cars}}$$ Cross-checking with the equivalent form $L_{q} = \rho^{2}/(1-\rho) = 0.7972/0.1071 = 7.44$ confirms the arithmetic.
3. Part (c) — Average number of cars in line. Reading “in line” as the whole system — those waiting plus the one being processed — the mean number in the system is $$L = \frac{\lambda}{\mu - \lambda} = \frac{25}{28 - 25} = \boxed{8.33\ \text{cars}}$$ The two answers are tied together by $L = L_{q} + \rho$: $7.44 + 0.893 = 8.33$, which is simply the statement that on average $\rho$ of a vehicle is at the booth. That identity is the cleanest way to keep parts (b) and (c) straight.
4. Part (d) — Waiting time and service time. Applying Little's law to the queue alone gives the mean wait before service, $$W_{q} = \frac{L_{q}}{\lambda} = \frac{7.44}{25} = 0.2976\ \text{h} = \boxed{17.9\ \text{min}}$$ The mean time actually spent being processed is the reciprocal of the service rate, $$\overline{t}_{s} = \frac{1}{\mu} = \frac{1}{28}\ \text{h} = \boxed{2.14\ \text{min}}$$ so the total time in the system is $W = W_{q} + 1/\mu = 20.0$ min, which also equals $L/\lambda = 8.33/25 = 0.3333$ h — a second independent check. Note the proportions: a driver spends about eight times as long waiting as being served, which is the practical argument for the second booth in part (e).
5. Part (e) — Probability a second booth is opened. A second booth is opened when the line exceeds five vehicles. Using the geometric state distribution, the probability of more than $n$ vehicles in the system is $$P(\gt n) = \rho^{\,n+1}$$ With $n = 5$, $$P(\gt 5) = \left(\frac{25}{28}\right)^{6} = (0.8929)^{6} = \boxed{0.507}$$ so the trigger is met roughly 50.7 per cent of the time. That is the substantive engineering conclusion: at 89 per cent utilisation a single booth is over the second-booth threshold for about half of every hour, so the “overflow” booth is not an exception but the normal operating state, and the border crossing should be staffed for two channels during this demand period.