Question 2 of 7: Single-Channel Queueing at a Drive-Thru

Question 2: Single-Channel Queueing at a Drive-Thru (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. A single-channel drive-thru with Poisson arrivals and negative-exponential service times, as tabulated above.

Find. The idle probability, the mean queue length and the mean system content, the mean waiting time and the mean time in system, and the probability that more than six vehicles are present.

Approach. Confirm stability from the traffic intensity, then apply the standard M/M/1 (single-channel, undersaturated) results in the order utilisation, queue content, time in queue, tail probability.

1. Establish the traffic intensity and hence stability. The traffic intensity is the ratio of demand to service capability, $$\rho = \frac{\lambda}{\mu} = \frac{125}{138} = 0.9058$$ Because $\rho \lt 1$ the queue does not grow without bound and the steady-state M/M/1 expressions may be used. It is worth noting how close to 1 this is: the worker is busy 90.6 per cent of the peak hour, which is exactly why the queue lengths below come out so large for such a modest arrival rate.
2. Part (a) — probability the worker is free. The probability of an empty system is the complement of the utilisation, $$P_0 = 1 - \rho = 1 - 0.9058 = \boxed{0.0942}$$ so the worker is free of any order for only about 9.4 per cent of the peak hour — roughly 5.7 minutes in 60.
3. Part (b) — average number waiting to be processed. The mean number in the queue, excluding the vehicle being served, is $$L_q = \frac{\lambda^{2}}{\mu(\mu - \lambda)} = \frac{125^{2}}{138(138 - 125)} = \frac{15\,625}{1794} = \boxed{8.71 \text{ veh}}$$ Equivalently $L_q = \rho^{2}/(1-\rho) = 0.8205/0.0942$, which returns the same 8.71 veh and is a useful independent check.
4. Part (c) — average number of cars in line. Taking “in line” as the whole system — the car at the window plus those behind it — the mean content is $$L = \frac{\lambda}{\mu - \lambda} = \frac{125}{13} = \boxed{9.62 \text{ veh}}$$ The two answers must differ by exactly the utilisation, and they do: $9.62 - 8.71 = 0.91 = \rho$, because the server is occupied a fraction $\rho$ of the time and holds one vehicle when occupied.
5. Part (d) — average wait, and average time in the system. Little's law converts each content into a time. The time spent queueing is $$W_q = \frac{L_q}{\lambda} = \frac{8.71}{125} = 0.06968 \text{ h} = \boxed{4.18 \text{ min}}$$ and the total time from joining the line to driving away with the order is $$W = \frac{L}{\lambda} = \frac{1}{\mu - \lambda} = \frac{1}{13} \text{ h} = \boxed{4.62 \text{ min}}$$ The difference, 0.43 min, is precisely the mean service time $1/\mu = 1/138$ h, which confirms both figures at once.
6. Part (e) — probability of backing up onto the street. For an M/M/1 queue the probability that more than $N$ vehicles are present is $$P(n \gt N) = \rho^{\,N+1}$$ The line backs onto Main Street once it is longer than six vehicles, so with $N = 6$, $$P(n \gt 6) = 0.9058^{7} = \boxed{0.500}$$ Half of the peak hour is spent with the line spilling onto the street. Summing $P_n = (1-\rho)\rho^{n}$ over $n = 0 \ldots 6$ and subtracting from unity reproduces the same 0.500, which is the cleanest check available on a tail probability.