Question 6 of 7: Single-channel queueing at a store cashier

Question 6 — Single-channel queueing at a store cashier (a) to (e), 4 marks each — 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.

Find. (a) the probability the cashier is idle; (b) the mean number of customers waiting; (c) the mean number in line; (d) the mean waiting time and the mean service time; (e) the probability that the trigger for opening a second cashier is met.

Approach. Confirm stability, then apply the standard M/M/1 results. Sub-parts (b) and (c) are read as the queue and the system respectively, following the order in which these papers set out the single-channel results.

1. Confirm the queue is stable before applying any formula. The traffic intensity is
   $$\rho=\frac{\lambda}{\mu}=\frac{5/10}{6/10}=\frac{5}{6}=0.8333$$
   Since $\rho<1$ the system reaches steady state and the M/M/1 results below are valid. At 0.833 the cashier is heavily loaded, which is why the queue lengths that follow are large for what sounds like a modest arrival rate — the classic non-linear behaviour of a single server near saturation.
2. Part (a) — probability the cashier is free. The cashier is free exactly when the system is empty, and the probability of the zero state is
   $$P_0=1-\rho=1-\frac{5}{6}=\frac{1}{6}=\boxed{0.167}$$
   So the cashier is idle 16.7 per cent of the time and busy 83.3 per cent of the time — the utilisation, which for a single server equals $\rho$ directly.
3. Part (b) — average number waiting to be processed. The mean queue length, counting only customers waiting and excluding the one at the till, is
   $$L_q=\frac{\rho^{2}}{1-\rho}=\frac{\left(5/6\right)^{2}}{1/6}=\frac{0.69444}{0.16667}=\boxed{4.17\ \text{customers}}$$
4. Part (c) — average number of customers in line. Counting the customer being served as part of the line gives the mean number in the system,
   $$L=\frac{\rho}{1-\rho}=\frac{5/6}{1/6}=\boxed{5.00\ \text{customers}}$$
   The relation between the two answers is a useful check: $L-L_q=\rho=0.833$, which is exactly the expected number in service, since the single server is busy a fraction $\rho$ of the time. Parts (b) and (c) therefore differ by less than one customer and any answer separating them by more has an arithmetic error.
5. Part (d) — average wait and average service time. The mean time spent waiting follows from Little's law applied to the queue,
   $$W_q=\frac{L_q}{\lambda}=\frac{4.1667}{0.5}=\boxed{8.33\ \text{min}}$$
   and the mean time actually being processed is the reciprocal of the service rate,
   $$\frac{1}{\mu}=\frac{1}{0.6}=\boxed{1.67\ \text{min}}$$
   Little's law closes the calculation: the total time in the system is $W=W_q+1/\mu=8.33+1.67=10.0$ min, and independently $W=L/\lambda=5.00/0.5=10.0$ min. The two agreeing confirms every preceding value at once. The engineering point is stark — a customer spends 100 s being served and 8.3 min waiting, so 83 per cent of the visit is queueing.
6. Part (e) — probability that the second cashier opens. The trigger is that the line is longer than three customers, so the second cashier opens whenever the system holds four or more. For M/M/1 the tail probability has the geometric closed form
   $$P\left(n\ge k\right)=\rho^{k}$$
   so with $k=4$,
   $$P\left(n\ge4\right)=\left(\frac{5}{6}\right)^{4}=\frac{625}{1296}=\boxed{0.482}$$
   Summing the individual state probabilities $(1-\rho)\rho^{n}$ from $n=4$ upwards gives the same 0.4823, which is the check on the tail formula. The second cashier is therefore needed almost half the time — a direct consequence of the 0.833 intensity, and a strong argument that the store is under-staffed at one till rather than occasionally busy.

Results.