Question 2 of 7: Single-server queueing at a cashier

Question 2 — Single-server queueing at a 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.

Approach. Treat the cashier as an M/M/1 system — Poisson arrivals, negative-exponential service, one server, infinite queue — compute the traffic intensity $\rho = \lambda/\mu$, and read every requested statistic off the standard steady-state relations.

1. Convert both rates to a common time base and form the traffic intensity. Rates quoted per 20 minutes are multiplied by three to give hourly rates:
   $$\lambda = 10 \times 3 = 30\ \text{cust}\,\text{h}^{-1}, \qquad \mu = 12 \times 3 = 36\ \text{cust}\,\text{h}^{-1}$$
   $$\rho = \frac{\lambda}{\mu} = \frac{30}{36} = 0.8333$$
   Because $\rho \lt 1$ the system is stable and a steady state exists; the cashier is busy 83.3 % of the time. (Working in customers per 20 minutes gives the identical $\rho = 10/12$; the hourly base is carried forward only so that the time answers come out in familiar units.)
2. (a) Probability that the cashier is free. The probability of an empty system is the complement of the utilisation:
   $$P_0 = 1 - \rho = 1 - 0.8333 = \boxed{0.1667}$$
   The cashier is idle about one-sixth of the time, or roughly 10 minutes in every hour.
3. (b) Average number of customers waiting to be processed. "Waiting to be processed" excludes the customer currently at the till, so this is the mean queue length $L_q$:
   $$L_q = \frac{\rho^{2}}{1-\rho} = \frac{(0.8333)^{2}}{0.1667} = \frac{0.6944}{0.1667} = \boxed{4.17\ \text{customers}}$$
4. (c) Average number of customers in line. Read as the mean number in the system — the queue plus the customer being served — this is $L$:
   $$L = \frac{\rho}{1-\rho} = \frac{0.8333}{0.1667} = \boxed{5.00\ \text{customers}}$$
   The two answers are consistent: $L = L_q + \rho = 4.17 + 0.83 = 5.00$, the extra $\rho$ being the expected number in service.
5. Establish the waiting and service times from Little's law. Little's law, $L = \lambda W$, applied to the queue and to the system gives the two time statistics directly:
   $$W_q = \frac{L_q}{\lambda}, \qquad W = \frac{L}{\lambda}$$
6. (d) Average wait time and average processing time. Substituting,
   $$W_q = \frac{4.1667}{30} = 0.1389\ \text{h} = \boxed{8.33\ \text{min}}$$
   and the average time actually spent being served is the reciprocal of the service rate,
   $$\frac{1}{\mu} = \frac{1}{36}\ \text{h} = \boxed{1.67\ \text{min}}$$
   For completeness, the total time in the system is $W = 5.00/30 = 0.1667$ h $= 10.0$ min, which checks against $W = W_q + 1/\mu = 8.33 + 1.67 = 10.0$ min. A customer therefore spends five times as long queueing as being served — the characteristic signature of an M/M/1 system running at $\rho = 0.83$.
7. (e) Probability that a second cashier is opened. A second cashier opens when the line exceeds five customers, that is when $n > 5$, equivalently $n \ge 6$. For M/M/1 the tail probability has the closed form
   $$P(n \ge k) = \rho^{k}$$
   so with $k = 6$:
   $$P(n > 5) = \rho^{6} = (0.8333)^{6} = \boxed{0.335}$$
   A second till is needed about one-third of the time, which is a strong operational argument for scheduling a second cashier through the peak rather than opening one reactively.