PCO analysis with M/M/1 (150 persons)

Kendall's notation

Kendall's notation A/B/C/K/N/D describes a queue: A = arrival process, B = service-time distribution, C = number of servers, K = system capacity, N = source population, D = queue discipline. M = Markovian (Poisson arrivals / exponential service), D = deterministic, G = general. Defaults are K = ∞, N = ∞, FCFS, so M/M/1 means Poisson arrivals, exponential holding time, one server, unlimited queue, first come first served.

PCO analysis with M/M/1

Assumption: the PCO (public call office) is open 24 hours and the 150 calls arrive at random (Poisson) over the day; holding time is exponential. (A 12-hour working day is checked afterwards.)

Parameters

$$\begin{aligned} \lambda &= \frac{150}{24} = 6.25\ \text{calls/h} \\ \mu &= \frac{60}{1.5} = 40\ \text{calls/h} \\ \rho &= \frac{\lambda}{\mu} = \frac{6.25}{40} = 0.15625 \end{aligned}$$

M/M/1 results

$$\begin{aligned} P(\text{wait}) &= \rho = 0.156 \\ L &= \frac{\rho}{1-\rho} = 0.185\ \text{persons in system} \\ L_q &= \frac{\rho^2}{1-\rho} = 0.0289\ \text{persons waiting} \\ W_q &= \frac{\rho}{\mu-\lambda} = \frac{0.15625}{33.75}\ \text{h} = 0.278\ \text{min} \approx 16.7\ \text{s} \\ W &= \frac{1}{\mu-\lambda} = \frac{1}{33.75}\ \text{h} = 1.78\ \text{min} \end{aligned}$$

Check for a 12-hour day ($\lambda = 12.5$/h):

Conclusion

The booth is busy only 16–31% of the time and the average wait is under one minute, with on average far less than one person in the queue. So, on the basis of average daily traffic, a second PCO is not justified; the complaint arises from short busy-hour peaks when many people arrive together. A second booth would be needed only if busy-hour measurements show $\rho$ approaching about 0.7–0.8 (for example, if more than about 28–32 calls arrive in one hour, where $W_q$ rises to several minutes).