PCO analysis with M/M/1 (300 persons)

Model: the PCO is a single server with a queue (M/M/1): Poisson arrivals, exponential holding times, one booth, people wait in line.

Given: 300 users per day, mean holding time $h = 5$ min.

Assumption: the booth is in use 24 hours a day and calls are spread evenly over the day. This is the most favourable case; a shorter working day or a busy-hour peak only makes the load higher.

Step 1: Arrival rate and offered traffic

$$\begin{aligned} \lambda &= \frac{300}{24 \times 60} = 0.2083\ \text{calls/min} \\ \mu &= \frac{1}{h} = 0.2\ \text{calls/min} \\ \rho = A &= \lambda h = 0.2083 \times 5 = 1.042\ \text{E} \end{aligned}$$

Step 2: Stability check

For an M/M/1 queue the mean queue length and waiting time are

$$L = \frac{\rho}{1-\rho}, \qquad W_q = \frac{\rho\,h}{1-\rho}$$

These are finite only when $\rho < 1$. Here $\rho = 1.042 > 1$: people arrive faster than one booth can serve them. The queue grows without limit and the waiting time becomes very long. If the booth is used only 12 hours a day, $\rho = \frac{300 \times 5}{720} = 2.08$ E, which is far worse.

Step 3: With a second PCO

Assume the users split equally between two booths (two M/M/1 queues):

$$\begin{aligned} \rho_1 &= \frac{1.042}{2} = 0.521 \\ L &= \frac{\rho_1}{1-\rho_1} = \frac{0.521}{0.479} = 1.087\ \text{persons} \\ L_q &= \frac{\rho_1^2}{1-\rho_1} = 0.566\ \text{persons} \\ W_q &= \frac{\rho_1 h}{1-\rho_1} = \frac{0.521 \times 5}{0.479} = 5.43\ \text{min} \\ W &= \frac{h}{1-\rho_1} = 10.43\ \text{min} \end{aligned}$$

Answer: with one PCO the utilisation is $\rho = 1.042 > 1$, so the M/M/1 queue is unstable and waiting times are unduly long, even with traffic spread over 24 hours. With two PCOs each booth has $\rho = 0.521$ and the mean wait drops to about 5.4 min. The public's suggestion is justified and deserves serious consideration.