---
title: "PCO analysis with M/M/1 (300 persons)"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:pco-300-persons-mm1"
source: https://www.bctengineer.com/q/semester_8/telecommunications/pco-300-persons-mm1
site: BCT Engineering
---

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

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2070 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/pco-300-persons-mm1
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

A public call office (PCO) is installed in a busy part of a town. 300 persons use the booth everyday. The average holding time for a call is 5 minutes. There is a suggestion from the public that another PCO is required in the same locality as the waiting times are unduly long. Analyse the situation using M/M/1 queue and determine if the suggestion deserves serious consideration.

## Short answer

Compute arrival rate 300/(24×60)=0.2083 calls/min, service rate 0.2 calls/min, so ρ=1.042>1; the M/M/1 queue is unstable. With two booths each handling half, ρ=0.521, queue wait ~5.4 min. The public's suggestion is justified.

## Answer

**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):

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

| Case | Load per booth | Mean wait in queue |
|---|---|---|
| One PCO | 1.042 E (> 1) | grows without limit |
| Two PCOs | 0.521 E | about 5.4 min |

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. 300 users per day are spread evenly over 24 hours. What is the arrival rate lambda?
   A. 300/24 = 12.5 calls/min
   B. 300/(24 x 60) = 0.2083 calls/min
   C. 300 x 5 = 1500 calls/min
   D. 5/300 = 0.0167 calls/min
   Answer: B. The day has 24 x 60 = 1440 minutes, so lambda = 300/1440 = 0.2083 calls per minute. (Key point: Arrival rate)

2. With h = 5 min, what is the traffic intensity (utilisation) of the single booth?
   A. 0.2 E
   B. 0.521 E
   C. 1.042 E
   D. 2.08 E
   Answer: C. rho = lambda x h = 0.2083 x 5 = 1.042 E. (Key point: Offered traffic)

3. Why does rho = 1.042 mean the waiting time is unduly long for one PCO?
   A. rho > 1, so arrivals exceed service capacity and the queue grows without limit
   B. rho > 1 only means the booth is well used, so waits are short
   C. rho is close to 0, so the booth is idle
   D. rho > 1 means calls are blocked and cleared, not queued
   Answer: A. M/M/1 formulas are finite only for rho < 1; with rho > 1 people arrive faster than they are served. (Key point: Stability condition)

4. If users split equally between two booths, what is the load per booth and the mean wait in queue?
   A. 1.042 E per booth, wait unchanged
   B. 0.521 E per booth, wait about 5.4 min
   C. 0.521 E per booth, wait about 10.4 min
   D. 0.2 E per booth, wait zero
   Answer: B. rho1 = 1.042/2 = 0.521 and Wq = rho1 h/(1 - rho1) = 5.43 min, so the suggestion is justified. (Key point: Effect of a second PCO)

## Related in Telecommunications

- [PCO analysis with M/M/1 (150 persons)](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Akendall-pco-150-persons) — Telecommunications, chapter 6