---
title: "PCO analysis with M/M/1 (150 persons)"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:kendall-pco-150-persons"
source: https://www.bctengineer.com/q/semester_8/telecommunications/kendall-pco-150-persons
site: BCT Engineering
---

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

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2079 Chaitra, 3+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/kendall-pco-150-persons
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

Explain Kendall's notation. A PCO is installed in the busy part of the town. 150 persons use the booth every day. The average holding time for the call is 1.5 minutes. There is a suggestion from the public that the waiting period is very long and they need another PCO in the same locality. Analyze using M/M/1 queue.

## Short answer

Kendall’s notation A/B/C/K/N/D describes queue: arrival process, service distribution, servers, capacity, population, discipline. M/M/1 analysis: λ = 150/24 = 6.25 calls/h, μ = 40/h, ρ = 0.156, W_q = 0.278 min, L_q = 0.029 persons. The booth is lightly loaded; a second PCO is not justified by average daily traffic.

## Answer

#### 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):
| Quantity | 24 h day | 12 h day |
|---|---|---|
| $\rho$ (booth busy) | 0.156 | 0.3125 |
| P(wait) | 15.6% | 31.3% |
| $L_q$ (persons) | 0.029 | 0.142 |
| $W_q$ (all callers) | 0.28 min | 0.68 min |
| $W$ (in system) | 1.78 min | 2.18 min |

#### 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).

## Check yourself

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

1. M/M/1 means:
   A. Poisson arrivals, exponential service, one server
   B. Deterministic arrivals, one server
   C. General arrivals, two servers
   D. Poisson arrivals, fixed service, one server
   Answer: A. Defaults are infinite capacity and population, FCFS. (Key point: M/M/1 meaning)

2. For 150 calls over 24 h and h = 1.5 min, λ and μ are:
   A. 6.25/h and 40/h
   B. 150/h and 1.5/h
   C. 6.25/h and 0.67/h
   D. 12.5/h and 40/h
   Answer: A. λ = 150/24 = 6.25; μ = 60/1.5 = 40. (Key point: Rates)

3. What is ρ for the 24 h case?
   A. 0.156
   B. 0.3125
   C. 0.5
   D. 0.85
   Answer: A. ρ = 6.25/40 = 0.156. (Key point: Utilisation)

4. What does the analysis conclude about a second PCO?
   A. Needed, since ρ is high
   B. Not justified on average traffic; complaints come from busy-hour peaks
   C. Needed because W > 10 min
   D. Needed because queue is infinite
   Answer: B. The booth is busy only 16-31% of the time with under one minute average wait. (Key point: Conclusion)

## Related in Telecommunications

- [PCO analysis with M/M/1 (300 persons)](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Apco-300-persons-mm1) — Telecommunications, chapter 6

## Related in other subjects

- [Factors of tariff; two-part and time of day tariff](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Atariff-factors-two-part-tod) — Utilization of Electrical Energy, chapter 6