---
title: "Device holding time and calls per hour"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:device-holding-time-calls"
source: https://www.bctengineer.com/q/semester_8/telecommunications/device-holding-time-calls
site: BCT Engineering
---

# Device holding time and calls per hour

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2076 Asoj, 2+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/device-holding-time-calls
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

A device in a telephone exchange is required to commence operation within an average period of 10 milliseconds after receiving a calling signal. (i) If the device is held, on average for 50 milliseconds per call, how many calls can it handle per hour? (ii) If the device is required to handle 18,000 calls per hour, what is the maximum permissible average holding time?

## Short answer

For an M/M/1 delay system with average wait W=10 ms. (i) h=50 ms, W = (A h)/(1-A) yields A=1/6; calls/h = A T/h = (1/6)×3600/0.05 = 12,000 calls/h. (ii) λ=5 calls/s, W=0.01 s, solve 5h^2/(1-5h)=0.01 gives h=0.04 s = 40 ms maximum.

## Answer

The device is a **single server** and calls that find it busy **wait** (delay system). With random (Poisson) calls and exponential holding time this is an **M/M/1 (Erlang delay, N = 1)** system. The "10 ms" is the mean delay before a call is served (averaged over all calls):
$$
W = \frac{A\,h}{1 - A}
$$
where $h$ = mean holding time and $A = \lambda h$ = traffic (occupancy) of the device.

#### (i) h = 50 ms, W = 10 ms
$$
\begin{aligned}
\frac{A \times 50}{1-A} &= 10 \\
50A &= 10 - 10A \\
A &= \frac{10}{60} = \frac{1}{6}\ \text{E}
\end{aligned}
$$
Calls per hour:
$$
\begin{aligned}
C &= \frac{A\,T}{h} = \frac{(1/6) \times 3600}{0.05} \\
&= 12000\ \text{calls/h}
\end{aligned}
$$
(Without the delay limit, a fully occupied device could take $3600/0.05 = 72000$ calls/h, but delay would then be infinite.)

#### (ii) 18000 calls/h, W = 10 ms
$\lambda = 18000/3600 = 5$ calls/s, so $A = 5h$ (h in seconds).
$$
\begin{aligned}
\frac{\lambda h \cdot h}{1 - \lambda h} &= 0.01 \\
5h^2 &= 0.01(1 - 5h) \\
5h^2 + 0.05h - 0.01 &= 0 \\
h &= \frac{-0.05 + \sqrt{0.05^2 + 4(5)(0.01)}}{2 \times 5} \\
&= \frac{-0.05 + 0.45}{10} = 0.04\ \text{s}
\end{aligned}
$$
Check: $A = 5 \times 0.04 = 0.2$; $W = 0.2 \times 40/(1-0.2) = 10$ ms.

**Answer:** (i) **12,000 calls per hour**; (ii) maximum average holding time **h = 40 ms**.

## Check yourself

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

1. Which model describes a single device where waiting calls queue?
   A. Erlang B with N = 1
   B. Engset
   C. Binomial
   D. M/M/1 (Erlang delay with N = 1)
   Answer: D. Calls wait for the single server, so it is a delay system. (Key point: Model choice)

2. What is the mean delay of all calls for the device?
   A. W = A h / (1 - A)
   B. W = A h
   C. W = h / (1 + A)
   D. W = (1 - A) / (A h)
   Answer: A. For M/M/1 W = A h/(1-A). (Key point: Mean delay formula)

3. With h = 50 ms and W = 10 ms, what is A?
   A. 1/5 E
   B. 1/6 E
   C. 1 E
   D. 10/50 E = 0.2 E
   Answer: B. 50A = 10(1-A) gives 60A = 10, A = 1/6. (Key point: Occupancy)

4. Why is the answer 12,000 calls/h rather than 72,000?
   A. 72,000 ignores holding time
   B. The device works only half the hour
   C. 72,000 would saturate the device and give infinite delay
   D. Delay does not depend on load
   Answer: C. At A = 1 the delay is infinite; the 10 ms limit requires A = 1/6, so C = (1/6)(3600)/0.05 = 12,000. (Key point: Calls per hour)

## Related in Telecommunications

- [Traffic numerical: 800 calls, 50 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-800-calls-50-lost) — Telecommunications, chapter 6, asked 6×
- [Working principle of a digital telephone exchange](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adigital-telephone-exchange-principle) — Telecommunications, chapter 4, asked 4×
- [Telephone exchange and switching hierarchy](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aexchange-switching-hierarchy) — Telecommunications, chapter 1, asked 3×
- [Telephone exchange structure; TST vs STS](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Aexchange-structure-tst-sts) — Telecommunications, chapter 4
- [Traffic numerical: 600 calls, 40 lost, 120 s](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-600-calls-120s) — Telecommunications, chapter 6, asked 5×
- [Time switch in digital exchange](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atime-switch-digital-exchange) — Telecommunications, chapter 4

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