---
title: "Trunk group numerical: 40 E, 20 s, 4 lost"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:trunks-40e-20s-four-lost"
source: https://www.bctengineer.com/q/semester_8/telecommunications/trunks-40e-20s-four-lost
site: BCT Engineering
---

# Trunk group numerical: 40 E, 20 s, 4 lost

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2075 Chaitra, 3+3+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/trunks-40e-20s-four-lost
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

During the busy hour, on an average, 40E is offered to a group of trunks and on average the total period during which all trunks are busy is 20s and four calls are lost. (i) Find the average number of calls carried by the group (ii) Find the average call duration (iii) Show that the average number of calls offered to the group during a period equal to the average call duration is 40.

## Short answer

Blocking probability B = 20s/3600s = 1/180. Calls offered = 4 / (1/180) = 720, so calls carried = 716. Average call duration h = (40E × 3600s) / 720 = 200s (3.33 min). During one holding time, calls offered = (720/3600)×200 = 40, matching A=40E.

## Answer

**Given:** offered traffic A = 40 E, total time all trunks busy = 20 s in the busy hour (3600 s), calls lost = 4.

For pure-chance traffic, the probability that a call is lost (call congestion) equals the fraction of time all trunks are busy (time congestion):
$$
\begin{aligned}
B &= \frac{20}{3600} = \frac{1}{180}
\end{aligned}
$$

#### (i) Average number of calls carried
Calls lost = B × calls offered, so
$$
\begin{aligned}
C_o &= \frac{\text{calls lost}}{B} = 4 \times 180 = 720\ \text{calls} \\
C_c &= C_o - C_l = 720 - 4 = 716\ \text{calls}
\end{aligned}
$$

#### (ii) Average call duration
$$
\begin{aligned}
A &= \frac{C_o\,h}{T} \Rightarrow h = \frac{A\,T}{C_o} \\
h &= \frac{40 \times 3600}{720} = 200\ \text{s} = 3.33\ \text{min}
\end{aligned}
$$

#### (iii) Calls offered in a period equal to h
Call arrival rate $= 720/3600$ calls per second. In 200 s:
$$
\begin{aligned}
n &= \frac{720}{3600} \times 200 = 40\ \text{calls}
\end{aligned}
$$
This equals the offered traffic in Erlangs, which is the definition of the Erlang: traffic in E = mean number of calls arriving during one mean holding time ($A = \lambda h$).

**Answer:** (i) **716 calls carried** (of 720 offered), (ii) **h = 200 s (3.33 min)**, (iii) calls offered in one holding time = **40** = A, as required.

## Check yourself

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

1. How is the loss probability B found?
   A. 4/40 = 0.1
   B. 20 x 4 = 80
   C. 40/20 = 2
   D. Time all trunks busy / 3600 = 1/180, using time congestion = call congestion
   Answer: D. For Poisson traffic call congestion equals time congestion. (Key point: Time and call congestion)

2. How many calls were offered, and carried?
   A. 720 offered, 716 carried
   B. 180 offered, 176 carried
   C. 800 offered, 796 carried
   D. 4 offered, 0 carried
   Answer: A. Offered = 4 / (1/180) = 720; carried = 716. (Key point: Calls offered)

3. Mean call duration from A = C h / T?
   A. 90 s
   B. 200 s
   C. 40 s
   D. 720 s
   Answer: B. h = 40 x 3600 / 720 = 200 s. (Key point: Holding time)

4. Why are 40 calls offered in one holding time?
   A. Because there are 40 trunks
   B. Because 40 calls are always lost
   C. A = lambda h, the calls arriving during one mean holding time
   D. Because T = 40 s
   Answer: C. Erlangs equal the number of arrivals during one mean holding time. (Key point: Meaning of Erlang)

## Close variants set in other exams

- [Trunk group numerical: 30 E, 12 s, 2 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atrunks-30e-12s-two-lost) — Telecommunications, chapter 6, asked 2×

## Related in Telecommunications

- [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×
- [Traffic numerical: 1000 calls, 5 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atraffic-1000-calls-5-lost) — Telecommunications, chapter 6, asked 6×
- [GOS numerical: 1400 calls, 14 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-busy-hour-1400-calls) — Telecommunications, chapter 6, asked 5×
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