---
title: "Trunk group numerical: 30 E, 12 s, 2 lost"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:trunks-30e-12s-two-lost"
source: https://www.bctengineer.com/q/semester_8/telecommunications/trunks-30e-12s-two-lost
site: BCT Engineering
---

# Trunk group numerical: 30 E, 12 s, 2 lost

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2075 Asoj, 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-30e-12s-two-lost
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

During the busy hour, on an average, 30 E is offered to a group of trunks and on average the total period during which all trunks are busy is 12 sec and two 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 30

## Short answer

B = 12s/3600s = 1/300. Calls offered = 2 × 300 = 600; carried = 598. Average call duration h = (30×3600)/600 = 180s (3 min). Calls arriving in one holding time = (600/3600)×180 = 30E, consistent with offered traffic.

## Answer

**Given:** offered traffic A = 30 E, total time all trunks busy = 12 s in the busy hour (3600 s), calls lost = 2.

For pure-chance traffic, the probability that a call is lost (call congestion) equals the fraction of time all trunks are busy (time congestion):
$$
\begin{aligned}
B &= \frac{12}{3600} = \frac{1}{300}
\end{aligned}
$$

#### (i) Average number of calls carried
Calls lost = B × calls offered, so
$$
\begin{aligned}
C_o &= \frac{\text{calls lost}}{B} = 2 \times 300 = 600\ \text{calls} \\
C_c &= C_o - C_l = 600 - 2 = 598\ \text{calls}
\end{aligned}
$$

#### (ii) Average call duration
$$
\begin{aligned}
A &= \frac{C_o\,h}{T} \Rightarrow h = \frac{A\,T}{C_o} \\
h &= \frac{30 \times 3600}{600} = 180\ \text{s} = 3\ \text{min}
\end{aligned}
$$

#### (iii) Calls offered in a period equal to h
Call arrival rate $= 600/3600$ calls per second. In 180 s:
$$
\begin{aligned}
n &= \frac{600}{3600} \times 180 = 30\ \text{calls}
\end{aligned}
$$
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) **598 calls carried** (of 600 offered), (ii) **h = 180 s (3 min)**, (iii) calls offered in one holding time = **30** = A, as required.

## Check yourself

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

1. How is B found?
   A. 2/30
   B. 12 x 2
   C. 30/12
   D. 12/3600 = 1/300 as time congestion equals call congestion
   Answer: D. Time congestion = call congestion for Poisson traffic. (Key point: Loss probability)

2. How many calls were offered and carried?
   A. 600 offered, 598 carried
   B. 300 offered, 298 carried
   C. 60 offered, 58 carried
   D. 3600 offered, 3598 carried
   Answer: A. Offered = 2 x 300 = 600; carried = 598. (Key point: Calls carried)

3. Average call duration?
   A. 120 s
   B. 180 s (3 min)
   C. 300 s
   D. 30 s
   Answer: B. h = A T / C = 30 x 3600 / 600 = 180 s. (Key point: Holding time)

4. Arrivals during one holding time are?
   A. 60
   B. 600
   C. 30, equal to A
   D. 180
   Answer: C. (600/3600) x 180 = 30 = lambda h = A. (Key point: Check)

## Close variants set in other exams

- [Trunk group numerical: 40 E, 20 s, 4 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atrunks-40e-20s-four-lost) — Telecommunications, chapter 6, asked 2×

## Related in Telecommunications

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- [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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