---
title: "Traffic numerical: 1000 calls, 5 lost"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:traffic-1000-calls-5-lost"
source: https://www.bctengineer.com/q/semester_8/telecommunications/traffic-1000-calls-5-lost
site: BCT Engineering
---

# Traffic numerical: 1000 calls, 5 lost

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2080 Baisakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/traffic-1000-calls-5-lost
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

During the busy hour, 1000 calls were offered to a group of trunks and 5 calls were lost. The average call duration was 4 minutes. Find: a) The traffic offered b) The traffic carried c) The traffic lost d) The grade of service e) The total duration of the period of congestion

## Short answer

Method: Traffic(erlangs) = calls × holding time (hours). Offered: 1000×(4/60)h = 66.67 E; carried: 995×(4/60)h = 66.33 E; lost: 5×(4/60)h = 0.33 E; GoS = 5/1000 = 0.005; congestion duration = GoS × 3600 s = 18 s.

## Answer

**Given:** calls offered $C_o = 1000$, calls lost $C_l = 5$, mean holding time $h = 4\ \text{min} = \tfrac{1}{15}$ h, period T = 1 h.

Traffic in Erlangs $= \dfrac{\text{calls} \times h}{T}$.

#### (a) Traffic offered
$$
\begin{aligned}
A_o &= \frac{C_o\,h}{T} = 1000 \times \tfrac{1}{15} \\
&= 66.67\ \text{E}
\end{aligned}
$$

#### (b) Traffic carried
Calls carried $= 1000 - 5 = 995$
$$
\begin{aligned}
A_c &= 995 \times \tfrac{1}{15} = 66.33\ \text{E}
\end{aligned}
$$

#### (c) Traffic lost
$$
\begin{aligned}
A_l &= A_o - A_c = 5 \times \tfrac{1}{15} = 0.33\ \text{E}
\end{aligned}
$$

#### (d) Grade of service
$$
\begin{aligned}
GoS &= \frac{C_l}{C_o} = \frac{5}{1000} = 0.005
\end{aligned}
$$

#### (e) Duration of congestion
For pure-chance (Poisson) traffic the proportion of time all trunks are busy (time congestion) equals the proportion of calls lost (call congestion), so congestion time in the busy hour is
$$
\begin{aligned}
T_c &= GoS \times 3600\ \text{s} = \frac{5}{1000} \times 3600 \\
&= 18\ \text{s}
\end{aligned}
$$

**Answer:** (a) offered = **66.67 E** (66 2/3 E), (b) carried = **66.33 E** (66 1/3 E), (c) lost = **0.33 E** (1/3 E), (d) GoS = **0.005**, (e) congestion period = **18 s** in the busy hour.

## Check yourself

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

1. How is traffic in Erlangs computed from calls?
   A. Number of calls x mean holding time / observation period
   B. Number of calls / mean holding time
   C. Number of calls x observation period / holding time
   D. Mean holding time / number of calls
   Answer: A. A = C h / T. (Key point: Erlang formula)

2. 1000 calls of 4 min mean duration in one hour give what offered traffic?
   A. 250 E
   B. 66.67 E
   C. 16.67 E
   D. 4000 E
   Answer: B. 1000 x 4 / 60 = 66.67 E. (Key point: Offered traffic)

3. How is the lost traffic found?
   A. Lost calls divided by offered calls = 0.005 E
   B. Lost calls x 3600 = 18000 E
   C. Offered minus carried, which equals lost calls x h / T = 0.33 E
   D. Carried divided by offered = 0.995 E
   Answer: C. Lost traffic is the traffic of the 5 lost calls, 5 x (1/15) = 0.33 E; the ratio 0.005 is GoS, not traffic. (Key point: Lost traffic)

4. Why is the congestion duration GoS x 3600 s = 18 s?
   A. Because each lost call lasts 3.6 s
   B. Because carried traffic is 18 E
   C. Because the holding time is 18 s
   D. For Poisson traffic time congestion equals call congestion
   Answer: D. For pure-chance traffic the fraction of time all trunks are busy equals the fraction of calls lost. (Key point: Congestion time)

## Close variants set in other exams

- [Traffic numerical: 1200 calls, 6 lost](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atraffic-1200-calls-6-lost) — Telecommunications, chapter 6, asked 6×
- [GOS vs blocking; 600 calls, 2.5 min](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-blocking-600-calls-2-5min) — Telecommunications, chapter 6, asked 6×
- [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×
- [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×
- [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×

## Related in Telecommunications

- [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×
- [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×
- [Traffic intensity, GOS and call duration probability](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atraffic-intensity-call-duration-probability) — Telecommunications, chapter 6