---
title: "Traffic numerical: 600 calls, 40 lost, 120 s"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:gos-600-calls-120s"
source: https://www.bctengineer.com/q/semester_8/telecommunications/gos-600-calls-120s
site: BCT Engineering
---

# Traffic numerical: 600 calls, 40 lost, 120 s

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2080 Chaitra, 2+6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/gos-600-calls-120s
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

Define lost calls and grade of service in telephone traffic engineering. During a busy hour, 600 calls were offered to a group of trunks and 40 calls were lost. If the average call duration was 120 seconds, find the traffic offered, carried traffic, lost traffic, GoS and duration of congestion.

## Short answer

Lost calls are blocked calls when all trunks busy; Grade of Service (GoS) is the fraction lost in the busy hour. With 600 offered calls, 40 lost, average holding 120 s, traffic offered = 20 E, carried = 18.67 E, lost = 1.33 E, GoS = 0.0667, congestion duration = 240 s.

## Answer

#### Lost calls and grade of service
- **Lost call:** a call attempt that cannot be connected because all servers (trunks/switch paths) are busy; in a loss system it is cleared and not served.
- **Grade of Service (GoS):** the proportion of calls lost in the busy hour, $GoS = \dfrac{\text{calls lost}}{\text{calls offered}} = \dfrac{A_l}{A_o}$. A lower GoS means better service (e.g. 0.01 = 1 call in 100 lost).

**Given:** calls offered $C_o = 600$, calls lost $C_l = 40$, mean holding time $h = 120\ \text{s} = \tfrac{1}{30}$ h, period T = 1 h.

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

#### (a) Traffic offered
$$
\begin{aligned}
A_o &= \frac{C_o\,h}{T} = 600 \times \tfrac{1}{30} \\
&= 20\ \text{E}
\end{aligned}
$$

#### (b) Traffic carried
Calls carried $= 600 - 40 = 560$
$$
\begin{aligned}
A_c &= 560 \times \tfrac{1}{30} = 18.67\ \text{E}
\end{aligned}
$$

#### (c) Traffic lost
$$
\begin{aligned}
A_l &= A_o - A_c = 40 \times \tfrac{1}{30} = 1.33\ \text{E}
\end{aligned}
$$

#### (d) Grade of service
$$
\begin{aligned}
GoS &= \frac{C_l}{C_o} = \frac{40}{600} = 0.0667
\end{aligned}
$$

#### (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
$$
\begin{aligned}
T_c &= GoS \times 3600\ \text{s} = \frac{40}{600} \times 3600 \\
&= 240\ \text{s}
\end{aligned}
$$

**Answer:** offered traffic = **20 E**, carried traffic = **18.67 E** (= 18 2/3 E), lost traffic = **1.33 E**, GoS = **0.0667** (about 1 in 15), congestion duration = **240 s = 4 min**.

## Check yourself

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

1. 600 calls with h = 120 s in one hour: offered traffic?
   A. 10 E
   B. 20 E
   C. 40 E
   D. 72 E
   Answer: B. A = 600 × 120 / 3600 = 20 E. (Key point: Offered traffic)

2. With 40 of 600 calls lost, carried traffic is:
   A. 18.67 E
   B. 19.33 E
   C. 20 E
   D. 1.33 E
   Answer: A. 560 × 1/30 = 18.67 E. (Key point: Carried traffic)

3. What is the GoS?
   A. 0.040
   B. 0.0667
   C. 0.0714
   D. 0.933
   Answer: B. GoS = 40/600 = 0.0667. (Key point: GoS)

4. How is the duration of congestion found?
   A. GoS × 3600 s = 240 s
   B. Lost traffic × 3600 s
   C. Carried traffic × 3600
   D. h × calls lost = 4800 s
   Answer: A. For Poisson traffic time congestion equals call congestion: 0.0667 × 3600 = 240 s. (Key point: Congestion)

## Close variants set in other exams

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

## Related in Telecommunications

- [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×
- [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 engineering goals; E.600 busy hours](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Atraffic-engineering-busy-hour-e600) — Telecommunications, chapter 6
- [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
- [E.600 busy hour, NTA duties and ITU sectors](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Ae600-busy-hour-nta-itu) — Telecommunications, chapter 7