---
title: "Infinite-source blocking; 40 subscribers"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:infinite-source-blocking-40-subscribers"
source: https://www.bctengineer.com/q/semester_8/telecommunications/infinite-source-blocking-40-subscribers
site: BCT Engineering
---

# Infinite-source blocking; 40 subscribers

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2079 Bhadra, 4+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/infinite-source-blocking-40-subscribers
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

Describe the blocking formulas used in infinite source. Over a 20-minute observation interval, 40 subscribers initiate calls. Total duration of the calls is 4800 seconds. Calculate the load offered to the network by the subscribers and the average subscriber traffic.

## Short answer

For infinite sources, Erlang B formula gives blocking: B = (A^N/N!)/Σ A^k/k!. Poisson lost-calls-held: P = e^{-A} Σ_{k=N}^∞ A^k/k!. Erlang C for delay: C(N,A). With 40 subscribers, 20 min, total call duration 4800 s, load = 4800/1200 = 4 E; average subscriber traffic = 4/40 = 0.1 E.

## Answer

#### Blocking formulas for infinite sources
With a very large number of sources, calls arrive as a Poisson process of constant rate and offered traffic A is independent of the number of busy servers. With N servers:

**1. Lost calls cleared – Erlang B formula** (blocked calls disappear):
$$
B = E_{1,N}(A) = \frac{A^N/N!}{\sum_{k=0}^{N}A^k/k!}
$$
Used for trunk groups; recursive form $E_{1,N} = \dfrac{A\,E_{1,N-1}}{N + A\,E_{1,N-1}}$, $E_{1,0}=1$.

**2. Lost calls held – Poisson formula** (blocked callers keep trying for a time equal to the holding time):
$$
P = e^{-A}\sum_{k=N}^{\infty}\frac{A^k}{k!}
$$
Gives slightly higher blocking than Erlang B; used in North American trunk design.

**3. Lost calls delayed – Erlang C formula** (blocked calls wait in an infinite queue):
$$
C(N,A) = \frac{\dfrac{A^N}{N!}\dfrac{N}{N-A}}{\sum_{k=0}^{N-1}\dfrac{A^k}{k!} + \dfrac{A^N}{N!}\dfrac{N}{N-A}}, \quad A < N
$$
Mean delay of all calls $= C(N,A)\,h/(N-A)$.

#### Numerical
**Given:** observation period T = 20 min = 1200 s, number of subscribers = 40 (each initiates a call), total call duration = 4800 s.

Offered load = total holding time / observation time:
$$
\begin{aligned}
A &= \frac{\sum h_i}{T} = \frac{4800}{1200} = 4\ \text{E}
\end{aligned}
$$

Average traffic per subscriber:
$$
\begin{aligned}
a &= \frac{A}{40} = \frac{4}{40} = 0.1\ \text{E}
\end{aligned}
$$

(Supporting values: call rate $\lambda = 40/20 = 2$ calls/min; mean holding time $h = 4800/40 = 120$ s $= 2$ min; $A = \lambda h = 2 \times 2 = 4$ E.)

**Answer:** offered load = **4 E**; average subscriber traffic = **0.1 E** (each subscriber busy 10% of the time).

## Check yourself

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

1. Which blocking formula applies to infinite sources when blocked calls are cleared?
   A. Erlang B
   B. Erlang C
   C. Engset
   D. Binomial
   Answer: A. Lost calls cleared with Poisson arrivals gives Erlang B. (Key point: Erlang B)

2. What happens to blocked calls in the Erlang C model?
   A. They vanish
   B. They wait in an infinite queue
   C. They keep retrying for one holding time
   D. They are rerouted to another trunk group
   Answer: B. Erlang C is lost calls delayed. (Key point: Erlang C)

3. 40 subscribers, total call time 4800 s in 1200 s. What is the offered load?
   A. 40 E
   B. 0.1 E
   C. 4 E
   D. 24 E
   Answer: C. A = total holding time / observation time = 4800/1200 = 4 E. (Key point: Offered load)

4. What is the average traffic per subscriber?
   A. 4 E
   B. 0.4 E
   C. 10 E
   D. 0.1 E
   Answer: D. 4 E / 40 subscribers = 0.1 E, i.e. each is busy 10% of the time. (Key point: Per-subscriber traffic)

## Related in Telecommunications

- [Decision tree; infinite source blocking formulas](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adecision-tree-infinite-sources) — Telecommunications, chapter 6, asked 2×
- [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×
- [Decision tree; finite source blocking formulas](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adecision-tree-finite-sources) — Telecommunications, chapter 6, asked 2×
- [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×
- [Decision tree formulas; 20 servers, 10 E](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Adecision-tree-20-servers) — Telecommunications, chapter 6
- [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×

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