---
title: "Pure loss system and binomial model"
subject: "Telecommunications"
chapter: 6
key: "semester_8/telecommunications:pure-loss-binomial-model"
source: https://www.bctengineer.com/q/semester_8/telecommunications/pure-loss-binomial-model
site: BCT Engineering
---

# Pure loss system and binomial model

Telecommunications · Chapter 6: Telephone Traffic

- Asked in: 2072 Chaitra, 2+6 marks
- Video: [TELE -TRAFFIC SYSTEM MODEL](https://www.youtube.com/watch?v=vWzan9bdon8) — Prof. Barapate's Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/pure-loss-binomial-model
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-6

## Question

What is pure loss system? Describe the teletraffic Binomial model.

## Short answer

A pure loss system has no waiting room; blocked calls are cleared. The Binomial model applies when N≤n sources are independent, each busy probability a = β/(1+β), with P(k) = C(N,k)a^k(1-a)^{N-k}. Traffic is smooth (variance < mean), and because N≤n, no call is blocked.

## Answer

#### Pure loss system

A **pure loss system** (lost-calls-cleared system) has a fixed number of servers (trunks) and **no waiting room**. A call that arrives when all servers are busy is rejected at once and disappears; it does not wait and is not retried. The performance measure is the **grade of service** (blocking probability). An ordinary circuit-switched trunk group giving "all lines busy" tone is the common example; Erlang B, Engset and the binomial model are pure loss models.

#### Teletraffic binomial model

The binomial model applies when the number of sources $N$ is small and **not larger than the number of servers** ($N \le n$), and the sources behave independently of each other.

**Assumptions**

1. $N$ independent sources; each source is either idle or busy.
2. Mean idle time $1/\gamma$ and mean holding time $h = 1/\mu$ (exponential).
3. Offered traffic per idle source $\beta = \gamma/\mu$.
4. Because $N \le n$, every call finds a free server, so no call is ever blocked.

**Probability of a source being busy.** Each source alternates between idle and busy, so the fraction of time one source is busy is

$$
a = \frac{h}{h + 1/\gamma} = \frac{\beta}{1+\beta}
$$

**State probabilities.** Since the sources are independent, the number of busy sources $k$ follows a binomial distribution:

$$
P(k) = \binom{N}{k}\, a^k (1-a)^{N-k}, \qquad k = 0,1,\dots,N
$$

This is obtained from the birth–death (cut) equations with arrival rate $(N-k)\gamma$ in state $k$ and departure rate $k\mu$:

$$
(N-k)\gamma\,P(k) = (k+1)\mu\,P(k+1)
$$

which gives $P(k) = \binom{N}{k}\beta^k P(0)$ and $P(0) = (1+\beta)^{-N}$.

**Traffic characteristics**

| Quantity | Value |
|---|---|
| Offered traffic | $A = N a$ |
| Carried traffic | $Y = N a$ (equal to offered) |
| Mean busy servers | $N a$ |
| Variance | $N a (1-a)$ (smaller than mean: smooth traffic) |
| Time congestion | $E = 0$ if $N < n$; $E = a^N$ if $N = n$ |
| Call congestion | $B = 0$ |

Because the variance is less than the mean, binomial traffic is called **smooth** traffic, unlike Poisson (random) traffic, where the variance equals the mean.

**Use of the model.** If $N > n$ and blocked calls are held, the same distribution is used: a call is blocked when $n$ or more of the other $N-1$ sources are busy, so the congestion is $\sum_{k=n}^{N-1}\binom{N-1}{k}a^k(1-a)^{N-1-k}$. Example: 4 subscribers, each busy 0.2 E, on 4 lines: the probability that all 4 lines are busy $= 0.2^4 = 0.0016$, but no call is lost since every subscriber has a line.

## Check yourself

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

1. What is a pure loss system?
   A. Servers with infinite queue
   B. Unlimited servers
   C. Blocked calls retry until served
   D. Fixed servers, no waiting room; blocked calls are cleared
   Answer: D. Blocked calls disappear immediately. (Key point: Pure loss)

2. Probability a source is busy?
   A. a = beta/(1+beta)
   B. a = beta
   C. a = 1/(1+beta)
   D. a = 1 - beta
   Answer: A. a = h/(h+1/gamma) = beta/(1+beta). (Key point: Busy probability)

3. Why is binomial traffic called smooth?
   A. Variance exceeds mean
   B. Variance Na(1-a) is less than mean Na
   C. Variance equals mean
   D. It has no variance
   Answer: B. Poisson has variance equal to mean. (Key point: Smooth traffic)

4. When N <= n, call congestion B is?
   A. a^N
   B. 1
   C. 0
   D. beta
   Answer: C. Every call finds a free server. (Key point: Congestion)

## Related in Telecommunications

- [Pure loss system and Engset model](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Apure-loss-engset-model) — Telecommunications, chapter 6
- [Infinite and pure loss systems; Engset model](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Ainfinite-pure-loss-engset) — Telecommunications, chapter 6
- [GOS and blocking probability in a loss system](https://www.bctengineer.com/question.md?key=semester_8%2Ftelecommunications%3Agos-blocking-loss-system) — Telecommunications, chapter 6

## Related in other subjects

- [Log-distance path loss model](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Alog-distance-path-loss) — Wireless Communication, chapter 3
- [Two-ray model advantages and path loss derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atwo-ray-advantages-path-loss) — Wireless Communication, chapter 3
- [Okumura model and median path loss numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-model-median-path-loss) — Wireless Communication, chapter 3
- [Define system, model, simulation](https://www.bctengineer.com/question.md?key=semester_8%2Fsimulation_and_modeling%3Asystem-model-simulation-define) — Simulation and Modeling, chapter 1