Pure loss system and binomial model

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

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.