Infinite-source blocking; 40 subscribers

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).