Trunk group numerical: 30 E, 12 s, 2 lost

Given: offered traffic A = 30 E, total time all trunks busy = 12 s in the busy hour (3600 s), calls lost = 2.

For pure-chance traffic, the probability that a call is lost (call congestion) equals the fraction of time all trunks are busy (time congestion):

$$\begin{aligned} B &= \frac{12}{3600} = \frac{1}{300} \end{aligned}$$

(i) Average number of calls carried

Calls lost = B × calls offered, so

$$\begin{aligned} C_o &= \frac{\text{calls lost}}{B} = 2 \times 300 = 600\ \text{calls} \\ C_c &= C_o - C_l = 600 - 2 = 598\ \text{calls} \end{aligned}$$

(ii) Average call duration

$$\begin{aligned} A &= \frac{C_o\,h}{T} \Rightarrow h = \frac{A\,T}{C_o} \\ h &= \frac{30 \times 3600}{600} = 180\ \text{s} = 3\ \text{min} \end{aligned}$$

(iii) Calls offered in a period equal to h

Call arrival rate $= 600/3600$ calls per second. In 180 s:

$$\begin{aligned} n &= \frac{600}{3600} \times 180 = 30\ \text{calls} \end{aligned}$$

This equals the offered traffic in Erlangs, which is the definition of the Erlang: traffic in E = mean number of calls arriving during one mean holding time ($A = \lambda h$).

Answer: (i) 598 calls carried (of 600 offered), (ii) h = 180 s (3 min), (iii) calls offered in one holding time = 30 = A, as required.