Traffic numerical: 1200 calls, 6 lost

Given: calls offered $C_o = 1200$ in the busy hour, calls lost $C_l = 6$, mean holding time $h = 3$ min, observation period $T = 60$ min.

Traffic intensity (in erlang) is the number of calls in the period multiplied by the mean holding time, divided by the period:

$$A = \frac{C\,h}{T}$$

a) Traffic offered

$$\begin{aligned} A_o &= \frac{1200 \times 3}{60} \\ &= 60\ \text{E} \end{aligned}$$

b) Traffic carried

Calls carried $= 1200 - 6 = 1194$.

$$\begin{aligned} A_c &= \frac{1194 \times 3}{60} \\ &= 59.7\ \text{E} \end{aligned}$$

c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 60 - 59.7 \\ &= 0.3\ \text{E} \end{aligned}$$

(Check: $6 \times 3 / 60 = 0.3$ E.)

d) Grade of service

GoS is the fraction of offered calls (or offered traffic) that is lost:

$$\begin{aligned} B &= \frac{\text{calls lost}}{\text{calls offered}} = \frac{6}{1200} \\ &= 0.005\ (\text{i.e. } 0.5\%) \end{aligned}$$

e) Total duration of the periods of congestion

For random (Poisson) traffic, the call congestion equals the time congestion, so the trunk group is fully busy for a fraction $B$ of the hour:

$$\begin{aligned} t_{cong} &= B \times T = 0.005 \times 3600\ \text{s} \\ &= 18\ \text{s} \end{aligned}$$

Answer: traffic offered = 60 E, traffic carried = 59.7 E, traffic lost = 0.3 E, GoS = 0.005 (0.5%), total congestion time = 18 s in the busy hour.