Traffic intensity, GOS and call duration probability

Traffic intensity

Average number of simultaneous calls in a period, in Erlangs: $A = \dfrac{C\,h}{T}$ (C calls, mean holding time h, period T). 1 E = one circuit busy all the time.

Grade of Service

Proportion of calls lost (or excessively delayed) in the busy hour: $GoS = \dfrac{\text{calls lost}}{\text{calls offered}}$; smaller is better.

Numerical

Call durations are taken as negative exponential with mean $h = 2$ min:

$$P(T > t) = e^{-t/h}$$

The exponential distribution is memoryless: the remaining duration of a call does not depend on how long it has already lasted. So the fact that the call has lasted 4 min does not change the answer.

(a) Lasts at least another 4 min

$$\begin{aligned} P(T > 8 \mid T > 4) &= P(T > 4) = e^{-4/2} \\ &= e^{-2} = 0.1353 \end{aligned}$$

(b) Ends within the next 4 min

$$\begin{aligned} P(T \le 8 \mid T > 4) &= 1 - e^{-2} \\ &= 0.8647 \end{aligned}$$

Answer: (a) 0.1353 (about 13.5%), (b) 0.8647 (about 86.5%).