Traffic numerical: 1000 calls, 5 lost

Given: calls offered $C_o = 1000$, calls lost $C_l = 5$, mean holding time $h = 4\ \text{min} = \tfrac{1}{15}$ h, period T = 1 h.

Traffic in Erlangs $= \dfrac{\text{calls} \times h}{T}$.

(a) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = 1000 \times \tfrac{1}{15} \\ &= 66.67\ \text{E} \end{aligned}$$

(b) Traffic carried

Calls carried $= 1000 - 5 = 995$

$$\begin{aligned} A_c &= 995 \times \tfrac{1}{15} = 66.33\ \text{E} \end{aligned}$$

(c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 5 \times \tfrac{1}{15} = 0.33\ \text{E} \end{aligned}$$

(d) Grade of service

$$\begin{aligned} GoS &= \frac{C_l}{C_o} = \frac{5}{1000} = 0.005 \end{aligned}$$

(e) Duration of congestion

For pure-chance (Poisson) traffic the proportion of time all trunks are busy (time congestion) equals the proportion of calls lost (call congestion), so congestion time in the busy hour is

$$\begin{aligned} T_c &= GoS \times 3600\ \text{s} = \frac{5}{1000} \times 3600 \\ &= 18\ \text{s} \end{aligned}$$

Answer: (a) offered = 66.67 E (66 2/3 E), (b) carried = 66.33 E (66 1/3 E), (c) lost = 0.33 E (1/3 E), (d) GoS = 0.005, (e) congestion period = 18 s in the busy hour.