Traffic numerical: 600 calls, 40 lost, 120 s

Lost calls and grade of service

• Lost call: a call attempt that cannot be connected because all servers (trunks/switch paths) are busy; in a loss system it is cleared and not served.
• Grade of Service (GoS): the proportion of calls lost in the busy hour, $GoS = \dfrac{\text{calls lost}}{\text{calls offered}} = \dfrac{A_l}{A_o}$. A lower GoS means better service (e.g. 0.01 = 1 call in 100 lost).

Given: calls offered $C_o = 600$, calls lost $C_l = 40$, mean holding time $h = 120\ \text{s} = \tfrac{1}{30}$ h, period T = 1 h.

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

(a) Traffic offered

$$\begin{aligned} A_o &= \frac{C_o\,h}{T} = 600 \times \tfrac{1}{30} \\ &= 20\ \text{E} \end{aligned}$$

(b) Traffic carried

Calls carried $= 600 - 40 = 560$

$$\begin{aligned} A_c &= 560 \times \tfrac{1}{30} = 18.67\ \text{E} \end{aligned}$$

(c) Traffic lost

$$\begin{aligned} A_l &= A_o - A_c = 40 \times \tfrac{1}{30} = 1.33\ \text{E} \end{aligned}$$

(d) Grade of service

$$\begin{aligned} GoS &= \frac{C_l}{C_o} = \frac{40}{600} = 0.0667 \end{aligned}$$

(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

$$\begin{aligned} T_c &= GoS \times 3600\ \text{s} = \frac{40}{600} \times 3600 \\ &= 240\ \text{s} \end{aligned}$$

Answer: offered traffic = 20 E, carried traffic = 18.67 E (= 18 2/3 E), lost traffic = 1.33 E, GoS = 0.0667 (about 1 in 15), congestion duration = 240 s = 4 min.