Device holding time and calls per hour

The device is a single server and calls that find it busy wait (delay system). With random (Poisson) calls and exponential holding time this is an M/M/1 (Erlang delay, N = 1) system. The "10 ms" is the mean delay before a call is served (averaged over all calls):

$$W = \frac{A\,h}{1 - A}$$

where $h$ = mean holding time and $A = \lambda h$ = traffic (occupancy) of the device.

(i) h = 50 ms, W = 10 ms

$$\begin{aligned} \frac{A \times 50}{1-A} &= 10 \\ 50A &= 10 - 10A \\ A &= \frac{10}{60} = \frac{1}{6}\ \text{E} \end{aligned}$$

Calls per hour:

$$\begin{aligned} C &= \frac{A\,T}{h} = \frac{(1/6) \times 3600}{0.05} \\ &= 12000\ \text{calls/h} \end{aligned}$$

(Without the delay limit, a fully occupied device could take $3600/0.05 = 72000$ calls/h, but delay would then be infinite.)

(ii) 18000 calls/h, W = 10 ms

$\lambda = 18000/3600 = 5$ calls/s, so $A = 5h$ (h in seconds).

$$\begin{aligned} \frac{\lambda h \cdot h}{1 - \lambda h} &= 0.01 \\ 5h^2 &= 0.01(1 - 5h) \\ 5h^2 + 0.05h - 0.01 &= 0 \\ h &= \frac{-0.05 + \sqrt{0.05^2 + 4(5)(0.01)}}{2 \times 5} \\ &= \frac{-0.05 + 0.45}{10} = 0.04\ \text{s} \end{aligned}$$

Check: $A = 5 \times 0.04 = 0.2$; $W = 0.2 \times 40/(1-0.2) = 10$ ms.

Answer: (i) 12,000 calls per hour; (ii) maximum average holding time h = 40 ms.