Question

A communications circuit is known to have an availability of 0.99 (that is, 99% of the time, the circuit is operational). A total of n such circuits are going to be set up by the FAA between San Francisco and Los Angeles in such a way that the circuits will fail indepen- dently of each other. How many such parallel circuits must be set up to attain an overall availabililty of 0.99999

Answer 1

Answer:

Assume that for the communication to be available means that at least one of the $n$ circuits is operational. It would take at least 3 circuits to achieve a $0.99999$ overall availability.

Step-by-step explanation:

The probability that one circuit is not working is $1 - 0.99 = 0.01$.

Since the circuits here are all independent of each other, the probability that none of them is working would be $\displaystyle \underbrace{0.01 \times 0.01 \times \cdots \times 0.01}_{\text{ n times}}$. That's the same as $0.01^n$.

The event that at least one of the $n$ circuits is working is the complement of the event that none of them is working. To find the probability that at least one of the $n$ circuits is working, simply subtract the probability that none of the circuit is working from one. That is:

$\begin{aligned}&P(\text{At least one working}) \cr &= 1 - P(\text{None is working}) \cr &= 1- 0.01^n\end{aligned}$.

The question requests that

$P(\text{At least one working}) \ge 0.99999$.

In other words,

$1- 0.01^n \ge 0.99999$.

$0.01^n \le 1 - 0.99999 = 0.000001 = 10^{-6}$.

Note that $0.01 = 10^{-2}$. Hence, the inequality becomes

$\left(10^{-2}\right)^n \le 10^{-6}$.

$10^{-2\,n} \le 10^{-6}$

Take the natural log of both sides of the equation:

$\ln\left(10^{-2\, n}\right) \le \ln \left(10^{-6}\right)$.

$(-2\, n)\ln\left(10\right) \le (-6) \ln\left(10\right)$.

$10 > 1$, hence $\ln(10) > 0$. Divide both sides by $\ln(10)$:

$-2\,n \le -6$.

$n \ge 3$.

In other words, at least three parallel circuits must be set up to achieve that availability.