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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

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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.

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 * 0.01 * \cdots * 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.

User Alexander Volkov
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