Answer:
Step-by-step explanation:
We are given that:
The failure rate = 0.1 failure/year
i.e. = [tex]\dfrac{0.1}{365\times 24}= 0.00001 \ failure /hour[/tex]
The repair rate = 365 repair/year, if we convert this to repair/hour; we get:
[tex]= \dfrac{365}{365\times 24} \\ \\ = 0.04 \ repair/ hour[/tex]
Time (t) = 1 hour
P(failure) = 1 - R ; if (t=1)
where:
[tex]Reliability (R ) = e^{-\lambda t}(1+ \lambda t) \\ \\ = e^{-0.00001*1}(1+0.00001 *1) \\ \\ \simeq 0.9999[/tex]
P (failure) = 1 - R
= 1 - 0.9999
= 0.0001
≅ 0
The frequency meantime of failure = [tex]\dfrac{1}{\lambda}[/tex]
[tex]\dfrac{1}{0.00001}[/tex]
= 100000
The average downtime of repair for the system = [tex]\dfrac{1}{\mu}[/tex]
[tex]=\dfrac{1}{0.04}[/tex]
= 25
