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Intelligent-Management-Of-Electrical-Systems-in-Industri-->View question

Asked On2019-04-07 17:20:53 by:Rohit498

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Reliability of a system with the components connected serially is given by the product of their individual reliabilities
i.e, initially, the reliability of the machine is:
R(M)=R(P).R(Q).R(R)
= 0.97*0.86*0.93
Then, 2 stand-by units of Q are added, now the total Q units become 3, where these are arranged parallel.
Reliability of a system with the components connected parallel is given by 1 - product of their 1- individual reliabilities
i.e, now at Q:
R(Q)=1-{[1-R(Q1)].[1-R(Q2)].[1-R(Q3)]}
=1-{[1-0.86]*[1-0.86]*[1-0.86]}
=0.997
Since P and R are connected serially to this, the total reliability becomes:
R(M)=R(P).R(Q).R(R)
= 0.97*0.997*0.93
= 0.899

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