IV. A. 2. d. Reliability Block Diagrams and Models – Complex

(C) 0.918

If you see this type of problem, while it’s easy to solve, it takes time. Hold till the end after you have answered all the quick one you know.

The approach to solve for the system reliability value is to reduce the parallel elements to form series elements. The equation to reduce two elements in parallel to determine the reliability of the parallel elements is

$$ {{R}_{a||b}}=1-\left( 1-{{R}_{a}} \right)\left( 1-{{R}_{b}} \right)$$

Looking at the diagram lets start by reducing the block R3 and R4

$$ \begin{array}{l}{{R}_{3||4}}=1-\left( 1-{{R}_{3}} \right)\left( 1-{{R}_{4}} \right)\\{{R}_{3||4}}=1-\left( 1-0.75 \right)\left( 1-0.7 \right)\\{{R}_{3||4}}=0.925\end{array}$$

Next, let’s reduce the parallel structure involving R5 and R6

$$ \begin{array}{l}{{R}_{5||6}}=1-\left( 1-{{R}_{5}} \right)\left( 1-{{R}_{6}} \right)\\{{R}_{5||6}}=1-\left( 1-0.6 \right)\left( 1-0.5 \right)\\{{R}_{5||6}}=0.8\end{array}$$

Then reduce the series string of R2 and R3||4

$$ \begin{array}{l}{{R}_{2-3||4}}={{R}_{2}}\times {{R}_{3||4}}\\{{R}_{_{2-3||4}}}=0.9\times 0.925\\{{R}_{_{2-3||4}}}=0.8325\end{array}$$

That sets up another simple parallel structure to reduce

$$ \begin{array}{l}{{R}_{(2-3||4)||(5||6)}}=1-\left( 1-{{R}_{2-3||4}} \right)\left( 1-{{R}_{5||6}} \right)\\{{R}_{(2-3||4)||(5||6)}}=1-0.8325\times 0.8\\{{R}_{(2-3||4)||(5||6)}}=0.9665\end{array}$$

Which leaves two elements in series

$$ \begin{array}{l}{{R}_{sys}}={{R}_{1}}\times {{R}_{(2-3||4)||(5||6)}}\\{{R}_{sys}}=0.95\times 0.9665\\{{R}_{sys}}=0.918\end{array}$$

(A) Take the reliability for A and combine it with the reliability for C and then combine that result with the reliability for E.

The simplest approach is to simplify the parallel elements, A and C, into a single block, which is then in series with E