Question

given the following diagram, determine the total system reliability if the individual component reliabilities are a = 0.91, b = 0.80, and c = 0.86. (hint: use equations 5.2 and 5.3, and note that the reliabilities of the parallel components are different.) round your answer to four decimal places. reliability:

Explanation:

Step1: Calculate parallel reliability (B and C)

The formula for parallel components (both must fail to system fail) is \( R_{parallel} = 1 - (1 - R_B)(1 - R_C) \).
Substitute \( R_B = 0.80 \), \( R_C = 0.86 \):
\( 1 - (1 - 0.80)(1 - 0.86) = 1 - (0.20)(0.14) = 1 - 0.028 = 0.972 \)

Step2: Calculate series reliability (A and parallel)

The formula for series components (all must work) is \( R_{series} = R_A \times R_{parallel} \).
Substitute \( R_A = 0.91 \), \( R_{parallel} = 0.972 \):
\( 0.91 \times 0.972 = 0.88452 \)

Step3: Round to four decimals

Round \( 0.88452 \) to four decimal places: \( 0.8845 \)

Answer:

0.8845