5) In a complex manufacturing process, three operations are performed in series.

Question

5) In a complex manufacturing process, three operations are performed in series. Because of the nature of the process, machines frequently fall out of adjustment and must be repaired. To keep the system going, two identical machines are used at each stage; thus, if one fails, the other can be used while the first is repaired.

The reliabilities of the machines are as follows:

Machine Reliability

A 0.85

B 0.90

C 0.75

a) What would be the system reliability if there were only one machine at each stage?

b) What is the system reliability with two machines at each stage?

Explanation / Answer

BACK UP THEORY

Machine reliability = 1 – P(Failure)

System reliability with 3 components in series = (reliability of component 1) x (reliability of component 2) x (reliability of component 3)

Machine reliability with a stand-by(two machines in place of one machine) = P(at least one will function) = 1 – P(Neither functions) = 1 – P(Both fails) = 1 – {(1- R1)(1 – R2)}, where R represents reliability.

SOLUTION FOR PART (a)

Since there is only 1 machine of each type, System reliability = (reliability of Machine A) x (reliability of Machine B) x (reliability of Machine C) = 0.85 x 0.90 x 0.75 = 0.57375 i.e., 57.375% reliability ANSWER

SOLUTION FOR PART (b)

Since there are 2 machines of each type, Machine reliability for A = 1 – {(1- 0.85)(1 – 0.85)}

= 1 – 0.152 = 0.9775. Similarly, Machine reliability for B = 1 – {(1- 0.9)(1 – 0.9)}

= 1 – 0.12 = 0.99 and Machine reliability for C = 1 – {(1- 0.75)(1 – 0.75)}

= 1 – 0.252 = 0.9375

System reliability = (reliability of Machine A) x (reliability of Machine B) x (reliability of Machine C) = 0.9775 x 0.99 x 0.9375 = 0.9072 i.e., 90.72% reliability ANSWER

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