Consider the following sys what is the probability that the system operates? Wha

Question

Consider the following sys what is the probability that the system operates? What is the probability that the system fails due to the components in series? Assume parallel components do not fail. what is the probability that the system fails due to the components in parallel? Assume series components do not fail Compute and compare the probabilities that the system fails when the probability that component C1 functions is improved to a value of 0.95 and when the probability that component C2 functions is improved to a value of 0.85. Which improvement increases the system reliability more? tem made up of functional components in parallel and series

Explanation / Answer

1. Probability that system operates

=P(All components work)+P(component two fails)+P(Component 3 fails)

= 0.9*0.8*0.85*0.95+0.9*0.2*0.85*0.95+0.9*0.15*0.8*0.95

=0.82935

2. P(System fails due to component in series)

= P(Component 1 fails)+P(Component 4 fails)+P(Components 1 and 4 fail)

=0.1*0.8*0.85*0.95+0.9*0.8*0.85*0.05+0.1*0.8*0.85*0.05

=0.0986

3. P(System fails due to components in parallel)

=P(Both components 2 and 3 fail)

=0.9*0.2*0.15*0.95

=0.02565

4.

When P(C1)=0.95

P(System fails)=1-P(System operates)

=1-P(All components work)+P(component two fails)+P(Component 3 fails)

= 1-0.95*0.8*0.85*0.95+0.95*0.2*0.85*0.95+0.95*0.15*0.8*0.95

=1-0.875425

=0.124575

When P(C2)=0.85

P(System fails)=1-P(System operates)

=1-P(All components work)+P(component two fails)+P(Component 3 fails)

= 1-0.9*0.85*0.85*0.95+0.9*0.15*0.85*0.95+0.9*0.15*0.85*0.95

=1-0.8357625

=0.1642375

As P(System fails) is more in case 2 than case 1, the reliability of case 1 is more than case 2.

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