Question

A bank loan processing system has three components with individual reliabilities as shown:

R₁ = 0.84
R₂ = 0.950
R₃ = 0.98

Round answers 3 decimal places

What is the reliability of the bank loan processing system?  .782

What would be the reliability of the bank system above if each of the three components had a backup with a reliability of 0.80?

How would the total reliability be different?

Answer 1

Question – a:

As all are in series (as question mentions that all are independent), the system reliability

R = R1*R2*R3

= 0.84*0.95*0.98

= 0.78204

Answer is: 0.782

Question – b:

Now each component has a backup with reliability 0.80

So first component combined reliability (with backup) = 1 – (1-R1)*(1-Rbackup) = 1 – (0.16)*(0.20)

= 1 – 0.032

= 0.968

Similarly, the combined reliability of second component = 1 – (0.05)*(0.2) = 1 – 0.01 = 0.99

The combined reliability of third component = 1 – (0.02)*(0.2) = 1 – 0.004 = 0.996

Now, the combined reliability of the whole system = 0.968*0.99*0.996 = 0.954

Answer is: 0.954

Question – c:

The reliabilities are different as in Q-b, there exists a backup component to work when the main component is in failure state, which eventually increases the reliability to a large extent.

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