please, I need clear writing if you choose to do it by hand.
thanks,
(a)
W1 & W2 are in parallel with each p = 0.9
Reliability of W1 W2 = 1 - (1 - 0.9)2 = 0.99
W3 & W4 are in parallel with each p = 0.9
Reliability of W3 W4 = 1 - (1 - 0.9)2 = 0.99
W1W2 and W3W4 are in series.
So,
Overall probability the device functions correctly = 0.99 X 0.99 = 0.9801
(b)
Case 1:
W1 = 0.99, W2 =0.99, W3 =0.9, W4 = 0.9
Reliability of W1 W2 = 1 - (1 - 0.99)2 = 0.9999
Reliability of W3 W4 = 1 - (1 - 0.9)2 = 0.99
So,
Overall probability the device functions correctly = 0.9999 X 0.99 = 0.9899
Case 2:
W1 = 0.9, W2 =0.9, W3 =0.99, W4 = 0.99
Reliability of W1 W2 = 1 - (1 - 0.9)2 = 0.99
Reliability of W3 W4 = 1 - (1 - 0.99)2 = 0.9999
So,
Overall probability the device functions correctly = 0.99 X 0.9999 = 0.9899
Case 3:
W1 = 0.99, W2 =0.9, W3 =0.99, W4 = 0.9
Reliability of W1 W2 = 1 - (1 - 0.99)(1-0.9) = 0.999
Reliability of W3 W4 = 1 - (1 - 0.99) (1-0.9)= 0.999
So,
Overall probability the device functions correctly = 0.999 X 0.999= 0.9980
Case 4:
W1 = 0.9, W2 =0.99, W3 =0.9, W4 = 0.99
Reliability of W1 W2 = 1 - (1 - 0.9)(1-0.99) = 0.999
Reliability of W3 W4 = 1 - (1 - 0.9) (1-0.99)= 0.999
So,
Overall probability the device functions correctly = 0.999 X 0.999= 0.9980
We note from the above:
Cases 3 & 4 each give overall reliability = 0.9980 > 0.9899 for Cases 1 & 2.
So,
we should replace parts 1 & 3 each with p = 0.99 in order to maximize device reliability
OR
we should replace parts 2 & 4 each with p = 0.99 in order to maximize device reliability
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