3. A device consists of two redundant subunits; only one needs to be working for
ID: 3352123 • Letter: 3
Question
3. A device consists of two redundant subunits; only one needs to be working for the device to function. The first subunit has two parts Wi and W2, and the second has two parts W3 and W4. In each subunit, both parts must function for the subunit to work. The four parts are identical and equally reliable with probability p 0.9 of functioning, independent of each other. (a) What is the overall probability the device functions correctly? (b) Suppose you have available two ultra-reliable parts, with p 0.99 that each functions, which two parts in the device should you replace with these in order to maximize device reliability? Justify your answer by calculating the resulting reliability and explain your choice.Explanation / Answer
(a)
Given:
R(W1) = R(W2) = R(W3) = R(W4) = 0.9
W1 & W2 are in series.
R(Subunit 1) = 0.9 X 0.9 = 0.81
W3 & W4 are in series.
R(Subunit 2) = 0.9 X 0.9 = 0.81
Subunit 1 & Subunit 2 are in parallel.
R(Device) = 1 - (1-0.81)2 = 0.9639
So,
Overall probability that the device functions correctly = 0.9639
(b)
The various options are:
OPTION 1:
W1 = W2 = 0.99
W3 = W4 = 0.9
R(Subunit 1) = 0.99 X 0.99 = 0.9801
R(Subunit 2) = 0.9 X 0.9 = 0.81
R(Device) = 1 - (1-0.9801) (1-0.81) = 1 - (0.0199 X 0.19) = 0.9962
OPTION 2:
W1 = W2 = 0.9
W3 = W4 = 0.99
R(Subunit 1) = 0.9 X 0.9 = 0.81
R(Subunit 2) = 0.99 X 0.99 = 0.9801
R(Device) = 1 - (1 - 0.81) (1 - 0.9801) = 0.9962
OPTION 3:
W1 = 0.9
W2 = 0.99
W3 = 0.9
W4 = 0.99
R(Subunit 1) = 0.9 X 0.99 = 0.891
R(Subunit 2) = 0.9 X 0.99 = 0.891
R(Device) = 1 - (1 - 0.891)2 = 0.9881
OPTION 4:
W1 = 0.99
W2 = 0.9
X3 = 0.99
W4 = 0.9
Similar to OPTION 3:
R(Device) = 0.9881
Comparing OPTIONS 1 to 4, we note:
OPTIONS 1 & 2, each with Reliability 0.9962 :
either replace parts 1 & 2
OR
replace parts 3 & 4
in order to maximize device reliability.
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