A design engineer in an electronics company is faced with a reliability-cost dec
ID: 3309597 • Letter: A
Question
A design engineer in an electronics company is faced with a reliability-cost decision regarding an electronic assembly that can be designed as a parallel sub- system. The mission reliability requirement for the device is at least 0.98. There are three components available to the engineer. Component A has 0.90 reliablity. a space requirement of 0.5 × 0.5 × 0.5 in, and a price of $38. Component B has 0.85 reliability, with a space requirement of 0.7x 0.7x 0.7 in, and a price of $20. Component C has 0.80 reliability, and a space requirement of 1 × 1 × 1 in, and a price of $15. The components can be mixed. For example, we can placc A, B, and/or C in parallel, or A and A in parallel, and so forth. [TY] a If minimum cost with acceptable reliability is the selection criterion, what is the best configuration? b If the design space is limited to 1.5 × 1.5 × 1.5 in3 and cost is not a consideration. 'what is the best configuration? c If space is not a limitation, and the configuration must yield a reliability of at least 0.99, what arrangement would you recommend and how much would it cost?Explanation / Answer
(a) So, number of opton avaiable here for minimum reliability criteria
Reliability can be calculated for parellal compnennt are R = 1 - (1 - Ri ) ( 1- Rj)
where i and j are components.
(i) A and A in parellel : R = 0.99 and Cost = 38 * 2 = $76
(ii) A and B in parellel : R = 0.985 and cost = $ 58
(iii) A and C in parellal : R = 0.98 and cost = $ 53
(iv) C and C and C in parellal R = 1 - (1 - 0.80)3 = 0.992 and Cost = $ 45
If minimum cost with acceptable reliability is the selection criteria, so the best configuration would be A and C in parallal with two components and if three components are acceptable then C-C-C are the best configuration.
(b) design space available = 1.5 x 1.5 x 1.5 in3
cost is not consideration.
So, acceptable combinations are
(i) A - A - A ; Reliability R = 1 - (1 - 0.9)3 = 0.999
(ii) A - A ; reliability R = 1 - (1- 0.9)2 = 0.99
(ii) A - B ; reliability R = 1 - (1 -0.9) * (1- 0.85) = 0.985
(iii) A - C ; Reliability = 0.98
so if two comonents are acceptable only we will use A - A combination otherwise we will use A-A -A for three components.
(c) If space is not a limitations, but the minimum reliability required is 0.99
then acceptable combinations are
(i) A-A ; reliability : 0.99; cost = $ 76
(ii) C-C-C in parellal ; reliaibility = 0.992 ; cost = $ 45
so we get other combination too but these would ccost more so we will use C-C-C is parellal.