Question

4. A piece of electronic equipment used for aviation has three elements connected in series or sequence. The reliability of each of the three elements is as follows:

Element A: 0.92                Element B: 0.94                 Element C: 0.91

a) Draw how these three elements are connected.

b) What is the reliability of this system?

c) Suppose we wanted the reliability to be at least 0.85 by upgrading Element C with a higher reliability. What would the reliability of Element C need to be?

d) Instead of increasing the reliability of Element C, each of the elements is supported by an identical element in parallel as a back-up. What is the reliability of this system?

Answer 1

a) Since Element A, B and C are connected in series or sequence, we draw them as shown below:

b) Reliability of system = R_s = R_A x R_B x R_C

R_A = Reliability of Element A

R_B = Reliability of Element B

R_C = Reliability of Element C

Hence, R_s = 0.92 x 0.94 x 0.91 = 0.787

c) If we want R_sn = 0.85, we get R_Cn = R_sn / (R_A x R_B) = 0.85 / (0.92 x 0.94) = 0.983

Hence, R_Cn = 0.983

d) If we support each element in parallel with its identical element, new reliability of each element can be calculated as

R_An = 1 - (1-R_A) x (1-R_A) = 1 - (1-0.92) x (1-0.92) = 0.9936

R_Bn = 1 - (1-R_B) x (1-R_B) = 1 - (1-0.94) x (1-0.94) = 0.9964

R_Cn = 1 - (1-R_C) x (1-R_C) = 1 - (1-0.91) x (1-0.91) = 0.9919

Reliability of system = R_An x R_Bn x R_Cn = 0.9936 x 0.9964 x 0.9919 = 0.982