Question

One generator is placed in standby redundancy to the main generator. The faliure rate of each generator is estimated to be λ = 0.05/hr. Compute the reliability of the system for 10 hrs and its MTBF assuming that the sensing and switching device is 100% reliable. If the reliability of this device is only 80%, how are the results modified?

Answer 1

Let A be an event in which main generator will not fail and B be an event in which standby generator will not fail.

Here,

P(A) = (1 - 0.05 per hour x 10 hours) = (1 - 0.5) = 0.5 and,

P(B) = 0.5 also.

Hence,

P(AUB)

= P(A) + P(B) - P(A∩B)

= P(A) + P(B) - P(A) x P(B) [As A & B are independent events]

= 0.5 + 0.5 - 0.5 x 0.5

= 1 - 0.25

= 0.75

Hence, probability that at least one of the two generators shall be in working condition for 10 hrs = 0.75

Thus, reliability of the system for 10 hrs

= 0.75 x 100 %

= 75 %

Thus failure rate = 1 - 75 % = 25 % = 0.25

Hence mean time between failures = MTBF = 1 / 0.25 = 4 hours [when switching devices are 100% reliable]

If reliability of switching devices is only 80 %, let C be an event in which switching devices work and D be an event in which at least one of the two generators shall be in working condition for 10 hrs, then,

P(C) = 80 % = 0.8

P(D) = P(AUB) = 0.75

Thus, P(A∩D) = P(A) x P(D) [ Independent Events] = 0.8 x 0.75 = 0.6

& Revised reliability of the system for 10 hrs

= 0.6 x 100 %

= 60%

Thus failure rate = 1 - 60 % = 40 % = 0.4

Hence mean time between failures = MTBF = 1 / 0.4 = 2.5 hours [when switching devices are 80% reliable]