Question

1. A system consists of 5 identical components, all of which must operate for system success. If the failure rate and average repair time of each component is 0.01 failures/year and 40 hours respectively, calculate the system failure rate, average down time and unavailability.

2. The system in Problem 1 is reinforced by connecting a second identical system in parallel with the first. Evaluate the new system failure rate, average down time and unavailability.

Answer 1

Explanation:

Given;

If the failure rate and average repair time of each component is 0.01 failures/year and 40 hours respectively,

Failure Rate for each component = ( Number of Failures / Hours ) = 0.01 / (365 x 24 ) = 1.14 x 10⁻⁶ failures per hour.

Failure Rate for system = 5 x 1.14 x 10⁻⁶ = 5.7 x 10⁻⁶ failures per hour.

Average Down Time = Repair Time / Failures in an year = 40 / 0.05 = 800 hours.

Mean time to Failure = 1 / Failure Rate = (5.7 x 10⁻⁶)-1= 175438.59 hours

Unavailability, U = Average Repair Time / ( Average Repair Time + MTTF )

= 800 / ( 800 + 175438.59 ) = 0.0045