SYSTEM WITH COMPONENTS IN SERIES
A system consists of three type A components and two type B components, all arranged
in series. These components have negative-exponential life spans, and the mean life span
is 30 h for type A and 36 h for type B. Find the reliability of the system for 9 h of opera-
tion.
Calculation Procedure:
- Compute the reliability of each component for 9 h
Let the subscripts A and B refer to the type of component, and take 1 h as the unit of time.
Then aA = I/HA = 1/30, and RA(f) = e'
t/30
9 or RA(9) = e'
9/30
= 0.7408. Similarly, RB(9) =
e-9/36 = 0.7788.
- Compute the reliability of the system for 9 h
Apply Eq. 17: Rs(9) = (0.7408)^3 (0.7788)^2 = 0.247. - Compute the reliability of the system
by an alternative method
Apply Eq. 17 to prove that the system also has a negative-exponential life span. Now as-
sume that when a component fails, it is instantly replaced with one of identical type, thus
maintaining continuous operation. In 18Oh of operation, the mean number of failures of
an individual component is 180/30 = 6 for type A and 180/36 = 5 for type B. Since failure
of any component causes failure of the system, the mean number of failures of the system
in 180 h is 3 x 6 + 2 x 5 = 28. Thus, the mean life span of the system is JJLS = 180/28, and
as = 28/180. Then ast = (28/180)9 = 1.4, and R£9) = e~lA = 0.247.
In the foregoing calculations, 18Oh was selected for convenience, since 180 is the
lowest common multiple of 30 and 36. However, a period of any length whatever can be
selected.
SYSTEM WITH COMPONENTS IN PARALLEL
With reference to the system in Fig. 30, the reliability of each component for a 60-day pe-
riod is as indicated. For example, the reliability of C 1 is 0.18. Determine the probability
FIGURE 30