that the system will be operating at the expiration of 60 days as a result of each of the fol-
lowing causes: (a) only one component survives; (b) only two components survive; (c) all
three components survive.
Calculation Procedure:
- Compute the probability that a component fails during
the 60-day period
The probability of failure is: for C 1 , 1 - 0.18 = 0.82; for C 2 , 1 - 0.23 = 0.77; for C 3 , 1 -
0.15 = 0.85.
- Compute the probability that the system survives because
one and only one component survives
Multiply the probabilities of the individual events. Thus, P(only C 1 survives) =
(0.18)(0.77)(0.85) - 0.11781; P(only C 2 survives) = (0.82)(0.23)(0.85) = 0.16031; P(omy
C 3 survives) = (0.82)(0.77)(0.15) = 0.09471. Sum the results: 0.11781 + 0.16031 +
0.09471 = 0.37283. This is the probability that only one component survives.
- Compute the probability that the system survives because two
and only two components survive
Proceed as in step 2. Thus, P(on\y C 1 and C 2 survive) = (0.18)(0.23)(0.85) = 0.03519;
P(only C 1 and C 3 survive) = (0.18)(0.77)(0.15) = 0.02079; P(only C 2 and C 3 survive) =
(0.82)(0.23)(0.15) - 0.02829. Sum the results: 0.03519 + 0.02079 + 0.02829 = 0.08427 =
the probability that only two components survive.
- Compute the probability that the system survives because all
three components survive
P(all survive) = (0.18)(0.23)(0.15) = 0.00621.
- Verify the foregoing results
Sum the results in steps 2, 3, and 4: Rs(60) = 0.37283 + 0.08427 + 0.00621 = 0.46331.
Now apply Eq. 18, giving tfs(60) = 1 - (0.82)(0.77)(0.85) - 0.46331. The equality of the
two values confirms the results obtained in the previous steps.
SYSTEM WITH IDENTICAL COMPONENTS
IN PARALLEL
A certain type of component has a reliability of 0.12 for 20 days. How many such compo-
nents must be connected in parallel if the reliability of the system is to be at least 0.49 for
20 days?
Calculation Procedure:
- Write the equation for the reliability of the system
Let n = number of components required. Apply Eq. 18, giving ^(2O) = 1 — (1 — 0.12)" =
1 - (0.88)", Eq. a.
- Determine the number of components
Set RS(2Q) = 0.49 and solve Eq. a for /i, giving n = (log 0.51)/(log 0.88) = 5.3. Use six
components.
