IV. Reliability Modeling and Prediction
A. Reliability Modeling
2. Reliability block diagrams and models (Create)
Generate and analyze various types of block diagrams and models, including series, parallel, partial redundancy, time-dependent, etc.
A parallel structure improves the system’s ability to withstand some failures.
Additional References
Parallel Systems (article)
k out of n (article)
Quick Quiz
1-51. Consider the following logic diagram with reliabilities as shown:
Assuming statistical independence of the three components, calculate the system reliability.
(A) 0.412
(B) 0.922
(C) 0.986
(D) 0.994
(D) 0.994
The formula to determine reliability of a parallel structure is
$$ {{R}_{sys}}\left( t \right)=1-\prod\limits_{i=1}^{n}{\left( 1-{{R}_{i}}\left( t \right) \right)}$$
This formula is multiplying the unreliabilities then the last 1 minus converts back to reliability. This works for active redundancy when only one element in parallel is necessary for the system to operate. This is a 1-out-of-n system.
Inserting the reliability values (assuming t is the the same for each value) for this problem we are able to quickly calculate the result
$$ {{R}_{sys}}=1-\left( \left( 1-0.92 \right)\left( 1-0.83 \right)\left( 1-0.54 \right) \right)=0.994$$
1-58. Consider the following system:
The following component failure data are available: failure rate of A = 0.00005 failures/hour, reliability of B = 0.91, MTTF of C = 12,000 hours, and reliability of D = 0.88.
(A) 0.9997
(B) 0.99997
(C) 0.99997
(D) 2630 hours
Answer
(C) 0.99997
One way to solve this is to solve each component for reliability at 900 hours. For component A given a failure rate of 0.0005, we solve for R(900)
Calculate this system’s reliability at 900 hours.
$$ \begin{array}{l}{{R}_{A}}\left( t \right)={{e}^{-\lambda t}}\\{{R}_{A}}\left( 900 \right)={{e}^{-\left( 0.0005 \right)\times 900}}=0.956\end{array}$$
And for component C with 12,000 hours MTBF we have
$$ \begin{array}{l}{{R}_{A}}\left( t \right)={{e}^{-\frac{t}{\theta }}}\\{{R}_{A}}\left( 900 \right)={{e}^{-\frac{900}{12,000}}}=0.928\end{array}$$
The reliability of the system is now
$$ \begin{array}{l}{{R}_{sys}}=1-\left( 1-{{R}_{A}} \right)\left( 1-{{R}_{B}} \right)\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{D}} \right)\\{{R}_{sys}}=1-\left( 1-0.956 \right)\left( 1-0.91 \right)\left( 1-0.928 \right)\left( 1-0.88 \right)\\{{R}_{sys}}=0.99997\end{array}$$
1-59. Two components operate in parallel. If the mean time to failure is 300 hours for each component, calculate the probability of system failure after one hour of operation/
(A) 0.9933
(B) 0.9967
(C) 0.99999
(D) 300 hours
(C) 0.99999
An easy approach is to determine the reliability of one component with t = 1.
$$ \begin{array}{*{35}{l}}{{R}_{A}}\left( t \right)={{e}^{-\frac{t}{\theta }}} \\{{R}_{A}}\left( 1 \right)={{e}^{-\frac{1}{300}}}=0.928 \\\end{array}$$
Then determine the system reliability with the two components in parallel.
$$ \begin{array}{l}{{R}_{sys}}=1-\left( 1-{{R}_{A}} \right)\left( 1-{{R}_{A}} \right)\\{{R}_{sys}}=1-\left( 1-0.99667 \right)\left( 1-0.99667 \right)\\{{R}_{sys}}=0.99999\end{array}$$
Hi Fred,
1-58 I think the question is not stated correctly.
1-59 The probability of failure in one hour should be quite low. I think the answers are the probability of survival.