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>>p = 0.01*[13 37 12 56 33 71 22 43 57 31];>>k = [2 5 7];>>P = ikn(p,k) P = 0.1401 0.1845 0.0225 % individual probabilities>>Pc = ckn(p,k) Pc = 0.9516 0.2921 0.0266 % cumulative probabilities
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Reliability of systems with independent components

Suppose a system has n components which fail independently. Let E i be the event the i th component survives the designated time period. Then R i = P ( E i ) is defined to be the reliability of that component. The reliability R of the complete system is a function of the component reliabilities. There are three basic configurations.General systems may be decomposed into subsystems of these types. The subsystems become components in the larger configuration. The three fundamental configurationsare:

  1. Series . The system operates iff all n components operate : R = i = 1 n R i
  2. Parallel . The system operates iff not all components fail : R = 1 - i = 1 n ( 1 - R i )
  3. k of n . The system operates iff k or more components operate. R may be calculated with the m-function ckn. If the component probabilities are all the same, itis more efficient to use the m-function cbinom (see Bernoulli trials and the binomial distribution, below).

MATLAB solution . Put the component reliabilities in matrix R C = [ R 1 R 2 R n ]

  1. Series Configuration >>R = prod(RC) % prod is a built in MATLAB function
  2. Parallel Configuration >>R = parallel(RC) % parallel is a user defined function
  3. k of n Configuration >>R = ckn(RC,k) % ckn is a user defined function (in file ckn.m).

There are eight components, numbered 1 through 8. Component 1 is in series with a parallel combination of components 2 and 3, followed bya 3 of 5 combination of components 4 through 8 (see Figure 1 for a schematic representation). Probabilities of the components in order are

0 . 95 0 . 90 0 . 92 0 . 80 0 . 83 0 . 91 0 . 85 0 . 85

The second and third probabilities are for the parallel pair, and the last five probabilities are for the 3 of 5 combination.

>>RC = 0.01*[95 90 92 80 83 91 85 85]; % Component reliabilities>>Ra = RC(1)*parallel(RC(2:3))*ckn(RC(4:8),3) % Solution Ra = 0.9172
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a representation of a system with 8 components a representation of a system with 8 components
Schematic representation of the system in [link]
>>RC = 0.01*[95 90 92 80 83 91 85 85]; % Component reliabilities 1--8>>Rb = prod(RC(1:2))*parallel([RC(3),ckn(RC(4:8),3)]) % SolutionRb = 0.8532
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Another 8 component system schematic Another 8 component system schematic
Schematic representation of the system in [link]

A test for independence

It is difficult to look at a list of minterm probabilities and determine whether or not the generating events form an independent class.The m-function imintest has as argument a vector of minterm probabilities. It checks for feasible size, determines the number of variables, and performs acheck for independence.

>>pm = 0.01*[15 5 2 18 25 5 18 12]; % An arbitrary class>>disp(imintest(pm)) The class is NOT independentMinterms for which the product rule fails 1 11 0 1 1 1 0
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>>pm = [0.10 0.15 0.20 0.25 0.30]: %An improper number of probabilities>>disp(imintest(pm)) The number of minterm probabilities incorrect
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>>pm = minprob([0.5 0.3 0.7]);>>disp(imintest(pm)) The class is independent
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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