K-out-of-n Systems: Difference between revisions
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Since the time to failure is exponential, system reliability is calculated via Equation 4.23 on page 78 by substituting the given data as: | Since the time to failure is exponential, system reliability is calculated via Equation 4.23 on page 78 by substituting the given data as: | ||
::<math>R(t) = \sum^{n}_{i=k}C^{i}_{ | ::<math>R(t) = \sum^{n}_{i=k}C^{i}_{n} e^{-\lambda i t} (1-e^{-\lambda t})^{n-i}\,\!</math> | ||
::<math>R(8760) = \sum^{5}_{i=3}C^{i}_{5} e^{-2.7 \times 10^{-5} \times 8760i} (1-e^{-2.7 \times 10^{-5} \times 8760})^{5-i} = 0.9336\,\!</math> | ::<math>R(8760) = \sum^{5}_{i=3}C^{i}_{5} e^{-2.7 \times 10^{-5} \times 8760i} (1-e^{-2.7 \times 10^{-5} \times 8760})^{5-i} = 0.9336\,\!</math> | ||
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Each server is modeled with an exponential distribution with a lambda of 2.7 × 10-5 failures per hour. | Each server is modeled with an exponential distribution with a lambda of 2.7 × 10<sup>-5</sup> failures per hour. | ||
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The mean time between failures (MTBF) is estimated to be 29,004 hours. | |||
[[Image:kn_qcpmtbf.png|center|500px]] | [[Image:kn_qcpmtbf.png|center|500px]] | ||
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The point reliability after 8760 hours of use is estimated to be 93. | The point reliability after 8760 hours of use is estimated to be 93.45%. | ||
[[Image:kn_simqcprel.png|center|500px]] | [[Image:kn_simqcprel.png|center|500px]] | ||
And the mean time to failure (MTTF) is estimated to be 29, | And the mean time to failure (MTTF) is estimated to be 29,048 hours. | ||
[[Image:kn_simqcpmttf.png|center|500px]] | [[Image:kn_simqcpmttf.png|center|500px]] |
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