Predicting Outcomes of Investments in Maintenance and Repair of Federal Facilities (2012) / Chapter Skim
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Appendix C: Some Fundamentals of the Risk-Based Approach
Appendix C: Some Fundamentals of the Risk-Based Approach
Pages 125-135
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From page 125... ...
Copyright 2012 by the National Academy of Sciences. All rights reserved
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From page 126... ...
The normal distribution, whose range of possible values is −∞ to +∞ is denoted as N(μ, ı) where μ is its mean value and ı is its standard deviation.
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From page 127... ...
where ȜX and ȗX are, respectively, the mean and standard deviation of lnX and they are the parameters of the lognormal distribution. These parameters are related to the mean and standard deviation of X as follows: λ = ln μ X − 1 ζ 2 2 and ª §σX · 2º » = ln(1 + δ X )
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From page 128... ...
ILLUSTRATIVE APPLICATIONS TO SPECIFIC OUTCOMES Described below are the numerical calculations of the risk or probability of "negative benefits" of three specific outcomesņaccident rates and types, deferred maintenance, and energy use. Accident Rates and Types In this example, let X = recordable incident rate (RIR)
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From page 129... ...
Assume that the variables are independent normal random variables; the means and standard deviations of each of the variables are: X = N(4, 1.2)
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From page 130... ...
Assume that the operational life T of a typical A/C unit can be described with the lognormal distribution, with a median life of tm months or years, and a COV of įT (or a standard deviation of σ T ≈ δ T × tm )
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From page 131... ...
If the agency has 1,000 A/C units, 230 of them are likely to fail within 2 years beyond the scheduled maintenance. If the average repair cost is $1,500 per unit, the deferred maintenance cost will be 230 × 1500 = $345,000.
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From page 132... ...
2 + (3σ Y ) 2 For numerical illustrations, assume hypothetically the following: Current average gasoline consumption, is ȝX = 10 million gallons, with a standard deviation of ıX = 2 million gallons.
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From page 133... ...
The corresponding probability of performance is then pS = 1 − p F Consider a system in which the available supply, X, is a Gaussian or normal random variable N(ȝX, ıX) and the demand is also a Gaussian random variable N(ȝY, ıY)
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From page 134... ...
as follows: pF β pF β 0.5 0 0.01 2.33 0.25 0.67 10−3 3.10 -4 0.16 1.00 10 3.72 -5 0.10 1.28 10 4.25 -6 0.05 1.65 10 4.75 The First-Order Reliability Method (FORM) Engineers are, traditionally, reluctant to admit a probability of failure; for this reason, a good alternative strategy is to use an equivalent measure, the safety index ȕ – which is a complete measure of the safety or performance of an engineered system.
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From page 135... ...
becomes σ X X '− σ Y Y '+ μ X − μY = 0 From the above figure in the reduced variates, we can clearly distinguish the failure region from the safe region, and distinguish the limit state equation (or failure surface) that separates the two regions.
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