U12, Time-Dependent Failure, 100218.pdf

C r10000036000 d t 095 51 lognormal student exercise

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c. R(100,000|36,000) d. t 0.95 51
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Lognormal Student Exercise 4 Solution a. R(36,000) = 1 – F[(1/0.7)ln(36,000/100,000] = 1 4 F[ 41.46] = 0.92786 b. MTTF = 100,000 e 0.49/2 = 127,762 mi. Var = 100,000 2 e 0.49 [ e 0.49 41] = 1.032 x 10 10 Std Dev = 101,594 mi. c. R(64,000|36,000) = R(100,000)/R(36,000) = 0.5/ 0.92786 = 0.54 52
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Student Exercise . Solution d. R(t 0.95 ) = 0.95 1 . F[ (1/0.7) ln(t 0.95 /100,000) ] = 0.95 (1/.7) ln(t 0.95 /100,000) = .1.645 t 0.95 = 100,000 e .1.645 x 0.7 = 31,616 mi. general approach: 53 R t = med t sz 1 R e
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The Gamma as a Failure Distribution Density function, pdf with 2 parameters and relation to previous notation for Erlang Gamma with a positive integer shape parameter, k ! (= k), shape parameter; α (= 1/λ), scale parameter 54 f(t) = t γ− 1 e t/ α α γ Γ ( γ ) γ , α > 0 and t 0 F(t) = f(t) = t γ− 1 e t/ α α γ Γ ( γ ) dt 0 t 0 t = I t α , γ Γ ( γ ) R(t) = 1 F(t) = 1 I t α , γ Γ ( γ ) , I t α , γ the incomplete gamma function Mean = ! α = MTTF Variance = ! α 2 t mode = α( ! K1) for ! > 1 0 otherwise The Chapter 4 Excel workbook template and the Ebeling software with the IRME text calculates values of the incomplete Gamma function. Values in tables or calculate using software Exponential for ! = 1, CFR
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Gamma Distribution Behavior The Gamma distribution is similar to the Weibull by providing a wide range of shapes. When shape parameter ! = 1, the density function is Exponential with mean = α = 1/λ This is the basic Gamma, not the Erlang Gamma version with integer values of the shape parameter, k, studied earlier. As shown below, the Gamma conditional failure rate function λ(t) behavior depends on the shape parameter ! and approaches 1/α as t ∞. Gamma Weibull λ(t) 0 < ! < 1 0 < β < 1 DFR ! = 1 β = 1 CFR ! > 1 β > 1 IFR 55
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Summary 56 R(t) = e −λ t R(t) = e t θ Β R(t) = 1 − Φ t μ σ R(t) = 1 − Φ 1 s ln t t med Exponential Weibull Normal Lognormal MTTF 1/ λ θΓ 1 + 1 β μ t med e s 2 /2 Reliability Gamma R(t) = 1 I t α , γ Γ ( γ ) ! α Variance 1/ λ 2 σ 2 ! α 2 t med 2 e s 2 [e s 2 1] 2 σ = 2 θ Γ 1+ 2 β - 2 Γ 1+ 1 β
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Gamma Failure Distribution Example Failures of a critical machine part due to cyclical vibration has a Gamma distribution with a shape parameter of ! = 2.3 and a scale parameter of α = 2000 operation hours. MTTF = ! α = (2.3)(2000) = 4600 hr Median = 3953 hr (note Median < MTTF due to the skewed distribution tail to large values of t) R(t = 1000 hr) = 0.946 (R(t) and median calculated using the Excel Chap 4 template for the Gamma distribution or in AgenaRisk) 57 σ = γ α 2 = 2.3 ( ) 2000 ( ) 2 = 3033 hr t mode = α ( γ − 1) = 2000(2.3 1) = 2600 hr
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