# 6 (2) - Constant Failure Rate Model(Cont Exponential...

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MFG5350 1 Constant Failure Rate Model (Cont’) Exponential Probability Distribution

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MFG5350 2 The Bathtub Curve
MFG5350 3 The Exponential Distribution 0 t , = (t) 0 t , e = e = R(t) t - dt - t 0 ' Then Let

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MFG5350 4 0.0000 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000 0 2 4 6 8 10 12 14 16 time lambda =.25 lambda =.50 lambda =1.0
MFG5350 5 The CDF and PDF e - 1 = F(t) t - e = dt F(t) d = f(t) t -

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MFG5350 6 Probability Density Function (PDF)
MFG5350 7 The MTTF 1 = - e = dt e = MTTF t - t - 0 0 | or 0 1 te t

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MFG5350 8 The Standard Deviation 2 t - 2 0 2 1 = dt e 1 - t = and MTTF 1
MFG5350 9 The Median Time to Failure .69315 = .5 1 - = t med ln = .69315 MTTF R t e t ( ) . 5

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MFG5350 10 The Design Life R = e = ) t R( t - R R For a given reliability R, let R 1 - = t then R ln
MFG5350 11 Memoryless Property e e = ) T R( ) T + R(t = ) T | T - ) T + (t - 0 0 0 0 0 R(t) = e = e e e = t - T - T - t - 0 0

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MFG5350 12 Failure Modes R(t) = n = R (t) i i 1 R i (t) is the reliability function for the ith failure mode, then, assuming independence among the failure modes, the system reliability, R(t) is found from
MFG5350 13 More on Failure Modes t i dt t i e t R 0 ' ' ) ( ) ( Let n i t t i e t R 1 ' ) ' ( 0 ) ( Then t n i i t 0 1 ' ) ' ( exp t t 0 ' ) ' ( exp (t) = (t) =1 n i i where

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MFG5350 14 Failure Modes and CFR i n =1 i = = (t) If a system consists of n independent, serially related components each with CFR, then e = e = R(t) t - dt - t 0 ' and R 1 R 2
MFG5350 15 Example #2 - Failure Modes An engine tune-up kit consists of 3 parts each having CFRs (in failures per mile) of .000034, .000017, and .0000086.

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## This note was uploaded on 04/25/2010 for the course IE 654 taught by Professor Smith during the Spring '10 term at 카이스트, 한국과학기술원.

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6 (2) - Constant Failure Rate Model(Cont Exponential...

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