315712858-Weibull -2.ppt

Weibull Analysisfor Life Data Jang JuSu Historical Background 1. Waloddi Weibull (1887-1979) invented the Weibull distribution in 1937. 2. His 1951 paper represents the culmination of his work in reliability analysis. 3. The U.S.Air Force recognized the merit of Weibull’s methods and funded his research to 1975. 4. Leonard Johnson at General Motors, improved Weibull’s methods. (Weibull used mean rank values for plotting but Johnson suggested the use of median rank values) 5. E.J.Gumbel proved that the Weibull distribution and the smallest extreme value distributions(Type III) are same. 6. The engineers at Pratt & Whitney found that the Weibull method worked well with extremely small samples, even 2 or 3 failures. MOASOFT Training Course for Reliability 2 Advantages of Weibull Analysis 1. The primary advantage of Weibull analysis is the ability to provide reasonably accurate failure analysis and failure forecast with extremely small samples. 2. Another advantage is a simple and useful graphical methods. MOASOFT Training Course for Reliability 3 Application of Weibull Analysis Life Data or Accelerated Life Data Weibull Analysis Failure Analysis and Forecasting Schedule Maintenance Application to Project Management for a System or Plant … Warranty and Cost Analysis MOASOFT Training Course for Reliability 4 . t   0     2. Weibull Distribution 1. Cumulative Distribution Function  t         F (t )  1  e  . t   0  : Shape parameter  : Scale parameter  : Location parameter MOASOFT Training Course for Reliability 5 . Probability Density Function  1  t     t        f (t )    e  . 5 0.4 0.3  2   2.1 5 10 15 20 25 30 5 10 15 20 25 30 0.3 0.2 0.2 0.3 0.1 5 10 15 20 25 30 5 10 15 20 25 30 MOASOFT Training Course for Reliability 6 .2 0.4 0. 0.4 0.1 0.3   0 .5  1 0. Weibull Distribution Shape parameter variations with   10 .1 0.4 0.2 0. 4 0.0002 50 100 150 200 250 300 500 1000 1500 2000 2500 3000 MOASOFT Training Course for Reliability 7 .0004 0.006   100 0.8 0.002 0.0008 0.004 0.2 0.08 0.04 0. 0.02 0.5 1 1. Weibull Distribution Scale parameter variations with  2 .008 0.06   10 0.0006   1000 0.5 2 2.6  1 0.5 3 5 10 15 20 25 30 0. Weibull Distribution Location parameter shift the curve along x-axis. CDF : shape parameter variation MOASOFT Training Course for Reliability 8 . Weibull Distribution  1  1  1 MOASOFT Training Course for Reliability 9 . 632 . Weibull Distribution  t       F (t )  1  e   From the above equation. So we can guess that  is defined as the age at which 63. MOASOFT Training Course for Reliability 10 . If we set the value of time to t  then F ( )  0.632 .2% of the units will fail.        F ( )  1  e 1  e 1  0. 2. Interpretation Assume that we have 2-types data. will be increased. failure and suspended.  .  . For shape parameter  .   1 : Implies Infant mortality and we can suspect MOASOFT Training Course for Reliability 11 . For scale parameter  . 1. hardly change but the scale parameter . In general the more we have suspended data the shape parameter .   1 : Implies random failures Maintenance errors. human errors. Interpretation Inadequate burn-in test or screening. misassembled. quality control. Production problems. MOASOFT Training Course for Reliability 12 . Failures due to nature. Overhaul problems. Solid state electronic failure.   4 : Implies ageing effects MOASOFT Training Course for Reliability 13 . 1    4 : Implies early wear out Low cycle fatigue. Interpretation Failures due to nature. Corrosion or erosion. Mixtures of data from 3 or more failure modes or different  s. Types of Data 0 Time 0 Time 0 Time 1 Complete 1 Type II 1 Type I Data 2 2 2 3 3 3 Unit Unit Unit 4 4 4 5 5 5 6 6 6 0 Time 0 Time 0 Time 1 Multiply Failure 1 Random 1 Interval Censored Censored (Grouped) 2 2 2 Data 3 3 3 Unit Unit Unit 4 4 4 5 5 5 6 6 6 MOASOFT Training Course for Reliability 14 . 8537 0.3664 -0.4099 8.4887 MOASOFT Training Course for Reliability 15 .3132 -2.7187 -1.2523 7.2353 4000 F 0. Median Rank Regression Failure time Status Median Rank Y X 1500 F 0.8041 8.2940 -0.5676 8.1763 5000 S 7000 F 0.0946 7.3088 1750 S 2250 F 0.6397 4300 F 0. . . 2. Median Rank Regression 1.4 n 1. Benard’s Approximation formula ( Rn  0. . 3 . 2 . 3.. 2 . .3) BRn  N  0. n  1 . Adjusted Rank formula reverse rank  Rn 1  ( N  1) Rn  reverse rank  1 R0  0 . MOASOFT Training Course for Reliability 16 .. Median Rank Regression  t  t       F (t )  1  e   1  F (t )  e  t    t  1 /(1  F (t ))  e   ln[ 1 /(1  F (t ))]      ln ln[ 1 /(1  F (t ))]   ln( t )   ln(  ) y  Bx  A MOASOFT Training Course for Reliability 17 . Median Rank Regression y  ln( failure time) x  ln ln( 1 /(1  Median Rank of Y ))  1      e A B MOASOFT Training Course for Reliability 18 . . .. Median Rank Regression Y e XC CX t (Y  XC )  0  y1   x1 ..1       y2   x2 .1 A     y   x .1  C  ( X t X ) 1 X tY  n  n  MOASOFT Training Course for Reliability 19 .1  B  X Y  X XC t t Y   X  C    . Median Rank Regression MOASOFT Training Course for Reliability 20 . . t  0  k L   f (ti )  f (t1 ) f (t 2 ). Maximum Likelihood Estimation t   1   f (t )  t e . f (t k ) i 1 ti  k     ti e  1  i 1  MOASOFT Training Course for Reliability 21 .. Maximum Likelihood Estimation  1  k 1 k  ln L  k ln     ln ti   ti    i 1  i 1 Maximizing Logarithm Likelihood Function (ln L) k 1    ln ti   ti ln ti  0     (ln L) k   2 0 i t     MOASOFT Training Course for Reliability 22 . Maximum Likelihood Estimation k  i ln ti t  1 r 1 G(  )  i 1   ln ti   0 k  i  r i 1 t i 1 1  k       t i     i 1   r      MOASOFT Training Course for Reliability 23 . Maximum Likelihood Estimation G(  n )  n1   n  G`(  n ) We so as to use Newton method have to find the derivative of G( ) MOASOFT Training Course for Reliability 24 . Maximum Likelihood Estimation k k t i  (ln ti ) 2 (t i   ln t ) i 2 1 G`(  )  i 1  i 1  2 k k  t i 1 i  ( t ) i 1 i  2 MOASOFT Training Course for Reliability 25 . Maximum Likelihood Estimation  (ln L) 2  1  t   t  2    t   t  2       i  (ln  i )      i  (ln  i )   2               n     r        (ln L) 2    t        t               i   (   1)    i   (   1)  2   2      2     2       n   r     (ln L) 2 1  ti   1   ti             ln    1  n            t    1   t       i     ln  i   1                r  MOASOFT Training Course for Reliability 26 . Maximum Likelihood Estimation         Z / 2 var(  )        exp           Z / 2 var(  )    exp               (ln t  ln  )        Z / 2 var( )     exp           Z / 2 var( )    exp         MOASOFT Training Course for Reliability 27 . Maximum Likelihood Estimation MOASOFT Training Course for Reliability 28 . Conclusion In case of small sample -> Monte Carlo Simulation! this will make an analysis enhanced with small sample. MLE preferred! MOASOFT Training Course for Reliability 29 . Topics MLE Rank Regression Graphical None OK Small sample Bad( more bias) more accurate Forecast Bad more accurate Risk Analysis ? ? Confidence Rigorous None Convergence ? Always Mixed mode Only one Available But large sample case. specimen size. MOASOFT Training Course for Reliability 30 .6171x10-5 per °C. Arrhenius Model Arrhenius Law E  A e kT where E : activation energy : electron-volts. and other factors. k : Boltzmann’s constant : 8. T : Kelvin temperature A : a constant that depends on product geometry. 303 k  1 MOASOFT Training Course for Reliability 31 .4343 E / k E  2. Arrhenius Model Linearized relationship log    0  ( 1 / T ) where  1  log( e) ( E / k )  0. T = 453 °K(180 °C).65 eV.6171x10-5 )[1/453 – 1/533]} = 12 MOASOFT Training Course for Reliability 32 .65/ 8. T’ = 533 °K(260 °C). K = exp{(0. Arrhenius Model Arrhenius acceleration factor K 1 1 K  /  '  exp{( E / k )[  ]} T T' Example : E = 0. Arrhenius Model MOASOFT Training Course for Reliability 33 .