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==The Lognormal Distribution==
#REDIRECT [[Distributions_Used_in_Accelerated_Testing#The_Lognormal_Distribution]]
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The lognormal distribution is commonly used for general reliability analysis, cycles-to-failure in fatigue, material strengths and loading variables in probabilistic design. A random variable is lognormally distributed if the logarithm of the random variable is normally distributed. Since the logarithms of a lognormally distributed random variable are normally distributed, the lognormal distribution is given by:
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::<math>f({T}')=\frac{1}{{{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{{T}'-\bar{{T}'}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}</math>
 
<br>
 
:where:
:• <math>{T}'=\ln T</math> , and where the  <math>T</math> s are the times-to-failure.
:• <math>\bar{{T}'}=</math> mean of the natural logarithms of the times to failure.
:• <math>{{\sigma }_{{{T}'}}}=</math> standard deviation of the natural logarithms of the times to failure.
 
The lognormal  <math>pdf</math>  can be obtained, realizing that for equal probabilities under the normal and lognormal  <math>pdf</math> s incremental areas should also be equal, or:
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::<math>f(T)dT=f({T}')d{T}'</math>
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Taking the derivative yields:
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::<math>d{T}'=\frac{dT}{T}</math>
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Substitution yields:
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::<math>\begin{align}
  & f(T)= & \frac{f({T}')}{T} \\
& = & \frac{1}{T\cdot {{\sigma }_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{{T}'-\bar{{T}'}}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}} 
\end{align}</math>
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:where:
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::<math>f(T)\ge 0,T>0,-\infty <\bar{{T}'}<\infty ,{{\sigma }_{{{T}'}}}>0</math>
 
{{alta ld statistical prop func}}
 
{{ald characteristics}}
 
===Parameter Estimation===
The estimate of the parameters of the lognormal distribution can be found graphically on probability plotting paper or analytically using either least squares or maximum likelihood. (Parameter estimation methods are presented in detail in Appendix B.)
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====Probability Plotting====
One method of calculating the parameter of the lognormal distribution is by using probability plotting. To better illustrate this procedure, consider the following example.
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====Example 5====
Let's assume six identical units are being reliability tested at the same application and operation stress levels. All of these units fail during the test after operating the following times (in hours),  <math>{{T}_{i}}</math> : 144, 385, 747, 1,144, 1,576 and 2,616.
The steps for determining the parameters of the lognormal  <math>pdf</math>  representing the data, using probability plotting, are as follows:
<br>
:• Rank the times-to-failure in ascending order as shown next.
<center><math>\begin{matrix}
  \text{Time-to-} & \text{Failure Order Number}  \\
  \text{failure, hrs} & \text{out of a Sample Size of 6}  \\
  \text{144} & \text{1}  \\
  \text{385} & \text{2}  \\
  \text{747} & \text{3}  \\
  \text{1,144} & \text{4}  \\
  \text{1,576} & \text{5}  \\
  \text{2,616} & \text{6}  \\
\end{matrix}</math></center>
:• Obtain their median rank plotting positions. The times-to-failure, with their corresponding median ranks, are shown next:
<br>
 
<center><math>\begin{matrix}
  \text{Time-to-} & \text{Median}  \\
  \text{failure, hr} & \text{Rank, }%  \\
  \text{144} & \text{10}\text{.91}  \\
  \text{385} & \text{26}\text{.44}  \\
  \text{747} & \text{42}\text{.14}  \\
  \text{1,144} & \text{57}\text{.86}  \\
  \text{1,576} & \text{73}\text{.56}  \\
  \text{2,616} & \text{89}\text{.09}  \\
\end{matrix}</math></center>
:• On a lognormal probability paper, plot the times and their corresponding rank value. Fig. 16 displays an example of a lognormal probability paper. The paper is simply a log-log paper. (The solution is given in Fig. 17.)
 
<br>
[[Image:ALTA4.10.gif|thumb|center|300px|Sample lognormal probability plotting paper.]]
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:• Draw the best possible straight line that goes through the  <math>t=0</math> 
and  <math>R(t)=100%</math>  point and through these points (as shown in Fig. 17).
:• At the  <math>Q(t)=50%</math>  ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the median. For this case, <math>\breve{T}=760</math> hr which means that <math>{{\bar{T}}^{\prime }}=\ln(\breve{T})=6.633</math>(see Eqn. Median).
<br>
::<math></math>
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[[Image:ALTA4.11.gif|thumb|center|300px|Probability plot for Example 5.]]
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<br>
:• The standard deviation,  <math>{{\sigma }_{{{T}'}}},</math>  can be found using the following equation:
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::<math>\begin{align}
  & {{\sigma }_{{{T}'}}}= & \frac{\ln \left[ T(Q=97.7%) \right]-\ln \left[ T(Q=2.3%) \right]}{4} \\
& = & \frac{\ln (5100)-\ln (120)}{4} \\
& = & 0.937376 
\end{align}</math>
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Now any reliability value for any mission time  <math>t</math>  can be obtained. For example, the reliability for a mission of 200 hr, or any other time, can now be obtained either from the plot or analytically.
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To obtain the value from the plot, draw a vertical line from the abscissa, at  <math>t=200</math>  hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read  <math>Q(t)</math> . In this case,  <math>R(t=200)=1-Q(t=200)=92%</math> . This can also be obtained analytically, from the lognormal reliability function. However, standard normal tables (or the Quick Statistical Reference in ALTA) must be used.
 
====MLE Parameter Estimation====
The parameters of the lognormal distribution can also be estimated using Maximum Likelihood Estimation (MLE). This general log-likelihood function is:
 
<br>
::<math>\begin{align}
  & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}{{{\sigma }_{{{T}'}}}{{T}_{i}}}\phi \left( \frac{\ln \left( {{T}_{i}} \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right) \right] \\
&  & \text{ }+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ln \left[ 1-\Phi \left( \frac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right) \right]+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime })] 
\end{align}</math>
 
<br>
:where:
<br>
::<math>z_{Li}^{\prime \prime }=\frac{\ln T_{Li}^{\prime \prime }-{\mu }'}{\sigma _{T}^{\prime }}</math>
<br>
::<math>z_{Ri}^{\prime \prime }=\frac{\ln T_{Ri}^{\prime \prime }-{\mu }'}{\sigma _{T}^{\prime }}</math>
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:and:
<br>
<br>
:• <math>{{F}_{e}}</math>  is the number of groups of times-to-failure data points.
:• <math>{{N}_{i}}</math>  is the number of times-to-failure in the  <math>{{i}^{th}}</math>  time-to-failure data group.
:• <math>{\mu }'</math>  is the mean of the natural logarithms of the times-to-failure (unknown a priori, the first of two parameters to be found).
:• <math>{{\sigma }_{{{T}'}}}</math>  is the standard deviation of the natural logarithms of the times-to-failure (unknown a priori, the second of two parameters to be found).
:• <math>{{T}_{i}}</math>  is the time of the  <math>{{i}^{th}}</math>  group of time-to-failure data.
:• <math>S</math>  is the number of groups of suspension data points.
:• <math>N_{i}^{\prime }</math>  is the number of suspensions in  <math>{{i}^{th}}</math>  group of suspension data points.
:• <math>T_{i}^{\prime }</math>  is the time of the  <math>{{i}^{th}}</math>  suspension data group.
:• <math>FI</math>  is the number of interval data groups.
:• <math>N_{i}^{\prime \prime }</math>  is the number of intervals in the i <math>^{th}</math>  group of data intervals.
:• <math>T_{Li}^{\prime \prime }</math>  is the beginning of the i <math>^{th}</math>  interval.
:• <math>T_{Ri}^{\prime \prime }</math>  is the ending of the i <math>^{th}</math>  interval.
The solution will be found by solving for a pair of parameters  <math>\left( {\mu }',{{\sigma }_{{{T}'}}} \right)</math>  so that  <math>\tfrac{\partial \Lambda }{\partial {\mu }'}=0</math>  and  <math>\tfrac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}=0,</math>  where:
<br>
::<math>\begin{align}
  & \frac{\partial \Lambda }{\partial {\mu }'}= & \frac{1}{\sigma _{{{T}'}}^{2}}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}(\ln ({{T}_{i}})-{\mu }') \\
&  & +\frac{1}{{{\sigma }_{{{T}'}}}}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}\overset{FI}{\mathop{\underset{i=1}{\mathop{-\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{\varphi (z_{Ri}^{\prime \prime })-\varphi (z_{Li}^{\prime \prime })}{\sigma _{T}^{\prime }(\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime }))} \\
&  &  \\
& \frac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}= & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{{{\left( \ln ({{T}_{i}})-{\mu }' \right)}^{2}}}{\sigma _{{{T}'}}^{3}}-\frac{1}{{{\sigma }_{{{T}'}}}} \right) \\
&  & +\frac{1}{{{\sigma }_{{{T}'}}}}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}\overset{FI}{\mathop{\underset{i=1}{\mathop{-\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{z_{Ri}^{\prime \prime }\varphi (z_{Ri}^{\prime \prime })-z_{Li}^{\prime \prime }\varphi (z_{Li}^{\prime \prime })}{\sigma _{T}^{\prime }(\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime }))} 
\end{align}</math>
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:and:
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<br>
::<math>\phi \left( x \right)=\frac{1}{\sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( x \right)}^{2}}}}</math>
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::<math>\Phi (x)=\frac{1}{\sqrt{2\pi }}\mathop{}_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dx</math>
<br>
====Example 6====
Using the same data as in the probability plotting example (Example 5), and assuming a lognormal distribution, estimate the parameters using the MLE method.
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''Solution''
<br>
<br>
In this example we have non-grouped data without suspensions. Thus, the partials reduce to:
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::<math>\begin{align}
  & \frac{\partial \Lambda }{\partial {\mu }'}= & \frac{1}{\sigma _{{{T}'}}^{2}}\cdot \underset{i=1}{\overset{14}{\mathop \sum }}\,\ln ({{T}_{i}})-{\mu }'=0 \\
& \frac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{T}_{i}})-{\mu }'}{\sigma _{{{T}'}}^{3}}-\frac{1}{{{\sigma }_{{{T}'}}}} \right)=0 
\end{align}</math>
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Substituting the values of  <math>{{T}_{i}}</math>  and solving the above system simultaneously, we get:
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::<math>\begin{align}
  & {{\sigma }_{{{T}'}}}= & 0.9537 \\
& {\mu }'= & 6.6356 
\end{align}</math>
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The mean and standard deviation of the times-to-failure can be estimated using Eqns. (mean) and (sdv):
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::<math>\overline{T}=\mu =1,200.31\text{ }hr</math>
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:and:
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::<math>{{\sigma }_{T}}=1,461.78\text{ }hr</math>
 
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[[Category:Acclerated_Testing_Reference]

Latest revision as of 03:23, 16 August 2012

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