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== The Lognormal Distribution ==
#REDIRECT [[Distributions_Used_in_Accelerated_Testing#The_Lognormal_Distribution]]
 
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: <br>
 
<br>
 
::<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:
 
:* <span class="texhtml">''T''' = ln ''T''</span>, and where the <span class="texhtml">''T''</span>s are the failure times.
:* <math>\bar{{T}'}=</math> mean of the natural logarithms of the times to failure.
:* <span class="texhtml">σ<sub>''T'''</sub> = </span> standard deviation of the natural logarithms of the failure times.
 
<br> The lognormal <span class="texhtml">''pdf''</span> can be obtained, realizing that for equal probabilities under the normal and lognormal <span class="texhtml">''pdf''</span> s incremental areas should also be equal, or:
 
<br>
 
::<math>\begin{align}
  f(T)dT = f(T')dT'
\end{align}</math>
 
<br> Taking the derivative yields:
 
<br>
 
::<math>d{T}'=\frac{dT}{T}</math>
 
<br> Substitution yields:
 
<br>
 
::<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>
 
<br>
 
where: <br>
 
::<math>f(T)\ge 0,T>0,-\infty <\bar{{T}'}<\infty ,{{\sigma }_{{{T}'}}}>0</math>
 
<br>
 
{{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: Parameter Estimation|Appendix B]].) <br>
 
{{ald pp}}
 
===== 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), <span class="texhtml">''T''<sub>''i''</sub></span>&nbsp;: 144, 385, 747, 1,144, 1,576 and 2,616. The steps for determining the parameters of the lognormal <span class="texhtml">''pdf''</span> representing the data, using probability plotting, are as follows: <br>
 
:* Rank the failure times in ascending order as shown next.
<center><math>\begin{matrix}
  \text{Failure} & \text{Failure Order Number}  \\
  \text{Time (Hr)} & \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>
<br>
 
:* Obtain their median rank plotting positions. The failure times, with their corresponding median ranks, are shown next:
 
<br>
<center><math>\begin{matrix}
  \text{Failure} & \text{Median}  \\
  \text{Time (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>
<br>
 
:* On a lognormal probability paper, plot the times and their corresponding rank value. The next figure displays an example of a lognormal probability paper. The paper is simply a log-log paper.
 
<br> [[Image:ALTA4.10.png|center|400px]] <br>
 
:* Draw the best possible straight line that goes through the <span class="texhtml">''t'' = 0</span> and <span class="texhtml">''R''(''t'') = 100%</span> point and through these points.
:* At the <span class="texhtml">''Q''(''t'') = 50%</span> 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> hours which means that <math>{{\bar{T}}^{\prime }}=\ln(\breve{T})=6.633</math>.
 
<br>
 
::
 
<br> [[Image:ALTA4.11.png|center|400px]] <br> <br>
 
:* The standard deviation, <span class="texhtml">σ<sub>''T'''</sub>,</span> can be found using the following equation:
 
<br>
 
::<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>
 
<br> Now any reliability value for any mission time <span class="texhtml">''t''</span> can be obtained. For example, the reliability for a mission of 200 hours, or any other time, can now be obtained either from the plot or analytically. <br> To obtain the value from the plot, draw a vertical line from the abscissa, at <span class="texhtml">''t'' = 200</span> hours, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read <span class="texhtml">''Q''(''t'')</span> . In this case, <span class="texhtml">''R''(''t'' = 200) = 1 − ''Q''(''t'' = 200) = 92%</span> . 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.
 
{{ald mle}}
 
===== 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.
 
'''Solution '''
 
In this example we have non-grouped data without suspensions. Thus, the partials reduce to:<br>
 
<br>
 
::<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>
 
<br> Substituting the values of <span class="texhtml">''T''<sub>''i''</sub></span> and solving the above system simultaneously, we get: <br>
 
<br>
 
::<math>\begin{align}
  & {{\sigma }_{{{T}'}}}= & 0.9537 \\
& {\mu }'= & 6.6356 
\end{align}</math>
 
<br> The mean and standard deviation of the times-to-failure can be estimated by: <br>
 
<br>
 
::<math>\overline{T}=\mu =1,200.31\text{ }hr</math>
 
<br> and: <br>
 
<br>
 
::<math>\begin{align}
  & {{\sigma }_{T}}= 1,461.78\text{ }hr
\end{align}</math>
 
<br>
{{RS Copyright}}
[[Category:Accelerated_Life_Data_Analysis_Reference]]

Latest revision as of 03:23, 16 August 2012

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