Template:Ipl ex mle: Difference between revisions

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====Maximum Likelihood Parameter Estimation====
#REDIRECT [[Inverse_Power_Law_(IPL)_Relationship#IPL-Exponential]]
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Substituting the inverse power law relationship into the exponential log-likelihood equation yields:
 
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::<math>\begin{align}
  & \ln (L)= \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ KV_{i}^{n}{{e}^{-KV_{i}^{n}{{T}_{i}}}} \right] -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }KV_{i}^{n}T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }] 
\end{align}</math>
 
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where:
 
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::<math>R_{Li}^{\prime \prime }={{e}^{-T_{Li}^{\prime \prime }KV_{i}^{n}}}</math>
 
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::<math>R_{Ri}^{\prime \prime }={{e}^{-T_{Ri}^{\prime \prime }KV_{i}^{n}}}</math>
 
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and:
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• <math>{{F}_{e}}</math>  is the number of groups of exact times-to-failure data points.
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• <math>{{N}_{i}}</math>  is the number of times-to-failure in the  <math>{{i}^{th}}</math>  time-to-failure data group.
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• <math>{{V}_{i}}</math>  is the stress level of the  <math>{{i}^{th}}</math>  group.
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• <math>K</math>  is the IPL parameter (unknown, the first of two parameters to be estimated).
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• <math>n</math>  is the second IPL parameter (unknown, the second of two parameters to be estimated).
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• <math>{{T}_{i}}</math>  is the exact failure time of the  <math>{{i}^{th}}</math>  group.
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• <math>S</math>  is the number of groups of suspension data points.
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• <math>N_{i}^{\prime }</math>  is the number of suspensions in the  <math>{{i}^{th}}</math>  group of suspension data points.
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• <math>T_{i}^{\prime }</math>  is the running time of the  <math>{{i}^{th}}</math> suspension data group.
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• <math>FI</math>  is the number of interval data groups.
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• <math>N_{i}^{\prime \prime }</math>  is the number of intervals in the i <math>^{th}</math>  group of data intervals.
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• <math>T_{Li}^{\prime \prime }</math>  is the beginning of the i <math>^{th}</math>  interval.
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• <math>T_{Ri}^{\prime \prime }</math>  is the ending of the i <math>^{th}</math>  interval.
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The solution (parameter estimates) will be found by solving for the parameters  <math>\widehat{K},</math>  <math>\widehat{n}</math>  so that  <math>\tfrac{\partial \Lambda }{\partial K}=0</math>  and  <math>\tfrac{\partial \Lambda }{\partial n}=0</math> , where:
 
 
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::<math>\begin{align}
  \frac{\partial \Lambda }{\partial K}=\ & \frac{1}{K}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}-\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}V_{i}^{n}{{T}_{i}}-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }V_{i}^{n}T_{i}^{\prime } \overset{FI}{\mathop{\underset{i=1}{\mathop{-\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{\left( T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime } \right)V_{i}^{n}}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} 
\end{align}</math>
 
 
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::<math>\begin{align}
  \frac{\partial \Lambda }{\partial n}=\ & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln ({{V}_{i}})-K\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}V_{i}^{n}\ln ({{V}_{i}}){{T}_{i}} -K\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }V_{i}^{n}\ln ({{V}_{i}})T_{i}^{\prime } \overset{FI}{\mathop{\underset{i=1}{\mathop{-\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{KV_{i}^{n}\ln ({{V}_{i}})\left( T_{Li}{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime } \right)}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} 
\end{align}</math>
 
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Latest revision as of 23:14, 15 August 2012

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