Template:Eyring-ex mle: Difference between revisions

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• <math>FI</math>  is the number of interval data groups.
• <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 <math>{{T}_{i}}</math>  group of data intervals.
• <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 <math>{{T}_{i}}</math>  interval.
• <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 <math>{{T}_{i}}</math> interval.
• <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{A}</math>  and  <math>\widehat{B}</math>  so that  <math>\tfrac{\partial \Lambda }{\partial A}=0</math>  and  <math>\tfrac{\partial \Lambda }{\partial B}=0</math>  where:
The solution (parameter estimates) will be found by solving for the parameters  <math>\widehat{A}</math>  and  <math>\widehat{B}</math>  so that  <math>\tfrac{\partial \Lambda }{\partial A}=0</math>  and  <math>\tfrac{\partial \Lambda }{\partial B}=0</math>  where:

Revision as of 17:57, 9 March 2012

Maximum Likelihood Estimation Method


The complete exponential log-likelihood function of the Eyring model is composed of two summation portions:


[math]\displaystyle{ \begin{align} & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ {{V}_{i}}\cdot {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}{{e}^{-{{V}_{i}}\cdot {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}\cdot {{T}_{i}}}} \right] -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\cdot {{V}_{i}}\cdot {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}\cdot 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]


where:


[math]\displaystyle{ R_{Li}^{\prime \prime }={{e}^{-T_{Li}^{\prime \prime }{{V}_{i}}{{e}^{A-\tfrac{B}{{{V}_{i}}}}}}} }[/math]


[math]\displaystyle{ R_{Ri}^{\prime \prime }={{e}^{-T_{Ri}^{\prime \prime }{{V}_{i}}{{e}^{A-\tfrac{B}{{{V}_{i}}}}}}} }[/math]


and:

[math]\displaystyle{ {{F}_{e}} }[/math] is the number of groups of exact times-to-failure data points.
[math]\displaystyle{ {{N}_{i}} }[/math] is the number of times-to-failure in the [math]\displaystyle{ {{i}^{th}} }[/math] time-to-failure data group.
[math]\displaystyle{ {{V}_{i}} }[/math] is the stress level of the [math]\displaystyle{ {{i}^{th}} }[/math] group.
[math]\displaystyle{ A }[/math] is the Eyring parameter (unknown, the first of two parameters to be estimated).
[math]\displaystyle{ B }[/math] is the second Eyring parameter (unknown, the second of two parameters to be estimated).
[math]\displaystyle{ {{T}_{i}} }[/math] is the exact failure time of the [math]\displaystyle{ {{i}^{th}} }[/math] group.
[math]\displaystyle{ S }[/math] is the number of groups of suspension data points.
[math]\displaystyle{ N_{i}^{\prime } }[/math] is the number of suspensions in the [math]\displaystyle{ {{i}^{th}} }[/math] group of suspension data points.
[math]\displaystyle{ T_{i}^{\prime } }[/math] is the running time of the [math]\displaystyle{ {{i}^{th}} }[/math] suspension data group.
[math]\displaystyle{ FI }[/math] is the number of interval data groups.
[math]\displaystyle{ N_{i}^{\prime \prime } }[/math] is the number of intervals in the i [math]\displaystyle{ ^{th} }[/math] group of data intervals.
[math]\displaystyle{ T_{Li}^{\prime \prime } }[/math] is the beginning of the i [math]\displaystyle{ ^{th} }[/math] interval.
[math]\displaystyle{ T_{Ri}^{\prime \prime } }[/math] is the ending of the i [math]\displaystyle{ ^{th} }[/math] interval.
The solution (parameter estimates) will be found by solving for the parameters [math]\displaystyle{ \widehat{A} }[/math] and [math]\displaystyle{ \widehat{B} }[/math] so that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial A}=0 }[/math] and [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial B}=0 }[/math] where:



[math]\displaystyle{ \begin{align} & \frac{\partial \Lambda }{\partial A}= & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( 1-{{V}_{i}}\cdot {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }{{V}_{i}}\cdot {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}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}}{{e}^{A-\tfrac{B}{{{V}_{i}}}}}}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} \end{align} }[/math]


[math]\displaystyle{ \begin{align} & \frac{\partial \Lambda }{\partial B}= & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left[ {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}{{T}_{i}}-\frac{1}{{{V}_{i}}} \right]+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\cdot {{e}^{\left( A-\tfrac{B}{{{V}_{i}}} \right)}}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){{e}^{A-\tfrac{B}{{{V}_{i}}}}}}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} \end{align} }[/math]