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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}
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| & \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] \\
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| & & -\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 }]
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| \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}
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| & \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 } \\
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| & & \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 }}
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| \end{align}</math>
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| ::<math>\begin{align}
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| & \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}} \\
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| & & -K\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }V_{i}^{n}\ln ({{V}_{i}})T_{i}^{\prime } \\
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| & & \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 }}
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| \end{align}</math>
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| <br>
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