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| ====MLE Parameter Estimation====
| | #REDIRECT [[The_Exponential_Distribution]] |
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| <br>
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| Let's assume six identical units are reliability tested at the same application and operation
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| stress levels. All of these units fail during the test after operating for the following times (in hours), <math>{{T}_{i}}</math> : 96, 257, 498, 763, 1051 and 1744.
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| The parameter of the exponential distribution can also be estimated using the maximum likelihood estimation (MLE) method.
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| This log-likelihood function is:
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| <br>
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| ::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda 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 }]</math>
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| <br>
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| where:
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| <br>
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| ::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>
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| <br>
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| ::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math>
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| <br>
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| and
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| <br>
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| ::• <math>{{F}_{e}}</math> is the number of groups of 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>\lambda </math> is the failure rate parameter (unknown a priori, the only parameter to be found).
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| ::• <math>{{T}_{i}}</math> is the time of the <math>{{i}^{th}}</math> group of time-to-failure data.
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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 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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| <br>
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| The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where:
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| <br>
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| ::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime }-\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}</math>
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| <br>
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| For the given data then the equation is
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| <br>
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| ::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=0</math>
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| <br>
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| Substituting the values for <math>T</math> we get:
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| <br>
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| ::<math>\begin{align}
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| \frac{6}{\lambda }= & 4409,\text{ or:} \\
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| \lambda = & 0.00136\text{ failure/hr}
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| \end{align}</math>
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