Template:Eyring-ex mle

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


 * $$\begin{align}

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


 * where:


 * $$R_{Li}^{\prime \prime }={{e}^{-T_{Li}^{\prime \prime }{{V}_{i}}{{e}^{A-\tfrac{B}}}}}$$


 * $$R_{Ri}^{\prime \prime }={{e}^{-T_{Ri}^{\prime \prime }{{V}_{i}}{{e}^{A-\tfrac{B}}}}}$$


 * and:

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


 * $$\begin{align}

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


 * $$\begin{align}

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