Template:Ipl ex mle

Maximum Likelihood Parameter Estimation
Substituting the inverse power law relationship into the exponential log-likelihood equation yields:


 * $$\begin{align}

& \ln (L)= \Lambda =\underset{i=1}{\overset{\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}$$

where:


 * $$R_{Li}^{\prime \prime }={{e}^{-T_{Li}^{\prime \prime }KV_{i}^{n}}}$$


 * $$R_{Ri}^{\prime \prime }={{e}^{-T_{Ri}^{\prime \prime }KV_{i}^{n}}}$$

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. •	 $$K$$ is the IPL parameter (unknown, the first of two parameters to be estimated). •	 $$n$$ is the second IPL 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 $$\widehat{K},$$   $$\widehat{n}$$  so that  $$\tfrac{\partial \Lambda }{\partial K}=0$$  and  $$\tfrac{\partial \Lambda }{\partial n}=0$$, where:


 * $$\begin{align}

\frac{\partial \Lambda }{\partial K}=\ & \frac{1}{K}\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}-\underset{i=1}{\overset{\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}$$


 * $$\begin{align}

\frac{\partial \Lambda }{\partial n}=\ & \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln ({{V}_{i}})-K\underset{i=1}{\overset{\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}$$