Mixed Weibull Log-Likelihood Functions and their Partials

Mixed Weibull Log-Likelihood Functions and their Partials
The log-likelihood function (without the constant) is composed of three summation portions:
 * $$\begin{align}

\frac{\partial \Lambda }{\partial {{\sigma }_}}= & \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\left( \frac{\sigma _^{3}}-\frac{1} \right) \\ & +\frac{1}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'} \right)\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'} \right)} \\ & -\frac{1}\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\frac{\left( \tfrac{\ln \left( T_{Ri}^{\prime \prime } \right)-{\mu }'} \right)\phi \left( \tfrac{\ln \left( T_{Ri}^{\prime \prime } \right)-{\mu }'} \right)-\left( \tfrac{\ln \left( T_{Li}^{\prime \prime } \right)-{\mu }'} \right)\phi \left( \tfrac{\ln \left( T_{Li}^{\prime \prime } \right)-{\mu }'} \right)}{\Phi \left( \tfrac{\ln \left( T_{Ri}^{\prime \prime } \right)-{\mu }'} \right)-\Phi \left( \tfrac{\ln \left( T_{Li}^{\prime \prime } \right)-{\mu }'} \right)} \end{align}$$


 * $$\begin{align}

\ln (L)= & \Lambda =\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \underset{k=1}{\overset{Q}{\mathop \sum }}\,{{\rho }_{k}}\frac{{\left( \frac{{{T}_{i}}} \right)}^{{{\beta }_{k}}-1}}{{e}^{-{{\left( \tfrac{{{T}_{i}}} \right)}^}}} \right] \\ & \text{ }+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ln \left[ \underset{k=1}{\overset{Q}{\mathop \sum }}\,{{\rho }_{k}}{{e}^{-{{\left( \tfrac{T_{i}^{\prime }} \right)}^}}} \right] \\ & \text{ }+\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\ln \left[ \underset{k=1}{\overset{Q}{\mathop \sum }}\,{{\rho }_{k}}\frac{{{\eta }_{k}}}{{\left( \frac{T_{Li}^{\prime \prime }+T_{Ri}^{\prime \prime }}{2{{\eta }_{k}}} \right)}^{{{\beta }_{k}}-1}}{{e}^{-{{\left( \tfrac{T_{Li}^{\prime \prime }+T_{Ri}^{\prime \prime }}{2{{\eta }_{k}}} \right)}^}}} \right] \end{align}$$
 * where:
 * •	$${{F}_{e}}$$ is the number of groups of times-to-failure data points
 * •	$${{N}_{i}}$$ is the number of times-to-failure in the $${{i}^{th}}$$ time-to-failure data group
 * •	$$Q$$ is the number of subpopulations
 * •	$${{\rho }_{k}}$$ is the proportionality of the $${{k}^{th}}$$ subpopulation (unknown a priori, the first set of three sets of parameters to be found)
 * •	$${{\beta }_{k}}$$ is the Weibull shape parameter of the $${{k}^{th}}$$ subpopulation (unknown a priori, the second set of three sets of parameters to be found)
 * •	$${{\eta }_{k}}$$ is the Weibull scale parameter (unknown a priori, the third set of three sets of parameters to be found)
 * •	$${{T}_{i}}$$ is the time of the $${{i}^{th}}$$ group of time-to-failure data
 * •	$$S$$ is the number of groups of suspension data points
 * •	$$N_{i}^{\prime }$$ is the number of suspensions in $${{i}^{th}}$$ group of suspension data points
 * •	$$T_{i}^{\prime }$$ is the time of the $${{i}^{th}}$$ suspension data group
 * •	$$FI$$ is the number of groups of interval data points
 * •	$$N_{i}^{\prime \prime }$$ is the number of intervals in $${{i}^{th}}$$ group of data intervals
 * •	$$T_{Li}^{\prime \prime }$$ is the beginning of the $${{i}^{th}}$$ interval
 * •	and $$T_{Ri}^{\prime \prime }$$ is the ending of the $${{i}^{th}}$$ interval

The solution will be found by solving for a group of parameters:


 * $$\left( \widehat\widehat,\widehat,\widehat\widehat,\widehat,...,\widehat\widehat,\widehat \right)$$


 * so that:


 * $$\begin{align}

\frac{\partial \Lambda }{\partial {{\rho }_{1}}}= & 0,\frac{\partial \Lambda }{\partial {{\beta }_{1}}}=0,\frac{\partial \Lambda }{\partial {{\eta }_{1}}}=0 \\ \frac{\partial \Lambda }{\partial {{\rho }_{2}}}= & 0,\frac{\partial \Lambda }{\partial {{\beta }_{2}}}=0,\frac{\partial \Lambda }{\partial {{\eta }_{2}}}=0 \\ \vdots \\ \frac{\partial \Lambda }{\partial {{\rho }_{Q-1}}}= & 0,\frac{\partial \Lambda }{\partial {{\beta }_{Q-1}}}=0,\frac{\partial \Lambda }{\partial {{\eta }_{Q-1}}}=0 \\ \frac{\partial \Lambda }{\partial {{\beta }_{Q}}}= & 0,\text{ and }\frac{\partial \Lambda }{\partial {{\eta }_{Q}}}=0 \end{align}$$