Template:Acb-w on time

Confidence Bounds on Time
The bounds around time, for a given lognormal percentile (unreliability), are estimated by first solving the reliability equation with respect to time, as follows:


 * $${T}'(V;\widehat{B},\widehat{C},{{\widehat{\sigma }}_})=\ln (\widehat{C})+\frac{\widehat{B}}{V}+z\cdot {{\widehat{\sigma }}_}$$

where:


 * $$\begin{align}

{T}'(V;\widehat{B},\widehat{C},{{\widehat{\sigma }}_})=&\ \ln (T) \\ z= & \ {{\Phi }^{-1}}\left[ F({T}') \right] \end{align}$$

and:


 * $$\Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z({T}')}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz$$

The next step is to calculate the variance of $${T}'(V;\widehat{B},\widehat{C},{{\widehat{\sigma }}_}):$$


 * $$\begin{align}

Var({T}')= & {{\left( \frac{\partial {T}'}{\partial B} \right)}^{2}}Var(\widehat{B})+{{\left( \frac{\partial {T}'}{\partial C} \right)}^{2}}Var(\widehat{C})+{{\left( \frac{\partial {T}'}{\partial {{\sigma }_}} \right)}^{2}}Var({{\widehat{\sigma }}_}) +2\left( \frac{\partial {T}'}{\partial B} \right)\left( \frac{\partial {T}'}{\partial C} \right)Cov\left( \widehat{B},\widehat{C} \right) \\ & +2\left( \frac{\partial {T}'}{\partial B} \right)\left( \frac{\partial {T}'}{\partial {{\sigma }_}} \right)Cov\left( \widehat{B},{{\widehat{\sigma }}_} \right) +2\left( \frac{\partial {T}'}{\partial C} \right)\left( \frac{\partial {T}'}{\partial {{\sigma }_}} \right)Cov\left( \widehat{C},{{\widehat{\sigma }}_} \right) \end{align}$$

or:


 * $$\begin{align}

& Var({T}')= \frac{1}Var(\widehat{B})+\frac{1}Var(\widehat{C})+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_}) +\frac{2}{B\cdot C}Cov\left( \widehat{B},\widehat{C} \right) +\frac{2\widehat{z}}{V}Cov\left( \widehat{B},{{\widehat{\sigma }}_} \right) +\frac{2\widehat{z}}{C}Cov\left( \widehat{C},{{\widehat{\sigma }}_} \right) \end{align}$$

The upper and lower bounds are then found by:


 * $$\begin{align}

& T_{U}^{\prime }= & \ln {{T}_{U}}={T}'+{{K}_{\alpha }}\sqrt{Var({T}')} \\ & T_{L}^{\prime }= & \ln {{T}_{L}}={T}'-{{K}_{\alpha }}\sqrt{Var({T}')} \end{align}$$

Solving for $${{T}_{U}}$$  and  $${{T}_{L}}$$  yields:


 * $$\begin{align}

& {{T}_{U}}= & {{e}^{T_{U}^{\prime }}}\text{ (Upper bound)} \\ & {{T}_{L}}= & {{e}^{T_{L}^{\prime }}}\text{ (Lower bound)} \end{align}$$