Power Law Model Confidence Bounds Example
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Using the data from the Power Law Model Parameter Estimation Example, calculate the mission reliability at [math]\displaystyle{ t=2000\,\! }[/math] hours and mission time [math]\displaystyle{ d=40\,\! }[/math] hours along with the confidence bounds at the 90% confidence level.
Solution
The maximum likelihood estimates of [math]\displaystyle{ \widehat{\lambda }\,\! }[/math] and [math]\displaystyle{ \widehat{\beta }\,\! }[/math] from the example are:
- [math]\displaystyle{ \begin{align} \widehat{\beta }= & 0.45300 \\ \widehat{\lambda }= & 0.36224 \end{align}\,\! }[/math]
The mission reliability at [math]\displaystyle{ t=2000\,\! }[/math] for mission time [math]\displaystyle{ d=40\,\! }[/math] is:
- [math]\displaystyle{ \begin{align} \widehat{R}(t)= & {{e}^{-\left[ \lambda {{\left( t+d \right)}^{\beta }}-\lambda {{t}^{\beta }} \right]}} \\ = & 0.90292 \end{align}\,\! }[/math]
At the 90% confidence level and [math]\displaystyle{ T=2000\,\! }[/math] hours, the Fisher matrix confidence bounds for the mission reliability for mission time [math]\displaystyle{ d=40\,\! }[/math] are given by:
- [math]\displaystyle{ CB=\frac{\widehat{R}(t)}{\widehat{R}(t)+(1-\widehat{R}(t)){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{R}(t))}/\left[ \widehat{R}(t)(1-\widehat{R}(t)) \right]}}}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{[\widehat{R}(t)]}_{L}}= & 0.83711 \\ {{[\widehat{R}(t)]}_{U}}= & 0.94392 \end{align}\,\! }[/math]
The Crow confidence bounds for the mission reliability are:
- [math]\displaystyle{ \begin{align} {{[\widehat{R}(t)]}_{L}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{1}}}}} \\ = & {{[0.90292]}^{\tfrac{1}{0.71440}}} \\ = & 0.86680 \\ {{[\widehat{R}(t)]}_{U}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{2}}}}} \\ = & {{[0.90292]}^{\tfrac{1}{1.6051}}} \\ = & 0.93836 \end{align}\,\! }[/math]
The next two figures show the Fisher matrix and Crow confidence bounds on mission reliability for mission time [math]\displaystyle{ d=40\,\! }[/math].
