Cumulative Damage-Power-Weibull Example

Using the simple step-stress data given here, one would define [math]\displaystyle{ x(t) }[/math] as:

[math]\displaystyle{ \begin{align} x(t)=\ & 2,\text{ }0\lt t\le 250 \\ =\ & 3,\text{ }250\lt t\le 350 \\ =\ & 4,\text{ }350\lt t\le 370 \\ =\ & 5,\text{ }370\lt t\le 380 \\ =\ & 6,\text{ }380\lt t\le 390 \\ =\ & 7,\text{ }390\lt t\le +\infty \end{align} }[/math]

Assuming a power relation as the underlying life-stress relationship and the Weibull distribution as the underlying life distribution, one can then formulate the log-likelihood function for the above data set as,

[math]\displaystyle{ \begin{align} & \ln (L)= & \Lambda =\overset{F}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,\ln \left\{ \beta {{\left[ \frac{x(t)}{a} \right]}^{n}}{{\left[ \mathop{}_{0}^{{{t}_{i}}}{{\left[ \frac{\left[ x(u) \right]}{a} \right]}^{n}}du \right]}^{\beta -1}} \right\} -\overset{F}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,\left\{ {{\left[ \mathop{}_{0}^{{{t}_{i}}}{{\left[ \frac{\left[ x(u) \right]}{a} \right]}^{n}}du \right]}^{\beta }} \right\} \end{align} }[/math]

where:
• [math]\displaystyle{ F }[/math] is the number of exact time-to-failure data points.
• [math]\displaystyle{ \beta }[/math] is the Weibull shape parameter.
• [math]\displaystyle{ a }[/math] and [math]\displaystyle{ n }[/math] are the IPL parameters.
• [math]\displaystyle{ x(t) }[/math] is the stress profile function.
• [math]\displaystyle{ {{t}_{i}} }[/math] is the [math]\displaystyle{ {{i}^{th}} }[/math] time to failure.

The parameter estimates for [math]\displaystyle{ \hat{\beta } }[/math] , [math]\displaystyle{ \hat{a} }[/math] and [math]\displaystyle{ \hat{n} }[/math] can be obtained by simultaneously solving, [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial a}=0 }[/math] and [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial n}=0 }[/math] . Using ALTA, the parameter estimates for this data set are:

[math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 2.67829 \\ \widehat{\alpha }=\ & 9.842122 \\ \widehat{n}=\ & -3.998466 \end{align} }[/math]

Once the parameters are obtained, one can now determine the reliability for these units at any time [math]\displaystyle{ t }[/math] and stress [math]\displaystyle{ x(t) }[/math] from:

[math]\displaystyle{ R\left( t,x\left( t \right) \right)={{e}^{-{{\left[ \int_{0}^{t}{{\left[ \tfrac{x(u)}{a} \right]}^{n}}du \right]}^{\beta }}}} }[/math]

or at a fixed stress level [math]\displaystyle{ x(t)=2V }[/math] and [math]\displaystyle{ t=300 }[/math] ,

[math]\displaystyle{ R\left( t=300,x(t)=2 \right)={{e}^{-{{\left[ \mathop{}_{0}^{t}{{\left[ \tfrac{x(u)}{a} \right]}^{n}}du \right]}^{\beta }}}}=97.5% }[/math]

The mean time to failure (MTTF) at any stress [math]\displaystyle{ x(t) }[/math] can be determined by:

[math]\displaystyle{ MTTF\left( x\left( t \right) \right)=\int_{0}^{\infty }t\left[ \left\{ \beta {{\left[ \frac{x\left( t \right)}{a} \right]}^{n}}{{\left[ \int_{0}^{t}{{\left[ \frac{x\left( u \right)}{a} \right]}^{n}}du \right]}^{\beta -1}} \right\}{{e}^{-{{\left[ \int_{0}^{t}{{\left[ \tfrac{x(u)}{a} \right]}^{n}}du \right]}^{\beta }}}} \right]dt }[/math]

or at a fixed stress level [math]\displaystyle{ x\left( t \right)=2V }[/math] ,

[math]\displaystyle{ MTTF\left( x\left( t \right) \right)=1046.3hrs }[/math]

Any other metric of interest (e.g. failure rate, conditional reliability etc.) can also be determined using the basic definitions given in Appendix A and calculated automatically with ALTA.