Cumulative Damage-Power-Weibull Example

Using the simple step-stress data given here, one would define $$x(t)$$   as:

$$\begin{align} x(t)=\ & 2,\text{   }0<t\le 250 \\ =\ & 3,\text{   }250<t\le 350 \\ =\ & 4,\text{   }350<t\le 370 \\ =\ & 5,\text{   }370<t\le 380 \\ =\ & 6,\text{   }380<t\le 390 \\ =\ & 7,\text{   }390<t\le +\infty \end{align}$$

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,


 * $$\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}$$

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

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


 * $$\begin{align}

\widehat{\beta }=\ & 2.67829 \\ \widehat{\alpha }=\ & 9.842122 \\ \widehat{n}=\ & -3.998466 \end{align}$$

Once the parameters are obtained, one can now determine the reliability for these units at any time $$t$$  and stress  $$x(t)$$  from:


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

or at a fixed stress level $$x(t)=2V$$  and  $$t=300$$ ,


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

The mean time to failure (MTTF) at any stress $$x(t)$$  can be determined by:


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

or at a fixed stress level $$x\left( t \right)=2V$$ ,


 * $$MTTF\left( x\left( t \right) \right)=1046.3hrs$$

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.