# Appendix C: Benchmark Examples

In this section, five published examples are presented for comparison purposes. ReliaSoft's R&D validated the ALTA software with hundreds of data sets and methods. ALTA also cross-validates each provided solution by independently re-evaluating the second partial derivatives based on the estimated parameters each time a calculation is performed. These partials will be equal to zero when a solution is reached. Double precision is used throughout ALTA.

# Example 1

From Wayne Nelson [28, p. 135].

## Published Results for Example 1

• Published Results:
\begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 0.59673 \\ \widehat{B}=\ & 9920.195 \\ \widehat{C}=\ & 9.69517\cdot {{10}^{-7}} \\ \end{align}\,\!

## Computed Results for Example 1

This same data set can be entered into ALTA by selecting the data sheet for grouped times-to-failure data with suspensions and using the Arrhenius model, the lognormal distribution, and MLE. ALTA computed parameters for maximum likelihood are:

\begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 0.59678 \\ \widehat{B}=\ & 9924.804 \\ \widehat{C}=\ & 9.58978\cdot {{10}^{-7}} \\ \end{align}\,\!

# Example 2

From Wayne Nelson [28, p. 453], time to breakdown of a transformer oil, tested at 26kV, 28kV, 30kV, 32kV, 34kV, 36kV and 38kV.

## Published Results for Example 2

• Published Results:
\begin{align} \widehat{\beta }=\ & 0.777 \\ \widehat{K}=\ & 6.8742\cdot {{10}^{-29}} \\ \widehat{n}=\ & 17.72958 \\ \end{align}\,\!
• Published 95% confidence limits on $\beta \,\!$:
$\begin{matrix} \left\{ 0.653,0.923 \right\} \\ \end{matrix}\,\!$

## Computed Results for Example 2

Use the inverse power law model and Weibull as the underlying life distribution. ALTA computed parameters are:

\begin{align} \widehat{\beta }=\ & 0.7765, \\ \widehat{K}=\ & 6.8741\cdot {{10}^{-29}} \\ \widehat{n}=\ & 17.7296 \\ \end{align}\,\!
• ALTA computed 95% confidence limits on the parameters:
$\left\{ 0.6535,0.9228 \right\}\text{ for }\widehat{\beta }\,\!$

# Example 3

From Wayne Nelson [28, p. 157], forty bearings were tested to failure at four different test loads. The data were analyzed using the inverse power law Weibull model.

## Published Results for Example 3

Nelson's [28, p. 306] IPL-Weibull parameter estimates:

\begin{align} \widehat{\beta }=\ & 1.243396 \\ \widehat{K}=\ & 0.4350735 \\ \widehat{n}=\ & 13.8528 \\ \end{align}\,\!
• The 95% 2-sided confidence bounds on the parameters:
\begin{align} & \left\{ 0.9746493,1.586245 \right\}\text{ for }\widehat{\beta } \\ & \left\{ 0.332906,0.568596 \right\}\text{ for }\widehat{K } \\ & \left\{ 11.43569,16.26991 \right\}\text{ for }\widehat{n } \end{align}\,\!
• Percentile estimates at a stress of 0.87, with 95% 2-sided confidence bounds:
Percentile Life Estimate 95% Lower 95% Upper
1% 0.3913096 0.1251383 1.223632
10% 2.589731 1.230454 5.450596
90% 30.94404 19.41020 49.33149
99% 54.03563 33.02691 88.40821

## Computed Results for Example 3

Use the inverse power law model and Weibull as the underlying life distribution.

• ALTA computed parameters are:
\begin{align} \widehat{\beta }=\ & 1.243375 \\ \widehat{K}=\ & 0.4350548 \\ \widehat{n}=\ & 13.8529 \\ \end{align}\,\!
• The 95% 2-sided confidence bounds on the parameters:
\begin{align} & \left\{ 0.9745811,1.586303 \right\}\text{ for }\widehat{\beta } \\ & \left\{ 0.330007,0.573542 \right\}\text{ for }\widehat{K } \\ & \left\{ 11.43510,16.27079 \right\}\text{ for }\widehat{n } \end{align}\,\!
• Percentile estimates at a stress of 0.87, with 95% 2-sided confidence bounds:
Percentile Life Estimate 95% Lower 95% Upper
1% 0.3913095 0.1251097 1.223911
10% 2.589814 1.230384 5.451588
90% 30.94632 19.40876 49.34240
99% 54.04012 33.02411 88.43039

# Example 4

From Meeker and Escobar [26, p. 504], Mylar-Polyurethane Insulating Structure data using the inverse power law lognormal model.

## Published Results for Example 4

• Published Results:
\begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 1.05, \\ \widehat{K}=\ & 1.14\cdot {{10}^{-12}}, \\ \widehat{n}=\ & 4.28. \\ \end{align}\,\!
• The 95% 2-sided confidence bounds on the parameters:
\begin{align} & \left\{ 0.83,1.32 \right\}\text{ for }{{\widehat{\sigma }}_{{{T}'}}} \\ & \left\{ 3.123\cdot {{10}^{-15}},4.16\cdot {{10}^{-10}} \right\}\text{ for }\widehat{K } \\ & \left\{ 3.11,5.46 \right\}\text{ for }\widehat{n } \end{align}\,\!

## Computed Results for Example 4

Use the inverse power law lognormal.

• ALTA computed parameters are:
\begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 1.04979 \\ \widehat{K}=\ & 1.15\cdot {{10}^{-12}} \\ \widehat{n}=\ & 4.289 \\ \end{align}\,\!
• ALTA computed 95% confidence limits on the parameters:
\begin{align} & \left\{ 0.833,1.323 \right\}\text{ for }{{\widehat{\sigma }}_{{{T}'}}} \\ & \left\{ 3.227\cdot {{10}^{-15}},4.095\cdot {{10}^{-10}} \right\}\text{ for }\widehat{K } \\ & \left\{ 3.115,5.464 \right\}\text{ for }\widehat{n } \end{align}\,\!

# Example 5

From Meeker and Escobar [26, p. 515], Tantalum capacitor data using the combination (Temperature-NonThermal) Weibull model.

## Published Results for Example 5

• Published Results:
\begin{align} \widehat{\beta }=\ & 0.4292 \\ \widehat{B}=\ & 3829.468 \\ \widehat{C}=\ & 4.513\cdot {{10}^{36}} \\ \widehat{n}=\ & 20.1 \\ \end{align}\,\!
• The 95% 2-sided confidence bounds on the parameters:
\begin{align} & \left\{ 0.3165,0.58 \right\}\text{ for }\widehat{\beta } \\ & \left\{ -464.177,8007.069 \right\}\text{ for }\widehat{B } \\ & \left\{ 1.265\cdot {{10}^{25}},1.609\cdot {{10}^{48}} \right\}\text{ for }\widehat{C } \\ & \left\{ 11.4,28.8 \right\}\text{ for }\widehat{n } \end{align}\,\!

## Computed Results for Example 5

Use the Temperature-NonThermal model and Weibull as the underlying life distribution.

• ALTA computed parameters are:
\begin{align} \widehat{\beta }=\ & 0.4287 \\ \widehat{B}=\ & 3780.298 \\ \widehat{C}=\ & 4.772\cdot {{10}^{36}} \\ \widehat{n}=\ & 20.09 \\ \end{align}\,\!
• ALTA computed 95% confidence limits on the parameters:
\begin{align} & \left\{ 0.3169,0.5799 \right\}\text{ for }\widehat{\beta } \\ & \left\{ -483.83,8044.426 \right\}\text{ for }\widehat{B } \\ & \left\{ 1.268\cdot {{10}^{25}},1.796\cdot {{10}^{48}} \right\}\text{ for }\widehat{C } \\ & \left\{ 11.37,28.8 \right\}\text{ for }\widehat{n } \end{align}\,\!