Appendix C: Benchmark Examples: Difference between revisions
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{{template:ALTABOOK | {{template:ALTABOOK|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. | 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= | =Example 1= | ||
From Wayne Nelson [[ | From Wayne Nelson [[Appendix_E:_References|[28, p. 135]]]. | ||
==Published Results for Example 1== | ==Published Results for Example 1== | ||
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\widehat{B}=\ & 9920.195 \\ | \widehat{B}=\ & 9920.195 \\ | ||
\widehat{C}=\ & 9.69517\cdot {{10}^{-7}} \\ | \widehat{C}=\ & 9.69517\cdot {{10}^{-7}} \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
==Computed Results for Example 1== | ==Computed Results for Example 1== | ||
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\widehat{B}=\ & 9924.804 \\ | \widehat{B}=\ & 9924.804 \\ | ||
\widehat{C}=\ & 9.58978\cdot {{10}^{-7}} \\ | \widehat{C}=\ & 9.58978\cdot {{10}^{-7}} \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
=Example 2= | =Example 2= | ||
From Wayne Nelson [[ | From Wayne Nelson [[Appendix_E:_References|[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 for Example 2== | ||
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\widehat{K}=\ & 6.8742\cdot {{10}^{-29}} \\ | \widehat{K}=\ & 6.8742\cdot {{10}^{-29}} \\ | ||
\widehat{n}=\ & 17.72958 \\ | \widehat{n}=\ & 17.72958 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*Published 95% confidence limits on | *Published 95% confidence limits on <math>\beta \,\!</math>: | ||
::<math>\begin{matrix} | ::<math>\begin{matrix} | ||
\left\{ 0.653,0.923 \right\} \\ | \left\{ 0.653,0.923 \right\} \\ | ||
\end{matrix}</math> | \end{matrix}\,\!</math> | ||
==Computed Results for Example 2== | ==Computed Results for Example 2== | ||
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\widehat{K}=\ & 6.8741\cdot {{10}^{-29}} \\ | \widehat{K}=\ & 6.8741\cdot {{10}^{-29}} \\ | ||
\widehat{n}=\ & 17.7296 \\ | \widehat{n}=\ & 17.7296 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*ALTA computed 95% confidence limits on the parameters: | *ALTA computed 95% confidence limits on the parameters: | ||
::<math>\left\{ 0.6535,0.9228 \right\}\text{ for }\widehat{\beta }</math> | ::<math>\left\{ 0.6535,0.9228 \right\}\text{ for }\widehat{\beta }\,\!</math> | ||
=Example 3= | =Example 3= | ||
From Wayne Nelson [[ | From Wayne Nelson [[Appendix_E:_References|[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== | ==Published Results for Example 3== | ||
Nelson's [[ | Nelson's [[Appendix_E:_References|[28, p. 306]]] IPL-Weibull parameter estimates: | ||
::<math>\begin{align} | ::<math>\begin{align} | ||
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\widehat{K}=\ & 0.4350735 \\ | \widehat{K}=\ & 0.4350735 \\ | ||
\widehat{n}=\ & 13.8528 \\ | \widehat{n}=\ & 13.8528 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*The 95% 2-sided confidence bounds on the parameters: | *The 95% 2-sided confidence bounds on the parameters: | ||
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& \left\{ 0.332906,0.568596 \right\}\text{ for }\widehat{K } \\ | & \left\{ 0.332906,0.568596 \right\}\text{ for }\widehat{K } \\ | ||
& \left\{ 11.43569,16.26991 \right\}\text{ for }\widehat{n } | & \left\{ 11.43569,16.26991 \right\}\text{ for }\widehat{n } | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*Percentile estimates at a stress of 0.87, with 95% 2-sided confidence bounds: | *Percentile estimates at a stress of 0.87, with 95% 2-sided confidence bounds: | ||
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\widehat{K}=\ & 0.4350548 \\ | \widehat{K}=\ & 0.4350548 \\ | ||
\widehat{n}=\ & 13.8529 \\ | \widehat{n}=\ & 13.8529 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*The 95% 2-sided confidence bounds on the parameters: | *The 95% 2-sided confidence bounds on the parameters: | ||
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& \left\{ 0.330007,0.573542 \right\}\text{ for }\widehat{K } \\ | & \left\{ 0.330007,0.573542 \right\}\text{ for }\widehat{K } \\ | ||
& \left\{ 11.43510,16.27079 \right\}\text{ for }\widehat{n } | & \left\{ 11.43510,16.27079 \right\}\text{ for }\widehat{n } | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*Percentile estimates at a stress of 0.87, with 95% 2-sided confidence bounds: | *Percentile estimates at a stress of 0.87, with 95% 2-sided confidence bounds: | ||
Line 133: | Line 133: | ||
=Example 4= | =Example 4= | ||
From Meeker and Escobar [[ | From Meeker and Escobar [[Appendix_E:_References|[26, p. 504]]], Mylar-Polyurethane Insulating Structure data using the inverse power law lognormal model. | ||
==Published Results for Example 4== | ==Published Results for Example 4== | ||
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\widehat{K}=\ & 1.14\cdot {{10}^{-12}}, \\ | \widehat{K}=\ & 1.14\cdot {{10}^{-12}}, \\ | ||
\widehat{n}=\ & 4.28. \\ | \widehat{n}=\ & 4.28. \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*The 95% 2-sided confidence bounds on the parameters: | *The 95% 2-sided confidence bounds on the parameters: | ||
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& \left\{ 3.123\cdot {{10}^{-15}},4.16\cdot {{10}^{-10}} \right\}\text{ for }\widehat{K } \\ | & \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 } | & \left\{ 3.11,5.46 \right\}\text{ for }\widehat{n } | ||
\end{align}</math> | \end{align}\,\!</math> | ||
==Computed Results for Example 4== | ==Computed Results for Example 4== | ||
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\widehat{K}=\ & 1.15\cdot {{10}^{-12}} \\ | \widehat{K}=\ & 1.15\cdot {{10}^{-12}} \\ | ||
\widehat{n}=\ & 4.289 \\ | \widehat{n}=\ & 4.289 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*ALTA computed 95% confidence limits on the parameters: | *ALTA computed 95% confidence limits on the parameters: | ||
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& \left\{ 3.227\cdot {{10}^{-15}},4.095\cdot {{10}^{-10}} \right\}\text{ for }\widehat{K } \\ | & \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 } | & \left\{ 3.115,5.464 \right\}\text{ for }\widehat{n } | ||
\end{align}</math> | \end{align}\,\!</math> | ||
=Example 5= | =Example 5= | ||
From Meeker and Escobar [[ | From Meeker and Escobar [[Appendix_E:_References|[26, p. 515]]], Tantalum capacitor data using the combination (Temperature-NonThermal) Weibull model. | ||
==Published Results for Example 5== | ==Published Results for Example 5== | ||
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\widehat{C}=\ & 4.513\cdot {{10}^{36}} \\ | \widehat{C}=\ & 4.513\cdot {{10}^{36}} \\ | ||
\widehat{n}=\ & 20.1 \\ | \widehat{n}=\ & 20.1 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*The 95% 2-sided confidence bounds on the parameters: | *The 95% 2-sided confidence bounds on the parameters: | ||
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& \left\{ 1.265\cdot {{10}^{25}},1.609\cdot {{10}^{48}} \right\}\text{ for }\widehat{C } \\ | & \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 } | & \left\{ 11.4,28.8 \right\}\text{ for }\widehat{n } | ||
\end{align}</math> | \end{align}\,\!</math> | ||
==Computed Results for Example 5== | ==Computed Results for Example 5== | ||
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\widehat{C}=\ & 4.772\cdot {{10}^{36}} \\ | \widehat{C}=\ & 4.772\cdot {{10}^{36}} \\ | ||
\widehat{n}=\ & 20.09 \\ | \widehat{n}=\ & 20.09 \\ | ||
\end{align}</math> | \end{align}\,\!</math> | ||
*ALTA computed 95% confidence limits on the parameters: | *ALTA computed 95% confidence limits on the parameters: | ||
Line 212: | Line 212: | ||
& \left\{ 1.268\cdot {{10}^{25}},1.796\cdot {{10}^{48}} \right\}\text{ for }\widehat{C } \\ | & \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 } | & \left\{ 11.37,28.8 \right\}\text{ for }\widehat{n } | ||
\end{align}</math> | \end{align}\,\!</math> |
Latest revision as of 19:18, 15 September 2023
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:
- [math]\displaystyle{ \begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 0.59673 \\ \widehat{B}=\ & 9920.195 \\ \widehat{C}=\ & 9.69517\cdot {{10}^{-7}} \\ \end{align}\,\! }[/math]
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:
- [math]\displaystyle{ \begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 0.59678 \\ \widehat{B}=\ & 9924.804 \\ \widehat{C}=\ & 9.58978\cdot {{10}^{-7}} \\ \end{align}\,\! }[/math]
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:
- [math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 0.777 \\ \widehat{K}=\ & 6.8742\cdot {{10}^{-29}} \\ \widehat{n}=\ & 17.72958 \\ \end{align}\,\! }[/math]
- Published 95% confidence limits on [math]\displaystyle{ \beta \,\! }[/math]:
- [math]\displaystyle{ \begin{matrix} \left\{ 0.653,0.923 \right\} \\ \end{matrix}\,\! }[/math]
Computed Results for Example 2
Use the inverse power law model and Weibull as the underlying life distribution. ALTA computed parameters are:
- [math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 0.7765, \\ \widehat{K}=\ & 6.8741\cdot {{10}^{-29}} \\ \widehat{n}=\ & 17.7296 \\ \end{align}\,\! }[/math]
- ALTA computed 95% confidence limits on the parameters:
- [math]\displaystyle{ \left\{ 0.6535,0.9228 \right\}\text{ for }\widehat{\beta }\,\! }[/math]
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:
- [math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 1.243396 \\ \widehat{K}=\ & 0.4350735 \\ \widehat{n}=\ & 13.8528 \\ \end{align}\,\! }[/math]
- The 95% 2-sided confidence bounds on the parameters:
- [math]\displaystyle{ \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}\,\! }[/math]
- 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:
- [math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 1.243375 \\ \widehat{K}=\ & 0.4350548 \\ \widehat{n}=\ & 13.8529 \\ \end{align}\,\! }[/math]
- The 95% 2-sided confidence bounds on the parameters:
- [math]\displaystyle{ \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}\,\! }[/math]
- 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:
- [math]\displaystyle{ \begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 1.05, \\ \widehat{K}=\ & 1.14\cdot {{10}^{-12}}, \\ \widehat{n}=\ & 4.28. \\ \end{align}\,\! }[/math]
- The 95% 2-sided confidence bounds on the parameters:
- [math]\displaystyle{ \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}\,\! }[/math]
Computed Results for Example 4
Use the inverse power law lognormal.
- ALTA computed parameters are:
- [math]\displaystyle{ \begin{align} {{\widehat{\sigma }}_{{{T}'}}}=\ & 1.04979 \\ \widehat{K}=\ & 1.15\cdot {{10}^{-12}} \\ \widehat{n}=\ & 4.289 \\ \end{align}\,\! }[/math]
- ALTA computed 95% confidence limits on the parameters:
- [math]\displaystyle{ \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}\,\! }[/math]
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:
- [math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 0.4292 \\ \widehat{B}=\ & 3829.468 \\ \widehat{C}=\ & 4.513\cdot {{10}^{36}} \\ \widehat{n}=\ & 20.1 \\ \end{align}\,\! }[/math]
- The 95% 2-sided confidence bounds on the parameters:
- [math]\displaystyle{ \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}\,\! }[/math]
Computed Results for Example 5
Use the Temperature-NonThermal model and Weibull as the underlying life distribution.
- ALTA computed parameters are:
- [math]\displaystyle{ \begin{align} \widehat{\beta }=\ & 0.4287 \\ \widehat{B}=\ & 3780.298 \\ \widehat{C}=\ & 4.772\cdot {{10}^{36}} \\ \widehat{n}=\ & 20.09 \\ \end{align}\,\! }[/math]
- ALTA computed 95% confidence limits on the parameters:
- [math]\displaystyle{ \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}\,\! }[/math]