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{{template:ALTABOOK|C}}
{{template:ALTABOOK|Appendix C|Benchmark Examples}}
=Reference Appendix C: Benchmark Examples=
<br>
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.
<br>
 
=Example 1=
=Example 1=
<br>
 
From Wayne Nelson [28, p. 135].
From Wayne Nelson [[Appendix_E:_References|[28, p. 135]]].
<br>
 
==Published Results for Example 1==
==Published Results for Example 1==
Published Results:   
*Published Results:   
::<math>\begin{matrix}
 
   {{\widehat{\sigma }}_{{{T}'}}}=0.59673  \\
::<math>\begin{align}
   \widehat{B}=9920.195  \\
   {{\widehat{\sigma }}_{{{T}'}}}=\ & 0.59673  \\
   \widehat{C}=9.69517\cdot {{10}^{-7}}  \\
   \widehat{B}=\ & 9920.195  \\
\end{matrix}</math>
   \widehat{C}=\ & 9.69517\cdot {{10}^{-7}}  \\
<br>
\end{align}\,\!</math>


==Computed Results for Example 1==
==Computed Results for Example 1==
<br>
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:
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>\begin{align}
  {{\widehat{\sigma }}_{{{T}'}}}=\ & 0.59678  \\
  \widehat{B}=\ & 9924.804  \\
  \widehat{C}=\ & 9.58978\cdot {{10}^{-7}}  \\
\end{align}\,\!</math>


::<math>\begin{matrix}
=Example 2=
  {{\widehat{\sigma }}_{{{T}'}}}=0.59678  \\
  \widehat{B}=9924.804  \\
  \widehat{C}=9.58978\cdot {{10}^{-7}}  \\
\end{matrix}</math>


<br>
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.


=Example 2=
<br>
From Wayne Nelson [28, p. 453], time to breakdown of a transformer oil, tested at 26kV, 28kV, 30kV, 32kV, 34kV, 36kV and 38kV.
<br>
<br>
==Published Results for Example 2==
==Published Results for Example 2==
<br>
*Published Results:
• Published Results:
 
::<math>\begin{align}
  \widehat{\beta }=\ & 0.777  \\
  \widehat{K}=\ & 6.8742\cdot {{10}^{-29}}  \\
  \widehat{n}=\ & 17.72958  \\
\end{align}\,\!</math>


::<math>\begin{matrix}
*Published 95% confidence limits on <math>\beta \,\!</math>:
  \widehat{\beta }=0.777  \\
  \widehat{K}=6.8742\cdot {{10}^{-29}}  \\
  \widehat{n}=17.72958  \\
\end{matrix}</math>


• Published 95% confidence limits on  <math>\beta </math> :
<br>
::<math>\begin{matrix}
::<math>\begin{matrix}
   \left\{ 0.653,0.923 \right\}  \\
   \left\{ 0.653,0.923 \right\}  \\
\end{matrix}</math>
\end{matrix}\,\!</math>


<br>
==Computed Results for Example 2==


==Computed Results for Example 2==
Use the inverse power law model and Weibull as the underlying life distribution. ALTA computed parameters are:  
<br>
 
Use the inverse power law model and Weibull as the underlying life distribution.  
::<math>\begin{align}
<br>
   \widehat{\beta }=\ & 0.7765,  \\
ALTA computed parameters are:  
   \widehat{K}=\ & 6.8741\cdot {{10}^{-29}}  \\
<br>
   \widehat{n}=\ & 17.7296  \\
::<math>\begin{matrix}
\end{align}\,\!</math>
   \widehat{\beta }=0.7765,  \\
   \widehat{K}=6.8741\cdot {{10}^{-29}}  \\
   \widehat{n}=17.7296  \\
\end{matrix}</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>
<br>


=Example 3=
=Example 3=
<br>
 
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.
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.
<br>
 
<br>
==Published Results for Example 3==
==Published Results for Example 3==
Nelson's [28, p. 306] IPL-Weibull parameter estimates:
Nelson's [[Appendix_E:_References|[28, p. 306]]] IPL-Weibull parameter estimates:
 
::<math>\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>\begin{matrix}
::<math>\begin{align}
  \widehat{\beta }=1.243396  \\
& \left\{ 0.9746493,1.586245 \right\}\text{ for }\widehat{\beta } \\
  \widehat{K}=0.4350735  \\
& \left\{ 0.332906,0.568596 \right\}\text{ for }\widehat{K } \\
  \widehat{n}=13.8528  \\
& \left\{ 11.43569,16.26991 \right\}\text{ for }\widehat{n }
\end{matrix}</math>
\end{align}\,\!</math>
<br>
• The 95% 2-sided confidence bounds on the parameters:


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:


{|align="center" border="1"
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
!Percentile
!Percentile
!Life Estimate
!Life Estimate
Line 107: Line 98:
Use the inverse power law model and Weibull as the underlying life distribution.
Use the inverse power law model and Weibull as the underlying life distribution.


ALTA computed parameters are:
*ALTA computed parameters are:
<br>
 
::<math>\begin{matrix}
::<math>\begin{align}
   \widehat{\beta }=1.243375  \\
   \widehat{\beta }=\ & 1.243375  \\
   \widehat{K}=0.4350548  \\
   \widehat{K}=\ & 0.4350548  \\
   \widehat{n}=13.8529  \\
   \widehat{n}=\ & 13.8529  \\
\end{matrix}</math>
\end{align}\,\!</math>
 
*The 95% 2-sided confidence bounds on the parameters:
 
::<math>\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>
   
   
• The 95% 2-sided confidence bounds on the parameters:
*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:


{|align="center" border="1"
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
!Percentile
!Percentile
!Life Estimate
!Life Estimate
Line 134: Line 131:
|}
|}


<br>
=Example 4=
 
From Meeker and Escobar [[Appendix_E:_References|[26, p. 504]]], Mylar-Polyurethane Insulating Structure data using the inverse power law lognormal model.


=Example 4=
<br>
From Meeker and Escobar [26, p. 504], Mylar-Polyurethane Insulating Structure data using the inverse power law lognormal model.
<br>
<br>
==Published Results for Example 4==
==Published Results for Example 4==
<br>
*Published Results:
• Published Results:
 
::<math>\begin{align}
  {{\widehat{\sigma }}_{{{T}'}}}=\ & 1.05,  \\
  \widehat{K}=\ & 1.14\cdot {{10}^{-12}},  \\
  \widehat{n}=\ & 4.28.  \\
\end{align}\,\!</math>


::<math>\begin{matrix}
*The 95% 2-sided confidence bounds on the parameters:
  {{\widehat{\sigma }}_{{{T}'}}}=1.05,  \\
  \widehat{K}=1.14\cdot {{10}^{-12}},  \\
  \widehat{n}=4.28.  \\
\end{matrix}</math>


• The 95% 2-sided confidence bounds on the parameters:
::<math>\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===
==Computed Results for Example 4==
Use the inverse power law lognormal.
Use the inverse power law lognormal.
• ALTA computed parameters are:


::<math>\begin{matrix}
*ALTA computed parameters are:
   {{\widehat{\sigma }}_{{{T}'}}}=1.04979  \\
 
   \widehat{K}=1.15\cdot {{10}^{-12}}  \\
::<math>\begin{align}
   \widehat{n}=4.289  \\
   {{\widehat{\sigma }}_{{{T}'}}}=\ & 1.04979  \\
\end{matrix}</math>
   \widehat{K}=\ & 1.15\cdot {{10}^{-12}}  \\
   \widehat{n}=\ & 4.289  \\
\end{align}\,\!</math>
 
*ALTA computed 95% confidence limits on the parameters:


• ALTA computed 95% confidence limits on the parameters:
::<math>\begin{align}
<br>
& \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=
=Example 5=
From Meeker and Escobar [26, p. 515], Tantalum Capacitor data using the combination (Temperature-NonThermal) Weibull model.
From Meeker and Escobar [[Appendix_E:_References|[26, p. 515]]], Tantalum capacitor data using the combination (Temperature-NonThermal) Weibull model.
<br>
 
<br>
==Published Results for Example 5==
==Published Results for Example 5==
<br>
*Published Results:
Published Results:
 
 
<math>\begin{matrix}
  \widehat{\beta }=0.4292  \\
  \widehat{B}=3829.468  \\
  \widehat{C}=4.513\cdot {{10}^{36}}  \\
  \widehat{n}=20.1  \\
\end{matrix}</math>


• The 95% 2-sided confidence bounds on the parameters:
::<math>\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>\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>


<br>
==Computed Results for Example 5==
==Computed Results for Example 5==
<br>
Use the Temperature-NonThermal model and Weibull as the underlying life distribution.
Use the Temperature-NonThermal model and Weibull as the underlying life distribution.
<br>
• ALTA computed parameters are:


*ALTA computed parameters are:
::<math>\begin{align}
  \widehat{\beta }=\ & 0.4287  \\
  \widehat{B}=\ & 3780.298  \\
  \widehat{C}=\ & 4.772\cdot {{10}^{36}}  \\
  \widehat{n}=\ & 20.09  \\
\end{align}\,\!</math>


<math>\begin{matrix}
*ALTA computed 95% confidence limits on the parameters:
  \widehat{\beta }=0.4287  \\
  \widehat{B}=3780.298  \\
  \widehat{C}=4.772\cdot {{10}^{36}}  \\
  \widehat{n}=20.09  \\
\end{matrix}</math>


• ALTA computed 95% confidence limits on the parameters:
::<math>\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>

Latest revision as of 19:18, 15 September 2023

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Chapter Appendix C: Appendix C: Benchmark Examples


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Chapter Appendix C  
Appendix C: Benchmark Examples  

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Available Software:
ALTA

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More Resources:
ALTA 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:
[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]