2P Exponential Example: Difference between revisions

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Construct the following table, as shown next.
Construct the following table, as shown next.


<center><math>\overset{{}}{\mathop{\text{Table}\text{- Least Squares Analysis}}}\,</math></center>
<center><math>\overset{{}}{\mathop{\text{Table}\text{- Least Squares Analysis}}}\,\,\!</math></center>


<center><math>\begin{matrix}
<center><math>\begin{matrix}
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   \text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324}  \\
   \text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324}  \\
   \sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899}  \\
   \sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899}  \\
\end{matrix}</math></center>
\end{matrix}\,\!</math></center>




The median rank values ( <math>F({{t}_{i}})</math> ) can be found in rank tables or they can be estimated using the '''Quick Statistical Reference''' in Weibull++.
The median rank values ( <math>F({{t}_{i}})\,\!</math> ) can be found in rank tables or they can be estimated using the '''Quick Statistical Reference''' in Weibull++.
Given the values in the table above, calculate <math>\hat{a}</math> and <math>\hat{b}</math>:
Given the values in the table above, calculate <math>\hat{a}\,\!</math> and <math>\hat{b}\,\!</math>:




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   \\  
   \\  
   \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14}   
   \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14}   
\end{align}</math>
\end{align}\,\!</math>


or:
or:


::<math>\hat{b}=-0.02711</math>
::<math>\hat{b}=-0.02711\,\!</math>


and:
and:


::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}</math>
::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\!</math>


or:
or:


::<math>\hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748</math>
::<math>\hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748\,\!</math>




Therefore:
Therefore:


::<math>\hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour}</math>
::<math>\hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour}\,\!</math>




and:
and:


::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711}</math>
::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711}\,\!</math>


or:
or:


::<math>\hat{\gamma }=10.1365\text{ hours}</math>
::<math>\hat{\gamma }=10.1365\text{ hours}\,\!</math>


Then:
Then:


::<math>f(t)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}}</math>
::<math>f(t)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}}\,\!</math>




The correlation coefficient can be estimated using equation for calculating the correlation coefficient:
The correlation coefficient can be estimated using equation for calculating the correlation coefficient:


::<math>\hat{\rho }=-0.9679</math>
::<math>\hat{\rho }=-0.9679\,\!</math>




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The estimated parameters and the correlation coefficient using Weibull++ were found to be:  
The estimated parameters and the correlation coefficient using Weibull++ were found to be:  


::<math>\hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679</math>
::<math>\hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679\,\!</math>


Please note that the user must deselect the '''Reset if location parameter > T1 on Exponential''' option on the Calculations page of the Application Setup window.  
Please note that the user must deselect the '''Reset if location parameter > T1 on Exponential''' option on the Calculations page of the Application Setup window.  

Revision as of 18:10, 25 September 2012

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This example also appears in the Life Data Analysis Reference book.

2-Parameter Exponential RRY Example
14 units were being reliability tested and the following life test data were obtained. Assuming that the data follow a 2-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, [math]\displaystyle{ \rho \,\! }[/math], using rank regression on Y (RRY).

Life Test Data
Data point index Time-to-failure
1 5
2 10
3 15
4 20
5 25
6 30
7 35
8 40
9 50
10 60
11 70
12 80
13 90
14 100

Solution

Construct the following table, as shown next.

[math]\displaystyle{ \overset{{}}{\mathop{\text{Table}\text{- Least Squares Analysis}}}\,\,\! }[/math]
[math]\displaystyle{ \begin{matrix} N & t_{i} & F(t_{i}) & y_{i} & t_{i}^{2} & y_{i}^{2} & t_{i} y_{i} \\ \text{1} & \text{5} & \text{0}\text{.0483} & \text{-0}\text{.0495} & \text{25} & \text{0}\text{.0025} & \text{-0}\text{.2475} \\ \text{2} & \text{10} & \text{0}\text{.1170} & \text{-0}\text{.1244} & \text{100} & \text{0}\text{.0155} & \text{-1}\text{.2443} \\ \text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.2064} & \text{225} & \text{0}\text{.0426} & \text{-3}\text{.0961} \\ \text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.2958} & \text{400} & \text{0}\text{.0875} & \text{-5}\text{.9170} \\ \text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.3942} & \text{625} & \text{0}\text{.1554} & \text{-9}\text{.8557} \\ \text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.5032} & \text{900} & \text{0}\text{.2532} & \text{-15}\text{.0956} \\ \text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.6257} & \text{1225} & \text{0}\text{.3915} & \text{-21}\text{.8986} \\ \text{8} & \text{40} & \text{0}\text{.5349} & \text{-0}\text{.7655} & \text{1600} & \text{0}\text{.5860} & \text{-30}\text{.6201} \\ \text{9} & \text{50} & \text{0}\text{.6046} & \text{-0}\text{.9279} & \text{2500} & \text{0}\text{.8609} & \text{-46}\text{.3929} \\ \text{10} & \text{60} & \text{0}\text{.6742} & \text{-1}\text{.1215} & \text{3600} & \text{1}\text{.2577} & \text{-67}\text{.2883} \\ \text{11} & \text{70} & \text{0}\text{.7439} & \text{-1}\text{.3622} & \text{4900} & \text{1}\text{.8456} & \text{-95}\text{.3531} \\ \text{12} & \text{80} & \text{0}\text{.8135} & \text{-1}\text{.6793} & \text{6400} & \text{2}\text{.8201} & \text{-134}\text{.3459} \\ \text{13} & \text{90} & \text{0}\text{.8830} & \text{-2}\text{.1456} & \text{8100} & \text{4}\text{.6035} & \text{-193}\text{.1023} \\ \text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324} \\ \sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899} \\ \end{matrix}\,\! }[/math]


The median rank values ( [math]\displaystyle{ F({{t}_{i}})\,\! }[/math] ) can be found in rank tables or they can be estimated using the Quick Statistical Reference in Weibull++. Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:


[math]\displaystyle{ \begin{align} \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/14} \\ \\ \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.02711\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748\,\! }[/math]


Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour}\,\! }[/math]


and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }=10.1365\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(t)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}}\,\! }[/math]


The correlation coefficient can be estimated using equation for calculating the correlation coefficient:

[math]\displaystyle{ \hat{\rho }=-0.9679\,\! }[/math]


This example can be repeated using Weibull++, choosing 2-parameter exponential and rank regression on Y (RRY), as shown next.

Exponential Example 2 Data Folio.png

The estimated parameters and the correlation coefficient using Weibull++ were found to be:

[math]\displaystyle{ \hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679\,\! }[/math]

Please note that the user must deselect the Reset if location parameter > T1 on Exponential option on the Calculations page of the Application Setup window.

The probability plot can be obtained simply by clicking the Plot icon.

Exponential Example 2 Plot.png