Grouped per Configuration Data - Logistic Model: Difference between revisions
Kate Racaza (talk | contribs) (Created page with '<noinclude>{{Banner RGA Examples}} ''This example appears in the Reliability Growth and Repairable System Analysis Reference book''. </noinclude> Some equipment und…') |
No edit summary |
||
| Line 1: | Line 1: | ||
<noinclude>{{Banner RGA Examples}} | <noinclude>{{Banner RGA Examples}} | ||
''This example appears in the [[Logistic|Reliability Growth and Repairable System Analysis Reference | ''This example appears in the [[Logistic|Reliability Growth and Repairable System Analysis Reference]]''. | ||
</noinclude> | </noinclude> | ||
| Line 30: | Line 30: | ||
|10|| 1|| 8|| 0.9000 | |10|| 1|| 8|| 0.9000 | ||
|} | |} | ||
'''Solution''' | '''Solution''' | ||
| Line 36: | Line 35: | ||
The observed reliability is <math>1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\!</math> and the last column of the table above shows the values for each group. With <math>N=9\,\!</math>, the least square estimator <math>\overline{Y}</math> becomes: | The observed reliability is <math>1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\!</math> and the last column of the table above shows the values for each group. With <math>N=9\,\!</math>, the least square estimator <math>\overline{Y}</math> becomes: | ||
:<math>\begin{align} | |||
\bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ | \bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ | ||
= & -1.4036 | = & -1.4036 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
and: | and: | ||
:<math>\begin{align} | |||
\bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\ | \bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\ | ||
= & 4 \\ | = & 4 \\ | ||
| Line 50: | Line 48: | ||
\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33 | \underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Now from the least squares estimators, <math>\hat{b_{i}}\,\!</math> and <math>\hat{b_{0}}\,\!</math>, we have: | Now from the least squares estimators, <math>\hat{b_{i}}\,\!</math> and <math>\hat{b_{0}}\,\!</math>, we have: | ||
:<math>\begin{align} | |||
{{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ | {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ | ||
= & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\ | = & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\ | ||
| Line 63: | Line 60: | ||
= & -0.2168 | = & -0.2168 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Therefore: | Therefore: | ||
:<math>\begin{align} | |||
\widehat{b}= & {{e}^{-0.2168}} \\ | \widehat{b}= & {{e}^{-0.2168}} \\ | ||
= & 0.8051 \\ | = & 0.8051 \\ | ||
| Line 73: | Line 69: | ||
= & 0.2967 | = & 0.2967 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
The Logistic reliability model that best fits the data is given by: | The Logistic reliability model that best fits the data is given by: | ||
:<math>R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\!</math> | |||
The figure below shows the Reliability vs. Time plot. | The figure below shows the Reliability vs. Time plot. | ||
[[Image:rga8.5.png|center|400px|Logistic Reliability vs. Time plot displaying the intervals.]] | [[Image:rga8.5.png|center|400px|Logistic Reliability vs. Time plot displaying the intervals.]] | ||
Revision as of 18:19, 29 May 2014
 |
New format available! This reference is now available in a new format that offers faster page load, improved display for calculations and images and more targeted search.
As of January 2024, this Reliawiki page will not continue to be updated. Please update all links and bookmarks to the latest references at RGA examples and RGA reference examples.
This example appears in the Reliability Growth and Repairable System Analysis Reference.
Some equipment underwent testing in different stages. The testing may have been performed in subsequent days, weeks or months with an unequal number of units tested every day. Each group was tested and several failures occurred. The data set is given in columns 1 and 2 of the following table. Find the Logistic model that best fits the data, and plot it along with the reliability observed from the raw data.
| Number of Units | Number of Failures | [math]\displaystyle{ T_i\,\! }[/math] | Observed Reliability |
|---|---|---|---|
| 10 | 5 | 0 | 0.5000 |
| 8 | 3 | 1 | 0.6250 |
| 9 | 3 | 2 | 0.6667 |
| 9 | 2 | 3 | 0.7778 |
| 10 | 2 | 4 | 0.8000 |
| 10 | 1 | 5 | 0.9000 |
| 10 | 1 | 6 | 0.9000 |
| 10 | 1 | 7 | 0.9000 |
| 10 | 1 | 8 | 0.9000 |
Solution
The observed reliability is [math]\displaystyle{ 1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\! }[/math] and the last column of the table above shows the values for each group. With [math]\displaystyle{ N=9\,\! }[/math], the least square estimator [math]\displaystyle{ \overline{Y} }[/math] becomes:
- [math]\displaystyle{ \begin{align} \bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ = & -1.4036 \end{align}\,\! }[/math]
and:
- [math]\displaystyle{ \begin{align} \bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\ = & 4 \\ \underset{i=0}{\overset{8}{\mathop \sum }}\,T_{i}^{2}= & 204 \\ \underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33 \end{align}\,\! }[/math]
Now from the least squares estimators, [math]\displaystyle{ \hat{b_{i}}\,\! }[/math] and [math]\displaystyle{ \hat{b_{0}}\,\! }[/math], we have:
- [math]\displaystyle{ \begin{align} {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ = & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\ = & -0.2967 \\ & \\ {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ = & \left( -1.4036 \right)-\left( -0.2967 \right)\cdot 4.0 \\ = & -0.2168 \end{align}\,\! }[/math]
Therefore:
- [math]\displaystyle{ \begin{align} \widehat{b}= & {{e}^{-0.2168}} \\ = & 0.8051 \\ \widehat{k}= & -(-0.2967) \\ = & 0.2967 \end{align}\,\! }[/math]
The Logistic reliability model that best fits the data is given by:
- [math]\displaystyle{ R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\! }[/math]
The figure below shows the Reliability vs. Time plot.
