Duane Confidence Bounds Example: Difference between revisions
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Using the values of <math>\widehat{b}\,\!</math> and <math>\widehat{\alpha }\,\!</math> estimated from the least squares analysis in <noinclude>[[Duane Linear Regression Examples| | Using the values of <math>\widehat{b}\,\!</math> and <math>\widehat{\alpha }\,\!</math> estimated from the least squares analysis in <noinclude>[[Duane Linear Regression Examples| | ||
Least Squares Example 2]]</noinclude><includeonly>Least Squares Example 2</includeonly>: | Least Squares Example 2]]</noinclude><includeonly>Least Squares Example 2</includeonly>: | ||
:<math>\widehat{b}=1.9453\,\!</math> | |||
:<math>\widehat{\alpha}=0.6133\,\!</math> | |||
calculate the 90% confidence bounds for: | calculate the 90% confidence bounds for: | ||
| Line 15: | Line 14: | ||
#The cumulative and instantaneous failure intensity. | #The cumulative and instantaneous failure intensity. | ||
#The cumulative and instantaneous MTBF. | #The cumulative and instantaneous MTBF. | ||
'''Solution''' | '''Solution''' | ||
1. | 1. Use the values of <math>\widehat{b}\,\!</math> and <math>\widehat{\alpha }\,\!</math> estimated from the least squares analysis. Then: | ||
:<math>\begin{align} | |||
{{S}_{xx}}&=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{(\ln {{t}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{t}_{i}}) \right)}^{2}} \\ | {{S}_{xx}}&=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{(\ln {{t}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{t}_{i}}) \right)}^{2}} \\ | ||
& = 1400.9084-1301.4545 \\ | & = 1400.9084-1301.4545 \\ | ||
| Line 27: | Line 25: | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
SE(\hat{\alpha })= & \frac{\sigma }{\sqrt{{{S}_{xx}}}} \\ | SE(\hat{\alpha })= & \frac{\sigma }{\sqrt{{{S}_{xx}}}} \\ | ||
= & \frac{0.08428}{9.9727} \\ | = & \frac{0.08428}{9.9727} \\ | ||
| Line 34: | Line 31: | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
SE(\ln \hat{b})= & \sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{(\ln {{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}} \\ | SE(\ln \hat{b})= & \sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{(\ln {{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}} \\ | ||
= & 0.065960 | = & 0.065960 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Thus, the 90% confidence bounds on parameter <math>\alpha \,\!</math> are: | Thus, the 90% confidence bounds on parameter <math>\alpha \,\!</math> are: | ||
:<math>C{{B}_{\alpha }}=\hat{\alpha }\pm {{t}_{n-2,\alpha /2}}SE(\hat{\alpha })\,\!</math> | |||
:<math>\begin{align} | |||
{{\alpha }_{L}}= & 0.602050 \\ | {{\alpha }_{L}}= & 0.602050 \\ | ||
{{\alpha }_{U}}= & 0.624417 | {{\alpha }_{U}}= & 0.624417 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
And 90% confidence bounds on parameter <math>b\,\!</math> are: | And 90% confidence bounds on parameter <math>b\,\!</math> are: | ||
:<math>C{{B}_{b}}=\hat{b}{{e}^{\pm {{t}_{n-2,\alpha /2}}SE\left[ \ln (\hat{b}) \right]}}\,\!</math> | |||
:<math>\begin{align} | |||
{{b}_{L}}= & 1.7831 \\ | {{b}_{L}}= & 1.7831 \\ | ||
{{b}_{U}}= & 2.1231 | {{b}_{U}}= & 2.1231 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
2. The cumulative failure intensity is: | |||
:<math>\begin{align} | |||
{{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ | {{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ | ||
= & 0.00111689 | = & 0.00111689 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
And the instantaneous failure intensity is equal to: | And the instantaneous failure intensity is equal to: | ||
:<math>\begin{align} | |||
{{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ | {{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ | ||
= & 0.00043198 | = & 0.00043198 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
So, at the 90% confidence level and for <math>T=22,000\,\!</math> hours, the confidence bounds on cumulative failure intensity are: | So, at the 90% confidence level and for <math>T=22,000\,\!</math> hours, the confidence bounds on cumulative failure intensity are: | ||
:<math>\begin{align} | |||
{{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00100254 \\ | {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00100254 \\ | ||
{{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00124429 | {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00124429 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
For the instantaneous failure intensity: | For the instantaneous failure intensity: | ||
:<math>\begin{align} | |||
{{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00038775 \\ | {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00038775 \\ | ||
{{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00048125 | {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00048125 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
The following figures show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds. | The following figures show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds. | ||
[[Image:rga4.7.png|center|400px|Cumulative Failure Intensity plot with 2-sided 90% confidence bounds.]] | [[Image:rga4.7.png|center|400px|Cumulative Failure Intensity plot with 2-sided 90% confidence bounds.]] | ||
[[Image:rga4.8.png|center|400px|Instantaneous Failure Intensity plot with 2-sided 90% confidence bounds.]] | [[Image:rga4.8.png|center|400px|Instantaneous Failure Intensity plot with 2-sided 90% confidence bounds.]] | ||
3. | 3. The cumulative MTBF is: | ||
:<math>\begin{align} | |||
{{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ | {{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ | ||
= & 895.3395 | = & 895.3395 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
And the instantaneous MTBF is: | And the instantaneous MTBF is: | ||
:<math>\begin{align} | |||
{{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ | {{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ | ||
= & 2314.9369 | = & 2314.9369 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
So, at 90% confidence level and for <math>T=22,000\,\!</math> hours, the confidence bounds on the cumulative MTBF are: | So, at 90% confidence level and for <math>T=22,000\,\!</math> hours, the confidence bounds on the cumulative MTBF are: | ||
:<math>\begin{align} | |||
{{m}_{c}}{{(t)}_{l}}= & 803.6695 \\ | {{m}_{c}}{{(t)}_{l}}= & 803.6695 \\ | ||
{{m}_{c}}{{(t)}_{u}}= & 997.4658 | {{m}_{c}}{{(t)}_{u}}= & 997.4658 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
The confidence bounds for the instantaneous MTBF are: | The confidence bounds for the instantaneous MTBF are: | ||
:<math>\begin{align} | |||
{{m}_{i}}{{(t)}_{l}}= & 2077.9204 \\ | {{m}_{i}}{{(t)}_{l}}= & 2077.9204 \\ | ||
{{m}_{i}}{{(t)}_{u}}= & 2578.9886 | {{m}_{i}}{{(t)}_{u}}= & 2578.9886 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
The figure below displays the cumulative MTBF. | The figure below displays the cumulative MTBF. | ||
Revision as of 22:50, 30 January 2014
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This example appears in the Reliability Growth and Repairable System Analysis Reference book.
Using the values of [math]\displaystyle{ \widehat{b}\,\! }[/math] and [math]\displaystyle{ \widehat{\alpha }\,\! }[/math] estimated from the least squares analysis in Least Squares Example 2:
- [math]\displaystyle{ \widehat{b}=1.9453\,\! }[/math]
- [math]\displaystyle{ \widehat{\alpha}=0.6133\,\! }[/math]
calculate the 90% confidence bounds for:
- The parameters [math]\displaystyle{ \alpha\,\! }[/math] and [math]\displaystyle{ b\,\! }[/math].
- The cumulative and instantaneous failure intensity.
- The cumulative and instantaneous MTBF.
Solution
1. Use the values of [math]\displaystyle{ \widehat{b}\,\! }[/math] and [math]\displaystyle{ \widehat{\alpha }\,\! }[/math] estimated from the least squares analysis. Then:
- [math]\displaystyle{ \begin{align} {{S}_{xx}}&=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{(\ln {{t}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{t}_{i}}) \right)}^{2}} \\ & = 1400.9084-1301.4545 \\ & = 99.4539 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} SE(\hat{\alpha })= & \frac{\sigma }{\sqrt{{{S}_{xx}}}} \\ = & \frac{0.08428}{9.9727} \\ = & 0.008452 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} SE(\ln \hat{b})= & \sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{(\ln {{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}} \\ = & 0.065960 \end{align}\,\! }[/math]
Thus, the 90% confidence bounds on parameter [math]\displaystyle{ \alpha \,\! }[/math] are:
- [math]\displaystyle{ C{{B}_{\alpha }}=\hat{\alpha }\pm {{t}_{n-2,\alpha /2}}SE(\hat{\alpha })\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{\alpha }_{L}}= & 0.602050 \\ {{\alpha }_{U}}= & 0.624417 \end{align}\,\! }[/math]
And 90% confidence bounds on parameter [math]\displaystyle{ b\,\! }[/math] are:
- [math]\displaystyle{ C{{B}_{b}}=\hat{b}{{e}^{\pm {{t}_{n-2,\alpha /2}}SE\left[ \ln (\hat{b}) \right]}}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{b}_{L}}= & 1.7831 \\ {{b}_{U}}= & 2.1231 \end{align}\,\! }[/math]
2. The cumulative failure intensity is:
- [math]\displaystyle{ \begin{align} {{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ = & 0.00111689 \end{align}\,\! }[/math]
And the instantaneous failure intensity is equal to:
- [math]\displaystyle{ \begin{align} {{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ = & 0.00043198 \end{align}\,\! }[/math]
So, at the 90% confidence level and for [math]\displaystyle{ T=22,000\,\! }[/math] hours, the confidence bounds on cumulative failure intensity are:
- [math]\displaystyle{ \begin{align} {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00100254 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00124429 \end{align}\,\! }[/math]
For the instantaneous failure intensity:
- [math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00038775 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00048125 \end{align}\,\! }[/math]
The following figures show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds.
3. The cumulative MTBF is:
- [math]\displaystyle{ \begin{align} {{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ = & 895.3395 \end{align}\,\! }[/math]
And the instantaneous MTBF is:
- [math]\displaystyle{ \begin{align} {{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ = & 2314.9369 \end{align}\,\! }[/math]
So, at 90% confidence level and for [math]\displaystyle{ T=22,000\,\! }[/math] hours, the confidence bounds on the cumulative MTBF are:
- [math]\displaystyle{ \begin{align} {{m}_{c}}{{(t)}_{l}}= & 803.6695 \\ {{m}_{c}}{{(t)}_{u}}= & 997.4658 \end{align}\,\! }[/math]
The confidence bounds for the instantaneous MTBF are:
- [math]\displaystyle{ \begin{align} {{m}_{i}}{{(t)}_{l}}= & 2077.9204 \\ {{m}_{i}}{{(t)}_{u}}= & 2578.9886 \end{align}\,\! }[/math]
The figure below displays the cumulative MTBF.
The next figure displays the instantaneous MTBF. Both are plotted with confidence bounds.
