Duane Confidence Bounds Example: Difference between revisions
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<noinclude>{{Banner RGA Examples}} | <noinclude>{{Banner RGA Examples}} | ||
''This example appears in the [ | ''This example appears in the [https://help.reliasoft.com/reference/reliability_growth_and_repairable_system_analysis Reliability growth reference]''. | ||
</noinclude> | </noinclude> | ||
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:<math>\hat{\alpha}=0.6133\,\!</math> | :<math>\hat{\alpha}=0.6133\,\!</math> | ||
Calculate the 90% confidence bounds for: | Calculate the 90% confidence bounds for the following: | ||
#The parameters <math>\alpha\,\!</math> and <math>b\,\!</math>. | #The parameters <math>\alpha\,\!</math> and <math>b\,\!</math>. | ||
Line 72: | Line 72: | ||
:<math>\begin{align} | :<math>\begin{align} | ||
{{[{{\lambda }_{c}}(t)]}_{L}}= & 0. | {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00106780 \\ | ||
{{[{{\lambda }_{c}}(t)]}_{U}}= & 0. | {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00116825 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Line 79: | Line 79: | ||
:<math>\begin{align} | :<math>\begin{align} | ||
{{[{{\lambda }_{i}}(t)]}_{L}}= & 0. | {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00041299 \\ | ||
{{[{{\lambda }_{c}}(t)]}_{U}}= & 0. | {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00045184 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Line 106: | Line 106: | ||
:<math>\begin{align} | :<math>\begin{align} | ||
{{m}_{c}}{{(t)}_{l}}= & | {{m}_{c}}{{(t)}_{l}}= & 855.9815 \\ | ||
{{m}_{c}}{{(t)}_{u}}= & | {{m}_{c}}{{(t)}_{u}}= & 936.5071 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Line 113: | Line 113: | ||
:<math>\begin{align} | :<math>\begin{align} | ||
{{m}_{i}}{{(t)}_{l}}= & | {{m}_{i}}{{(t)}_{l}}= & 2213.1753 \\ | ||
{{m}_{i}}{{(t)}_{u}}= & | {{m}_{i}}{{(t)}_{u}}= & 2421.3776 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Latest revision as of 21:23, 18 September 2023
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This example appears in the Reliability growth reference.
Using the values of [math]\displaystyle{ \hat{b}\,\! }[/math] and [math]\displaystyle{ \hat{\alpha }\,\! }[/math] estimated from the least squares analysis in Least Squares Example 2:
- [math]\displaystyle{ \hat{b}=1.9453\,\! }[/math]
- [math]\displaystyle{ \hat{\alpha}=0.6133\,\! }[/math]
Calculate the 90% confidence bounds for the following:
- The parameters [math]\displaystyle{ \alpha\,\! }[/math] and [math]\displaystyle{ b\,\! }[/math].
- The cumulative and instantaneous failure intensity.
- The cumulative and instantaneous MTBF.
Solution
- Use the values of [math]\displaystyle{ \hat{b}\,\! }[/math] and [math]\displaystyle{ \hat{\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]
- [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]
- [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]
- The cumulative failure intensity is:
- [math]\displaystyle{ \begin{align} {{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ = & 0.00111689 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ = & 0.00043198 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00106780 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00116825 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00041299 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00045184 \end{align}\,\! }[/math]
- The cumulative MTBF is:
- [math]\displaystyle{ \begin{align} {{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ = & 895.3395 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ = & 2314.9369 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{m}_{c}}{{(t)}_{l}}= & 855.9815 \\ {{m}_{c}}{{(t)}_{u}}= & 936.5071 \end{align}\,\! }[/math]
- [math]\displaystyle{ \begin{align} {{m}_{i}}{{(t)}_{l}}= & 2213.1753 \\ {{m}_{i}}{{(t)}_{u}}= & 2421.3776 \end{align}\,\! }[/math]
The next figure displays the instantaneous MTBF. Both are plotted with confidence bounds.