Duane Confidence Bounds Example

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This example appears in the Reliability Growth and Repairable System Analysis Reference book.

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:

1. The parameters [math]\displaystyle{ \alpha\,\! }[/math] and [math]\displaystyle{ b\,\! }[/math].
2. The cumulative and instantaneous failure intensity.
3. The cumulative and instantaneous MTBF.

Solution

1. 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]

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.