Duane Confidence Bounds Example

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

Using the values of $$\hat{b}\,\!$$ and $$\hat{\alpha }\,\!$$ estimated from the least squares analysis in Least Squares Example 2  Least Squares Example 2 :
 * $$\hat{b}=1.9453\,\!$$
 * $$\hat{\alpha}=0.6133\,\!$$

Calculate the 90% confidence bounds for the following:


 * 1) The parameters $$\alpha\,\!$$ and $$b\,\!$$.
 * 2) The cumulative and instantaneous failure intensity.
 * 3) The cumulative and instantaneous MTBF.

Solution

 Use the values of $$\hat{b}\,\!$$ and $$\hat{\alpha }\,\!$$ estimated from the least squares analysis. Then:


 * $$\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}\,\!$$


 * $$\begin{align}

SE(\hat{\alpha })= & \frac{\sigma }{\sqrt} \\ = & \frac{0.08428}{9.9727} \\ = & 0.008452 \end{align}\,\!$$


 * $$\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}\,\!$$

Thus, the 90% confidence bounds on parameter $$\alpha \,\!$$ are:


 * $$C{{B}_{\alpha }}=\hat{\alpha }\pm {{t}_{n-2,\alpha /2}}SE(\hat{\alpha })\,\!$$


 * $$\begin{align}

{{\alpha }_{L}}= & 0.602050 \\ {{\alpha }_{U}}= & 0.624417 \end{align}\,\!$$

And 90% confidence bounds on parameter $$b\,\!$$ are:


 * $$C{{B}_{b}}=\hat{b}{{e}^{\pm {{t}_{n-2,\alpha /2}}SE\left[ \ln (\hat{b}) \right]}}\,\!$$


 * $$\begin{align}

{{b}_{L}}= & 1.7831 \\ {{b}_{U}}= & 2.1231 \end{align}\,\!$$  The cumulative failure intensity is:


 * $$\begin{align}

{{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\ = & 0.00111689 \end{align}\,\!$$

And the instantaneous failure intensity is equal to:


 * $$\begin{align}

{{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\ = & 0.00043198 \end{align}\,\!$$

So, at the 90% confidence level and for $$T=22,000\,\!$$ hours, the confidence bounds on cumulative failure intensity are:


 * $$\begin{align}

{{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00106780 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00116825 \end{align}\,\!$$

For the instantaneous failure intensity:


 * $$\begin{align}

{{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00041299 \\ {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00045184 \end{align}\,\!$$

The following figures show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds.



 The cumulative MTBF is:


 * $$\begin{align}

{{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\ = & 895.3395 \end{align}\,\!$$

And the instantaneous MTBF is:


 * $$\begin{align}

{{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\ = & 2314.9369 \end{align}\,\!$$

So, at 90% confidence level and for $$T=22,000\,\!$$ hours, the confidence bounds on the cumulative MTBF are:


 * $$\begin{align}

{{m}_{c}}{{(t)}_{l}}= & 855.9815 \\ {{m}_{c}}{{(t)}_{u}}= & 936.5071 \end{align}\,\!$$

The confidence bounds for the instantaneous MTBF are:


 * $$\begin{align}

{{m}_{i}}{{(t)}_{l}}= & 2213.1753 \\ {{m}_{i}}{{(t)}_{u}}= & 2421.3776 \end{align}\,\!$$

The figure below displays the cumulative MTBF.



The next figure displays the instantaneous MTBF. Both are plotted with confidence bounds.

 