Failure Discounting Example

This example appears in the article Failure Discounting.

Assume that during the 22 launches given in the first table below, the first failure was caused by Mode 1, the second and fourth failures were caused by Mode 2, the third and fifth failures were caused by Mode 3, the sixth failure was caused by Mode 4 and the seventh failure was caused by Mode 5.


 * 1)	Find the standard Gompertz reliability growth curve using the results of the first 15 launches.
 * 2)	Find the predicted reliability after launch 22.
 * 3)	Calculate the reliability after launch 22 based on the full data set from the second table, and compare it with the estimate obtained for question 2.

Solution

1)	The first table above is organized as follows:


 * The failures are represented by columns "Failure 1", "Failure 2", etc. The "Result/Mode" column shows whether each launch is a failure (indicated by the failure modes F1, F2, etc.) or a success (S).


 * The values of failure are based on $$CL=0.90\,\!$$ and are calculated from:


 * $$f=1-{{(1-CL)}^{\tfrac{1}}}\,\!$$


 * These values are summed and the reliability is calculated from:


 * $$R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%\,\!$$


 * where $$N\,\!$$ is the number of failures and $$n\,\!$$ is the number of events, tests, runs or launches.


 * Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process. So at launch 3, $${{S}_{n}}=1\,\!$$, and so on.


 * Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore, $${{S}_{n}}=1\,\!$$ at launch 4 and at launch 7, and so on.


 * Failure 3 is Mode 3; it occurs at launch 3 and it recurs at launch 6. Therefore, $${{S}_{n}}=1\,\!$$ at launch 5 and at launch 8, and so on.


 * Failure 6 is Mode 4; it occurs at launch 17 and it does not recur throughout the process. So at launch 19, $${{S}_{n}}=1\,\!$$, and so on.


 * Failure 7 is Mode 5; it occurs at launch 19 and it does not recur throughout the process. So at launch 21, $${{S}_{n}}=1\,\!$$, and so on.


 * For launch 3 and failure 1, $${{S}_{n}}=1\,\!$$.


 * $$\begin{align}

{{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900 \end{align}\,\!$$


 * For launch 4 and failure 1, $${{S}_{n}}=2\,\!$$.


 * $$\begin{align}

{{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684 \end{align}\,\!$$


 * And so on.


 * Calculate the initial values of the Gompertz parameters using the second table above. Based on the equations from the Gompertz Models chapter, the initial values are:


 * $$\begin{align}

c &= \left ( \frac{S_{3}-S_{2}}{S_{2}-S_{1}} \right )^\frac{1}{n\cdot I} \\ &= \left [ \frac{26.946-26.776}{26.776-22.218} \right ]^\frac{1}{6} \\ &= 0.578 \\ \end{align}\,\!$$


 * $$\begin{align}

a &= e^\left [\frac{1}{n}\left (S_{1} + \frac {S_{2}-S_{1}}{1-e^{n\cdot I}} \right )\right ] \\ &= e^\left [\frac{1}{6}\left (22.218 + \frac{26.776 - 22.218}{1-0.578^{6}}\right ) \right ] \\ &= 89.31% \\ \end{align}\,\!$$


 * $$\begin{align}

b &= e^\left [\frac{(S_{2}-S_{1})(c-1)}{(1-c^{n})^{2}} \right ] \\ &= e^\left [\frac{(26.776-22.218)(0.578-1)}{(1-0.578^{6})^{2}} \right ] \\ &= 0.127 \\ \end{align}\,\!$$


 * Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting $${{Y}_{i}}={{R}_{i}},\,\!$$ $$g_{1}^{(0)}=89.31,\,\!$$ $$g_{2}^{(0)}=0.127,\,\!$$ $$g_{3}^{(0)}=0.578\,\!$$. The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:


 * $$\begin{align}

\widehat{a}&= 0.9299 \\ \widehat{b} &= 0.0943 \\ \widehat{c} &= 0.7170 \end{align}\,\!$$


 * The next figure shows the entered data and the estimated parameters.




 * The Gompertz reliability growth curve may now be written as follows where $${{L}_{G}}\,\!$$ is the number of launches, with the first successful launch being counted as $${{L}_{G}}=1\,\!$$. Therefore, $${{L}_{G}}\,\!$$ is equal to 19, since reliability growth starts with launch 4.


 * $$R=0.9299{{(0.0943)}^}\,\!$$

2)	The predicted reliability after launch 22 is therefore:


 * $$\begin{align}

R &= 0.9299{{(0.0943)}^} \\ &= 0.9260 \end{align}\,\!$$


 * The predicted reliability after launch 22 is calculated using the Quick Calculation Pad (QCP), as shown next.



3)	In the second table, the predicted reliability values are compared with the reliabilities that are calculated from the raw data using failure discounting. It can be seen in the table, and in the following figure, that the Gompertz curve appears to provide a good fit to the actual data.