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{{template:RGA BOOK|AA|Failure Discounting}}
{{Template:RGA_BOOK|Appendix A|Failure Discounting}}
During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [[RGA_References|[4]]]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. However, certain questions arise, such as to what extent should the failure value be diminished or discounted, and how should the failure value be defined? One answer would be to use engineering judgment (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore, that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis needs to be selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, logistic, and the standard and modified Gompertz models.


=Failure Discounting=
The value of the failure, <math>f\,\!</math>, is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:  
<br>
During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished [4]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. But certain questions arise, such as, To what extent should the failure value be diminished or discounted? and How should the failure value be defined? One answer would be to use engineering judgment, e.g. a panel of specialists would agree that the probability of failure has been reduced by  <math>50%</math>  or  <math>90%</math>  and therefore that failure should be given a value of 0.5 or 0.9. The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis is selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, Logistic and the Standard and Modified Gompertz models.
The value of the failure, <math>f</math> , is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:  


::<math>f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}</math>
:<math>f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\!</math>


where:  
where:  


::<math>\begin{align}
*<math>CL\,\!</math> is the confidence level.
  & CL= & \text{the confidence level}\text{.} \\
*<math>{{S}_{n}}\,\!</math> is the number of successful tests after the first success following the corrective action.  
& {{S}_{n}}= & \text{the number of successful tests after the first success } \\
&  & \text{following the corrective action}\text{.
\end{align}</math>


For example, after one successful test following a corrective action, <math>{{S}_{n}}=1\,\!</math>, the failure is given a value of 0.9 based on a 90% confidence level. After two successful tests, <math>{{S}_{n}}=2\,\!</math>, the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.


For example, after one successful test following a corrective action,  <math>{{S}_{n}}=1</math> , the failure is given a value of 0.9 based on a  <math>90%</math>  confidence level. After two successful tests,  <math>{{S}_{n}}=2</math> , the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.
<br>
<br>
'''Example'''
'''Example'''
<br>
{{:Failure_Discounting_Example}}
Use failure discounting to answer the questions below. Assume that during the 22 launches given in Table A.1, 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.
<br>
<br>
1) Find the Standard Gompertz reliability growth curve using the results of the first 15 launches.
<br>
2) Find the predicted reliability after launch 22.
<br>
3) Calculate the reliability after launch 22 based on the full data set from Table A.2 and compare with the estimate obtained for question 2.
 
<br>
<br>
Table A.1 - Launch sequence with failure modes and failure values
 
Launch Result/ Sum of
Number Mode Failure 1 Failure 2 Failure 3 Failure 4 Failure 5 Failure 6 Failure 7 Failures
1 F1 1.000 1.000
2 F2 1.000 1.000 2.000
3 F3 0.900 1.000 1.000 2.900
4 S 0.684 0.900 1.000 2.584
5 F2 0.536 1.000 0.900 1.000 3.436
6 F3 0.438 1.000 1.000 1.000 1.000 4.438
7 S 0.369 0.900 1.000 0.900 1.000 4.169
8 S 0.319 0.684 0.900 0.684 0.900 3.486
9 S 0.280 0.536 0.684 0.536 0.684 2.720
10 S 0.250 0.438 0.536 0.438 0.536 2.197
11 S 0.226 0.369 0.438 0.369 0.438 1.839
12 S 0.206 0.319 0.369 0.319 0.369 1.581
13 S 0.189 0.280 0.319 0.280 0.319 1.387
14 S 0.175 0.250 0.280 0.250 0.280 1.235
15 S 0.162 0.226 0.250 0.226 0.250 1.114
16 S 0.152 0.206 0.226 0.206 0.226 1.014
17 F4 0.142 0.189 0.206 0.189 0.206 1.000 1.931
18 S 0.134 0.175 0.189 0.175 0.189 1.000 1.861
19 F5 0.127 0.162 0.175 0.162 0.175 0.900 1.000 2.701
20 S 0.120 0.152 0.162 0.152 0.162 0.684 1.000 2.432
21 S 0.114 0.142 0.152 0.142 0.152 0.536 0.900 2.138
22 S 0.109 0.134 0.142 0.134 0.142 0.438 0.684 1.783
 
 
 
<br>
Table A.2 - Comparison of the predicted reliability with the actual data
 
Launch Calculated Gompertz
Number Reliability (%) ln(R) Reliability (%)
1 0.000
2 0.000
3 3.333 1.204
4 35.406 3.567 16.426
5 31.283 3.443 26.691
6 26.039 3.260 37.858
7 40.442 3.670 48.691
8 56.422 4.033 58.363
9 69.783 4.245 66.496
<math>{{S}_{1}}</math>  = 22.218
10 78.029 4.357 73.044
11 83.281 4.422 78.155
12 86.824 4.464 82.055
13 89.331 4.492 84.983
14 91.175 4.513 87.155
15 92.573 4.528 88.754
<math>{{S}_{2}}</math>  = 26.776
16 93.660 4.540 89.923
17 88.639 4.484 90.774
18 89.661 4.496 91.392
19 85.787 4.452 91.839
20 87.841 4.476 92.163
21 89.820 4.498 92.396
<math>{{S}_{3}}</math> = 26.946
22 91.896 4.521 92.565
 
 
<br>
'''Solution'''
<br>
1) Table A.1 is organized as follows:
<br>
• Each failure is represented by a single column and named Failure 1, Failure 2 etc. The failure mode for each failure is shown in the Result/Mode column where is also incidated in the launch is a success.
<br>
• The values of failure are calculated from Eqn. (att4) and are based on  <math>CL=0.90</math> .
<br>
• These values are summed and the reliability is calculated from:
::<math>R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%</math>
where  <math>N</math>  is the number of failures and  <math>n</math>  is the number of events, tests, runs or launches.
<br>
• Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process. <br>
So at launch 3,  <math>{{S}_{n}}=1</math> , and so on.
<br>
• Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore,  <math>{{S}_{n}}=1</math>  at launch 4 and at launch 7, and so on.
<br>
• Failure 3 is Mode 3; it occurs at launch 3 and it recurs at launch 6. Therefore,  <math>{{S}_{n}}=1</math>  at launch 5 and at launch 8, and so on.
<br>
• Failure 6 is Mode 4; it occurs at launch 17 and it does not recur throughout the process. So at launch 19,  <math>{{S}_{n}}=1</math> , and so on.
<br>
• Failure 7 is Mode 5; it occurs at launch 19 and it does not recur throughout the process. So at launch 21,  <math>{{S}_{n}}=1</math> , and so on.
<br>
For launch 3 and failure 1,  <math>{{S}_{n}}=1</math>  in Eqn. (att4).
 
::<math>{{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900</math>
 
For launch 4 and failure 1,  <math>{{S}_{n}}=2</math>  in Eqn. (att4).
 
::<math>{{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684</math>
 
And so on.
 
Calculate the initial values of the Gompertz parameters using Table A.2. Based on the equations from Chapter 7, the initial values are:
 
Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting  <math>{{Y}_{i}}={{R}_{i}},</math>  <math>g_{1}^{(0)}=89.31,</math>  <math>g_{2}^{(0)}=0.127,</math>  <math>g_{3}^{(0)}=0.578</math> . The iterations are continued to solve for the parameters. Using RGA, the estimators of the parameters for the given example are:
 
::<math>\begin{align}
  & \widehat{a}= & 0.9299 \\
& \widehat{b}= & 0.0943 \\
& \widehat{c}= & 0.7170 
\end{align}</math>
 
 
Figure entered data Standard GMP shows the entered data and the estimated parameters.
The Gompertz reliability growth curve may now be written as follows where  <math>{{L}_{G}}</math>  is the number of launches with the first successful launch being counted as  <math>{{L}_{G}}=1</math> . Therefore,  <math>{{L}_{G}}</math>  is equal to 19, since reliability growth starts with launch 4.
 
 
::<math>R=0.9299{{(0.0943)}^{{{0.7170}^{{{L}_{G}}}}}}</math>
<br>
<br>
<br>
[[File:rgaA.1.png|center]]
<br>
::Figure A.1: Entered data and the estimated Standard Gompertz parameters.
<br>
 
 
2) Based on Eqn. (predictR), the predicted reliability after launch 22 is:
 
 
::<math>\begin{align}
  & R= & 0.9299{{(0.0943)}^{{{0.7170}^{19}}}} \\
& = & 0.9260 
\end{align}</math>
 
The predicted reliability after launch 22 is calculated using the Quick Calculation Pad and is shown in Figure A1.
<br>
<br>
3) In Table A.2, the predicted reliability values, as calculated from Eqn. (predictR), are compared with the reliabilities that are calculated from the raw data using failure discounting. It can be seen in Table A.2 and in Figure oldfig34 that the Gompertz curve appears to provide a good fit to the actual data.

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Chapter Appendix A: Failure Discounting


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Chapter Appendix A  
Failure Discounting  

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Available Software:
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More Resources:
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During a reliability growth test, once a failure has been analyzed and corrective actions for that specific failure mode have been implemented, the probability of its recurrence is diminished, as given in Lloyd [4]. Then for the success/failure data that follow, the value of the failure for which corrective actions have already been implemented should be subtracted from the total number of failures. However, certain questions arise, such as to what extent should the failure value be diminished or discounted, and how should the failure value be defined? One answer would be to use engineering judgment (e.g., a panel of specialists would agree that the probability of failure has been reduced by 50% or 90% and therefore, that failure should be given a value of 0.5 or 0.9). The obvious disadvantage of this approach is its arbitrariness and the difficulty of reaching an agreement. Therefore, a statistical basis needs to be selected, one that is repeatable and less arbitrary. Failure discounting is applied when using the Lloyd-Lipow, logistic, and the standard and modified Gompertz models.

The value of the failure, [math]\displaystyle{ f\,\! }[/math], is chosen to be the upper confidence limit on the probability of failure based on the number of successful tests following implementation of the corrective action. The failure value is given by the following equation:

[math]\displaystyle{ f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\! }[/math]

where:

  • [math]\displaystyle{ CL\,\! }[/math] is the confidence level.
  • [math]\displaystyle{ {{S}_{n}}\,\! }[/math] is the number of successful tests after the first success following the corrective action.

For example, after one successful test following a corrective action, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math], the failure is given a value of 0.9 based on a 90% confidence level. After two successful tests, [math]\displaystyle{ {{S}_{n}}=2\,\! }[/math], the failure is given a value of 0.684, and so on. The procedure for applying this method is illustrated in the next example.

Example


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.
Launch Sequence with Failure Modes and Failure Values
Launch Number Result/Mode Failure 1 Failure 2 Failure 3 Failure 4 Failure 5 Sum of Failures
1 F1 1.000 1.000
2 F2 1.000 1.000 2.000
3 F3 0.900 1.000 1.000 2.900
4 S 0.684 0.900 1.000 2.584
5 F2 0.536 1.000 0.900 2.436
6 F3 0.438 1.000 1.000 2.438
7 S 0.369 0.900 1.000 2.269
8 S 0.319 0.684 0.900 1.902
9 S 0.280 0.536 0.684 1.500
10 S 0.250 0.438 0.536 1.224
11 S 0.226 0.369 0.438 1.032
12 S 0.206 0.319 0.369 0.894
13 S 0.189 0.280 0.319 0.788
14 S 0.175 0.250 0.280 0.705
15 S 0.162 0.226 0.250 0.638
16 S 0.152 0.206 0.226 0.584
17 F4 0.142 0.189 0.206 1.000 1.537
18 S 0.134 0.175 0.189 1.000 1.498
19 F5 0.127 0.162 0.175 0.900 1.000 2.364
20 S 0.120 0.152 0.162 0.684 1.000 2.118
21 S 0.114 0.142 0.152 0.536 0.900 1.844
22 S 0.109 0.134 0.142 0.438 0.684 1.507


Comparison of the Predicted Reliability with the Actual Data
Launch Number Calculated Reliability (%) ln(R) Gompertz Reliability (%)
1 0.000
2 0.000
3 3.333 1.204
4 35.406 3.567 28.617
5 51.283 3.937 45.883
6 59.372 4.084 60.841
7 67.585 4.213 72.017
8 76.219 4.334 79.654
9 83.334 4.423 84.600
[math]\displaystyle{ {{S}_{1}}\,\! }[/math] = 24.558
10 87.764 4.475 87.701
11 90.614 4.507 89.609
12 92.555 4.528 90.769
13 93.939 4.543 91.469
14 94.964 4.553 91.891
15 95.746 4.562 92.143
[math]\displaystyle{ {{S}_{2}}\,\! }[/math] = 27.167
16 96.356 4.568 92.295
17 90.960 4.510 92.385
18 91.681 4.518 92.439
19 87.560 4.472 92.472
20 89.411 4.493 92.491
21 91.219 4.513 92.503
[math]\displaystyle{ {{S}_{3}}\,\! }[/math] = 27.076
22 93.152 4.534 92.510

Solution

  1. In the table above, 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 [math]\displaystyle{ CL=0.90\,\! }[/math] and are calculated from:
    [math]\displaystyle{ f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\! }[/math]
    These values are summed and the reliability is calculated from:
    [math]\displaystyle{ R=\left[ 1-\left( \frac{\mathop{}_{i=1}^{N}{{f}_{i}}}{n} \right) \right]\cdot 100\text{ }%\,\! }[/math]
    where [math]\displaystyle{ N\,\! }[/math] is the number of failures and [math]\displaystyle{ n\,\! }[/math] 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, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math], and so on.
    • Failure 2 is Mode 2; it occurs at launch 2 and it recurs at launch 5. Therefore, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math] 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, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math] 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, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math], 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, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math], and so on.
    For launch 3 and failure 1, [math]\displaystyle{ {{S}_{n}}=1\,\! }[/math].
    [math]\displaystyle{ \begin{align} {{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900 \end{align}\,\! }[/math]
    For launch 4 and failure 1, [math]\displaystyle{ {{S}_{n}}=2\,\! }[/math].
    [math]\displaystyle{ \begin{align} {{f}_{1/4}}=1-{{(1-0.90)}^{1/2}}=0.684 \end{align}\,\! }[/math]
    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:
    [math]\displaystyle{ \begin{align} c &= \left ( \frac{S_{2}-S_{3}}{S_{2}-S_{1}} \right )^\frac{1}{n\cdot I} \\ &= \left [ \frac{27.167-27.076}{27.167-24.558} \right ]^\frac{1}{6} \\ &= 0.572 \\ \end{align}\,\! }[/math]
    [math]\displaystyle{ \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 (24.558 + \frac{27.167 - 24.558}{1-0.572^{6}}\right ) \right ] \\ &= 94.024% \\ \end{align}\,\! }[/math]
    [math]\displaystyle{ \begin{align} b &= e^\left [\frac{(S_{2}-S_{1})(c-1)}{(1-c^{n})^{2}} \right ] \\ &= e^\left [\frac{(27.167-24.558)(0.572-1)}{(1-0.572^{6})^{2}} \right ] \\ &= 0.301 \\ \end{align}\,\! }[/math]
    Now, since the initial values have been determined, the Gauss-Newton method can be used. Substituting [math]\displaystyle{ {{Y}_{i}}={{R}_{i}},\,\! }[/math] [math]\displaystyle{ g_{1}^{(0)}=94.024,\,\! }[/math] [math]\displaystyle{ g_{2}^{(0)}=0.301,\,\! }[/math] [math]\displaystyle{ g_{3}^{(0)}=0.572\,\! }[/math]. The iterations are continued to solve for the parameters. Using the RGA software, the estimators of the parameters for the given example are:
    [math]\displaystyle{ \begin{align} \widehat{a}&= 0.9252 \\ \widehat{b} &= 0.1404 \\ \widehat{c} &= 0.5977 \end{align}\,\! }[/math]
    The next figure shows the entered data and the estimated parameters.
    RgaA.1.png

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

    [math]\displaystyle{ R=0.9299{{(0.0943)}^{{{0.7170}^{{{L}_{G}}}}}}\,\! }[/math]
  2. The predicted reliability after launch 22 is therefore:
    [math]\displaystyle{ \begin{align} R &= 0.9252{{(0.1404)}^{{{0.5977}^{19}}}} \\ &= 0.9251 \end{align}\,\! }[/math]
    The predicted reliability after launch 22 is calculated using the Quick Calculation Pad (QCP), as shown next.
    RgaA.2.png
  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.
    RgaA.3.png