Failure Discounting Example

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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.
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]{{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]{{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]{{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]CL=0.90\,\![/math] and are calculated from:
    [math]f=1-{{(1-CL)}^{\tfrac{1}{{{S}_{n}}}}}\,\![/math]
    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.
    • Failure 1 is Mode 1; it occurs at launch 1 and it does not recur throughout the process. So at launch 3, [math]{{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]{{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]{{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]{{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]{{S}_{n}}=1\,\![/math], and so on.
    For launch 3 and failure 1, [math]{{S}_{n}}=1\,\![/math].
    [math]\begin{align} {{f}_{1/3}}=1-{{(1-0.90)}^{1/1}}=0.900 \end{align}\,\![/math]
    For launch 4 and failure 1, [math]{{S}_{n}}=2\,\![/math].
    [math]\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]\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]\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]\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]{{Y}_{i}}={{R}_{i}},\,\![/math] [math]g_{1}^{(0)}=94.024,\,\![/math] [math]g_{2}^{(0)}=0.301,\,\![/math] [math]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]\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]{{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]
  2. The predicted reliability after launch 22 is therefore:
    [math]\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