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.
- Find the Standard Gompertz reliability growth curve using
the results of the first 15 launches.
- Find the predicted reliability after launch 22.
- 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.
Solution to Appendix A Example
- Table A.1 is organized as follows:
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, Sn = 1, and so on.
- Failure 2 is Mode 2; it occurs at launch 2 and it recurs
at launch 5. Therefore, Sn = 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, Sn = 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, Sn = 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, Sn = 1, and so on.
For launch 3 and failure 1, Sn = 1 in Eqn.
(1).
For launch 4 and failure 1, Sn = 2 in Eqn.
(1).
And so on.
Table A.1 - Launch
sequence with failure modes and failure values |
Launch
Number |
Result/
Mode |
Failure 1 |
Failure 2 |
Failure 3 |
Failure 4 |
Failure 5 |
Failure 6 |
Failure 7 |
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 |
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 |
|
Table A.2 - 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 |
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 |
|
|
S1
= 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 |
|
|
S2
= 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 |
|
|
S3= 26.946 |
|
22 |
91.896 |
4.521 |
92.565 |
|
Calculate the initial values of the Gompertz parameters using Table
A.2. Based on the equations from the Gompertz
Models chapter of this reference, the initial values are:
Now, since the initial values have been determined, the Gauss-Newton
method can be used. Substituting Yi = Ri,
g1(0)
= 89.31, g2(0) = 0.127, g3(0) = 0.578. The iterations are continued to solve
for the parameters. Using RGA,
the estimators of the parameters for the given example are:
Figure A.1 shows the entered data and the estimated parameters.
Figure
A.1: Entered data and the estimated Standard Gompertz parameters |
The Gompertz reliability growth curve may now be written as follows
where LG
is the number of launches with the first successful launch being counted
as LG = 1. Therefore, LG
is equal to 19, since reliability growth starts with launch 4.
(3)
- Based on Eqn. (3), the predicted reliability after launch
22 is:
The predicted reliability after launch 22 is calculated using the Quick
Calculation Pad and is shown in Figure A.2.
- In Table A.2, the predicted reliability values, as calculated
from Eqn. (3), 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 A.3 that the Gompertz curve appears to provide a
good fit to the actual data.
Figure
A.2: Predicted reliability after launch 22 |
Figure
A.3: Standard Gompertz reliability growth curve |
(2)