General Examples Using the Crow model

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General Examples Using the Crow-

AMSAA Model

Crow-AMSAA Example 11

Six systems were subjected to a reliability growth test and a total of 81 failures were

observed. Table 5.9 presents the start and end times, along with the times-to-failure for each system. Do the following:

1. Estimate the parameters of the Crow-AMSAA model using maximum likelihood

estimation.2. How many additional failures would be generated if testing continues until 3000

hours?

Table 5.9 - Multiple systems (concurrent operating times) data for Example 11

System # 1 2 3 4 5 6

Start Time 0 0 0 0 0 0

End Time 504 541 454 474 436 500

Times-to-Failure 21 83 26 36 23 7

29 83 26 306 46 13

43 83 57 306 127 13

43 169 64 334 166 31

43 213 169 354 169 31

66 299 213 395 213 82

115 375 231 403 213 109

159 431 231 448 255 137

199 231 456 369 166

202 231 461 374 200

222 304 380 210

248 383 415 220

248 422

255 437

286 469

286 469

304

320

348

364

404

410

429



Solution to Crow-AMSAA Example 11

1. Figure 5.28 shows the parameters estimated using RGA.

2. The number of failures can be estimated using the Quick Calculation Pad asshown in Figure 5.29. The estimated number of failures at 3000 hours is equal to

83.2451 and 81 failures were observed during testing. Therefore, the number of

additional failures generated if testing continues until 3000 hours is equal to83.2451 - 81 = 2.2451 ≈ 3.

Figure 5.28: Estimated parameters of the Crow-AMSAA model

http://rga.reliasoft.com/





Figure 5.29: Expected number of failures at 3000 hours


A prototype of a system was tested at the end of one of its design stages. The test was run

for a total of 300 hours and 27 failures were observed. Table 5.10 shows the collecteddata set. The prototype has a design specification of an MTBF equal to 10 hours with a

90% confidence level at 300 hours. Do the following:

1. Estimate the parameters of the Crow-AMSAA model using maximum likelihood

estimation.2. Does the prototype meet the specified goal?

Table 5.10 - Failure times data for Example 12

2.6 56.5 98.1 190.7

16.5 63.1 101.1 193

16.5 70.6 132 198.7

17 73 142.2 251.9



21.4 77.7 147.7 282.5

29.1 93.9 149 286.1

33.3 95.5 167.2



2. The instantaneous MTBF with one-sided 90% confidence bounds can becalculated using the Quick Calculation Pad (QCP) as shown in Figure 5.31. From

the QCP, it is estimated that the lower limit on the MTBF at 300 hours with a

90% confidence level is equal to 10.8170 hours. Therefore, the prototype has metthe specified goal.




Figure 5.31: Instantaneous MTBF with one-sided 90% confidence bounds


A one-shot system underwent reliability growth development for a total of 50 trials. The

test was performed as a combination of configuration in groups and individual trial bytrial. Table 5.11 shows the obtained test data set. The first column specifies the number of

failures that occurred in each interval and the second column the cumulative number of

trials in that interval. Do the following:

1. Estimate the parameters of the Crow-AMSAA model using maximum likelihoodestimators.

2. What are the instantaneous reliability and the 2-sided 90% confidence bounds atthe end of the test?

3. Plot the cumulative reliability with 2-sided 90% confidence bounds.

4. If the test was continued for another 25 trials what would the expected number of

additional failures be?

Table 5.10 - Mixed data for Example 13



Failures in Interval Cumulative Trials



3 4

0 5

4 9

1 12

0 13

1 15

2 19

1 20

1 22

0 24

1 25

1 28

0 32

2 37

0 39

1 40

1 44

0 46

1 49

0 50






2. Figure 5.33 shows the calculation of the instantaneous reliability with the 2-sided

90% confidence bounds. From the QCP it is estimated that the instantaneous

reliability at stage 50 (or at the end of the test) is 0.7487 with an upper and lower 2-sided 90% confidence bound of 0.8424 and 0.3815 respectively.

Figure 5.33: Instantaneous reliability with 2-sided 90% confidence bounds

3. Figure 5.34 shows the plot of the cumulative reliability with the 2-sided 90%

confidence bounds.



Figure 5.34: Plot of cumulative reliability with 2-sided 90% confidence bounds

4. Figure 5.35 shows the calculation of the expected number of failures after 75trials. From the QCP it is estimated that the cumulative number of failures after

75 trials is 26.3770 ≈ 27. Since 19 failures occurred in the first 50 trials, the

estimated number of additional failures is 6.



Figure 5.35: Cumulative number of failures after 75 trials

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General Examples Using the Crow model