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:

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

Figure 5.28 shows the parameters estimated using RGA.
The number of failures can be estimated using the Quick Calculation Pad as shown 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 to 83.2451 - 81 = 2.2451 ≈ 3.

Figure 5.28: Estimated parameters of the Crow-AMSAA model

Figure 5.29: Expected number of failures at 3000 hours

Crow-AMSAA Example 12

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 collected data 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:

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

Solution to Crow-AMSAA Example 12

Figure 5.30 shows the parameters estimated using RGA.
The instantaneous MTBF with one-sided 90% confidence bounds can be calculated 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 met the specified goal.

Figure 5.30: Estimated parameters of the Crow-AMSAA model

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

Crow-AMSAA Example 13

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 by trial. 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:

Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimators.
What are the instantaneous reliability and the 2-sided 90% confidence bounds at the end of the test?
Plot the cumulative reliability with 2-sided 90% confidence bounds.
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

Solution to Crow-AMSAA Example 13

Figure 5.32 shows the parameters estimated using RGA.

rga7_ch5amsaa_mixed_data.wmf

Figure 5.32: Estimated parameters of the Crow-AMSAA model

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
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
Figure 5.35 shows the calculation of the expected number of failures after 75 trials. 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