Concurrent Operating Times - Crow-AMSAA (NHPP) Example

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This example appears in the Reliability Growth and Repairable System Analysis Reference book.


Six systems were subjected to a reliability growth test, and a total of 82 failures were observed. Given the data in the table below, which presents the start/end times and times-to-failure for each system, do the following:

  1. Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
  2. Determine how many additional failures would be generated if testing continues until 3,000 hours.
Multiple Systems (Concurrent Operating Times) Data
System # 1 2 3 4 5 6
Start Time (Hr) 0 0 0 0 0 0
End Time (Hr) 504 541 454 474 436 500
Failure Times (Hr) 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
248 301
255 422
286 437
286 469
304 469
320
348
364
404
410
429

Solution

  1. To estimate the parameters [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda}\,\! }[/math], the equivalent single system (ESS) must first be determined. The ESS is given below:
Equivalent Single System
Row Time to Event (hr) Row Time to Event (hr) Row Time to Event (hr) Row Time to Event (hr)
1 42 22 498 43 1386 64 2214
2 78 23 654 44 1386 65 2244
3 78 24 690 45 1386 66 2250
4 126 25 762 46 1386 67 2280
5 138 26 822 47 1488 68 2298
6 156 27 954 48 1488 69 2370
7 156 28 996 49 1530 70 2418
8 174 29 996 50 1530 71 2424
9 186 30 1014 51 1716 72 2460
10 186 31 1014 52 1716 73 2490
11 216 32 1014 53 1794 74 2532
12 258 33 1194 54 1806 75 2574
13 258 34 1200 55 1824 76 2586
14 258 35 1212 56 1824 77 2621
15 276 36 1260 57 1836 78 2676
16 342 37 1278 58 1836 79 2714
17 384 38 1278 59 1920 80 2734
18 396 39 1278 60 2004 81 2766
19 492 40 1278 61 2088 82 2766
20 498 41 1320 62 2124
21 498 42 1332 63 2184

Given the ESS data, the value of [math]\displaystyle{ \hat{\beta }\,\! }[/math] is calculated using:

[math]\displaystyle{ \hat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}}\,\! }[/math]
[math]\displaystyle{ \hat{\beta }=0.8939\,\! }[/math]

where [math]\displaystyle{ n\,\! }[/math] is the number of failures and [math]\displaystyle{ T^*\,\! }[/math] is the termination time. The termination time is the sum of end times for each of the systems which equals 2909.

[math]\displaystyle{ \hat{\lambda}\,\! }[/math] is estimated with:

[math]\displaystyle{ \hat{\lambda }=\frac{n}{{{T}^{*}}^{\beta }} }[/math]
[math]\displaystyle{ \hat{\lambda }=0.0657\,\! }[/math]

The next figure shows the parameters estimated using RGA.

Estimated parameters of the Crow-AMSAA model


  1. The number of failures can be estimated using the Quick Calculation Pad, as shown next. The estimated number of failures at 3,000 hours is equal to 84.2892 and 82 failures were observed during testing. Therefore, the number of additional failures generated if testing continues until 3,000 hours is equal to [math]\displaystyle{ 84.2892-82=2.2892\approx 3\,\! }[/math]
Expected number of failures at 3000 hours