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Eleven systems from the field were chosen for the purposes of a fleet analysis. Each system had at least one failure. All of the systems had a start time equal to zero and the last failure for each system corresponds to the end time. Group the data based on a fixed interval of 3000 hours and assume a fixed effectiveness factor equal to 0.4. Do the following:
Table 13.9 - Fleet data for Example 5
System |
Times-to-Failure |
1 |
1137 BD1, 1268 BD2 |
2 |
682 BD3, 744 A, 1336 BD1 |
3 |
95 BD1, 1593 BD3 |
4 |
1421 A |
5 |
1091 A, 1574 BD2 |
6 |
1415 BD4 |
7 |
598 BD4, 1290 BD1 |
8 |
1556 BD5 |
9 |
55 BD4 |
10 |
730 BD1, 1124 BD3 |
11 |
1400 BD4, 1568 A |

Figure 13.8: Estimated Crow Extended parameters |

Figure 13.9: System Operation plot |
This case study is based on the data given in the article "Graphical Analysis of Repair Data" by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
Table 13.10 - Automatic transmission data
Car |
Mileage |
|
Car |
Mileage |
1 |
7068, 26744+ |
|
18 |
17955+ |
2 |
28, 13809+ |
|
19 |
19507+ |
3 |
48, 1440, 29834+ |
|
20 |
24177+ |
4 |
530, 25660+ |
|
21 |
22854+ |
5 |
21762+ |
|
22 |
17844+ |
6 |
14235+ |
|
23 |
22637+ |
7 |
1388, 18228+ |
|
24 |
375, 19607+ |
8 |
21401+ |
|
25 |
19403+ |
9 |
21876+ |
|
26 |
20997+ |
10 |
5094, 18228+ |
|
27 |
19175+ |
11 |
21691+ |
|
28 |
20425+ |
12 |
20890+ |
|
29 |
22149+ |
13 |
22486+ |
|
30 |
21144+ |
14 |
19321+ |
|
31 |
21237+ |
15 |
21585+ |
|
32 |
14281+ |
16 |
18676+ |
|
33 |
8250, 21974+ |
17 |
23520+ |
|
34 |
19250, 21888+ |

Figure 13.10: Entered transmission data and the estimated Power Law parameters |

Figure 13.11: Cumulative number of failures at 36,000 miles |
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
Table 13.11 - Field data
System 1 |
System 2 |
System 3 |
1006.3 |
722.7 |
619.1 |
2261.2 |
1950.9 |
1519.1 |
2367 |
3259.6 |
2956.6 |
2615.5 |
4733.9 |
3114.8 |
2848.1 |
5105.1 |
3657.9 |
4073 |
5624.1 |
4268.9 |
5708.1 |
5806.3 |
6690.2 |
6464.1 |
5855.6 |
6803.1 |
6519.7 |
6325.2 |
7323.9 |
6799.1 |
6999.4 |
7501.4 |
7342.9 |
7084.4 |
7641.2 |
7736 |
7105.9 |
7851.6 |
8246.1 |
7290.9 |
8147.6 |
|
7614.2 |
8221.9 |
|
8332.1 |
9560.5 |
|
8368.5 |
9575.4 |
|
8947.9 |
|
|
9012.3 |
|
|
9135.9 |
|
|
9147.5 |
|
|
9601 |
|

Figure 13.12: Entered data and the estimated Power Law parameters |

Figure 13.13: The optimum overhaul interval |
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:

Figure 13.14: Crow Extended model for repairable systems |

Figure 13.15: MTBFs from Crow Extended model |
Figure 13.16 shows the expected number of failures at time = 1,000.

Figure 13.16: Cumulative number of failures at time = 1,000 |