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Fielded Systems

Repairable Systems Analysis

Fleet Analysis

Fielded Systems Data

General Examples Using Fielded Systems

Fielded Systems Example 5 (Fleet Data Analysis)

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:

  1. Estimate the parameters of the Crow Extended model.
  2. Based on the analysis does it appear that the systems were randomly ordered?
  3. After the implementation of the delayed fixes, how many failures would you expect within the next 4000 hours of fleet operation.

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

Figure 13.8: Estimated Crow Extended parameters

 

Figure

Figure 13.9: System Operation plot

Solution to Fielded Systems Example 5
  1. Figure 13.8 shows the estimated Crow Extended parameters.
  2. Upon observing the estimated parameter β it does appear that the systems were randomly ordered since β = 0.8569. This value is close to 1. You can also verify that the confidence bounds on β include 1 by going to the QCP and calculating the parameter bounds or by viewing the Beta Bounds plot. However, you can also determine graphically if the systems were randomly ordered by using the System Operation plot as shown in Figure 13.9. Looking at the Cum. Time Line, it does not appear that the failures have a trend associated with them. Therefore, the systems can be assumed to be randomly ordered.
Fielded Systems Example 6 (Repairable Systems Data)

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:

  1. Estimate the parameters of the Power Law model.
  2. Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.

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

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

 

Figure

Figure 13.11: Cumulative number of failures at 36,000 miles

Solution to Fielded Systems Example 6
  1. The estimated Power Law parameters are shown in Figure 13.10.
  2. The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure 13.11. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
Fielded Systems Example 7 (Repairable System Data)

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:

  1. Estimate the parameters of the Power Law model.
  2. Determine the optimum overhaul interval.
  3. If β < 1, would it be cost-effective to implement an overhaul policy?

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

 

Solution to Fielded Systems Example 7
  1. Figure 13.12 shows the estimated Power Law parameters.
  2. The QCP can be used to calculate the optimum overhaul interval as shown in Figure 13.13.
  3. Since β < 1 then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.

Figure

Figure 13.12: Entered data and the estimated Power Law parameters

 

Figure

Figure 13.13: The optimum overhaul interval

Fielded Systems Example 8 (Repairable System Data)

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:

  1. Estimate the parameters of the Crow Extended model.
  2. Calculate the projected MTBF after the delayed fixes.
  3. What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?

Figure

Figure 13.14: Crow Extended model for repairable systems

Solution to Fielded Systems Example 8
  1. Figure 13.14 shows the estimated Crow Extended parameters.
  2. Figure 13.15 shows the projected MTBF at time = 541 (i.e. the age of the oldest system).

Figure

Figure 13.15: MTBFs from Crow Extended model


  1. Figure 13.16 shows the expected number of failures at time = 1,000.

Figure

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