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	Crow Extended Model

General Examples Using the Crow Extended Model

Crow Extended Example 8

Three systems were subjected to a reliability growth test to evaluate the prototype of a new product. Once the test was completed a failure analysis was done and, based on this, a management strategy was able to be defined. It was determined that all corrective actions will be delayed until after the test. The collected data set is shown in Table 9.6 and the associated effectiveness factors for each unique BD mode are presented in Table 9.6. The prototype is required to meet a projected MTBF goal of 55 hours. Do the following:

Estimate the parameters of the Crow Extended model.
Based on the current management strategy what is the projected MTBF?
If the projected MTBF goal is not met, alter the current management strategy to meet this requirement with as little adjustment as possible and without changing the EFs of the existing BD modes. Assume an EF = 0.7 for any newly assigned BD modes.

Table 9.6 - Multiple Systems (Concurrent Operating Times) data for Example 8

	System 1	System 2	System 3
Start Time	0	0	0
End Time	541	454	436
Times-to-Failure	83 BD37	26 BD25	23 BD30
	83 BD43	26 BD43	46 BD49
	83 BD46	57 BD37	127 BD47
	169 A45	64 BD19	166 A2
	213 A18	169 A45	169 BD23
	299 A42	213 A32	213 BD7
	375 A1	231 BD8	213 BD29
	431 BD16	231 BD25	255 BD26
		231 BD27	369 A33
		231 A28	374 BD29
		304 BD24	380 BD22
		383 BD40	415 BD7

Table 9.7 - Effectiveness factors for Example 8

BD Mode	Effectiveness Factor
30	0.75
43	0.5
25	0.5
49	0.75
37	0.9
19	0.75
46	0.75
47	0.25
23	0.5
7	0.25
29	0.25
8	0.5
27	0.5
26	0.75
24	0.5
22	0.5
40	0.75
16	0.75

Solution to Crow Extended Example 8

Figure 9.29 shows the estimated Crow Extended parameters.
There are a couple of ways to calculate the projected MTBF, but the easiest is via the Quick Calculation Pad (QCP), as shown in Figure 9.30.
From the previous question, the projected MTBF is estimated to be 53.9390 hours, which is below the goal of 55 hours. To reach our goal, or to see if we can even get there, the management strategy must be changed. The effectiveness factors for the existing BD modes cannot be changed, however it is possible to change an A mode to a BD mode. But which A mode(s) should be changed? To answer this question, you can view the Individual Mode Failure Intensity plot with just the A modes displayed as shown in Figure 9.31. As you can see from the plot, failure mode A45 has the highest failure intensity. Therefore, among the A modes this particular failure mode is having the greatest negative effect in regards to the system MTBF. So change A45 to BD45. Be sure to change all instances of A45 to a BD mode. Enter an effectiveness factor for BD45 equal to 0.7 and recalculate the parameters of the Crow Extended model. Now go back to the QCP to calculate the projected MTBF as shown in Figure 9.32. The projected MTBF is now estimated to be 55.5903 hours. Based on the change in the management strategy, the projected MTBF goal is now expected to be met.

Figure 9.29: Entered data and the estimated Crow Extended parameters

Figure 9.30: Calculate the projected MTBF

Figure 9.31: Individual Mode Failure Intensity chart

Figure 9.32: Calculate the projected MTBF based on the change to the management strategy

Crow Extended Example 9

A reliability growth test was conducted for 200 hours. Some of the corrective actions were applied during the test while others were delayed until after the test was completed. The data set is given in Table 9.8. The effectiveness factors for the BD modes are given in Table 9.9. Do the following:

Estimate the parameters of the Crow Extended model.
Determine the average effectiveness factor of the BC modes using the Function Wizard.
What percent of the failure intensity will be left in the system due to the BD modes after implementing the delayed fixes?

Table 9.8 - Grouped Failure Times data for Example 9

Number at Event	Time to Event	Classification	Mode
3	25	BC	1
1	25	BD	9
1	25	BC	2
1	50	BD	10
1	50	BD	11
1	75	BD	12
1	75	BC	3
2	75	BD	13
1	75	A
1	100	BC	4
1	100	BD	14
1	125	BD	15
1	125	A
1	125	A
1	125	BC	5
1	125	BD	10
1	125	BC	6
1	150	A
1	150	BD	16
1	175	BC	4
1	175	BC	8
1	175	A
1	175	BC	7
1	200	BD	16
1	200	BC	3
1	200	BD	17

Table 9.9 - Effectiveness factors for Example 9

BD Mode	Effectiveness Factor
9	0.75
10	0.5
11	0.9
12	0.6
13	0.8
14	0.8
15	0.25
16	0.75
17	0.8

Solution to Crow Extended Example 9

Figure 9.33 shows the estimated parameters of the Crow Extended model.
After inserting a General Spreadsheet into the Folio, the Function Wizard can be accessed via the Tools menu. Once the Function Wizard is loaded, select Average Effectiveness Factor from the list of available functions and under Avg. Eff. Factor select BC modes as shown in Figure 9.34. Click OK and the result will be placed into the General Spreadsheet. The average effectiveness factor for the BC modes is 0.6983.
The percent of the failure intensity left in the system due to the BD modes can be determined using the Failure Mode Strategy plot as shown in Figure 9.35. Therefore, the percent of the failure intensity left in the system due to the BD modes is 1.79%.

Figure 9.33: Entered data and the estimated Crow Extended parameters

Figure 9.34: Calculate the average effectiveness factor for the BC modes using the Function Wizard

Figure 9.35: Failure Mode Strategy plot

Crow Extended Example 10

Two prototypes of a new design are tested simultaneously. Whenever a failure was observed for one unit, the current operating time of the other unit was also recorded. The test was terminated after 300 hours. All of the design changes for the prototypes were delayed until after completing the test and the data set is given in Table 9.10. Assume a fixed effectiveness factor equal to 0.7. The MTBF goal for the new design is 30 hours. Do the following:

Estimate the parameters of the Crow Extended model.
What is the projected MTBF and growth potential?
Under the current management strategy, is it even possible to reach the MTBF goal of 30 hours?

Table 9.10 - Multiple Systems (Known Operating Times) data for Example 10

Failed Unit ID	Time Unit 1	Time Unit 2	Classification	Mode
1	16.5	0	BD	seal leak
1	16.5	0	BD	valve
1	17	0	A
2	20.5	0.9	A
2	25.3	3.8	BD	hose
2	28.7	4.6	BD	operator error
1	41.8	14.7	BD	bearing
1	45.5	17.6	A
2	48.6	22	A
2	49.6	23.4	BD	seal leak
1	51.4	26.3	A
1	58.2	35.7	BD	seal leak
2	59	36.5	A
2	60.6	37.6	BD	hose
1	61.9	39.1	BD	seal leak
1	76.6	55.4	BD	bearing
2	81.1	61.1	A
1	84.1	63.6	A
1	84.7	64.3	A
1	94.6	72.6	A
2	100	78.1	BD	valve
1	104	81.4	BD	bearing
2	104.8	85.9	BD	spring
2	105.9	87.1	BD	operator error
1	108.8	89.9	BD	hose
2	132.4	119.5	BD	spring
2	132.4	150.1	BD	operator error
2	132.4	153.7	A

Solution to Crow Extended Example 10

Figure 9.36 shows the estimated Crow Extended parameters.
One possible method to calculate the projected MTBF and growth potential is to use the Quick Calculation Pad, but you can also view these two values at the same time by viewing the Growth Potential MTBF plot, which is displayed in Figure 9.37. From the plot, the projected MTBF is equal to 16.87 hours and the growth potential is equal to 18.63 hours.
The current projected MTBF and growth potential MTBF are both well below the required goal of 30 hours. To check if this goal can even be reached, you can set the effectiveness factor equal to 1. In other words, if all of the corrective actions were to remove the failure modes completely then what would be the projected and growth potential MTBF? After changing the fixed effectiveness factor to 1, the parameters are recalculated and the Growth Potential plot is refreshed. The refreshed plot is shown in Figure 9.38. Even if you assume an effectiveness factor equal to 1, the growth potential is still only 27.27 hours. Based on the current design process, it will not be possible to reach the MTBF goal of 30 hours. Therefore, there are basically two options: start a new design stage or reduce the required MTBF goal.

Figure 9.36: Entered data and the estimated Crow Extended parameters

Figure 9.37: Growth Potential MTBF plot (EF = 0.7)

Figure 9.38: Growth Potential MTBF plot (EF = 1)