Applying the Crow Extended Model to Fleet Data

As it was mentioned previously, the main motivation of the fleet analysis is to apply the Crow Extended model for in-service reliability improvements. The methodology to be used is identical to the application of the Crow Extended model for Grouped Data described in the Grouped Data for the Crow Extended Model section of this on-line reference. Consider the fleet data in Table 13.4. In order to apply the Crow Extended model, put N = 37 failure times on a cumulative time scale over (0, T), where T = 52110. In the example, each Ti corresponds to a failure time Xij. This is often not the situation. However, in all cases the accumulated operating time Yq at a failure time Xir is:

And q indexes the successive order of the failures. Thus, in this example N = 37, Y1 = 1396, Y2 = 5893, Y3 = 6418, ..., Y37 = 52110. See Table 13.6.

q	Yq	Mode	q	Yq	Mode
1	1396	BD1	20	26361	BD1
2	5893	BD2	21	26392	A
3	6418	A	22	26845	BD8
4	7650	BD3	23	30477	BD1
5	7877	BD4	24	31500	A
6	8012	BD2	25	31661	BD3
7	8031	BD2	26	31697	BD2
8	8843	BD1	27	36428	BD1
9	10867	BD1	28	40223	BD1
10	11183	BD5	29	40803	BD9
11	11810	A	30	42656	BD1
12	11870	BD1	31	42724	BD10
13	16139	BD2	32	44554	BD1
14	16104	BD6	33	45795	BD11
15	18178	BD7	34	46666	BD12
16	18677	BD2	35	48368	BD1
17	20751	BD4	36	51924	BD13
18	20772	BD2	37	52110	BD2
19	25815	BD1

Each system failure time in Table 13.4 corresponds to a problem and a cause (failure mode). The management strategy can be to not fix the failure mode (A mode) or to fix the failure mode with a delayed corrective action (BD mode). There are NA = 4 failures due to A failure modes. There are NBD = 33 total failures due to M = 13 distinct BD failure modes. Some of the distinct BD modes had repeats of the same problem. For example, mode BD1 had 12 occurrences of the same problem. Therefore, in this example, there are 13 distinct corrective actions corresponding to 13 distinct BD failure modes.

The objective of the Crow Extended model is to estimate the impact of the 13 distinct corrective actions. The analyst will choose an average effectiveness factor (EF) based on the proposed corrective actions and historical experience. Historical industry and government data supports a typical average effectiveness factor $\overline{d}=.70$ for many systems. In this example, an average EF of $\ \overline{d}=0.4$ was assumed in order to be conservative regarding the impact of the proposed corrective actions. Since there are no BC failure modes (corrective actions applied during the test), the projected failure intensity is:

To estimate the last term $\overline{d}h(T)$ of the Crow Extended model, partition the data in Table 13.6 into intervals. This partition consists of D successive intervals. The length of the qth interval is Lq, q = 1, 2, ..., D. It is not required that the intervals be of the same length, but there should be several (e.g. at least 5) cycles per interval on average. Also, let S1 = L1, S2 = L1 + L2, ..., etc. be the accumulated time through the qth interval. For the qth interval note the number of distinct BD modes, MIq, appearing for the first time, q = 1, 2, ..., D. See Table 13.7.

Interval	No. of Distinct BD Mode Failures	Length	Accumulated Time
1	MI1	L1	S1
2	MI2	L2	S2
.	.	.	.
.	.	.	.
.	.	.	.
D	MID	LD	SD

The term $\widehat{h}(T)$ is calculated as

and the values $\widehat{\lambda }$ and $\widehat{\beta }$ satisfy Eqns. (22) and (23). This is the grouped data version of the Crow-AMSAA model applied only to the first occurrence of distinct BD modes.

For the data in Table 13.6 the first 4 intervals had a length of 10000 and the last interval was 12110. Therefore, D = 5. This choice gives an average of about 5 overhaul cycles per interval. See Table 13.8.

Interval	No. of Distinct BD Mode Failures	Length	Accumulated Time
1	4	10000	10000
2	3	10000	20000
3	1	10000	30000
4	0	10000	40000
5	5	12110	52110
Total	13

Consequently, for $\overline{d}=0.4$ the last term of the Crow Extended model is given by:

This estimates that the 13 proposed corrective actions will reduce the number of failures per cycle of operation hours from the current

The average time between failures is estimated to increase from the current 1408.38 hours to 1876.93 hours.