Template:Applying the crow extended model to fleet data

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 Chapter 9. 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 $${{T}_{i}}$$  corresponds to a failure time  $${{X}_{ij}}$$. This is often not the situation. However, in all cases the accumulated operating time $${{Y}_{q}}$$  at a failure time  $${{X}_{ir}}$$  is:


 * $$\begin{align}

& {{Y}_{q}}= & {{X}_{i,r}}+\underset{j=1}{\overset{r-1}{\mathop \sum }}\,{{T}_{j}},\ \ \ q=1,2,\ldots ,N \\ & N= & \underset{j=1}{\overset{K}{\mathop \sum }}\,{{N}_{j}} \end{align}$$

And $$q$$  indexes the successive order of the failures. Thus, in this example $$N=37,\,{{Y}_{1}}=1396,\,{{Y}_{2}}=5893,\,{{Y}_{3}}=6418,\ldots ,{{Y}_{37}}=52110$$. See Table 13.6.

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 $${{N}_{A}}=4$$  failures due to A failure modes. There are $${{N}_{BD}}=33$$  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 $$\bar{d}=0.4$$ 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:


 * $$\widehat{r}(T)=\left( \frac{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T} \right)+\overline{d}h(T)$$

The first term is estimated by:


 * $${{\widehat{\lambda }}_{A}}=\frac{T}=0.000077$$

The second term is:


 * $$\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T}=0.00038$$

This estimates the growth potential failure intensity:


 * $$\begin{align}

& {{\widehat{\gamma }}_{GP}}(T)= & \frac{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T} \\ & = & 0.00046 \end{align}$$

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 $${{q}^{th}}$$  interval is  $${{L}_{q}},$$   $$\,q=1,2,\ldots ,D$$. 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 $${{S}_{1}}={{L}_{1}},$$   $${{S}_{2}}={{L}_{1}}+{{L}_{2}},\ldots ,$$  etc. be the accumulated time through the  $${{q}^{th}}$$  interval. For the $${{q}^{th}}$$  interval note the number of distinct BD modes,  $$M{{I}_{q}}$$, appearing for the first time,  $$q=1,2,\ldots ,D$$. See Table 13.7. The term $$\widehat{h}(T)$$  is calculated as  $$\widehat{h}(T)=\widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}}$$ and the values  $$\widehat{\lambda }$$  and  $$\widehat{\beta }$$  satisfy Eqns. (cc1) and (cc2). 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.


 * Thus:


 * $$\begin{align}

& \widehat{\lambda }= & 0.00330 \\ & \widehat{\beta }= & 0.76219 \end{align}$$


 * This gives:


 * $$\begin{align}

& \widehat{h}(T)= & \widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}} \\ & = & 0.00019 \end{align}$$

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


 * $$\overline{d}h(T)=0.000076$$

The projected failure intensity is:


 * $$\begin{align}

& \widehat{r}(T)= & \frac{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{T}+\overline{d}h(T) \\ & = & 0.000077+0.6\times (0.00063)+0.4\times (0.00019) \\ & = & 0.000533 \end{align}$$

This estimates that the 13 proposed corrective actions will reduce the number of failures per cycle of operation hours from the current $$\widehat{r}(0)=\tfrac{{{N}_{A}}+{{N}_{BD}}}{T}=0.00071$$  to  $$\widehat{r}(T)=0.00053.$$  The average time between failures is estimated to increase from the current 1408.38 hours to 1876.93 hours.