Template:Crow-amsaa(nhpp)

=Crow-AMSAA (NHPP)= In Reliability Analysis for Complex, Repairable Systems (1974), Dr. Larry H. Crow noted that the Duane model could be stochastically represented as a Weibull process, allowing for statistical procedures to be used in the application of this model in reliability growth. This statistical extension became what is known as the Crow-AMSAA (NHPP) model. This method was first developed at the U.S. Army Materiel Systems Analysis Activity (AMSAA). It is frequently used on systems when usage is measured on a continuous scale. It can also be applied for the analysis of one shot items when there is high reliability and large number of trials. Test programs are generally conducted on a phase by phase basis. The Crow-AMSAA model is designed for tracking the reliability within a test phase and not across test phases. A development testing program may consist of several separate test phases. If corrective actions are introduced during a particular test phase then this type of testing and the associated data are appropriate for analysis by the Crow-AMSAA model. The model analyzes the reliability growth progress within each test phase and can aid in determining the following:

•	Reliability of the configuration currently on test •	Reliability of the configuration on test at the end of the test phase •	Expected reliability if the test time for the phase is extended •	Growth rate •	Confidence intervals •	Applicable goodness-of-fit tests

The reliability growth pattern for the Crow-AMSAA model is exactly the same pattern as for the Duane postulate (Chapter 4). That is, the cumulative number of failures is linear when plotted on ln-ln scale. Unlike the Duane postulate, the Crow-AMSAA model is statistically based. Under the Duane postulate, the failure rate is linear on ln-ln scale. However for the Crow-AMSAA model statistical structure, the failure intensity of the underlying non-homogeneous Poisson process (NHPP) is linear when plotted on ln-ln scale. Let $$N(t)$$  be the cumulative number of failures observed in cumulative test time  $$t$$  and let  $$\rho (t)$$  be the failure intensity for the Crow-AMSAA model. Under the NHPP model, $$\rho (t)\Delta t$$  is approximately the probably of a failure occurring over the interval  $$[t,t+\Delta t]$$  for small  $$\Delta t$$. In addition, the expected number of failures experienced over the test interval $$[0,T]$$  under the Crow-AMSAA model is given by:


 * $$E[N(T)]=\mathop{}_{0}^{T}\rho (t)dt$$

The Crow-AMSAA model assumes that $$\rho (T)$$  may be approximated by the Weibull failure rate function:


 * $$\rho (T)=\frac{\beta }{{T}^{\beta -1}}$$

Therefore, if $$\lambda =\tfrac{1},$$  the intensity function,  $$\rho (T),$$  or the instantaneous failure intensity,  $${{\lambda }_{i}}(T)$$, is defined as:


 * $${{\lambda }_{i}}(T)=\lambda \beta {{T}^{\beta -1}},\text{with }T>0,\text{ }\lambda >0\text{ and }\beta >0$$

In the special case of exponential failure times there is no growth and the failure intensity, $$\rho (t)$$, is equal to  $$\lambda $$. In this case, the expected number of failures is given by:


 * $$\begin{align}

& E[N(T)]= & \mathop{}_{0}^{T}\rho (t)dt \\ & = & \lambda T \end{align}$$

In order for the plot to be linear when plotted on ln-ln scale under the general reliability growth case, the following must hold true where the expected number of failures is equal to:


 * $$\begin{align}

& E[N(T)]= & \mathop{}_{0}^{T}\rho (t)dt \\ & = & \lambda {{T}^{\beta }} \end{align}$$

To put a statistical structure on the reliability growth process, consider again the special case of no growth. In this case the number of failures, $$N(T),$$  experienced during the testing over  $$[0,T]$$  is random. The expected number of failures, $$N(T),$$  is said to follow the homogeneous (constant) Poisson process with mean  $$\lambda T$$  and is given by:


 * $$\underset{}{\overset{}{\mathop{\Pr }}}\,[N(T)=n]=\frac{n!};\text{ }n=0,1,2,\ldots $$

The Crow-AMSAA generalizes this no growth case to allow for reliability growth due to corrective actions. This generalization keeps the Poisson distribution for the number of failures but allows for the expected number of failures, $$E[N(T)],$$  to be linear when plotted on ln-ln scale. The Crow-AMSAA model lets $$E[N(T)]=\lambda {{T}^{\beta }}$$. The probability that the number of failures, $$N(T),$$  will be equal to  $$n$$  under growth is then given by the Poisson distribution:


 * $$\underset{}{\overset{}{\mathop{\Pr }}}\,[N(T)=n]=\frac{n!};\text{ }n=0,1,2,\ldots $$

This is the general growth situation and the number of failures, $$N(T)$$, follows a non-homogeneous Poisson process. The exponential, "no growth" homogeneous Poisson process is a special case of the non-homogeneous Crow-AMSAA model. This is reflected in the Crow-AMSAA model parameter where $$\beta =1$$. The cumulative failure rate, $${{\lambda }_{c}}$$, is:


 * $${{\lambda }_{c}}=\lambda {{T}^{\beta -1}}$$

The cumulative $$MTB{{F}_{c}}$$  is:


 * $$MTB{{F}_{c}}=\frac{1}{\lambda }{{T}^{1-\beta }}$$

As mentioned above, the local pattern for reliability growth within a test phase is the same as the growth pattern observed by Duane, discussed in Chapter 4. The Duane $$MTB{{F}_{c}}$$  is equal to:


 * $$MTB{{F}_}=b{{T}^{\alpha }}$$

And the Duane cumulative failure rate, $${{\lambda }_{c}}$$, is:


 * $${{\lambda }_}=\frac{1}{b}{{T}^{-\alpha }}$$

Thus a relationship between Crow-AMSAA parameters and Duane parameters can be developed, such that:


 * $$\begin{align}

& {{b}_{DUANE}}= & \frac{1} \\ & {{\alpha }_{DUANE}}= & 1-{{\beta }_{AMSAA}} \end{align}$$

Note that these relationships are not absolute. They change according to how the parameters (slopes, intercepts, etc.) are defined when the analysis of the data is performed. For the exponential case, $$\beta =1$$, then  $${{\lambda }_{i}}(T)=\lambda $$ , a constant. For $$\beta >1$$,  $${{\lambda }_{i}}(T)$$  is increasing. This indicates a deterioration in system reliability. For $$\beta <1$$,  $${{\lambda }_{i}}(T)$$  is decreasing. This is indicative of reliability growth. Note that the model assumes a Poisson process with Weibull intensity function, not the Weibull distribution. Therefore, statistical procedures for the Weibull distribution do not apply for this model. The parameter $$\lambda $$  is called a scale parameter because it depends upon the unit of measurement chosen for  $$T$$. $$\beta $$ is the shape parameter that characterizes the shape of the graph of the intensity function. The total number of failures, $$N(T)$$, is a random variable with Poisson distribution. Therefore, the probability that exactly $$n$$  failures occur by time  $$T$$  is:


 * $$P[N(T)=n]=\frac{n!}$$

The number of failures occurring in the interval from $${{T}_{1}}$$  to  $${{T}_{2}}$$  is a random variable having a Poisson distribution with mean:


 * $$\theta ({{T}_{2}})-\theta ({{T}_{1}})=\lambda (T_{2}^{\beta }-T_{1}^{\beta })$$

The number of failures in any interval is statistically independent of the number of failures in any interval that does not overlap the first interval. At time $${{T}_{0}}$$, the failure intensity is  $${{\lambda }_{i}}({{T}_{0}})=\lambda \beta T_{0}^{\beta -1}$$. If improvements are not made to the system after time $${{T}_{0}}$$, it is assumed that failures would continue to occur at the constant rate  $${{\lambda }_{i}}({{T}_{0}})=\lambda \beta T_{0}^{\beta -1}$$. Future failures would then follow an exponential distribution with mean $$m({{T}_{0}})=\tfrac{1}{\lambda \beta T_{0}^{\beta -1}}$$. The instantaneous $$MTBF$$  of the system at time  $$T$$  is:


 * $$m(T)=\frac{1}{\lambda \beta {{T}^{\beta -1}}}$$

Example 11
Six systems were subjected to a reliability growth test and a total of 81 failures were observed. Table 5.9 presents the start and end times, along with the times-to-failure for each system. Do the following:
 * 1)	Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
 * 2)	How many additional failures would be generated if testing continues until 3000 hours?



Solution to Example 11

 * 1)	Figure ex9a shows the parameters estimated using RGA.
 * 2)	The number of failures can be estimated using the Quick Calculation Pad as shown in Figure ex9b. The estimated number of failures at 3000 hours is equal to $$83.2451$$  and 81 failures were observed during testing. Therefore, the number of additional failures generated if testing continues until 3000 hours is equal to  $$83.2451-81=2.2451\approx 3$$.

$$$$

Example 12
A prototype of a system was tested at the end of one of its design stages. The test was run for a total of 300 hours and 27 failures were observed. Table 5.10 shows the collected data set. The prototype has a design specification of an MTBF equal to 10 hours with a 90% confidence level at 300 hours. Do the following:
 * 1)	Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
 * 2)	Does the prototype meet the specified goal?

Table 5.10 - Failure times data for Example 12

Solution to Example 12

 * 1)	Figure ex10a shows the parameters estimated using RGA.
 * 2)	The instantaneous MTBF with one-sided 90% confidence bounds can be calculated using the

Quick Calculation Pad (QCP) as shown in Figure ex10b. From the QCP, it is estimated that the lower limit on the MTBF at 300 hours with a 90% confidence level is equal to 10.8170 hours. Therefore, the prototype has met the specified goal. $$$$

Example 13
A one-shot system underwent reliability growth development for a total of 50 trials. The test was performed as a combination of configuration in groups and individual trial by trial. Table 5.11 shows the obtained test data set. The first column specifies the number of failures that occurred in each interval and the second column the cumulative number of trials in that interval. Do the following:
 * 1)	Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimators.
 * 2)	What are the instantaneous reliability and the 2-sided 90% confidence bounds at the end of the test?
 * 3)	Plot the cumulative reliability with 2-sided 90% confidence bounds.
 * 4)	If the test was continued for another 25 trials what would the expected number of additional failures be?

Table 5.11 - Mixed data for Example 13

Solution to Example 13

 * 1)	Figure Mixedolio shows the parameters estimated using RGA.




 * 2)	Figure MixedCP shows the calculation of the instantaneous reliability with the 2-sided 90% confidence bounds. From the QCP it is estimated that the instantaneous reliability at stage 50 (or at the end of the test) is 72.6971% with an upper and lower 2-sided 90% confidence bound of 82.3627% and 39.5926% respectively.




 * 3)	Figure Mixedeliabilitylot shows the plot of the cumulative reliability with the 2-sided 90% confidence bounds.




 * 4)	Figure MixedCPumFailures shows the calculation of the expected number of failures after 75 trials. From the QCP it is estimated that the cumulative number of failures after 75 trials is $$26.3770\approx 27$$ . Since 20 failures occurred in the first 50 trials, the estimated number of additional failures is 7.