Crow-AMSAA (NHPP)

Dr. Larry H. Crow [17] 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 a 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

Crow-AMSAA (NHPP) Model
The reliability growth pattern for the Crow-AMSAA model is exactly the same pattern as for the Duane postulate, 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 model 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:


 * $$\begin{align}

{{\lambda }_{c}}=\lambda {{T}^{\beta -1}} \end{align}\,\!$$

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. 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 the 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 $$T\,\!$$, while $$\beta \,\!$$ 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:


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

Note About Applicability
The Duane and Crow-AMSAA models are the most frequently used reliability growth models. Their relationship comes from the fact that both make use of the underlying observed linear relationship between the logarithm of cumulative MTBF and cumulative test time. However, the Duane model does not provide a capability to test whether the change in MTBF observed over time is significantly different from what might be seen due to random error between phases. The Crow-AMSAA model allows for such assessments. Also, the Crow-AMSAA allows for development of hypothesis testing procedures to determine growth presence in the data (where $$\beta <1\,\!$$ indicates that there is growth in MTBF, $$\beta =1\,\!$$ indicates a constant MTBF and $$\beta >1\,\!$$ indicates a decreasing MTBF). Additionally, the Crow-AMSAA model views the process of reliability growth as probabilistic, while the Duane model views the process as deterministic.

Maximum Likelihood Estimators
The probability density function (pdf) of the $${{i}^{th}}\,\!$$ event given that the $${{(i-1)}^{th}}\,\!$$ event occurred at $${{T}_{i-1}}\,\!$$ is:


 * $$f({{T}_{i}}|{{T}_{i-1}})=\frac{\beta }{\eta }{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta -1}}\cdot {{e}^{-\tfrac{1}\left( T_{i}^{\beta }-T_{i-1}^{\beta } \right)}}\,\!$$

The likelihood function is:


 * $$L={{\lambda }^{n}}{{\beta }^{n}}{{e}^{-\lambda {{T}^{*\beta }}}}\underset{i=1}{\overset{n}{\mathop \prod }}\,T_{i}^{\beta -1}\,\!$$

where $${{T}^{*}}\,\!$$ is the termination time and is given by:


 * $${{T}^{*}}=\left\{ \begin{matrix}

{{T}_{n}}\text{ if the test is failure terminated} \\ T>{{T}_{n}}\text{ if the test is time terminated} \\ \end{matrix} \right\}\,\!$$

Taking the natural log on both sides:


 * $$\Lambda =n\ln \lambda +n\ln \beta -\lambda {{T}^{*\beta }}+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}\,\!$$

And differentiating with respect to $$\lambda \,\!$$ yields:


 * $$\frac{\partial \Lambda }{\partial \lambda }=\frac{n}{\lambda }-{{T}^{*\beta }}\,\!$$

Set equal to zero and solve for $$\lambda \,\!$$ :


 * $$\widehat{\lambda }=\frac{n}\,\!$$

Now differentiate with respect to $$\beta \,\!$$ :


 * $$\frac{\partial \Lambda }{\partial \beta }=\frac{n}{\beta }-\lambda {{T}^{*\beta }}\ln {{T}^{*}}+\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}\,\!$$

Set equal to zero and solve for $$\beta \,\!$$ :


 * $$\widehat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}}\,\!$$

Biasing and Unbiasing of Beta
The equation above returns the biased estimate of $$\beta \,\!$$. The unbiased estimate of $$\beta \,\!$$ can be calculated by using the following relationships. For time terminated data (meaning that the test ends after a specified number of failures):


 * $$\bar{\beta }=\frac{N-1}{N}\hat{\beta }\,\!$$

For failure terminated data (meaning that the test ends after a specified test time):


 * $$\bar{\beta }=\frac{N-2}{N}\hat{\beta }\,\!$$

Crow-AMSAA Model Example
Two prototypes of a system were tested simultaneously with design changes incorporated during the test. The following table presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.

Developmental test data for two identical systems Solution

For the failure terminated test, $${\beta}\,\!$$ is:


 * $$\begin{align}

\widehat{\beta }&=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}} \\ &=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}} \\ \end{align}\,\!$$


 * where:


 * $$\underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355\,\!$$


 * Then:


 * $$\widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142\,\!$$

And for $${\lambda}\,\!$$ :


 * $$\begin{align}

\widehat{\lambda }&=\frac{n} \\ & =\frac{22}=0.4239 \\ \end{align}\,\!$$

Therefore, $${{\lambda }_{i}}(T)\,\!$$ becomes:


 * $$\begin{align}

{{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\ = & 0.0217906\frac{\text{failures}}{\text{hr}} \end{align}\,\!$$

The next figure shows the plot of the failure rate. If no further changes are made, the estimated MTBF is $$\tfrac{1}{0.0217906}\,\!$$ or 46 hours.



Confidence Bounds
The RGA software provides two methods to estimate the confidence bounds for the Crow Extended model when applied to developmental testing data. The Fisher Matrix approach is based on the Fisher Information Matrix and is commonly employed in the reliability field. The Crow bounds were developed by Dr. Larry Crow. See the Crow-AMSAA Confidence Bounds chapter for details on how the confidence bounds are calculated.

Confidence Bounds Example
Calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity for the data from the Crow-AMSAA Model Example given above.

Solution

Fisher Matrix Bounds

Using $$\widehat{\beta }\,\!$$ and $$\widehat{\lambda }\,\!$$ estimated in the Crow-AMSAA example, the Fisher Matrix bounds on beta are:


 * $$\begin{align}

\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} = & -\frac{22}=-122.43 \\ \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}} = & -\frac{22}-0.4239\cdot {{620}^{0.6142}}{{(\ln 620)}^{2}}=-967.68 \\ \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta } = & -{{620}^{0.6142}}\ln 620=-333.64 \end{align}\,\!$$

The Fisher Matrix then becomes:


 * $$\begin{align}

\begin{bmatrix}122.43 & 333.64\\ 333.64 & 967.68\end{bmatrix}^{-1} & = \begin{bmatrix}Var(\hat{\lambda}) & Cov(\hat{\beta},\hat{\lambda})\\ Cov(\hat{\beta},\hat{\lambda}) & Var(\hat{\beta})\end{bmatrix} \\ & = \begin{bmatrix} 0.13519969 & -0.046614609\\ -0.046614609 & 0.017105343 \end{bmatrix} \end{align}\,\!$$

For $$T=620\,\!$$ hours, the partial derivatives of the cumulative and instantaneous failure intensities are:


 * $$\begin{align}

\frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \widehat{\lambda }{{T}^{\widehat{\beta }-1}}\ln (T) \\ = & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ = & 0.22811336 \\   \frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }=  & {{T}^{\widehat{\beta }-1}} \\ = & {{620}^{-0.3858}} \\   =  & 0.083694185  \end{align}\,\!$$


 * $$\begin{align}

\frac{\partial {{\lambda }_{i}}(T)}{\partial \beta }= & \widehat{\lambda }{{T}^{\widehat{\beta }-1}}+\widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}}\ln T \\ = & 0.4239\cdot {{620}^{-0.3858}}+0.4239\cdot 0.6142\cdot {{620}^{-0.3858}}\ln 620 \\ = & 0.17558519  \end{align}\,\!$$


 * $$\begin{align}

\frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \widehat{\beta }{{T}^{\widehat{\beta }-1}} \\ = & 0.6142\cdot {{620}^{-0.3858}} \\ = & 0.051404969  \end{align}\,\!$$

Therefore, the variances become:


 * $$\begin{align}

Var(\hat{\lambda_{c}}(T)) & = 0.22811336^{2}\cdot 0.017105343\ + 0.083694185^{2} \cdot 0.13519969\ -2\cdot 0.22811336\cdot 0.083694185\cdot 0.046614609 \\ & = 0.00005721408 \\ Var(\hat{\lambda_{i}}(T)) & = 0.17558519^{2}\cdot 0.01715343\ + 0.051404969^{2}\cdot 0.13519969\ -2\cdot 0.17558519\cdot 0.051404969\cdot 0.046614609 \\ &= 0.0000431393 \end{align}\,\!$$

The cumulative and instantaneous failure intensities at $$T=620\,\!$$ hours are:


 * $$\begin{align}

{{\lambda }_{c}}(T)= & 0.03548 \\ {{\lambda }_{i}}(T)= & 0.02179 \end{align}\,\!$$

So, at the 90% confidence level and for $$T=620\,\!$$ hours, the Fisher Matrix confidence bounds for the cumulative failure intensity are:


 * $$\begin{align}

{{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039 \end{align}\,\!$$

The confidence bounds for the instantaneous failure intensity are:


 * $$\begin{align}

{{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579 \end{align}\,\!$$

The following figures display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively.





Crow Bounds

The Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for $$T=620\,\!$$ hours are:


 * $$\begin{align}

{{[{{\lambda }_{c}}(T)]}_{L}} = & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ = & \frac{29.787476}{2*620} \\ = & 0.02402 \\  {{[{{\lambda }_{c}}(T)]}_{U}} = & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \\ = & \frac{62.8296}{2*620} \\ = & 0.05067 \end{align}\,\!$$

The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for $$T=620\,\!$$ hours are:


 * $$\begin{align}

{{[{{\lambda }_{i}}(t)]}_{L}} = & \frac{1} \\ = & \frac{1}{MTB{{F}_{i}}\cdot U} \\ = & 0.01179 \end{align}\,\!$$


 * $$\begin{align}

{{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1} \\ = & \frac{1}{MTB{{F}_{i}}\cdot L} \\ = & 0.03253  \end{align}\,\!$$

The following figures display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.





Another Confidence Bounds Example
Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data from the Crow-AMSAA Model Example given above.

Solution

Fisher Matrix Bounds

From the previous example:


 * $$\begin{align}

Var(\widehat{\lambda }) = & 0.13519969 \\ Var(\widehat{\beta }) = & 0.017105343 \\ Cov(\widehat{\beta },\widehat{\lambda }) = & -0.046614609 \end{align}\,\!$$

And for $$T=620\,\!$$ hours, the partial derivatives of the cumulative and instantaneous MTBF are:


 * $$\begin{align}

\frac{\partial {{m}_{c}}(T)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }}{{T}^{1-\widehat{\beta }}}\ln T \\ = & -\frac{1}{0.4239}{{620}^{0.3858}}\ln 620 \\ = & -181.23135 \\  \frac{\partial {{m}_{c}}(T)}{\partial \lambda } = & -\frac{1}{{T}^{1-\widehat{\beta }}} \\ = & -\frac{1}{{620}^{0.3858}} \\ = & -66.493299 \\  \frac{\partial {{m}_{i}}(T)}{\partial \beta } = & -\frac{1}{\widehat{\lambda }{{\widehat{\beta }}^{2}}}{{T}^{1-\beta }}-\frac{1}{\widehat{\lambda }\widehat{\beta }}{{T}^{1-\widehat{\beta }}}\ln T \\ = & -\frac{1}{0.4239\cdot {{0.6142}^{2}}}{{620}^{0.3858}}-\frac{1}{0.4239\cdot 0.6142}{{620}^{0.3858}}\ln 620 \\ = & -369.78634 \\  \frac{\partial {{m}_{i}}(T)}{\partial \lambda } = & -\frac{1}{{{\widehat{\lambda }}^{2}}\widehat{\beta }}{{T}^{1-\widehat{\beta }}} \\ = & -\frac{1}{{{0.4239}^{2}}\cdot 0.6142}\cdot {{620}^{0.3858}} \\ = & -108.26001 \end{align}\,\!$$

Therefore, the variances become:


 * $$\begin{align}

Var({{\widehat{m}}_{c}}(T)) = & {{\left( -181.23135 \right)}^{2}}\cdot 0.017105343+{{\left( -66.493299 \right)}^{2}}\cdot 0.13519969 \\ & -2\cdot \left( -181.23135 \right)\cdot \left( -66.493299 \right)\cdot 0.046614609 \\ = & 36.113376 \end{align}\,\!$$


 * $$\begin{align}

Var({{\widehat{m}}_{i}}(T)) = & {{\left( -369.78634 \right)}^{2}}\cdot 0.017105343+{{\left( -108.26001 \right)}^{2}}\cdot 0.13519969 \\ & -2\cdot \left( -369.78634 \right)\cdot \left( -108.26001 \right)\cdot 0.046614609 \\ = & 191.33709 \end{align}\,\!$$

So, at 90% confidence level and $$T=620\,\!$$ hours, the Fisher Matrix confidence bounds are:


 * $$\begin{align}

{{[{{m}_{c}}(T)]}_{L}} = & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ = & 19.84581 \\  {{[{{m}_{c}}(T)]}_{U}} = & {{{\hat{m}}}_{c}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ = & 40.01927 \end{align}\,\!$$


 * $$\begin{align}

{{[{{m}_{i}}(T)]}_{L}} = & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ = & 27.94261 \\ {{[{{m}_{i}}(T)]}_{U}} = & {{{\hat{m}}}_{i}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ = & 75.34193 \end{align}\,\!$$

The following two figures show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs.





Crow Bounds

The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for $$T=620\,\!$$ hours are:


 * $$\begin{align}

{{[{{m}_{c}}(T)]}_{L}} = & \frac{1} \\ = & 20.5023 \\  {{[{{m}_{c}}(T)]}_{U}}  = & \frac{1} \\ = & 41.6282 \end{align}\,\!$$


 * $$\begin{align}

{{[MTB{{F}_{i}}]}_{L}} = & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ = & 30.7445 \\  {{[MTB{{F}_{i}}]}_{U}}  = & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \\ = & 84.7972 \end{align}\,\!$$

The figures below show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF.



Confidence bounds can also be obtained on the parameters $$\widehat{\beta }\,\!$$ and $$\widehat{\lambda }\,\!$$. For Fisher Matrix confidence bounds:


 * $$\begin{align}

{{\beta }_{L}} = & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ = & 0.4325 \\ {{\beta }_{U}}  = & \hat{\beta }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ = & 0.8722 \end{align}\,\!$$


 * and:


 * $$\begin{align}

{{\lambda }_{L}} = & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ = & 0.1016 \\  {{\lambda }_{U}}  = & \hat{\lambda }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ = & 1.7691 \end{align}\,\!$$

For Crow confidence bounds:


 * $$\begin{align}

{{\beta }_{L}}= & 0.4527 \\ {{\beta }_{U}}= & 0.9350 \end{align}\,\!$$


 * and:


 * $$\begin{align}

{{\lambda }_{L}}= & 0.2870 \\ {{\lambda }_{U}}= & 0.5827 \end{align}\,\!$$

Grouped Data
For analyzing grouped data, we follow the same logic described previously for the Duane model. If the $$E[N(T)]\,\!$$ equation from the Crow-AMSAA (NHPP) Model section above is linearized:


 * $$\begin{align}

\ln [E(N(T))]=\ln \lambda +\beta \ln T \end{align}\,\!$$

According to Crow [9], the likelihood function for the grouped data case, (where $${{n}_{1}},\,\!$$ $${{n}_{2}},\,\!$$ $${{n}_{3}},\ldots ,\,\!$$ $${{n}_{k}}\,\!$$ failures are observed and $$k\,\!$$ is the number of groups), is:


 * $$\underset{i=1}{\overset{k}{\mathop \prod }}\,\underset{}{\overset{}{\mathop{\Pr }}}\,({{N}_{i}}={{n}_{i}})=\underset{i=1}{\overset{k}{\mathop \prod }}\,\frac{{{(\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })}^{{{n}_{i}}}}\cdot {{e}^{-(\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })}}}{{{n}_{i}}!}\,\!$$

And the MLE of $$\lambda \,\!$$ based on this relationship is:


 * $$\widehat{\lambda }=\frac{n}{T_{k}^{\widehat{\beta }}}\,\!$$

where $$n \,\!$$ is the total number of failures from all the groups.

The estimate of $$\beta \,\!$$ is the value $$\widehat{\beta }\,\!$$ that satisfies:


 * $$\underset{i=1}{\overset{k}{\mathop \sum }}\,{{n}_{i}}\left[ \frac{T_{i}^{\widehat{\beta }}\ln {{T}_{i}}-T_{i-1}^{\widehat{\beta }}\ln {{T}_{i-1}}}{T_{i}^{\widehat{\beta }}-T_{i-1}^{\widehat{\beta }}}-\ln {{T}_{k}} \right]=0\,\!$$

See the Crow-AMSAA Confidence Bounds for details on how confidence bounds for grouped data are calculated.

Grouped Data Example
Consider the grouped failure times data given in the following table. Solve for the Crow-AMSAA parameters using MLE.

Solution

Using RGA, the value of $$\widehat{\beta }\,\!$$, which must be solved numerically, is 0.6315. Using this value, the estimator of $$\lambda \,\!$$ is:


 * $$\begin{align}

\widehat{\lambda } = & \frac{11}{3,{{000}^{0.6315}}} \\ = & 0.0701 \end{align}\,\!$$

Therefore, the intensity function becomes:


 * $$\widehat{\rho }(T)=0.0701\cdot 0.6315\cdot {{T}^{-0.3685}}\,\!$$

Grouped Data Confidence Bounds Example
A new helicopter system is under development. System failure data has been collected on five helicopters during the final test phase. The actual failure times cannot be determined since the failures are not discovered until after the helicopters are brought into the maintenance area. However, total flying hours are known when the helicopters are brought in for service, and every 2 weeks each helicopter undergoes a thorough inspection to uncover any failures that may have occurred since the last inspection. Therefore, the cumulative total number of flight hours and the cumulative total number of failures for the 5 helicopters are known for each 2-week period. The total number of flight hours from the test phase is 500, which was accrued over a period of 12 weeks (six 2-week intervals). For each 2-week interval, the total number of flight hours and total number of failures for the 5 helicopters were recorded. The grouped data set is displayed in the following table.


 * 1)	Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
 * 2)	Calculate the confidence bounds on the cumulative and instantaneous MTBF using the Fisher Matrix and Crow methods.

Solution


 * 1)	Using RGA, the value of $$\widehat{\beta }\,\!$$, which must be solved numerically, is 0.81361. Using this value, $$\widehat{\lambda }\,\!$$ is:


 * $$\widehat{\lambda }=0.44585\,\!$$

The grouped Fisher Matrix confidence bounds can be obtained on the parameters $$\widehat{\beta }\,\!$$ and $$\widehat{\lambda }\,\!$$ at the 90% confidence level by:


 * $$\begin{align}

{{\beta }_{L}} = & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ = & 0.6546 \\  {{\beta }_{U}}  = & \hat{\beta }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ = & 1.0112 \end{align}\,\!$$


 * and:


 * $$\begin{align}

{{\lambda }_{L}} = & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ = & 0.14594 \\  {{\lambda }_{U}}  = & \hat{\lambda }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ = & 1.36207 \end{align}\,\!$$

Crow confidence bounds can also be obtained on the parameters $$\widehat{\beta }\,\!$$ and $$\widehat{\lambda }\,\!$$ at the 90% confidence level, as:


 * $$\begin{align}

{{\beta }_{L}} = & \hat{\beta }(1-S) \\ = & 0.63552 \\  {{\beta }_{U}}  = & \hat{\beta }(1+S) \\ = & 0.99170 \end{align}\,\!$$


 * and:


 * $$\begin{align}

{{\lambda }_{L}} = & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot T_{k}^{\beta }} \\ = & 0.36197 \\  {{\lambda }_{U}}  = & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot T_{k}^{\beta }} \\ = & 0.53697 \end{align}\,\!$$


 * 2)	The Fisher Matrix confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% 2-sided confidence level and for $$T=500\,\!$$ hour are:


 * $$\begin{align}

{{[{{m}_{c}}(T)]}_{L}} = & {{{\hat{m}}}_{c}}(t){{e}^{{{z}_{\alpha /2}}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ = & 5.8680 \\ {{[{{m}_{c}}(T)]}_{U}}  = & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha /2}}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ = & 8.6947 \end{align}\,\!$$


 * and:


 * $$\begin{align}

{{[MTB{{F}_{i}}]}_{L}} = & {{{\hat{m}}}_{i}}(t){{e}^{{{z}_{\alpha /2}}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ = & 6.6483 \\  {{[MTB{{F}_{i}}]}_{U}}  = & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha /2}}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ = & 11.5932 \end{align}\,\!$$

The next two figures show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBF.





The Crow confidence bounds for the cumulative and instantaneous MTBF at the 90% 2-sided confidence level and for $$T = 500\,\!$$hours are:


 * $$\begin{align}

{{[{{m}_{c}}(T)]}_{L}} = & \frac{1}{C{{(t)}_{U}}} \\ = & 5.85449 \\ {{[{{m}_{c}}(T)]}_{U}}  = & \frac{1}{C{{(t)}_{L}}} \\ = & 8.79822 \end{align}\,\!$$


 * and:


 * $$\begin{align}

{{[MTB{{F}_{i}}]}_{L}} = & {{\widehat{m}}_{i}}(1-W) \\ = & 6.19623 \\ {{[MTB{{F}_{i}}]}_{U}}  = & {{\widehat{m}}_{i}}(1+W) \\ = & 11.36223 \end{align}\,\!$$

The next two figures show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF.





Crow Discrete Reliability Growth Model
The Crow-AMSAA model can be adapted for the analysis of success/failure data (also called discrete or attribute data). Suppose system development is represented by $$i\,\!$$ configurations. This corresponds to $$i-1\,\!$$ configuration changes, unless fixes are applied at the end of the test phase, in which case there would be $$i\,\!$$ configuration changes. Let $${{N}_{i}}\,\!$$ be the number of trials during configuration $$i\,\!$$ and let $${{M}_{i}}\,\!$$ be the number of failures during configuration $$i\,\!$$. Then the cumulative number of trials through configuration $$i\,\!$$, namely $${{T}_{i}}\,\!$$, is the sum of the $${{N}_{i}}\,\!$$ for all $$i\,\!$$, or:


 * $${{T}_{i}}=\underset{}{\overset{}{\mathop \sum }}\,{{N}_{i}}\,\!$$

And the cumulative number of failures through configuration $$i\,\!$$, namely $${{K}_{i}}\,\!$$, is the sum of the $${{M}_{i}}\,\!$$ for all $$i\,\!$$, or:


 * $${{K}_{i}}=\underset{}{\overset{}{\mathop \sum }}\,{{M}_{i}}\,\!$$

The expected value of $${{K}_{i}}\,\!$$ can be expressed as $$E[{{K}_{i}}]\,\!$$ and defined as the expected number of failures by the end of configuration $$i\,\!$$. Applying the learning curve property to $$E[{{K}_{i}}]\,\!$$ implies:


 * $$E\left[ {{K}_{i}} \right]=\lambda T_{i}^{\beta }\,\!$$

Denote $${{f}_{1}}\,\!$$ as the probability of failure for configuration 1 and use it to develop a generalized equation for $${{f}_{i}}\,\!$$ in terms of the $${{T}_{i}}\,\!$$ and $${{N}_{i}}\,\!$$. From the equation above, the expected number of failures by the end of configuration 1 is:


 * $$E\left[ {{K}_{1}} \right]=\lambda T_{1}^{\beta }={{f}_{1}}{{N}_{1}}\,\!$$


 * $$\therefore {{f}_{1}}=\frac{\lambda T_{1}^{\beta }}\,\!$$

Applying the $$E\left[ {{K}_{i}}\right]\,\!$$ equation again and noting that the expected number of failures by the end of configuration 2 is the sum of the expected number of failures in configuration 1 and the expected number of failures in configuration 2:


 * $$\begin{align}

E\left[ {{K}_{2}} \right] = & \lambda T_{2}^{\beta } \\ = & {{f}_{1}}{{N}_{1}}+{{f}_{2}}{{N}_{2}} \\ = & \lambda T_{1}^{\beta }+{{f}_{2}}{{N}_{2}} \end{align}\,\!$$


 * $$\therefore {{f}_{2}}=\frac{\lambda T_{2}^{\beta }-\lambda T_{1}^{\beta }}\,\!$$

By this method of inductive reasoning, a generalized equation for the failure probability on a configuration basis, $${{f}_{i}}\,\!$$, is obtained, such that:


 * $${{f}_{i}}=\frac{\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta }}\,\!$$

For the special case where $${{N}_{i}}=1\,\!$$ for all $$i\,\!$$, the equation above becomes a smooth curve, $${{g}_{i}}\,\!$$, that represents the probability of failure for trial by trial data, or:


 * $${{g}_{i}}=\lambda \cdot {{i}^{\beta }}-\lambda \cdot {{\left( i-1 \right)}^{\beta }}\,\!$$

In this equation, $$i\,\!$$ represents the trial number. Thus an equation for the reliability (probability of success) for the $${{i}^{th}}\,\!$$ configuration is obtained:


 * $$\begin{align}

{{R}_{i}}=1-{{f}_{i}} \end{align}\,\!$$

The equation for the reliability for the $${{i}^{th}}\,\!$$ trial is:


 * $$\begin{align}

{{R}_{i}}=1-{{g}_{i}} \end{align}\,\!$$

Maximum Likelihood Estimators for Discrete Model
This section describes procedures for estimating the parameters of the Crow-AMSAA model for success/failure data. An example is presented illustrating these concepts. The estimation procedures provide maximum likelihood estimates (MLEs) for the model's two parameters, $$\lambda \,\!$$ and $$\beta \,\!$$. The MLEs for $$\lambda \,\!$$ and $$\beta \,\!$$ allow for point estimates for the probability of failure, given by:


 * $${{\hat{f}}_{i}}=\frac{\hat{\lambda }T_{i}^-\hat{\lambda }T_{i-1}^}=\frac{\hat{\lambda }\left( T_{i}^-T_{i-1}^ \right)}\,\!$$

And the probability of success (reliability) for each configuration $$i\,\!$$ is equal to:


 * $${{\hat{R}}_{i}}=1-{{\hat{f}}_{i}}\,\!$$

The likelihood function is:


 * $$\underset{i=1}{\overset{k}{\mathop \prod }}\,\left( \begin{matrix}

{{N}_{i}} \\ {{M}_{i}} \\ \end{matrix} \right){{\left( \frac{\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta }} \right)}^}{{\left( \frac{{{N}_{i}}-\lambda T_{i}^{\beta }+\lambda T_{i-1}^{\beta }} \right)}^{{{N}_{i}}-{{M}_{i}}}}\,\!$$

Taking the natural log on both sides yields:


 * $$\begin{align}

& \Lambda = & \underset{i=1}{\overset{K}{\mathop \sum }}\,\left[ \ln \left( \begin{matrix}  {{N}_{i}}  \\   {{M}_{i}}  \\ \end{matrix} \right)+{{M}_{i}}\left[ \ln (\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })-\ln {{N}_{i}} \right] \right] \\ & & +\underset{i=1}{\overset{K}{\mathop \sum }}\,\left[ ({{N}_{i}}-{{M}_{i}})\left[ \ln ({{N}_{i}}-\lambda T_{i}^{\beta }+\lambda T_{i-1}^{\beta })-\ln {{N}_{i}} \right] \right] \end{align}\,\!$$

Taking the derivative with respect to $$\lambda \,\!$$ and $$\beta \,\!$$ respectively, exact MLEs for $$\lambda \,\!$$ and $$\beta \,\!$$ are values satisfying the following two equations:


 * $$\begin{align}

& \underset{i=1}{\overset{K}{\mathop \sum }}\,{{H}_{i}}\times {{S}_{i}}= & 0 \\ & \underset{i=1}{\overset{K}{\mathop \sum }}\,{{U}_{i}}\times {{S}_{i}}= & 0 \end{align}\,\!$$


 * where:


 * $$\begin{align}

{{H}_{i}}= & \underset{i=1}{\overset{K}{\mathop \sum }}\,\left[ T_{i}^{\beta }\ln {{T}_{i}}-T_{i-1}^{\beta }\ln {{T}_{i-1}} \right] \\ {{S}_{i}}= & \frac{\left[ \lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta } \right]}-\frac{{{N}_{i}}-{{M}_{i}}}{\left[ {{N}_{i}}-\lambda T_{i}^{\beta }+\lambda T_{i-1}^{\beta } \right]} \\ {{U}_{i}}= & T_{i}^{\beta }-T_{i-1}^{\beta }\, \end{align}\,\!$$

Discrete Model Example
A one-shot system underwent reliability growth development testing for a total of 68 trials. Delayed corrective actions were incorporated after the 14th, 33rd and 48th trials. From trial 49 to trial 68, the configuration was not changed.


 * Configuration 1 experienced 5 failures,
 * Configuration 2 experienced 3 failures,
 * Configuration 3 experienced 4 failures and
 * Configuration 4 experienced 4 failures.

The objectives are:
 * 1)	Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
 * 2)	Estimate the unreliability and reliability by configuration.

Solution


 * 1)	The solution from the MLE equations for discrete models yield $$\lambda = 0.5954\,\!$$ and $$\beta =0.7801\,\!$$.
 * 2)	The following table displays the results for probability of failure and reliability, and these results are displayed in the next two plots.

Estimated Failure Probability and Reliability by Configuration





Discrete Model for Mixed Data
In the RGA software, the Discrete Data > Mixed Data option gives a data sheet that can have input data that is either a configuration in groups or individual trial by trial, or a mixed combination of individual trials and configurations of more than one trial. The calculations use the same mathematical methods described in the Grouped Data section.

Mixed Data Example
The table below shows the number of failures of each interval of trials and the cumulative number of trials in each interval for a reliability growth test. For example, the first row indicates that for an interval of 14 trials, 5 failures occurred.

Using the RGA software, the parameters of the Crow-AMSAA model are estimated as follows:


 * $$\widehat{\beta }=0.7950\,\!$$


 * and:


 * $$\widehat{\lambda }=0.5588\,\!$$

As we have seen, the Crow-AMSAA instantaneous failure intensity, $${{\lambda }_{i}}(T)\,\!$$, is defined as:


 * $$\begin{align}

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

Using the parameter estimates, we can calculate the instantaneous unreliability at the end of the test, or $$T=68.\,\!$$


 * $${{R}_{i}}(68)=0.5588\cdot 0.7950\cdot {{68}^{0.7950-1}}=0.1871\,\!$$

This result that can be obtained from the Quick Calculation Pad (QCP), for $$T=68,\,\!$$ as seen in the following picture.



The instantaneous reliability can then be calculated as:


 * $$\begin{align}

{{R}_{inst}}=1-0.1871=0.8129 \end{align}\,\!$$

The average unreliability is calculated as:


 * $$\text{Average Unreliability }({{t}_{1,}}{{t}_{2}})=\frac{\lambda t_{2}^{\beta }-\lambda t_{1}^{\beta }}{{{t}_{2}}-{{t}_{1}}}\,\!$$

and the average reliability is calculated as:


 * $$\text{Average Reliability }({{t}_{1,}}{{t}_{2}})=1-\frac{\lambda t_{2}^{\beta }-\lambda t_{1}^{\beta }}{{{t}_{2}}-{{t}_{1}}}\,\!$$

Mixed Data Confidence Bounds
Bounds on Average Failure Probability for Mixed Data The process to calculate the average unreliability confidence bounds for mixed data is as follows:
 * 1)	Calculate the average failure probability $$({{t}_{1}},{{t}_{2}})\,\!$$.
 * 2)	There will exist a $${{t}^{*}}\,\!$$ between $${{t}_{1}}\,\!$$ and $${{t}_{2}}\,\!$$ such that the instantaneous unreliability at $${{t}^{*}}\,\!$$ equals the average unreliability $$({{t}_{1}},{{t}_{2}})\,\!$$. The confidence intervals for the instantaneous unreliability at $${{t}^{*}}\,\!$$ are the confidence intervals for the average unreliability $$({{t}_{1}},{{t}_{2}})\,\!$$.

Bounds on Average Reliability for Mixed Data The process to calculate the average reliability confidence bounds for mixed data is as follows:
 * 1)	Calculate confidence bounds for average unreliability $$({{t}_{1}},{{t}_{2}})\,\!$$ as described above.
 * 2)	The confidence bounds for reliability are 1 minus these confidence bounds for average unreliability.

Estimating the Number of Failures if Testing Continues
Six systems were subjected to a reliability growth test and a total of 81 failures were observed. The following table 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 3,000 hours?

Solution


 * 1)	The next figure shows the parameters estimated using RGA.




 * 2)	The number of failures can be estimated using the Quick Calculation Pad, as shown next. The estimated number of failures at 3,000 hours is equal to 83.2451 and 81 failures were observed during testing. Therefore, the number of additional failures generated if testing continues until 3,000 hours is equal to $$83.2451-81=2.2451\approx 3\,\!$$.



Determining Whether a Design Meets the MTBF Goal
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. The table below 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?

Failure Times Data

Solution


 * 1)	The next figure 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 next. 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.



Analyzing Mixed Data for a One-Shot System
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. The table below shows the data set obtained from the test. The first column specifies the number of failures that occurred in each interval, and the second column shows 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?

Mixed Data Solution


 * 1)	The next figure shows the parameters estimated using RGA.




 * 2)	The figure below 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)	The following plot shows the cumulative reliability with the 2-sided 90% confidence bounds.




 * 4)	The last figure 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.