# Crow-AMSAA Grouped Data Examples

## Contents

These examples appear in the Reliability Growth and Repairable System Analysis Reference book.

## Grouped Data Example 1

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

Grouped Failure Times Data
Run Number Cumulative Failures End Time(hours) $\ln{(T_i)}\,\!$ $\ln{(T_i)^2}\,\!$ $\ln{(\theta_i)}\,\!$ $\ln{(T_i)}\cdot\ln{(\theta_i)}\,\!$
1 2 200 5.298 28.072 0.693 3.673
2 3 400 5.991 35.898 1.099 6.582
3 4 600 6.397 40.921 1.386 8.868
4 11 3000 8.006 64.102 2.398 19.198
Sum = 25.693 168.992 5.576 38.321

Solution

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

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

Therefore, the intensity function becomes:

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

## Grouped Data Example 2

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.

Grouped Data for a New Helicopter System
Interval Interval Length Failures in Interval
1 0 - 62 12
2 62 -100 6
3 100 - 187 15
4 187 - 210 3
5 210 - 350 18
6 350 - 500 16

Do the following:

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 $\hat{\beta }\,\!$, must be solved numerically. Once $\hat{\beta }\,\!$ has been estimated then the value of $\hat{\lambda }\,\!$ can be determined. The parameter values are displayed below:
$\hat{\beta }= 0.81361\,\!$
$\hat{\lambda }= 0.44585\,\!$
The grouped Fisher Matrix confidence bounds can be obtained on the parameters $\hat{\beta }\,\!$ and $\hat{\lambda }\,\!$ 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 $\hat{\beta }\,\!$ and $\hat{\lambda }\,\!$ 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}} = & {{\hat{m}}_{i}}(1-W) \\ = & 6.19623 \\ {{[MTB{{F}_{i}}]}_{U}} = & {{\hat{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.