Crow-AMSAA Model Examples

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

Parameter Estimation Examples

Parameter Estimation for Failure Times Data

A prototype of a system was tested 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

Row	Time to Event (hr)	$ln{(T)}\,\!$
1	2.7	0.99325
2	10.3	2.33214
3	12.5	2.52573
4	30.6	3.42100
5	57.0	4.04305
6	61.3	4.11578
7	80.0	4.38203
8	109.5	4.69592
9	125.0	4.82831
10	128.6	4.85671
11	143.8	4.96842
12	167.9	5.12337
13	229.2	5.43459
14	296.7	5.69272
15	320.6	5.77019
16	328.2	5.79362
17	366.2	5.90318
18	396.7	5.98318
19	421.1	6.04287
20	438.2	6.08268
21	501.2	6.21701
22	620.0	6.42972

Solution

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

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

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 ${\lambda }\,\!$ :

${\begin{aligned}{\widehat {\lambda }}&={\frac {n}{{T}^{*\beta }}}\\&={\frac {22}{{620}^{0.6142}}}=0.4239\\\end{aligned}}\,\!$

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

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

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.

Example - Confidence Bounds on Failure Intensity

Using the values of ${\hat {\beta }}\,\!$ and ${\hat {\lambda }}\,\!$ estimated in the example given above, calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity.

Solution

Fisher Matrix Bounds

The partial derivatives for the Fisher Matrix confidence bounds are:

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

The Fisher Matrix then becomes:

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

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

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

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

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

Therefore, the variances become:

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

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

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

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

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

The confidence bounds for the instantaneous failure intensity are:

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

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

Crow Bounds

Given that the data is failure terminated, the Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for $T=620\,\!$ hours are:

${\begin{aligned}{{[{{\lambda }_{c}}(T)]}_{L}}=&{\frac {\chi _{{\tfrac {\alpha }{2}},2N}^{2}}{2\cdot t}}\\=&{\frac {29.787476}{2*620}}\\=&0.02402\\{{[{{\lambda }_{c}}(T)]}_{U}}=&{\frac {\chi _{1-{\tfrac {\alpha }{2}},2N}^{2}}{2\cdot t}}\\=&{\frac {60.48089}{2*620}}\\=&0.048775\end{aligned}}\,\!$

The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for $T=620\,\!$ hours are calculated by first estimating the bounds on the instantaneous MTBF. Once these are calculated, take the inverse as shown below. Details on the confidence bounds for instantaneous MTBF are presented here.

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

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

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

Example - Confidence Bounds on MTBF

Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data from the example given above.

Solution

Fisher Matrix Bounds

From the previous example:

${\begin{aligned}Var({\hat {\lambda }})=&0.13519969\\Var({\hat {\beta }})=&0.017105343\\Cov({\hat {\beta }},{\hat {\lambda }})=&-0.046614609\end{aligned}}\,\!$

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

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

Therefore, the variances become:

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

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

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

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

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

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:

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

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

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 ${\hat {\beta }}\,\!$ and ${\hat {\lambda }}\,\!$. For Fisher Matrix confidence bounds:

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

and:

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

For Crow confidence bounds:

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

and:

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

Parameter Estimation for Discrete Data

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.

Do the following:

1. Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
2. Estimate the unreliability and reliability by configuration.

Solution

1. The parameter estimates for the Crow-AMSAA model using the parameter estimation for discrete data methodology yields $\lambda =0.5954\,\!$ 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

   Configuration($i\,\!$)	Estimated Failure Probability	Estimated Reliability
   1	0.333	0.667
   2	0.234	0.766
   3	0.206	0.794
   4	0.190	0.810

   
   

Parameter Estimation for Discrete-Mixed Data

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.

Mixed Data

Failures in Interval	Cumulative Trials
5	14
3	33
4	48
0	52
1	53
0	57
1	58
0	62
1	63
0	67
1	68

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

${\hat {\beta }}=0.7950\,\!$

and:

${\hat {\lambda }}=0.5588\,\!$

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

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

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:

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

Failure Times Data Example

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

2.6	56.5	98.1	190.7
16.5	63.1	101.1	193
16.5	70.6	132	198.7
17	73	142.2	251.9
21.4	77.7	147.7	282.5
29.1	93.9	149	286.1
33.3	95.5	167.2

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.
   

Failure Times Grouped Data Examples

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:

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

Therefore, the intensity function becomes:

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

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:

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

   and:

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

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

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

   and:

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

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:

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

   and:

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

   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:

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

   and:

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

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

   
   

Discrete Mixed Data Example

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

Failures in Interval	Cumulative Trials	Failures in Interval	Cumulative Trials
3	4	1	25
0	5	1	28
4	9	0	32
1	12	2	37
0	13	0	39
1	15	1	40
2	19	1	44
1	20	0	46
1	22	1	49
0	24	0	50

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 72.70% with an upper and lower 2-sided 90% confidence bound of 82.36% and 39.59%, 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.
   

Multiple Systems: Concurrent Operating Times Example

Six systems were subjected to a reliability growth test, and a total of 82 failures were observed. Given the data in the table below, which presents the start/end times and times-to-failure for each system, do the following:

1. Estimate the parameters of the Crow-AMSAA model using maximum likelihood estimation.
2. Determine how many additional failures would be generated if testing continues until 3,000 hours.

Multiple Systems (Concurrent Operating Times) Data

System #	1	2	3	4	5	6
Start Time (Hr)	0	0	0	0	0	0
End Time (Hr)	504	541	454	474	436	500
Failure Times (Hr)	21	83	26	36	23	7
	29	83	26	306	46	13
	43	83	57	306	127	13
	43	169	64	334	166	31
	43	213	169	354	169	31
	66	299	213	395	213	82
	115	375	231	403	213	109
	159	431	231	448	255	137
	199		231	456	369	166
	202		231	461	374	200
	222		304		380	210
	248		383		415	220
	248					301
	255					422
	286					437
	286					469
	304					469
	320
	348
	364
	404
	410
	429

Solution

1. To estimate the parameters ${\hat {\beta }}\,\!$ and ${\hat {\lambda }}\,\!$, the equivalent single system (ESS) must first be determined. The ESS is given below: Equivalent Single System

   Row	Time to Event (hr)		Row	Time to Event (hr)		Row	Time to Event (hr)		Row	Time to Event (hr)
   1	42		22	498		43	1386		64	2214
   2	78		23	654		44	1386		65	2244
   3	78		24	690		45	1386		66	2250
   4	126		25	762		46	1386		67	2280
   5	138		26	822		47	1488		68	2298
   6	156		27	954		48	1488		69	2370
   7	156		28	996		49	1530		70	2418
   8	174		29	996		50	1530		71	2424
   9	186		30	1014		51	1716		72	2460
   10	186		31	1014		52	1716		73	2490
   11	216		32	1014		53	1794		74	2532
   12	258		33	1194		54	1806		75	2574
   13	258		34	1200		55	1824		76	2586
   14	258		35	1212		56	1824		77	2621
   15	276		36	1260		57	1836		78	2676
   16	342		37	1278		58	1836		79	2714
   17	384		38	1278		59	1920		80	2734
   18	396		39	1278		60	2004		81	2766
   19	492		40	1278		61	2088		82	2766
   20	498		41	1320		62	2124
   21	498		42	1332		63	2184

   Given the ESS data, the value of ${\hat {\beta }}\,\!$ is calculated using:

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

   ${\hat {\beta }}=0.8939\,\!$

   where $n\,\!$ is the number of failures and $T^{*}\,\!$ is the termination time. The termination time is the sum of end times for each of the systems, which equals 2,909.

   ${\hat {\lambda }}\,\!$ is estimated with:

   ${\hat {\lambda }}={\frac {n}{{{T}^{*}}^{\beta }}}$

   ${\hat {\lambda }}=0.0657\,\!$

   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 84.2892 and 82 failures were observed during testing. Therefore, the number of additional failures generated if testing continues until 3,000 hours is equal to $84.2892-82=2.2892\approx 3\,\!$