Template:Grouped data camsaa: Difference between revisions

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(Created page with '==Grouped Data== For analyzing grouped data, we follow the same logic described in Chapter 4 for the Duane model. If Eqn. (amsaa2a) is linearized: ::<math>\ln [E(N(T))]=\ln \la…')
 
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==Grouped Data==
#REDIRECT [[Crow-AMSAA - NHPP]]
For analyzing grouped data, we follow the same logic described in Chapter 4 for the Duane model. If Eqn. (amsaa2a) is linearized:
 
::<math>\ln [E(N(T))]=\ln \lambda +\beta \ln T</math>
 
According to Crow [9], the likelihood function for the grouped data case, (where  <math>{{n}_{1}},</math>  <math>{{n}_{2}},</math>  <math>{{n}_{3}},\ldots ,</math>  <math>{{n}_{k}}</math>  failures are observed and  <math>k</math>  is the number of groups), is:
 
::<math>\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}}!}</math>
 
And the MLE of  <math>\lambda </math>  based on this relationship is:
 
::<math>\widehat{\lambda }=\frac{n}{T_{k}^{\widehat{\beta }}}</math>
 
And the estimate of  <math>\beta </math>  is the value  <math>\widehat{\beta }</math>  that satisfies:
 
::<math>\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</math>
 
'''Example 4'''
<br>
Consider the grouped failure times data given in Table 5.2. Solve for the Crow-AMSAA parameters using MLE.
<br>
{|style= align="center" border="1"
|+'''Table 5.2 - Grouped failure times data'''
!Run Number
!Cumulative Failures
!End Time(hr)
!<math>\ln{(T_i)}</math>
!<math>\ln{(T_i)^2}</math>
!<math>\ln{(\theta_i)}</math>
!<math>\ln{(T_i)}\cdot\ln{(\theta_i)}</math>
|-
|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
|-
| ||   ||<span style="color:blue">Sum =</span>|| <span style="color:blue">25.693</span> ||<span style="color:blue">168.992</span>||<span style="color:blue">5.576</span>||<span style="color:blue">38.321</span>
|}
 
'''Solution'''
To obtain the estimator of  <math>\beta </math> , Eqn. (vv) must be solved numerically for  <math>\beta </math> . Using RGA, the value of  <math>\widehat{\beta }</math>  is  <math>0.6315</math> . Now plugging this value into Eqn. (vv1), the estimator of  <math>\lambda </math>  is:
 
::<math>\begin{align}
  & \widehat{\lambda }= & \frac{11}{3,{{000}^{0.6315}}} \\
& = & 0.0701 
\end{align}</math>
 
Therefore, the intensity function becomes:
 
::<math>\widehat{\rho }(T)=0.0701\cdot 0.6315\cdot {{T}^{-0.3685}}</math>

Latest revision as of 12:09, 23 August 2012

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