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==
==Grouped Data== <!-- THIS SECTION HEADER IS LINKED FROM: Operational Mission Profile Testing. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->
For analyzing grouped data, we follow the same logic described in Chapter 4 for the Duane model. If Eqn. (amsaa2a) is linearized:  
For analyzing grouped data, we follow the same logic described previously for the [[Duane Model|Duane model]]. If Eqn. (amsaa2a) is linearized:  


::<math>\ln [E(N(T))]=\ln \lambda +\beta \ln T</math>
::<math>\ln [E(N(T))]=\ln \lambda +\beta \ln T</math>

Revision as of 00:57, 23 August 2012

Grouped Data

For analyzing grouped data, we follow the same logic described previously for the Duane model. If Eqn. (amsaa2a) is linearized:

[math]\displaystyle{ \ln [E(N(T))]=\ln \lambda +\beta \ln T }[/math]

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

[math]\displaystyle{ \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]\displaystyle{ \lambda }[/math] based on this relationship is:

[math]\displaystyle{ \widehat{\lambda }=\frac{n}{T_{k}^{\widehat{\beta }}} }[/math]

And the estimate of [math]\displaystyle{ \beta }[/math] is the value [math]\displaystyle{ \widehat{\beta } }[/math] that satisfies:

[math]\displaystyle{ \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
Consider the grouped failure times data given in Table 5.2. Solve for the Crow-AMSAA parameters using MLE.

Table 5.2 - Grouped failure times data
Run Number Cumulative Failures End Time(hr) [math]\displaystyle{ \ln{(T_i)} }[/math] [math]\displaystyle{ \ln{(T_i)^2} }[/math] [math]\displaystyle{ \ln{(\theta_i)} }[/math] [math]\displaystyle{ \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
Sum = 25.693 168.992 5.576 38.321

Solution To obtain the estimator of [math]\displaystyle{ \beta }[/math] , Eqn. (vv) must be solved numerically for [math]\displaystyle{ \beta }[/math] . Using RGA, the value of [math]\displaystyle{ \widehat{\beta } }[/math] is [math]\displaystyle{ 0.6315 }[/math] . Now plugging this value into Eqn. (vv1), the estimator of [math]\displaystyle{ \lambda }[/math] is:

[math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{11}{3,{{000}^{0.6315}}} \\ & = & 0.0701 \end{align} }[/math]

Therefore, the intensity function becomes:

[math]\displaystyle{ \widehat{\rho }(T)=0.0701\cdot 0.6315\cdot {{T}^{-0.3685}} }[/math]