Template:Grouped data camsaa - Revision history

Lisa Hacker: Redirected page to Crow-AMSAA - NHPP

2012-08-23T12:09:49Z

← Older revision		Revision as of 12:09, 23 August 2012
Line 1:		Line 1:
	~~==Grouped Data== <!-- THIS SECTION HEADER IS LINKED FROM: Operational Mission Profile Testing. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->~~		#REDIRECT [[Crow-AMSAA - NHPP]]
	~~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>~~

	~~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>~~

Lisa Hacker: /* Grouped Data */

2012-08-23T00:57:54Z

Grouped Data

← Older revision		Revision as of 00:57, 23 August 2012
Line 1:		Line 1:
	==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>

Nicolette Young: 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: :: $\ln [E(N(T))]=\ln \la…'$

2012-01-05T23:05:24Z

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…'

New page

==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 \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'''
 
Consider the grouped failure times data given in Table 5.2. Solve for the Crow-AMSAA parameters using MLE.
 
{|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
|-
| || ||Sum =|| 25.693 ||168.992||5.576||38.321
|}

'''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>

Template:Grouped data camsaa - Revision history

Lisa Hacker: Redirected page to Crow-AMSAA - NHPP

Lisa Hacker: /* Grouped Data */

Nicolette Young: 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: ::\ln [E(N(T))]=\ln \la…'

Nicolette Young: 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: :: $\ln [E(N(T))]=\ln \la…'$