Template:Parameter estimation camsaa - Revision history

Lisa Hacker: Redirected page to Crow-AMSAA - NHPP

2012-08-23T12:09:52Z

← Older revision		Revision as of 12:09, 23 August 2012
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			#REDIRECT [[Crow-AMSAA - NHPP]]
	~~==Parameter Estimation==~~
	~~{{maximum likelihood estimators camsaa~~-~~pe}}~~

	~~{{biasing and unbiasing of beta camsaa}}~~

Nicolette Young: /* Biasing and Unbiasing of Beta */

2012-01-05T21:04:48Z

Biasing and Unbiasing of Beta

← Older revision		Revision as of 21:04, 5 January 2012
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	{{maximum likelihood estimators camsaa-pe}}		{{maximum likelihood estimators camsaa-pe}}

	~~===Biasing and Unbiasing of Beta===~~		{{biasing and unbiasing of beta camsaa}}
	Eqn. (6) returns the biased estimate of <math>\beta </math> . The unbiased estimate of <math>\beta </math> can be calculated by using the following relationships. For time terminated data (meaning that the test ends after a specified number of failures):

	~~::<math>\bar{\beta }=\frac~~{~~N-1}~~{~~N}\hat{\beta }</math>~~

	~~<br>~~
	~~For failure terminated data (meaning that the test ends after a specified test time):~~

	~~<br>~~
	~~::<math>\bar{\beta }=\frac{N-2}{N}\hat{\beta }</math>~~

	~~<br>~~
	~~'''Example 1'''~~
	~~<br>~~
	~~Two prototypes~~ of a system were tested simultaneously with design changes incorporated during the test. Table 5.1 presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.
	~~<br>~~
	~~<br>~~
	~~<center>Table 5.1 - Developmental test data for two identical systems </center>~~
	~~{\|style= align="center" border="1"~~
	~~!Failure Number~~
	~~!Failed Unit~~
	~~!Test Time Unit 1(hr)~~
	~~!Test Time Unit 2(hr)~~
	~~!Total Test Time(hr)~~
	~~!<math>ln{(T)}</math>~~
	\|-
	~~\|1\|\| 1\|\| 1.0\|\| 1.7\|\| 2.7\|\| 0.99325~~
	\|-
	~~\|2\|\| 1\|\| 7.3\|\| 3.0\|\| 10.3\|\| 2.33214~~
	\|-
	~~\|3\|\| 2\|\| 8.7\|\| 3.8\|\| 12.5\|\| 2.52573~~
	\|-
	~~\|4\|\| 2\|\| 23.3\|\| 7.3\|\| 30.6\|\| 3.42100~~
	\|-
	~~\|5\|\| 2\|\| 46.4\|\| 10.6\|\| 57.0\|\| 4.04305~~
	\|-
	~~\|6\|\| 1\|\| 50.1\|\| 11.2\|\| 61.3\|\| 4.11578~~
	\|-
	~~\|7\|\| 1\|\| 57.8\|\| 22.2\|\| 80.0\|\| 4.38203~~
	\|-
	~~\|8\|\| 2\|\| 82.1\|\| 27.4\|\| 109.5\|\| 4.69592~~
	\|-
	~~\|9\|\| 2\|\| 86.6\|\| 38.4\|\| 125.0\|\|4.82831~~
	\|-
	~~\|10\|\| 1\|\| 87.0\|\| 41.6\|\| 128.6\|\| 4.85671~~
	\|-
	~~\|11\|\| 2\|\| 98.7\|\| 45.1\|\| 143.8\|\| 4.96842~~
	\|-
	~~\|12\|\| 1\|\| 102.2\|\| 65.7\|\| 167.9\|\| 5.12337~~
	\|-
	~~\|13\|\| 1\|\| 139.2 \|\|90.0\|\|229.2\|\| 5.43459~~
	\|-
	~~\|14\|\| 1\|\| 166.6\|\| 130.1\|\| 296.7\|\| 5.69272~~
	\|-
	~~\|15\|\| 2\|\| 180.8\|\| 139.8 \|\|320.6\|\|5.77019~~
	\|-
	~~\|16\|\| 1\|\| 181.3\|\| 146.9\|\| 328.2\|\| 5.79362~~
	\|-
	~~\|17\|\| 2\|\| 207.9\|\| 158.3 \|\|366.2\|\|5.90318~~
	\|-
	~~\|18\|\| 2\|\| 209.8\|\| 186.9\|\| 396.7\|\| 5.98318~~
	\|-
	~~\|19\|\| 2\|\| 226.9\|\| 194.2\|\| 421.1\|\| 6.04287~~
	\|-
	~~\|20\|\| 1\|\| 232.2\|\| 206.0\|\| 438.2\|\| 6.08268~~
	\|-
	~~\|21\|\| 2\|\| 267.5\|\| 233.7\|\| 501.2\|\| 6.21701~~
	\|-
	~~\|22\|\| 2\|\| 330.1\|\| 289.9\|\| 620.0\|\| 6.42972~~
	\|}

	~~'''Solution'''~~
	~~<br>~~
	~~For the failure terminated test, using Eqn. (amsaa6):~~

	~~::<math>\widehat{\~~beta }~~=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}}</math>~~

	~~:where:~~

	~~<br>~~
	~~::<math>\underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355</math>~~

	~~<br>~~
	~~:Then:~~

	~~<br>~~
	~~::<math>\widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142</math>~~

	~~<br>~~
	~~From Eqn. (amsaa5):~~

	~~<br>~~
	~~::<math>\widehat{\lambda }=\frac{22}{{{620}^{0.6142}}}=0.4239</math>~~

	~~<br>~~
	~~Therefore, <math>{{\lambda }_{i}}(T)</math> becomes:~~

	~~<br>~~
	~~::<math>\begin{align}~~
	~~& {{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\~~
	~~& = & 0.0217906\frac{\text{failures}}{\text{hr}}~~
	~~\end{align}</math>~~

	~~<br>~~
	~~Figure 4fig81 shows the plot of the failure rate. If no further changes are made, the estimated MTBF is <math>\tfrac{1}{0.0217906~~}~~</math> or 46 hr.~~
	~~<br>~~
	~~<br>~~
	~~[[Image:rga5.1.png\|thumb\|center\|400px\|Failure rate plot for Example 5-1 using Maximum Likelihood Estimation.]]~~
	~~<br>~~

Nicolette Young: /* Maximum Likelihood Estimators */

2012-01-05T21:04:09Z

Maximum Likelihood Estimators

@@ Line 1: / Line 1: @@
 ==Parameter Estimation==
-===Maximum Likelihood Estimators===
+{{maximum likelihood estimators camsaa-pe}}
 ===Biasing and Unbiasing of Beta===

Nicolette Young: Created page with ' ==Parameter Estimation== ===Maximum Likelihood Estimators=== The probability density function ( $pdf$ ) of the ${{i}^{th}}$ event given that the ${…'$

2012-01-05T21:03:24Z

Created page with ' ==Parameter Estimation== ===Maximum Likelihood Estimators=== The probability density function ( <math>pdf</math> ) of the <math>{{i}^{th}}</math> event given that the <math>{…'

New page

==Parameter Estimation==
===Maximum Likelihood Estimators===
The probability density function ( <math>pdf</math> ) of the <math>{{i}^{th}}</math> event given that the <math>{{(i-1)}^{th}}</math> event occurred at <math>{{T}_{i-1}}</math> is:

 
::<math>f({{T}_{i}}|{{T}_{i-1}})=\frac{\beta }{\eta }{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta -1}}\cdot {{e}^{-\tfrac{1}{{{\eta }^{\beta }}}\left( T_{i}^{\beta }-T_{i-1}^{\beta } \right)}}</math>

The likelihood function is:

 
::<math>L={{\lambda }^{n}}{{\beta }^{n}}{{e}^{-\lambda {{T}^{*\beta }}}}\underset{i=1}{\overset{n}{\mathop \prod }}\,T_{i}^{\beta -1}</math>

where <math>{{T}^{*}}</math> is the termination time and is given by:

 
::<math>{{T}^{*}}=\left\{ \begin{matrix}
{{T}_{n}}\text{ if the test is failure terminated} \\
T>{{T}_{n}}\text{ if the test is time terminated} \\
\end{matrix} \right\}</math>

Taking the natural log on both sides:

 
::<math>\Lambda =n\ln \lambda +n\ln \beta -\lambda {{T}^{*\beta }}+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}</math>

 
And differentiating with respect to <math>\lambda </math> yields:

 
::<math>\frac{\partial \Lambda }{\partial \lambda }=\frac{n}{\lambda }-{{T}^{*\beta }}</math>

 
Set equal to zero and solve for <math>\lambda </math> :

 
::<math>\widehat{\lambda }=\frac{n}{{{T}^{*\beta }}}</math>

 
Now differentiate Eqn. (amsaa4) with respect to <math>\beta </math> :

 
::<math>\frac{\partial \Lambda }{\partial \beta }=\frac{n}{\beta }-\lambda {{T}^{*\beta }}\ln {{T}^{*}}+\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}</math>

 
Set equal to zero and solve for <math>\beta </math> :

 
::<math>\widehat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}}</math>

===Biasing and Unbiasing of Beta===
Eqn. (6) returns the biased estimate of <math>\beta </math> . The unbiased estimate of <math>\beta </math> can be calculated by using the following relationships. For time terminated data (meaning that the test ends after a specified number of failures):

::<math>\bar{\beta }=\frac{N-1}{N}\hat{\beta }</math>

 
For failure terminated data (meaning that the test ends after a specified test time):

 
::<math>\bar{\beta }=\frac{N-2}{N}\hat{\beta }</math>

 
'''Example 1'''
 
Two prototypes of a system were tested simultaneously with design changes incorporated during the test. Table 5.1 presents the data collected over the entire test. Find the Crow-AMSAA parameters and the intensity function using maximum likelihood estimators.
 
 
<center>Table 5.1 - Developmental test data for two identical systems </center>
{|style= align="center" border="1"
!Failure Number
!Failed Unit
!Test Time Unit 1(hr)
!Test Time Unit 2(hr)
!Total Test Time(hr)
!<math>ln{(T)}</math>
|-
|1|| 1|| 1.0|| 1.7|| 2.7|| 0.99325
|-
|2|| 1|| 7.3|| 3.0|| 10.3|| 2.33214
|-
|3|| 2|| 8.7|| 3.8|| 12.5|| 2.52573
|-
|4|| 2|| 23.3|| 7.3|| 30.6|| 3.42100
|-
|5|| 2|| 46.4|| 10.6|| 57.0|| 4.04305
|-
|6|| 1|| 50.1|| 11.2|| 61.3|| 4.11578
|-
|7|| 1|| 57.8|| 22.2|| 80.0|| 4.38203
|-
|8|| 2|| 82.1|| 27.4|| 109.5|| 4.69592
|-
|9|| 2|| 86.6|| 38.4|| 125.0||4.82831
|-
|10|| 1|| 87.0|| 41.6|| 128.6|| 4.85671
|-
|11|| 2|| 98.7|| 45.1|| 143.8|| 4.96842
|-
|12|| 1|| 102.2|| 65.7|| 167.9|| 5.12337
|-
|13|| 1|| 139.2 ||90.0||229.2|| 5.43459
|-
|14|| 1|| 166.6|| 130.1|| 296.7|| 5.69272
|-
|15|| 2|| 180.8|| 139.8 ||320.6||5.77019
|-
|16|| 1|| 181.3|| 146.9|| 328.2|| 5.79362
|-
|17|| 2|| 207.9|| 158.3 ||366.2||5.90318
|-
|18|| 2|| 209.8|| 186.9|| 396.7|| 5.98318
|-
|19|| 2|| 226.9|| 194.2|| 421.1|| 6.04287
|-
|20|| 1|| 232.2|| 206.0|| 438.2|| 6.08268
|-
|21|| 2|| 267.5|| 233.7|| 501.2|| 6.21701
|-
|22|| 2|| 330.1|| 289.9|| 620.0|| 6.42972
|}

'''Solution'''
 
For the failure terminated test, using Eqn. (amsaa6):

::<math>\widehat{\beta }=\frac{22}{22\ln 620-\underset{i=1}{\overset{22}{\mathop{\sum }}}\,\ln {{T}_{i}}}</math>

:where:

 
::<math>\underset{i=1}{\overset{22}{\mathop \sum }}\,\ln {{T}_{i}}=105.6355</math>

 
:Then:

 
::<math>\widehat{\beta }=\frac{22}{22\ln 620-105.6355}=0.6142</math>

 
From Eqn. (amsaa5):

 
::<math>\widehat{\lambda }=\frac{22}{{{620}^{0.6142}}}=0.4239</math>

 
Therefore, <math>{{\lambda }_{i}}(T)</math> becomes:

 
::<math>\begin{align}
& {{\widehat{\lambda }}_{i}}(T)= & 0.4239\cdot 0.6142\cdot {{620}^{-0.3858}} \\
& = & 0.0217906\frac{\text{failures}}{\text{hr}}
\end{align}</math>

 
Figure 4fig81 shows the plot of the failure rate. If no further changes are made, the estimated MTBF is <math>\tfrac{1}{0.0217906}</math> or 46 hr.
 
 
[[Image:rga5.1.png|thumb|center|400px|Failure rate plot for Example 5-1 using Maximum Likelihood Estimation.]]

Template:Parameter estimation camsaa - Revision history

Lisa Hacker: Redirected page to Crow-AMSAA - NHPP

Nicolette Young: /* Biasing and Unbiasing of Beta */

Nicolette Young: /* Maximum Likelihood Estimators */

Nicolette Young: Created page with ' ==Parameter Estimation== ===Maximum Likelihood Estimators=== The probability density function ( pdf ) of the {{i}^{th}} event given that the {…'

Nicolette Young: Created page with ' ==Parameter Estimation== ===Maximum Likelihood Estimators=== The probability density function ( $pdf$ ) of the ${{i}^{th}}$ event given that the ${…'$