Template:Parameter estimation camsaa: Difference between revisions

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(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>{…')
 
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#REDIRECT [[Crow-AMSAA - NHPP]]
==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:
 
<br>
::<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:
 
<br>
::<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:
 
<br>
::<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:
 
<br>
::<math>\Lambda =n\ln \lambda +n\ln \beta -\lambda {{T}^{*\beta }}+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}</math>
 
<br>
And differentiating with respect to  <math>\lambda </math>  yields:
 
<br>
::<math>\frac{\partial \Lambda }{\partial \lambda }=\frac{n}{\lambda }-{{T}^{*\beta }}</math>
 
<br>
Set equal to zero and solve for  <math>\lambda </math> :
 
<br>
::<math>\widehat{\lambda }=\frac{n}{{{T}^{*\beta }}}</math>
 
<br>
Now differentiate Eqn. (amsaa4) with respect to  <math>\beta </math> :
 
<br>
::<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>
 
<br>
Set equal to zero and solve for  <math>\beta </math> :
 
<br>
::<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>
 
<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>

Latest revision as of 12:09, 23 August 2012

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