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== Estimation of the Weibull Parameters ==
#REDIRECT [[The Weibull Distribution]]
 
The estimates of the parameters of the Weibull distribution can be found graphically via probability plotting paper, or analytically, either using least squares or maximum likelihood.
 
{{weibull parameters probability plotting}}
 
===Example 2===
 
Six identical units are reliability tested under the same stresses and conditions. All units are tested to failure and the following times-to-failure are recorded: 48, 66, 85, 107, 125 and 152 hours. Find the parameters of the three-parameter Weibull distribution using probability plotting.
 
==== Solution to Example 2 ====
 
The following figure shows the results. Note that since the original data set was concave down, 17.26 was subtracted from all the times-to-failure and replotted, resulting in a straight line, thus <span class="texhtml">γ = 17.26</span>. (We used Weibull++ to get the results. To perform this by hand, one would attempt different values of <span class="texhtml">γ,</span> using a trial and error methodology, until an acceptable straight line is found. When performed manually, you do not expect decimal accuracy.)
<br>
[[Image:lda6.9.gif|thumb|center|500px|]]
 
=== Rank Regression on Y ===
Performing rank regression on Y requires that a straight line mathematically be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized. This is in essence the same methodology as the probability plotting method, except that we use the principle of least squares to determine the line through the points, as opposed to just eyeballing it. The first step is to bring our function into a linear form. For the two-parameter Weibull distribution, the  (cumulative density function) is:
 
::<math> F(T)=1-e^{-\left( \frac{T}{\eta }\right) ^{\beta }}  (Fw) </math>
 
Taking the natural logarithm of both sides of the equation yields:
 
::<math>\ln[ 1-F(T)] =-( \frac{T}{\eta }) ^{\beta } </math>
 
::<math> \ln{ -\ln[ 1-F(T)]} =\beta \ln ( \frac{T}{ \eta }) </math>
 
:or:
 
::<math> \ln \{ -\ln[ 1-F(T)]\} =-\beta \ln (\eta )+\beta \ln (T) EQNREF logw </math>
 
:Now let:
 
::<math> y = \ln \{ -\ln[ 1-F(T)]\}  ( yw )</math>
 
::<math> a = − βln(\eta) </math>  (aw)
 
:and:
 
::<math> b= \beta</math>  ( bw  )
 
which results in the linear equation of:
 
::<math>y=a+bx</math>
 
The least squares parameter estimation method (also known as regression analysis) was discussed in Chapter 3 and the following equations for regression on Y were derived in Appendix A:
 
::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} EQNREF aaw </math>
 
:and:
 
::<math> \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}x_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}x_{i}\right) ^{2}}{N}}} EQNREF bbw </math>
 
In this case the equations for <span class="texhtml">''y''<sub>''i''</sub></span> and <span class="texhtml">''x''<sub>''i''</sub></span> are:
 
::<math> y_{i}=\ln \left\{ -\ln [1-F(T_{i})]\right\} , </math>
 
:and:
 
::<span class="texhtml">''x''<sub>''i''</sub> = ln(''T''<sub>''i''</sub>).</span>
 
The <math> F(T_{i})^{\prime }s </math> are estimated from the median ranks.
 
Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, then <math> \hat{\beta } </math> and <math> \hat{\eta } </math> can easily be obtained from Eqns. (EQNREF aw ) and (\ref {bw}).
 
==== The Correlation Coefficient ====
The correlation coefficient is defined as follows:
 
::<math> \rho ={\frac{\sigma _{xy}}{\sigma _{x}\sigma _{y}}} </math>
 
where, <span class="texhtml">σ<sub>''x'' ''y''</sub> = </span>covariance of  and , <span class="texhtml">σ<sub>''x''</sub> = </span>standard deviation of , and <span class="texhtml">σ<sub>''y''</sub> = </span>standard deviation of . The estimator of <span class="texhtml">ρ</span> is the sample ''correlation coefficient'', <math> \hat{\rho} </math>, given by:
 
::<math> \hat{\rho}=\frac{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})(y_{i}-\overline{y} )}{\sqrt{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})^{2}\cdot \sum\limits_{i=1}^{N}(y_{i}-\overline{y})^{2}}} EQNREF RHOw </math>
<br>
 
===== Example 3 =====
Consider the data in Example 1, where six units were tested to failure and the following failure times were recorded: 16, 34, 53, 75, 93 and 120 hours. Estimate the parameters and the correlation coefficient using rank regression on Y, assuming that the data follow the two-parameter Weibull distribution.
 
===== Solution to Example 3 =====
Construct a table as shown below.
 
{|align="center" border=1 cellspacing=1
|-
|colspan="8" style="text-align:center"| Table 6.1 - Least Squares Analysis
|-
!<math>N</math>
!<math>T_{i}</math>
!<math>ln(T_{i})</math>
!<math>F(T_i)</math>
!<math>y_{i}</math>
!<math>(ln{T_i})^2</math>
!<math>{y_i}^2</math>
!<math>(ln{T_i})y_i</math>
|-
|1 ||16||2.7726||0.1091||-2.1583||7.6873||4.6582||-5.9840
|-
|2 ||34||3.5264||0.2645||-1.1802||12.4352||1.393||-4.1620
|-
|3 ||53||3.9703||0.4214||-0.6030||15.7632||0.3637||-2.3943
|-
|4 ||75||4.3175||0.5786||-0.146||18.6407||0.0213||-0.6303
|-
|5 ||93||4.5326||0.7355||0.2851||20.5445||0.0813||1.2923
|-
|6 ||120||4.7875||0.8909||0.7955||22.9201||0.6328||3.8083
|-
|<math>\sum</math>||  ||23.9068|| ||-3.007||97.9909||7.1502||-8.0699
|}
 
 
Utilizing the values from Table 6.1, calculate <math> \hat{a} </math> and <math> \hat{b} </math> using Eqns. (EQNREF aaw ) and (EQNREF bbw ):
::<math> \hat{b} =\frac{\sum\limits_{i=1}^{6}(\ln T_{i})y_{i}-(\sum\limits_{i=1}^{6}\ln T_{i})(\sum\limits_{i=1}^{6}y_{i})/6}{ \sum\limits_{i=1}^{6}(\ln T_{i})^{2}-(\sum\limits_{i=1}^{6}\ln T_{i})^{2}/6}
</math>
 
::<math> \hat{b}=\frac{-8.0699-(23.9068)(-3.0070)/6}{97.9909-(23.9068)^{2}/6} </math>
 
:or
 
::<math> \hat{b}=1.4301 </math>
 
:and:
 
::<math> \hat{a}=\overline{y}-\hat{b}\overline{T}=\frac{\sum \limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{\sum\limits_{i=1}^{N}\ln T_{i}}{N } </math>
 
:or:
 
::<math> \hat{a}=\frac{(-3.0070)}{6}-(1.4301)\frac{23.9068}{6}=-6.19935 </math>
 
Therefore, from Eqn. (EQNREF bw ):
 
::<math> \hat{\beta }=\hat{b}=1.4301 </math>
 
and from Eqn. (EQNREF aw ):
 
::<math> \hat{\eta }=e^{-\frac{\hat{a}}{\hat{b}}}=e^{-\frac{(-6.19935)}{ 1.4301}} </math>
 
:or:
 
::<math> \hat{\eta }=76.318\text{ hr} </math>
 
The correlation coefficient can be estimated using Eqn. (EQNREF RHOw ):
 
::<math> \hat{\rho }=0.9956 </math>
 
The above example can be repeated using Weibull++. Start Weibull++ and create a new ''Data Folio''.
 
[[Image:projectwizardweibull.png|thumb|center|400px|]]
 
Select the '' Times-to-failure data'' option.
 
[[Image:times--to-failure.png|thumb|center|400px| The effect of the Weibull shape parameter on the <math>pdf</math>. ]]
 
Enter the times-to-failure in the datasheet (ignore the Subset ID column), as shown next. The times-to-failure need not be sorted, Weibull++ will automatically sort the data.
<br>
[[Image:weibullfolio16.png|thumb|center|400px| ]]
 
Select the desired method of analysis. Note that we are assuming that the underlying distribution is the Weibull, so make sure that the Weibull distribution is selected. Under ''Parameters/Type'' on the Main page, select ''2''.
<br>
[[Image:parameterweibull.png|thumb|center|400px|]]
 
Also, so that you get the same results as this example, switch to the ''Analysis'' page and make sure you are using the ''Rank Regression on Y (RRY)'' calculation method with this example, as shown next.
 
[[Image:weibullanalysis.png|thumb|center|400px| ]]
 
Note that this can also be done from the ''Main'' page by clicking the left bottom box under the Results area. Each time you click that box you will see the method switch between MLE, RRX, and RRY. Click the ''Calculate'' icon,
 
[[Image:calculateicon.png|thumb|center|400px|]]
or select ''Calculate'' from the ''Data'' menu. The results will appear in the Data Folio's ''Results area''. The next figure shows the results for this example.
 
[[Image:weibullfolio16calculate.png|thumb|center|400px| ]]
 
<br>
You can now plot the results by clicking the ''Plot'' icon,
 
[[Image:ploticon.png|thumb|center|400px| ]]
 
or by selecting ''Plot Probability'' from the ''Data'' menu.
 
The Weibull probability plot for these data is shown next.
 
The confidence bounds, as determined from the Fisher matrix, can also be plotted. Select ''Confidence Bounds'' from the ''Plot'' menu, choose ''Two-Sided'' under ''Sides,'' ''Reliability (Type II)'' under ''Type'' and enter ''90'' for the ''Confidence level.
 
[[Image:confidenceboundsweibull.png|thumb|center|400px| ]]
 
[[Image:weibullconfidencebounds.png|thumb|center|400px| ]]
The plot will appear as follows,
 
[[Image:weibullfolio16plot.png|thumb|center|400px| ]]
 
If desired, the Weibull  <math>pdf</math> representing these data can be written as:
 
::<math> f(T)={\frac{\beta }{\eta }}\left( {\frac{T}{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{T}{\eta }}\right) ^{\beta }} </math>
 
:or:
 
::<math> f(T)={\frac{1.4302}{76.317}}\left( {\frac{T}{76.317}}\right) ^{0.4302}e^{-\left( {\frac{T}{76.317}}\right) ^{1.4302}} </math> You can also plot the Weibull  by selecting ''Pdf Plot'' from the ''Plot Type'' drop-down menu on the control panel to the right of the plot area.
 
[[Image:weibullpdffolio16.png|thumb|center|400px]]
 
From this point on, different results, reports and plots can be obtained.
 
=== Rank Regression on X ===
 
Performing a rank regression on X is similar to the process for rank regression on Y, with the difference being that the ''horizontal'' deviations from the points to the line are minimized rather than the vertical. Again, the first task is to bring our  function, Eqn. (EQNREF Fw ), into a linear form. This step is exactly the same as in the regression on Y analysis and Eqns. (EQNREF logw ), (EQNREF yw ), (EQNREF aw ) and (EQNREF bw ) apply in this case too. The derivation from the previous analysis begins on the least squares fit part, where in this case we treat  as the dependent variable and  as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter 3), is the straight line:
 
::<math> x= \hat{a}+\hat{b}y </math> EQNREF xlinew
 
The corresponding equations for <math> \hat{a} </math> and <math> \hat{b} </math> are:
 
::<math> \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\sum\limits_{i=1}^{N}x_{i}}{N} -\hat{b}\frac{\sum\limits_{i=1}^{N}y_{i}}{N} </math>
 
:and
 
::<math> \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}y_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}y_{i}\right) ^{2}}{N}}} </math>
 
:where:
 
::<math> y_{i}=\ln \left\{ -\ln [1-F(T_{i})]\right\} </math> and:
 
<span class="texhtml">''x''<sub>''i''</sub> = ln(''T''<sub>''i''</sub>)</span> and the <span class="texhtml">''F''(''T''<sub>''i''</sub>)</span> values are again obtained from the median ranks.
 
Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, solve Eqn. (EQNREF xlinew ) for <span class="texhtml">''y'',</span> which corresponds to:
 
::<math> y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}}x </math> Solving for the parameters from Eqns. (EQNREF aw ) and (EQNREF bw ) we get
 
::<math> a=-\frac{\hat{a}}{\hat{b}}=-\beta \ln (\eta )</math>  EQNREF awx
 
:and
 
::<math> b=\frac{1}{\hat{b}}=\beta EQNREF bwx </math> The correlation coefficient is evaluated as before using Eqn. (EQNREF RHOw ).
 
==== Example 4 ====
Repeat Example 1 using rank regression on X.
 
===== Solution to Example 4 =====
Solution to Example 4 Table 6.1, constructed in Example 3, can also be applied to this example.
 
Using the values from this table we get:
 
::<math> \hat{b} ={\frac{\sum\limits_{i=1}^{6}(\ln T_{i})y_{i}-\frac{ \sum\limits_{i=1}^{6}\ln T_{i}\sum\limits_{i=1}^{6}y_{i}}{6}}{ \sum\limits_{i=1}^{6}y_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{6}y_{i}\right) ^{2}}{6}}}
</math>
 
::<math>\hat{b} =\frac{-8.0699-(23.9068)(-3.0070)/6}{7.1502-(-3.0070)^{2}/6} </math>
 
:or:
 
::<math> \hat{b}=0.6931 </math>
 
:and:
 
::<math> \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\sum\limits_{i=1}^{6}\ln T_{i} }{6}-\hat{b}\frac{\sum\limits_{i=1}^{6}y_{i}}{6} </math>
 
:or:
 
::<math> \hat{a}=\frac{23.9068}{6}-(0.6931)\frac{(-3.0070)}{6}=4.3318 </math>
 
Therefore, from Eqn. (EQNREF bwx ):
 
::<math> \hat{\beta }=\frac{1}{\hat{b}}=\frac{1}{0.6931}=1.4428 </math>
 
:and from Eqn. (EQNREF awx )
 
::<math> \hat{\eta }=e^{\frac{\hat{a}}{\hat{b}}\cdot \frac{1}{\hat{ \beta }}}=e^{\frac{4.3318}{0.6931}\cdot \frac{1}{1.4428}}=76.0811\text{ hr} </math>
 
The correlation coefficient is found using Eqn. (EQNREF RHOw ):
 
::<math> \hat{\rho }=0.9956 </math>
 
The results and the associated graph using Weibull++ are given next. Note that the slight variation in the results is due to the number of significant figures used in the estimation of the median ranks. Weibull++ by default uses double precision accuracy when computing the median ranks.
 
[[Image:onevariableplot.png|thumb|center|400px| ]]
 
<br>
 
=== Three-Parameter Weibull Regression ===
 
When the MR versus <span class="texhtml">''T''<sub>''j''</sub></span> points plotted on the Weibull probability paper do not fall on a satisfactory straight line and the points fall on a curve,(Note that other shapes, particularly shapes, might suggest the existence of more than one population. In these cases, the multiple population, mixed Weibull distribution, may be more appropriate. Chapter 10 presents the mixed Weibull distribution.) then a location parameter, <span class="texhtml">γ</span>, might exist which may straighten out these points. The goal in this case is to fit a curve, instead of a line, through the data points using nonlinear regression. The Gauss-Newton method can be used to solve for the parameters, <span class="texhtml">β</span>, <span class="texhtml">η</span> and <span class="texhtml">γ</span>, by performing a Taylor series expansion on <span class="texhtml">''F''(''T''<sub>''i''</sub>;β,η,γ)</span>. Then the nonlinear model is approximated with linear terms and ordinary least squares are employed to estimate the parameters. This procedure is iterated until a satisfactory solution is reached. Weibull++ 7 calculates the value of <span class="texhtml">γ</span> by utilizing an optimized Nelder-Mead algorithm, and adjusts the points by this value of <span class="texhtml">γ</span> such that they fall on a straight line, and then plots both the adjusted and the original unadjusted points. To draw a curve through the original unadjusted points, if so desired, select Weibull 3P Line Unadjusted for Gamma from the ''Show Plot Line'' submenu under the ''Plot Options'' menu.  The returned estimations of the parameters are the same when selecting RRX or RRY. To display the unadjusted data points and line along with the adjusted data points and line, select ''Show/Hide Items'' under the ''Plot Options ''menu and include the unadjusted data points and line as follows:
 
[[Image:showhideplotitems.png|thumb|center|300px]]
 
[[Image:showhideplotwindow.png|thumb|center|300px]]
 
The results and the associated graph for the previous example using the three-parameter Weibull case are shown next:
 
[[Image:3parameterweibullplot.png|thumb|center|400px| ]]
<br>
 
=== Maximum Likelihood Estimation ===
 
As outlined in Chapter 3, maximum likelihood estimation works by developing a likelihood function based on the available data and finding the values of the parameter estimates that maximize the likelihood function.  This can be achieved by using iterative methods to determine the parameter estimate values that maximize the likelihood function, but this can be rather difficult and time-consuming, particularly when dealing with the three-parameter distribution.  Another method of finding the parameter estimates involves taking the partial derivatives of the likelihood function with respect to the parameters, setting the resulting equations equal to zero and solving simultaneously to determine the values of the parameter estimates. ( Note that MLE asymptotic properties do not hold when estimating <span class="texhtml">γ</span> using MLE [27].) The log-likelihood functions and associated partial derivatives used to determine maximum likelihood estimates for the Weibull distribution are covered in Appendix C.
 
 
====Example 5====
Repeat Example 1 using maximum likelihood estimation.
 
 
=====Solution to Example 5=====
In this case, we have non-grouped data with no suspensions or intervals, i.e. complete data. The equations for the partial derivatives of the log-likelihood function are derived in Appendix C and given next:
::<math> \frac{\partial \Lambda }{\partial \beta }=\frac{6}{\beta } +\sum_{i=1}^{6}\ln \left( \frac{T_{i}}{\eta }\right) -\sum_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }\ln \left( \frac{T_{i}}{\eta }\right) =0
</math>
 
:and:
 
::<math> \frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{ \beta }{\eta }\sum\limits_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }=0 </math>
 
Solving the above equations simultaneously we get:
 
::<math> \hat{\beta }=1.933,</math> <math>\hat{\eta }=73.526 </math>
 
<br>
The variance/covariance matrix is found to be,
 
::<math> \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.4211 & \hat{Cov}( \hat{\beta },\hat{\eta })=3.272  \\
 
\hat{Cov}(\hat{\beta },\hat{\eta })=3.272 & \hat{Var} \left( \hat{\eta }\right) =266.646 \end{array} \right] </math>
 
The results and the associated graph using Weibull++ (MLE) are shown next.
 
[[Image:weibullfolio16plot.png|thumb|center|400px| ]]
 
You can view the variance/covariance matrix directly by clicking the ''Quick Calculation Pad ''(QCP) icon
 
[[Image:qcpicon.gif|thumb|center|400px| ]]
 
 
[[Image:qcpfolio16.png|thumb|center|400px| ]]
 
 
<br> Note that the decimal accuracy displayed and used is based on your individual User Setup.

Latest revision as of 09:07, 3 August 2012

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