Reliability Data - Logistic Model: Difference between revisions

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<noinclude>{{Banner RGA Examples}}
<noinclude>{{Banner RGA Examples}}
''This example appears in the [[Logistic|Reliability Growth and Repairable System Analysis Reference book]]''.
''This example appears in the [[Logistic|Reliability Growth and Repairable System Analysis Reference]]''.
</noinclude>
</noinclude>


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<li>The figure below shows the entered data and the estimated parameters using the standard Gompertz model.
<li>The figure below shows the entered data and the estimated parameters using the standard Gompertz model.


[[Image:rga8.1.png|center|400px|Estimated Standard Gompertz parameters for Example 1.]]
[[Image:rga8.1.png|center|650px|Estimated Standard Gompertz parameters for Example 1.]]


:Therefore:  
Therefore:  


::<math>\begin{align}
:<math>\begin{align}
   & \widehat{a}= & 0.9999 \\  
   & \widehat{a}= & 0.9999 \\  
  & \widehat{b}= & 0.2485 \\  
  & \widehat{b}= & 0.2485 \\  
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\end{align}\,\!</math>
\end{align}\,\!</math>


:<math>R=(0.9999){{(0.2485)}^{{{0.6858}^{T}}}}\,\!</math>


::<math>R=(0.9999){{(0.2485)}^{{{0.6858}^{T}}}}\,\!</math>
The values of the predicted reliabilities are plotted in the figure below.
 
 
:The values of the predicted reliabilities are plotted in the figure below.


[[Image:rga8.2.png|center|400px|Gompertz Reliability vs. Time plot.]]
[[Image:rga8.2.png|center|400px|Gompertz Reliability vs. Time plot.]]


 
Notice how the standard Gompertz model is not really capable of handling the S-shaped characteristics of this data.  
:Notice how the standard Gompertz model is not really capable of handling the S-shaped characteristics of this data.  


</li><li>The least squares estimators of the Logistic growth curve parameters are given by Crow [[RGA_References|[9]]]:  
</li><li>The least squares estimators of the Logistic growth curve parameters are given by Crow [[RGA_References|[9]]]:  


:where:  
where:  


::<math>\begin{align}
:<math>\begin{align}
   {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\
   {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\
   {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ \\
   {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ \\
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   \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}}   
   \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}}   
\end{align}\,\!</math>
\end{align}\,\!</math>


In this example <math>N=9\,\!</math>, which gives:  
In this example <math>N=9\,\!</math>, which gives:  


::<math>\begin{align}
:<math>\begin{align}
\overline{Y}&=\frac{1}{9}\sum_{i=0}^{8}ln\left (\frac{1}{R_{i}}-1  \right ) \\
\overline{Y}&=\frac{1}{9}\sum_{i=0}^{8}ln\left (\frac{1}{R_{i}}-1  \right ) \\
&= -1.7355 \\ \\
&= -1.7355 \\ \\
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\sum_{i=0}^{8}T_{i}Y_{i} &= -106.8630
\sum_{i=0}^{8}T_{i}Y_{i} &= -106.8630
\end{align}\,\!</math>
\end{align}\,\!</math>


From the equations for <math>b_{i}\,\!</math> and <math>\hat{b_{0}}\,\!</math>:
From the equations for <math>b_{i}\,\!</math> and <math>\hat{b_{0}}\,\!</math>:


::<math>\begin{align}
:<math>\begin{align}
\hat{b_{1}} &= \frac{-106.8630 - 9(4)(-1.7355)}{204-9(4)_{2}} \\
\hat{b_{1}} &= \frac{-106.8630 - 9(4)(-1.7355)}{204-9(4)_{2}} \\
& = 0.7398 \\
& = 0.7398 \\
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&= 1.2235
&= 1.2235
\end{align}\,\!</math>
\end{align}\,\!</math>


And from the least squares estimators for <math>\hat{b}\,\!</math> and <math>\hat{k}\,\!</math>:  
And from the least squares estimators for <math>\hat{b}\,\!</math> and <math>\hat{k}\,\!</math>:  


::<math>\begin{align}
:<math>\begin{align}
   \widehat{b}= & {{e}^{1.2235}} \\  
   \widehat{b}= & {{e}^{1.2235}} \\  
   = & 3.3991 \\  
   = & 3.3991 \\  
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   = & 0.7398   
   = & 0.7398   
\end{align}\,\!</math>
\end{align}\,\!</math>


Therefore, the Logistic reliability growth curve that represents this data set is given by:
Therefore, the Logistic reliability growth curve that represents this data set is given by:


::<math>R=\frac{1}{1+3.3991\,{{e}^{-0.7398\,T}}}\,\!</math>
:<math>R=\frac{1}{1+3.3991\,{{e}^{-0.7398\,T}}}\,\!</math>
 


The following figure shows the Reliability vs. Time plot. The plot shows that the observed data set is estimated well by the Logistic reliability growth curve, except in the region closely surrounding the inflection point of the observed reliability. This problem can be overcome by using the [[Gompertz_Models#The_Modified_Gompertz_Model|modified Gompertz model]].
The following figure shows the Reliability vs. Time plot. The plot shows that the observed data set is estimated well by the Logistic reliability growth curve, except in the region closely surrounding the inflection point of the observed reliability. This problem can be overcome by using the [[Gompertz_Models#The_Modified_Gompertz_Model|modified Gompertz model]].


[[Image:rga8.3.png|center|400px|Logistic Reliability vs. Time plot.]]</li></ol>
[[Image:rga8.3.png|center|400px|Logistic Reliability vs. Time plot.]]</li></ol>

Revision as of 18:24, 29 May 2014

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This example appears in the Reliability Growth and Repairable System Analysis Reference.


Using the reliability growth data given in the table below, do the following:

  1. Find a Gompertz curve that represents the data and plot it with the raw data.
  2. Find a Logistic reliability growth curve that represents the data and plot it with the raw data.
Development Time vs. Observed Reliability data and Predicted Reliabilities
Time, months Raw Data Reliability (%) Gompertz Reliability (%) Logistic Reliablity (%)
0 31.00 24.85 22.73
1 35.50 38.48 38.14
2 49.30 51.95 56.37
3 70.10 63.82 73.02
4 83.00 73.49 85.01
5 92.20 80.95 92.24
6 96.40 86.51 96.14
7 98.60 90.54 98.12
8 99.00 93.41 99.09

Solution

  1. The figure below shows the entered data and the estimated parameters using the standard Gompertz model.
    Estimated Standard Gompertz parameters for Example 1.

    Therefore:

    [math]\displaystyle{ \begin{align} & \widehat{a}= & 0.9999 \\ & \widehat{b}= & 0.2485 \\ & \widehat{c}= & 0.6858 \end{align}\,\! }[/math]
    [math]\displaystyle{ R=(0.9999){{(0.2485)}^{{{0.6858}^{T}}}}\,\! }[/math]

    The values of the predicted reliabilities are plotted in the figure below.

    Gompertz Reliability vs. Time plot.

    Notice how the standard Gompertz model is not really capable of handling the S-shaped characteristics of this data.

  2. The least squares estimators of the Logistic growth curve parameters are given by Crow [9]: where:
    [math]\displaystyle{ \begin{align} {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{N-1}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ \\ {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ \\ {{Y}_{i}}= & \ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ \\ \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \end{align}\,\! }[/math]
    In this example [math]\displaystyle{ N=9\,\! }[/math], which gives:
    [math]\displaystyle{ \begin{align} \overline{Y}&=\frac{1}{9}\sum_{i=0}^{8}ln\left (\frac{1}{R_{i}}-1 \right ) \\ &= -1.7355 \\ \\ \overline{T}&=\frac{1}{9}\sum_{i=0}^{8}T_{i} = 4 \\ \sum_{i=0}^{8}T_{i}^{2} &= 204 \\ \sum_{i=0}^{8}T_{i}Y_{i} &= -106.8630 \end{align}\,\! }[/math]
    From the equations for [math]\displaystyle{ b_{i}\,\! }[/math] and [math]\displaystyle{ \hat{b_{0}}\,\! }[/math]:
    [math]\displaystyle{ \begin{align} \hat{b_{1}} &= \frac{-106.8630 - 9(4)(-1.7355)}{204-9(4)_{2}} \\ & = 0.7398 \\ \hat{b_{0}} &= -1.7355 - (-0.7398)(4)\\ &= 1.2235 \end{align}\,\! }[/math]
    And from the least squares estimators for [math]\displaystyle{ \hat{b}\,\! }[/math] and [math]\displaystyle{ \hat{k}\,\! }[/math]:
    [math]\displaystyle{ \begin{align} \widehat{b}= & {{e}^{1.2235}} \\ = & 3.3991 \\ \widehat{k}= & -(-0.7398) \\ = & 0.7398 \end{align}\,\! }[/math]
    Therefore, the Logistic reliability growth curve that represents this data set is given by:
    [math]\displaystyle{ R=\frac{1}{1+3.3991\,{{e}^{-0.7398\,T}}}\,\! }[/math]
    The following figure shows the Reliability vs. Time plot. The plot shows that the observed data set is estimated well by the Logistic reliability growth curve, except in the region closely surrounding the inflection point of the observed reliability. This problem can be overcome by using the modified Gompertz model.
    Logistic Reliability vs. Time plot.