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{{template:RGA BOOK|7|Logistic}}
{{template:RGA BOOK|3.7|Logistic}}
The Logistic reliability growth model has an S-shaped curve and is given by [3]:  
The Logistic reliability growth model has an S-shaped curve and is given by Kececioglu [[RGA_References|[3]]]:  


<INSERT EQUATION>
:<math>R = \frac{1}{1+be^{-kt}},  b > 0, k > 0, T \simeq 0</math>
 
where <math>b</math> and <math>k</math> are parameters. Similar to the analysis given for [[Gompertz Models|the Gompertz curve]], the following may be concluded:
where <math>b\,\!</math> and <math>k\,\!</math> are parameters. Similar to the analysis given for the [[Gompertz Models|Gompertz curve]], the following may be concluded:
<br>
 
:1) The point of inflection is given by:  
<ol>
<li>The point of inflection is given by:  
::<math>{{T}_{i}}=\frac{\ln (b)}{k}</math>
:2) When <math>b>1</math> then <math>{{T}_{i}}>0</math> and an S-shaped curve will be generated. However, w<math>%</math>hen  <math>0<b\le 1</math> then <math>{{T}_{i}}\le 0</math> and the Logistic reliability growth model will not be described by an S-shaped curve.
:<math>{{T}_{i}}=\frac{\ln (b)}{k}\,\!</math>
<br>
</li>
:3) The value of <math>R</math> is equal to 0.5 at the inflection point.
<li>When <math>b>1\,\!</math>, then <math>{{T}_{i}}>0\,\!</math> and an S-shaped curve will be generated. However, when <math>0<b\le 1\,\!</math> , then <math>{{T}_{i}}\le 0\,\!</math> and the Logistic reliability growth model will not be described by an S-shaped curve.
</li>
<li>The value of <math>R\,\!</math> is equal to 0.5 at the inflection point.
</li>
</ol>


=Parameter Estimation=
==Parameter Estimation==
In this section, we will demonstrate the parameter estimation method for the Logistic model using three examples for different types of data.
In this section, we will demonstrate the parameter estimation method for the Logistic model using three examples for different types of data.
==Example: Logistic for Reliability Data==
==Example: Logistic for Reliability Data==
Using the reliability growth data given in the table below, do the following:
{{:Reliability_Data_-_Logistic_Model}}
<br>
:1) Find a Gompertz curve that represents the data and plot it with the raw data.
<br>
:2) Find a Logistic reliability growth curve that represents the data and plot it with the raw data.
<br>
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+Development time versus 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
|}
<br>
'''Solution'''
<br>
:1) Figure Logistic1 shows the entered data and the estimated parameters using the Standard Gompertz model.
<br>
:Therefore:
::<math>\begin{align}
  & \widehat{a}= & 0.9999 \\
& \widehat{b}= & 0.2485 \\
& \widehat{c}= & 0.6858 
\end{align}</math>
<br>
::<math>R=(0.9999){{(0.2485)}^{{{0.6858}^{T}}}}</math>
<br>
The values of predicted reliabilities are plotted in Figure Logig81.
<br>
[[Image:rga8.1.png|thumb|center|400px|Estimated Standard Gompertz parameters for Example 1.]]
<br>
[[Image:rga8.2.png|thumb|center|400px|Gompertz Reliability vs. Time plot.]]
<br>
Notice how the Standard Gompertz model is not really capable of handling the S-shaped characteristics of this data.
<br>
:2.  The least squares estimators of the Logistic growth curve parameters are [9]:
<br>
:where:
<br>
::<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}}}_{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>
<br>
In this example  <math>N=9</math> , which gives:
<br>
From Eqns. (eq37) and (eq38):
And from Eqns. (eq36a) and (eq36b):
<br>
::<math>\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>R=\frac{1}{1+3.3991\,{{e}^{-0.7398\,T}}}</math>
Figure Logig82 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|Modified Gompertz model]], which was presented in a previous chapter.
<br>
[[Image:rga8.3.png|thumb|center|400px|Logistic Reliability vs. Time plot.]]
<br>
==Example: Logistic for Sequential Success/Failure Data==
==Example: Logistic for Sequential Success/Failure Data==
A prototype was tested under a success/failure pattern. The test consisted of 15 runs and the following table presents the data from the test. Find the Logistic model that best fits the data set and plot it along with the reliability observed from the raw data.
{{:Sequential_Data_-_Logistic_Model}}
<br>
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+Sequential success/failure data with observed reliability values
!Time
!Result
!Observed Reliability
|-
|0|| F|| 0.5000
|-
|1|| F|| 0.3333
|-
|2|| S|| 0.5000
|-
|3|| S|| 0.6000
|-
|4|| F|| 0.5000
|-
|5|| S|| 0.5714
|-
|6|| S|| 0.6250
|-
|7|| S|| 0.6667
|-
|8|| S|| 0.7000
|-
|9|| F|| 0.6364
|-
|10|| S|| 0.6667
|-
|11|| S|| 0.6923
|-
|12|| S|| 0.7143
|-
|13|| S|| 0.7333
|}
 
'''Solution'''
<br>
The first run is ignored because it was a success and the reliability at that point was 100%. This failure will be ignored throughout the analysis because it is considered that the test starts when the reliability is not equal to zero or one. The test essentially begins at time 1, and is now considered as time 0 with  <math>N=14</math> . The observed reliability is shown in the last column of Table 8.2. Keep in mind that the observed reliability values still account for the initial suspension.
Eqn. (eq41a) becomes:
 
::<math>\begin{align}
  & \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \\
& = & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\
& = & -0.43163 
\end{align}</math>
<br>
:and:
<br>
::<math>\begin{align}
  & \bar{T}= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}} \\
& = & 6.5 \\
& \underset{i=0}{\overset{13}{\mathop \sum }}\,T_{i}^{2}= & 819.0 \\
& \underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -61.69 
\end{align}</math>
 
Now, from the least squares estimators, Eqns. (eq37) and (eq38) are:
:Therefore:
<br>
::<math>\begin{align}
  & \widehat{b}= & {{e}^{0.2087}} \\
& = & 1.2321 \\
& \widehat{k}= & -(-0.0985) \\
& = & 0.0985 
\end{align}</math>
 
The Logistic reliability model that best fits the data is given by:
<br>
::<math>R=\frac{1}{1+1.2321\cdot \ \,{{e}^{-0.0985T}}}</math>
 
Figure Logig83 shows the Reliability vs. Time plot.
 
==Example: Logistic for Grouped per Configuration Data==
Some equipment underwent testing in different stages. The testing may have been performed in subsequent days, weeks or months with an unequal number of units tested every day. Each group was tested and several failures occurred. The data set is given in columns 1 and 2 of the following table. Find the Logistic model that best fits the data and plot it along with the reliability observed from the raw data.
<br>
 
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|+Grouped per configuration data
!Number of Units
!Number of Failures
!<math>T_i</math>
!Observed Reliability
|-
|10|| 5|| 0|| 0.5000
|-
|8|| 3|| 1|| 0.6250
|-
|9|| 3|| 2|| 0.6667
|-
|9|| 2|| 3|| 0.7778
|-
|10|| 2|| 4|| 0.8000
|-
|10|| 1|| 5|| 0.9000
|-
|10|| 1|| 6|| 0.9000
|-
|10|| 1|| 7|| 0.9000
|-
|10|| 1|| 8|| 0.9000
|}
 
 
'''Solution'''
 
The observed reliability is  <math>1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}</math>  and the last column of Table 8.3 shows the values for each group. With  <math>N=9</math> , Eqn. (eq41a) becomes:
 
::<math>\begin{align}
  & \bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\
& = & -1.4036 
\end{align}</math>
 
:and:
 
::<math>\begin{align}
  & \bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\
& = & 4 \\
& \underset{i=0}{\overset{8}{\mathop \sum }}\,T_{i}^{2}= & 204 \\
& \underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33 
\end{align}</math>
 
Now from the least squares estimators, Eqns. (eq37) and (eq38) give:  
 
::<math>\begin{align}
  & {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\
& = & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\
& = & -0.2967 \\
&  &  \\
& {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\
& = & \left( -1.4036 \right)-\left( -0.2967 \right)\cdot 4.0 \\
& = & -0.2168 
\end{align}</math>
 
:Therefore:
 
::<math>\begin{align}
  & \widehat{b}= & {{e}^{-0.2168}} \\
& = & 0.8051 \\
& \widehat{k}= & -(-0.2967) \\
& = & 0.2967 
\end{align}</math>
 
The Logistic reliability model that best fits the data is given by:
 
::<math>R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}</math>
 
Figure Logig85 shows the Reliability vs. Time plot.
 
[[Image:rga8.4.png|thumb|center|400px|Logistic Reliability vs. Time plot.]]
 


[[Image:rga8.5.png|thumb|center|400px|Logistic Reliability vs. Time plot displaying the intervals.]]
===Example: Logistic for Grouped per Configuration Data===
{{:Grouped_per_Configuration_Data_-_Logistic_Model}}


=Confidence Bounds=
==Confidence Bounds==
Least squares is used to estimate the parameters of the following Logistic model.
Least squares is used to estimate the parameters of the following Logistic model.


::<math>\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}</math>
:<math>\ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}\,\!</math>
 
Thus the confidence bounds on the parameters are given by:
 
::<math>b=\hat{b}{{e}^{{{t}_{n-2,\alpha /2}}SE(\ln \hat{b})}}</math>
 
:where:
 
::<math>SE(\ln \hat{b})=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{({{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}},\ \ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}</math>
 
::<math>\sigma =\sqrt{SSE/(n-2)}</math>
 
:and:
 
::<math>k=\hat{k}\pm {{t}_{n-2,\alpha /2}}SE(\hat{k})</math>
 
:where:
 
::<math>SE(\hat{k})=\frac{\sigma }{\sqrt{{{S}_{xx}}}},\ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}</math>
 
Since the reliability is always between 0 and 1, the logit transformation is used to obtain the confidence bounds on reliability.
 
::<math>CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}</math>
 
==Example: Logistic Confidence Bounds==
For the data given above for the [[Logistic#Example:_Logistic_for_Reliability_Data|reliability data example]], calculate the 2-sided 90% confidence bounds under the Logistic model for the following:
<br>
<br>
:1) The parameters  <math>b</math>  and  <math>k</math> .
<br>
:2) Reliability at month 5.
 
<br>
'''Solution'''
<br>
:1) The values of  <math>\widehat{b}</math>  and  <math>\widehat{k}</math>  estimated from the least squares analysis in the [[Logistic#Example:_Logistic_for_Reliability_Data|reliability data example]] above:
 
::<math>\begin{align}
  & \widehat{b}= & 3.3991 \\
& \widehat{\alpha }= & 0.7398 
\end{align}</math>
 
Thus the 2-sided 90% confidence bounds on parameter  <math>b</math>  using Eqn. (LogCBb) are:
 
::<math>\begin{align}
  & {{b}_{lower}}= & 2.5547 \\
& {{b}_{upper}}= & 4.5225 
\end{align}</math>
 
The 2-sided 90% confidence bounds on parameter  <math>k</math>  using Eqn. (logCBk) are:
 
::<math>\begin{align}
  & {{k}_{lower}}= & 0.6798 \\
& {{k}_{upper}}= & 0.7997 
\end{align}</math>
 
 
:2) First calculate the reliability estimation at month 5:
::<math>\begin{align}
  & {{R}_{5}}= & \frac{1}{1+b{{e}^{-5k}}} \\
& = & 0.9224 
\end{align}</math>
Thus the 2-sided 90% confidence bounds on reliability at month 5 using Eqn. (LogCR) are:
 
::<math>\begin{align}
  & {{[{{R}_{5}}]}_{lower}}= & 0.8493 \\
& {{[{{R}_{5}}]}_{upper}}= & 0.9955 
\end{align}</math>
 
The next figure shows a graph of the reliability plotted with 2-sided 90% confidence bounds.
 
[[Image:rga8.6.png|thumb|center|400px|Logistic Reliability vs. Time plot with 2-sided 90% confidence bounds.]]


=More Examples=
Thus, the confidence bounds on the parameter <math>b\,\!</math> are given by:
==Auto Transmission Reliability Data==
The following table presents the reliabilities observed monthly for an automobile transmission that was tested for one year.


:1) Find a Logistic reliability growth curve that best represents the data.
:<math>b=\hat{b}{{e}^{{{t}_{n-2,\alpha /2}}SE(\ln \hat{b})}}\,\!</math>
:2) Plot it comparatively with the raw data.
:3) If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99% be achieved?
:4) If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of January the following year?


{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
where:  
|+Reliability data
!Month
!Observed Reliability(%)
|-
|June|| 22
|-
|July|| 26
|-
|August|| 30
|-
|September|| 34
|-
|October|| 45
|-
|November|| 58
|-
|December|| 68
|-
|January|| 79
|-
|February|| 85
|-
|March|| 89
|-
|April|| 92
|-
|May|| 95
|}


'''Solution '''
:<math>\begin{align}
<br>
SE(\ln \hat{b})&=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{({{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}},\ \ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}} \\ \\
:1) The next figure shows the estimated parameters.
\sigma &=\sqrt{SSE/(n-2)}
<br>
\end{align}\,\!</math>
[[Image:rga8.7.png|thumb|center|400px|Entered data and the estimated Logistic parameters.]]
<br>
:2) The next figure displays the Reliability vs. Time plot.
<br>
[[Image:rga8.8.png|thumb|center|400px|Reliability vs. Time plot.]]
<br>
:3) Using the QCP, the next figure displays when the reliability goal of 99% will be achieved.
<br>
[[Image:rga8.9.png|thumb|center|400px|When the reliability goal of 99% will be achieved.]]
<br>
:4) The last figure shows the reliability at the end of January the following year (i.e. after 20 months of testing and development).
<br>
[[Image:rga8.10.png|thumb|center|400px|The reliability at the end of the following January (month=20)]]
<br>


==Sequential Data from Missile Launch Test==
and the confidence bounds on the parameter <math>k\,\!</math> are:


The next table presents the results for a missile launch test. The test consisted of 20 attempts. If the missile launched, it was recorded as a success. If not, it was recorded as a failure. Note that, at this development stage, the test did not consider whether or not the target was destroyed.
:<math>k=\hat{k}\pm {{t}_{n-2,\alpha /2}}SE(\hat{k})\,\!</math>


:1) Find a Logistic reliability growth curve that best represents the data.
where:
:2) Plot it comparatively with the raw data.
:3) If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99.5% with a 90% confidence level be achieved?
:4) If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of the 35th launch?


{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
:<math>SE(\hat{k})=\frac{\sigma }{\sqrt{{{S}_{xx}}}},\ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}\,\!</math>
|+Sequential success/failure data
!Launch Number
!Result
|-
|1|| F
|-
|2|| F
|-
|3|| S
|-
|4|| F
|-
|5|| F
|-
|6|| S
|-
|7|| S
|-
|8|| S
|-
|9|| F
|-
|10|| S
|-
|11|| F
|-
|12|| S
|-
|13|| S
|-
|14|| S
|-
|15|| S
|-
|16|| S
|-
|17|| S
|-
|18|| S
|-
|19|| S
|-
|20|| S
|}


'''Solution'''
Since the reliability is always between 0 and 1, the logit transformation is used to obtain the confidence bounds on reliability, which is:  
<br>
:1) The next figure shows the entered data and the estimated parameters.
<br>
[[Image:rga8.11.png|thumb|center|400px|Entered data and the estimated logistic parameters.]]
<br>
:2) The next figure displays the Reliability vs. Time plot.
<br>
[[Image:rga8.12.png|thumb|center|400px|Reliability vs. Time plot.]]
<br>
:3) The next figure displays when the reliability goal of 99.5% will be achieved with a 90% confidence level.
<br>
[[Image:rga8.13.png|thumb|center|400px|When the reliability goal of 99.5% with a 90% confidence level will be achieved.]]
<br>
:4) The next figure displays the reliability achieved after the 35th launch.
<br>
[[Image:rga8.14.png|thumb|center|400px|The reliability at the end of the 35th launch.]]


==Sequential Data with Failure Modes==
:<math>CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}\,\!</math>
Consider the data given for the missile launch test in the [[Logistic#Sequential_Data_from_Missile_Launch_Test|previous example]]. Now suppose that the engineers assigned failure modes to each failure and that the appropriate corrective actions were taken.


The table below presents the data.
===Example: Logistic Confidence Bounds===
:1) Find the Logistic reliability growth curve that best represents the data.
{{:Logistic_Confidence_Bounds_Example}}
:2) Plot it comparatively with the raw data.
:3) If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99.50% be achieved?
:4) If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of the 35th launch?
<br>


{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
==More Examples==
|+Sequential success/failure data with modes
===Auto Transmission Reliability Data===
!Launch Number
{{:Logistic_Model_-_Auto_Transmission_Example}}
!Result
!Mode
|-
|1|| F|| 2
|-
|2|| F|| 1
|-
|3|| S||
|-
|4|| F|| 3
|-
|5|| F|| 3
|-
|6|| S||
|-
|7|| S||
|-
|8|| S||
|-
|9|| F|| 2
|-
|10|| S||
|-
|11|| F|| 1
|-
|12|| S
|-
|13|| S
|-
|14|| S
|-
|15|| S
|-
|16|| S
|-
|17|| S
|-
|18|| S
|-
|19|| S
|-
|20|| S
|}


'''Solution'''
===Sequential Data from Missile Launch Test===
<br>
{{:Logistic_Model_-_Missile_Launch_Test_Example}}
:1) Figure Loge51 shows the estimated parameters.
:2) Figure Loge52 displays the Reliability vs. Time plot.
:3) Figure Loge53 displays when the reliability goal of 99.5% will be achieved.
:4) Figure Loge54 displays the reliability after the 35th launch.


[[Image:rga8.15.png|thumb|center|400px|Entered data and the estimated Logistic parameters.]]
===Sequential Data with Failure Modes===
<br>
{{:Sequential_Data_with_Failure_Modes_-_Logistic_Model}}
<br>
[[Image:rga8.16.png|thumb|center|400px|Reliability vs. Time plot.]]
<br>
<br>
[[Image:rga8.17.png|thumb|center|400px|Calculate when the reliability goal of 99.5% will be achieved.]]
<br>
<br>
[[Image:rga8.18.png|thumb|center|400px|Calculate the reliability at the end of the 35th launch.]]

Latest revision as of 18:18, 29 May 2014

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Chapter 3.7: Logistic


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Chapter 3.7  
Logistic  

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The Logistic reliability growth model has an S-shaped curve and is given by Kececioglu [3]:

[math]\displaystyle{ R = \frac{1}{1+be^{-kt}}, b \gt 0, k \gt 0, T \simeq 0 }[/math]

where [math]\displaystyle{ b\,\! }[/math] and [math]\displaystyle{ k\,\! }[/math] are parameters. Similar to the analysis given for the Gompertz curve, the following may be concluded:

  1. The point of inflection is given by:
    [math]\displaystyle{ {{T}_{i}}=\frac{\ln (b)}{k}\,\! }[/math]
  2. When [math]\displaystyle{ b\gt 1\,\! }[/math], then [math]\displaystyle{ {{T}_{i}}\gt 0\,\! }[/math] and an S-shaped curve will be generated. However, when [math]\displaystyle{ 0\lt b\le 1\,\! }[/math] , then [math]\displaystyle{ {{T}_{i}}\le 0\,\! }[/math] and the Logistic reliability growth model will not be described by an S-shaped curve.
  3. The value of [math]\displaystyle{ R\,\! }[/math] is equal to 0.5 at the inflection point.

Parameter Estimation

In this section, we will demonstrate the parameter estimation method for the Logistic model using three examples for different types of data.

Example: Logistic for Reliability Data

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.

Example: Logistic for Sequential Success/Failure Data

A prototype was tested under a success/failure pattern. The test consisted of 15 runs. The following table presents the data from the test. Find the Logistic model that best fits the data set, and plot it along with the reliability observed from the raw data.

Sequential Success/Failure Data with Observed Reliability Values
Time Result Observed Reliability
0 F 0.5000
1 F 0.3333
2 S 0.5000
3 S 0.6000
4 F 0.5000
5 S 0.5714
6 S 0.6250
7 S 0.6667
8 S 0.7000
9 F 0.6364
10 S 0.6667
11 S 0.6923
12 S 0.7143
13 S 0.7333

Solution

The first run is ignored because it was a success, and the reliability at that point was 100%. This failure will be ignored throughout the analysis because it is considered that the test starts when the reliability is not equal to zero or one. The test essentially begins at time 1, and is now considered as time 0 with [math]\displaystyle{ N=14\,\! }[/math]. The observed reliability is shown in the last column of the table. Keep in mind that the observed reliability values still account for the initial suspension.

Therefore:

[math]\displaystyle{ \begin{align} \bar{Y}= & \frac{1}{N}\underset{i=0}{\overset{N-1}{\mathop \sum }}\,{{Y}_{i}} \\ = & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ = & -0.43163 \end{align}\,\! }[/math]

and:

[math]\displaystyle{ \begin{align} \bar{T}= & \frac{1}{14}\underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}} \\ = & 6.5 \\ \underset{i=0}{\overset{13}{\mathop \sum }}\,T_{i}^{2}= & 819.0 \\ \underset{i=0}{\overset{13}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -61.69 \end{align}\,\! }[/math]

Now, from the least squares estimators, the values are:

[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}}} \\ \\ &= \frac{-61.69-14 \cdot 6.5 \cdot (-.43163)}{819.0-14 \cdot 6.5^{2}} \\ &= -0.0985 \\ \\ \hat{b_{0}}&= \overline{Y} - \hat{b_{1}}\overline{T} \\ &= (-.043163)-(-0.0985) \cdot 6.5 \\ &= 0.2087 \end{align}\,\! }[/math]

Therefore:

[math]\displaystyle{ \begin{align} \widehat{b}= & {{e}^{0.2087}} \\ = & 1.2321 \\ \widehat{k}= & -(-0.0985) \\ = & 0.0985 \end{align}\,\! }[/math]

The Logistic reliability model that best fits the data is given by:

[math]\displaystyle{ R=\frac{1}{1+1.2321\cdot \ \,{{e}^{-0.0985T}}}\,\! }[/math]

The following figure shows the Reliability vs. Time plot.

Rga8.4.png

Example: Logistic for Grouped per Configuration Data

Some equipment underwent testing in different stages. The testing may have been performed in subsequent days, weeks or months with an unequal number of units tested every day. Each group was tested and several failures occurred. The data set is given in columns 1 and 2 of the following table. Find the Logistic model that best fits the data, and plot it along with the reliability observed from the raw data.

Grouped per Configuration Data
Number of Units Number of Failures [math]\displaystyle{ T_i\,\! }[/math] Observed Reliability
10 5 0 0.5000
8 3 1 0.6250
9 3 2 0.6667
9 2 3 0.7778
10 2 4 0.8000
10 1 5 0.9000
10 1 6 0.9000
10 1 7 0.9000
10 1 8 0.9000

Solution

The observed reliability is [math]\displaystyle{ 1-\tfrac{\#\text{ of failures}}{\#\text{ of units}}\,\! }[/math] and the last column of the table above shows the values for each group. With [math]\displaystyle{ N=9\,\! }[/math], the least square estimator [math]\displaystyle{ \overline{Y} }[/math] becomes:

[math]\displaystyle{ \begin{align} \bar{Y}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,\ln \left( \frac{1}{{{R}_{i}}}-1 \right) \\ = & -1.4036 \end{align}\,\! }[/math]

and:

[math]\displaystyle{ \begin{align} \bar{T}= & \frac{1}{9}\underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}} \\ = & 4 \\ \underset{i=0}{\overset{8}{\mathop \sum }}\,T_{i}^{2}= & 204 \\ \underset{i=0}{\overset{8}{\mathop \sum }}\,{{T}_{i}}{{Y}_{i}}= & -68.33 \end{align}\,\! }[/math]

Now from the least squares estimators, [math]\displaystyle{ \hat{b_{i}}\,\! }[/math] and [math]\displaystyle{ \hat{b_{0}}\,\! }[/math], we have:

[math]\displaystyle{ \begin{align} {{{\hat{b}}}_{1}}= & \frac{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,{{T}_{i}}{{Y}_{i}}-N\cdot \bar{T}\cdot \bar{Y}}{\underset{i=0}{\overset{8}{\mathop{\sum }}}\,T_{i}^{2}-N\cdot {{{\bar{T}}}^{2}}} \\ = & \frac{-68.33-9\cdot 4\cdot \left( -1.4036 \right)}{204-9\cdot {{4}^{2}}} \\ = & -0.2967 \\ & \\ {{{\hat{b}}}_{0}}= & \bar{Y}-{{{\hat{b}}}_{1}}\bar{T} \\ = & \left( -1.4036 \right)-\left( -0.2967 \right)\cdot 4.0 \\ = & -0.2168 \end{align}\,\! }[/math]

Therefore:

[math]\displaystyle{ \begin{align} \widehat{b}= & {{e}^{-0.2168}} \\ = & 0.8051 \\ \widehat{k}= & -(-0.2967) \\ = & 0.2967 \end{align}\,\! }[/math]

The Logistic reliability model that best fits the data is given by:

[math]\displaystyle{ R=\frac{1}{1+0.8051\cdot \ \,{{e}^{-0.2967T}}}\,\! }[/math]

The figure below shows the Reliability vs. Time plot.

Rga8.5.png

Confidence Bounds

Least squares is used to estimate the parameters of the following Logistic model.

[math]\displaystyle{ \ln (\frac{1}{{{{\hat{R}}}_{i}}}-1)=\ln (b)-k{{T}_{i}}\,\! }[/math]

Thus, the confidence bounds on the parameter [math]\displaystyle{ b\,\! }[/math] are given by:

[math]\displaystyle{ b=\hat{b}{{e}^{{{t}_{n-2,\alpha /2}}SE(\ln \hat{b})}}\,\! }[/math]

where:

[math]\displaystyle{ \begin{align} SE(\ln \hat{b})&=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{({{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}},\ \ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}} \\ \\ \sigma &=\sqrt{SSE/(n-2)} \end{align}\,\! }[/math]

and the confidence bounds on the parameter [math]\displaystyle{ k\,\! }[/math] are:

[math]\displaystyle{ k=\hat{k}\pm {{t}_{n-2,\alpha /2}}SE(\hat{k})\,\! }[/math]

where:

[math]\displaystyle{ SE(\hat{k})=\frac{\sigma }{\sqrt{{{S}_{xx}}}},\ \ {{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{({{T}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{T}_{i}} \right)}^{2}}\,\! }[/math]

Since the reliability is always between 0 and 1, the logit transformation is used to obtain the confidence bounds on reliability, which is:

[math]\displaystyle{ CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}\,\! }[/math]

Example: Logistic Confidence Bounds

For the data given above for the reliability data example, calculate the 2-sided 90% confidence bounds under the Logistic model for the following:

  1. The parameters [math]\displaystyle{ b\,\! }[/math] and [math]\displaystyle{ k\,\! }[/math].
  2. Reliability at month 5.

Solution

  1. The values of [math]\displaystyle{ \hat{b}\,\! }[/math] and [math]\displaystyle{ \hat{k}\,\! }[/math] that were estimated from the least squares analysis in the reliability data example are:
    [math]\displaystyle{ \begin{align} \widehat{b}= & 3.3991 \\ \widehat{\alpha }= & 0.7398 \end{align}\,\! }[/math]
    Thus, the 2-sided 90% confidence bounds on parameter [math]\displaystyle{ b\,\! }[/math] are:
    [math]\displaystyle{ \begin{align} {{b}_{lower}}= & 2.5547 \\ {{b}_{upper}}= & 4.5225 \end{align}\,\! }[/math]
    The 2-sided 90% confidence bounds on parameter [math]\displaystyle{ k\,\! }[/math] are:
    [math]\displaystyle{ \begin{align} {{k}_{lower}}= & 0.6798 \\ {{k}_{upper}}= & 0.7997 \end{align}\,\! }[/math]
  2. First, calculate the reliability estimation at month 5:
    [math]\displaystyle{ \begin{align} {{R}_{5}}= & \frac{1}{1+b{{e}^{-5k}}} \\ = & 0.9224 \end{align}\,\! }[/math]
    Thus, the 2-sided 90% confidence bounds on reliability at month 5 are:
    [math]\displaystyle{ \begin{align} {{[{{R}_{5}}]}_{lower}}= & 0.8493 \\ {{[{{R}_{5}}]}_{upper}}= & 0.9955 \end{align}\,\! }[/math]
    The next figure shows a graph of the reliability plotted with 2-sided 90% confidence bounds.
    Rga8.6.png

More Examples

Auto Transmission Reliability Data

The following table presents the reliabilities observed monthly for an automobile transmission that was tested for one year.

  1. Find a Logistic reliability growth curve that best represents the data.
  2. Plot it comparatively with the raw data.
  3. If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99% be achieved?
  4. If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of January the following year?
Reliability Data
Month Observed Reliability(%)
June 22
July 26
August 30
September 34
October 45
November 58
December 68
January 79
February 85
March 89
April 92
May 95

Solution

  1. The next figure shows the entered data and the estimated parameters.
    Rga8.7.png


  2. The next figure displays the Reliability vs. Time plot.
    Rga8.8.png


  3. Using the QCP, the next figure displays, in months, when the reliability goal of 99% will be achieved.
    Rga8.9.png


  4. The last figure shows the reliability at the end of January the following year (i.e., after 20 months of testing and development).
    Rga8.10.png


Sequential Data from Missile Launch Test

The following table presents the results for a missile launch test. The test consisted of 20 attempts. If the missile launched, it was recorded as a success. If not, it was recorded as a failure. Note that, at this development stage, the test did not consider whether or not the target was destroyed.

  1. Find a Logistic reliability growth curve that best represents the data.
  2. Plot it comparatively with the raw data.
  3. If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99.5% with a 90% confidence level be achieved?
  4. If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of the 35th launch?
Sequential Success/Failure Data
Launch Number Result
1 F
2 F
3 S
4 F
5 F
6 S
7 S
8 S
9 F
10 S
11 F
12 S
13 S
14 S
15 S
16 S
17 S
18 S
19 S
20 S

Solution

  1. The next figure shows the entered data and the estimated parameters.
    Rga8.11.png
  2. The next figure displays the Reliability vs. Time plot.
    Rga8.12.png
  3. The next figure displays the number of launches before the reliability goal of 99.5% will be achieved with a 90% confidence level.
    Rga8.13.png
  4. The next figure displays the reliability achieved after the 35th launch.
    Rga8.14.png

Sequential Data with Failure Modes

Consider the data given in the previous example. Now suppose that the engineers assigned failure modes to each failure and that the appropriate corrective actions were taken.

The table below presents the data.

  1. Find the Logistic reliability growth curve that best represents the data.
  2. Plot it comparatively with the raw data.
  3. If design changes continue to be incorporated and the testing continues, when will the reliability goal of 99.50% be achieved?
  4. If design changes continue to be incorporated and the testing continues, what will be the attainable reliability at the end of the 35th launch?
Sequential Success/Failure Data with Modes
Launch Number Result Mode
1 F 2
2 F 1
3 S
4 F 3
5 F 3
6 S
7 S
8 S
9 F 2
10 S
11 F 1
12 S
13 S
14 S
15 S
16 S
17 S
18 S
19 S
20 S

Solution

  1. The next figure shows the entered data and the estimated parameters.
    Rga8.15.png

  2. The next figure displays the Reliability vs. Time plot.
    Rga8.16.png

  3. The next figure displays the number of launches before the reliability goal of 99.5% will be achieved.
    Rga8.17.png

  4. The last figure displays the reliability after the 35th launch.
    Rga8.18.png