Logistic Model Examples
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This example appears in the Reliability Growth and Repairable System Analysis Reference book.
Parameter Estimation: Reliability Data
Using the reliability growth data given in the table below, do the following:
- Find a Gompertz curve that represents the data and plot it with the raw data.
- Find a Logistic reliability growth curve that represents the data and plot it with the raw data.
| 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
- The figure below shows the entered data and the estimated parameters using the standard Gompertz model.
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.
Notice how the standard Gompertz model is not really capable of handling the S-shaped characteristics of this data.
- 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]
- [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]
- [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]
- [math]\displaystyle{ \begin{align} \widehat{b}= & {{e}^{1.2235}} \\ = & 3.3991 \\ \widehat{k}= & -(-0.7398) \\ = & 0.7398 \end{align}\,\! }[/math]
- [math]\displaystyle{ R=\frac{1}{1+3.3991\,{{e}^{-0.7398\,T}}}\,\! }[/math]
Parameter Estimation: 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.
| 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.
Parameter Estimation: 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.
| 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.
