Non Parametric RDA MCF Example: Difference between revisions

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<noinclude>{{Banner Weibull Examples}}
<noinclude>{{Banner Weibull Examples}}
''This example appears in the [[Non-Parametric Recurrent Event Data Analysis]] article.''
''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''


</noinclude>
</noinclude>
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   \text{4} & \text{13, 15, 24+}  \\
   \text{4} & \text{13, 15, 24+}  \\
   \text{5} & \text{16, 22, 25, 28+}  \\
   \text{5} & \text{16, 22, 25, 28+}  \\
\end{matrix}</math></center>
\end{matrix}\,\!</math></center>


Estimate the MCF values, with 95% confidence bounds.  
Estimate the MCF values, with 95% confidence bounds.  
Line 28: Line 28:
   \text{1} & \text{10} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.40 + 0}\text{.20 = 0}\text{.60}  \\
   \text{1} & \text{10} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.40 + 0}\text{.20 = 0}\text{.60}  \\
   \text{3} & \text{12} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.60 + 0}\text{.20 = 0}\text{.80}  \\
   \text{3} & \text{12} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.60 + 0}\text{.20 = 0}\text{.80}  \\
   \text{2} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.80+0}\text{.20 =1}\text{.00}  \\
   \text{2} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.80 + 0}\text{.20 = 1}\text{.00}  \\
   \text{4} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.00 + 0}\text{.20 = 1}\text{.20}  \\
   \text{4} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.00 + 0}\text{.20 = 1}\text{.20}  \\
   \text{1} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.20 + 0}\text{.20 =1}\text{.40}  \\
   \text{1} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.20 + 0}\text{.20 = 1}\text{.40}  \\
   \text{4} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.40 + 0}\text{.20 = 1}\text{.60}  \\
   \text{4} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.40 + 0}\text{.20 = 1}\text{.60}  \\
   \text{5} & \text{16} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.60 + 0}\text{.20 = 1}\text{.80}  \\
   \text{5} & \text{16} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.60 + 0}\text{.20 = 1}\text{.80}  \\
   \text{2} & \text{17} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.80 + 0}\text{.20 = 2}\text{.0}  \\
   \text{2} & \text{17} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.80 + 0}\text{.20 = 2}\text{.00}  \\
   \text{1} & \text{17} & \text{S} & \text{4} & {} & {}  \\
   \text{1} & \text{17} & \text{S} & \text{4} & {} & {}  \\
   \text{2} & \text{19} & \text{S} & \text{3} & {} & {}  \\
   \text{2} & \text{19} & \text{S} & \text{3} & {} & {}  \\
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   \text{3} & \text{26} & \text{S} & \text{1} & {} & {}  \\
   \text{3} & \text{26} & \text{S} & \text{1} & {} & {}  \\
   \text{5} & \text{28} & \text{S} & \text{0} & {} & {}  \\
   \text{5} & \text{28} & \text{S} & \text{0} & {} & {}  \\
\end{matrix}</math></center>
\end{matrix}\,\!</math></center>


Using the MCF variance equation, the following table of variance values can be obtained:  
Using the MCF variance equation, the following table of variance values can be obtained:  
Line 52: Line 52:
! Months  
! Months  
! State  
! State  
! <span class="texhtml">''r''<sub>''i''</sub></span>  
! <math>{{r}_{i}}\,\!</math>  
! <span class="texhtml">''V''''a''''r''<sub>''i''</sub></span>
! <math>Va{{r}_{i}}\,\!</math>
|-
|-
| 1  
| 1  
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| F  
| F  
| 5  
| 5  
| <math>(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.032</math>
| <math>(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.032\,\!</math>
|-
|-
| 2  
| 2  
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| F  
| F  
| 5  
| 5  
| <math>0.032+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.064</math>
| <math>0.032+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.064\,\!</math>
|-
|-
| 1  
| 1  
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| F  
| F  
| 5  
| 5  
| <math>0.064+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.096</math>
| <math>0.064+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.096\,\!</math>
|-
|-
| 3  
| 3  
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| F  
| F  
| 5  
| 5  
| <math>0.096+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.128</math>
| <math>0.096+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.128\,\!</math>
|-
|-
| 2  
| 2  
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| F  
| F  
| 5  
| 5  
| <math>0.128+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.160</math>
| <math>0.128+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.160\,\!</math>
|-
|-
| 4  
| 4  
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| F  
| F  
| 5  
| 5  
| <math>0.160+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.192</math>
| <math>0.160+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.192\,\!</math>
|-
|-
| 1  
| 1  
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| F  
| F  
| 5  
| 5  
| <math>0.192+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.224</math>
| <math>0.192+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.224\,\!</math>
|-
|-
| 4  
| 4  
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| F  
| F  
| 5  
| 5  
| <math>0.224+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.256</math>
| <math>0.224+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.256\,\!</math>
|-
|-
| 5  
| 5  
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| F  
| F  
| 5  
| 5  
| <math>0.256+(\tfrac{1}{5})^2[3(0-\tfrac{1}{5})^2+2(1-\tfrac{1}{5})^2]=0.288</math>
| <math>0.256+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.288\,\!</math>
|-
|-
| 2  
| 2  
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| F  
| F  
| 5  
| 5  
| <math>0.288+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.320</math>
| <math>0.288+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.320\,\!</math>
|-
|-
| 1  
| 1  
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| F  
| F  
| 3  
| 3  
| <math>0.320+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+2(0-\tfrac{1}{5})^2]=0.394</math>
| <math>0.320+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.394\,\!</math>
|-
|-
| 5  
| 5  
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| F  
| F  
| 3  
| 3  
| <math>0.394+(\tfrac{1}{5})^2[2(0-\tfrac{1}{5})^2+4(1-\tfrac{1}{5})^2]=0.468</math>
| <math>0.394+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.468\,\!</math>
|-
|-
| 4  
| 4  
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| F  
| F  
| 2  
| 2  
| <math>0.468+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.593</math>
| <math>0.468+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.593\,\!</math>
|-
|-
| 5  
| 5  
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| F  
| F  
| 2  
| 2  
| <math>0.580+(\tfrac{1}{5})^2[(0-\tfrac{1}{5})^2+4(1-\tfrac{1}{5})^2]=0.718</math>
| <math>0.593+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.718\,\!</math>
|-
|-
| 3  
| 3  
Line 170: Line 170:
|}
|}


Using the equation for the MCF bounds and <span class="texhtml">''K''<sub>5</sub> = 1.644</span> for a 95% confidence level, the confidence bounds can be obtained as follows:  
Using the equation for the MCF bounds and <math>{{K}_{5}} = 1.644\,\!</math> for a 95% confidence level, the confidence bounds can be obtained as follows:  


<center><math>\begin{matrix}
<center><math>\begin{matrix}
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   \text{4} & \text{15} & \text{F} & \text{1}\text{.60} & \text{0}\text{.256} & 0.9511 & 2.6916  \\
   \text{4} & \text{15} & \text{F} & \text{1}\text{.60} & \text{0}\text{.256} & 0.9511 & 2.6916  \\
   \text{5} & \text{16} & \text{F} & \text{1}\text{.80} & \text{0}\text{.288} & 1.1023 & 2.9393  \\
   \text{5} & \text{16} & \text{F} & \text{1}\text{.80} & \text{0}\text{.288} & 1.1023 & 2.9393  \\
   \text{2} & \text{17} & \text{F} & \text{2}\text{.0} & \text{0}\text{.320} & 1.2560 & 3.1848  \\
   \text{2} & \text{17} & \text{F} & \text{2}\text{.00} & \text{0}\text{.320} & 1.2560 & 3.1848  \\
   \text{1} & \text{17} & \text{S} & {} & {} & {} & {}  \\
   \text{1} & \text{17} & \text{S} & {} & {} & {} & {}  \\
   \text{2} & \text{19} & \text{S} & {} & {} & {} & {}  \\
   \text{2} & \text{19} & \text{S} & {} & {} & {} & {}  \\
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   \text{3} & \text{26} & \text{S} & {} & {} & {} & {}  \\
   \text{3} & \text{26} & \text{S} & {} & {} & {} & {}  \\
   \text{5} & \text{28} & \text{S} & {} & {} & {} & {}  \\
   \text{5} & \text{28} & \text{S} & {} & {} & {} & {}  \\
\end{matrix}</math></center>
\end{matrix}\,\!</math></center>


The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.
The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.


[[Image:Recurrent Data Example 2 Data.png|center|550px]]  
[[Image:Recurrent Data Example 2 Data.png|center|650px]]  


Note: In the folio above, the <span class="texhtml">''F''</span> refers to failures and <span class="texhtml">''E''</span> refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.  
Note: In the folio above, the <math>F\,\!</math> refers to failures and <math>E\,\!</math> refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.  


[[Image:Recurrent Data Example 2 Result.png|center|550px]]  
[[Image:Recurrent Data Example 2 Result.png|center|650px]]  




[[Image:Recurrent Data Example 2 Plot.png|center|550px]]
[[Image:Recurrent Data Example 2 Plot.png|center|550px]]

Latest revision as of 18:53, 18 September 2023

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This example appears in the Life data analysis reference.


A health care company maintains five identical pieces of equipment used by a hospital. When a piece of equipment fails, the company sends a crew to repair it. The following table gives the failure and censoring ages for each machine, where the + sign indicates a censoring age.


[math]\displaystyle{ \begin{matrix} Equipment ID & Months \\ \text{1} & \text{5, 10 , 15, 17+} \\ \text{2} & \text{6, 13, 17, 19+} \\ \text{3} & \text{12, 20, 25, 26+} \\ \text{4} & \text{13, 15, 24+} \\ \text{5} & \text{16, 22, 25, 28+} \\ \end{matrix}\,\! }[/math]

Estimate the MCF values, with 95% confidence bounds.


Solution

The MCF estimates are obtained as follows:


[math]\displaystyle{ \begin{matrix} ID & Months ({{t}_{i}}) & State & {{r}_{i}} & 1/{{r}_{i}} & {{M}^{*}}({{t}_{i}}) \\ \text{1} & \text{5} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.20} \\ \text{2} & \text{6} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.20 + 0}\text{.20 = 0}\text{.40} \\ \text{1} & \text{10} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.40 + 0}\text{.20 = 0}\text{.60} \\ \text{3} & \text{12} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.60 + 0}\text{.20 = 0}\text{.80} \\ \text{2} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.80 + 0}\text{.20 = 1}\text{.00} \\ \text{4} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.00 + 0}\text{.20 = 1}\text{.20} \\ \text{1} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.20 + 0}\text{.20 = 1}\text{.40} \\ \text{4} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.40 + 0}\text{.20 = 1}\text{.60} \\ \text{5} & \text{16} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.60 + 0}\text{.20 = 1}\text{.80} \\ \text{2} & \text{17} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.80 + 0}\text{.20 = 2}\text{.00} \\ \text{1} & \text{17} & \text{S} & \text{4} & {} & {} \\ \text{2} & \text{19} & \text{S} & \text{3} & {} & {} \\ \text{3} & \text{20} & \text{F} & \text{3} & \text{0}\text{.33} & \text{2}\text{.00 + 0}\text{.33 = 2}\text{.33} \\ \text{5} & \text{22} & \text{F} & \text{3} & \text{0}\text{.33} & \text{2}\text{.33 + 0}\text{.33 = 2}\text{.66} \\ \text{4} & \text{24} & \text{S} & \text{2} & {} & {} \\ \text{3} & \text{25} & \text{F} & \text{2} & \text{0}\text{.50} & \text{2}\text{.66 + 0}\text{.50 = 3}\text{.16} \\ \text{5} & \text{25} & \text{F} & \text{2} & \text{0}\text{.50} & \text{3}\text{.16 + 0}\text{.50 = 3}\text{.66} \\ \text{3} & \text{26} & \text{S} & \text{1} & {} & {} \\ \text{5} & \text{28} & \text{S} & \text{0} & {} & {} \\ \end{matrix}\,\! }[/math]

Using the MCF variance equation, the following table of variance values can be obtained:

ID Months State [math]\displaystyle{ {{r}_{i}}\,\! }[/math] [math]\displaystyle{ Va{{r}_{i}}\,\! }[/math]
1 5 F 5 [math]\displaystyle{ (\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.032\,\! }[/math]
2 6 F 5 [math]\displaystyle{ 0.032+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.064\,\! }[/math]
1 10 F 5 [math]\displaystyle{ 0.064+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.096\,\! }[/math]
3 12 F 5 [math]\displaystyle{ 0.096+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.128\,\! }[/math]
2 13 F 5 [math]\displaystyle{ 0.128+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.160\,\! }[/math]
4 13 F 5 [math]\displaystyle{ 0.160+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.192\,\! }[/math]
1 15 F 5 [math]\displaystyle{ 0.192+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.224\,\! }[/math]
4 15 F 5 [math]\displaystyle{ 0.224+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.256\,\! }[/math]
5 16 F 5 [math]\displaystyle{ 0.256+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.288\,\! }[/math]
2 17 F 5 [math]\displaystyle{ 0.288+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.320\,\! }[/math]
1 17 S 4
2 19 S 3
3 20 F 3 [math]\displaystyle{ 0.320+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.394\,\! }[/math]
5 22 F 3 [math]\displaystyle{ 0.394+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.468\,\! }[/math]
4 24 S 2
3 25 F 2 [math]\displaystyle{ 0.468+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.593\,\! }[/math]
5 25 F 2 [math]\displaystyle{ 0.593+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.718\,\! }[/math]
3 26 S 1
5 28 S 0

Using the equation for the MCF bounds and [math]\displaystyle{ {{K}_{5}} = 1.644\,\! }[/math] for a 95% confidence level, the confidence bounds can be obtained as follows:

[math]\displaystyle{ \begin{matrix} ID & Months & State & MC{{F}_{i}} & Va{{r}_{i}} & MC{{F}_{{{L}_{i}}}} & MC{{F}_{{{U}_{i}}}} \\ \text{1} & \text{5} & \text{F} & \text{0}\text{.20} & \text{0}\text{.032} & 0.0459 & 0.8709 \\ \text{2} & \text{6} & \text{F} & \text{0}\text{.40} & \text{0}\text{.064} & 0.1413 & 1.1320 \\ \text{1} & \text{10} & \text{F} & \text{0}\text{.60} & \text{0}\text{.096} & 0.2566 & 1.4029 \\ \text{3} & \text{12} & \text{F} & \text{0}\text{.80} & \text{0}\text{.128} & 0.3834 & 1.6694 \\ \text{2} & \text{13} & \text{F} & \text{1}\text{.00} & \text{0}\text{.160} & 0.5179 & 1.9308 \\ \text{4} & \text{13} & \text{F} & \text{1}\text{.20} & \text{0}\text{.192} & 0.6582 & 2.1879 \\ \text{1} & \text{15} & \text{F} & \text{1}\text{.40} & \text{0}\text{.224} & 0.8028 & 2.4413 \\ \text{4} & \text{15} & \text{F} & \text{1}\text{.60} & \text{0}\text{.256} & 0.9511 & 2.6916 \\ \text{5} & \text{16} & \text{F} & \text{1}\text{.80} & \text{0}\text{.288} & 1.1023 & 2.9393 \\ \text{2} & \text{17} & \text{F} & \text{2}\text{.00} & \text{0}\text{.320} & 1.2560 & 3.1848 \\ \text{1} & \text{17} & \text{S} & {} & {} & {} & {} \\ \text{2} & \text{19} & \text{S} & {} & {} & {} & {} \\ \text{3} & \text{20} & \text{F} & \text{2}\text{.33} & \text{0}\text{.394} & 1.4990 & 3.6321 \\ \text{5} & \text{22} & \text{F} & \text{2}\text{.66} & \text{0}\text{.468} & 1.7486 & 4.0668 \\ \text{4} & \text{24} & \text{S} & {} & {} & {} & {} \\ \text{3} & \text{25} & \text{F} & \text{3}\text{.16} & \text{0}\text{.593} & 2.1226 & 4.7243 \\ \text{5} & \text{25} & \text{F} & \text{3}\text{.66} & \text{0}\text{.718} & 2.5071 & 5.3626 \\ \text{3} & \text{26} & \text{S} & {} & {} & {} & {} \\ \text{5} & \text{28} & \text{S} & {} & {} & {} & {} \\ \end{matrix}\,\! }[/math]

The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.

Recurrent Data Example 2 Data.png

Note: In the folio above, the [math]\displaystyle{ F\,\! }[/math] refers to failures and [math]\displaystyle{ E\,\! }[/math] refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.

Recurrent Data Example 2 Result.png


Recurrent Data Example 2 Plot.png
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