Non Parametric RDA MCF Example

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This example appears in the Non-Parametric Recurrent Event Data Analysis article.


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.


\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}\,\!

Estimate the MCF values, with 95% confidence bounds.


Solution

The MCF estimates are obtained as follows:


\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}\,\!

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

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

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

\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}\,\!

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

Note: In the folio above, the F\,\! refers to failures and E\,\! 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.


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