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.


[math]\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]\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]{{r}_{i}}\,\![/math] [math]Va{{r}_{i}}\,\![/math]
1 5 F 5 [math](\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.032\,\![/math]
2 6 F 5 [math]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]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]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]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]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]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]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]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]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]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]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]0.468+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.593\,\![/math]
5 25 F 2 [math]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]{{K}_{5}} = 1.644\,\![/math] for a 95% confidence level, the confidence bounds can be obtained as follows:

[math]\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]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.

Recurrent Data Example 2 Result.png


Recurrent Data Example 2 Plot.png