Non Parametric RDA MCF Example

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.

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

Estimate the MCF values, with 95% confidence bounds.

Solution

The MCF estimates are obtained as follows:

${\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}}\,\!}$

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

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

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

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

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 ${\displaystyle F\,\!}$ refers to failures and ${\displaystyle E\,\!}$ 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.