Kaplan-Meier Example
A group of 20 units are put on a life test with the following results.
[math]\displaystyle{ \begin{matrix}
Number & State & State \\
in State & (F or S) & End Time \\
3 & F & 9 \\
1 & S & 9 \\
1 & F & 11 \\
1 & S & 12 \\
1 & F & 13 \\
1 & S & 13 \\
1 & S & 15 \\
1 & F & 17 \\
1 & F & 21 \\
1 & S & 22 \\
1 & S & 24 \\
1 & S & 26 \\
1 & F & 28 \\
1 & F & 30 \\
1 & S & 32 \\
2 & S & 35 \\
1 & S & 39 \\
1 & S & 41 \\
\end{matrix} }[/math]
Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.
Solution
Using the data and the reliability equation of the Kaplan-Meier estimator, the following table can be constructed:
[math]\displaystyle{ \begin{matrix}
State & Number of & Number of & Available & {} & {} \\
End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} & \mathop{}_{}^{}\prod\tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} \\
9 & 3 & 1 & 20 & 0.850 & 0.850 \\
11 & 1 & 0 & 16 & 0.938 & 0.797 \\
12 & 0 & 1 & 15 & 1.000 & 0.797 \\
13 & 1 & 1 & 14 & 0.929 & 0.740 \\
15 & 0 & 1 & 12 & 1.000 & 0.740 \\
17 & 1 & 0 & 11 & 0.909 & 0.673 \\
21 & 1 & 0 & 10 & 0.900 & 0.605 \\
22 & 0 & 1 & 9 & 1.000 & 0.605 \\
24 & 0 & 1 & 8 & 1.000 & 0.605 \\
26 & 0 & 1 & 7 & 1.000 & 0.605 \\
28 & 1 & 0 & 6 & 0.833 & 0.505 \\
30 & 1 & 0 & 5 & 0.800 & 0.404 \\
32 & 0 & 1 & 4 & 1.000 & 0.404 \\
35 & 0 & 1 & 3 & 1.000 & 0.404 \\
39 & 0 & 1 & 2 & 1.000 & 0.404 \\
41 & 0 & 1 & 1 & 1.000 & 0.404 \\
\end{matrix} }[/math]
As can be determined from the preceding table, the reliability estimates for the failure times are:
[math]\displaystyle{ \begin{matrix}
Failure Time & Reliability Est. \\
9 & 85.0% \\
11 & 79.7% \\
13 & 74.0% \\
17 & 67.3% \\
21 & 60.5% \\
28 & 50.5% \\
30 & 40.4% \\
\end{matrix} }[/math]