Non-Parametric LDA Example (Kaplan-Meier)
Problem Statement(Kaplan-Meier)
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(Kaplan-Meier)
Using the data and Eqn. (kapmeier), 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{}_{}^{}\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]
Non-Parametric LDA Example (Simple Actuarial Method)
Problem Statement(Simple Actuarial Method)
A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:
[math]\displaystyle{ \begin{matrix}
Start & End & Number of & Number of \\
Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} \\
0 & 50 & 2 & 4 \\
50 & 100 & 0 & 5 \\
100 & 150 & 2 & 2 \\
150 & 200 & 3 & 5 \\
200 & 250 & 2 & 1 \\
250 & 300 & 1 & 2 \\
300 & 350 & 2 & 1 \\
350 & 400 & 3 & 3 \\
400 & 450 & 3 & 4 \\
450 & 500 & 1 & 2 \\
500 & 550 & 2 & 1 \\
550 & 600 & 1 & 0 \\
600 & 650 & 2 & 1 \\
\end{matrix} }[/math]
Solution (Simple Actuarial Method)
The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):
[math]\displaystyle{ \begin{matrix}
Start & End & Number of & Number of & Available & {} & {} \\
Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \\
0 & 50 & 2 & 4 & 55 & 0.964 & 0.964 \\
50 & 100 & 0 & 5 & 49 & 1.000 & 0.964 \\
100 & 150 & 2 & 2 & 44 & 0.955 & 0.920 \\
150 & 200 & 3 & 5 & 40 & 0.925 & 0.851 \\
200 & 250 & 2 & 1 & 32 & 0.938 & 0.798 \\
250 & 300 & 1 & 2 & 29 & 0.966 & 0.770 \\
300 & 350 & 2 & 1 & 26 & 0.923 & 0.711 \\
350 & 400 & 3 & 3 & 23 & 0.870 & 0.618 \\
400 & 450 & 3 & 4 & 17 & 0.824 & 0.509 \\
450 & 500 & 1 & 2 & 10 & 0.900 & 0.458 \\
500 & 550 & 2 & 1 & 7 & 0.714 & 0.327 \\
550 & 600 & 1 & 0 & 4 & 0.750 & 0.245 \\
600 & 650 & 2 & 1 & 3 & 0.333 & 0.082 \\
\end{matrix} }[/math]
As can be determined from the preceding table, the reliability estimates for the failure times are:
[math]\displaystyle{ \begin{matrix}
Failure Period & Reliability \\
End Time & Estimate \\
50 & 96.4% \\
150 & 92.0% \\
200 & 85.1% \\
250 & 79.8% \\
300 & 77.0% \\
350 & 71.1% \\
400 & 61.8% \\
450 & 50.9% \\
500 & 45.8% \\
550 & 32.7% \\
600 & 24.5% \\
650 & 8.2% \\
\end{matrix} }[/math]
Non-Parametric LDA Example (Standard Actuarial Method)
Problem Statement(Standard Actuarial Method)
Find reliability estimates for the data in Example for the Simple Actuarial Method using the standard actuarial method.
Solution(Standard Actuarial Method)
The solution to this example is similar to that of Example 10, with the exception of the inclusion of the [math]\displaystyle{ n_{i}^{\prime } }[/math] term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:
[math]\displaystyle{ \begin{matrix}
Start & End & Number of & Number of & Adjusted & {} & {} \\
Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\
0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\
50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\
100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\
150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\
200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\
250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\
300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\
350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\
400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\
450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\
500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\
550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\
600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\
\end{matrix} }[/math]
As can be determined from the preceding table, the reliability estimates for the failure times are: