Standard Actuarial Example
Find reliability estimates for the data in the Actuarial-Simple Example using the actuarial-standard method.
Solution
The solution to this example is similar to that in the Actuarial-Simple Example, with the exception of the inclusion of the [math]\displaystyle{ n_{i}^{\prime } }[/math] term, which is used in the equation for the actuarial-standard method. 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 }} & \prod\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:
[math]\displaystyle{ \begin{matrix}
Failure Period & Reliability \\
End Time & Estimate \\
50 & 96.2% \\
150 & 91.8% \\
200 & 84.4% \\
250 & 79.1% \\
300 & 76.2% \\
350 & 70.2% \\
400 & 60.4% \\
450 & 48.4% \\
500 & 43.0% \\
550 & 29.8% \\
600 & 22.3% \\
650 & 4.5% \\
\end{matrix} }[/math]