Standard Actuarial Example
Find reliability estimates for the data in Example 10 using the standard actuarial method.
Solution
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: