Template:Example: Standard Actuarial Example

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 $$n_{i}^{\prime }$$ term, which is used in the equation for the actuarial-standard method. Applying this equation to the data, we can generate the following table: $$\begin{matrix} Start & End & Number of & Number of & Adjusted & {} & {} \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{n_{j}^{\prime }} & \prod\mathop{}_{}^{}1-\tfrac{n_{j}^{\prime }} \\ 0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\   50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962  \\   100 & 150 & 2 & 2 & 43 & 0.953 & 0.918  \\   150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844  \\   200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791  \\   250 & 300 & 1 & 2 & 28 & 0.964 & 0.762  \\   300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702  \\   350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604  \\   400 & 450 & 3 & 4 & 15 & 0.800 & 0.484  \\   450 & 500 & 1 & 2 & 9 & 0.889 & 0.430  \\   500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298  \\   550 & 600 & 1 & 0 & 4 & 0.750 & 0.223  \\   600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045  \\ \end{matrix}$$ As can be determined from the preceding table, the reliability estimates for the failure times are: $$\begin{matrix} Failure Period & Reliability \\ End Time & Estimate \\ 50 & 96.2% \\   150 & 91.8%  \\   200 & 84.4%  \\   250 & 79.1%  \\   300 & 76.2%  \\   350 & 70.2%  \\   400 & 60.4%  \\   450 & 48.4%  \\   500 & 43.0%  \\   550 & 29.8%  \\   600 & 22.3%  \\   650 & 4.5%  \\ \end{matrix}$$