Template:Non-Parametric LDA Example (Standard Actuarial Method)

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: