Template:Standard actuarial method

Standard Actuarial Method
The standard actuarial model is a variation of the simple actuarial model that involves adjusting the value for the number of operating units in an interval. The Kaplan-Meier and simple actuarial methods assume that the suspensions in a time period or interval occur at the end of that interval, after the failures have occurred. The standard actuarial model assumes that the suspensions occur in the middle of the interval, which has the effect of reducing the number of available units in the interval by half of the suspensions in that interval or:


 * $$n_{i}^{\prime }={{n}_{i}}-\frac{2}$$

With this adjustment, the calculations are carried out just as they were for the simple actuarial model in Eqn. (simpact) or:


 * $$\widehat{R}({{t}_{i}})=\underset{j=1}{\overset{i}{\mathop \prod }}\,\left( 1-\frac{n_{j}^{\prime }} \right),\text{ }i=1,...,m$$

Example 11
Find reliability estimates for the data in Example 10 using the standard actuarial method.

Solution to Example 11
The solution to this example is similar to that of Example 10, with the exception of the inclusion of the $$n_{i}^{\prime }$$  term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:

$$\begin{matrix} Start & End & Number of & Number of & Adjusted & {} & {} \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{n_{j}^{\prime }} \\ 0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\   50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962  \\   100 & 150 & 2 & 2 & 43 & 0.953 & 0.918  \\   150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844  \\   200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791  \\   250 & 300 & 1 & 2 & 28 & 0.964 & 0.762  \\   300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702  \\   350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604  \\   400 & 450 & 3 & 4 & 15 & 0.800 & 0.484  \\   450 & 500 & 1 & 2 & 9 & 0.889 & 0.430  \\   500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298  \\   550 & 600 & 1 & 0 & 4 & 0.750 & 0.223  \\   600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045  \\ \end{matrix}$$

As can be determined from the preceding table, the reliability estimates for the failure times are: