Template:Non-parametric LDA Examples

Example 9
A group of 20 units are put on a life test with the following results.

$$\begin{matrix} Number & State & State \\ in State & (F or S) & End Time \\ 3 & F & 9 \\ 1 & S & 9 \\ 1 & F & 11 \\ 1 & S & 12 \\ 1 & F & 13 \\ 1 & S & 13 \\ 1 & S & 15 \\ 1 & F & 17 \\ 1 & F & 21 \\ 1 & S & 22 \\ 1 & S & 24 \\ 1 & S & 26 \\ 1 & F & 28 \\ 1 & F & 30 \\ 1 & S & 32 \\ 2 & S & 35 \\ 1 & S & 39 \\ 1 & S & 41 \\ \end{matrix}$$

Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.

Solution to Example 9
Using the data and Eqn. (kapmeier), the following table can be constructed:

$$\begin{matrix} State & Number of & Number of & Available & {} & {}  \\ End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}} \\ 9 & 3 & 1 & 20 & 0.850 & 0.850 \\   11 & 1 & 0 & 16 & 0.938 & 0.797  \\   12 & 0 & 1 & 15 & 1.000 & 0.797  \\   13 & 1 & 1 & 14 & 0.929 & 0.740  \\   15 & 0 & 1 & 12 & 1.000 & 0.740  \\   17 & 1 & 0 & 11 & 0.909 & 0.673  \\   21 & 1 & 0 & 10 & 0.900 & 0.605  \\   22 & 0 & 1 & 9 & 1.000 & 0.605  \\   24 & 0 & 1 & 8 & 1.000 & 0.605  \\   26 & 0 & 1 & 7 & 1.000 & 0.605  \\   28 & 1 & 0 & 6 & 0.833 & 0.505  \\   30 & 1 & 0 & 5 & 0.800 & 0.404  \\   32 & 0 & 1 & 4 & 1.000 & 0.404  \\   35 & 0 & 1 & 3 & 1.000 & 0.404  \\   39 & 0 & 1 & 2 & 1.000 & 0.404  \\   41 & 0 & 1 & 1 & 1.000 & 0.404  \\ \end{matrix}$$

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

$$\begin{matrix} Failure Time & Reliability Est. \\ 9 & 85.0% \\   11 & 79.7%  \\   13 & 74.0%  \\   17 & 67.3%  \\   21 & 60.5%  \\   28 & 50.5%  \\   30 & 40.4%  \\ \end{matrix}$$

Example 10
A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:

$$\begin{matrix} Start & End & Number of & Number of \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} \\ 0 & 50 & 2 & 4 \\   50 & 100 & 0 & 5  \\   100 & 150 & 2 & 2  \\   150 & 200 & 3 & 5  \\   200 & 250 & 2 & 1  \\   250 & 300 & 1 & 2  \\   300 & 350 & 2 & 1  \\   350 & 400 & 3 & 3  \\   400 & 450 & 3 & 4  \\   450 & 500 & 1 & 2  \\   500 & 550 & 2 & 1  \\   550 & 600 & 1 & 0  \\   600 & 650 & 2 & 1  \\ \end{matrix}$$

Solution to Example 10
The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):

$$\begin{matrix} Start & End & Number of & Number of & Available & {} & {} \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & 1-\tfrac & \mathop{}_{}^{}1-\tfrac \\ 0 & 50 & 2 & 4 & 55 & 0.964 & 0.964 \\   50 & 100 & 0 & 5 & 49 & 1.000 & 0.964  \\   100 & 150 & 2 & 2 & 44 & 0.955 & 0.920  \\   150 & 200 & 3 & 5 & 40 & 0.925 & 0.851  \\   200 & 250 & 2 & 1 & 32 & 0.938 & 0.798  \\   250 & 300 & 1 & 2 & 29 & 0.966 & 0.770  \\   300 & 350 & 2 & 1 & 26 & 0.923 & 0.711  \\   350 & 400 & 3 & 3 & 23 & 0.870 & 0.618  \\   400 & 450 & 3 & 4 & 17 & 0.824 & 0.509  \\   450 & 500 & 1 & 2 & 10 & 0.900 & 0.458  \\   500 & 550 & 2 & 1 & 7 & 0.714 & 0.327  \\   550 & 600 & 1 & 0 & 4 & 0.750 & 0.245  \\   600 & 650 & 2 & 1 & 3 & 0.333 & 0.082  \\ \end{matrix}$$

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

$$\begin{matrix} Failure Period & Reliability \\ End Time & Estimate \\ 50 & 96.4% \\   150 & 92.0%  \\   200 & 85.1%  \\   250 & 79.8%  \\   300 & 77.0%  \\   350 & 71.1%  \\   400 & 61.8%  \\   450 & 50.9%  \\   500 & 45.8%  \\   550 & 32.7%  \\   600 & 24.5%  \\   650 & 8.2%  \\ \end{matrix}$$

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: