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{{Non-Parametric LDA Example (Kaplan-Meier)}}
#REDIRECT [[Non-Parametric_Life_Data_Analysis]]
 
====Non-Parametric LDA Example (Simple Actuarial Method)====
=====Problem Statement(Simple Actuarial Method)=====
 
A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:
 
<center><math>\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}</math></center>
 
 
=====Solution (Simple Actuarial Method)=====
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):
 
<center><math>\begin{matrix}
  Start & End & Number of & Number of & Available & {} & {}  \\
  Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{{{n}_{j}}}  \\
  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}</math></center>
 
As can be determined from the preceding table, the reliability estimates for the failure times are:
 
<center><math>\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}</math></center>
 
====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>n_{i}^{\prime }</math>  term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:
 
<center><math>\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></center>
 
 
As can be determined from the preceding table, the reliability estimates for the failure times are:

Latest revision as of 10:04, 9 August 2012

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