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====Example 9====
#REDIRECT [[Non-Parametric_Life_Data_Analysis]]
A group of 20 units are put on a life test with the following results.
 
<center><math>\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}</math></center>
 
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:
 
<center><math>\begin{matrix}
  State & Number of & Number of & Available  & {} & {}  \\
  End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{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}</math></center>
 
As can be determined from the preceding table, the reliability estimates for the failure times are:
 
<center><math>\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}</math></center>
====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:
 
<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 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):
 
<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>
====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  <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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