Template:Kaplan-meier estimator

Kaplan-Meier Estimator

The Kaplan-Meier estimator, also known as the product limit estimator, can be used to calculate values for nonparametric reliability for data sets with multiple failures and suspensions. The equation of the estimator is given by:

[math]\displaystyle{ \widehat{R}({{t}_{i}})=\underset{j=1}{\overset{i}{\mathop \prod }}\,\frac{{{n}_{j}}-{{r}_{j}}}{{{n}_{j}}},\text{ }i=1,...,m }[/math]

where:

[math]\displaystyle{ \begin{align} & m= & \text{the total number of data points} \\ & n= & \text{the total number of units} \end{align} }[/math]

The variable [math]\displaystyle{ {{n}_{i}} }[/math] is defined by:

[math]\displaystyle{ {{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m }[/math]

where:

[math]\displaystyle{ \begin{align} & {{r}_{j}}= & \text{the number of failures in the }{{j}^{th}}\text{ data group} \\ & {{s}_{j}}= & \text{the number of suspensions in the }{{j}^{th}}\text{ data group} \end{align} }[/math]

Note that the reliability estimate is only calculated for times at which one or more failures occurred. For the sake of calculating the value of [math]\displaystyle{ {{n}_{j}} }[/math] at time values that have failures and suspensions, it is assumed that the suspensions occur slightly after the failures, so that the suspended units are considered to be operating and included in the count of [math]\displaystyle{ {{n}_{j}} }[/math] .

Example 9

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

[math]\displaystyle{ \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]

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:

[math]\displaystyle{ \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]

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

[math]\displaystyle{ \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]