Non-Parametric Life Data Analysis

Nonparametric analysis allows the user to analyze data without assuming an underlying distribution. This can have certain advantages as well as disadvantages. The ability to analyze data without assuming an underlying life distribution avoids the potentially large errors brought about by making incorrect assumptions about the distribution. On the other hand, the confidence bounds associated with nonparametric analysis are usually much wider than those calculated via parametric analysis, and predictions outside the range of the observations are not possible. Some practitioners recommend that any set of life data should first be subjected to a nonparametric analysis before moving on to the assumption of an underlying distribution. There are several methods for conducting a nonparametric analysis, including the Kaplan-Meier, simple actuarial, and standard actuarial methods. A method for attaching confidence bounds to the results of these nonparametric analysis techniques can also be developed. The basis of nonparametric life data analysis is the empirical [math]\displaystyle{ cdf }[/math] function, which is given by:

[math]\displaystyle{ \widehat{F}(t)=\frac{observations\le t}{n} }[/math]

Note that this is similar to the Bernard's approximation of the median ranks, as discussed in Chapter 3. The following nonparametric analysis methods are essentially variations of this concept.

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]

Simple Actuarial Method

The simple actuarial method is an easy-to-use form of nonparametric data analysis that can be used for multiply censored data that are arranged in intervals. This method is based on calculating the number of failures in a time interval, [math]\displaystyle{ {{r}_{j}}, }[/math] versus the number of operating units in that time period, [math]\displaystyle{ {{n}_{j}} }[/math] . The equation for the reliability estimator for the standard actuarial method is given by:

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

where:

[math]\displaystyle{ \begin{align} & m= & \text{the total number of intervals} \\ & 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 interval }j \\ & {{s}_{j}}= & \text{the number of suspensions in interval }j \end{align} }[/math]

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:

[math]\displaystyle{ \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]

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):

[math]\displaystyle{ \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]

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

[math]\displaystyle{ \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]

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:

[math]\displaystyle{ n_{i}^{\prime }={{n}_{i}}-\frac{{{s}_{i}}}{2} }[/math]

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

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

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]\displaystyle{ n_{i}^{\prime } }[/math] term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:

[math]\displaystyle{ \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] As can be determined from the preceding table, the reliability estimates for the failure times are:

Nonparametric Confidence Bounds

Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for nonparametric data, Weibull++ uses Greenwood's formula [27]:

[math]\displaystyle{ \widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right]}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)} }[/math]

where:

[math]\displaystyle{ \begin{align} & m= & \text{ the total number of intervals} \\ & 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 interval }j \\ & {{s}_{j}}= & \text{the number of suspensions in interval }j \end{align} }[/math]

Once the variance has been calculated, the standard error can be determined by taking the square root of the variance:

[math]\displaystyle{ {{\widehat{se}}_{\widehat{R}}}=\sqrt{\widehat{Var}(\widehat{R}({{t}_{i}}))} }[/math]

This information can then be applied to determine the confidence bounds:

[math]\displaystyle{ \left[ LC{{B}_{\widehat{R}}},\text{ }UC{{B}_{\widehat{R}}} \right]=\left[ \frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})\cdot w},\text{ }\frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})/w} \right] }[/math]

where:

[math]\displaystyle{ w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}} }[/math]

and [math]\displaystyle{ \alpha }[/math] is the desired confidence level for the 1-sided confidence bounds.

Example 12

Determine the 1-sided confidence bounds for the reliability estimates in Example 11, with a 95% confidence level.

Solution to Example 12

Once again, this type of problem is most readily solved by constructing a table similar to the following:

The following plot illustrates these results graphically: