Template:Weibull++ Examples and Case Studies

Confidence Bound Examples

Likelihood Ratio Bounds on Parameters

Likelihood Ratio Bounds on Parameters

Five units were put on a reliability test and experienced failures at 10, 20, 30, 40 and 50 hours. Assuming a Weibull distribution, the MLE parameter estimates are calculated to be [math]\displaystyle{ \widehat{\beta }=2.2938\,\! }[/math] and [math]\displaystyle{ \widehat{\eta }=33.9428.\,\! }[/math] Calculate the 90% two-sided confidence bounds on these parameters using the likelihood ratio method.

Solution

The first step is to calculate the likelihood function for the parameter estimates:

[math]\displaystyle{ \begin{align} L(\widehat{\beta },\widehat{\eta })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\widehat{\beta },\widehat{\eta })=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\widehat{\beta }}{\widehat{\eta }}\cdot {{\left( \frac{{{x}_{i}}}{\widehat{\eta }} \right)}^{\widehat{\beta }-1}}\cdot {{e}^{-{{\left( \tfrac{{{x}_{i}}}{\widehat{\eta }} \right)}^{\widehat{\beta }}}}} \\ \\ L(\widehat{\beta },\widehat{\eta })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{2.2938}{33.9428}\cdot {{\left( \frac{{{x}_{i}}}{33.9428} \right)}^{1.2938}}\cdot {{e}^{-{{\left( \tfrac{{{x}_{i}}}{33.9428} \right)}^{2.2938}}}} \\ \\ L(\widehat{\beta },\widehat{\eta })= & 1.714714\times {{10}^{-9}} \end{align}\,\! }[/math]

where [math]\displaystyle{ {{x}_{i}}\,\! }[/math] are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:

[math]\displaystyle{ L(\beta ,\eta )-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0\,\! }[/math]

Since our specified confidence level, [math]\displaystyle{ \delta\,\! }[/math], is 90%, we can calculate the value of the chi-squared statistic, [math]\displaystyle{ \chi _{0.9;1}^{2}=2.705543.\,\! }[/math] We then substitute this information into the equation:

[math]\displaystyle{ \begin{align} L(\beta ,\eta )-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\ L(\beta ,\eta )-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ \\ L(\beta ,\eta )-4.432926\cdot {{10}^{-10}}= & 0 \end{align}\,\! }[/math]

The next step is to find the set of values of [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \eta\,\! }[/math] that satisfy this equation, or find the values of [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \eta\,\! }[/math] such that [math]\displaystyle{ L(\beta ,\eta )=4.432926\cdot {{10}^{-10}}.\,\! }[/math]

The solution is an iterative process that requires setting the value of [math]\displaystyle{ \beta\,\! }[/math] and finding the appropriate values of [math]\displaystyle{ \eta\,\! }[/math], and vice versa. The following table gives values of [math]\displaystyle{ \beta\,\! }[/math] based on given values of [math]\displaystyle{ \eta\,\! }[/math].

These data are represented graphically in the following contour plot:

(Note that this plot is generated with degrees of freedom [math]\displaystyle{ k = 1\,\! }[/math], as we are only determining bounds on one parameter. The contour plots generated in Weibull++ are done with degrees of freedom [math]\displaystyle{ k = 2\,\! }[/math], for use in comparing both parameters simultaneously.) As can be determined from the table, the lowest calculated value for [math]\displaystyle{ \beta\,\! }[/math] is 1.142, while the highest is 3.950. These represent the two-sided 90% confidence limits on this parameter. Since solutions for the equation do not exist for values of [math]\displaystyle{ \eta\,\! }[/math] below 23 or above 50, these can be considered the 90% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on [math]\displaystyle{ \eta\,\! }[/math], we can perform the same procedure as before, but finding the two values of [math]\displaystyle{ \eta\,\! }[/math] that correspond with a given value of [math]\displaystyle{ \beta\,\! }[/math] Using this method, we find that the 90% confidence limits on [math]\displaystyle{ \eta\,\! }[/math] are 22.474 and 49.967, which are close to the initial estimates of 23 and 50.

Note that the points where [math]\displaystyle{ \beta\,\! }[/math] are maximized and minimized do not necessarily correspond with the points where [math]\displaystyle{ \eta\,\! }[/math] are maximized and minimized. This is due to the fact that the contour plot is not symmetrical, so that the parameters will have their extremes at different points.

Likelihood Ratio Bounds on Time (Type I)

Likelihood Ratio Bounds on Time (Type I)

For the data given in Example 1, determine the 90% two-sided confidence bounds on the time estimate for a reliability of 50%. The ML estimate for the time at which [math]\displaystyle{ R(t)=50%\,\! }[/math] is 28.930.

Solution

In this example, we are trying to determine the 90% two-sided confidence bounds on the time estimate of 28.930. As was mentioned, we need to rewrite the likelihood ratio equation so that it is in terms of [math]\displaystyle{ t\,\! }[/math] and [math]\displaystyle{ \beta .\,\! }[/math] This is accomplished by using a form of the Weibull reliability equation, [math]\displaystyle{ R={{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}.\,\! }[/math] This can be rearranged in terms of [math]\displaystyle{ \eta \,\! }[/math], with [math]\displaystyle{ R\,\! }[/math] being considered a known variable or:

[math]\displaystyle{ \eta =\frac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}}\,\! }[/math]

This can then be substituted into the [math]\displaystyle{ \eta \,\! }[/math] term in the likelihood ratio equation to form a likelihood equation in terms of [math]\displaystyle{ t\,\! }[/math] and [math]\displaystyle{ \beta \,\! }[/math] or:

[math]\displaystyle{ \begin{align} & L(\beta ,t)= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\beta ,t,R) \\ & & \end{align}\,\! }[/math]

[math]\displaystyle{ =\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\beta }{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)}\cdot {{\left( \frac{{{x}_{i}}}{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)} \right)}^{\beta -1}}\cdot \text{exp}\left[ -{{\left( \frac{{{x}_{i}}}{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)} \right)}^{\beta }} \right]\,\! }[/math]

where [math]\displaystyle{ {{x}_{i}}\,\! }[/math] are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:

[math]\displaystyle{ L(\beta ,t)-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0\,\! }[/math]

Since our specified confidence level, [math]\displaystyle{ \delta \,\! }[/math], is 90%, we can calculate the value of the chi-squared statistic, [math]\displaystyle{ \chi _{0.9;1}^{2}=2.705543.\,\! }[/math] We can now substitute this information into the equation:

[math]\displaystyle{ \begin{align} L(\beta ,t)-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\ L(\beta ,t)-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ & \\ L(\beta ,t)-4.432926\cdot {{10}^{-10}}= & 0 \end{align}\,\! }[/math]

Note that the likelihood value for [math]\displaystyle{ L(\widehat{\beta },\widehat{\eta })\,\! }[/math] is the same as it was for Example 1. This is because we are dealing with the same data and parameter estimates or, in other words, the maximum value of the likelihood function did not change. It now remains to find the values of [math]\displaystyle{ \beta \,\! }[/math] and [math]\displaystyle{ t\,\! }[/math] which satisfy this equation. This is an iterative process that requires setting the value of [math]\displaystyle{ \beta \,\! }[/math] and finding the appropriate values of [math]\displaystyle{ t\,\! }[/math]. The following table gives the values of [math]\displaystyle{ t\,\! }[/math] based on given values of [math]\displaystyle{ \beta \,\! }[/math].

These points are represented graphically in the following contour plot:

As can be determined from the table, the lowest calculated value for [math]\displaystyle{ t\,\! }[/math] is 17.389, while the highest is 41.714. These represent the 90% two-sided confidence limits on the time at which reliability is equal to 50%.

Likelihood Ratio Bounds on Reliability (Type 2)

Likelihood Ratio Bounds on Reliability (Type 2)

For the data given in Example 1, determine the 90% two-sided confidence bounds on the reliability estimate for [math]\displaystyle{ t=45\,\! }[/math]. The ML estimate for the reliability at [math]\displaystyle{ t=45\,\! }[/math] is 14.816%.

Solution

In this example, we are trying to determine the 90% two-sided confidence bounds on the reliability estimate of 14.816%. As was mentioned, we need to rewrite the likelihood ratio equation so that it is in terms of [math]\displaystyle{ R\,\! }[/math] and [math]\displaystyle{ \beta .\,\! }[/math] This is again accomplished by substituting the Weibull reliability equation into the [math]\displaystyle{ \eta \,\! }[/math] term in the likelihood ratio equation to form a likelihood equation in terms of [math]\displaystyle{ R\,\! }[/math] and [math]\displaystyle{ \beta \,\! }[/math]:

[math]\displaystyle{ \begin{align} & L(\beta ,R)= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\beta ,t,R) \\ & & \end{align}\,\! }[/math]

[math]\displaystyle{ =\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\beta }{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)}\cdot {{\left( \frac{{{x}_{i}}}{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)} \right)}^{\beta -1}}\cdot \text{exp}\left[ -{{\left( \frac{{{x}_{i}}}{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)} \right)}^{\beta }} \right]\,\! }[/math]

where [math]\displaystyle{ {{x}_{i}}\,\! }[/math] are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:

[math]\displaystyle{ L(\beta ,R)-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0\,\! }[/math]

Since our specified confidence level, [math]\displaystyle{ \delta \,\! }[/math], is 90%, we can calculate the value of the chi-squared statistic, [math]\displaystyle{ \chi _{0.9;1}^{2}=2.705543.\,\! }[/math] We can now substitute this information into the equation:

[math]\displaystyle{ \begin{align} L(\beta ,R)-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\ L(\beta ,R)-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ \\ L(\beta ,R)-4.432926\cdot {{10}^{-10}}= & 0 \end{align}\,\! }[/math]

It now remains to find the values of [math]\displaystyle{ \beta \,\! }[/math] and [math]\displaystyle{ R\,\! }[/math] that satisfy this equation. This is an iterative process that requires setting the value of [math]\displaystyle{ \beta \,\! }[/math] and finding the appropriate values of [math]\displaystyle{ R\,\! }[/math]. The following table gives the values of [math]\displaystyle{ R\,\! }[/math] based on given values of [math]\displaystyle{ \beta \,\! }[/math].

These points are represented graphically in the following contour plot:

As can be determined from the table, the lowest calculated value for [math]\displaystyle{ R\,\! }[/math] is 2.38%, while the highest is 44.26%. These represent the 90% two-sided confidence limits on the reliability at [math]\displaystyle{ t=45\,\! }[/math].

Comparing Parameter Estimation Methods Using Simulation Based Bounds

Comparing Parameter Estimation Methods Using Simulation Based Bounds

The purpose of this example is to determine the best parameter estimation method for a sample of ten units with complete time-to-failure data for each unit (i.e., no censoring). The data set follows a Weibull distribution with [math]\displaystyle{ \beta =2\,\! }[/math] and [math]\displaystyle{ \eta =100\,\! }[/math] hours.

The confidence bounds for the data set could be obtained by using Weibull++'s SimuMatic utility. To obtain the results, use the following settings in SimuMatic.

1. On the Main tab, choose the 2P-Weibull distribution and enter the given parameters (i.e., [math]\displaystyle{ \beta =2\,\! }[/math] and [math]\displaystyle{ \eta =100\,\! }[/math] hours)
2. On the Censoring tab, select the No censoring option.
3. On the Settings tab, set the number of data sets to 1,000 and the number of data points to 10.
4. On the Analysis tab, choose the RRX analysis method and set the confidence bounds to 90.

The following plot shows the simulation-based confidence bounds for the RRX parameter estimation method, as well as the expected variation due to sampling error.

Create another SimuMatic folio and generate a second data using the same settings, but this time, select the RRY analysis method on the Analysis tab. The following plot shows the result.

The following plot shows the results using the MLE analysis method.

The results clearly demonstrate that the median RRX estimate provides the least deviation from the truth for this sample size and data type. However, the MLE outputs are grouped more closely together, as evidenced by the bounds.

This experiment can be repeated in SimuMatic using multiple censoring schemes (including Type I and Type II right censoring as well as random censoring) with various distributions. Multiple experiments can be performed with this utility to evaluate assumptions about the appropriate parameter estimation method to use for data sets.

Exponential Disribution Examples

One Parameter Exponential Probability Plot Example

1-Parameter Exponential Probability Plot Example

6 units are put on a life test and tested to failure. The failure times are 7, 12, 19, 29, 41, and 67 hours. Estimate the failure rate for a 1-parameter exponential distribution using the probability plotting method.

In order to plot the points for the probability plot, the appropriate reliability estimate values must be obtained. These will be equivalent to [math]\displaystyle{ 100%-MR\,\! }[/math] since the y-axis represents the reliability and the [math]\displaystyle{ MR\,\! }[/math] values represent unreliability estimates.

[math]\displaystyle{ \begin{matrix} \text{Time-to-failure, hr} & \text{Reliability Estimate, }% \\ 7 & 100-10.91=89.09% \\ 12 & 100-26.44=73.56% \\ 19 & 100-42.14=57.86% \\ 29 & 100-57.86=42.14% \\ 41 & 100-73.56=26.44% \\ 67 & 100-89.09=10.91% \\ \end{matrix}\,\! }[/math]

Next, these points are plotted on an exponential probability plotting paper. A sample of this type of plotting paper is shown next, with the sample points in place. Notice how these points describe a line with a negative slope.

Once the points are plotted, draw the best possible straight line through these points. The time value at which this line intersects with a horizontal line drawn at the 36.8% reliability mark is the mean life, and the reciprocal of this is the failure rate [math]\displaystyle{ \lambda\,\! }[/math]. This is because at [math]\displaystyle{ t=m=\tfrac{1}{\lambda }\,\! }[/math]:

[math]\displaystyle{ \begin{align} R(t)= & {{e}^{-\lambda \cdot t}} \\ R(t)= & {{e}^{-\lambda \cdot \tfrac{1}{\lambda }}} \\ R(t)= & {{e}^{-1}}=0.368=36.8%. \end{align}\,\! }[/math]

The following plot shows that the best-fit line through the data points crosses the [math]\displaystyle{ R=36.8%\,\! }[/math] line at [math]\displaystyle{ t=33\,\! }[/math] hours. And because [math]\displaystyle{ \tfrac{1}{\lambda }=33\,\! }[/math] hours, [math]\displaystyle{ \lambda =0.0303\,\! }[/math] failures/hour.

2 Parameter Exponential Distribution RRY

The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where [math]\displaystyle{ \beta =1\,\! }[/math]. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).

Exponential Probability Density Function

The 2-Parameter Exponential Distribution

The 2-parameter exponential pdf is given by:

[math]\displaystyle{ f(t)=\lambda {{e}^{-\lambda (t-\gamma )}},f(t)\ge 0,\lambda \gt 0,t\ge \gamma \,\! }[/math]

where [math]\displaystyle{ \gamma \,\! }[/math] is the location parameter. Some of the characteristics of the 2-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{t}-\gamma =m-\gamma \,\! }[/math].
• The exponential pdf has no shape parameter, as it has only one shape.
• The distribution starts at [math]\displaystyle{ t=\gamma \,\! }[/math] at the level of [math]\displaystyle{ f(t=\gamma )=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases beyond [math]\displaystyle{ \gamma \,\! }[/math] and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The 1-Parameter Exponential Distribution

The 1-parameter exponential pdf is obtained by setting [math]\displaystyle{ \gamma =0\,\! }[/math], and is given by:

[math]\displaystyle{ \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, & t\ge 0, \lambda \gt 0,m\gt 0 \end{align} \,\! }[/math]

where:

[math]\displaystyle{ \lambda \,\! }[/math] = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)

[math]\displaystyle{ \lambda =\frac{1}{m}\,\! }[/math]

[math]\displaystyle{ m\,\! }[/math] = mean time between failures, or to failure

[math]\displaystyle{ t\,\! }[/math] = operating time, life, or age, in hours, cycles, miles, actuations, etc.

This distribution requires the knowledge of only one parameter, [math]\displaystyle{ \lambda \,\! }[/math], for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], is zero.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=m\,\! }[/math].
• As [math]\displaystyle{ \lambda \,\! }[/math] is decreased in value, the distribution is stretched out to the right, and as [math]\displaystyle{ \lambda \,\! }[/math] is increased, the distribution is pushed toward the origin.
• This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, [math]\displaystyle{ \lambda \,\! }[/math]).
• The distribution starts at [math]\displaystyle{ t=0\,\! }[/math] at the level of [math]\displaystyle{ f(t=0)=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases, and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].
• The pdf can be thought of as a special case of the Weibull pdf with [math]\displaystyle{ \gamma =0\,\! }[/math] and [math]\displaystyle{ \beta =1\,\! }[/math].

Exponential Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T},\,\! }[/math] or mean time to failure (MTTF) is given by:

[math]\displaystyle{ \begin{align} \bar{T}= & \int_{\gamma }^{\infty }t\cdot f(t)dt \\ = & \int_{\gamma }^{\infty }t\cdot \lambda \cdot {{e}^{-\lambda t}}dt \\ = & \gamma +\frac{1}{\lambda }=m \end{align}\,\! }[/math]

Note that when [math]\displaystyle{ \gamma =0\,\! }[/math], the MTTF is the inverse of the exponential distribution's constant failure rate. This is only true for the exponential distribution. Most other distributions do not have a constant failure rate. Consequently, the inverse relationship between failure rate and MTTF does not hold for these other distributions.

The Median

The median, [math]\displaystyle{ \breve{T}, \,\! }[/math] is:

[math]\displaystyle{ \breve{T}=\gamma +\frac{1}{\lambda}\cdot 0.693 \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T},\,\! }[/math] is:

[math]\displaystyle{ \tilde{T}=\gamma \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ {\sigma }_{T}\,\! }[/math], is:

[math]\displaystyle{ {\sigma}_{T}=\frac{1}{\lambda }=m\,\! }[/math]

The Exponential Reliability Function

The equation for the 2-parameter exponential cumulative density function, or cdf, is given by:

[math]\displaystyle{ \begin{align} F(t)=Q(t)=1-{{e}^{-\lambda (t-\gamma )}} \end{align}\,\! }[/math]

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function of the 2-parameter exponential distribution is given by:

[math]\displaystyle{ R(t)=1-Q(t)=1-\int_{0}^{t-\gamma }f(x)dx\,\! }[/math]

[math]\displaystyle{ R(t)=1-\int_{0}^{t-\gamma }\lambda {{e}^{-\lambda x}}dx={{e}^{-\lambda (t-\gamma )}}\,\! }[/math]

The 1-parameter exponential reliability function is given by:

[math]\displaystyle{ R(t)={{e}^{-\lambda t}}={{e}^{-\tfrac{t}{m}}}\,\! }[/math]

The Exponential Conditional Reliability Function

The exponential conditional reliability equation gives the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration, having already successfully accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation up to the start of this new mission. The exponential conditional reliability function is:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{{{e}^{-\lambda (T+t-\gamma )}}}{{{e}^{-\lambda (T-\gamma )}}}={{e}^{-\lambda t}}\,\! }[/math]

which says that the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration undertaken after the component or equipment has already accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. This is referred to as the memoryless property.

The Exponential Reliable Life Function

The reliable life, or the mission duration for a desired reliability goal, [math]\displaystyle{ {{t}_{R}}\,\! }[/math], for the 1-parameter exponential distribution is:

[math]\displaystyle{ R({{t}_{R}})={{e}^{-\lambda ({{t}_{R}}-\gamma )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \ln[R({{t}_{R}})]=-\lambda({{t}_{R}}-\gamma ) \end{align}\,\! }[/math]

or:

[math]\displaystyle{ {{t}_{R}}=\gamma -\frac{\ln [R({{t}_{R}})]}{\lambda }\,\! }[/math]

The Exponential Failure Rate Function

The exponential failure rate function is:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\lambda {{e}^{-\lambda (t-\gamma )}}}{{{e}^{-\lambda (t-\gamma )}}}=\lambda =\text{constant}\,\! }[/math]

Once again, note that the constant failure rate is a characteristic of the exponential distribution, and special cases of other distributions only. Most other distributions have failure rates that are functions of time.

Characteristics of the Exponential Distribution

The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. Unfortunately, this fact also leads to the use of this model in situations where it is not appropriate. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The constant failure rate of the exponential distribution would require the assumption that the automobile would be just as likely to experience a breakdown during the first mile as it would during the one-hundred-thousandth mile. Clearly, this is not a valid assumption. However, some inexperienced practitioners of reliability engineering and life data analysis will overlook this fact, lured by the siren-call of the exponential distribution's relatively simple mathematical models.

The Effect of lambda and gamma on the Exponential pdf

• The exponential pdf has no shape parameter, as it has only one shape.
• The exponential pdf is always convex and is stretched to the right as [math]\displaystyle{ \lambda \,\! }[/math] decreases in value.
• The value of the pdf function is always equal to the value of [math]\displaystyle{ \lambda \,\! }[/math] at [math]\displaystyle{ t=0\,\! }[/math] (or [math]\displaystyle{ t=\gamma \,\! }[/math]).
• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before this time.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{T}-\gamma =m-\gamma \,\! }[/math].
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The Effect of lambda and gamma on the Exponential Reliability Function

• The 1-parameter exponential reliability function starts at the value of 100% at [math]\displaystyle{ t=0\,\! }[/math], decreases thereafter monotonically and is convex.
• The 2-parameter exponential reliability function remains at the value of 100% for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and decreases thereafter monotonically and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ R(t\to \infty )\to 0\,\! }[/math].
• The reliability for a mission duration of [math]\displaystyle{ t=m=\tfrac{1}{\lambda }\,\! }[/math], or of one MTTF duration, is always equal to [math]\displaystyle{ 0.3679\,\! }[/math] or 36.79%. This means that the reliability for a mission which is as long as one MTTF is relatively low and is not recommended because only 36.8% of the missions will be completed successfully. In other words, of the equipment undertaking such a mission, only 36.8% will survive their mission.

The Effect of lambda and gamma on the Failure Rate Function

• The 1-parameter exponential failure rate function is constant and starts at [math]\displaystyle{ t=0\,\! }[/math].
• The 2-parameter exponential failure rate function remains at the value of 0 for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and then keeps at the constant value of [math]\displaystyle{ \lambda\,\! }[/math].

Exponential Distribution Examples

Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
Number of Units in Group	Time-to-Failure
7	100
5	200
3	300
2	400
1	500
2	600

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the pdf plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

Solution

1. For the 2-parameter exponential distribution and for [math]\displaystyle{ \hat{\gamma }=100\,\! }[/math] hours (first failure), the partial of the log-likelihood function, [math]\displaystyle{ \lambda\,\! }[/math], becomes:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\ \Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)] + \ldots +2[\frac{1}{\lambda}-(600-100)]\\ = & 0\\ \Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr} \end{align} \,\! }[/math]

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the pdf plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, [math]\displaystyle{ \gamma \,\! }[/math], at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time [math]\displaystyle{ {{T}_{i}}\,\! }[/math] where [math]\displaystyle{ i\,\! }[/math] indicates the group number. In this example, the total number of groups is [math]\displaystyle{ N=6\,\! }[/math] and the total number of units is [math]\displaystyle{ {{N}_{T}}=20\,\! }[/math]. Thus, the median rank values will be estimated for 20 units and for the total failed units ([math]\displaystyle{ {{N}_{{{F}_{i}}}}\,\! }[/math]) up to the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] group, for the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the [math]\displaystyle{ {{17}^{th}}\,\! }[/math] rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

[math]\displaystyle{ \begin{matrix} N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\ \text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\ \text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\ \text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\ \text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\ \text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\ \text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\ \sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\ \end{matrix}\,\! }[/math]

Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:

[math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\ & & \\ & \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.005392\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\! }[/math]

Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\! }[/math]

and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }\simeq 51.8\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\! }[/math]

Using Weibull++, the estimated parameters are:

[math]\displaystyle{ \begin{align} \hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}\,\! }[/math]

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the [math]\displaystyle{ 15^{th}\,\! }[/math] decimal point.

Using Auto Batch Run

A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].

Table - Leukemia Treatment Results
Time (weeks)	Number of Patients	Treament	Comments
1	2	placebo
2	2	placebo
3	1	placebo
4	2	placebo
5	2	placebo
6	4	6MP	3 patients completed
7	1	6MP
8	4	placebo
9	1	6MP	Not completed
10	2	6MP	1 patient completed
11	2	placebo
11	1	6MP	Not completed
12	2	placebo
13	1	6MP
15	1	placebo
16	1	6MP
17	1	placebo
17	1	6MP	Not completed
19	1	6MP	Not completed
20	1	6MP	Not completed
22	1	placebo
22	1	6MP
23	1	placebo
23	1	6MP
25	1	6MP	Not completed
32	2	6MP	Not completed
34	1	6MP	Not completed
35	1	6MP	Not completed

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.

2 Parameter Exponential Distribution RRX

The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where [math]\displaystyle{ \beta =1\,\! }[/math]. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).

Exponential Probability Density Function

The 2-Parameter Exponential Distribution

The 2-parameter exponential pdf is given by:

[math]\displaystyle{ f(t)=\lambda {{e}^{-\lambda (t-\gamma )}},f(t)\ge 0,\lambda \gt 0,t\ge \gamma \,\! }[/math]

where [math]\displaystyle{ \gamma \,\! }[/math] is the location parameter. Some of the characteristics of the 2-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{t}-\gamma =m-\gamma \,\! }[/math].
• The exponential pdf has no shape parameter, as it has only one shape.
• The distribution starts at [math]\displaystyle{ t=\gamma \,\! }[/math] at the level of [math]\displaystyle{ f(t=\gamma )=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases beyond [math]\displaystyle{ \gamma \,\! }[/math] and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The 1-Parameter Exponential Distribution

The 1-parameter exponential pdf is obtained by setting [math]\displaystyle{ \gamma =0\,\! }[/math], and is given by:

[math]\displaystyle{ \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, & t\ge 0, \lambda \gt 0,m\gt 0 \end{align} \,\! }[/math]

where:

[math]\displaystyle{ \lambda \,\! }[/math] = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)

[math]\displaystyle{ \lambda =\frac{1}{m}\,\! }[/math]

[math]\displaystyle{ m\,\! }[/math] = mean time between failures, or to failure

[math]\displaystyle{ t\,\! }[/math] = operating time, life, or age, in hours, cycles, miles, actuations, etc.

This distribution requires the knowledge of only one parameter, [math]\displaystyle{ \lambda \,\! }[/math], for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], is zero.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=m\,\! }[/math].
• As [math]\displaystyle{ \lambda \,\! }[/math] is decreased in value, the distribution is stretched out to the right, and as [math]\displaystyle{ \lambda \,\! }[/math] is increased, the distribution is pushed toward the origin.
• This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, [math]\displaystyle{ \lambda \,\! }[/math]).
• The distribution starts at [math]\displaystyle{ t=0\,\! }[/math] at the level of [math]\displaystyle{ f(t=0)=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases, and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].
• The pdf can be thought of as a special case of the Weibull pdf with [math]\displaystyle{ \gamma =0\,\! }[/math] and [math]\displaystyle{ \beta =1\,\! }[/math].

Exponential Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T},\,\! }[/math] or mean time to failure (MTTF) is given by:

[math]\displaystyle{ \begin{align} \bar{T}= & \int_{\gamma }^{\infty }t\cdot f(t)dt \\ = & \int_{\gamma }^{\infty }t\cdot \lambda \cdot {{e}^{-\lambda t}}dt \\ = & \gamma +\frac{1}{\lambda }=m \end{align}\,\! }[/math]

Note that when [math]\displaystyle{ \gamma =0\,\! }[/math], the MTTF is the inverse of the exponential distribution's constant failure rate. This is only true for the exponential distribution. Most other distributions do not have a constant failure rate. Consequently, the inverse relationship between failure rate and MTTF does not hold for these other distributions.

The Median

The median, [math]\displaystyle{ \breve{T}, \,\! }[/math] is:

[math]\displaystyle{ \breve{T}=\gamma +\frac{1}{\lambda}\cdot 0.693 \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T},\,\! }[/math] is:

[math]\displaystyle{ \tilde{T}=\gamma \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ {\sigma }_{T}\,\! }[/math], is:

[math]\displaystyle{ {\sigma}_{T}=\frac{1}{\lambda }=m\,\! }[/math]

The Exponential Reliability Function

The equation for the 2-parameter exponential cumulative density function, or cdf, is given by:

[math]\displaystyle{ \begin{align} F(t)=Q(t)=1-{{e}^{-\lambda (t-\gamma )}} \end{align}\,\! }[/math]

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function of the 2-parameter exponential distribution is given by:

[math]\displaystyle{ R(t)=1-Q(t)=1-\int_{0}^{t-\gamma }f(x)dx\,\! }[/math]

[math]\displaystyle{ R(t)=1-\int_{0}^{t-\gamma }\lambda {{e}^{-\lambda x}}dx={{e}^{-\lambda (t-\gamma )}}\,\! }[/math]

The 1-parameter exponential reliability function is given by:

[math]\displaystyle{ R(t)={{e}^{-\lambda t}}={{e}^{-\tfrac{t}{m}}}\,\! }[/math]

The Exponential Conditional Reliability Function

The exponential conditional reliability equation gives the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration, having already successfully accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation up to the start of this new mission. The exponential conditional reliability function is:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{{{e}^{-\lambda (T+t-\gamma )}}}{{{e}^{-\lambda (T-\gamma )}}}={{e}^{-\lambda t}}\,\! }[/math]

which says that the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration undertaken after the component or equipment has already accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. This is referred to as the memoryless property.

The Exponential Reliable Life Function

The reliable life, or the mission duration for a desired reliability goal, [math]\displaystyle{ {{t}_{R}}\,\! }[/math], for the 1-parameter exponential distribution is:

[math]\displaystyle{ R({{t}_{R}})={{e}^{-\lambda ({{t}_{R}}-\gamma )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \ln[R({{t}_{R}})]=-\lambda({{t}_{R}}-\gamma ) \end{align}\,\! }[/math]

or:

[math]\displaystyle{ {{t}_{R}}=\gamma -\frac{\ln [R({{t}_{R}})]}{\lambda }\,\! }[/math]

The Exponential Failure Rate Function

The exponential failure rate function is:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\lambda {{e}^{-\lambda (t-\gamma )}}}{{{e}^{-\lambda (t-\gamma )}}}=\lambda =\text{constant}\,\! }[/math]

Once again, note that the constant failure rate is a characteristic of the exponential distribution, and special cases of other distributions only. Most other distributions have failure rates that are functions of time.

Characteristics of the Exponential Distribution

The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. Unfortunately, this fact also leads to the use of this model in situations where it is not appropriate. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The constant failure rate of the exponential distribution would require the assumption that the automobile would be just as likely to experience a breakdown during the first mile as it would during the one-hundred-thousandth mile. Clearly, this is not a valid assumption. However, some inexperienced practitioners of reliability engineering and life data analysis will overlook this fact, lured by the siren-call of the exponential distribution's relatively simple mathematical models.

The Effect of lambda and gamma on the Exponential pdf

• The exponential pdf has no shape parameter, as it has only one shape.
• The exponential pdf is always convex and is stretched to the right as [math]\displaystyle{ \lambda \,\! }[/math] decreases in value.
• The value of the pdf function is always equal to the value of [math]\displaystyle{ \lambda \,\! }[/math] at [math]\displaystyle{ t=0\,\! }[/math] (or [math]\displaystyle{ t=\gamma \,\! }[/math]).
• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before this time.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{T}-\gamma =m-\gamma \,\! }[/math].
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The Effect of lambda and gamma on the Exponential Reliability Function

• The 1-parameter exponential reliability function starts at the value of 100% at [math]\displaystyle{ t=0\,\! }[/math], decreases thereafter monotonically and is convex.
• The 2-parameter exponential reliability function remains at the value of 100% for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and decreases thereafter monotonically and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ R(t\to \infty )\to 0\,\! }[/math].
• The reliability for a mission duration of [math]\displaystyle{ t=m=\tfrac{1}{\lambda }\,\! }[/math], or of one MTTF duration, is always equal to [math]\displaystyle{ 0.3679\,\! }[/math] or 36.79%. This means that the reliability for a mission which is as long as one MTTF is relatively low and is not recommended because only 36.8% of the missions will be completed successfully. In other words, of the equipment undertaking such a mission, only 36.8% will survive their mission.

The Effect of lambda and gamma on the Failure Rate Function

• The 1-parameter exponential failure rate function is constant and starts at [math]\displaystyle{ t=0\,\! }[/math].
• The 2-parameter exponential failure rate function remains at the value of 0 for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and then keeps at the constant value of [math]\displaystyle{ \lambda\,\! }[/math].

Exponential Distribution Examples

Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
Number of Units in Group	Time-to-Failure
7	100
5	200
3	300
2	400
1	500
2	600

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the pdf plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

Solution

1. For the 2-parameter exponential distribution and for [math]\displaystyle{ \hat{\gamma }=100\,\! }[/math] hours (first failure), the partial of the log-likelihood function, [math]\displaystyle{ \lambda\,\! }[/math], becomes:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\ \Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)] + \ldots +2[\frac{1}{\lambda}-(600-100)]\\ = & 0\\ \Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr} \end{align} \,\! }[/math]

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the pdf plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, [math]\displaystyle{ \gamma \,\! }[/math], at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time [math]\displaystyle{ {{T}_{i}}\,\! }[/math] where [math]\displaystyle{ i\,\! }[/math] indicates the group number. In this example, the total number of groups is [math]\displaystyle{ N=6\,\! }[/math] and the total number of units is [math]\displaystyle{ {{N}_{T}}=20\,\! }[/math]. Thus, the median rank values will be estimated for 20 units and for the total failed units ([math]\displaystyle{ {{N}_{{{F}_{i}}}}\,\! }[/math]) up to the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] group, for the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the [math]\displaystyle{ {{17}^{th}}\,\! }[/math] rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

[math]\displaystyle{ \begin{matrix} N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\ \text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\ \text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\ \text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\ \text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\ \text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\ \text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\ \sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\ \end{matrix}\,\! }[/math]

Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:

[math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\ & & \\ & \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.005392\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\! }[/math]

Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\! }[/math]

and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }\simeq 51.8\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\! }[/math]

Using Weibull++, the estimated parameters are:

[math]\displaystyle{ \begin{align} \hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}\,\! }[/math]

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the [math]\displaystyle{ 15^{th}\,\! }[/math] decimal point.

Using Auto Batch Run

A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].

Table - Leukemia Treatment Results
Time (weeks)	Number of Patients	Treament	Comments
1	2	placebo
2	2	placebo
3	1	placebo
4	2	placebo
5	2	placebo
6	4	6MP	3 patients completed
7	1	6MP
8	4	placebo
9	1	6MP	Not completed
10	2	6MP	1 patient completed
11	2	placebo
11	1	6MP	Not completed
12	2	placebo
13	1	6MP
15	1	placebo
16	1	6MP
17	1	placebo
17	1	6MP	Not completed
19	1	6MP	Not completed
20	1	6MP	Not completed
22	1	placebo
22	1	6MP
23	1	placebo
23	1	6MP
25	1	6MP	Not completed
32	2	6MP	Not completed
34	1	6MP	Not completed
35	1	6MP	Not completed

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.

MLE for Exponential Distribution

The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where [math]\displaystyle{ \beta =1\,\! }[/math]. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).

Exponential Probability Density Function

The 2-Parameter Exponential Distribution

The 2-parameter exponential pdf is given by:

[math]\displaystyle{ f(t)=\lambda {{e}^{-\lambda (t-\gamma )}},f(t)\ge 0,\lambda \gt 0,t\ge \gamma \,\! }[/math]

where [math]\displaystyle{ \gamma \,\! }[/math] is the location parameter. Some of the characteristics of the 2-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{t}-\gamma =m-\gamma \,\! }[/math].
• The exponential pdf has no shape parameter, as it has only one shape.
• The distribution starts at [math]\displaystyle{ t=\gamma \,\! }[/math] at the level of [math]\displaystyle{ f(t=\gamma )=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases beyond [math]\displaystyle{ \gamma \,\! }[/math] and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The 1-Parameter Exponential Distribution

The 1-parameter exponential pdf is obtained by setting [math]\displaystyle{ \gamma =0\,\! }[/math], and is given by:

[math]\displaystyle{ \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, & t\ge 0, \lambda \gt 0,m\gt 0 \end{align} \,\! }[/math]

where:

[math]\displaystyle{ \lambda \,\! }[/math] = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)

[math]\displaystyle{ \lambda =\frac{1}{m}\,\! }[/math]

[math]\displaystyle{ m\,\! }[/math] = mean time between failures, or to failure

[math]\displaystyle{ t\,\! }[/math] = operating time, life, or age, in hours, cycles, miles, actuations, etc.

This distribution requires the knowledge of only one parameter, [math]\displaystyle{ \lambda \,\! }[/math], for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], is zero.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=m\,\! }[/math].
• As [math]\displaystyle{ \lambda \,\! }[/math] is decreased in value, the distribution is stretched out to the right, and as [math]\displaystyle{ \lambda \,\! }[/math] is increased, the distribution is pushed toward the origin.
• This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, [math]\displaystyle{ \lambda \,\! }[/math]).
• The distribution starts at [math]\displaystyle{ t=0\,\! }[/math] at the level of [math]\displaystyle{ f(t=0)=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases, and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].
• The pdf can be thought of as a special case of the Weibull pdf with [math]\displaystyle{ \gamma =0\,\! }[/math] and [math]\displaystyle{ \beta =1\,\! }[/math].

Exponential Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T},\,\! }[/math] or mean time to failure (MTTF) is given by:

[math]\displaystyle{ \begin{align} \bar{T}= & \int_{\gamma }^{\infty }t\cdot f(t)dt \\ = & \int_{\gamma }^{\infty }t\cdot \lambda \cdot {{e}^{-\lambda t}}dt \\ = & \gamma +\frac{1}{\lambda }=m \end{align}\,\! }[/math]

Note that when [math]\displaystyle{ \gamma =0\,\! }[/math], the MTTF is the inverse of the exponential distribution's constant failure rate. This is only true for the exponential distribution. Most other distributions do not have a constant failure rate. Consequently, the inverse relationship between failure rate and MTTF does not hold for these other distributions.

The Median

The median, [math]\displaystyle{ \breve{T}, \,\! }[/math] is:

[math]\displaystyle{ \breve{T}=\gamma +\frac{1}{\lambda}\cdot 0.693 \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T},\,\! }[/math] is:

[math]\displaystyle{ \tilde{T}=\gamma \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ {\sigma }_{T}\,\! }[/math], is:

[math]\displaystyle{ {\sigma}_{T}=\frac{1}{\lambda }=m\,\! }[/math]

The Exponential Reliability Function

The equation for the 2-parameter exponential cumulative density function, or cdf, is given by:

[math]\displaystyle{ \begin{align} F(t)=Q(t)=1-{{e}^{-\lambda (t-\gamma )}} \end{align}\,\! }[/math]

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function of the 2-parameter exponential distribution is given by:

[math]\displaystyle{ R(t)=1-Q(t)=1-\int_{0}^{t-\gamma }f(x)dx\,\! }[/math]

[math]\displaystyle{ R(t)=1-\int_{0}^{t-\gamma }\lambda {{e}^{-\lambda x}}dx={{e}^{-\lambda (t-\gamma )}}\,\! }[/math]

The 1-parameter exponential reliability function is given by:

[math]\displaystyle{ R(t)={{e}^{-\lambda t}}={{e}^{-\tfrac{t}{m}}}\,\! }[/math]

The Exponential Conditional Reliability Function

The exponential conditional reliability equation gives the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration, having already successfully accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation up to the start of this new mission. The exponential conditional reliability function is:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{{{e}^{-\lambda (T+t-\gamma )}}}{{{e}^{-\lambda (T-\gamma )}}}={{e}^{-\lambda t}}\,\! }[/math]

which says that the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration undertaken after the component or equipment has already accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. This is referred to as the memoryless property.

The Exponential Reliable Life Function

The reliable life, or the mission duration for a desired reliability goal, [math]\displaystyle{ {{t}_{R}}\,\! }[/math], for the 1-parameter exponential distribution is:

[math]\displaystyle{ R({{t}_{R}})={{e}^{-\lambda ({{t}_{R}}-\gamma )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \ln[R({{t}_{R}})]=-\lambda({{t}_{R}}-\gamma ) \end{align}\,\! }[/math]

or:

[math]\displaystyle{ {{t}_{R}}=\gamma -\frac{\ln [R({{t}_{R}})]}{\lambda }\,\! }[/math]

The Exponential Failure Rate Function

The exponential failure rate function is:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\lambda {{e}^{-\lambda (t-\gamma )}}}{{{e}^{-\lambda (t-\gamma )}}}=\lambda =\text{constant}\,\! }[/math]

Once again, note that the constant failure rate is a characteristic of the exponential distribution, and special cases of other distributions only. Most other distributions have failure rates that are functions of time.

Characteristics of the Exponential Distribution

The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. Unfortunately, this fact also leads to the use of this model in situations where it is not appropriate. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The constant failure rate of the exponential distribution would require the assumption that the automobile would be just as likely to experience a breakdown during the first mile as it would during the one-hundred-thousandth mile. Clearly, this is not a valid assumption. However, some inexperienced practitioners of reliability engineering and life data analysis will overlook this fact, lured by the siren-call of the exponential distribution's relatively simple mathematical models.

The Effect of lambda and gamma on the Exponential pdf

• The exponential pdf has no shape parameter, as it has only one shape.
• The exponential pdf is always convex and is stretched to the right as [math]\displaystyle{ \lambda \,\! }[/math] decreases in value.
• The value of the pdf function is always equal to the value of [math]\displaystyle{ \lambda \,\! }[/math] at [math]\displaystyle{ t=0\,\! }[/math] (or [math]\displaystyle{ t=\gamma \,\! }[/math]).
• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before this time.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{T}-\gamma =m-\gamma \,\! }[/math].
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The Effect of lambda and gamma on the Exponential Reliability Function

• The 1-parameter exponential reliability function starts at the value of 100% at [math]\displaystyle{ t=0\,\! }[/math], decreases thereafter monotonically and is convex.
• The 2-parameter exponential reliability function remains at the value of 100% for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and decreases thereafter monotonically and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ R(t\to \infty )\to 0\,\! }[/math].
• The reliability for a mission duration of [math]\displaystyle{ t=m=\tfrac{1}{\lambda }\,\! }[/math], or of one MTTF duration, is always equal to [math]\displaystyle{ 0.3679\,\! }[/math] or 36.79%. This means that the reliability for a mission which is as long as one MTTF is relatively low and is not recommended because only 36.8% of the missions will be completed successfully. In other words, of the equipment undertaking such a mission, only 36.8% will survive their mission.

The Effect of lambda and gamma on the Failure Rate Function

• The 1-parameter exponential failure rate function is constant and starts at [math]\displaystyle{ t=0\,\! }[/math].
• The 2-parameter exponential failure rate function remains at the value of 0 for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and then keeps at the constant value of [math]\displaystyle{ \lambda\,\! }[/math].

Exponential Distribution Examples

Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
Number of Units in Group	Time-to-Failure
7	100
5	200
3	300
2	400
1	500
2	600

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the pdf plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

Solution

1. For the 2-parameter exponential distribution and for [math]\displaystyle{ \hat{\gamma }=100\,\! }[/math] hours (first failure), the partial of the log-likelihood function, [math]\displaystyle{ \lambda\,\! }[/math], becomes:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\ \Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)] + \ldots +2[\frac{1}{\lambda}-(600-100)]\\ = & 0\\ \Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr} \end{align} \,\! }[/math]

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the pdf plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, [math]\displaystyle{ \gamma \,\! }[/math], at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time [math]\displaystyle{ {{T}_{i}}\,\! }[/math] where [math]\displaystyle{ i\,\! }[/math] indicates the group number. In this example, the total number of groups is [math]\displaystyle{ N=6\,\! }[/math] and the total number of units is [math]\displaystyle{ {{N}_{T}}=20\,\! }[/math]. Thus, the median rank values will be estimated for 20 units and for the total failed units ([math]\displaystyle{ {{N}_{{{F}_{i}}}}\,\! }[/math]) up to the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] group, for the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the [math]\displaystyle{ {{17}^{th}}\,\! }[/math] rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

[math]\displaystyle{ \begin{matrix} N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\ \text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\ \text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\ \text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\ \text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\ \text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\ \text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\ \sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\ \end{matrix}\,\! }[/math]

Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:

[math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\ & & \\ & \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.005392\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\! }[/math]

Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\! }[/math]

and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }\simeq 51.8\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\! }[/math]

Using Weibull++, the estimated parameters are:

[math]\displaystyle{ \begin{align} \hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}\,\! }[/math]

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the [math]\displaystyle{ 15^{th}\,\! }[/math] decimal point.

Using Auto Batch Run

A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].

Table - Leukemia Treatment Results
Time (weeks)	Number of Patients	Treament	Comments
1	2	placebo
2	2	placebo
3	1	placebo
4	2	placebo
5	2	placebo
6	4	6MP	3 patients completed
7	1	6MP
8	4	placebo
9	1	6MP	Not completed
10	2	6MP	1 patient completed
11	2	placebo
11	1	6MP	Not completed
12	2	placebo
13	1	6MP
15	1	placebo
16	1	6MP
17	1	placebo
17	1	6MP	Not completed
19	1	6MP	Not completed
20	1	6MP	Not completed
22	1	placebo
22	1	6MP
23	1	placebo
23	1	6MP
25	1	6MP	Not completed
32	2	6MP	Not completed
34	1	6MP	Not completed
35	1	6MP	Not completed

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.

Likelihood Ratio Bound on Lambda

The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where [math]\displaystyle{ \beta =1\,\! }[/math]. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).

Exponential Probability Density Function

The 2-Parameter Exponential Distribution

The 2-parameter exponential pdf is given by:

[math]\displaystyle{ f(t)=\lambda {{e}^{-\lambda (t-\gamma )}},f(t)\ge 0,\lambda \gt 0,t\ge \gamma \,\! }[/math]

where [math]\displaystyle{ \gamma \,\! }[/math] is the location parameter. Some of the characteristics of the 2-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{t}-\gamma =m-\gamma \,\! }[/math].
• The exponential pdf has no shape parameter, as it has only one shape.
• The distribution starts at [math]\displaystyle{ t=\gamma \,\! }[/math] at the level of [math]\displaystyle{ f(t=\gamma )=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases beyond [math]\displaystyle{ \gamma \,\! }[/math] and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The 1-Parameter Exponential Distribution

The 1-parameter exponential pdf is obtained by setting [math]\displaystyle{ \gamma =0\,\! }[/math], and is given by:

[math]\displaystyle{ \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, & t\ge 0, \lambda \gt 0,m\gt 0 \end{align} \,\! }[/math]

where:

[math]\displaystyle{ \lambda \,\! }[/math] = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)

[math]\displaystyle{ \lambda =\frac{1}{m}\,\! }[/math]

[math]\displaystyle{ m\,\! }[/math] = mean time between failures, or to failure

[math]\displaystyle{ t\,\! }[/math] = operating time, life, or age, in hours, cycles, miles, actuations, etc.

This distribution requires the knowledge of only one parameter, [math]\displaystyle{ \lambda \,\! }[/math], for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], is zero.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=m\,\! }[/math].
• As [math]\displaystyle{ \lambda \,\! }[/math] is decreased in value, the distribution is stretched out to the right, and as [math]\displaystyle{ \lambda \,\! }[/math] is increased, the distribution is pushed toward the origin.
• This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, [math]\displaystyle{ \lambda \,\! }[/math]).
• The distribution starts at [math]\displaystyle{ t=0\,\! }[/math] at the level of [math]\displaystyle{ f(t=0)=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases, and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].
• The pdf can be thought of as a special case of the Weibull pdf with [math]\displaystyle{ \gamma =0\,\! }[/math] and [math]\displaystyle{ \beta =1\,\! }[/math].

Exponential Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T},\,\! }[/math] or mean time to failure (MTTF) is given by:

[math]\displaystyle{ \begin{align} \bar{T}= & \int_{\gamma }^{\infty }t\cdot f(t)dt \\ = & \int_{\gamma }^{\infty }t\cdot \lambda \cdot {{e}^{-\lambda t}}dt \\ = & \gamma +\frac{1}{\lambda }=m \end{align}\,\! }[/math]

Note that when [math]\displaystyle{ \gamma =0\,\! }[/math], the MTTF is the inverse of the exponential distribution's constant failure rate. This is only true for the exponential distribution. Most other distributions do not have a constant failure rate. Consequently, the inverse relationship between failure rate and MTTF does not hold for these other distributions.

The Median

The median, [math]\displaystyle{ \breve{T}, \,\! }[/math] is:

[math]\displaystyle{ \breve{T}=\gamma +\frac{1}{\lambda}\cdot 0.693 \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T},\,\! }[/math] is:

[math]\displaystyle{ \tilde{T}=\gamma \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ {\sigma }_{T}\,\! }[/math], is:

[math]\displaystyle{ {\sigma}_{T}=\frac{1}{\lambda }=m\,\! }[/math]

The Exponential Reliability Function

The equation for the 2-parameter exponential cumulative density function, or cdf, is given by:

[math]\displaystyle{ \begin{align} F(t)=Q(t)=1-{{e}^{-\lambda (t-\gamma )}} \end{align}\,\! }[/math]

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function of the 2-parameter exponential distribution is given by:

[math]\displaystyle{ R(t)=1-Q(t)=1-\int_{0}^{t-\gamma }f(x)dx\,\! }[/math]

[math]\displaystyle{ R(t)=1-\int_{0}^{t-\gamma }\lambda {{e}^{-\lambda x}}dx={{e}^{-\lambda (t-\gamma )}}\,\! }[/math]

The 1-parameter exponential reliability function is given by:

[math]\displaystyle{ R(t)={{e}^{-\lambda t}}={{e}^{-\tfrac{t}{m}}}\,\! }[/math]

The Exponential Conditional Reliability Function

The exponential conditional reliability equation gives the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration, having already successfully accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation up to the start of this new mission. The exponential conditional reliability function is:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{{{e}^{-\lambda (T+t-\gamma )}}}{{{e}^{-\lambda (T-\gamma )}}}={{e}^{-\lambda t}}\,\! }[/math]

which says that the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration undertaken after the component or equipment has already accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. This is referred to as the memoryless property.

The Exponential Reliable Life Function

The reliable life, or the mission duration for a desired reliability goal, [math]\displaystyle{ {{t}_{R}}\,\! }[/math], for the 1-parameter exponential distribution is:

[math]\displaystyle{ R({{t}_{R}})={{e}^{-\lambda ({{t}_{R}}-\gamma )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \ln[R({{t}_{R}})]=-\lambda({{t}_{R}}-\gamma ) \end{align}\,\! }[/math]

or:

[math]\displaystyle{ {{t}_{R}}=\gamma -\frac{\ln [R({{t}_{R}})]}{\lambda }\,\! }[/math]

The Exponential Failure Rate Function

The exponential failure rate function is:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\lambda {{e}^{-\lambda (t-\gamma )}}}{{{e}^{-\lambda (t-\gamma )}}}=\lambda =\text{constant}\,\! }[/math]

Once again, note that the constant failure rate is a characteristic of the exponential distribution, and special cases of other distributions only. Most other distributions have failure rates that are functions of time.

Characteristics of the Exponential Distribution

The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. Unfortunately, this fact also leads to the use of this model in situations where it is not appropriate. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The constant failure rate of the exponential distribution would require the assumption that the automobile would be just as likely to experience a breakdown during the first mile as it would during the one-hundred-thousandth mile. Clearly, this is not a valid assumption. However, some inexperienced practitioners of reliability engineering and life data analysis will overlook this fact, lured by the siren-call of the exponential distribution's relatively simple mathematical models.

The Effect of lambda and gamma on the Exponential pdf

• The exponential pdf has no shape parameter, as it has only one shape.
• The exponential pdf is always convex and is stretched to the right as [math]\displaystyle{ \lambda \,\! }[/math] decreases in value.
• The value of the pdf function is always equal to the value of [math]\displaystyle{ \lambda \,\! }[/math] at [math]\displaystyle{ t=0\,\! }[/math] (or [math]\displaystyle{ t=\gamma \,\! }[/math]).
• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before this time.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{T}-\gamma =m-\gamma \,\! }[/math].
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The Effect of lambda and gamma on the Exponential Reliability Function

• The 1-parameter exponential reliability function starts at the value of 100% at [math]\displaystyle{ t=0\,\! }[/math], decreases thereafter monotonically and is convex.
• The 2-parameter exponential reliability function remains at the value of 100% for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and decreases thereafter monotonically and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ R(t\to \infty )\to 0\,\! }[/math].
• The reliability for a mission duration of [math]\displaystyle{ t=m=\tfrac{1}{\lambda }\,\! }[/math], or of one MTTF duration, is always equal to [math]\displaystyle{ 0.3679\,\! }[/math] or 36.79%. This means that the reliability for a mission which is as long as one MTTF is relatively low and is not recommended because only 36.8% of the missions will be completed successfully. In other words, of the equipment undertaking such a mission, only 36.8% will survive their mission.

The Effect of lambda and gamma on the Failure Rate Function

• The 1-parameter exponential failure rate function is constant and starts at [math]\displaystyle{ t=0\,\! }[/math].
• The 2-parameter exponential failure rate function remains at the value of 0 for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and then keeps at the constant value of [math]\displaystyle{ \lambda\,\! }[/math].

Exponential Distribution Examples

Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
Number of Units in Group	Time-to-Failure
7	100
5	200
3	300
2	400
1	500
2	600

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the pdf plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

Solution

1. For the 2-parameter exponential distribution and for [math]\displaystyle{ \hat{\gamma }=100\,\! }[/math] hours (first failure), the partial of the log-likelihood function, [math]\displaystyle{ \lambda\,\! }[/math], becomes:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\ \Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)] + \ldots +2[\frac{1}{\lambda}-(600-100)]\\ = & 0\\ \Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr} \end{align} \,\! }[/math]

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the pdf plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, [math]\displaystyle{ \gamma \,\! }[/math], at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time [math]\displaystyle{ {{T}_{i}}\,\! }[/math] where [math]\displaystyle{ i\,\! }[/math] indicates the group number. In this example, the total number of groups is [math]\displaystyle{ N=6\,\! }[/math] and the total number of units is [math]\displaystyle{ {{N}_{T}}=20\,\! }[/math]. Thus, the median rank values will be estimated for 20 units and for the total failed units ([math]\displaystyle{ {{N}_{{{F}_{i}}}}\,\! }[/math]) up to the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] group, for the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the [math]\displaystyle{ {{17}^{th}}\,\! }[/math] rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

[math]\displaystyle{ \begin{matrix} N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\ \text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\ \text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\ \text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\ \text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\ \text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\ \text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\ \sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\ \end{matrix}\,\! }[/math]

Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:

[math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\ & & \\ & \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.005392\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\! }[/math]

Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\! }[/math]

and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }\simeq 51.8\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\! }[/math]

Using Weibull++, the estimated parameters are:

[math]\displaystyle{ \begin{align} \hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}\,\! }[/math]

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the [math]\displaystyle{ 15^{th}\,\! }[/math] decimal point.

Using Auto Batch Run

A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].

Table - Leukemia Treatment Results
Time (weeks)	Number of Patients	Treament	Comments
1	2	placebo
2	2	placebo
3	1	placebo
4	2	placebo
5	2	placebo
6	4	6MP	3 patients completed
7	1	6MP
8	4	placebo
9	1	6MP	Not completed
10	2	6MP	1 patient completed
11	2	placebo
11	1	6MP	Not completed
12	2	placebo
13	1	6MP
15	1	placebo
16	1	6MP
17	1	placebo
17	1	6MP	Not completed
19	1	6MP	Not completed
20	1	6MP	Not completed
22	1	placebo
22	1	6MP
23	1	placebo
23	1	6MP
25	1	6MP	Not completed
32	2	6MP	Not completed
34	1	6MP	Not completed
35	1	6MP	Not completed

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.

Likelihood Ratio Bound on Time

The exponential distribution is a commonly used distribution in reliability engineering. Mathematically, it is a fairly simple distribution, which many times leads to its use in inappropriate situations. It is, in fact, a special case of the Weibull distribution where [math]\displaystyle{ \beta =1\,\! }[/math]. The exponential distribution is used to model the behavior of units that have a constant failure rate (or units that do not degrade with time or wear out).

Exponential Probability Density Function

The 2-Parameter Exponential Distribution

The 2-parameter exponential pdf is given by:

[math]\displaystyle{ f(t)=\lambda {{e}^{-\lambda (t-\gamma )}},f(t)\ge 0,\lambda \gt 0,t\ge \gamma \,\! }[/math]

where [math]\displaystyle{ \gamma \,\! }[/math] is the location parameter. Some of the characteristics of the 2-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{t}-\gamma =m-\gamma \,\! }[/math].
• The exponential pdf has no shape parameter, as it has only one shape.
• The distribution starts at [math]\displaystyle{ t=\gamma \,\! }[/math] at the level of [math]\displaystyle{ f(t=\gamma )=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases beyond [math]\displaystyle{ \gamma \,\! }[/math] and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The 1-Parameter Exponential Distribution

The 1-parameter exponential pdf is obtained by setting [math]\displaystyle{ \gamma =0\,\! }[/math], and is given by:

[math]\displaystyle{ \begin{align}f(t)= & \lambda {{e}^{-\lambda t}}=\frac{1}{m}{{e}^{-\tfrac{1}{m}t}}, & t\ge 0, \lambda \gt 0,m\gt 0 \end{align} \,\! }[/math]

where:

[math]\displaystyle{ \lambda \,\! }[/math] = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.)

[math]\displaystyle{ \lambda =\frac{1}{m}\,\! }[/math]

[math]\displaystyle{ m\,\! }[/math] = mean time between failures, or to failure

[math]\displaystyle{ t\,\! }[/math] = operating time, life, or age, in hours, cycles, miles, actuations, etc.

This distribution requires the knowledge of only one parameter, [math]\displaystyle{ \lambda \,\! }[/math], for its application. Some of the characteristics of the 1-parameter exponential distribution are discussed in Kececioglu [19]:

• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], is zero.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=m\,\! }[/math].
• As [math]\displaystyle{ \lambda \,\! }[/math] is decreased in value, the distribution is stretched out to the right, and as [math]\displaystyle{ \lambda \,\! }[/math] is increased, the distribution is pushed toward the origin.
• This distribution has no shape parameter as it has only one shape, (i.e., the exponential, and the only parameter it has is the failure rate, [math]\displaystyle{ \lambda \,\! }[/math]).
• The distribution starts at [math]\displaystyle{ t=0\,\! }[/math] at the level of [math]\displaystyle{ f(t=0)=\lambda \,\! }[/math] and decreases thereafter exponentially and monotonically as [math]\displaystyle{ t\,\! }[/math] increases, and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].
• The pdf can be thought of as a special case of the Weibull pdf with [math]\displaystyle{ \gamma =0\,\! }[/math] and [math]\displaystyle{ \beta =1\,\! }[/math].

Exponential Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T},\,\! }[/math] or mean time to failure (MTTF) is given by:

[math]\displaystyle{ \begin{align} \bar{T}= & \int_{\gamma }^{\infty }t\cdot f(t)dt \\ = & \int_{\gamma }^{\infty }t\cdot \lambda \cdot {{e}^{-\lambda t}}dt \\ = & \gamma +\frac{1}{\lambda }=m \end{align}\,\! }[/math]

Note that when [math]\displaystyle{ \gamma =0\,\! }[/math], the MTTF is the inverse of the exponential distribution's constant failure rate. This is only true for the exponential distribution. Most other distributions do not have a constant failure rate. Consequently, the inverse relationship between failure rate and MTTF does not hold for these other distributions.

The Median

The median, [math]\displaystyle{ \breve{T}, \,\! }[/math] is:

[math]\displaystyle{ \breve{T}=\gamma +\frac{1}{\lambda}\cdot 0.693 \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T},\,\! }[/math] is:

[math]\displaystyle{ \tilde{T}=\gamma \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ {\sigma }_{T}\,\! }[/math], is:

[math]\displaystyle{ {\sigma}_{T}=\frac{1}{\lambda }=m\,\! }[/math]

The Exponential Reliability Function

The equation for the 2-parameter exponential cumulative density function, or cdf, is given by:

[math]\displaystyle{ \begin{align} F(t)=Q(t)=1-{{e}^{-\lambda (t-\gamma )}} \end{align}\,\! }[/math]

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function of the 2-parameter exponential distribution is given by:

[math]\displaystyle{ R(t)=1-Q(t)=1-\int_{0}^{t-\gamma }f(x)dx\,\! }[/math]

[math]\displaystyle{ R(t)=1-\int_{0}^{t-\gamma }\lambda {{e}^{-\lambda x}}dx={{e}^{-\lambda (t-\gamma )}}\,\! }[/math]

The 1-parameter exponential reliability function is given by:

[math]\displaystyle{ R(t)={{e}^{-\lambda t}}={{e}^{-\tfrac{t}{m}}}\,\! }[/math]

The Exponential Conditional Reliability Function

The exponential conditional reliability equation gives the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration, having already successfully accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation up to the start of this new mission. The exponential conditional reliability function is:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{{{e}^{-\lambda (T+t-\gamma )}}}{{{e}^{-\lambda (T-\gamma )}}}={{e}^{-\lambda t}}\,\! }[/math]

which says that the reliability for a mission of [math]\displaystyle{ t\,\! }[/math] duration undertaken after the component or equipment has already accumulated [math]\displaystyle{ T\,\! }[/math] hours of operation from age zero is only a function of the mission duration, and not a function of the age at the beginning of the mission. This is referred to as the memoryless property.

The Exponential Reliable Life Function

The reliable life, or the mission duration for a desired reliability goal, [math]\displaystyle{ {{t}_{R}}\,\! }[/math], for the 1-parameter exponential distribution is:

[math]\displaystyle{ R({{t}_{R}})={{e}^{-\lambda ({{t}_{R}}-\gamma )}}\,\! }[/math]

[math]\displaystyle{ \begin{align} \ln[R({{t}_{R}})]=-\lambda({{t}_{R}}-\gamma ) \end{align}\,\! }[/math]

or:

[math]\displaystyle{ {{t}_{R}}=\gamma -\frac{\ln [R({{t}_{R}})]}{\lambda }\,\! }[/math]

The Exponential Failure Rate Function

The exponential failure rate function is:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\lambda {{e}^{-\lambda (t-\gamma )}}}{{{e}^{-\lambda (t-\gamma )}}}=\lambda =\text{constant}\,\! }[/math]

Once again, note that the constant failure rate is a characteristic of the exponential distribution, and special cases of other distributions only. Most other distributions have failure rates that are functions of time.

Characteristics of the Exponential Distribution

The primary trait of the exponential distribution is that it is used for modeling the behavior of items with a constant failure rate. It has a fairly simple mathematical form, which makes it fairly easy to manipulate. Unfortunately, this fact also leads to the use of this model in situations where it is not appropriate. For example, it would not be appropriate to use the exponential distribution to model the reliability of an automobile. The constant failure rate of the exponential distribution would require the assumption that the automobile would be just as likely to experience a breakdown during the first mile as it would during the one-hundred-thousandth mile. Clearly, this is not a valid assumption. However, some inexperienced practitioners of reliability engineering and life data analysis will overlook this fact, lured by the siren-call of the exponential distribution's relatively simple mathematical models.

The Effect of lambda and gamma on the Exponential pdf

• The exponential pdf has no shape parameter, as it has only one shape.
• The exponential pdf is always convex and is stretched to the right as [math]\displaystyle{ \lambda \,\! }[/math] decreases in value.
• The value of the pdf function is always equal to the value of [math]\displaystyle{ \lambda \,\! }[/math] at [math]\displaystyle{ t=0\,\! }[/math] (or [math]\displaystyle{ t=\gamma \,\! }[/math]).
• The location parameter, [math]\displaystyle{ \gamma \,\! }[/math], if positive, shifts the beginning of the distribution by a distance of [math]\displaystyle{ \gamma \,\! }[/math] to the right of the origin, signifying that the chance failures start to occur only after [math]\displaystyle{ \gamma \,\! }[/math] hours of operation, and cannot occur before this time.
• The scale parameter is [math]\displaystyle{ \tfrac{1}{\lambda }=\bar{T}-\gamma =m-\gamma \,\! }[/math].
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ f(t)\to 0\,\! }[/math].

The Effect of lambda and gamma on the Exponential Reliability Function

• The 1-parameter exponential reliability function starts at the value of 100% at [math]\displaystyle{ t=0\,\! }[/math], decreases thereafter monotonically and is convex.
• The 2-parameter exponential reliability function remains at the value of 100% for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and decreases thereafter monotonically and is convex.
• As [math]\displaystyle{ t\to \infty \,\! }[/math], [math]\displaystyle{ R(t\to \infty )\to 0\,\! }[/math].
• The reliability for a mission duration of [math]\displaystyle{ t=m=\tfrac{1}{\lambda }\,\! }[/math], or of one MTTF duration, is always equal to [math]\displaystyle{ 0.3679\,\! }[/math] or 36.79%. This means that the reliability for a mission which is as long as one MTTF is relatively low and is not recommended because only 36.8% of the missions will be completed successfully. In other words, of the equipment undertaking such a mission, only 36.8% will survive their mission.

The Effect of lambda and gamma on the Failure Rate Function

• The 1-parameter exponential failure rate function is constant and starts at [math]\displaystyle{ t=0\,\! }[/math].
• The 2-parameter exponential failure rate function remains at the value of 0 for [math]\displaystyle{ t=0\,\! }[/math] up to [math]\displaystyle{ t=\gamma \,\! }[/math], and then keeps at the constant value of [math]\displaystyle{ \lambda\,\! }[/math].

Exponential Distribution Examples

Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
Number of Units in Group	Time-to-Failure
7	100
5	200
3	300
2	400
1	500
2	600

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the pdf plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

Solution

1. For the 2-parameter exponential distribution and for [math]\displaystyle{ \hat{\gamma }=100\,\! }[/math] hours (first failure), the partial of the log-likelihood function, [math]\displaystyle{ \lambda\,\! }[/math], becomes:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\ \Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)] + \ldots +2[\frac{1}{\lambda}-(600-100)]\\ = & 0\\ \Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr} \end{align} \,\! }[/math]

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the pdf plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, [math]\displaystyle{ \gamma \,\! }[/math], at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time [math]\displaystyle{ {{T}_{i}}\,\! }[/math] where [math]\displaystyle{ i\,\! }[/math] indicates the group number. In this example, the total number of groups is [math]\displaystyle{ N=6\,\! }[/math] and the total number of units is [math]\displaystyle{ {{N}_{T}}=20\,\! }[/math]. Thus, the median rank values will be estimated for 20 units and for the total failed units ([math]\displaystyle{ {{N}_{{{F}_{i}}}}\,\! }[/math]) up to the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] group, for the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the [math]\displaystyle{ {{17}^{th}}\,\! }[/math] rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

[math]\displaystyle{ \begin{matrix} N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\ \text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\ \text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\ \text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\ \text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\ \text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\ \text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\ \sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\ \end{matrix}\,\! }[/math]

Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:

[math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\ & & \\ & \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.005392\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\! }[/math]

Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\! }[/math]

and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }\simeq 51.8\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\! }[/math]

Using Weibull++, the estimated parameters are:

[math]\displaystyle{ \begin{align} \hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}\,\! }[/math]

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the [math]\displaystyle{ 15^{th}\,\! }[/math] decimal point.

Using Auto Batch Run

A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].

Table - Leukemia Treatment Results
Time (weeks)	Number of Patients	Treament	Comments
1	2	placebo
2	2	placebo
3	1	placebo
4	2	placebo
5	2	placebo
6	4	6MP	3 patients completed
7	1	6MP
8	4	placebo
9	1	6MP	Not completed
10	2	6MP	1 patient completed
11	2	placebo
11	1	6MP	Not completed
12	2	placebo
13	1	6MP
15	1	placebo
16	1	6MP
17	1	placebo
17	1	6MP	Not completed
19	1	6MP	Not completed
20	1	6MP	Not completed
22	1	placebo
22	1	6MP
23	1	placebo
23	1	6MP
25	1	6MP	Not completed
32	2	6MP	Not completed
34	1	6MP	Not completed
35	1	6MP	Not completed

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.

Likelihood Ratio Bound on Reliability

Likelihood Ratio Bound on Reliability

For the data given in Example 5, determine the 85% two-sided confidence bounds on the reliability estimate for a [math]\displaystyle{ t=50 }[/math]. The ML estimate for the time at [math]\displaystyle{ t=50 }[/math] is [math]\displaystyle{ \hat{R}=50.881% }[/math].

Solution

In this example, we are trying to determine the 85% two-sided confidence bounds on the reliability estimate of 50.881%. This is accomplished by substituting [math]\displaystyle{ t=50 }[/math] and [math]\displaystyle{ \alpha =0.85 }[/math] into the likelihood ratio bound equation. It now remains to find the values of [math]\displaystyle{ R }[/math] which satisfy this equation. Since there is only one parameter, there are only two values of [math]\displaystyle{ t }[/math] that will satisfy the equation. These values represent the [math]\displaystyle{ \delta =85% }[/math] two-sided confidence limits of the reliability estimate [math]\displaystyle{ \hat{R} }[/math]. For our problem, the confidence limits are:

[math]\displaystyle{ {{\hat{R}}_{t=50}}=(29.861%,71.794%) }[/math]

Exponential Dstribution for Grouped Data

20 units were reliability tested with the following results:

Table - Life Test Data
Number of Units in Group	Time-to-Failure
7	100
5	200
3	300
2	400
1	500
2	600

1. Assuming a 2-parameter exponential distribution, estimate the parameters by hand using the MLE analysis method.

2. Repeat the above using Weibull++. (Enter the data as grouped data to duplicate the results.)

3. Show the Probability plot for the analysis results.

4. Show the Reliability vs. Time plot for the results.

5. Show the pdf plot for the results.

6. Show the Failure Rate vs. Time plot for the results.

7. Estimate the parameters using the rank regression on Y (RRY) analysis method (and using grouped ranks).

Solution

1. For the 2-parameter exponential distribution and for [math]\displaystyle{ \hat{\gamma }=100\,\! }[/math] hours (first failure), the partial of the log-likelihood function, [math]\displaystyle{ \lambda\,\! }[/math], becomes:

[math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial \lambda }= &\underset{i=1}{\overset{6}{\mathop \sum }}\,{N_i} \left[ \frac{1}{\lambda }-\left( {{T}_{i}}-100 \right) \right]=0\\ \Rightarrow & 7[\frac{1}{\lambda }-(100-100)]+5[\frac{1}{\lambda}-(200-100)] + \ldots +2[\frac{1}{\lambda}-(600-100)]\\ = & 0\\ \Rightarrow & \hat{\lambda}=\frac{20}{3100}=0.0065 \text{fr/hr} \end{align} \,\! }[/math]

2. Enter the data in a Weibull++ standard folio and calculate it as shown next.

3. On the Plot page of the folio, the exponential Probability plot will appear as shown next.

4. View the Reliability vs. Time plot.

5. View the pdf plot.

6. View the Failure Rate vs. Time plot.

Note that, as described at the beginning of this chapter, the failure rate for the exponential distribution is constant. Also note that the Failure Rate vs. Time plot does show values for times before the location parameter, [math]\displaystyle{ \gamma \,\! }[/math], at 100 hours.

7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. This is because the median rank values are determined from the total number of failures observed by time [math]\displaystyle{ {{T}_{i}}\,\! }[/math] where [math]\displaystyle{ i\,\! }[/math] indicates the group number. In this example, the total number of groups is [math]\displaystyle{ N=6\,\! }[/math] and the total number of units is [math]\displaystyle{ {{N}_{T}}=20\,\! }[/math]. Thus, the median rank values will be estimated for 20 units and for the total failed units ([math]\displaystyle{ {{N}_{{{F}_{i}}}}\,\! }[/math]) up to the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] group, for the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference tool.

For example, the median rank value of the fourth group will be the [math]\displaystyle{ {{17}^{th}}\,\! }[/math] rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed.

[math]\displaystyle{ \begin{matrix} N & {{N}_{F}} & {{N}_{{{F}_{i}}}} & {{T}_{i}} & F({{T}_{i}}) & {{y}_{i}} & T_{i}^{2} & y_{i}^{2} & {{T}_{i}}{{y}_{i}} \\ \text{1} & \text{7} & \text{7} & \text{100} & \text{0}\text{.32795} & \text{-0}\text{.3974} & \text{10000} & \text{0}\text{.1579} & \text{-39}\text{.7426} \\ \text{2} & \text{5} & \text{12} & \text{200} & \text{0}\text{.57374} & \text{-0}\text{.8527} & \text{40000} & \text{0}\text{.7271} & \text{-170}\text{.5402} \\ \text{3} & \text{3} & \text{15} & \text{300} & \text{0}\text{.72120} & \text{-1}\text{.2772} & \text{90000} & \text{1}\text{.6313} & \text{-383}\text{.1728} \\ \text{4} & \text{2} & \text{17} & \text{400} & \text{0}\text{.81945} & \text{-1}\text{.7117} & \text{160000} & \text{2}\text{.9301} & \text{-684}\text{.6990} \\ \text{5} & \text{1} & \text{18} & \text{500} & \text{0}\text{.86853} & \text{-2}\text{.0289} & \text{250000} & \text{4}\text{.1166} & \text{-1014}\text{.4731} \\ \text{6} & \text{2} & \text{20} & \text{600} & \text{0}\text{.96594} & \text{-3}\text{.3795} & \text{360000} & \text{11}\text{.4211} & \text{-2027}\text{.7085} \\ \sum_{}^{} & {} & {} & \text{2100} & {} & \text{-9}\text{.6476} & \text{910000} & \text{20}\text{.9842} & \text{-4320}\text{.3362} \\ \end{matrix}\,\! }[/math]

Given the values in the table above, calculate [math]\displaystyle{ \hat{a}\,\! }[/math] and [math]\displaystyle{ \hat{b}\,\! }[/math]:

[math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{y}_{i}})/6}{\underset{i=1}{\overset{6}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/6} \\ & & \\ & \hat{b}= & \frac{-4320.3362-(2100)(-9.6476)/6}{910,000-{{(2100)}^{2}}/6} \end{align}\,\! }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.005392\,\! }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}\,\! }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793\,\! }[/math]

Therefore:

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}\,\! }[/math]

and:

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}\,\! }[/math]

or:

[math]\displaystyle{ \hat{\gamma }\simeq 51.8\text{ hours}\,\! }[/math]

Then:

[math]\displaystyle{ f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}\,\! }[/math]

Using Weibull++, the estimated parameters are:

[math]\displaystyle{ \begin{align} \hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}\,\! }[/math]

The small difference in the values from Weibull++ is due to rounding. In the application, the calculations and the rank values are carried out up to the [math]\displaystyle{ 15^{th}\,\! }[/math] decimal point.

Exponential Distribution Auto Batch Run Example

A number of leukemia patients were treated with either drug 6MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately [21, p.175].

Table - Leukemia Treatment Results
Time (weeks)	Number of Patients	Treament	Comments
1	2	placebo
2	2	placebo
3	1	placebo
4	2	placebo
5	2	placebo
6	4	6MP	3 patients completed
7	1	6MP
8	4	placebo
9	1	6MP	Not completed
10	2	6MP	1 patient completed
11	2	placebo
11	1	6MP	Not completed
12	2	placebo
13	1	6MP
15	1	placebo
16	1	6MP
17	1	placebo
17	1	6MP	Not completed
19	1	6MP	Not completed
20	1	6MP	Not completed
22	1	placebo
22	1	6MP
23	1	placebo
23	1	6MP
25	1	6MP	Not completed
32	2	6MP	Not completed
34	1	6MP	Not completed
35	1	6MP	Not completed

Create a new Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions. In the first column, enter the number of patients. Whenever there are uncompleted tests, enter the number of patients who completed the test separately from the number of patients who did not (e.g., if 4 patients had symptoms return after 6 weeks and only 3 of them completed the test, then enter 1 in one row and 3 in another). In the second column enter F if the patients completed the test and S if they didn't. In the third column enter the time, and in the fourth column (Subset ID) specify whether the 6MP drug or a placebo was used.

Next, open the Batch Auto Run utility and select to separate the 6MP drug from the placebo, as shown next.

The software will create two data sheets, one for each subset ID, as shown next.

Calculate both data sheets using the 2-parameter exponential distribution and the MLE analysis method, then insert an additional plot and select to show the analysis results for both data sheets on that plot, which will appear as shown next.

Weibull Distribution Examples

Probability Plotting Example

The Weibull distribution is one of the most widely used lifetime distributions in reliability engineering. It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, [math]\displaystyle{ {\beta} \,\! }[/math]. This chapter provides a brief background on the Weibull distribution, presents and derives most of the applicable equations and presents examples calculated both manually and by using ReliaSoft's Weibull++ software.

Weibull Probability Density Function

The 3-Parameter Weibull

The 3-parameter Weibull pdf is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

where:

[math]\displaystyle{ f(t)\geq 0,\text{ }t\geq \gamma \,\! }[/math]

[math]\displaystyle{ \beta\gt 0\ \,\! }[/math]

[math]\displaystyle{ \eta \gt 0 \,\! }[/math]

[math]\displaystyle{ -\infty \lt \gamma \lt +\infty \,\! }[/math]

and:

[math]\displaystyle{ \eta= \,\! }[/math] scale parameter, or characteristic life

[math]\displaystyle{ \beta= \,\! }[/math] shape parameter (or slope)

[math]\displaystyle{ \gamma= \,\! }[/math] location parameter (or failure free life)

The 2-Parameter Weibull

The 2-parameter Weibull pdf is obtained by setting [math]\displaystyle{ \gamma=0 \,\! }[/math], and is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( { \frac{t}{\eta }}\right) ^{\beta }} \,\! }[/math]

The 1-Parameter Weibull

The 1-parameter Weibull pdf is obtained by again setting [math]\displaystyle{ \gamma=0 \,\! }[/math] and assuming [math]\displaystyle{ \beta=C=Constant \,\! }[/math] assumed value or:

[math]\displaystyle{ f(t)={ \frac{C}{\eta }}\left( {\frac{t}{\eta }}\right) ^{C-1}e^{-\left( {\frac{t}{ \eta }}\right) ^{C}} \,\! }[/math]

where the only unknown parameter is the scale parameter, [math]\displaystyle{ \eta\,\! }[/math].

Note that in the formulation of the 1-parameter Weibull, we assume that the shape parameter [math]\displaystyle{ \beta \,\! }[/math] is known a priori from past experience with identical or similar products. The advantage of doing this is that data sets with few or no failures can be analyzed.

Weibull Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T} \,\! }[/math], (also called MTTF) of the Weibull pdf is given by:

[math]\displaystyle{ \overline{T}=\gamma +\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

where

[math]\displaystyle{ \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

is the gamma function evaluated at the value of:

[math]\displaystyle{ \left( { \frac{1}{\beta }}+1\right) \,\! }[/math]

The gamma function is defined as:

[math]\displaystyle{ \Gamma (n)=\int_{0}^{\infty }e^{-x}x^{n-1}dx \,\! }[/math]

For the 2-parameter case, this can be reduced to:

[math]\displaystyle{ \overline{T}=\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

Note that some practitioners erroneously assume that [math]\displaystyle{ \eta \,\! }[/math] is equal to the MTTF, [math]\displaystyle{ \overline{T}\,\! }[/math]. This is only true for the case of: [math]\displaystyle{ \beta=1 \,\! }[/math] or:

[math]\displaystyle{ \begin{align} \overline{T} &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {2}\right) \\ &= \eta \cdot 1\\ &= \eta \end{align} \,\! }[/math]

The Median

The median, [math]\displaystyle{ \breve{T}\,\! }[/math], of the Weibull distribution is given by:

[math]\displaystyle{ \breve{T}=\gamma +\eta \left( \ln 2\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T} \,\! }[/math], is given by:

[math]\displaystyle{ \tilde{T}=\gamma +\eta \left( 1-\frac{1}{\beta }\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ \sigma _{T}\,\! }[/math], is given by:

[math]\displaystyle{ \sigma _{T}=\eta \cdot \sqrt{\Gamma \left( {\frac{2}{\beta }}+1\right) -\Gamma \left( {\frac{1}{ \beta }}+1\right) ^{2}} \,\! }[/math]

The Weibull Reliability Function

The equation for the 3-parameter Weibull cumulative density function, cdf, is given by:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t-\gamma }{\eta }\right) ^{\beta }} \,\! }[/math]

This is also referred to as unreliability and designated as [math]\displaystyle{ Q(t) \,\! }[/math] by some authors.

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function for the 3-parameter Weibull distribution is then given by:

[math]\displaystyle{ R(t)=e^{-\left( { \frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

The Weibull Conditional Reliability Function

The 3-parameter Weibull conditional reliability function is given by:

[math]\displaystyle{ R(t|T)={ \frac{R(T+t)}{R(T)}}={\frac{e^{-\left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }}}{e^{-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }}}} \,\! }[/math]

or:

[math]\displaystyle{ R(t|T)=e^{-\left[ \left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }\right] } \,\! }[/math]

These give the reliability for a new mission of [math]\displaystyle{ t \,\! }[/math] duration, having already accumulated [math]\displaystyle{ T \,\! }[/math] time of operation up to the start of this new mission, and the units are checked out to assure that they will start the next mission successfully. It is called conditional because you can calculate the reliability of a new mission based on the fact that the unit or units already accumulated hours of operation successfully.

The Weibull Reliable Life

The reliable life, [math]\displaystyle{ T_{R}\,\! }[/math], of a unit for a specified reliability, [math]\displaystyle{ R\,\! }[/math], starting the mission at age zero, is given by:

[math]\displaystyle{ T_{R}=\gamma +\eta \cdot \left\{ -\ln ( R ) \right\} ^{ \frac{1}{\beta }} \,\! }[/math]

This is the life for which the unit/item will be functioning successfully with a reliability of [math]\displaystyle{ R\,\! }[/math]. If [math]\displaystyle{ R = 0.50\,\! }[/math], then [math]\displaystyle{ T_{R}=\breve{T} \,\! }[/math], the median life, or the life by which half of the units will survive.

The Weibull Failure Rate Function

The Weibull failure rate function, [math]\displaystyle{ \lambda(t) \,\! }[/math], is given by:

[math]\displaystyle{ \lambda \left( t\right) = \frac{f\left( t\right) }{R\left( t\right) }=\frac{\beta }{\eta }\left( \frac{ t-\gamma }{\eta }\right) ^{\beta -1} \,\! }[/math]

Characteristics of the Weibull Distribution

The Weibull distribution is widely used in reliability and life data analysis due to its versatility. Depending on the values of the parameters, the Weibull distribution can be used to model a variety of life behaviors. We will now examine how the values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], and the scale parameter, [math]\displaystyle{ \eta\,\! }[/math], affect such distribution characteristics as the shape of the curve, the reliability and the failure rate. Note that in the rest of this section we will assume the most general form of the Weibull distribution, (i.e., the 3-parameter form). The appropriate substitutions to obtain the other forms, such as the 2-parameter form where [math]\displaystyle{ \gamma = 0,\,\! }[/math] or the 1-parameter form where [math]\displaystyle{ \beta = C = \,\! }[/math] constant, can easily be made.

Effects of the Shape Parameter, beta

The Weibull shape parameter, [math]\displaystyle{ \beta\,\! }[/math], is also known as the slope. This is because the value of [math]\displaystyle{ \beta\,\! }[/math] is equal to the slope of the regressed line in a probability plot. Different values of the shape parameter can have marked effects on the behavior of the distribution. In fact, some values of the shape parameter will cause the distribution equations to reduce to those of other distributions. For example, when [math]\displaystyle{ \beta = 1\,\! }[/math], the pdf of the 3-parameter Weibull distribution reduces to that of the 2-parameter exponential distribution or:

[math]\displaystyle{ f(t)={\frac{1}{\eta }}e^{-{\frac{t-\gamma }{\eta }}} \,\! }[/math]

where [math]\displaystyle{ \frac{1}{\eta }=\lambda = \,\! }[/math] failure rate. The parameter [math]\displaystyle{ \beta\,\! }[/math] is a pure number, (i.e., it is dimensionless). The following figure shows the effect of different values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], on the shape of the pdf. As you can see, the shape can take on a variety of forms based on the value of [math]\displaystyle{ \beta\,\! }[/math].

For [math]\displaystyle{ 0\lt \beta \leq 1 \,\! }[/math]:

• As [math]\displaystyle{ t \rightarrow 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]), [math]\displaystyle{ f(t)\rightarrow \infty.\,\! }[/math]
• As [math]\displaystyle{ t\rightarrow \infty\,\! }[/math], [math]\displaystyle{ f(t)\rightarrow 0\,\! }[/math].
• [math]\displaystyle{ f(t)\,\! }[/math] decreases monotonically and is convex as it increases beyond the value of [math]\displaystyle{ \gamma\,\! }[/math].
• The mode is non-existent.

For [math]\displaystyle{ \beta \gt 1 \,\! }[/math]:

• [math]\displaystyle{ f(t) = 0\,\! }[/math] at [math]\displaystyle{ t = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]).
• [math]\displaystyle{ f(t)\,\! }[/math] increases as [math]\displaystyle{ t\rightarrow \tilde{T} \,\! }[/math] (the mode) and decreases thereafter.
• For [math]\displaystyle{ \beta \lt 2.6\,\! }[/math] the Weibull pdf is positively skewed (has a right tail), for [math]\displaystyle{ 2.6 \lt \beta \lt 3.7\,\! }[/math] its coefficient of skewness approaches zero (no tail). Consequently, it may approximate the normal pdf, and for [math]\displaystyle{ \beta \gt 3.7\,\! }[/math] it is negatively skewed (left tail). The way the value of [math]\displaystyle{ \beta\,\! }[/math] relates to the physical behavior of the items being modeled becomes more apparent when we observe how its different values affect the reliability and failure rate functions. Note that for [math]\displaystyle{ \beta = 0.999\,\! }[/math], [math]\displaystyle{ f(0) = \infty\,\! }[/math], but for [math]\displaystyle{ \beta = 1.001\,\! }[/math], [math]\displaystyle{ f(0) = 0.\,\! }[/math] This abrupt shift is what complicates MLE estimation when [math]\displaystyle{ \beta\,\! }[/math] is close to 1.

The Effect of beta on the cdf and Reliability Function

The above figure shows the effect of the value of [math]\displaystyle{ \beta\,\! }[/math] on the cdf, as manifested in the Weibull probability plot. It is easy to see why this parameter is sometimes referred to as the slope. Note that the models represented by the three lines all have the same value of [math]\displaystyle{ \eta\,\! }[/math]. The following figure shows the effects of these varied values of [math]\displaystyle{ \beta\,\! }[/math] on the reliability plot, which is a linear analog of the probability plot.

• [math]\displaystyle{ R(t)\,\! }[/math] decreases sharply and monotonically for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases monotonically but less sharply than for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases as increases. As wear-out sets in, the curve goes through an inflection point and decreases sharply.

The Effect of beta on the Weibull Failure Rate

The value of [math]\displaystyle{ \beta\,\! }[/math] has a marked effect on the failure rate of the Weibull distribution and inferences can be drawn about a population's failure characteristics just by considering whether the value of [math]\displaystyle{ \beta\,\! }[/math] is less than, equal to, or greater than one.

As indicated by above figure, populations with [math]\displaystyle{ \beta \lt 1\,\! }[/math] exhibit a failure rate that decreases with time, populations with [math]\displaystyle{ \beta = 1\,\! }[/math] have a constant failure rate (consistent with the exponential distribution) and populations with [math]\displaystyle{ \beta \gt 1\,\! }[/math] have a failure rate that increases with time. All three life stages of the bathtub curve can be modeled with the Weibull distribution and varying values of [math]\displaystyle{ \beta\,\! }[/math]. The Weibull failure rate for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] is unbounded at [math]\displaystyle{ T = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\!)\,\! }[/math]. The failure rate, [math]\displaystyle{ \lambda(t),\,\! }[/math] decreases thereafter monotonically and is convex, approaching the value of zero as [math]\displaystyle{ t\rightarrow \infty\,\! }[/math] or [math]\displaystyle{ \lambda (\infty) = 0\,\! }[/math]. This behavior makes it suitable for representing the failure rate of units exhibiting early-type failures, for which the failure rate decreases with age. When encountering such behavior in a manufactured product, it may be indicative of problems in the production process, inadequate burn-in, substandard parts and components, or problems with packaging and shipping. For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] yields a constant value of [math]\displaystyle{ { \frac{1}{\eta }} \,\! }[/math] or:

[math]\displaystyle{ \lambda (t)=\lambda ={\frac{1}{\eta }} \,\! }[/math]

This makes it suitable for representing the failure rate of chance-type failures and the useful life period failure rate of units.

For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] increases as [math]\displaystyle{ t\,\! }[/math] increases and becomes suitable for representing the failure rate of units exhibiting wear-out type failures. For [math]\displaystyle{ 1 \lt \beta \lt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is concave, consequently the failure rate increases at a decreasing rate as [math]\displaystyle{ t\,\! }[/math] increases.

For [math]\displaystyle{ \beta = 2\,\! }[/math] there emerges a straight line relationship between [math]\displaystyle{ \lambda(t)\,\! }[/math] and [math]\displaystyle{ t\,\! }[/math], starting at a value of [math]\displaystyle{ \lambda(t) = 0\,\! }[/math] at [math]\displaystyle{ t = \gamma\,\! }[/math], and increasing thereafter with a slope of [math]\displaystyle{ { \frac{2}{\eta ^{2}}} \,\! }[/math]. Consequently, the failure rate increases at a constant rate as [math]\displaystyle{ t\,\! }[/math] increases. Furthermore, if [math]\displaystyle{ \eta = 1\,\! }[/math] the slope becomes equal to 2, and when [math]\displaystyle{ \gamma = 0\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] becomes a straight line which passes through the origin with a slope of 2. Note that at [math]\displaystyle{ \beta = 2\,\! }[/math], the Weibull distribution equations reduce to that of the Rayleigh distribution.

When [math]\displaystyle{ \beta \gt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is convex, with its slope increasing as [math]\displaystyle{ t\,\! }[/math] increases. Consequently, the failure rate increases at an increasing rate as [math]\displaystyle{ t\,\! }[/math] increases, indicating wearout life.

Effects of the Scale Parameter, eta

A change in the scale parameter [math]\displaystyle{ \eta\,\! }[/math] has the same effect on the distribution as a change of the abscissa scale. Increasing the value of [math]\displaystyle{ \eta\,\! }[/math] while holding [math]\displaystyle{ \beta\,\! }[/math] constant has the effect of stretching out the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of the pdf curve will also decrease with the increase of [math]\displaystyle{ \eta\,\! }[/math], as indicated in the above figure.

• If [math]\displaystyle{ \eta\,\! }[/math] is increased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets stretched out to the right and its height decreases, while maintaining its shape and location.
• If [math]\displaystyle{ \eta\,\! }[/math] is decreased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets pushed in towards the left (i.e., towards its beginning or towards 0 or [math]\displaystyle{ \gamma\,\! }[/math]), and its height increases.
• [math]\displaystyle{ \eta\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Effects of the Location Parameter, gamma

The location parameter, [math]\displaystyle{ \gamma\,\! }[/math], as the name implies, locates the distribution along the abscissa. Changing the value of [math]\displaystyle{ \gamma\,\! }[/math] has the effect of sliding the distribution and its associated function either to the right (if [math]\displaystyle{ \gamma \gt 0\,\! }[/math]) or to the left (if [math]\displaystyle{ \gamma \lt 0\,\! }[/math]).

• When [math]\displaystyle{ \gamma = 0,\,\! }[/math] the distribution starts at [math]\displaystyle{ t=0\,\! }[/math] or at the origin.
• If [math]\displaystyle{ \gamma \gt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the right of the origin.
• If [math]\displaystyle{ \gamma \lt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the left of the origin.
• [math]\displaystyle{ \gamma\,\! }[/math] provides an estimate of the earliest time-to-failure of such units.
• The life period 0 to [math]\displaystyle{ + \gamma\,\! }[/math] is a failure free operating period of such units.
• The parameter [math]\displaystyle{ \gamma\,\! }[/math] may assume all values and provides an estimate of the earliest time a failure may be observed. A negative [math]\displaystyle{ \gamma\,\! }[/math] may indicate that failures have occurred prior to the beginning of the test, namely during production, in storage, in transit, during checkout prior to the start of a mission, or prior to actual use.
• [math]\displaystyle{ \gamma\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Weibull Distribution Examples

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

2P Weibull Distribution RRY Example

The Weibull distribution is one of the most widely used lifetime distributions in reliability engineering. It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, [math]\displaystyle{ {\beta} \,\! }[/math]. This chapter provides a brief background on the Weibull distribution, presents and derives most of the applicable equations and presents examples calculated both manually and by using ReliaSoft's Weibull++ software.

Weibull Probability Density Function

The 3-Parameter Weibull

The 3-parameter Weibull pdf is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

where:

[math]\displaystyle{ f(t)\geq 0,\text{ }t\geq \gamma \,\! }[/math]

[math]\displaystyle{ \beta\gt 0\ \,\! }[/math]

[math]\displaystyle{ \eta \gt 0 \,\! }[/math]

[math]\displaystyle{ -\infty \lt \gamma \lt +\infty \,\! }[/math]

and:

[math]\displaystyle{ \eta= \,\! }[/math] scale parameter, or characteristic life

[math]\displaystyle{ \beta= \,\! }[/math] shape parameter (or slope)

[math]\displaystyle{ \gamma= \,\! }[/math] location parameter (or failure free life)

The 2-Parameter Weibull

The 2-parameter Weibull pdf is obtained by setting [math]\displaystyle{ \gamma=0 \,\! }[/math], and is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( { \frac{t}{\eta }}\right) ^{\beta }} \,\! }[/math]

The 1-Parameter Weibull

The 1-parameter Weibull pdf is obtained by again setting [math]\displaystyle{ \gamma=0 \,\! }[/math] and assuming [math]\displaystyle{ \beta=C=Constant \,\! }[/math] assumed value or:

[math]\displaystyle{ f(t)={ \frac{C}{\eta }}\left( {\frac{t}{\eta }}\right) ^{C-1}e^{-\left( {\frac{t}{ \eta }}\right) ^{C}} \,\! }[/math]

where the only unknown parameter is the scale parameter, [math]\displaystyle{ \eta\,\! }[/math].

Note that in the formulation of the 1-parameter Weibull, we assume that the shape parameter [math]\displaystyle{ \beta \,\! }[/math] is known a priori from past experience with identical or similar products. The advantage of doing this is that data sets with few or no failures can be analyzed.

Weibull Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T} \,\! }[/math], (also called MTTF) of the Weibull pdf is given by:

[math]\displaystyle{ \overline{T}=\gamma +\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

where

[math]\displaystyle{ \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

is the gamma function evaluated at the value of:

[math]\displaystyle{ \left( { \frac{1}{\beta }}+1\right) \,\! }[/math]

The gamma function is defined as:

[math]\displaystyle{ \Gamma (n)=\int_{0}^{\infty }e^{-x}x^{n-1}dx \,\! }[/math]

For the 2-parameter case, this can be reduced to:

[math]\displaystyle{ \overline{T}=\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

Note that some practitioners erroneously assume that [math]\displaystyle{ \eta \,\! }[/math] is equal to the MTTF, [math]\displaystyle{ \overline{T}\,\! }[/math]. This is only true for the case of: [math]\displaystyle{ \beta=1 \,\! }[/math] or:

[math]\displaystyle{ \begin{align} \overline{T} &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {2}\right) \\ &= \eta \cdot 1\\ &= \eta \end{align} \,\! }[/math]

The Median

The median, [math]\displaystyle{ \breve{T}\,\! }[/math], of the Weibull distribution is given by:

[math]\displaystyle{ \breve{T}=\gamma +\eta \left( \ln 2\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T} \,\! }[/math], is given by:

[math]\displaystyle{ \tilde{T}=\gamma +\eta \left( 1-\frac{1}{\beta }\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ \sigma _{T}\,\! }[/math], is given by:

[math]\displaystyle{ \sigma _{T}=\eta \cdot \sqrt{\Gamma \left( {\frac{2}{\beta }}+1\right) -\Gamma \left( {\frac{1}{ \beta }}+1\right) ^{2}} \,\! }[/math]

The Weibull Reliability Function

The equation for the 3-parameter Weibull cumulative density function, cdf, is given by:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t-\gamma }{\eta }\right) ^{\beta }} \,\! }[/math]

This is also referred to as unreliability and designated as [math]\displaystyle{ Q(t) \,\! }[/math] by some authors.

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function for the 3-parameter Weibull distribution is then given by:

[math]\displaystyle{ R(t)=e^{-\left( { \frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

The Weibull Conditional Reliability Function

The 3-parameter Weibull conditional reliability function is given by:

[math]\displaystyle{ R(t|T)={ \frac{R(T+t)}{R(T)}}={\frac{e^{-\left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }}}{e^{-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }}}} \,\! }[/math]

or:

[math]\displaystyle{ R(t|T)=e^{-\left[ \left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }\right] } \,\! }[/math]

These give the reliability for a new mission of [math]\displaystyle{ t \,\! }[/math] duration, having already accumulated [math]\displaystyle{ T \,\! }[/math] time of operation up to the start of this new mission, and the units are checked out to assure that they will start the next mission successfully. It is called conditional because you can calculate the reliability of a new mission based on the fact that the unit or units already accumulated hours of operation successfully.

The Weibull Reliable Life

The reliable life, [math]\displaystyle{ T_{R}\,\! }[/math], of a unit for a specified reliability, [math]\displaystyle{ R\,\! }[/math], starting the mission at age zero, is given by:

[math]\displaystyle{ T_{R}=\gamma +\eta \cdot \left\{ -\ln ( R ) \right\} ^{ \frac{1}{\beta }} \,\! }[/math]

This is the life for which the unit/item will be functioning successfully with a reliability of [math]\displaystyle{ R\,\! }[/math]. If [math]\displaystyle{ R = 0.50\,\! }[/math], then [math]\displaystyle{ T_{R}=\breve{T} \,\! }[/math], the median life, or the life by which half of the units will survive.

The Weibull Failure Rate Function

The Weibull failure rate function, [math]\displaystyle{ \lambda(t) \,\! }[/math], is given by:

[math]\displaystyle{ \lambda \left( t\right) = \frac{f\left( t\right) }{R\left( t\right) }=\frac{\beta }{\eta }\left( \frac{ t-\gamma }{\eta }\right) ^{\beta -1} \,\! }[/math]

Characteristics of the Weibull Distribution

The Weibull distribution is widely used in reliability and life data analysis due to its versatility. Depending on the values of the parameters, the Weibull distribution can be used to model a variety of life behaviors. We will now examine how the values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], and the scale parameter, [math]\displaystyle{ \eta\,\! }[/math], affect such distribution characteristics as the shape of the curve, the reliability and the failure rate. Note that in the rest of this section we will assume the most general form of the Weibull distribution, (i.e., the 3-parameter form). The appropriate substitutions to obtain the other forms, such as the 2-parameter form where [math]\displaystyle{ \gamma = 0,\,\! }[/math] or the 1-parameter form where [math]\displaystyle{ \beta = C = \,\! }[/math] constant, can easily be made.

Effects of the Shape Parameter, beta

The Weibull shape parameter, [math]\displaystyle{ \beta\,\! }[/math], is also known as the slope. This is because the value of [math]\displaystyle{ \beta\,\! }[/math] is equal to the slope of the regressed line in a probability plot. Different values of the shape parameter can have marked effects on the behavior of the distribution. In fact, some values of the shape parameter will cause the distribution equations to reduce to those of other distributions. For example, when [math]\displaystyle{ \beta = 1\,\! }[/math], the pdf of the 3-parameter Weibull distribution reduces to that of the 2-parameter exponential distribution or:

[math]\displaystyle{ f(t)={\frac{1}{\eta }}e^{-{\frac{t-\gamma }{\eta }}} \,\! }[/math]

where [math]\displaystyle{ \frac{1}{\eta }=\lambda = \,\! }[/math] failure rate. The parameter [math]\displaystyle{ \beta\,\! }[/math] is a pure number, (i.e., it is dimensionless). The following figure shows the effect of different values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], on the shape of the pdf. As you can see, the shape can take on a variety of forms based on the value of [math]\displaystyle{ \beta\,\! }[/math].

For [math]\displaystyle{ 0\lt \beta \leq 1 \,\! }[/math]:

• As [math]\displaystyle{ t \rightarrow 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]), [math]\displaystyle{ f(t)\rightarrow \infty.\,\! }[/math]
• As [math]\displaystyle{ t\rightarrow \infty\,\! }[/math], [math]\displaystyle{ f(t)\rightarrow 0\,\! }[/math].
• [math]\displaystyle{ f(t)\,\! }[/math] decreases monotonically and is convex as it increases beyond the value of [math]\displaystyle{ \gamma\,\! }[/math].
• The mode is non-existent.

For [math]\displaystyle{ \beta \gt 1 \,\! }[/math]:

• [math]\displaystyle{ f(t) = 0\,\! }[/math] at [math]\displaystyle{ t = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]).
• [math]\displaystyle{ f(t)\,\! }[/math] increases as [math]\displaystyle{ t\rightarrow \tilde{T} \,\! }[/math] (the mode) and decreases thereafter.
• For [math]\displaystyle{ \beta \lt 2.6\,\! }[/math] the Weibull pdf is positively skewed (has a right tail), for [math]\displaystyle{ 2.6 \lt \beta \lt 3.7\,\! }[/math] its coefficient of skewness approaches zero (no tail). Consequently, it may approximate the normal pdf, and for [math]\displaystyle{ \beta \gt 3.7\,\! }[/math] it is negatively skewed (left tail). The way the value of [math]\displaystyle{ \beta\,\! }[/math] relates to the physical behavior of the items being modeled becomes more apparent when we observe how its different values affect the reliability and failure rate functions. Note that for [math]\displaystyle{ \beta = 0.999\,\! }[/math], [math]\displaystyle{ f(0) = \infty\,\! }[/math], but for [math]\displaystyle{ \beta = 1.001\,\! }[/math], [math]\displaystyle{ f(0) = 0.\,\! }[/math] This abrupt shift is what complicates MLE estimation when [math]\displaystyle{ \beta\,\! }[/math] is close to 1.

The Effect of beta on the cdf and Reliability Function

The above figure shows the effect of the value of [math]\displaystyle{ \beta\,\! }[/math] on the cdf, as manifested in the Weibull probability plot. It is easy to see why this parameter is sometimes referred to as the slope. Note that the models represented by the three lines all have the same value of [math]\displaystyle{ \eta\,\! }[/math]. The following figure shows the effects of these varied values of [math]\displaystyle{ \beta\,\! }[/math] on the reliability plot, which is a linear analog of the probability plot.

• [math]\displaystyle{ R(t)\,\! }[/math] decreases sharply and monotonically for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases monotonically but less sharply than for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases as increases. As wear-out sets in, the curve goes through an inflection point and decreases sharply.

The Effect of beta on the Weibull Failure Rate

The value of [math]\displaystyle{ \beta\,\! }[/math] has a marked effect on the failure rate of the Weibull distribution and inferences can be drawn about a population's failure characteristics just by considering whether the value of [math]\displaystyle{ \beta\,\! }[/math] is less than, equal to, or greater than one.

As indicated by above figure, populations with [math]\displaystyle{ \beta \lt 1\,\! }[/math] exhibit a failure rate that decreases with time, populations with [math]\displaystyle{ \beta = 1\,\! }[/math] have a constant failure rate (consistent with the exponential distribution) and populations with [math]\displaystyle{ \beta \gt 1\,\! }[/math] have a failure rate that increases with time. All three life stages of the bathtub curve can be modeled with the Weibull distribution and varying values of [math]\displaystyle{ \beta\,\! }[/math]. The Weibull failure rate for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] is unbounded at [math]\displaystyle{ T = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\!)\,\! }[/math]. The failure rate, [math]\displaystyle{ \lambda(t),\,\! }[/math] decreases thereafter monotonically and is convex, approaching the value of zero as [math]\displaystyle{ t\rightarrow \infty\,\! }[/math] or [math]\displaystyle{ \lambda (\infty) = 0\,\! }[/math]. This behavior makes it suitable for representing the failure rate of units exhibiting early-type failures, for which the failure rate decreases with age. When encountering such behavior in a manufactured product, it may be indicative of problems in the production process, inadequate burn-in, substandard parts and components, or problems with packaging and shipping. For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] yields a constant value of [math]\displaystyle{ { \frac{1}{\eta }} \,\! }[/math] or:

[math]\displaystyle{ \lambda (t)=\lambda ={\frac{1}{\eta }} \,\! }[/math]

This makes it suitable for representing the failure rate of chance-type failures and the useful life period failure rate of units.

For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] increases as [math]\displaystyle{ t\,\! }[/math] increases and becomes suitable for representing the failure rate of units exhibiting wear-out type failures. For [math]\displaystyle{ 1 \lt \beta \lt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is concave, consequently the failure rate increases at a decreasing rate as [math]\displaystyle{ t\,\! }[/math] increases.

For [math]\displaystyle{ \beta = 2\,\! }[/math] there emerges a straight line relationship between [math]\displaystyle{ \lambda(t)\,\! }[/math] and [math]\displaystyle{ t\,\! }[/math], starting at a value of [math]\displaystyle{ \lambda(t) = 0\,\! }[/math] at [math]\displaystyle{ t = \gamma\,\! }[/math], and increasing thereafter with a slope of [math]\displaystyle{ { \frac{2}{\eta ^{2}}} \,\! }[/math]. Consequently, the failure rate increases at a constant rate as [math]\displaystyle{ t\,\! }[/math] increases. Furthermore, if [math]\displaystyle{ \eta = 1\,\! }[/math] the slope becomes equal to 2, and when [math]\displaystyle{ \gamma = 0\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] becomes a straight line which passes through the origin with a slope of 2. Note that at [math]\displaystyle{ \beta = 2\,\! }[/math], the Weibull distribution equations reduce to that of the Rayleigh distribution.

When [math]\displaystyle{ \beta \gt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is convex, with its slope increasing as [math]\displaystyle{ t\,\! }[/math] increases. Consequently, the failure rate increases at an increasing rate as [math]\displaystyle{ t\,\! }[/math] increases, indicating wearout life.

Effects of the Scale Parameter, eta

A change in the scale parameter [math]\displaystyle{ \eta\,\! }[/math] has the same effect on the distribution as a change of the abscissa scale. Increasing the value of [math]\displaystyle{ \eta\,\! }[/math] while holding [math]\displaystyle{ \beta\,\! }[/math] constant has the effect of stretching out the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of the pdf curve will also decrease with the increase of [math]\displaystyle{ \eta\,\! }[/math], as indicated in the above figure.

• If [math]\displaystyle{ \eta\,\! }[/math] is increased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets stretched out to the right and its height decreases, while maintaining its shape and location.
• If [math]\displaystyle{ \eta\,\! }[/math] is decreased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets pushed in towards the left (i.e., towards its beginning or towards 0 or [math]\displaystyle{ \gamma\,\! }[/math]), and its height increases.
• [math]\displaystyle{ \eta\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Effects of the Location Parameter, gamma

The location parameter, [math]\displaystyle{ \gamma\,\! }[/math], as the name implies, locates the distribution along the abscissa. Changing the value of [math]\displaystyle{ \gamma\,\! }[/math] has the effect of sliding the distribution and its associated function either to the right (if [math]\displaystyle{ \gamma \gt 0\,\! }[/math]) or to the left (if [math]\displaystyle{ \gamma \lt 0\,\! }[/math]).

• When [math]\displaystyle{ \gamma = 0,\,\! }[/math] the distribution starts at [math]\displaystyle{ t=0\,\! }[/math] or at the origin.
• If [math]\displaystyle{ \gamma \gt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the right of the origin.
• If [math]\displaystyle{ \gamma \lt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the left of the origin.
• [math]\displaystyle{ \gamma\,\! }[/math] provides an estimate of the earliest time-to-failure of such units.
• The life period 0 to [math]\displaystyle{ + \gamma\,\! }[/math] is a failure free operating period of such units.
• The parameter [math]\displaystyle{ \gamma\,\! }[/math] may assume all values and provides an estimate of the earliest time a failure may be observed. A negative [math]\displaystyle{ \gamma\,\! }[/math] may indicate that failures have occurred prior to the beginning of the test, namely during production, in storage, in transit, during checkout prior to the start of a mission, or prior to actual use.
• [math]\displaystyle{ \gamma\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Weibull Distribution Examples

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

2P Weibull Distribution RRX Example

The Weibull distribution is one of the most widely used lifetime distributions in reliability engineering. It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, [math]\displaystyle{ {\beta} \,\! }[/math]. This chapter provides a brief background on the Weibull distribution, presents and derives most of the applicable equations and presents examples calculated both manually and by using ReliaSoft's Weibull++ software.

Weibull Probability Density Function

The 3-Parameter Weibull

The 3-parameter Weibull pdf is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

where:

[math]\displaystyle{ f(t)\geq 0,\text{ }t\geq \gamma \,\! }[/math]

[math]\displaystyle{ \beta\gt 0\ \,\! }[/math]

[math]\displaystyle{ \eta \gt 0 \,\! }[/math]

[math]\displaystyle{ -\infty \lt \gamma \lt +\infty \,\! }[/math]

and:

[math]\displaystyle{ \eta= \,\! }[/math] scale parameter, or characteristic life

[math]\displaystyle{ \beta= \,\! }[/math] shape parameter (or slope)

[math]\displaystyle{ \gamma= \,\! }[/math] location parameter (or failure free life)

The 2-Parameter Weibull

The 2-parameter Weibull pdf is obtained by setting [math]\displaystyle{ \gamma=0 \,\! }[/math], and is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( { \frac{t}{\eta }}\right) ^{\beta }} \,\! }[/math]

The 1-Parameter Weibull

The 1-parameter Weibull pdf is obtained by again setting [math]\displaystyle{ \gamma=0 \,\! }[/math] and assuming [math]\displaystyle{ \beta=C=Constant \,\! }[/math] assumed value or:

[math]\displaystyle{ f(t)={ \frac{C}{\eta }}\left( {\frac{t}{\eta }}\right) ^{C-1}e^{-\left( {\frac{t}{ \eta }}\right) ^{C}} \,\! }[/math]

where the only unknown parameter is the scale parameter, [math]\displaystyle{ \eta\,\! }[/math].

Note that in the formulation of the 1-parameter Weibull, we assume that the shape parameter [math]\displaystyle{ \beta \,\! }[/math] is known a priori from past experience with identical or similar products. The advantage of doing this is that data sets with few or no failures can be analyzed.

Weibull Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T} \,\! }[/math], (also called MTTF) of the Weibull pdf is given by:

[math]\displaystyle{ \overline{T}=\gamma +\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

where

[math]\displaystyle{ \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

is the gamma function evaluated at the value of:

[math]\displaystyle{ \left( { \frac{1}{\beta }}+1\right) \,\! }[/math]

The gamma function is defined as:

[math]\displaystyle{ \Gamma (n)=\int_{0}^{\infty }e^{-x}x^{n-1}dx \,\! }[/math]

For the 2-parameter case, this can be reduced to:

[math]\displaystyle{ \overline{T}=\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

Note that some practitioners erroneously assume that [math]\displaystyle{ \eta \,\! }[/math] is equal to the MTTF, [math]\displaystyle{ \overline{T}\,\! }[/math]. This is only true for the case of: [math]\displaystyle{ \beta=1 \,\! }[/math] or:

[math]\displaystyle{ \begin{align} \overline{T} &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {2}\right) \\ &= \eta \cdot 1\\ &= \eta \end{align} \,\! }[/math]

The Median

The median, [math]\displaystyle{ \breve{T}\,\! }[/math], of the Weibull distribution is given by:

[math]\displaystyle{ \breve{T}=\gamma +\eta \left( \ln 2\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T} \,\! }[/math], is given by:

[math]\displaystyle{ \tilde{T}=\gamma +\eta \left( 1-\frac{1}{\beta }\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ \sigma _{T}\,\! }[/math], is given by:

[math]\displaystyle{ \sigma _{T}=\eta \cdot \sqrt{\Gamma \left( {\frac{2}{\beta }}+1\right) -\Gamma \left( {\frac{1}{ \beta }}+1\right) ^{2}} \,\! }[/math]

The Weibull Reliability Function

The equation for the 3-parameter Weibull cumulative density function, cdf, is given by:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t-\gamma }{\eta }\right) ^{\beta }} \,\! }[/math]

This is also referred to as unreliability and designated as [math]\displaystyle{ Q(t) \,\! }[/math] by some authors.

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function for the 3-parameter Weibull distribution is then given by:

[math]\displaystyle{ R(t)=e^{-\left( { \frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

The Weibull Conditional Reliability Function

The 3-parameter Weibull conditional reliability function is given by:

[math]\displaystyle{ R(t|T)={ \frac{R(T+t)}{R(T)}}={\frac{e^{-\left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }}}{e^{-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }}}} \,\! }[/math]

or:

[math]\displaystyle{ R(t|T)=e^{-\left[ \left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }\right] } \,\! }[/math]

These give the reliability for a new mission of [math]\displaystyle{ t \,\! }[/math] duration, having already accumulated [math]\displaystyle{ T \,\! }[/math] time of operation up to the start of this new mission, and the units are checked out to assure that they will start the next mission successfully. It is called conditional because you can calculate the reliability of a new mission based on the fact that the unit or units already accumulated hours of operation successfully.

The Weibull Reliable Life

The reliable life, [math]\displaystyle{ T_{R}\,\! }[/math], of a unit for a specified reliability, [math]\displaystyle{ R\,\! }[/math], starting the mission at age zero, is given by:

[math]\displaystyle{ T_{R}=\gamma +\eta \cdot \left\{ -\ln ( R ) \right\} ^{ \frac{1}{\beta }} \,\! }[/math]

This is the life for which the unit/item will be functioning successfully with a reliability of [math]\displaystyle{ R\,\! }[/math]. If [math]\displaystyle{ R = 0.50\,\! }[/math], then [math]\displaystyle{ T_{R}=\breve{T} \,\! }[/math], the median life, or the life by which half of the units will survive.

The Weibull Failure Rate Function

The Weibull failure rate function, [math]\displaystyle{ \lambda(t) \,\! }[/math], is given by:

[math]\displaystyle{ \lambda \left( t\right) = \frac{f\left( t\right) }{R\left( t\right) }=\frac{\beta }{\eta }\left( \frac{ t-\gamma }{\eta }\right) ^{\beta -1} \,\! }[/math]

Characteristics of the Weibull Distribution

The Weibull distribution is widely used in reliability and life data analysis due to its versatility. Depending on the values of the parameters, the Weibull distribution can be used to model a variety of life behaviors. We will now examine how the values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], and the scale parameter, [math]\displaystyle{ \eta\,\! }[/math], affect such distribution characteristics as the shape of the curve, the reliability and the failure rate. Note that in the rest of this section we will assume the most general form of the Weibull distribution, (i.e., the 3-parameter form). The appropriate substitutions to obtain the other forms, such as the 2-parameter form where [math]\displaystyle{ \gamma = 0,\,\! }[/math] or the 1-parameter form where [math]\displaystyle{ \beta = C = \,\! }[/math] constant, can easily be made.

Effects of the Shape Parameter, beta

The Weibull shape parameter, [math]\displaystyle{ \beta\,\! }[/math], is also known as the slope. This is because the value of [math]\displaystyle{ \beta\,\! }[/math] is equal to the slope of the regressed line in a probability plot. Different values of the shape parameter can have marked effects on the behavior of the distribution. In fact, some values of the shape parameter will cause the distribution equations to reduce to those of other distributions. For example, when [math]\displaystyle{ \beta = 1\,\! }[/math], the pdf of the 3-parameter Weibull distribution reduces to that of the 2-parameter exponential distribution or:

[math]\displaystyle{ f(t)={\frac{1}{\eta }}e^{-{\frac{t-\gamma }{\eta }}} \,\! }[/math]

where [math]\displaystyle{ \frac{1}{\eta }=\lambda = \,\! }[/math] failure rate. The parameter [math]\displaystyle{ \beta\,\! }[/math] is a pure number, (i.e., it is dimensionless). The following figure shows the effect of different values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], on the shape of the pdf. As you can see, the shape can take on a variety of forms based on the value of [math]\displaystyle{ \beta\,\! }[/math].

For [math]\displaystyle{ 0\lt \beta \leq 1 \,\! }[/math]:

• As [math]\displaystyle{ t \rightarrow 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]), [math]\displaystyle{ f(t)\rightarrow \infty.\,\! }[/math]
• As [math]\displaystyle{ t\rightarrow \infty\,\! }[/math], [math]\displaystyle{ f(t)\rightarrow 0\,\! }[/math].
• [math]\displaystyle{ f(t)\,\! }[/math] decreases monotonically and is convex as it increases beyond the value of [math]\displaystyle{ \gamma\,\! }[/math].
• The mode is non-existent.

For [math]\displaystyle{ \beta \gt 1 \,\! }[/math]:

• [math]\displaystyle{ f(t) = 0\,\! }[/math] at [math]\displaystyle{ t = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]).
• [math]\displaystyle{ f(t)\,\! }[/math] increases as [math]\displaystyle{ t\rightarrow \tilde{T} \,\! }[/math] (the mode) and decreases thereafter.
• For [math]\displaystyle{ \beta \lt 2.6\,\! }[/math] the Weibull pdf is positively skewed (has a right tail), for [math]\displaystyle{ 2.6 \lt \beta \lt 3.7\,\! }[/math] its coefficient of skewness approaches zero (no tail). Consequently, it may approximate the normal pdf, and for [math]\displaystyle{ \beta \gt 3.7\,\! }[/math] it is negatively skewed (left tail). The way the value of [math]\displaystyle{ \beta\,\! }[/math] relates to the physical behavior of the items being modeled becomes more apparent when we observe how its different values affect the reliability and failure rate functions. Note that for [math]\displaystyle{ \beta = 0.999\,\! }[/math], [math]\displaystyle{ f(0) = \infty\,\! }[/math], but for [math]\displaystyle{ \beta = 1.001\,\! }[/math], [math]\displaystyle{ f(0) = 0.\,\! }[/math] This abrupt shift is what complicates MLE estimation when [math]\displaystyle{ \beta\,\! }[/math] is close to 1.

The Effect of beta on the cdf and Reliability Function

The above figure shows the effect of the value of [math]\displaystyle{ \beta\,\! }[/math] on the cdf, as manifested in the Weibull probability plot. It is easy to see why this parameter is sometimes referred to as the slope. Note that the models represented by the three lines all have the same value of [math]\displaystyle{ \eta\,\! }[/math]. The following figure shows the effects of these varied values of [math]\displaystyle{ \beta\,\! }[/math] on the reliability plot, which is a linear analog of the probability plot.

• [math]\displaystyle{ R(t)\,\! }[/math] decreases sharply and monotonically for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases monotonically but less sharply than for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases as increases. As wear-out sets in, the curve goes through an inflection point and decreases sharply.

The Effect of beta on the Weibull Failure Rate

The value of [math]\displaystyle{ \beta\,\! }[/math] has a marked effect on the failure rate of the Weibull distribution and inferences can be drawn about a population's failure characteristics just by considering whether the value of [math]\displaystyle{ \beta\,\! }[/math] is less than, equal to, or greater than one.

As indicated by above figure, populations with [math]\displaystyle{ \beta \lt 1\,\! }[/math] exhibit a failure rate that decreases with time, populations with [math]\displaystyle{ \beta = 1\,\! }[/math] have a constant failure rate (consistent with the exponential distribution) and populations with [math]\displaystyle{ \beta \gt 1\,\! }[/math] have a failure rate that increases with time. All three life stages of the bathtub curve can be modeled with the Weibull distribution and varying values of [math]\displaystyle{ \beta\,\! }[/math]. The Weibull failure rate for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] is unbounded at [math]\displaystyle{ T = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\!)\,\! }[/math]. The failure rate, [math]\displaystyle{ \lambda(t),\,\! }[/math] decreases thereafter monotonically and is convex, approaching the value of zero as [math]\displaystyle{ t\rightarrow \infty\,\! }[/math] or [math]\displaystyle{ \lambda (\infty) = 0\,\! }[/math]. This behavior makes it suitable for representing the failure rate of units exhibiting early-type failures, for which the failure rate decreases with age. When encountering such behavior in a manufactured product, it may be indicative of problems in the production process, inadequate burn-in, substandard parts and components, or problems with packaging and shipping. For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] yields a constant value of [math]\displaystyle{ { \frac{1}{\eta }} \,\! }[/math] or:

[math]\displaystyle{ \lambda (t)=\lambda ={\frac{1}{\eta }} \,\! }[/math]

This makes it suitable for representing the failure rate of chance-type failures and the useful life period failure rate of units.

For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] increases as [math]\displaystyle{ t\,\! }[/math] increases and becomes suitable for representing the failure rate of units exhibiting wear-out type failures. For [math]\displaystyle{ 1 \lt \beta \lt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is concave, consequently the failure rate increases at a decreasing rate as [math]\displaystyle{ t\,\! }[/math] increases.

For [math]\displaystyle{ \beta = 2\,\! }[/math] there emerges a straight line relationship between [math]\displaystyle{ \lambda(t)\,\! }[/math] and [math]\displaystyle{ t\,\! }[/math], starting at a value of [math]\displaystyle{ \lambda(t) = 0\,\! }[/math] at [math]\displaystyle{ t = \gamma\,\! }[/math], and increasing thereafter with a slope of [math]\displaystyle{ { \frac{2}{\eta ^{2}}} \,\! }[/math]. Consequently, the failure rate increases at a constant rate as [math]\displaystyle{ t\,\! }[/math] increases. Furthermore, if [math]\displaystyle{ \eta = 1\,\! }[/math] the slope becomes equal to 2, and when [math]\displaystyle{ \gamma = 0\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] becomes a straight line which passes through the origin with a slope of 2. Note that at [math]\displaystyle{ \beta = 2\,\! }[/math], the Weibull distribution equations reduce to that of the Rayleigh distribution.

When [math]\displaystyle{ \beta \gt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is convex, with its slope increasing as [math]\displaystyle{ t\,\! }[/math] increases. Consequently, the failure rate increases at an increasing rate as [math]\displaystyle{ t\,\! }[/math] increases, indicating wearout life.

Effects of the Scale Parameter, eta

A change in the scale parameter [math]\displaystyle{ \eta\,\! }[/math] has the same effect on the distribution as a change of the abscissa scale. Increasing the value of [math]\displaystyle{ \eta\,\! }[/math] while holding [math]\displaystyle{ \beta\,\! }[/math] constant has the effect of stretching out the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of the pdf curve will also decrease with the increase of [math]\displaystyle{ \eta\,\! }[/math], as indicated in the above figure.

• If [math]\displaystyle{ \eta\,\! }[/math] is increased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets stretched out to the right and its height decreases, while maintaining its shape and location.
• If [math]\displaystyle{ \eta\,\! }[/math] is decreased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets pushed in towards the left (i.e., towards its beginning or towards 0 or [math]\displaystyle{ \gamma\,\! }[/math]), and its height increases.
• [math]\displaystyle{ \eta\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Effects of the Location Parameter, gamma

The location parameter, [math]\displaystyle{ \gamma\,\! }[/math], as the name implies, locates the distribution along the abscissa. Changing the value of [math]\displaystyle{ \gamma\,\! }[/math] has the effect of sliding the distribution and its associated function either to the right (if [math]\displaystyle{ \gamma \gt 0\,\! }[/math]) or to the left (if [math]\displaystyle{ \gamma \lt 0\,\! }[/math]).

• When [math]\displaystyle{ \gamma = 0,\,\! }[/math] the distribution starts at [math]\displaystyle{ t=0\,\! }[/math] or at the origin.
• If [math]\displaystyle{ \gamma \gt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the right of the origin.
• If [math]\displaystyle{ \gamma \lt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the left of the origin.
• [math]\displaystyle{ \gamma\,\! }[/math] provides an estimate of the earliest time-to-failure of such units.
• The life period 0 to [math]\displaystyle{ + \gamma\,\! }[/math] is a failure free operating period of such units.
• The parameter [math]\displaystyle{ \gamma\,\! }[/math] may assume all values and provides an estimate of the earliest time a failure may be observed. A negative [math]\displaystyle{ \gamma\,\! }[/math] may indicate that failures have occurred prior to the beginning of the test, namely during production, in storage, in transit, during checkout prior to the start of a mission, or prior to actual use.
• [math]\displaystyle{ \gamma\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Weibull Distribution Examples

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Maximum Likelihood Estimation Example

The Weibull distribution is one of the most widely used lifetime distributions in reliability engineering. It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, [math]\displaystyle{ {\beta} \,\! }[/math]. This chapter provides a brief background on the Weibull distribution, presents and derives most of the applicable equations and presents examples calculated both manually and by using ReliaSoft's Weibull++ software.

Weibull Probability Density Function

The 3-Parameter Weibull

The 3-parameter Weibull pdf is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

where:

[math]\displaystyle{ f(t)\geq 0,\text{ }t\geq \gamma \,\! }[/math]

[math]\displaystyle{ \beta\gt 0\ \,\! }[/math]

[math]\displaystyle{ \eta \gt 0 \,\! }[/math]

[math]\displaystyle{ -\infty \lt \gamma \lt +\infty \,\! }[/math]

and:

[math]\displaystyle{ \eta= \,\! }[/math] scale parameter, or characteristic life

[math]\displaystyle{ \beta= \,\! }[/math] shape parameter (or slope)

[math]\displaystyle{ \gamma= \,\! }[/math] location parameter (or failure free life)

The 2-Parameter Weibull

The 2-parameter Weibull pdf is obtained by setting [math]\displaystyle{ \gamma=0 \,\! }[/math], and is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( { \frac{t}{\eta }}\right) ^{\beta }} \,\! }[/math]

The 1-Parameter Weibull

The 1-parameter Weibull pdf is obtained by again setting [math]\displaystyle{ \gamma=0 \,\! }[/math] and assuming [math]\displaystyle{ \beta=C=Constant \,\! }[/math] assumed value or:

[math]\displaystyle{ f(t)={ \frac{C}{\eta }}\left( {\frac{t}{\eta }}\right) ^{C-1}e^{-\left( {\frac{t}{ \eta }}\right) ^{C}} \,\! }[/math]

where the only unknown parameter is the scale parameter, [math]\displaystyle{ \eta\,\! }[/math].

Note that in the formulation of the 1-parameter Weibull, we assume that the shape parameter [math]\displaystyle{ \beta \,\! }[/math] is known a priori from past experience with identical or similar products. The advantage of doing this is that data sets with few or no failures can be analyzed.

Weibull Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T} \,\! }[/math], (also called MTTF) of the Weibull pdf is given by:

[math]\displaystyle{ \overline{T}=\gamma +\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

where

[math]\displaystyle{ \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

is the gamma function evaluated at the value of:

[math]\displaystyle{ \left( { \frac{1}{\beta }}+1\right) \,\! }[/math]

The gamma function is defined as:

[math]\displaystyle{ \Gamma (n)=\int_{0}^{\infty }e^{-x}x^{n-1}dx \,\! }[/math]

For the 2-parameter case, this can be reduced to:

[math]\displaystyle{ \overline{T}=\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

Note that some practitioners erroneously assume that [math]\displaystyle{ \eta \,\! }[/math] is equal to the MTTF, [math]\displaystyle{ \overline{T}\,\! }[/math]. This is only true for the case of: [math]\displaystyle{ \beta=1 \,\! }[/math] or:

[math]\displaystyle{ \begin{align} \overline{T} &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {2}\right) \\ &= \eta \cdot 1\\ &= \eta \end{align} \,\! }[/math]

The Median

The median, [math]\displaystyle{ \breve{T}\,\! }[/math], of the Weibull distribution is given by:

[math]\displaystyle{ \breve{T}=\gamma +\eta \left( \ln 2\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T} \,\! }[/math], is given by:

[math]\displaystyle{ \tilde{T}=\gamma +\eta \left( 1-\frac{1}{\beta }\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ \sigma _{T}\,\! }[/math], is given by:

[math]\displaystyle{ \sigma _{T}=\eta \cdot \sqrt{\Gamma \left( {\frac{2}{\beta }}+1\right) -\Gamma \left( {\frac{1}{ \beta }}+1\right) ^{2}} \,\! }[/math]

The Weibull Reliability Function

The equation for the 3-parameter Weibull cumulative density function, cdf, is given by:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t-\gamma }{\eta }\right) ^{\beta }} \,\! }[/math]

This is also referred to as unreliability and designated as [math]\displaystyle{ Q(t) \,\! }[/math] by some authors.

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function for the 3-parameter Weibull distribution is then given by:

[math]\displaystyle{ R(t)=e^{-\left( { \frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

The Weibull Conditional Reliability Function

The 3-parameter Weibull conditional reliability function is given by:

[math]\displaystyle{ R(t|T)={ \frac{R(T+t)}{R(T)}}={\frac{e^{-\left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }}}{e^{-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }}}} \,\! }[/math]

or:

[math]\displaystyle{ R(t|T)=e^{-\left[ \left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }\right] } \,\! }[/math]

These give the reliability for a new mission of [math]\displaystyle{ t \,\! }[/math] duration, having already accumulated [math]\displaystyle{ T \,\! }[/math] time of operation up to the start of this new mission, and the units are checked out to assure that they will start the next mission successfully. It is called conditional because you can calculate the reliability of a new mission based on the fact that the unit or units already accumulated hours of operation successfully.

The Weibull Reliable Life

The reliable life, [math]\displaystyle{ T_{R}\,\! }[/math], of a unit for a specified reliability, [math]\displaystyle{ R\,\! }[/math], starting the mission at age zero, is given by:

[math]\displaystyle{ T_{R}=\gamma +\eta \cdot \left\{ -\ln ( R ) \right\} ^{ \frac{1}{\beta }} \,\! }[/math]

This is the life for which the unit/item will be functioning successfully with a reliability of [math]\displaystyle{ R\,\! }[/math]. If [math]\displaystyle{ R = 0.50\,\! }[/math], then [math]\displaystyle{ T_{R}=\breve{T} \,\! }[/math], the median life, or the life by which half of the units will survive.

The Weibull Failure Rate Function

The Weibull failure rate function, [math]\displaystyle{ \lambda(t) \,\! }[/math], is given by:

[math]\displaystyle{ \lambda \left( t\right) = \frac{f\left( t\right) }{R\left( t\right) }=\frac{\beta }{\eta }\left( \frac{ t-\gamma }{\eta }\right) ^{\beta -1} \,\! }[/math]

Characteristics of the Weibull Distribution

The Weibull distribution is widely used in reliability and life data analysis due to its versatility. Depending on the values of the parameters, the Weibull distribution can be used to model a variety of life behaviors. We will now examine how the values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], and the scale parameter, [math]\displaystyle{ \eta\,\! }[/math], affect such distribution characteristics as the shape of the curve, the reliability and the failure rate. Note that in the rest of this section we will assume the most general form of the Weibull distribution, (i.e., the 3-parameter form). The appropriate substitutions to obtain the other forms, such as the 2-parameter form where [math]\displaystyle{ \gamma = 0,\,\! }[/math] or the 1-parameter form where [math]\displaystyle{ \beta = C = \,\! }[/math] constant, can easily be made.

Effects of the Shape Parameter, beta

The Weibull shape parameter, [math]\displaystyle{ \beta\,\! }[/math], is also known as the slope. This is because the value of [math]\displaystyle{ \beta\,\! }[/math] is equal to the slope of the regressed line in a probability plot. Different values of the shape parameter can have marked effects on the behavior of the distribution. In fact, some values of the shape parameter will cause the distribution equations to reduce to those of other distributions. For example, when [math]\displaystyle{ \beta = 1\,\! }[/math], the pdf of the 3-parameter Weibull distribution reduces to that of the 2-parameter exponential distribution or:

[math]\displaystyle{ f(t)={\frac{1}{\eta }}e^{-{\frac{t-\gamma }{\eta }}} \,\! }[/math]

where [math]\displaystyle{ \frac{1}{\eta }=\lambda = \,\! }[/math] failure rate. The parameter [math]\displaystyle{ \beta\,\! }[/math] is a pure number, (i.e., it is dimensionless). The following figure shows the effect of different values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], on the shape of the pdf. As you can see, the shape can take on a variety of forms based on the value of [math]\displaystyle{ \beta\,\! }[/math].

For [math]\displaystyle{ 0\lt \beta \leq 1 \,\! }[/math]:

• As [math]\displaystyle{ t \rightarrow 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]), [math]\displaystyle{ f(t)\rightarrow \infty.\,\! }[/math]
• As [math]\displaystyle{ t\rightarrow \infty\,\! }[/math], [math]\displaystyle{ f(t)\rightarrow 0\,\! }[/math].
• [math]\displaystyle{ f(t)\,\! }[/math] decreases monotonically and is convex as it increases beyond the value of [math]\displaystyle{ \gamma\,\! }[/math].
• The mode is non-existent.

For [math]\displaystyle{ \beta \gt 1 \,\! }[/math]:

• [math]\displaystyle{ f(t) = 0\,\! }[/math] at [math]\displaystyle{ t = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]).
• [math]\displaystyle{ f(t)\,\! }[/math] increases as [math]\displaystyle{ t\rightarrow \tilde{T} \,\! }[/math] (the mode) and decreases thereafter.
• For [math]\displaystyle{ \beta \lt 2.6\,\! }[/math] the Weibull pdf is positively skewed (has a right tail), for [math]\displaystyle{ 2.6 \lt \beta \lt 3.7\,\! }[/math] its coefficient of skewness approaches zero (no tail). Consequently, it may approximate the normal pdf, and for [math]\displaystyle{ \beta \gt 3.7\,\! }[/math] it is negatively skewed (left tail). The way the value of [math]\displaystyle{ \beta\,\! }[/math] relates to the physical behavior of the items being modeled becomes more apparent when we observe how its different values affect the reliability and failure rate functions. Note that for [math]\displaystyle{ \beta = 0.999\,\! }[/math], [math]\displaystyle{ f(0) = \infty\,\! }[/math], but for [math]\displaystyle{ \beta = 1.001\,\! }[/math], [math]\displaystyle{ f(0) = 0.\,\! }[/math] This abrupt shift is what complicates MLE estimation when [math]\displaystyle{ \beta\,\! }[/math] is close to 1.

The Effect of beta on the cdf and Reliability Function

The above figure shows the effect of the value of [math]\displaystyle{ \beta\,\! }[/math] on the cdf, as manifested in the Weibull probability plot. It is easy to see why this parameter is sometimes referred to as the slope. Note that the models represented by the three lines all have the same value of [math]\displaystyle{ \eta\,\! }[/math]. The following figure shows the effects of these varied values of [math]\displaystyle{ \beta\,\! }[/math] on the reliability plot, which is a linear analog of the probability plot.

• [math]\displaystyle{ R(t)\,\! }[/math] decreases sharply and monotonically for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases monotonically but less sharply than for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases as increases. As wear-out sets in, the curve goes through an inflection point and decreases sharply.

The Effect of beta on the Weibull Failure Rate

The value of [math]\displaystyle{ \beta\,\! }[/math] has a marked effect on the failure rate of the Weibull distribution and inferences can be drawn about a population's failure characteristics just by considering whether the value of [math]\displaystyle{ \beta\,\! }[/math] is less than, equal to, or greater than one.

As indicated by above figure, populations with [math]\displaystyle{ \beta \lt 1\,\! }[/math] exhibit a failure rate that decreases with time, populations with [math]\displaystyle{ \beta = 1\,\! }[/math] have a constant failure rate (consistent with the exponential distribution) and populations with [math]\displaystyle{ \beta \gt 1\,\! }[/math] have a failure rate that increases with time. All three life stages of the bathtub curve can be modeled with the Weibull distribution and varying values of [math]\displaystyle{ \beta\,\! }[/math]. The Weibull failure rate for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] is unbounded at [math]\displaystyle{ T = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\!)\,\! }[/math]. The failure rate, [math]\displaystyle{ \lambda(t),\,\! }[/math] decreases thereafter monotonically and is convex, approaching the value of zero as [math]\displaystyle{ t\rightarrow \infty\,\! }[/math] or [math]\displaystyle{ \lambda (\infty) = 0\,\! }[/math]. This behavior makes it suitable for representing the failure rate of units exhibiting early-type failures, for which the failure rate decreases with age. When encountering such behavior in a manufactured product, it may be indicative of problems in the production process, inadequate burn-in, substandard parts and components, or problems with packaging and shipping. For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] yields a constant value of [math]\displaystyle{ { \frac{1}{\eta }} \,\! }[/math] or:

[math]\displaystyle{ \lambda (t)=\lambda ={\frac{1}{\eta }} \,\! }[/math]

This makes it suitable for representing the failure rate of chance-type failures and the useful life period failure rate of units.

For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] increases as [math]\displaystyle{ t\,\! }[/math] increases and becomes suitable for representing the failure rate of units exhibiting wear-out type failures. For [math]\displaystyle{ 1 \lt \beta \lt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is concave, consequently the failure rate increases at a decreasing rate as [math]\displaystyle{ t\,\! }[/math] increases.

For [math]\displaystyle{ \beta = 2\,\! }[/math] there emerges a straight line relationship between [math]\displaystyle{ \lambda(t)\,\! }[/math] and [math]\displaystyle{ t\,\! }[/math], starting at a value of [math]\displaystyle{ \lambda(t) = 0\,\! }[/math] at [math]\displaystyle{ t = \gamma\,\! }[/math], and increasing thereafter with a slope of [math]\displaystyle{ { \frac{2}{\eta ^{2}}} \,\! }[/math]. Consequently, the failure rate increases at a constant rate as [math]\displaystyle{ t\,\! }[/math] increases. Furthermore, if [math]\displaystyle{ \eta = 1\,\! }[/math] the slope becomes equal to 2, and when [math]\displaystyle{ \gamma = 0\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] becomes a straight line which passes through the origin with a slope of 2. Note that at [math]\displaystyle{ \beta = 2\,\! }[/math], the Weibull distribution equations reduce to that of the Rayleigh distribution.

When [math]\displaystyle{ \beta \gt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is convex, with its slope increasing as [math]\displaystyle{ t\,\! }[/math] increases. Consequently, the failure rate increases at an increasing rate as [math]\displaystyle{ t\,\! }[/math] increases, indicating wearout life.

Effects of the Scale Parameter, eta

A change in the scale parameter [math]\displaystyle{ \eta\,\! }[/math] has the same effect on the distribution as a change of the abscissa scale. Increasing the value of [math]\displaystyle{ \eta\,\! }[/math] while holding [math]\displaystyle{ \beta\,\! }[/math] constant has the effect of stretching out the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of the pdf curve will also decrease with the increase of [math]\displaystyle{ \eta\,\! }[/math], as indicated in the above figure.

• If [math]\displaystyle{ \eta\,\! }[/math] is increased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets stretched out to the right and its height decreases, while maintaining its shape and location.
• If [math]\displaystyle{ \eta\,\! }[/math] is decreased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets pushed in towards the left (i.e., towards its beginning or towards 0 or [math]\displaystyle{ \gamma\,\! }[/math]), and its height increases.
• [math]\displaystyle{ \eta\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Effects of the Location Parameter, gamma

The location parameter, [math]\displaystyle{ \gamma\,\! }[/math], as the name implies, locates the distribution along the abscissa. Changing the value of [math]\displaystyle{ \gamma\,\! }[/math] has the effect of sliding the distribution and its associated function either to the right (if [math]\displaystyle{ \gamma \gt 0\,\! }[/math]) or to the left (if [math]\displaystyle{ \gamma \lt 0\,\! }[/math]).

• When [math]\displaystyle{ \gamma = 0,\,\! }[/math] the distribution starts at [math]\displaystyle{ t=0\,\! }[/math] or at the origin.
• If [math]\displaystyle{ \gamma \gt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the right of the origin.
• If [math]\displaystyle{ \gamma \lt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the left of the origin.
• [math]\displaystyle{ \gamma\,\! }[/math] provides an estimate of the earliest time-to-failure of such units.
• The life period 0 to [math]\displaystyle{ + \gamma\,\! }[/math] is a failure free operating period of such units.
• The parameter [math]\displaystyle{ \gamma\,\! }[/math] may assume all values and provides an estimate of the earliest time a failure may be observed. A negative [math]\displaystyle{ \gamma\,\! }[/math] may indicate that failures have occurred prior to the beginning of the test, namely during production, in storage, in transit, during checkout prior to the start of a mission, or prior to actual use.
• [math]\displaystyle{ \gamma\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Weibull Distribution Examples

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Weibull-Bayesian with Lognormal Prior Example

The Weibull distribution is one of the most widely used lifetime distributions in reliability engineering. It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, [math]\displaystyle{ {\beta} \,\! }[/math]. This chapter provides a brief background on the Weibull distribution, presents and derives most of the applicable equations and presents examples calculated both manually and by using ReliaSoft's Weibull++ software.

Weibull Probability Density Function

The 3-Parameter Weibull

The 3-parameter Weibull pdf is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

where:

[math]\displaystyle{ f(t)\geq 0,\text{ }t\geq \gamma \,\! }[/math]

[math]\displaystyle{ \beta\gt 0\ \,\! }[/math]

[math]\displaystyle{ \eta \gt 0 \,\! }[/math]

[math]\displaystyle{ -\infty \lt \gamma \lt +\infty \,\! }[/math]

and:

[math]\displaystyle{ \eta= \,\! }[/math] scale parameter, or characteristic life

[math]\displaystyle{ \beta= \,\! }[/math] shape parameter (or slope)

[math]\displaystyle{ \gamma= \,\! }[/math] location parameter (or failure free life)

The 2-Parameter Weibull

The 2-parameter Weibull pdf is obtained by setting [math]\displaystyle{ \gamma=0 \,\! }[/math], and is given by:

[math]\displaystyle{ f(t)={ \frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( { \frac{t}{\eta }}\right) ^{\beta }} \,\! }[/math]

The 1-Parameter Weibull

The 1-parameter Weibull pdf is obtained by again setting [math]\displaystyle{ \gamma=0 \,\! }[/math] and assuming [math]\displaystyle{ \beta=C=Constant \,\! }[/math] assumed value or:

[math]\displaystyle{ f(t)={ \frac{C}{\eta }}\left( {\frac{t}{\eta }}\right) ^{C-1}e^{-\left( {\frac{t}{ \eta }}\right) ^{C}} \,\! }[/math]

where the only unknown parameter is the scale parameter, [math]\displaystyle{ \eta\,\! }[/math].

Note that in the formulation of the 1-parameter Weibull, we assume that the shape parameter [math]\displaystyle{ \beta \,\! }[/math] is known a priori from past experience with identical or similar products. The advantage of doing this is that data sets with few or no failures can be analyzed.

Weibull Distribution Functions

The Mean or MTTF

The mean, [math]\displaystyle{ \overline{T} \,\! }[/math], (also called MTTF) of the Weibull pdf is given by:

[math]\displaystyle{ \overline{T}=\gamma +\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

where

[math]\displaystyle{ \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

is the gamma function evaluated at the value of:

[math]\displaystyle{ \left( { \frac{1}{\beta }}+1\right) \,\! }[/math]

The gamma function is defined as:

[math]\displaystyle{ \Gamma (n)=\int_{0}^{\infty }e^{-x}x^{n-1}dx \,\! }[/math]

For the 2-parameter case, this can be reduced to:

[math]\displaystyle{ \overline{T}=\eta \cdot \Gamma \left( {\frac{1}{\beta }}+1\right) \,\! }[/math]

Note that some practitioners erroneously assume that [math]\displaystyle{ \eta \,\! }[/math] is equal to the MTTF, [math]\displaystyle{ \overline{T}\,\! }[/math]. This is only true for the case of: [math]\displaystyle{ \beta=1 \,\! }[/math] or:

[math]\displaystyle{ \begin{align} \overline{T} &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {\frac{1}{1}}+1\right) \\ &= \eta \cdot \Gamma \left( {2}\right) \\ &= \eta \cdot 1\\ &= \eta \end{align} \,\! }[/math]

The Median

The median, [math]\displaystyle{ \breve{T}\,\! }[/math], of the Weibull distribution is given by:

[math]\displaystyle{ \breve{T}=\gamma +\eta \left( \ln 2\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Mode

The mode, [math]\displaystyle{ \tilde{T} \,\! }[/math], is given by:

[math]\displaystyle{ \tilde{T}=\gamma +\eta \left( 1-\frac{1}{\beta }\right) ^{\frac{1}{\beta }} \,\! }[/math]

The Standard Deviation

The standard deviation, [math]\displaystyle{ \sigma _{T}\,\! }[/math], is given by:

[math]\displaystyle{ \sigma _{T}=\eta \cdot \sqrt{\Gamma \left( {\frac{2}{\beta }}+1\right) -\Gamma \left( {\frac{1}{ \beta }}+1\right) ^{2}} \,\! }[/math]

The Weibull Reliability Function

The equation for the 3-parameter Weibull cumulative density function, cdf, is given by:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t-\gamma }{\eta }\right) ^{\beta }} \,\! }[/math]

This is also referred to as unreliability and designated as [math]\displaystyle{ Q(t) \,\! }[/math] by some authors.

Recalling that the reliability function of a distribution is simply one minus the cdf, the reliability function for the 3-parameter Weibull distribution is then given by:

[math]\displaystyle{ R(t)=e^{-\left( { \frac{t-\gamma }{\eta }}\right) ^{\beta }} \,\! }[/math]

The Weibull Conditional Reliability Function

The 3-parameter Weibull conditional reliability function is given by:

[math]\displaystyle{ R(t|T)={ \frac{R(T+t)}{R(T)}}={\frac{e^{-\left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }}}{e^{-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }}}} \,\! }[/math]

or:

[math]\displaystyle{ R(t|T)=e^{-\left[ \left( {\frac{T+t-\gamma }{\eta }}\right) ^{\beta }-\left( {\frac{T-\gamma }{\eta }}\right) ^{\beta }\right] } \,\! }[/math]

These give the reliability for a new mission of [math]\displaystyle{ t \,\! }[/math] duration, having already accumulated [math]\displaystyle{ T \,\! }[/math] time of operation up to the start of this new mission, and the units are checked out to assure that they will start the next mission successfully. It is called conditional because you can calculate the reliability of a new mission based on the fact that the unit or units already accumulated hours of operation successfully.

The Weibull Reliable Life

The reliable life, [math]\displaystyle{ T_{R}\,\! }[/math], of a unit for a specified reliability, [math]\displaystyle{ R\,\! }[/math], starting the mission at age zero, is given by:

[math]\displaystyle{ T_{R}=\gamma +\eta \cdot \left\{ -\ln ( R ) \right\} ^{ \frac{1}{\beta }} \,\! }[/math]

This is the life for which the unit/item will be functioning successfully with a reliability of [math]\displaystyle{ R\,\! }[/math]. If [math]\displaystyle{ R = 0.50\,\! }[/math], then [math]\displaystyle{ T_{R}=\breve{T} \,\! }[/math], the median life, or the life by which half of the units will survive.

The Weibull Failure Rate Function

The Weibull failure rate function, [math]\displaystyle{ \lambda(t) \,\! }[/math], is given by:

[math]\displaystyle{ \lambda \left( t\right) = \frac{f\left( t\right) }{R\left( t\right) }=\frac{\beta }{\eta }\left( \frac{ t-\gamma }{\eta }\right) ^{\beta -1} \,\! }[/math]

Characteristics of the Weibull Distribution

The Weibull distribution is widely used in reliability and life data analysis due to its versatility. Depending on the values of the parameters, the Weibull distribution can be used to model a variety of life behaviors. We will now examine how the values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], and the scale parameter, [math]\displaystyle{ \eta\,\! }[/math], affect such distribution characteristics as the shape of the curve, the reliability and the failure rate. Note that in the rest of this section we will assume the most general form of the Weibull distribution, (i.e., the 3-parameter form). The appropriate substitutions to obtain the other forms, such as the 2-parameter form where [math]\displaystyle{ \gamma = 0,\,\! }[/math] or the 1-parameter form where [math]\displaystyle{ \beta = C = \,\! }[/math] constant, can easily be made.

Effects of the Shape Parameter, beta

The Weibull shape parameter, [math]\displaystyle{ \beta\,\! }[/math], is also known as the slope. This is because the value of [math]\displaystyle{ \beta\,\! }[/math] is equal to the slope of the regressed line in a probability plot. Different values of the shape parameter can have marked effects on the behavior of the distribution. In fact, some values of the shape parameter will cause the distribution equations to reduce to those of other distributions. For example, when [math]\displaystyle{ \beta = 1\,\! }[/math], the pdf of the 3-parameter Weibull distribution reduces to that of the 2-parameter exponential distribution or:

[math]\displaystyle{ f(t)={\frac{1}{\eta }}e^{-{\frac{t-\gamma }{\eta }}} \,\! }[/math]

where [math]\displaystyle{ \frac{1}{\eta }=\lambda = \,\! }[/math] failure rate. The parameter [math]\displaystyle{ \beta\,\! }[/math] is a pure number, (i.e., it is dimensionless). The following figure shows the effect of different values of the shape parameter, [math]\displaystyle{ \beta\,\! }[/math], on the shape of the pdf. As you can see, the shape can take on a variety of forms based on the value of [math]\displaystyle{ \beta\,\! }[/math].

For [math]\displaystyle{ 0\lt \beta \leq 1 \,\! }[/math]:

• As [math]\displaystyle{ t \rightarrow 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]), [math]\displaystyle{ f(t)\rightarrow \infty.\,\! }[/math]
• As [math]\displaystyle{ t\rightarrow \infty\,\! }[/math], [math]\displaystyle{ f(t)\rightarrow 0\,\! }[/math].
• [math]\displaystyle{ f(t)\,\! }[/math] decreases monotonically and is convex as it increases beyond the value of [math]\displaystyle{ \gamma\,\! }[/math].
• The mode is non-existent.

For [math]\displaystyle{ \beta \gt 1 \,\! }[/math]:

• [math]\displaystyle{ f(t) = 0\,\! }[/math] at [math]\displaystyle{ t = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\! }[/math]).
• [math]\displaystyle{ f(t)\,\! }[/math] increases as [math]\displaystyle{ t\rightarrow \tilde{T} \,\! }[/math] (the mode) and decreases thereafter.
• For [math]\displaystyle{ \beta \lt 2.6\,\! }[/math] the Weibull pdf is positively skewed (has a right tail), for [math]\displaystyle{ 2.6 \lt \beta \lt 3.7\,\! }[/math] its coefficient of skewness approaches zero (no tail). Consequently, it may approximate the normal pdf, and for [math]\displaystyle{ \beta \gt 3.7\,\! }[/math] it is negatively skewed (left tail). The way the value of [math]\displaystyle{ \beta\,\! }[/math] relates to the physical behavior of the items being modeled becomes more apparent when we observe how its different values affect the reliability and failure rate functions. Note that for [math]\displaystyle{ \beta = 0.999\,\! }[/math], [math]\displaystyle{ f(0) = \infty\,\! }[/math], but for [math]\displaystyle{ \beta = 1.001\,\! }[/math], [math]\displaystyle{ f(0) = 0.\,\! }[/math] This abrupt shift is what complicates MLE estimation when [math]\displaystyle{ \beta\,\! }[/math] is close to 1.

The Effect of beta on the cdf and Reliability Function

The above figure shows the effect of the value of [math]\displaystyle{ \beta\,\! }[/math] on the cdf, as manifested in the Weibull probability plot. It is easy to see why this parameter is sometimes referred to as the slope. Note that the models represented by the three lines all have the same value of [math]\displaystyle{ \eta\,\! }[/math]. The following figure shows the effects of these varied values of [math]\displaystyle{ \beta\,\! }[/math] on the reliability plot, which is a linear analog of the probability plot.

• [math]\displaystyle{ R(t)\,\! }[/math] decreases sharply and monotonically for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases monotonically but less sharply than for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] and is convex.
• For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ R(t)\,\! }[/math] decreases as increases. As wear-out sets in, the curve goes through an inflection point and decreases sharply.

The Effect of beta on the Weibull Failure Rate

The value of [math]\displaystyle{ \beta\,\! }[/math] has a marked effect on the failure rate of the Weibull distribution and inferences can be drawn about a population's failure characteristics just by considering whether the value of [math]\displaystyle{ \beta\,\! }[/math] is less than, equal to, or greater than one.

As indicated by above figure, populations with [math]\displaystyle{ \beta \lt 1\,\! }[/math] exhibit a failure rate that decreases with time, populations with [math]\displaystyle{ \beta = 1\,\! }[/math] have a constant failure rate (consistent with the exponential distribution) and populations with [math]\displaystyle{ \beta \gt 1\,\! }[/math] have a failure rate that increases with time. All three life stages of the bathtub curve can be modeled with the Weibull distribution and varying values of [math]\displaystyle{ \beta\,\! }[/math]. The Weibull failure rate for [math]\displaystyle{ 0 \lt \beta \lt 1\,\! }[/math] is unbounded at [math]\displaystyle{ T = 0\,\! }[/math] (or [math]\displaystyle{ \gamma\,\!)\,\! }[/math]. The failure rate, [math]\displaystyle{ \lambda(t),\,\! }[/math] decreases thereafter monotonically and is convex, approaching the value of zero as [math]\displaystyle{ t\rightarrow \infty\,\! }[/math] or [math]\displaystyle{ \lambda (\infty) = 0\,\! }[/math]. This behavior makes it suitable for representing the failure rate of units exhibiting early-type failures, for which the failure rate decreases with age. When encountering such behavior in a manufactured product, it may be indicative of problems in the production process, inadequate burn-in, substandard parts and components, or problems with packaging and shipping. For [math]\displaystyle{ \beta = 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] yields a constant value of [math]\displaystyle{ { \frac{1}{\eta }} \,\! }[/math] or:

[math]\displaystyle{ \lambda (t)=\lambda ={\frac{1}{\eta }} \,\! }[/math]

This makes it suitable for representing the failure rate of chance-type failures and the useful life period failure rate of units.

For [math]\displaystyle{ \beta \gt 1\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] increases as [math]\displaystyle{ t\,\! }[/math] increases and becomes suitable for representing the failure rate of units exhibiting wear-out type failures. For [math]\displaystyle{ 1 \lt \beta \lt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is concave, consequently the failure rate increases at a decreasing rate as [math]\displaystyle{ t\,\! }[/math] increases.

For [math]\displaystyle{ \beta = 2\,\! }[/math] there emerges a straight line relationship between [math]\displaystyle{ \lambda(t)\,\! }[/math] and [math]\displaystyle{ t\,\! }[/math], starting at a value of [math]\displaystyle{ \lambda(t) = 0\,\! }[/math] at [math]\displaystyle{ t = \gamma\,\! }[/math], and increasing thereafter with a slope of [math]\displaystyle{ { \frac{2}{\eta ^{2}}} \,\! }[/math]. Consequently, the failure rate increases at a constant rate as [math]\displaystyle{ t\,\! }[/math] increases. Furthermore, if [math]\displaystyle{ \eta = 1\,\! }[/math] the slope becomes equal to 2, and when [math]\displaystyle{ \gamma = 0\,\! }[/math], [math]\displaystyle{ \lambda(t)\,\! }[/math] becomes a straight line which passes through the origin with a slope of 2. Note that at [math]\displaystyle{ \beta = 2\,\! }[/math], the Weibull distribution equations reduce to that of the Rayleigh distribution.

When [math]\displaystyle{ \beta \gt 2,\,\! }[/math] the [math]\displaystyle{ \lambda(t)\,\! }[/math] curve is convex, with its slope increasing as [math]\displaystyle{ t\,\! }[/math] increases. Consequently, the failure rate increases at an increasing rate as [math]\displaystyle{ t\,\! }[/math] increases, indicating wearout life.

Effects of the Scale Parameter, eta

A change in the scale parameter [math]\displaystyle{ \eta\,\! }[/math] has the same effect on the distribution as a change of the abscissa scale. Increasing the value of [math]\displaystyle{ \eta\,\! }[/math] while holding [math]\displaystyle{ \beta\,\! }[/math] constant has the effect of stretching out the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of the pdf curve will also decrease with the increase of [math]\displaystyle{ \eta\,\! }[/math], as indicated in the above figure.

• If [math]\displaystyle{ \eta\,\! }[/math] is increased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets stretched out to the right and its height decreases, while maintaining its shape and location.
• If [math]\displaystyle{ \eta\,\! }[/math] is decreased while [math]\displaystyle{ \beta\,\! }[/math] and [math]\displaystyle{ \gamma\,\! }[/math] are kept the same, the distribution gets pushed in towards the left (i.e., towards its beginning or towards 0 or [math]\displaystyle{ \gamma\,\! }[/math]), and its height increases.
• [math]\displaystyle{ \eta\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Effects of the Location Parameter, gamma

The location parameter, [math]\displaystyle{ \gamma\,\! }[/math], as the name implies, locates the distribution along the abscissa. Changing the value of [math]\displaystyle{ \gamma\,\! }[/math] has the effect of sliding the distribution and its associated function either to the right (if [math]\displaystyle{ \gamma \gt 0\,\! }[/math]) or to the left (if [math]\displaystyle{ \gamma \lt 0\,\! }[/math]).

• When [math]\displaystyle{ \gamma = 0,\,\! }[/math] the distribution starts at [math]\displaystyle{ t=0\,\! }[/math] or at the origin.
• If [math]\displaystyle{ \gamma \gt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the right of the origin.
• If [math]\displaystyle{ \gamma \lt 0,\,\! }[/math] the distribution starts at the location [math]\displaystyle{ \gamma\,\! }[/math] to the left of the origin.
• [math]\displaystyle{ \gamma\,\! }[/math] provides an estimate of the earliest time-to-failure of such units.
• The life period 0 to [math]\displaystyle{ + \gamma\,\! }[/math] is a failure free operating period of such units.
• The parameter [math]\displaystyle{ \gamma\,\! }[/math] may assume all values and provides an estimate of the earliest time a failure may be observed. A negative [math]\displaystyle{ \gamma\,\! }[/math] may indicate that failures have occurred prior to the beginning of the test, namely during production, in storage, in transit, during checkout prior to the start of a mission, or prior to actual use.
• [math]\displaystyle{ \gamma\,\! }[/math] has the same units as [math]\displaystyle{ t\,\! }[/math], such as hours, miles, cycles, actuations, etc.

Weibull Distribution Examples

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Weibull-Bayesian with Lognormal Prior Example

A manufacturer has tested prototypes of a modified product. The test was terminated at 2,000 hours, with only 2 failures observed from a sample size of 18. The following table shows the data.

Number of State	State of F or S	State End Time
1	F	1180
1	F	1842
16	S	2000

Because of the lack of failure data in the prototype testing, the manufacturer decided to use information gathered from prior tests on this product to increase the confidence in the results of the prototype testing. This decision was made because failure analysis indicated that the failure mode of the two failures is the same as the one that was observed in previous tests. In other words, it is expected that the shape of the distribution (beta) hasn't changed, but hopefully the scale (eta) has, indicating longer life. The 2-parameter Weibull distribution was used to model all prior tests results. The estimated beta ([math]\displaystyle{ \beta\,\! }[/math]) parameters of the prior test results are as follows:

Betas Obtained for Similar Mode
1.7
2.1
2.4
3.1
3.5

Solution

First, in order to fit the data to a Bayesian-Weibull model, a prior distribution for beta needs to be determined. Based on the beta values in the prior tests, the prior distribution for beta is found to be a lognormal distribution with [math]\displaystyle{ \mu = 0.9064\,\! }[/math], [math]\displaystyle{ \sigma = 0.3325\,\! }[/math]. (The values of the parameters can be obtained by entering the beta values into a Weibull++ standard folio and analyzing it using the lognormal distribution and the RRX analysis method.)

Next, enter the data from the prototype testing into a standard folio. On the control panel, choose the Bayesian-Weibull > B-W Lognormal Prior distribution. Click Calculate and enter the parameters of the lognormal distribution, as shown next.

Click OK. The result is Beta (Median) = 2.361219 and Eta (Median) = 5321.631912 (by default Weibull++ returns the median values of the posterior distribution). Suppose that the reliability at 3,000 hours is the metric of interest in this example. Using the QCP, the reliability is calculated to be 76.97% at 3,000 hours. The following picture depicts the posterior pdf plot of the reliability at 3,000, with the corresponding median value as well as the 10th percentile value. The 10th percentile constitutes the 90% lower 1-sided bound on the reliability at 3,000 hours, which is calculated to be 50.77%.

The pdf of the times-to-failure data can be plotted in Weibull++, as shown next:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Weibull Disribution Unreliability RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is:

Median Rank Plot Example

In this example, we will determine the median rank value used for plotting the 6th failure from a sample size of 10. This example will use Weibull++'s Quick Statistical Reference (QSR) tool to show how the points in the plot of the following example are calculated.

First, open the Quick Statistical Reference tool and select the Inverse F-Distribution Values option.

In this example, n1 = 10, j = 6, m = 2(10 - 6 + 1) = 10, and n2 = 2 x 6 = 12.

Thus, from the F-distribution rank equation:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{10-6+1}{6} \right){{F}_{0.5;10;12}}}\,\! }[/math]

Use the QSR to calculate the value of F_0.5;10;12 = 0.9886, as shown next:

Consequently:

[math]\displaystyle{ MR=\frac{1}{1+\left( \frac{5}{6} \right)\times 0.9886}=0.5483=54.83%\,\! }[/math]

Another method is to use the Median Ranks option directly, which yields MR(%) = 54.8305%, as shown next:

Complete Data Example

Assume that 10 identical units (N = 10) are being reliability tested at the same application and operation stress levels. 6 of these units fail during this test after operating the following numbers of hours, [math]\displaystyle{ {T}_{j}\,\! }[/math]: 150, 105, 83, 123, 64 and 46. The test is stopped at the 6th failure. Find the parameters of the Weibull pdf that represents these data.

Solution

Create a new Weibull++ standard folio that is configured for grouped times-to-failure data with suspensions.

Enter the data in the appropriate columns. Note that there are 4 suspensions, as only 6 of the 10 units were tested to failure (the next figure shows the data as entered). Use the 3-parameter Weibull and MLE for the calculations.

Plot the data.

Note that the original data points, on the curved line, were adjusted by subtracting 30.92 hours to yield a straight line as shown above.

Suspension Data Example

ACME company manufactures widgets, and it is currently engaged in reliability testing a new widget design. 19 units are being reliability tested, but due to the tremendous demand for widgets, units are removed from the test whenever the production cannot cover the demand. The test is terminated at the 67th day when the last widget is removed from the test. The following table contains the collected data.

Widget Test Data

Data Point Index	State (F/S)	Time to Failure
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Solution

In this example, we see that the number of failures is less than the number of suspensions. This is a very common situation, since reliability tests are often terminated before all units fail due to financial or time constraints. Furthermore, some suspensions will be recorded when a failure occurs that is not due to a legitimate failure mode, such as operator error. In cases such as this, a suspension is recorded, since the unit under test cannot be said to have had a legitimate failure.

Enter the data into a Weibull++ standard folio that is configured for times-to-failure data with suspensions. The folio will appear as shown next:

We will use the 2-parameter Weibull to solve this problem. The parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=1.145 \\ & \hat{\eta }=65.97 \\ \end{align}\,\! }[/math]

Using RRX:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.914\\ & \hat{\eta }=79.38 \\ \end{align}\,\! }[/math]

Using RRY:

[math]\displaystyle{ \begin{align} & \hat{\beta }=0.895\\ & \hat{\eta }=82.02 \\ \end{align}\,\! }[/math]

Interval Data Example

Suppose we have run an experiment with 8 units tested and the following is a table of their last inspection times and failure times:

Data Point Index	Last Inspection	Failure Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

Analyze the data using several different parameter estimation techniques and compare the results.

Solution

Enter the data into a Weibull++ standard folio that is configured for interval data. The data is entered as follows:

The computed parameters using maximum likelihood are:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.76 \\ & \hat{\eta }=44.68 \\ \end{align}\,\! }[/math]

Using RRX or rank regression on X:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.70 \\ & \hat{\eta }=44.54 \\ \end{align}\,\! }[/math]

Using RRY or rank regression on Y:

[math]\displaystyle{ \begin{align} & \hat{\beta }=5.41 \\ & \hat{\eta }=44.76 \\ \end{align}\,\! }[/math]

The plot of the MLE solution with the two-sided 90% confidence bounds is:

Mixed Data Types Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 406. [20].

Estimate the parameters for the 3-parameter Weibull, for a sample of 10 units that are all tested to failure. The recorded failure times are 200; 370; 500; 620; 730; 840; 950; 1,050; 1,160 and 1,400 hours.

Published Results:

Published results (using probability plotting):

[math]\displaystyle{ {\widehat{\beta}} = 3.0\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1,220\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -300\,\! }[/math]

Computed Results in Weibull++

Weibull++ computed parameters for rank regression on X are:

[math]\displaystyle{ {\widehat{\beta}} = 2.9013\,\! }[/math], [math]\displaystyle{ {\widehat{\eta}} = 1195.5009\,\! }[/math], [math]\displaystyle{ {\widehat{\gamma}} = -279.000\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ are due to the difference in the estimation method. In the publication the parameters were estimated using probability plotting (i.e., the fitted line was "eye-balled"). In Weibull++, the parameters were estimated using non-linear regression (a more accurate, mathematically fitted line). Note that γ in this example is negative. This means that the unadjusted for γ line is concave up, as shown next.

Weibull Distribution RRX Example

Assume that 6 identical units are being tested. The failure times are: 93, 34, 16, 120, 53 and 75 hours.

1. What is the unreliability of the units for a mission duration of 30 hours, starting the mission at age zero?

2. What is the reliability for a mission duration of 10 hours, starting the new mission at the age of T = 30 hours?

3. What is the longest mission that this product should undertake for a reliability of 90%?

Solution

1. First, we use Weibull++ to obtain the parameters using RRX.

Then, we investigate several methods of solution for this problem. The first, and more laborious, method is to extract the information directly from the plot. You may do this with either the screen plot in RS Draw or the printed copy of the plot. (When extracting information from the screen plot in RS Draw, note that the translated axis position of your mouse is always shown on the bottom right corner.)

Using this first method, enter either the screen plot or the printed plot with T = 30 hours, go up vertically to the straight line fitted to the data, then go horizontally to the ordinate, and read off the result. A good estimate of the unreliability is 23%. (Also, the reliability estimate is 1.0 - 0.23 = 0.77 or 77%.)

The second method involves the use of the Quick Calculation Pad (QCP).

Select the Prob. of Failure calculation option and enter 30 hours in the Mission End Time field.

Note that the results in QCP vary according to the parameter estimation method used. The above results are obtained using RRX.

2. The conditional reliability is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}\,\! }[/math]

or:

[math]\displaystyle{ \hat{R}(10hr|30hr)=\frac{\hat{R}(10+30)}{\hat{R}(30)}=\frac{\hat{R}(40)}{\hat{R}(30)}\,\! }[/math]

Again, the QCP can provide this result directly and more accurately than the plot.

3. To use the QCP to solve for the longest mission that this product should undertake for a reliability of 90%, choose Reliable Life and enter 0.9 for the required reliability. The result is 15.9933 hours.

Benchmark with Published Examples

The following examples compare published results to computed results obtained with Weibull++.

Complete Data RRY Example

From Dimitri Kececioglu, Reliability & Life Testing Handbook, Page 418 [20].

Sample of 10 units, all tested to failure. The failures were recorded at 16, 34, 53, 75, 93, 120, 150, 191, 240 and 339 hours.

Published Results

Published Results (using Rank Regression on Y):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.20 \\ & \widehat{\eta} = 146.2 \\ & \hat{\rho }=0.998703\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard data sheet. Use RRY for the estimation method.

Weibull++ computed parameters for RRY are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.1973 \\ & \widehat{\eta} = 146.2545 \\ & \hat{\rho }=0.9999\\ \end{align}\,\! }[/math]

The small difference between the published results and the ones obtained from Weibull++ is due to the difference in the median rank values between the two (in the publication, median ranks are obtained from tables to 3 decimal places, whereas in Weibull++ they are calculated and carried out up to the 15th decimal point).

You will also notice that in the examples that follow, a small difference may exist between the published results and the ones obtained from Weibull++. This can be attributed to the difference between the computer numerical precision employed by Weibull++ and the lower number of significant digits used by the original authors. In most of these publications, no information was given as to the numerical precision used.

Suspension Data MLE Example

From Wayne Nelson, Fan Example, Applied Life Data Analysis, page 317 [30].

70 diesel engine fans accumulated 344,440 hours in service and 12 of them failed. A table of their life data is shown next (+ denotes non-failed units or suspensions, using Dr. Nelson's nomenclature). Evaluate the parameters with their two-sided 95% confidence bounds, using MLE for the 2-parameter Weibull distribution.

Published Results:

Weibull parameters (2P-Weibull, MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,296 \\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Note that Nelson expresses the results as multiples of 1,000 (or = 26.297, etc.). The published results were adjusted by this factor to correlate with Weibull++ results.

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio, using 2-parameter Weibull and MLE to calculate the parameter estimates.

You can also enter the data as given in table without grouping them by opening a data sheet configured for suspension data. Then click the Group Data icon and chose Group exactly identical values.

The data will be automatically grouped and put into a new grouped data sheet.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.0584 \\ & \widehat{\eta} = 26,297 \\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 0.6441, \text{ }1.7394\rbrace \\ & \widehat{\eta} = \lbrace 10,522, \text{ }65,532\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

The two-sided 95% bounds on the parameters can be determined from the QCP. Calculate and then click Report to see the results.

Interval Data MLE Example

From Wayne Nelson, Applied Life Data Analysis, Page 415 [30]. 167 identical parts were inspected for cracks. The following is a table of their last inspection times and times-to-failure:

Published Results:

Published results (using MLE):

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.486 \\ & \widehat{\eta} = 71.687\\ \end{align}\,\! }[/math]

Published 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.962, \text{ }82.938\rbrace \\ \end{align}\,\! }[/math]

Published variance/covariance matrix:

Computed Results in Weibull++

This same data set can be entered into a Weibull++ standard folio that's configured for grouped times-to-failure data with suspensions and interval data.

Weibull++ computed parameters for maximum likelihood are:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=1.485 \\ & \widehat{\eta} = 71.690\\ \end{align}\,\! }[/math]

Weibull++ computed 95% FM confidence limits on the parameters:

[math]\displaystyle{ \begin{align} & \widehat{\beta }=\lbrace 1.224, \text{ }1.802\rbrace \\ & \widehat{\eta} = \lbrace 61.961, \text{ }82.947\rbrace \\ \end{align}\,\! }[/math]

Weibull++ computed/variance covariance matrix:

Grouped Suspension MLE Example

From Dallas R. Wingo, IEEE Transactions on Reliability Vol. R-22, No 2, June 1973, Pages 96-100.

Wingo uses the following times-to-failure: 37, 55, 64, 72, 74, 87, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 102, 102, 105, 105, 107, 113, 117, 120, 120, 120, 122, 124, 126, 130, 135, 138, 182. In addition, the following suspensions are used: 4 at 70, 5 at 80, 4 at 99, 3 at 121 and 1 at 150.

Published Results (using MLE)

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Computed Results in Weibull++

[math]\displaystyle{ \begin{align} & \widehat{\beta }=3.7596935\\ & \widehat{\eta} = 106.49758 \\ & \hat{\gamma }=14.451684\\ \end{align}\,\! }[/math]

Note that you must select the Use True 3-P MLEoption in the Weibull++ Application Setup to replicate these results.

3-P Probability Plot Example

Suppose we want to model a left censored, right censored, interval, and complete data set, consisting of 274 units under test of which 185 units fail. The following table contains the data.

The Test Data

Data Point Index	Number in State	Last Inspection	State (S or F)	State End Time
1	2	5	F	5
2	23	5	S	5
3	28	0	F	7
4	4	10	F	10
5	7	15	F	15
6	8	20	F	20
7	29	20	S	20
8	32	0	F	22
9	6	25	F	25
10	4	27	F	30
11	8	30	F	35
12	5	30	F	40
13	9	27	F	45
14	7	25	F	50
15	5	20	F	55
16	3	15	F	60
17	6	10	F	65
18	3	5	F	70
19	37	100	S	100
20	48	0	F	102

Solution

Since standard ranking methods for dealing with these different data types are inadequate, we will want to use the ReliaSoft ranking method. This option is the default in Weibull++ when dealing with interval data. The filled-out standard folio is shown next:

The computed parameters using MLE are:

[math]\displaystyle{ \hat{\beta }=0.748;\text{ }\hat{\eta }=44.38\,\! }[/math]

Using RRX:

[math]\displaystyle{ \hat{\beta }=1.057;\text{ }\hat{\eta }=36.29\,\! }[/math]

Using RRY:

[math]\displaystyle{ \hat{\beta }=0.998;\text{ }\hat{\eta }=37.16\,\! }[/math]

The plot with the two-sided 90% confidence bounds for the rank regression on X solution is: