Weibull++ Standard Folio Data 2P-Exponential

Reliability Web Notes
Standard Folio Data 2P-Exponential
Weibull++
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.
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