The Exponential Distribution

=The 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 $$\beta =1$$. 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).

Example 8
Twenty units were reliability tested with the following results:

8-1. Assuming a two-parameter exponential distribution, estimate the parameters analytically using the MLE method.

8-2. Repeat part 8-1 using Weibull++ (enter the data as grouped data to duplicate the results of 8-1).

8-3. Plot the exponential probability vs. time-to-failure using Weibull++.

8-4. Plot $$R(t)$$ vs. time using Weibull++.

8-5. Plot the $$pdf$$ using Weibull++.

8-6. Plot the failure rate vs. time using Weibull++.

8-7. Estimate the parameters analytically using the RRY method (using grouped ranks).

Solution To Example 8
8-1. For the two-parameter exponential distribution and for $$\hat{\gamma }=100$$ hours (first failure), the partial of the log-likelihood function, $$\Lambda $$, becomes:


 * $$ \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} $$

8-2. The data as entered in Weibull++ along with results are shown next.





Select Reliability vs. Time.



The exponential $$pdf$$ plot is shown next.



The exponential failure rate plot is shown next.



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 plot does not exist for times before the location parameter, $$\gamma $$, at 100 hours.

8-7. In the case of grouped data, one must be cautious when estimating the parameters using a rank regression method. That is because the median rank values are determined from the total number of failures observed by time $${{T}_{i}}$$ where $$i$$ indicates the group number. In this example the total number of groups is $$N=6$$ and the total number of units is $${{N}_{T}}=20$$. Thus, the median rank values will be estimated for twenty units and for the total failed units ($${{N}_}$$) up to the $${{i}^{th}}$$ group, for the $${{i}^{th}}$$ rank value. The median ranks values can be found from rank tables or they can be estimated using ReliaSoft's Quick Statistical Reference.

For example, the median rank value of the fourth group will be the $${{17}^{th}}$$ rank out of a sample size of twenty units (or 81.945%).

The following table is then constructed (as in Example 2).

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

Given the values in the table above, calculate $$\hat{a}$$ and $$\hat{b}$$ using Eqns. (aae) and (bbe):


 * $$\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}$$


 * or:


 * $$\hat{b}=-0.005392$$


 * and:


 * $$\hat{a}=\overline{y}-\hat{b}\overline{T}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{T}_{i}}}{N}$$


 * or:


 * $$\hat{a}=\frac{-9.6476}{6}-(-0.005392)\frac{2100}{6}=0.2793$$


 * Therefore, from Eqn. (be):


 * $$\hat{\lambda }=-\hat{b}=-(-0.005392)=0.05392\text{ failures/hour}$$


 * and from Eqn. (ae):


 * $$\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2793}{0.005392}$$


 * or:


 * $$\hat{\gamma }\simeq 51.8\text{ hours}$$


 * Then:


 * $$f(T)=(0.005392){{e}^{-0.005392(T-51.8)}}$$

Using Weibull++, the estimated parameters are:


 * $$\begin{align}

\hat{\lambda }= & 0.0054\text{ failures/hour} \\ \hat{\gamma }= & 51.82\text{ hours} \end{align}$$

The small difference in the values from Weibull++ is due to rounding. In Weibull++ the calculations and the rank values are carried out up to the $${{15}^{th}}$$ decimal point.

Example 9
A number of leukemia patients were treated with either drug 6 MP or a placebo, and the times in weeks until cancer symptoms returned were recorded. Analyze each treatment separately. [21, p.175]

Solution to Example 9
Enter the data into Weibull++, by selecting Times to Failure, with Right Censored Data (Suspensions) and 'with Grouped Observations. 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. In the second column enter F for completed tests and S for uncompleted. In the third column enter the time. In the fourth column (Subset ID) enter the name of the treatment. The title of each column can be changed by double-clicking it and typing the desired name.



Now click the Batch Auto Run icon




 * or:



and click Select All Available> > to separate the 6 MP drug from the placebo as shown next.



Click OK and you will get a new Data Sheet for each treatment with the corresponding results, as shown next.

Make sure both data sheets are calculated, then from the Life Data tab, click on Insert Additional Plot. Click on the Select Data Sheets button.





and check the two data sheets under the Folio1 project.

The plot is shown next,