Calculating the Parameters of the Weibull Distribution

Parameter Estimation

The estimates of the parameters of the Weibull distribution can be found graphically on probability plotting paper or analytically using either least squares or maximum likelihood. Parameter estimation methods are presented in detail in Appendix B: Parameter Estimations.

Probability Plotting

One method of calculating the parameters of the Weibull distribution is by using probability plotting. To better illustrate this procedure, consider the following example [19].

Example 3

Let's assume six identical units are being reliability tested at the same application and operation stress levels. All of these units fail during the test after operating the following times (in hours), Ti: 93, 34, 16, 120, 53 and 75.

The steps for determining the parameters of the Weibull pdf representing the data, using probability plotting, are as follows:

Rank the times-to-failure in ascending order as shown next.
Obtain their median rank plotting positions. The times-to-failure, with their corresponding median ranks, are shown next.
On a Weibull probability paper, plot the times and their corresponding ranks. Figure 8 displays an example of a Weibull probability paper (the solution is given in Figure 9).

Fig. 8: Sample Weibull probability plotting paper

Draw the best possible straight line through the plotted points (as shown in Figure 9).
Obtain the slope of this line by drawing a line, parallel to the one just obtained, through the slope indicator. This value is the estimate of the shape parameter . In this case = 1.4.
At the Q(t) = 63.2% ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of . For this case = 76 hr. (This is always at 63.2% since Q(t) = = 1 - e-1 = 0.632 = 63.2%).

Fig. 9: Probability Plot for Example 3

Now any reliability value for any mission time t can be obtained. For example the reliability for a mission of 15 hr, or any other time, can now be obtained either from the plot or analytically (i.e. using the equations given in the Weibull Distribution section).

To obtain the value from the plot, draw a vertical line from the abscissa, at t = 15 hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read Q(t), in this case Q(t = 15) = 9.8%. Thus, R(t = 15) = 1 - Q(t) = 90.2%. This can also be obtained analytically, from the Weibull reliability function, since both of the parameters are known or:

MLE Parameter Estimation

The parameters of the 2-parameter Weibull distribution can also be estimated using maximum likelihood estimation (MLE). This log-likelihood function is:

where:

and:

Fe is the number of groups of times-to-failure data points.
Ni is the number of times-to-failure in the ith time-to-failure data group.
β is the Weibull shape parameter (unknown a priori, the first of two parameters to be found).
η is the Weibull scale parameter (unknown a priori, the second of two parameters to be found).
Ti is the time of the ith group of time-to-failure data.
S is the number of groups of suspension data points.
is the number of suspensions in ith group of suspension data points.
is the time of the ith suspension data group.
FI is the number of interval data groups.
is the number of intervals in the ith group of data intervals.
is the beginning of the ith interval.
is the ending of the ith interval.

The solution will be found by solving for a pair of parameters so that = 0 and = 0. (Other methods can also be used, such as direct maximization of the likelihood function, without having to compute the derivatives.)

(10)

(11)

Example 4

Using the same data as in the probability plotting example (Example 3) and assuming a 2-parameter Weibull distribution, estimate the parameter using the MLE method.

Solution

In this case we have non-grouped data with no suspensions, thus Eqns. (10) and (11) become:

and:

Solving the above equations simultaneously we get: