Repairable systems like production equipment more so than parts fail and are then repaired and restored to service only to fail again. Failures occurring in repairable items are examples of discrete events happening randomly over time and are called stochastic point processes. When you analyze data like this it is important to determine if there is a trend, meaning failures come slower (improvement) or they come faster (deterioration).
The Laplace test can be used for this purpose.
Where T(n) equals total operating time or, X(i) is the age at the ith successive failure and n is the number of failures.
In case the period analyzed ends at an event replace n with n-1 and exclude the time to the last event from the summation. This is the formula used since you calculate every time at the completion of a work order.
An example will likely give a better picture. Assume you have the same failure times as before in our first Weibull example in the following sequence 16, 34, 53, 75, 93, and 120. After each failure, the equipment is restored and brought back in service. If x(i) represents the time between failures and X(i) represents the cumulative time to failure for the successive events, then:
X(1) = 16
X(2) = 16+34 = 50
X(3) = 16+34+53 = 103
X(4) = 16+34+53+75 = 178
X(5) = 16+34+53+75+93 = 271
X(6) = 16+34+53+75+93+120 = 391
T(n) = 16+34+53+75+93+120 = 391
With these values U = -1.424386203
Now sort these same failure occurrences descending. The sequence then becomes 120, 93, 75, 53, 34, and 16. Similarly:
X(1) = 120
X(2) = 120+93 = 213
X(3) = 120+93+75 = 288
X(4) = 120+93+75+53 = 341
X(5) = 120+93+75+53+34 = 375
X(6) = 120+93+75+53+34+16 = 391
T(n) = 120+93+75+53+34+16 = 391
With these values U = +1.424386203
If U > 0 the time between failures is becoming shorter and the failure rate is increasing.
If U < 0 the time between failures is becoming longer and the failure rate is decreasing.
If U = 0 the time between failures is random.
Please note that the Weibull distribution would have generated the same values for β (= 1.42696711) and η (= 76.3454156), because the Weibull distribution requires us to sort the failures by failure time in order to determine the distribution.