Statistics for Maintenance Modelling: Repairable Systems

# Exercise 4¤

Use the data set from screen 16 of successive failures of a piece of software developed as part of a large data system:
9, 12, 11, 4, 7, 2, 5, 8, 5, 7, 1, 6, 1, 9, 4, 1, 2, 2, 6, 1, 11, 33, 7, 91, 2, 1, 87, 47, 12, 9, 135, 258, 16, 35. Compute Ln L by substituting 1(t) = and 2(t) = into the log likelihood as discussed on screens 13 and 15 to give:

resp. Which model gives a better result? Select the one you think, and then click the "Check Answer" button when you've finished.

ln L = n 0 + 1 ti -
ln L = n(ln + ln ) + ( - 1) ln ti -

Yes, that's right, 2(t) is better choice. For 1(t) the log likelihood is -728, and for 2(t) the log likelihood is -163.34. Since 2(t) creates a ln L with a higher value, this choice is better.

No, in fact 2(t) is better choice. For 1(t) the log likelihood is -728, and for 2(t) the log likelihood is -163.34. Since 2(t) creates a ln L with a higher value, this choice is better.