OPS.15.1:5.1 - A window cuts through unfinished work
Observe an operation from hour zero to hour eight. Case A arrived at minus three and leaves at two. B arrives at one and leaves at four. C arrives at six and is still present at eight.
| Case | Full known residence | Residence inside [0,8] |
|---|---|---|
| A | 5 hours | 2 hours |
| B | 3 hours | 3 hours |
| C | At least 2 hours; unfinished | 2 hours |
The population-time area is seven case-hours, so mean population is 7/8. Counting departures gives two in eight hours. Their full residence mean is four hours; multiplying 2/8 by four gives one, not 7/8.
There is no contradiction. The departure cohort includes three hours of A before the observation window and excludes C’s two observed hours. Computing the clipped intervals answers the window question. C’s eventual completion is unnecessary for that result.
Now the question changes to the mean full residence of all three cases. C’s endpoint matters. If no later information is available, the sum is at least ten hours and the mean at least 10/3, with no supported finite upper bound. The earlier mean population does not fill that missing endpoint.