OPS.8.1:5.2 - A free place and a full receiving resource

A manufacturing resource S is available continuously from 0 to 5. Job A is already running there from 0 to 2. A part X undergoing inspection elsewhere may need one additional two-hour visit to S; the answer becomes available at hour one, and any such visit must be complete by five.

A new job Y would require two uninterrupted hours at S and is also wanted by five. Its upstream activity has space and emits a release signal. There are no other operations needed for these completions.

Protecting the stated return and promising Y requires 2 + 2 + 2 = 6 hours in a five-hour window. The free upstream place does not remove this obstruction.

At hour one, one hour of A remains. If X is accepted, the remaining requirement is one hour of A plus two of Y; A finishes at two and Y runs 2–4. If X returns, its visit can run 2–4, but Y cannot also finish by five. Replacing the two-hour return allowance with X’s known visit preserves this result; adding both would falsely claim eight original hours of demand.

This example uses full occupied hours and a finite deadline. A workload norm based on a different weighted quantity would need its own interpretation. If Y instead used an independent resource, holding it because of S’s protected hours would require another reason.