OPS.10.2:5.3 - Shared attendance still permits machine overlap
Five orders arrive at zero. Each needs A for two hours and then B for three, with order 1-5 on each station. Each station has one continuously available machine; there are no setups, returns or other work. B completion makes an order ready for its customer. No operation is interrupted. Admit the first two at zero and the next order immediately when a B completion frees one of the two places.
With independent station resources, earliest A intervals are 0-2, 2-4, 5-7, 8-10 and 11-13. B finishes at 5, 8, 11, 14 and 17.
Now both stations use one operator. A requires continuous attendance; B requires attendance only in its first hour and holds its machine for all three hours. The old A intervals conflict with B attendance. Recovering the phases yields this revised plan:
| Order | Admission | A with operator | B with operator | B unattended | Completion |
|---|---|---|---|---|---|
| 1 | 0 | 0-2 | 2-3 | 3-5 | 5 |
| 2 | 0 | 3-5 | 5-6 | 6-8 | 8 |
| 3 | 5 | 6-8 | 8-9 | 9-11 | 11 |
| 4 | 8 | 9-11 | 11-12 | 12-14 | 14 |
| 5 | 11 | 12-14 | 14-15 | 15-17 | 17 |
At hour two, start B1 before A2; then place A2 inside B1’s unattended phase. Repeating that choice keeps B working continuously. The operator’s intervals do not overlap, and no more than two orders are admitted and unfinished.
B cannot begin before hour two and must process five orders sequentially for three hours each. Every completion therefore satisfies f_Bi >= 2+3i; the plan attains all five bounds. The shared operator changes starts at A without delaying any recipient result.
Customer times average eleven hours. Internal times are 5, 8, 6, 6 and 6 hours, averaging 6.2; the remaining average 4.8 hours occurs before admission. These quantities retain their different event boundaries.