Scheduling Two Machines Around a Shared Operator
A small print shop runs two jobs on two machines that share one trained operator. Job A must run on Machine 1 and Job B on Machine 2. Machine 1 needs 45 minutes of setup, then runs 3.5 hours of production, but the operator must be present for setup and for the final 30 minutes of production only. Machine 2 needs 20 minutes of setup and 2.25 hours of production, and the operator must be present the entire time. The shop opens at 8:00 a.m. and the operator has a mandatory 30-minute lunch that must start no earlier than 11:30 a.m. and must end by 1:00 p.m. Both jobs must finish the same day. The operator cannot be in two places at once.
1. Can both jobs be completed before the operator's lunch if lunch starts at 11:30 a.m.? Show the timeline and the constraint that decides it.
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Machine 1 operator time: 45 min setup + 30 min final production = 75 min. Machine 2 operator time: 20 min setup + 135 min production = 155 min. Total operator time = 230 min = 3 h 50 min. From 8:00 a.m. to 11:30 a.m. is 3 h 30 min = 210 min. Since 230 > 210, no - both cannot finish before 11:30. Even ignoring sequencing conflicts, the operator's total required minutes exceed the window by 20 minutes.
2. Find a sequence that finishes both jobs as early as possible, and state the completion time of the later job.
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Run Machine 2 first because it is operator-bound the whole time. 8:00–8:20 setup, 8:20–10:35 production (2 h 15 min). Operator is free at 10:35. Now run Machine 1: 10:35–11:20 setup (45 min). Production runs 11:20–2:50 (3.5 h), and the operator is only needed for the last 30 min, 2:20–2:50. Lunch must be 30 min ending by 1:00, so place lunch 12:30–1:00. The operator is free during that block because Machine 1 production needs no attention then. Later job (Machine 1) finishes at 2:50 p.m. Check: lunch ends by 1:00 ✓; operator not double-booked ✓.
3. A rush order adds 40 minutes of operator-attended work to Machine 2, all of it during production. Recompute whether the 2:50 p.m. finish still holds, and if not, give the new finish.
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Machine 2 operator time becomes 20 + 135 + 40 = 195 min, so Machine 2 now ends at 8:00 + 195 min = 11:15 a.m. Machine 1 setup then runs 11:15–12:00, production 12:00–3:30, with operator needed 3:00–3:30. Lunch fits 12:30–1:00 (operator idle on Machine 1 then). New later finish = 3:30 p.m., a 40-minute slip. The added operator minutes push everything back one-for-one because the operator is the bottleneck.