Two Theatres, One Surgeon's Window, One Emergency
The district hospital runs two operating theatres, OT-1 and OT-2, on a single 08:00–16:00 list (480 minutes each). Each theatre must take one uninterrupted 30-minute lunch break starting between 12:30 and 14:00, and a 25-minute changeover is required between consecutive cases in the same theatre, whether the next case is elective or emergency. Six elective cases are booked: A = 90 min, B = 60 min, C = 120 min, D = 75 min, E = 45 min, F = 105 min. Dr Rao is the only surgeon for cases A and C and is available only between 08:00 and 12:00. Any case may be done in either theatre, but A and C must both be completed by Dr Rao inside his window. At 11:00 an emergency laparotomy needing 70 minutes arrives and must be done the same day. The theatre manager must publish a schedule that keeps every elective case and also fits the emergency, and must be able to show that no theatre is double-booked at any moment.
1. Calculate the total theatre time the day requires, including changeovers and lunch, and decide whether the day is feasible in aggregate. Then explain why aggregate feasibility alone is not enough to publish the list.
Read the worked answer
Case minutes: 90 + 60 + 120 + 75 + 45 + 105 = 495 elective, plus 70 emergency = 565 minutes of operating.
Changeovers: 7 cases across 2 theatres means a split of 4 and 3, giving 3 + 2 = 5 changeovers = 125 minutes. This is fixed: any split of 7 cases into two non-empty lists gives 5 changeovers.
Lunch: 2 × 30 = 60 minutes.
Total requirement = 565 + 125 + 60 = 750 minutes against 480 × 2 = 960 minutes available. Slack = 210 minutes, so the day is feasible in aggregate.
Why that is not enough: the aggregate hides the binding constraint. Dr Rao has 240 minutes (08:00–12:00) and must personally do A + C = 210 minutes, with at least one 25-minute changeover between them: 235 minutes. That leaves 5 minutes of slack. So the whole day turns on sequencing A and C back-to-back at 08:00 in the same theatre, with nothing inserted between them. A manager who publishes the list from the aggregate figure alone will produce a schedule that fails clinically even though the arithmetic adds up.
2. Produce a timetable that satisfies every constraint and places the emergency as early as possible. Show the arithmetic for each theatre.
Read the worked answer
OT-1 – Dr Rao list first
- A: 08:00–09:30
- Changeover: 09:30–09:55
- C: 09:55–11:55 (Rao uses 235 of his 240 minutes)
- Lunch: 12:30–13:00 (inside the permitted window)
- E: 13:00–13:45
OT-2 – general list with emergency inserted at 11:00
- B: 08:00–09:00
- Changeover: 09:00–09:25
- D: 09:25–10:40
- Changeover: 10:40–11:05
- Emergency: 11:05–12:15 (only a 5-minute wait, because OT-2 is already clean and staffed)
- Lunch: 12:15–12:45
- Changeover: 12:45–13:10
- F: 13:10–14:55
Checks: OT-1 finishes at 13:45, OT-2 at 14:55, both before 16:00. Rao's two cases sit inside his window with 5 minutes to spare. No theatre is double-booked, and no case runs across a lunch break.
The judgement point: running F at 11:05 and pushing the emergency to 13:45 is arithmetically identical - same total minutes, same finish times. It is clinically indefensible: it would delay a laparotomy by about 2 hours 40 minutes to protect a routine case. When two schedules are equally valid on minutes, the tie-break is always patient risk, not list order.
3. On the day, OT-1 is unusable because of a steriliser failure. Can the whole list be delivered from OT-2 alone? If not, quantify the shortfall and state which cases you would move.
Read the worked answer
Available in one theatre: 480 − 30 (lunch) = 450 minutes of usable time.
Required for all 7 cases in one theatre: 565 minutes of operating + 6 changeovers × 25 = 150 + 30 lunch = 745 minutes. Shortfall = 745 − 450 = 295 minutes. The list cannot be delivered; at least 295 minutes of case time plus its associated changeover must leave the day.
Defensible plan: move F (105 min) and D (75 min) to the next list = 180 minutes of operating, plus the 2 changeovers they generate = 230 minutes removed, leaving 515 minutes required - still 65 over. So a third case must move; E (45 min) with its changeover (25) removes 70, leaving 445 ≤ 450.
Feasible single-theatre schedule: A 08:00–09:30, changeover to 09:55, C 09:55–11:55, changeover to 12:20, B 12:20–13:20, lunch 13:20–13:50, E 13:50–14:35.
Why these cases and not others: A and C are locked to Dr Rao's 4-hour window and travel together. B is short. F and D are the longest and most movable if the patients are stable and consented for a short postponement. If F is a cancer case, the choice must be made by the surgeon on clinical grounds, and the postponement must be recorded and communicated to the patient the same morning - not left for the patient to discover on arrival.