Five dermatology slots: a scheduling judgement call
A district hospital's dermatology clinic runs five 20-minute slots starting at 08:00 - 08:00, 08:20, 08:40, 09:00 and 09:20. Five patients, Asha, Bikash, Chandani, Dipesh and Elina, are booked one per slot. The receptionist's rules are: (i) Chandani must take the third slot because she has a bus to her village at 09:30; (ii) Asha must be seen immediately before Elina because they share a vehicle; (iii) Dipesh cannot take the first or the last slot, since his wound dressing needs the room when the steriliser has finished cycling; (iv) Bikash must be seen after Dipesh, because he cannot sit in the waiting room for long. At 08:40 the doctor is running 15 minutes late, and a walk-in patient with an itchy rash arrives asking to be fitted in.
1. Deduce the unique order of the five patients, showing each elimination step.
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Fixed point. Rule (i) puts Chandani in slot 3. Slots 1, 2, 4 and 5 remain for Asha, Bikash, Dipesh, Elina.
Where can Dipesh go? Rule (iii) removes slots 1 and 5, and slot 3 is taken, so Dipesh is in slot 2 or slot 4.
Test Dipesh = 2. Rule (iv) requires Bikash after slot 2, so Bikash is in slot 4 or 5. Asha and Elina need consecutive slots among those left (1, 4, 5) - only (4, 5) is consecutive, so Asha/Elina occupy 4 and 5. That leaves slot 1 for Bikash, contradicting rule (iv). Dipesh ≠ 2.
Therefore Dipesh = 4. Bikash must follow, so with slot 5 the only remaining place, Bikash = 5. Slots 1 and 2 are left for Asha and Elina, and rule (ii) forces Asha = 1, Elina = 2.
Final schedule: 08:00 Asha, 08:20 Elina, 08:40 Chandani, 09:00 Dipesh, 09:20 Bikash. The key move is testing the impossible branch first; a single contradiction collapses half the possibilities.
2. A colleague claims rule (iii) is unnecessary because rule (iv) already settles Dipesh's position. Evaluate this claim.
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Remove rule (iii) and test. With Chandani fixed in slot 3 and Asha immediately before Elina, the pair can occupy (1, 2) or (4, 5).
- Asha 1, Elina 2: slots 4 and 5 remain for Bikash and Dipesh. Rule (iv) forces Dipesh 4, Bikash 5. Valid.
- Asha 4, Elina 5: slots 1 and 2 remain. Rule (iv) forces Dipesh 1, Bikash 2. Also consistent with rules (i), (ii) and (iv).
Two schedules satisfy every remaining rule, so the claim is false: without rule (iii) the arrangement is not unique. Dipesh can indeed be pushed into slot 1.
Why the error arises. Rule (iv) constrains only the relative order of Bikash and Dipesh; it says nothing about absolute position. Only rule (iii) - an absolute constraint tied to the steriliser cycle - pins Dipesh away from the first slot. The lesson is that ordering constraints and positional constraints are logically different and cannot substitute for one another; a plausible-sounding inference is not a proof of uniqueness.
3. The clinic is 15 minutes behind from Chandani's slot onward. State when Bikash is actually seen, and assess Chandani's bus.
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Revised timings. Chandani's slot starts on time at 08:40 but finishes at 09:15 instead of 09:00, so every later slot shifts by 15 minutes: Dipesh 09:15–09:35, Bikash 09:35–09:55.
Chandani's bus. She needs to be out of the room by about 09:20 to reach a 09:30 departure. She is free at 09:15, leaving roughly five minutes - survivable but with no margin; if the overrun stretches to 25 minutes she finishes at 09:25 and misses the bus. The correct response is to tell her at once and offer the next day's first slot, not to let her discover it in the queue.
The walk-in. The absolute constraints still hold - Bikash is still after Dipesh. The judgement is queue discipline: an itchy rash is rarely an emergency, so the walk-in should be placed after 09:55 unless red flags (airway swelling, fever with a spreading rash, mucosal involvement) are present. Squeezing the walk-in between booked patients would cascade delays onto people who have already rearranged travel, and would punish the patients who followed the rules.