A Five-Day Induction That Will Not Reshuffle
A five-day induction week runs Monday to Friday, one 90-minute session per day, one module per day. The modules are Compliance (C), Data protection (D), Ethics (E), Fire-safety (F) and Grievance handling (G).
The coordinator works to five constraints:
- Grievance handling is on Tuesday.
- Compliance is exactly two days before Data protection.
- Fire-safety is on the day immediately before Ethics.
- Data protection is not on Friday.
- No two modules share a day.
The Fire-safety trainer has now emailed that he can come only on the Monday or the Friday of that week. The coordinator says this is easy to accommodate because "the rest of the plan can simply be reshuffled". The HR head is not convinced and wants a definite answer before she confirms his slot, because moving him later costs Rs 12,000 in trainer fees and pushes the induction back a week.
1. Derive the day-wise schedule from the five constraints, showing why the arrangement is the only one possible.
Read the worked answer
Step 1 - fix G. Constraint 1 puts G on Tuesday (day 2). Days 1, 3, 4 and 5 remain for C, D, F, E.
Step 2 - where can the C-D pair go? C is exactly two days before D, so the legal pairs are (1,3), (2,4) and (3,5).
- (2,4) is blocked: day 2 is G.
- (3,5) would leave days 1 and 4 for F and E, which are not consecutive - breaks constraint 3.
- So C = Monday, D = Wednesday.
Step 3 - place F and E. Days 4 and 5 remain; F must be immediately before E, so F = Thursday, E = Friday.
Result: Mon C, Tue G, Wed D, Thu F, Fri E.
Check constraint 4: D is Wednesday, not Friday - satisfied (and in fact redundant once the rest is fixed). No other arrangement survives, so the schedule is unique. Note the practical point: constraint 2 and constraint 3 are both rigid relations, and together they leave almost no slack - which is why the coordinator's "simply reshuffle" instinct is unsafe.
2. Can the Fire-safety trainer's Monday-or-Friday restriction be met? If not, how many of the five constraints must be given up, and which?
Read the worked answer
Friday is impossible outright. F on Friday forces E on Saturday, outside the working week - constraint 3 cannot be satisfied.
Monday is impossible under the current set. F on Monday forces E on Tuesday, but Tuesday is G. Move G, and the pair C-D (exactly two days apart) can only be (3,5), i.e. C Wednesday, D Friday - which breaches constraint 4. Try G on Monday: days 2-5 remain for C, D, F, E, and the C-D pair is either (2,4) or (3,5); in both cases the leftover two days are non-consecutive, so F and E cannot sit next to each other. G on Wednesday or Friday fails for the same reason.
Conclusion: no schedule exists with F on Monday or Friday. To place F on Monday you must give up both constraint 1 (G moves to Thursday) and constraint 4 (D ends up on Friday): F Mon, E Tue, C Wed, G Thu, D Fri.
The coordinator's assumption is wrong, and the HR head should confirm the Thursday slot now rather than pay Rs 12,000 later.
3. A venue problem forces Grievance handling onto Thursday instead. Show whether any schedule exists, and name one single constraint that, if deleted entirely, would make Thursday work.
Read the worked answer
Test G = Thursday (day 4). Days 1, 2, 3 and 5 remain for C, D, F, E. Legal C-D pairs from those days: (1,3) and (3,5) - (2,4) is blocked by G.
- C = 1, D = 3: days 2 and 5 remain for F and E, which are not consecutive. Fails constraint 3.
- C = 3, D = 5: D lands on Friday. Fails constraint 4.
So no schedule exists with G on Thursday. G is effectively pinned to Tuesday by the interaction of constraints 2, 3 and 4 - a rigidity the coordinator has not noticed.
Single deletion that rescues it: delete constraint 4 (D not on Friday). Then C Wed, D Fri, F Mon, E Tue, G Thu is valid: F immediately precedes E, C is exactly two days before D, G is Thursday. Deleting constraint 3 also works if "immediately before" is dropped to "sometime before": with G on Thursday, C Mon, D Wed, F Tue, E Fri satisfies everything else.
The examinable lesson: before proposing a "small change", test whether the constraint set is over-determined. A set with two rigid relational constraints often has exactly one solution, and then any local change cascades.