Scheduling five pending inspections under fixed conditions
A branch has to complete five pending inspections - files A, B, C, D and E - one on each working day from Monday to Friday, one file per day. The supervisor's note records four conditions: (i) file C is taken up exactly two days after file A (if A is on Monday, C is on Wednesday); (ii) file E is taken up on Thursday; (iii) file B is taken up immediately before file D; (iv) no file is repeated and no working day is left vacant. A newly posted assistant claims the note is over-specified, arguing that condition (ii) contributes nothing because the leftover day would take E anyway. Before accepting that argument, the supervisor wants the roster verified. Separately, a clerk reports that the officer holding file E will be on leave on Thursday, so E would have to move to Saturday, which is not a working day in this office.
1. Work out the day on which each file is taken up, showing how the conditions interact.
Read the worked answer
Start from the most rigid condition.
Step 1 - E is fixed on Thursday.
Step 2 - B immediately before D. The available consecutive pairs are Monday–Tuesday and Tuesday–Wednesday (Wednesday–Thursday and Thursday–Friday are blocked because Thursday is taken by E).
Step 3 - test Tuesday–Wednesday for B–D. That leaves Monday and Friday for A and C. But C must be exactly two days after A: A on Monday would put C on Wednesday, which is taken; A on Friday has no day two days later within the week. So this branch fails.
Step 4 - B on Monday, D on Tuesday. The free days are Wednesday and Friday, and A on Wednesday puts C on Friday, satisfying the two-day gap.
Final roster: Monday – B; Tuesday – D; Wednesday – A; Thursday – E; Friday – C.
The key is that condition (iii) plus the fixed Thursday collapses the possibilities to one; the two-day gap then has only one place to sit.
2. Test the assistant's claim that condition (ii) is redundant.
Read the worked answer
Drop condition (ii) and enumerate every placement of the consecutive block B–D in Monday–Friday:
| B–D | Days left for A, C, E | Outcome |
|---|---|---|
| Mon–Tue | Wed, Thu, Fri | A Wed, C Fri, E Thu - valid |
| Tue–Wed | Mon, Thu, Fri | A Mon needs C on Wed (taken); A Thu needs C on Sat (outside week) - invalid |
| Wed–Thu | Mon, Tue, Fri | A Mon needs C on Wed (taken); A Tue needs C on Thu (taken) - invalid |
| Thu–Fri | Mon, Tue, Wed | A Mon, C Wed, E Tue - valid |
Two schedules survive, not one. So condition (ii) does real work: it eliminates the second schedule (A Mon, E Tue, C Wed, B Thu, D Fri). The assistant has confused 'E ends up on Thursday in the answer' with 'E is on Thursday as a premise'. A clue is redundant only if removing it leaves the solution unchanged, which is not the case here.
3. E must move to Saturday. What are the consequences, and what should the supervisor do?
Read the worked answer
Re-solve with E removed from Monday–Friday and no spare day permitted for the others. The B–D block can sit on Mon–Tue (A Wed, C Fri, Thursday idle) or on Thu–Fri (A Mon, C Wed, Tuesday idle); the Tue–Wed and Wed–Thu placements both fail the two-day rule. Two outcomes, each with one working day vacant.
So a straight 'move E to Saturday' breaks condition (iv): the week can no longer carry five inspections on five days, only four, and an idle day appears. Moving E out of the week is therefore not a neutral adjustment; it reduces the week's throughput by one inspection.
Recommended handling: (a) if the deadline is firm, depute another officer to hold file E on Thursday so the roster stands; (b) if E truly cannot be heard before Saturday, accept the idle day but record in writing which day is vacant and why, and re-plan the following week's load; (c) do not silently leave a day blank, because the vacancy will later be read as a missed inspection rather than a leave adjustment.