Slotting 3,750 Candidates into a Five-Lab Test Centre
A recruitment board must conduct a computer-based test for 3,750 candidates at one Lucknow centre. The centre has 5 laboratories with 60 working machines each, and four sessions per day are possible at 09:00, 11:30, 14:00 and 16:30. The board's standard operating procedure requires a 5% spare-seat buffer per session so that a candidate can be shifted if a machine fails mid-test. Each candidate must report 45 minutes before the session, and each session runs for 90 minutes. The vendor's contract permits a maximum of 4 test days at this centre, and no session may start after 16:30 because the building's security cover ends at 20:00. However, on day 3 the district has requisitioned two of the five laboratories from 09:00 to 13:00 for a departmental examination, so only three labs can be used in the 09:00 and 11:30 sessions that day. The board also knows from experience that about 8% of candidates do not turn up.
1. Calculate the usable seats per session and per day after the 5% buffer, and determine whether three full days would be enough for 3,750 candidates. Show the arithmetic.
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Raw capacity: 5 labs × 60 machines = 300 seats per session.
Usable after the buffer: the 5% buffer is not extra capacity, it is capacity that must be held back. So usable = 300 × 0.95 = 285 seats per session. (Equivalently, keep 15 machines free per session.)
Per full day: 285 × 4 sessions = 1,140 candidates.
Three full days: 1,140 × 3 = 3,420 candidates.
Conclusion: 3,420 < 3,750, so three full days are short by 330 candidates even before the day-3 lab restriction is considered. With the restriction the shortfall is larger still, so the answer to 'can we finish in three days?' is no, not at this centre with 5 labs and 4 sessions.
The 8% absenteeism figure - roughly 300 of 3,750 - must not be used to close this gap. Seats are allotted on the assumption that every candidate may appear; if attendance is higher than expected, candidates who travelled to the centre would have to be turned away, which is a far worse outcome than an extra day of testing. Absenteeism reduces the load, not the planned capacity, and the 5% buffer exists for machine failure, not for no-shows.
2. Now build a feasible four-day plan that respects the day-3 laboratory restriction and state the total capacity and slack.
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Day 3 first, because it is the constrained day.
- 09:00 and 11:30 sessions: 3 labs × 60 = 180 machines → usable 180 × 0.95 = 171 seats each.
- 14:00 and 16:30 sessions: all 5 labs available → 285 seats each.
- Day 3 total = 171 + 171 + 285 + 285 = 912 candidates.
Days 1, 2 and 4 (unrestricted): 1,140 each = 3,420.
Grand total capacity over 4 days = 3,420 + 912 = 4,332 seats against a demand of 3,750.
Slack = 4,332 − 3,750 = 582 seats, which is comfortable: the plan is feasible with roughly 13% headroom.
A balanced loading could be:
- Day 1: 1,140
- Day 2: 1,140
- Day 3: 912 (the maximum the restricted day allows)
- Day 4: 558
- Total: 3,750.
Two practical points. First, the smallest unit of scheduling is a session, so the last day's 558 must be filled session by session (for example, 285 + 273 in the first two sessions, or four lighter sessions), not as one block. Second, the last session of the final day starts at 16:30, so candidates must report at 15:45, and the session ends at 18:00 - well inside the 20:00 security cut-off. No session may be pushed to 19:00 because the machine reset and candidate exit would overrun the security cover.
3. If the board refuses a fourth day, what options are genuinely available and which would you reject and why? Also state the effect if only two labs (not three) were available on day 3.
Read the worked answer
Options if the fourth day is refused:
- Add laboratories. 2 more labs → 7 × 60 = 420 raw, 399 usable per session, 1,596 per day, 4,788 in three days. This works, but only if 120 additional working machines, network points and invigilators exist and are certified.
- Add a fifth session. Rejected here: the last permitted start is 16:30 and security cover ends at 20:00, so a 19:00 session cannot be held.
- Shift about 330 candidates to another centre in the same city. Workable if the other centre has verified spare capacity and the candidates' admit cards can be reissued in time; otherwise it creates travel and grievance problems.
- Reduce the 5% buffer to fit everyone in three days. Reject this first. The buffer is what allows a candidate whose machine dies mid-test to be reseated; removing it means a single hardware failure leaves a candidate with no option but cancellation and re-test. The saving would be 15 seats per session, and the arithmetic still does not give three comfortable days, so the board would be trading a real risk for no benefit.
If only two labs were free on day 3: 09:00 and 11:30 sessions give 2 × 60 × 0.95 = 114 seats each, so day 3 becomes 114 + 114 + 285 + 285 = 798, and the four-day total falls to 3,420 + 798 = 4,218 - still above 3,750, so the plan survives with 468 seats of slack. It would not survive if the day-3 disruption spread to a second day.