Ten-Day Shutdown Window with One Crew and Two Blocked Days
One line crew is available at a 110 kV substation for a ten-day window, Days 1 to 10, and five shutdown jobs are pending: P (2 days), Q (1 day), R (3 days), S (2 days) and T (1 day). Constraints are: R must be completed before S starts; P must be over before Day 5 to satisfy a fertilizer consumer's supply agreement; no shutdown of any kind is permitted on Days 6 and 7 because of a local festival and a planned 220 kV bus outage; and the crew can work on only one job at a time. A contractor offers a second independent crew at ₹25,000 per day, but that crew cannot work on Q or T because those two jobs need the same 110 kV bus-coupler isolation, which can be issued to only one crew at a time. The utility estimates that every day the whole package runs beyond its own completion target costs ₹40,000 in continued operation of an aged breaker that is due for replacement.
1. Is it possible to complete all five jobs by Day 10? Show the arithmetic and give either a valid schedule or the earliest feasible finish.
Read the worked answer
Work content. Total crew-days = 2 + 1 + 3 + 2 + 1 = 9 crew-days.
Capacity. Usable days are 1, 2, 3, 4, 5, 8, 9, 10 = 8 days, since Days 6 and 7 are embargoed and one crew can do only one job at a time.
Verdict. 9 crew-days of work in 8 available days is impossible regardless of sequencing. The earliest finish is the 9th working day, which is Day 11.
A valid schedule finishing Day 11
- P: Days 1–2 (completes before Day 5)
- R: Days 3–5
- S: Days 8–9 (after R, precedence satisfied)
- Q: Day 10
- T: Day 11
Days 6 and 7 are idle, as required.
Why P must be early. Try R first (Days 1–3) and P at Days 4–5: P then finishes on Day 5, breaching the 'over before Day 5' condition. Any sequence that puts P after the festival also breaches it. So the deadline on P is a real sequencing constraint, not a formality, and it should be written into the permit plan rather than left to the crew.
What to report upward. State plainly that Day 10 is not achievable with one crew; do not accept a target that the arithmetic rules out. Offer Day 11 with the sequence above, or the second-crew option below.
2. The fertilizer consumer demands P complete by the end of Day 3. Is that compatible with the rest of the work, and what exactly would you commit to in writing?
Read the worked answer
P by Day 3 is achievable at no cost. In the schedule above P runs Days 1–2 and is already complete by Day 3, with the rest of the package finishing on Day 11.
But do not promise P by Day 3 and everything by Day 10. If P occupies Days 1–2, the remaining work is 7 crew-days (Q, R, S, T) and only Days 3, 4, 5, 8, 9, 10 = 6 days remain. That combination is infeasible and would be a commitment the utility cannot keep.
Commitment to issue in writing
- P will be charged and energised by the end of Day 3, subject to the shutdown being availed on Day 1 and no forced outage on the 110 kV bus.
- Q, R, S and T will be completed by Day 11, with Days 6 and 7 excluded for the festival and the 220 kV bus outage.
- If the second contractor crew is engaged, all five jobs will be completed by Day 5 - but this must be confirmed after the commercial decision, not assumed.
Practical points. Tell the consumer about the Days 6–7 embargo so it can plan its load, and confirm that P's bay can be restored and loaded without the work on R, since R also touches the same station. A short written note now is cheaper than an unplanned outage dispute later.
3. Evaluate the second crew at ₹25,000 per day against the ₹40,000 per day deferral cost, and state whether your answer changes if the deferral cost is ₹15,000 per day.
Read the worked answer
Longest useful hire (5 days). Crew 2 takes R (Days 1–3) then S (Days 4–5). Crew 1 takes P (Days 1–2), Q (Day 3) and T (Day 4). Everything finishes on Day 5. Precedence R before S holds, P is complete by Day 3, and only one crew is ever on the bus coupler for Q and T, so the isolation rule is respected. Cost = 5 × ₹25,000 = ₹1,25,000. Saving = 11 − 5 = 6 days × ₹40,000 = ₹2,40,000. Net benefit ≈ ₹1,15,000. Engage the crew.
Shorter hire (3 days). Crew 2 takes only R (Days 1–3); crew 1 does P (1–2), S (4–5), Q (Day 8), T (Day 9). Finish Day 9. Cost = ₹75,000; saving = 2 days × ₹40,000 = ₹80,000; net ≈ ₹5,000. Technically positive but too thin to be worth the coordination risk.
If the deferral cost is ₹15,000/day. The 5-day hire saves 6 × ₹15,000 = ₹90,000 against ₹1,25,000 → net loss ₹35,000. The 3-day hire saves ₹30,000 against ₹75,000 → net loss ₹45,000. Recommendation reverses: use one crew and finish on Day 11, and spend the money instead on condition monitoring of the aged breaker.
Conclusion. The correct choice is driven by the marginal value of early completion, not by the contractor's rate card. Fix the ₹40,000/day estimate with the asset manager in writing before signing, since the whole decision turns on it.