Crashing a shutdown overhaul with two critical paths
A fabrication shop plans a 17-day shutdown overhaul. The activity list is: A (1–2) 6 days; B (2–3) 4 days; C (2–4) 3 days; D (3–5) 5 days; E (4–5) 6 days; F (5–6) 2 days. Crash limits and costs per day are: A 2 days at Rs 400/day; B 2 days at Rs 300/day; C 1 day at Rs 200/day; D 2 days at Rs 500/day; E 2 days at Rs 350/day; F 1 day at Rs 600/day. Indirect shutdown cost is Rs 1,000 per day. The shop has booked a crane for day 18, so any duration beyond 17 days is impossible; the maintenance manager wants the cheapest duration between 13 and 17 days. The production manager argues that crashing A is pointless because A is common to both paths and therefore does not shorten either path.
1. Identify the critical path or paths and the normal project duration.
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There are two paths from 1 to 6:
Path 1: A–B–D–F = 6 + 4 + 5 + 2 = 17 days. Path 2: A–C–E–F = 6 + 3 + 6 + 2 = 17 days.
Both paths are critical, and the normal project duration is 17 days. Any shortening must reduce both paths; crashing an activity on only one path does not shorten the project while the other path remains at 17 days.
2. Find the least-cost way to reach 14 days. What is the cost and the resulting activity durations?
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To reduce the project by 3 days, both paths must each fall by 3 days. Crashing A is attractive because A lies on both paths: crash A by its full 2 days at 2×400 = Rs 800. Both paths become 15 days. One more day is needed on each path. The cheapest pair that covers both paths is B (path 1) and C (path 2): 300 + 200 = Rs 500. Total crash cost = 800 + 500 = Rs 1,300.
Resulting durations: A = 4 days, B = 3 days, C = 2 days, D = 5 days, E = 6 days, F = 2 days. Path 1 = 4+3+5+2 = 14; path 2 = 4+2+6+2 = 14. The 14-day plan is feasible and costs Rs 1,300 in direct crash cost, against 3×1,000 = Rs 3,000 saved in shutdown indirect cost.
3. Is 13 days economically justified? Compute the marginal cost and evaluate the production manager's claim about activity A.
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From the 14-day plan, one more day requires a further reduction on both paths. The cheapest single common activity is F, which is uncrashed and can save 1 day at Rs 600. Crashing B and E instead would cost 300 + 350 = Rs 650, so F is cheaper. The 13-day total crash cost is 1,300 + 600 = Rs 1,900. The extra day costs Rs 600 and saves Rs 1,000, giving a net benefit of Rs 400, so 13 days is justified.
The production manager's claim is wrong. A is common to both paths, so crashing A reduces both path lengths simultaneously. That is exactly why A was crashed first: 2 days of A bought 2 days of project time at Rs 800, cheaper than attacking each path separately. The limitation is A's 2-day crash limit, not its commonality.