Crashing a 16-Day Polytechnic Block to 13 Days Under Liquidated Damages
An additional classroom block for a government polytechnic must be handed over in 13 working days to avoid a liquidated-damages clause of ₹10,000 per day. The approved network has five activities: A (site clearance, 3 days) precedes B (foundation, 5 days) and C (plinth beam and columns, 4 days); B and C both precede D (roof slab, 2 days); D precedes E (finishing, 6 days). The contractor's planning cell reports these crash options within the same sequence: A can be shortened by 1 day at ₹6,000 per day, B by 2 days at ₹4,000 per day, C by 1 day at ₹3,000 per day, D by 1 day at ₹5,000 per day, and E by 2 days at ₹8,000 per day. Normal completion is 16 days. The executive engineer must approve the least-cost crash plan that delivers the block in 13 days, and must also answer the contractor's alternative proposal to crash E by two days instead of touching the structural activities.
1. Identify the critical path, the project duration and the float available, and state how much time must be recovered.
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Two paths exist:
- Path 1: A–B–D–E = 3 + 5 + 2 + 6 = 16 days
- Path 2: A–C–D–E = 3 + 4 + 2 + 6 = 15 days
Critical path is A–B–D–E at 16 days. A, B, D and E have zero total float; C has 16 − 15 = 1 day of float.
Time to recover = 16 − 13 = 3 days, to avoid 3 × ₹10,000 = ₹30,000 of liquidated damages.
The trap in this network. Crashing an activity shortens the project only while it lies on the critical path. Path 2 is only 1 day behind, so more than one day of crashing B alone buys nothing beyond the first day - the moment path 1 drops to 15, C becomes critical and further work on B is wasted money. Every crash decision must be re-checked against both paths, which is why a simple 'cheapest activity first' answer is wrong.
2. Work out the least-cost plan to reach 13 days and show its economics.
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Cheapest combination. Let reductions be a, b, c, d, e. The constraints are:
- Path 1: a + b + d + e ≥ 3
- Path 2: a + c + d + e ≥ 2 with limits a ≤ 1, b ≤ 2, c ≤ 1, d ≤ 1, e ≤ 2.
A, D and E appear on both paths (shared); B helps only path 1, C only path 2. Cheapest solution: a = 1, b = 1, d = 1.
Cost = 6,000 + 4,000 + 5,000 = ₹15,000. Check: Path 1 = 2 + 4 + 1 + 6 = 13 ✓; Path 2 = 2 + 4 + 1 + 6 = 13 ✓ (C stays at 4).
Economics. LD avoided = 3 × ₹10,000 = ₹30,000, crash cost ₹15,000 → net benefit ₹15,000.
Alternative for comparison. Crash B by 2 (₹8,000) → path 1 = 14, path 2 = 15, but the project is now 15, so 2 days of B bought 1 day. Then C by 1 (₹3,000) → both 14; then D by 1 (₹5,000) → 13. Total ₹16,000 - feasible but ₹1,000 dearer. Recommend A + B + D at ₹15,000.
3. The contractor proposes crashing E by two days instead of touching structural activities. Evaluate the cost and the engineering judgement.
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Arithmetic. Crash E by 2 days at ₹8,000 = ₹16,000. Path 1 = 16 − 2 = 14; path 2 = 15 − 2 = 13. The project becomes 14 days, not 13 - E is on both paths but path 1 still governs, so 2 days of crashing yields only 2 days of time.
Value: LD avoided 2 × ₹10,000 = ₹20,000 against ₹16,000 → net ₹4,000, compared with ₹15,000 net for the A + B + D plan. On money alone it loses.
Engineering judgement. E is 6 days of finishing compressed to 4, at the highest rate in the table. Finishing is labour-bound and moisture-bound: adding shifts to plaster, curing and painting is precisely where rework, shrinkage cracks and leakage complaints come from, and the ₹8,000/day rate understates that risk. E is worth using only when cheaper, shared activities are exhausted - its cost, ₹8,000, is still below the ₹10,000 LD rate, so it is not forbidden, just last in the queue.
Recommendation. Approve A + B + D for ₹15,000; hold E in reserve only if the structural activities turn out to be un-crashable on site.