- 2079 Shrawan · 8 marks
TCTO numerical: seven activities, Rs 2000/day
Short answer
Compute cost slopes. Crash critical activities one day at a time by cheapest slope while saving Rs 2000/day indirect cost. Normal 21 days at Rs 88,000. Optimum: 13 days, total cost Rs 77,000. Minimum duration: 11 days at Rs 77,800.
From the given data in table below. Calculate minimum cost and minimum duration schedule using time cost trade off analysis. Indirect cost is Rs. 2000/day.
| Activity | Predecessor | Normal Duration (days) | Crash Duration (days) | Normal cost (Rs.) | Crash Cost (Rs.) |
|---|---|---|---|---|---|
| A | - | 5 | 2 | 5000 | 6200 |
| B | A | 4 | 3 | 6000 | 7000 |
| C | A | 3 | 2 | 5000 | 6500 |
| D | A | 6 | 4 | 4000 | 4400 |
| E | B | 5 | 3 | 15000 | 18000 |
| F | D | 4 | 2 | 3000 | 4800 |
| G | C, E, F | 6 | 3 | 8000 | 10400 |
Given data and method
Indirect cost = 2,000 Rs/day. Time-cost trade-off (TCTO) crashes the activities on the critical path, one time-unit at a time, always choosing the cheapest cost slope, as long as the saving in indirect cost is more than the extra direct cost.
| Activity | Normal dur | Crash dur | Normal cost | Crash cost | Slope |
|---|---|---|---|---|---|
| A | 5 | 2 | 5,000 | 6,200 | 400 |
| B | 4 | 3 | 6,000 | 7,000 | 1,000 |
| C | 3 | 2 | 5,000 | 6,500 | 1,500 |
| D | 6 | 4 | 4,000 | 4,400 | 200 |
| E | 5 | 3 | 15,000 | 18,000 | 1,500 |
| F | 4 | 2 | 3,000 | 4,800 | 900 |
| G | 6 | 3 | 8,000 | 10,400 | 800 |
Paths at normal duration
- A-B-E-G = 20 days
- A-C-G = 14 days
- A-D-F-G = 21 days
Critical activities (on the longest path or paths): A, D, F, G; normal duration = 21 days; normal direct cost = 46,000 Rs.
Crashing, step by step
| Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total |
|---|---|---|---|---|---|---|
| 0 | - | - | 21 | 46,000 | 42,000 | 88,000 |
| 1 | D | 200 | 20 | 46,200 | 40,000 | 86,200 |
| 2 | A | 400 | 19 | 46,600 | 38,000 | 84,600 |
| 3 | A | 400 | 18 | 47,000 | 36,000 | 83,000 |
| 4 | A | 400 | 17 | 47,400 | 34,000 | 81,400 |
| 5 | G | 800 | 16 | 48,200 | 32,000 | 80,200 |
| 6 | G | 800 | 15 | 49,000 | 30,000 | 79,000 |
| 7 | G | 800 | 14 | 49,800 | 28,000 | 77,800 |
| 8 | B+D | 1,200 | 13 | 51,000 | 26,000 | 77,000 |
| 9 | E+F | 2,400 | 12 | 53,400 | 24,000 | 77,400 |
| 10 | E+F | 2,400 | 11 | 55,800 | 22,000 | 77,800 |
(Rows with several activities show activities crashed together because two paths are critical.)
Result
- Optimum (minimum-cost) schedule: duration 13 days, direct cost 51,000 + indirect 26,000 = 77,000 Rs.
- Activities crashed at the optimum: A by 3, B by 1, D by 2, G by 3. Durations: A=2, B=3, C=3, D=4, E=5, F=4, G=3.
- Minimum possible duration: 11 days, with total cost 77,800 Rs (direct 55,800 + indirect 22,000).
Answer: minimum-cost schedule = 13 days, total cost Rs 77,000; minimum-duration schedule = 11 days, total cost Rs 77,800. Crashing beyond 13 days costs more (Rs 2,400 per day) than the Rs 2,000/day saved.
✓ Answer checked by Claude Opus 5.5 (high effort)
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