- 2070 Magh · 6 marks
TCTO numerical: seven activities, Rs 1000/day
Short answer
Compute cost slopes. Crash cheapest critical activities one day at a time while saving Rs 1000/day indirect cost. Normal 21 days at Rs 67,000. Optimum: 14 days, total cost Rs 63,800. Minimum duration: 11 days at Rs 66,800.
Calculate minimum cost and minimum duration schedule using time cost trade-off analysis of the project with following detail. Indirect cost is Rs. 1000 per day.
| Activity | Predecessor | Normal Duration | Crash Duration | Normal Cost | Crash Cost |
|---|---|---|---|---|---|
| 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 = 1,000 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 | 21,000 | 67,000 |
| 1 | D | 200 | 20 | 46,200 | 20,000 | 66,200 |
| 2 | A | 400 | 19 | 46,600 | 19,000 | 65,600 |
| 3 | A | 400 | 18 | 47,000 | 18,000 | 65,000 |
| 4 | A | 400 | 17 | 47,400 | 17,000 | 64,400 |
| 5 | G | 800 | 16 | 48,200 | 16,000 | 64,200 |
| 6 | G | 800 | 15 | 49,000 | 15,000 | 64,000 |
| 7 | G | 800 | 14 | 49,800 | 14,000 | 63,800 |
| 8 | B+D | 1,200 | 13 | 51,000 | 13,000 | 64,000 |
| 9 | E+F | 2,400 | 12 | 53,400 | 12,000 | 65,400 |
| 10 | E+F | 2,400 | 11 | 55,800 | 11,000 | 66,800 |
(Rows with several activities show activities crashed together because two paths are critical.)
Result
- Optimum (minimum-cost) schedule: duration 14 days, direct cost 49,800 + indirect 14,000 = 63,800 Rs.
- Activities crashed at the optimum: A by 3, D by 1, G by 3. Durations: A=2, B=4, C=3, D=5, E=5, F=4, G=3.
- Minimum possible duration: 11 days, with total cost 66,800 Rs (direct 55,800 + indirect 11,000).
Answer: minimum-cost schedule = 14 days, total cost Rs 63,800; minimum-duration schedule = 11 days, total cost Rs 66,800.
✓ Answer checked by Claude Opus 5.5 (high effort)
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