- 2078 Kartik · 8 marks
TCTO numerical: seven activities, Rs 225/day
Short answer
With indirect cost Rs 225/day, cost slopes (cheapest Rs 200/day) exceed the saving. Optimum is normal 19 days, total cost Rs 13,475. Minimum possible time is 12 days (total Rs 15,466.67).
Calculate the minimum cost, optimum duration using time cost trade off analysis for the following project schedule consisting of seven activities. Also find the possible minimum time to complete the project. Indirect cost of the project is Rs. 225 per day.
| S.N. | Project Activity | Immediate Predecessor | Normal Duration (days) | Crash Duration (days) | Normal Cost (NPR) | Crash Cost (NPR) |
|---|---|---|---|---|---|---|
| 1 | A | - | 6 | 4 | 1400 | 1900 |
| 2 | B | - | 8 | 5 | 2000 | 2800 |
| 3 | C | A | 4 | 2 | 1100 | 1500 |
| 4 | D | A | 3 | 2 | 800 | 1400 |
| 5 | E | B, C | 6 | 3 | 900 | 1600 |
| 6 | F | D | 10 | 6 | 2500 | 3500 |
| 7 | G | E | 3 | 2 | 500 | 800 |
Given data and method
Indirect cost = 225 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 | 6 | 4 | 1,400 | 1,900 | 250 |
| B | 8 | 5 | 2,000 | 2,800 | 266.67 |
| C | 4 | 2 | 1,100 | 1,500 | 200 |
| D | 3 | 2 | 800 | 1,400 | 600 |
| E | 6 | 3 | 900 | 1,600 | 233.33 |
| F | 10 | 6 | 2,500 | 3,500 | 250 |
| G | 3 | 2 | 500 | 800 | 300 |
Paths at normal duration
- A-C-E-G = 19 days
- A-D-F = 19 days
- B-E-G = 17 days
Critical activities (on the longest path or paths): A, C, D, E, F, G; normal duration = 19 days; normal direct cost = 9,200 Rs.
Crashing, step by step
| Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total |
|---|---|---|---|---|---|---|
| 0 | - | - | 19 | 9,200 | 4,275 | 13,475 |
| 1 | A | 250 | 18 | 9,450 | 4,050 | 13,500 |
| 2 | A | 250 | 17 | 9,700 | 3,825 | 13,525 |
| 3 | E+F | 483.33 | 16 | 10,183.33 | 3,600 | 13,783.33 |
| 4 | E+F | 483.33 | 15 | 10,666.67 | 3,375 | 14,041.67 |
| 5 | E+F | 483.33 | 14 | 11,150 | 3,150 | 14,300 |
| 6 | F+G | 550 | 13 | 11,700 | 2,925 | 14,625 |
| 7 | B+C+D | 1,066.67 | 12 | 12,766.67 | 2,700 | 15,466.67 |
(Rows with several activities show activities crashed together because two or more paths are critical. After step 2, all three paths, including B-E-G, are 17 days long, so a cut must shorten all three: E+F (Rs 483.33) is the cheapest such combination; C+F would leave B-E-G at 17 days.)
Result
- Optimum (minimum-cost) schedule: duration 19 days, direct cost 9,200 + indirect 4,275 = 13,475 Rs.
- Activities crashed at the optimum: none. Durations: A=6, B=8, C=4, D=3, E=6, F=10, G=3.
- Minimum possible duration: 12 days, with total cost 15,466.67 Rs (direct 12,766.67 + indirect 2,700).
Answer: minimum cost Rs 13,475 at the optimum (normal) duration of 19 days, since the cheapest way to shorten both critical paths (A, Rs 250/day) costs more than the Rs 225/day indirect cost saved; the minimum possible time to complete the project is 12 days (total cost Rs 15,466.67).
✓ Answer checked by Claude Opus 5.5 (high effort)
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