- Asked 2/19
- 2078 Chaitra · 7 marks
TCTO numerical: optimal crashed time, Rs 2500/week
Short answer
Crash critical activities with smallest cost slope as long as indirect saving (Rs 2500/week) exceeds slope. Normal 22 weeks at Rs 79,000. Optimum: 19 weeks, total cost Rs 71,950. No further crashing reduces total cost.
Consider the data of a project as shown in table below. If the indirect cost per week is Rs. 2500, find the optimal crashed project completion time and corresponding minimum cost.
| Activity | Immediate Predecessor | Normal Time (Week) | Normal Cost (Rs.) | Crash Duration (Week) | Crash Cost (Rs.) |
|---|---|---|---|---|---|
| P | - | 7 | 2500 | 4 | 2200 |
| Q | - | 9 | 4500 | 7 | 4200 |
| R | P | 5 | 3500 | 4 | 3000 |
| S | P | 8 | 5500 | 7 | 5100 |
| T | Q | 9 | 3000 | 8 | 2500 |
| U | Q | 12 | 5500 | 12 | - |
| V | S, T | 4 | 1500 | 2 | 1800 |
Given data and method
Indirect cost = 2,500 Rs/week. 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.
Assumption: for P, Q, R, S and T the table lists the crash cost below the normal cost, which is impossible for a crash. These columns are taken as interchanged, i.e. the smaller value is the normal cost and the larger is the crash cost. U cannot be crashed (crash duration equals normal duration).
| Activity | Normal dur | Crash dur | Normal cost | Crash cost | Slope |
|---|---|---|---|---|---|
| P | 7 | 4 | 2,200 | 2,500 | 100 |
| Q | 9 | 7 | 4,200 | 4,500 | 150 |
| R | 5 | 4 | 3,000 | 3,500 | 500 |
| S | 8 | 7 | 5,100 | 5,500 | 400 |
| T | 9 | 8 | 2,500 | 3,000 | 500 |
| U | 12 | 12 | 5,500 | 5,500 | cannot crash |
| V | 4 | 2 | 1,500 | 1,800 | 150 |
Paths at normal duration
- P-R = 12 weeks
- P-S-V = 19 weeks
- Q-T-V = 22 weeks
- Q-U = 21 weeks
Critical activities (on the longest path or paths): Q, T, V; normal duration = 22 weeks; normal direct cost = 24,000 Rs.
Crashing, step by step
| Step | Crash (by 1 unit) | Slope/cost | Duration | Direct | Indirect | Total |
|---|---|---|---|---|---|---|
| 0 | - | - | 22 | 24,000 | 55,000 | 79,000 |
| 1 | Q | 150 | 21 | 24,150 | 52,500 | 76,650 |
| 2 | Q | 150 | 20 | 24,300 | 50,000 | 74,300 |
| 3 | V | 150 | 19 | 24,450 | 47,500 | 71,950 |
Result
- Optimum (minimum-cost) schedule: duration 19 weeks, direct cost 24,450 + indirect 47,500 = 71,950 Rs.
- Activities crashed at the optimum: Q by 2, V by 1. Durations: P=7, Q=7, R=5, S=8, T=9, U=12, V=3.
- Minimum possible duration: 19 weeks, with total cost 71,950 Rs (direct 24,450 + indirect 47,500).
Answer: optimal crashed project completion time = 19 weeks; minimum total cost = Rs 71,950.
✓ Answer checked by Claude Opus 5.5 (high effort)
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