Skip to main content

TCTO numerical: crashing critical path, Rs 30/day

Short answer

Lowest cost schedule: crash 1-2 by 1 day (slope Rs 25/day) and 3-5 by 3 days (slope Rs 26.67/day). Resulting duration 16 days, total cost Rs1,165 (direct Rs685 + overhead Rs480). No other crashing is economical against Rs30/day indirect saving.

For the activities lying on the critical path of a network, the normal duration and the crash duration along with their respective direct cost are given in the table.

ActivityNormal durationNormal cost (Rs)Crash durationCrash costs (Rs)
1-22 days501 day75
2-34 days1603 days225
3-59 days2706 days350
5-65 days1004 days250

The overhead indirect cost is Rs.30/- per day. Crashing is possible for 1 day for each of the activities 1-2, 2-3, and 5-6 and 3 days for activity 3-5. Find the lowest cost schedule by crashing step by step assuming no fresh critical path is developed on crashing a project.

Given data and method

Indirect cost = 30 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.

The four activities lie in series on the critical path (normal duration 2 + 4 + 9 + 5 = 20 days). Crashing limits: 1-2, 2-3 and 5-6 by 1 day each; 3-5 by 3 days. No new critical path appears.

Cost slope=Crash cost−Normal costNormal duration−Crash duration\text{Cost slope}=\frac{\text{Crash cost}-\text{Normal cost}}{\text{Normal duration}-\text{Crash duration}}
ActivityNormal durCrash durNormal costCrash costSlope
1-221507525
2-34316022565
3-59627035026.67
5-654100250150

Paths at normal duration

  • 1-2 + 2-3 + 3-5 + 5-6 = 20 days

Critical activities (on the longest path or paths): 1-2, 2-3, 3-5, 5-6; normal duration = 20 days; normal direct cost = 580 Rs.

Crashing, step by step

StepCrash (by 1 unit)Slope/costDurationDirectIndirectTotal
0--205806001,180
11-225196055701,175
23-526.6718631.675401,171.67
33-526.6717658.335101,168.33
43-526.67166854801,165
52-365157504501,200
65-6150149004201,320

Result

  • Optimum (minimum-cost) schedule: duration 16 days, direct cost 685 + indirect 480 = 1,165 Rs.
  • Activities crashed at the optimum: 1-2 by 1, 3-5 by 3. Durations: 1-2=1, 2-3=4, 3-5=6, 5-6=5.
  • Minimum possible duration: 14 days, with total cost 1,320 Rs (direct 900 + indirect 420).

Answer: the lowest-cost schedule is obtained by crashing 1-2 by 1 day and 3-5 by 3 days: duration 16 days, total cost Rs 1,165 (direct Rs 685 + overhead Rs 480), against Rs 1,180 at the normal duration. Crashing 2-3 (Rs 65/day) or 5-6 (Rs 150/day) costs more than the Rs 30/day saved.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Construction Management chapter 2 questions →