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Time cost trade-off and steps in planning

Short answer

Steps: draw network, find critical path, compute each activity’s cost slope, crash critical activity with lowest slope one unit at a time, recompute duration and costs, repeat until slope exceeds indirect cost saved. In the example, crashing both P and Q fully (slopes Rs 400/day and Rs 300/day, below Rs 500/day indirect) gives optimum duration 4 days and total cost Rs 6,600.

Discuss on time cost trade-off with examples and steps in planning.

Time-cost trade-off is the study of the relation between project duration and cost, in order to find the duration with minimum total cost. Direct cost rises when activities are crashed; indirect cost falls when the project is shorter.

Steps in planning

  1. Draw the network and find the critical path and normal duration.
  2. Find the normal and crash time and cost of every activity.
  3. Compute the cost slope = (crash cost - normal cost)/(normal time - crash time).
  4. Crash the critical activity with the lowest slope by one unit (not below its crash time). If there are several critical paths, crash a combination that cuts all of them.
  5. Recompute project duration, direct cost, indirect cost and total cost.
  6. Repeat until the slope exceeds the indirect cost saved per unit time, or all critical activities are at crash limit.
  7. Choose the duration with the lowest total cost as the optimum.

Example

Activities in series: P (normal 4 days, Rs 2,000; crash 2 days, Rs 2,800) and Q (normal 3 days, Rs 1,500; crash 2 days, Rs 1,800). Indirect cost Rs 500/day.

Slopes: P = (2800 - 2000)/2 = Rs 400/day; Q = 300/1 = Rs 300/day.

StepCrashDurationDirectIndirectTotal
0-73,5003,5007,000
1Q63,8003,0006,800
2P54,2002,5006,700
3P44,6002,0006,600

Answer: all slopes are below the indirect cost of Rs 500/day, so the project is crashed fully: optimum duration 4 days, total cost Rs 6,600.

✓ Answer checked by Claude Opus 5.5 (high effort)

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