Shorten a project by recalculating the critical path
In project scheduling, crashing means paying an additional direct cost to reduce an activity’s duration within its allowed limit. Under a linear cost assumption, slope = (crash cost − normal cost)/(normal duration − crash duration). Only shortening activities that constrain the finish can shorten the whole project. Recalculate all relevant path lengths after each reduction because a second path may become critical.
Worked practice example
Two paths run from start to finish: A–B takes 4 + 6 = 10 days; C–D takes 3 + 5 = 8 days. A can be shortened from 4 to 3 days for KSh 500. B can be shortened from 6 to 4 days for KSh 1,600 total. Find the cheapest way to finish in 8 days under these linear assumptions.
| Step | Working |
|---|---|
| Critical path initially | A–B is 10 days; C–D is 8 days. |
| Crash slopes | A: 500/(4 − 3) = KSh 500/day. B: 1600/(6 − 4) = KSh 800/day. |
| First reduction | Shorten A by one day: A–B becomes 9 days, extra cost KSh 500. |
| Second reduction | Shorten B by one day: both paths become 8 days. Total extra cost = 500 + 800 = KSh 1,300. |
Check your method
Crashing B by two days would also meet 8 days but cost KSh 1,600. Crashing C or D initially would not shorten the 10-day path. Once both paths are critical, shortening only A–B cannot reduce the project below 8 days.
Try it yourself
Would one more day taken from B alone reduce the project duration to 7 days?
Show the answer and method
No. A–B would become 7 days but C–D would remain 8 days, so project duration would remain 8 days. Every critical path must be shortened to reduce the finish time further.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.