ITPEC IP April 2022 Question 52
ITPEC IP April 2022 — Question 52 of 100
The whole project is delayed by one day. The original path durations are A–D = 2 + 4 = 6 days, B–E = 4 + 3 = 7 days, and C–F = 7 + 1 = 8 days, so C–F is initially the critical path. A and D are delayed by three days in total, making their path 6 + 3 = 9 days. The project duration therefore changes from 8 to 9 days, a delay of one day.
Answer (a)
Why not others:
- the full three-day activity delay is partly absorbed by the A–D path's original two days of float
- delays of two, three, or four days do not match the change in the longest path
Key rule: A project delay equals the increase in the longest path, not necessarily the full delay of an activity on a noncritical path.
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