A project has the following activities, durations and predecessors: A: 4 days (no predecessor); B: 2 days (none); C: 3 days (A); D: 2 days (B); E: 5 days (C); F: 4 days (C and D); G: 2 days (F). If all other activities start as early as possible, what is the maximum number of days by which the start of activity B can be delayed without extending the total project duration?

Answer: C. 3 days

The earliest start is day 4 for C, day 7 for E and F (for F, the later of day 7 when C ends and day 4 when D ends) and day 11 for G, so the whole project takes A → C → F → G = 4 + 3 + 4 + 2 = 13 days. Working backward for the latest starts gives day 11 for G, day 7 for F, 7 − 2 = day 5 for D and 5 − 2 = day 3 for B, so B's total float is 3 − 0 = 3 days. Delaying B also delays the start of D, but D has the same float, so up to 3 days does not extend the project. 0 days is B's free float (the delay possible without delaying the earliest start of its successor D), and 1 day is E's float. In the critical path method (CPM), the floats are found in this way, and the path connecting activities with zero float is the critical path.

• A. This is B's free float. Delaying B delays the start of its successor D, but D also has float, so the project is not extended right away.
• B. This is the float of activity E (latest start day 8 − earliest start day 7).
• C. Correct. B's latest start is day 3 and its earliest start is day 0, so its total float is 3 days.