The method: forward pass, backward pass, float
The critical path method finds the longest chain of dependent activities. That chain sets the shortest time the whole project can take. This calculator counts elapsed days from 0, the convention the Bid Bond lesson uses.
- Forward pass: early start (ES) is the largest early finish (EF) of the predecessors, or 0 with none. EF = ES + duration.
- Backward pass: late finish (LF) is the smallest late start (LS) of the successors, or the project finish with none. LS = LF − duration.
- Total float = LS − ES (or LF − EF): how long an activity can slip without moving the project finish.
- Free float = earliest ES of the successors − EF: how long it can slip without moving any later activity.
- Activities with zero total float are critical. Free float is never more than total float.
Worked example 1: six activities
A takes 4 days. B (6 days) and C (3 days) follow A. D (5 days) follows B. E (4 days) follows C. F (2 days) follows both D and E. These are invented study numbers and the calculator's starting example.
- Forward pass: A 0 to 4, B 4 to 10, C 4 to 7, D 10 to 15, E 7 to 11. F starts at the later of 15 and 11, so F runs 15 to 17.
- Backward pass from 17: F 15 to 17, D 10 to 15, E 11 to 15, B 4 to 10. C must finish by E's late start of 11, so C runs 8 to 11. A must finish by the earlier of B's 4 and C's 8, so A runs 0 to 4.
- Total float: C = 8 − 4 = 4 days, E = 11 − 7 = 4 days, and A, B, D and F have 0.
- Free float: C = 7 − 7 = 0, because any slip in C delays E; E = 15 − 11 = 4.
- Critical path: A-B-D-F, 17 days.
Check by paths: A-B-D-F is 4 + 6 + 5 + 2 = 17 days and A-C-E-F is 4 + 3 + 4 + 2 = 13 days. The difference, 17 − 13 = 4, is the float shared by C and E.
Worked example 2: shorter float
The same logic with new durations: A 3, B 5, C 2, D 4, E 6 and F 3 days.
- Forward pass: A 0 to 3, B 3 to 8, C 3 to 5, D 8 to 12, E 5 to 11, F 12 to 15.
- Backward pass from 15: F 12 to 15, D 8 to 12, E 6 to 12, B 3 to 8, C 4 to 6, A 0 to 3.
- Total float: C = 4 − 3 = 1 day and E = 6 − 5 = 1 day. Free float: C = 5 − 5 = 0 and E = 12 − 11 = 1.
Check by paths: A-B-D-F is 3 + 5 + 4 + 3 = 15 and A-C-E-F is 3 + 2 + 6 + 3 = 14, a difference of 1 day. C and E share that one day: if C slips a day, E has none left. These are original arithmetic examples, not exam questions. The CPM scheduling and float guide walks through the same network.
CPM float questions
How do you find total float and free float?
Run a forward pass for the early dates and a backward pass for the late dates. Total float is late start minus early start. Free float is the earliest early start of the following activities minus this activity's early finish. In the worked example, activity C has 4 days of total float but 0 free float, because any slip in C delays E.
How do I find the critical path?
The critical path is the chain of activities with zero total float, and it is the longest path through the network. In the worked example A-B-D-F takes 4 + 6 + 5 + 2 = 17 days and A-C-E-F takes 4 + 3 + 4 + 2 = 13, so A-B-D-F is critical and the project takes 17 days.
Why does the first activity start at day 0?
This calculator, like the Bid Bond lesson, counts elapsed days from 0, so early finish = early start + duration and a successor starts the day its predecessor finishes. Some books count the first day as day 1. The floats and the critical path come out the same either way; only the start and finish labels shift.
Where is this on the NASCLA exam?
The Bid Bond course teaches CPM scheduling and float in its General Requirements lesson. General Requirements is one of the twelve subject areas on the NASCLA outline, with 25 of the 115 scored questions.
See the NASCLA exam outline for all twelve subject areas and their question counts.