ITPEC IP October 2018 Question 52

4 days — the float of C→F: the latest time at F, minus the earliest time at C, minus the 6 days of C→F.

Earliest time of each node (the longest path into it):

• •A 0, B 3, C 3, E 7
• •D: the later of 3 + 4 = 7 via B and 3 + 3 = 6 via C, so 7
• •F: the latest of 3 + 6 = 9 via C, 7 + 6 = 13 via D and 7 + 4 = 11 via E, so 13
• •G: the later of 13 + 4 = 17 via F and 7 + 7 = 14 via E, so 17

The critical path is A→B→D→F→G, 17 days, so F must be reached by day 13. C→F can start on day 3 and takes 6 days, so it has 13 − 3 − 6 = 4 days to spare. The same result from paths: A→C→F→G takes 3 + 6 + 4 = 13 days against 17.

Answer (d)

Why not others:

• •(a) 1 — the slack of path A→C→D→F→G (3 + 3 + 6 + 4 = 16 days), which limits A→C and C→D; C→F is not on that path
• •(b) 2 — the slack of path A→E→F→G (7 + 4 + 4 = 15 days)
• •(c) 3 — the slack of path A→E→G (7 + 7 = 14 days)

Key rule: An activity's total float is the latest time of its end node minus the earliest time of its start node minus its duration; another activity leaving the same node does not limit it.

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