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3.8c · Critical Path Analysis · Sub-skill
Find the critical path, calculate float, read Gantt charts, and work out the effect of delays and crashing.
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Float is the spare time an activity has: float = late time − early time − duration. An activity can be delayed by up to its float without delaying the project.
An activity with zero float is critical. Join the critical activities from start to finish to get the critical path; its length is the minimum completion time. There can be more than one critical path. Any delay to a critical activity delays the whole project by the same amount.
A Gantt chart (or cascade diagram) draws each activity as a bar on a time axis, starting at its early time. The float is shown as a dashed box after the bar. Critical activities have no dashed box. You can read off which activities must be happening at a given time (the critical ones) and which could be.
If an activity is delayed by more than its float, the project is delayed by (delay − float). Crashing means shortening an activity (usually at extra cost): it only helps if the activity is critical — and once a path has been shortened enough, another path may become critical. A resource histogram shows how many workers are needed each day; non-critical activities can be moved within their float to level the peaks.
For the network A (3), B (5), C (4, after A), D (3, after B), E (3, after C and D), find the float of each activity and say what happens if C is delayed by 3 days.
Floats (late − early − duration): A: 4−0−3 = 1; B: 5−0−5 = 0; C: 8−3−4 = 1; D: 8−5−3 = 0; E: 11−8−3 = 0. Critical path: B → D → E (11 days).
C has float 1, so a 3-day delay uses up its 1 day of float and delays the project by 3 − 1 = 2 days.
Test yourself
The activities of a project have early and late times (E, L) as follows, with durations: A (4): E 0, L 4; B (6): E 0, L 7; C (5): E 4, L 9; D (2): E 6, L 9; E (7): E 4, L 12; F (3): E 9, L 12. The minimum completion time is 12 days.
(a) Find the float of each activity and state the critical path.
(b) Activity E is delayed by 3 days. What is the effect on the project?
(c) Activity C is crashed so it takes 3 days. Find the new minimum completion time (the longest other route is A, E = 11 days and B, D, F = 11 days). [6 marks]
(a) Float = L − E − duration. A: 4−0−4 = 0; B: 7−0−6 = 1; C: 9−4−5 = 0; D: 9−6−2 = 1; E: 12−4−7 = 1; F: 12−9−3 = 0. Critical path: A → C → F (4+5+3 = 12). (3 marks)
(b) E has float 1, so a 3-day delay delays the project by 3 − 1 = 2 days (completion 14 days). (1 mark)
(c) A → C → F becomes 4+3+3 = 10, but the other routes are still 11, so the new minimum is 11 days — crashing C saves only 1 day because another route becomes critical. (2 marks)
Practice
Five short questions. Work through them, then reveal the mark scheme to check.