The Critical Path Method (CPM) answers: "What is the shortest time to finish the project, and which activities cannot be delayed?" A project is a network of activities where some must finish before others can start. The critical path is the longest chain from start to finish — it sets the project duration. Activities on the critical path have zero float (any delay delays the whole project). Activities not on the critical path have float (some slack): Total Float = LS − ES = LF − EF. Forward pass (ES/EF) finds the earliest possible times; backward pass (LS/LF) finds the latest allowable times without delaying the project.
$$ES_j=\max(EF_i\text{ of all predecessors}), \quad EF=ES+\text{duration}$$
$$LF_i=\min(LS_j\text{ of all successors}), \quad LS=LF-\text{duration}$$
$$TF=LS-ES=LF-EF \quad (\text{Critical if }TF=0)$$
Parallel Activity Critical Path
Activity A takes 4 days. Then B takes 6 days and C takes 5 days in parallel. D takes 3 days after both finish. Find the project duration.
$$A-B-D=4+6+3=13\text{ days}$$
$$A-C-D=4+5+3=12\text{ days}$$
Final answer: 13 days, controlled by path A-B-D.
★ Total Float of an Activity
Activity E has Early Start = 5, Late Start = 9, duration = 4 days. Find its total float and determine if it is on the critical path.
$$TF=LS-ES=9-5=4\text{ days}$$
Since TF = 4 ≠ 0, Activity E is not on the critical path. It can be delayed up to 4 days without affecting the project finish date.
★★ Forward and Backward Pass
Three activities: A (duration 5, start of project), B (duration 3, starts after A), C (duration 7, starts after A). The project ends when both B and C are complete. Find: project duration, TF for B and C, and identify the critical path.
Critical path: A → C (both have TF=0). B has 4 days of float.
★★★ Five-Activity Network
A project has these precedence relationships and durations: A(3), B(5, after A), C(4, after A), D(6, after B and C), E(2, after B). Find the project duration and critical path.
Critical path: A → B → D. Activity C has 1 day float; E has 4 days float.
★★★ Crashing a Project
The critical path duration is 20 days. Activity A (critical, normal duration 6 days, crash duration 4 days, crash cost P5,000/day) and Activity B (critical, normal 8 days, crash 6 days, crash cost P3,000/day). The client offers P4,000/day for every day the project finishes early. Should you crash, and which activity first?
Crash cost per day: A = P5,000/day, B = P3,000/day. Benefit of crashing = P4,000/day saved.
B costs less to crash (P3,000 < P4,000 benefit) → crash B first.
Crash B by 2 days: cost = 2 × P3,000 = P6,000. Benefit = 2 × P4,000 = P8,000. Net gain = P2,000.
A costs P5,000/day to crash but earns only P4,000/day — not economical. Do not crash A.
$$\text{Net benefit of crashing B by 2 days}=8000-6000=\text{P}2{,}000$$
Final answer: Crash Activity B by 2 days for a net saving of P2,000.
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Exam Generator Problems
Additional board-style practice items for this topic.
Which of the following is/are main drawback(s) in adopting bar charts? I. All the activities are shown as being independent of each other. II. The sequence of activities is not defined at all. III. It is difficult to judge whether an activity is completed or not.
I only
II only
III only
I, II, and III
Bar charts are easy to read, but they do not show dependency relationships among activities, do not define the sequence of activities, and make it difficult to judge completion and detect schedule slippage in sufficient detail. Therefore, I, II, and III are drawbacks.
If you are preparing a report on PERT CPM network in a construction job, what time scheduling should you consider in the preparation of the PERT CPM network diagram?
pessimistic time
optimistic time
probable time
all of the above
A PERT CPM network should consider pessimistic time, optimistic time, and probable time. Considering all three estimates gives a more comprehensive schedule that accounts for both ideal and delayed conditions. Therefore, the answer is all of the above.
