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.
