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Strategic Cost Management · Network Analysis - PERT, CPM

Critical Path Method (CPM) and How to Find the Critical Path

Updated 11 October 2026 · Fact-checked

The critical path method finds the longest chain of activities in a project network. Do a forward pass to get earliest times, then a backward pass from the project end date to get latest times. Activities with zero total float form the critical path, and its length is the minimum project duration.

Understand Critical Path Method (CPM) and Critical Path

A project is a set of activities with a fixed order. Some activities cannot start until others finish. CPM draws this order as a network and asks one question: what is the shortest time in which the whole project can finish?

The answer is the critical path. It is the longest path from the start event to the end event. Every other path is shorter, so those paths have spare time. The critical path has none. If a critical activity runs late by one day, the whole project runs late by one day.

To find it, you make two passes through the network. The forward pass moves left to right and gives the earliest event time (E) for each event. The backward pass moves right to left and gives the latest event time (L). The latest time is the last moment an event can occur without delaying the project.

An event with E = L has no slack and is a critical event. Be careful, though. A path is critical only if its activities also pass the activity test: the activity duration must equal the gap between the L of its head event and the E of its tail event. Linking critical events alone is not enough.

CPM uses a single fixed time for each activity. This is different from PERT, which uses three time estimates. In the exam, CPM is the base on which float, crashing and PERT questions are built, so get this right first.

Key rules to remember

Earliest event time (forward pass)
E(j) = maximum of [E(i) + duration(i, j)] over all activities ending at event j
Start with E(1) = 0. Where several activities merge into an event, take the largest value.
Latest event time (backward pass)
L(i) = minimum of [L(j) − duration(i, j)] over all activities starting at event i
Start with L(last event) = E(last event). Where several activities leave an event, take the smallest value.
Earliest start and finish of an activity
ES = E(tail event); EF = ES + duration
In activity-on-node tables, ES = largest EF among all predecessors.
Latest finish and start of an activity
LF = L(head event); LS = LF − duration
In activity-on-node tables, LF = smallest LS among all successors.
Total float of an activity
Total float = LS − ES = LF − EF = L(j) − E(i) − duration(i, j)
Total float is zero for every critical activity.
Critical path rule
Critical activity: E(i) = L(i), E(j) = L(j) and L(j) − E(i) = duration(i, j)
The critical path is the chain of critical activities from start to end. Project duration = E of the last event.

How to solve Critical Path Method (CPM) and Critical Path questions

Use this order for any CPM question. It works for both arrow networks and predecessor tables.

  1. 1List every activity with its duration and its predecessors. If a table is given, draw or at least sketch the network so merge and burst points are clear.
  2. 2Forward pass: set the earliest time of the start event to 0. Move left to right. At each event, take the largest of (earlier event time + activity duration) across all incoming activities.
  3. 3Read the project duration from the earliest time of the last event.
  4. 4Backward pass: set the latest time of the last event equal to its earliest time. Move right to left. At each event, take the smallest of (later event latest time − activity duration) across all outgoing activities.
  5. 5Mark each event with its two times, E and L. Events with E = L are candidates for the critical path.
  6. 6Test every activity: it is critical only if E(i) = L(i), E(j) = L(j) and duration = L(j) − E(i). Equivalently, its total float is zero.
  7. 7Trace the critical path from start to end through critical activities only. State the path and the project duration clearly.
  8. 8If asked, compute total float for non-critical activities as L(j) − E(i) − duration.

Quickest way: Table method: ES, EF, LS, LF with float

When to use it: Use this when activities and predecessors are given in a table. It is fast, and the float column checks your work.

  1. Write one row per activity: duration and predecessors.
  2. Forward pass down the rows: ES = largest EF of predecessors (0 if none). EF = ES + duration.
  3. Project duration = largest EF.
  4. Backward pass from the last row up: LF = smallest LS of successors (project duration if none). LS = LF − duration.
  5. Float = LS − ES. Every row with float 0 is critical.
  6. Check: the durations of the zero-float activities, added along the chain, must equal the project duration. If they do not, recheck a merge point.

Common mistakes in Critical Path Method (CPM) and Critical Path

  • Taking the smaller value at a merge event in the forward pass

    Students carry the habit of 'shortest route' from other network problems.

    Fix: In the forward pass always take the maximum, because the event cannot occur until all incoming activities are done. In the backward pass always take the minimum.

  • Marking a path critical just because its events have E = L

    Two critical events can be joined by a short activity that still has float.

    Fix: Also check that the activity duration equals L(j) − E(i). If not, the activity has float and is not critical.

  • Starting the backward pass with a wrong latest time for the last event

    Students use a deadline or forget to copy the forward-pass result.

    Fix: Unless the question gives a target completion date, set L of the last event equal to its E.

  • Ignoring dummy activities or giving them a duration

    Dummies look like extra arrows and are skipped, or are treated as real work.

    Fix: A dummy has zero duration but still carries the dependency. Include it in both passes.

  • Listing more than one critical path as an error, or giving only one when two exist

    Students stop after finding the first zero-float chain.

    Fix: Several paths can tie for the longest length. Scan all zero-float activities and write every chain that runs start to end.

  • Stating the critical path without the project duration, or the duration without the path

    Students rush the last line.

    Fix: Write both in the final answer: the path in activity letters and the total time with its unit.

Worked examples

Example 1

A project has these activities (event numbers in brackets) and durations in days: A (1-2) 4; B (1-3) 3; C (2-4) 5; D (3-4) 6; E (2-5) 7; F (4-6) 2; G (5-6) 3. Find the earliest and latest event times, the critical path and the project duration.

Show the solution
  1. Forward pass: E1 = 0. E2 = 0 + 4 = 4. E3 = 0 + 3 = 3.
  2. E4 = maximum of (4 + 5 = 9) and (3 + 6 = 9) = 9. E5 = 4 + 7 = 11.
  3. E6 = maximum of (9 + 2 = 11) and (11 + 3 = 14) = 14. Project duration = 14 days.
  4. Backward pass: L6 = 14. L5 = 14 − 3 = 11. L4 = 14 − 2 = 12.
  5. L3 = 12 − 6 = 6. L2 = minimum of (12 − 5 = 7) and (11 − 7 = 4) = 4.
  6. L1 = minimum of (4 − 4 = 0) and (6 − 3 = 3) = 0, which matches E1 = 0.
  7. Events with E = L: 1 (0, 0), 2 (4, 4), 5 (11, 11), 6 (14, 14). Events 3 and 4 have 3 days of slack each (E3 = 3, L3 = 6; E4 = 9, L4 = 12).
  8. Activity test: A: 4 = 4 − 0, critical. E: 7 = 11 − 4, critical. G: 3 = 14 − 11, critical.
  9. Other activities each have total float 3 days, for example C: 12 − 4 − 5 = 3.

Answer: Critical path is 1-2-5-6, that is A → E → G, and the project duration is 4 + 7 + 3 = 14 days.

Example 2

Activities, durations (weeks) and predecessors: A 3 (none); B 5 (none); C 4 (A); D 6 (A, B); E 2 (C); F 5 (C, D); G 4 (E, F). Find ES, EF, LS, LF, total float, the critical path and the project duration.

Show the solution
  1. Forward pass: A: ES 0, EF 3. B: ES 0, EF 5. C: ES 3, EF 7.
  2. D: ES = maximum of (3, 5) = 5, EF 11. E: ES 7, EF 9.
  3. F: ES = maximum of (EF of C = 7, EF of D = 11) = 11, EF 16.
  4. G: ES = maximum of (9, 16) = 16, EF 20. Project duration = 20 weeks.
  5. Backward pass: G: LF 20, LS 16. F: LF 16, LS 11. E: LF 16, LS 14.
  6. D: LF = LS of F = 11, LS 5. C: LF = minimum of (LS of E = 14, LS of F = 11) = 11, LS 7.
  7. B: LF = LS of D = 5, LS 0. A: LF = minimum of (LS of C = 7, LS of D = 5) = 5, LS 2.
  8. Total float = LS − ES: A = 2; B = 0; C = 4; D = 0; E = 7; F = 0; G = 0.
  9. Zero-float chain: B → D → F → G = 5 + 6 + 5 + 4 = 20 weeks, which equals the project duration.

Answer: Critical path is B → D → F → G and the project duration is 20 weeks. Float: A 2, C 4, E 7 weeks; the critical activities have zero float.

Exam tips

  • Draw a small box at each event with E on the left and L on the right. It keeps the two passes visible and earns method marks even if one number slips.
  • Always finish with a one-line answer: the critical path in letters and the duration with its unit.
  • Check every merge and burst point twice. Almost all lost marks come from taking the wrong maximum or minimum there.
  • Questions often continue into float, crashing or PERT probability. A correct critical path is the base for all of them, so do not rush it.
  • In MCQs, you often only need the project duration. Do the forward pass first and read the answer before attempting the backward pass.

Practice questions from Network Analysis - PERT, CPM

Critical Path Method (CPM) and Critical Path: frequently asked questions

What is the critical path in CPM?

It is the longest path through the project network from start to end. It decides the minimum project duration. Activities on it have zero total float, so any delay in them delays the project.

Why do we take the maximum in the forward pass and the minimum in the backward pass?

An event can occur only after all incoming activities finish, so its earliest time is the largest of the incoming finish times. In the backward pass, an event must occur early enough for every outgoing activity, so its latest time is the smallest of the allowed values.

Can a project have more than one critical path?

Yes. If two or more paths tie for the longest length, each is critical. All activities on these paths have zero total float, and a delay on any of them delays the project.

What is the difference between CPM and PERT?

CPM uses one fixed duration per activity and is used where times are well known. PERT uses three time estimates to get an expected time and a variance, so it allows probability statements about completion. The forward and backward pass method is the same in both.