Critical Path Calculator

From the order and durations of a project's tasks, this tool calculates how many days the whole project takes at minimum and which tasks delay the whole project if they slip (the critical path), using the critical path method (CPM). You can check the earliest start, latest start and float used in arrow diagram (PERT chart) problems in a table.

Tasks

Task names can be up to 8 letters, digits or underscores; the duration is a number of 0 or more; list predecessors separated by spaces or commas, and omit them if there are none. Up to 26 tasks.

Results

Total project duration
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Critical path
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Task on the critical path Task with float Float (days the task can be delayed)

Days are counted from day 0 at the start of the project. ES, EF, LS and LF are the earliest start, earliest finish, latest start and latest finish. Total float is the number of days a task can be delayed without extending the total project duration, and free float is the number of days it can be delayed without delaying the earliest start of any following task.

What is the critical path?

Project tasks have order relations: some cannot start until earlier tasks have finished. Of all the chains of tasks from start to finish, the one with the longest total duration is the critical path, and its length is the shortest possible duration of the whole project. A delay of even one day in a task on the critical path delays the whole project. In the "Six tasks" example, B (5 days) → D (4 days) → E (3 days), at 12 days, is the critical path, and A, C and F have float.

Finding the earliest and latest start

The earliest start is the latest day on which all predecessors have finished, found from the start as "the maximum of the predecessors' earliest finishes" (the forward pass). The latest start is the latest day a task can begin without extending the total project duration, found backward from the end as "the minimum of the successors' latest starts minus the task's own duration" (the backward pass). The difference between the latest start and the earliest start is the float (total float), and the tasks with zero float connected together form the critical path.

Arrow diagrams and PERT

A diagram that shows the order relations of tasks with arrows is called an arrow diagram, and it is used in the technique called PERT (Program Evaluation and Review Technique). In an arrow diagram, tasks are drawn as arrows and the points between them as circles (nodes), and dotted arrows that represent only an order relation (dummy activities, with a duration of 0) are sometimes used. This tool takes input as a list of predecessors for each task (the same idea as a precedence diagram), so dummy activities are not needed.

Shortening the schedule

To finish a project earlier, only shortening tasks on the critical path has any effect; shortening tasks that have float does not change the total duration. Methods include crashing, which shortens tasks by adding people, and fast tracking, which overlaps tasks that would otherwise run in sequence. Shortening a task on the critical path can make another chain the new critical path, so check the table as you go.

Practice related questions

In an arrow diagram with nodes ① to ⑥, the activities, written as 'name: start node → end node, duration', are A: ① → ②, 6 days; B: ① → ③, 4 days; C: ② → ④, 4 days; D: ③ → ⑤, 4 days; E: ④ → ⑥, 2 days; F: ⑤ → ⑥, 3 days; and a dummy activity: ② → ③ (duration 0). What is the shortest duration of the whole project?

  1. A11 days
  2. B12 days
  3. C13 days
Show answer and explanation

Answer: C. 13 days

The dummy activity ② → ③ means that activity D cannot start until A, as well as B, has finished. The earliest event times are 6 days at ②, max(4, 6 + 0) = 6 days at ③, 6 + 4 = 10 days at ④, 6 + 4 = 10 days at ⑤ and max(10 + 2, 10 + 3) = 13 days at ⑥, so the whole project takes 13 days and the critical path is A → dummy → D → F. Overlooking the dummy activity would let D start on day 4, making A → C → E, at 12 days, the longest path. 11 days is the length of the B → D → F path alone.

A project has the following activities, durations and predecessors: A: 4 days (no predecessor); B: 2 days (none); C: 3 days (A); D: 2 days (B); E: 5 days (C); F: 4 days (C and D); G: 2 days (F). If all other activities start as early as possible, what is the maximum number of days by which the start of activity B can be delayed without extending the total project duration?

  1. A0 days
  2. B1 day
  3. C3 days
Show answer and explanation

Answer: C. 3 days

The earliest start is day 4 for C, day 7 for E and F (for F, the later of day 7 when C ends and day 4 when D ends) and day 11 for G, so the whole project takes A → C → F → G = 4 + 3 + 4 + 2 = 13 days. Working backward for the latest starts gives day 11 for G, day 7 for F, 7 − 2 = day 5 for D and 5 − 2 = day 3 for B, so B's total float is 3 − 0 = 3 days. Delaying B also delays the start of D, but D has the same float, so up to 3 days does not extend the project. 0 days is B's free float (the delay possible without delaying the earliest start of its successor D), and 1 day is E's float. In the critical path method (CPM), the floats are found in this way, and the path connecting activities with zero float is the critical path.

A project has the following activities, durations and predecessors: A: 6 days (none); B: 4 days (none); C: 4 days (A); D: 4 days (A and B); E: 2 days (C); F: 3 days (D), and its total duration is 13 days. The extra cost of shortening each activity by one day is ¥300,000 for A, ¥50,000 for B, ¥100,000 for C, ¥200,000 for D, ¥80,000 for E and ¥250,000 for F. Which is the cheapest way to shorten the total project duration by one day?

  1. AShorten activity B by 1 day (¥50,000)
  2. BShorten activity D by 1 day (¥200,000)
  3. CShorten activity A by 1 day (¥300,000)
Show answer and explanation

Answer: B. Shorten activity D by 1 day (¥200,000)

The critical path is A → D → F (6 + 4 + 3 = 13 days); the other paths are A → C → E at 12 days and B → D → F at 11 days. To shorten the project, an activity on the critical path must be shortened, and the cheapest of them is D at ¥200,000, after which A → D → F and A → C → E both take 12 days. B is the cheapest of all, but it is not on the critical path, so shortening it leaves the project at 13 days. Shortening A also gives 12 days, but it costs ¥300,000. Shortening activities by spending money in this way is called crashing.

FAQ

What is the difference between total float and free float?

Total float is the number of days a task can be delayed without delaying the whole project. Free float is the number of days it can be delayed without delaying the earliest start of the next task, and it is never larger than the total float. Using up the total float also reduces the float of later tasks.

Can there be two or more critical paths?

Yes. If several chains share the longest total duration, all of them are critical paths. In that case, all the critical paths must be shortened to shorten the whole project.

What if the durations are uncertain?

PERT estimates three values, an optimistic value a, a most likely value m and a pessimistic value b, and uses (a + 4m + b) ÷ 6 as the expected value. Calculate the expected value of each task before entering it into this tool.

How does this appear in exams?

Japan's Fundamental and Applied Information Technology Engineer Examinations often ask you to find the total project duration, the critical path, or a task's earliest start, latest start and float from an arrow diagram, or how many days the whole project is shortened when a task is shortened.

Related terms

Critical path method (CPM)

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