Scheduling and the Critical Path: Lesson, Examples and Practice

1Learning goal: what you will be able to do

A budget tells you what a project costs. A schedule tells you when it will be done, and which delays you can afford. By the end of this lesson you will be able to turn a list of tasks into a network, calculate the earliest finish date, and find the one chain of tasks that decides it, the critical path.

Time55 to 60 minutes

You needPen, paper and a calculator

AssumedAddition and finding the larger of two numbers. Helpful but not required: the previous lesson on scope and work breakdown

Learning outcomes

FoundationDefine an activity, a dependency, a milestone and a schedule network, and read a simple network diagram.
CoreCarry out a forward pass and a backward pass to find earliest and latest start and finish times for every activity.
CoreCalculate total float, identify the critical path, and explain why delays on it delay the whole project.
ExtensionChoose between crashing and fast-tracking to shorten a schedule, and explain how risk and uncertainty affect a deadline.

Where this lesson fits

This is the second lesson of Module 2 in Project Management Essentials. It builds on the work breakdown structure from the first lesson: once you know what the work packages are, scheduling decides in what order and by when. The other lessons shown are planned and are not yet published.

Module 1, What a project is and how it is organisedPlanned
Module 2, Lesson 1: Defining project scope and managing riskOpen
Module 2, Lesson 2: Scheduling and the critical pathThis lesson
Module 3, Budgeting and tracking costPlanned
Module 4, Leading the team and reporting progressPlanned

Where this topic appears in practice

The critical path method was developed in the late 1950s and is now a standard part of the major project management frameworks, including the PMI Guide to the Project Management Body of Knowledge (PMBOK Guide), PRINCE2, and the ISO standard on project management (ISO 21502). Planning software such as Microsoft Project and Primavera calculates it automatically, and certification syllabuses test it by hand. Vocabulary differs a little between sources, so use the terms your own organisation or syllabus uses. This lesson is general education, supports and does not replace a certification course, and does not award any certificate.

2A familiar example: cooking a roast dinner

You are cooking a roast dinner and the guests arrive at 7 pm. You want everything hot at the same moment. Think about the tasks: prepare the vegetables (20 minutes), roast the chicken (75 minutes), roast the potatoes (45 minutes, after they are peeled and parboiled, which takes 25), make the gravy (10 minutes, using juices from the chicken), and lay the table (15 minutes).

Some tasks can happen at the same time. You can lay the table while the chicken roasts. Others have to wait: the gravy cannot start until the chicken has produced juices. Some tasks have slack. Laying the table takes 15 minutes, but the chicken takes 75, so you could start laying it up to an hour late and the dinner would not be delayed.

Predict first: which task decides when you must start cooking?

Every task matters to the meal, but only one chain of tasks sets the start time. Decide which before you open this.

The chicken, then the gravy. The chicken (75 minutes) followed by the gravy (10 minutes) is the longest chain: 85 minutes. The potatoes chain (25 + 45 = 70 minutes) is shorter, and the table (15 minutes) far shorter. So the dinner is ready 85 minutes after you start, and any delay to the chicken or the gravy delays the meal minute for minute. A delay of ten minutes in laying the table delays nothing. That longest chain is the critical path, and the spare time on the other tasks is called float.

You have just done the central job of scheduling: you found the longest chain of dependent tasks, which sets the earliest finish, and you noticed which tasks have room to move. Project managers do exactly this with dozens or hundreds of tasks. Try the explorer below, then read the method.

Try it: the critical path explorer

This is a small building job with six activities. Move a slider to change a duration in days and watch the earliest and latest times, the float and the critical path update. Activities in red are critical. Try lengthening C (prepare site) and see when it becomes critical.

Schedule table
Activity	Days	ES	EF	LS	LF	Float

ES and EF are the earliest start and finish. LS and LF are the latest start and finish that do not delay the project. The thin line after a blue bar shows its float.

3Clear explanation

Activities, dependencies and milestones

Start from the work packages of your work breakdown structure. Each one becomes one or more activities: pieces of work with a duration. A milestone is a point in time with no duration, such as "design approved". A dependency says one activity cannot start or finish until another has. The common type is finish-to-start: B cannot start until A finishes. Others exist (start-to-start, finish-to-finish), but this lesson uses finish-to-start throughout, which covers most real cases and every exam-style problem here.

Drawing the activities as boxes joined by arrows gives a schedule network diagram. An activity with no predecessor starts the project. An activity with no successor ends it. Where two arrows join, the activity must wait for both.

Estimating durations

Durations are estimates, and they are usually wrong in a predictable direction: people forget interruptions and rework. Ask the people who will do the work, estimate from similar past tasks, and state assumptions. For uncertain tasks, three-point estimating gives a more honest figure. With an optimistic time O, a most likely time M and a pessimistic time P, the PERT estimate is

E = (O + 4M + P) / 6

It weights the most likely value but is pulled towards the pessimistic one when the downside is larger.

The forward pass: earliest times

Work from the start of the network to the end.

An activity with no predecessor has earliest start ES = 0.
Otherwise ES is the largest earliest finish among its predecessors, because it must wait for the slowest one.
Earliest finish EF = ES + duration.
The project duration is the largest EF of all the activities at the end.

The backward pass: latest times

Now work from the end back to the start, to find how late each activity can be without delaying the project.

For the final activities, latest finish LF = the project duration.
Otherwise LF is the smallest latest start among its successors, because it must finish in time for the most demanding one.
Latest start LS = LF − duration.

Float and the critical path

Total float = LS − ES = LF − EF

Total float (or slack) is how long an activity can slip without delaying the project end date. An activity with zero float is critical. The critical path is the chain of critical activities running from start to finish; it is the longest path through the network, and its length equals the project duration. Three facts follow:

Any delay to a critical activity delays the project, day for day.
Speeding up a non-critical activity does not shorten the project.
A network can have more than one critical path, and non-critical activities can become critical if they use up their float.

Shortening a schedule

If the date is too late, only changes on the critical path help. Two standard techniques:

Technique	What it does	Typical cost
Crashing	Adds resources (people, overtime, equipment) to critical activities	More money, with diminishing returns
Fast-tracking	Runs activities in parallel that were planned in sequence	More risk of rework and mistakes

Crash the activity that saves a day at the lowest extra cost, and recheck the critical path after each change, because it can switch to a different chain. Each choice is a trade-off, a form of opportunity cost: what you spend to save time cannot be spent elsewhere.

From plan to Gantt chart

A Gantt chart shows each activity as a bar on a calendar, as in the explorer above. It is the easiest view to share with a team or a client. The network diagram shows the logic and the Gantt chart shows the timing; good planning software shows both from the same data. A schedule is only a plan: compare actual progress with it at every review and update it. The lesson on risk showed that a risk register belongs beside the schedule, because most delays are risks that were never priced.

Common mistakes

Mistake	Why it hurts	Fix
Adding durations of parallel tasks	Overstates the project length	Only tasks in sequence add; for parallel paths take the longest
Using the smaller EF in the forward pass	Starts an activity before a predecessor is done	Forward pass: take the largest EF
Using the larger LS in the backward pass	Lets a predecessor finish too late for a successor	Backward pass: take the smallest LS
Forgetting a dependency	The schedule looks faster than it can be	Ask for every activity: what must be finished first?
Assuming float belongs to you	Float is shared along a chain; using it all makes the path critical	Treat float as a reserve to be managed, and agree who may use it
Never updating the schedule	The critical path moves and nobody notices	Recalculate at every review

4Worked examples

Read each example with a pencil in hand. Cover the steps, draw the network yourself, then compare. Examples 2 to 5 use the same six-activity project as the explorer: A (3 days), B (4, after A), C (2, after A), D (5, after B), E (3, after B and C) and F (4, after D and E).

Example 1 Foundation 3 marks: the longest path

A project has four activities: P (4 days, no predecessor), Q (3 days, after P), R (5 days, after P) and S (2 days, after Q and R). Find every path, the project duration and the critical path.

Path P → Q → S: 4 + 3 + 2 = 9 days.
Path P → R → S: 4 + 5 + 2 = 11 days.
The project cannot finish until the longer path is complete, so the duration is 11 days and the critical path is P → R → S.

Answer: 11 days, critical path P → R → S. Q is not critical: it has 5 − 3 = 2 days of float.

Example 2 Core 5 marks: the forward pass

Find the earliest start and finish of every activity in the six-activity project, and the project duration.

A has no predecessor: ES = 0, EF = 0 + 3 = 3.
B and C both follow A: B has ES = 3, EF = 7; C has ES = 3, EF = 5.
D follows B: ES = 7, EF = 12.
E follows B and C, so ES = max(7, 5) = 7, EF = 10.
F follows D and E, so ES = max(12, 10) = 12, EF = 16.

Answer: the project duration is 16 days (EF of F). Notice E is held back by B, the slower predecessor.

Example 3 Core 6 marks: the backward pass, float and the critical path

Continue Example 2: find the latest times, the float of each activity and the critical path.

F is last: LF = 16, LS = 16 − 4 = 12.
D and E both feed F: LF = 12 for each. D: LS = 12 − 5 = 7. E: LS = 12 − 3 = 9.
B feeds D (LS 7) and E (LS 9), so LF = min(7, 9) = 7, LS = 7 − 4 = 3.
C feeds E only: LF = 9, LS = 9 − 2 = 7.
A feeds B (LS 3) and C (LS 7): LF = min(3, 7) = 3, LS = 0.
Float = LS − ES: A 0, B 0, C 7 − 3 = 4, D 0, E 9 − 7 = 2, F 0.

Answer: the zero-float activities are A, B, D and F, so the critical path is A → B → D → F (3 + 4 + 5 + 4 = 16 days). C has 4 days of float and E has 2.

Example 4 Core 4 marks: what a delay does

Using the results above, what happens to the finish date if (a) C takes 3 days longer, (b) C takes 6 days longer, (c) E takes 3 days longer?

(a) C has 4 days of float and uses 3, so the project still finishes on day 16.
(b) C uses its 4 days and exceeds it by 2, so the project is delayed by 2 days, to day 18.
(c) E has 2 days of float and overruns by 1, so the project is delayed by 1 day, to day 17.

Answer: (a) no delay, (b) 18 days, (c) 17 days. The rule: a delay moves the end date only by the amount that exceeds the activity’s float. Check (b) with the explorer.

Example 5 Core 5 marks: crashing the schedule

The sponsor wants the project done in 14 days. Crashing D (build structure) costs an extra $800 per day saved, and D can be shortened by at most 3 days. Is crashing D enough, and what does it cost?

D is critical, so shortening it can help. Crash D by 2 days: D = 3 days.
Recalculate. A → B → D → F = 3 + 4 + 3 + 4 = 14. But A → B → E → F = 3 + 4 + 3 + 4 = 14 as well.
Both paths are now critical, so the project is 14 days. Cost: 2 × $800 = $1,600.
Crashing D by a third day would give no gain: the path through E is still 14 days.

Answer: yes, crash D by 2 days for $1,600. Past this point you must shorten activities on both critical paths (for example B or F, which are on both), so the next saving costs more. Always recalculate after each change.

Example 6 Extension 4 marks: when the critical path moves

Suppose C (prepare site) takes 8 days instead of 2. Recalculate the project duration and the critical path.

A: EF = 3. C: ES = 3, EF = 11. B: ES = 3, EF = 7. D: ES = 7, EF = 12.
E follows B and C: ES = max(7, 11) = 11, EF = 14.
F follows D and E: ES = max(12, 14) = 14, EF = 18.
The longest path is A → C → E → F = 3 + 8 + 3 + 4 = 18.

Answer: 18 days, with critical path A → C → E → F. Activities B and D, once critical, now have float. The critical path is not fixed; it must be recalculated whenever durations change.

Example 7 Extension 3 marks: a three-point estimate

For an uncertain task the team estimates an optimistic time of 4 days, a most likely time of 6 days and a pessimistic time of 14 days. What duration should go in the schedule?

E = (O + 4M + P)/6 = (4 + 4 × 6 + 14)/6.
= (4 + 24 + 14)/6 = 42/6 = 7.

Answer: 7 days. The pessimistic tail pulls the estimate above the most likely value of 6. Tasks with large downside risk are good candidates for the risk register.

6Answer explanations

Attempt every question before you open these. A right answer with a wrong method is luck, so read each explanation even when you were correct. Questions 3 and 6 show their own worked solution as soon as you answer.

Question 1: the critical path

Answer: the longest path through the network, whose activities have zero float.

Cost (a), number of activities (b) and risk score (d) do not decide the end date. Duration does. The longest chain of dependent activities is the shortest time in which the project can possibly finish.

Question 2: parallel paths

Path A → B → D = 5 + 7 + 3 = 15. Path A → C → D = 5 + 9 + 3 = 17. The project takes the longer: 17 days.

The slip to avoid is adding all four durations (24), which treats parallel work as if it were done one task after another.

Question 3: generated earliest finish

Take the later of the two predecessor finishes as the earliest start, then add the duration. Use New numbers to practise until the rule is automatic.

Question 4: latest start

LS = LF − duration = 20 − 6 = 14. If the activity starts any later than day 14 it cannot finish by day 20 and will delay its successors.

Question 5: the forward pass rule

Answer: the larger of the two earliest finishes.

The activity needs both predecessors finished, so it must wait for the slower one. Taking the smaller (a) would start it before a predecessor is done. Sums and averages (c, d) have no meaning in the method.

Question 6: generated float

Find EF = ES + duration, then float = LF − EF. Zero means critical. Use New numbers to practise, including the zero-float cases.

Question 7: speeding up a non-critical activity

Answer: none; the end date is unchanged.

The end date is set by the critical path. An activity with 5 days of float is already finishing earlier than the project needs, so making it faster only adds cost. Effort to save time belongs on critical activities.

Question 8: the PERT estimate

E = (2 + 4 × 5 + 14)/6 = (2 + 20 + 14)/6 = 36/6 = 6. The long pessimistic tail lifts the estimate above the most likely value of 5.

Question 9: fast-tracking

Answer: running activities in parallel that were planned in sequence, saving time but raising the risk of rework.

Adding people (a) is crashing. Removing scope (b) shortens the schedule by changing the deliverable and needs a change request. Cutting every duration by a percentage (c) is not a technique, because only critical activities help.

7Short recap

A schedule turns work packages into activities with durations and dependencies, drawn as a network.
Forward pass: ES is the largest predecessor EF; EF = ES + duration. The largest EF at the end is the project duration.
Backward pass: LF is the smallest successor LS; LS = LF − duration.
Total float = LS − ES = LF − EF. Zero float means critical.
The critical path is the longest path. A delay to a critical activity delays the project day for day; a delay to another activity matters only by the amount it exceeds its float.
To shorten a schedule, work on critical activities: crash them (more resources, more cost) or fast-track them (parallel work, more risk). Recalculate after each change.
Estimates are uncertain. The PERT formula (O + 4M + P)/6 gives a considered estimate, and large downsides belong in the risk register.
A schedule must be updated at every review, because the critical path can move.

Tick what you can now do

I can define activities, dependencies and milestones, and read a simple network diagram.
I can carry out a forward pass and a backward pass.
I can calculate float, find the critical path and predict the effect of a delay.
I can compare crashing and fast-tracking and calculate a three-point estimate.

If you ticked fewer than four, return to the matching worked example, redo it without looking, then try the generated questions again. Next, take a project you know, list ten activities and find its critical path by hand.

Related lessons and Encyclopedia entries

Defining Project Scope and Managing Risk: the previous lesson, which produces the work breakdown structure this schedule starts from.
Critical path method: the Encyclopedia entry on this technique.
Risk register, Risk scoring and Risk assessment matrix: record and rank the threats to your dates.
Stakeholder engagement: who needs to hear about a slipping critical path, and when.
Opportunity cost: the trade-off behind every decision to crash a schedule.
Agile software development: an alternative approach to planning when the work cannot be fully known in advance.

References and further reading

Titles and publishers only. Search each site for the title to reach the exact page.

Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK Guide), the sections on schedule management. https://www.pmi.org/
AXELOS, Managing Successful Projects with PRINCE2, the chapters on plans and progress. https://www.axelos.com/
International Organization for Standardization, ISO 21502: Project, programme and portfolio management, guidance on project management. https://www.iso.org/
Kerzner, H., Project Management: A Systems Approach to Planning, Scheduling, and Controlling. Wiley.