The Critical Path Method (CPM) and Float | Free Project Management Course

What you'll learn

Draw a simple network diagram from an activity list.
Run the forward pass (early start and finish) and the backward pass (late start and finish).
Calculate total float and free float, and explain the difference.
Identify the critical path and predict which delays move the finish date.
Shorten a schedule by crashing or fast-tracking without wasting money.

The critical path is the longest chain of dependent activities in a project, and it sets the earliest possible finish date. Any delay on it delays the whole project, one day for one day. Activities off the critical path have float (also called slack): time they can slip without moving the finish. The Critical Path Method (CPM) is the calculation that finds both. In this lesson you'll run it step by step on a real-world construction schedule in Dayton, Ohio, then use it to decide where extra money actually buys time.

40workdays: the length of the critical path in this lesson's example

7 of 10activities are critical. Only three have any float.

2passes through the network: forward for early dates, backward for late dates

Draw the network
Forward pass
Backward pass
Calculate float
Compress the schedule

Step 1: the activities and the network

We're back at the garden center addition for Miller Family Hardware in Dayton (the WBS lesson). The contractor's PM has turned the work packages into ten activities, with durations in workdays:

ID	Activity	Duration	Predecessors
A	Permits	5	–
B	Site preparation	3	A
C	Foundation	7	B
D	Order steel and lumber	5	A
E	Deliver and stage materials	3	D
F	Framing and roof	8	C, E
G	Electrical rough-in	6	F
H	Plumbing rough-in	4	F
I	Drywall and finishes	9	G, H
J	Final inspection	2	I

A network diagram draws each activity as a box and each dependency as an arrow. Here it is with the results you're about to calculate:

Orange boxes and arrows form the critical path (A-B-C-F-G-I-J). ES/EF are early start and finish; LS/LF are late start and finish, all in workdays from day 0.

Step 2: the forward pass (early dates)

Go left to right. The first activity starts at day 0. For every other activity:

Early start (ES)  = the LATEST early finish among its predecessors
Early finish (EF) = ES + duration

A: ES 0, EF 0 + 5 = 5.
B and D both follow A, so both start at 5. B finishes at 8, D at 10.
C: 8 + 7 = 15. E: 10 + 3 = 13.
F needs both C and E. It must wait for the later one: ES = max(15, 13) = 15, EF = 23.
G and H start at 23; G finishes at 29, H at 27. I waits for both: max(29, 27) = 29, EF 38.
J: 38 + 2 = 40. The project needs 40 workdays.

1. An activity has three predecessors with early finishes of 12, 9 and 14. What is its early start?

An activity can't start until all its predecessors finish, so you take the largest early finish: max(12, 9, 14) = 14.

Step 3: the backward pass (late dates)

Now go right to left, starting from the finish (40). For every activity:

Late finish (LF) = the EARLIEST late start among its successors
Late start (LS)  = LF − duration

J: LF 40, LS 38. I: LF 38, LS 29.
G and H both feed I, so both have LF 29. G: LS 23. H: LS 29 − 4 = 25.
F feeds G (LS 23) and H (LS 25). Take the smaller: LF = 23, LS = 15.
C and E both feed F, so both have LF 15. C: LS 8. E: LS 12. Then D: LF 12, LS 7. B: LF 8, LS 5.
A feeds B (LS 5) and D (LS 7): LF = 5, LS = 0. Back to day 0, which confirms the math.

Step 4: calculate float

Two kinds of float matter:

Total float = LS − ES  (or LF − EF)
  How long an activity can slip without delaying the PROJECT.

Free float  = earliest ES of its successors − its EF
  How long it can slip without delaying ANY successor.

Activity	ES	EF	LS	LF	Total float	Free float
A Permits	0	5	0	5	0	0
B Site prep	5	8	5	8	0	0
C Foundation	8	15	8	15	0	0
D Order materials	5	10	7	12	2	0
E Deliver and stage	10	13	12	15	2	2
F Framing and roof	15	23	15	23	0	0
G Electrical	23	29	23	29	0	0
H Plumbing	23	27	25	29	2	2
I Drywall and finishes	29	38	29	38	0	0
J Final inspection	38	40	38	40	0	0

The activities with zero total float form the critical path: A-B-C-F-G-I-J. Now look at D. It has 2 days of total float but zero free float. If ordering slips one day, delivery (E) is pushed one day too, but the project still finishes on day 40. D and E share the same 2 days: if D uses one, only one is left for E. That's why you should never promise the same float to two people on the same chain.

Another way to see the critical path is to add up every path through the network:

Length of each path (workdays)

A-B-C-F-G-I-J40

A-B-C-F-H-I-J38

A-D-E-F-G-I-J38

A-D-E-F-H-I-J36

The longest path (40) is critical. The two 38-day paths are "near-critical": a 2-day slip on either one makes it critical too. Watch near-critical paths almost as closely as the critical one.

2. The steel delivery (E, total float 2) arrives 3 days late. What happens to the project finish?

Nothing: E isn't on the critical path
It moves from day 40 to day 41
It moves from day 40 to day 43

E can absorb 2 days of delay. The third day pushes F, which is critical, so the project slips by 3 − 2 = 1 day, to day 41. A-D-E-F-G-I-J becomes the new critical path.

3. The plumber (H) needs one extra day. Does the project finish later?

No. H has 2 days of total float, so one day is absorbed.
Yes, by one day
Yes, by two days

H's total float is LS − ES = 25 − 23 = 2. A one-day slip leaves one day of float and the finish stays at day 40.

Step 5: compress the schedule

The owner wants to open 3 workdays earlier to catch a spring promotion. You have two tools:

CrashingAdd resources to critical activities

How: Overtime, an extra crew, rented equipment.
Cost: More money, usually per day saved.
Rule: Only crash critical activities, cheapest first.

Trades money for time

Fast-trackingOverlap activities planned in sequence

How: Start framing on one side while the foundation cures on the other.
Cost: More risk and possible rework.
Rule: Only overlap where the dependency is soft, not physical.

Trades risk for time

The contractor prices the options:

Activity	Critical?	Max days saved	Cost per day saved
B Site prep (second excavator)	Yes	1	$800
C Foundation (extra crew)	Yes	2	$1,200
E Deliver and stage (expedited freight)	No	2	$400
G Electrical (weekend overtime)	Yes	2	$700
I Drywall and finishes (second crew)	Yes	3	$1,000

The cheapest line, expedited freight for E, is a trap: E isn't critical, so paying $400 a day saves nothing. Here's the right sequence:

Crash G by 2 days ($700 × 2 = $1,400). The finish moves from 40 to 38. But now H (plumbing) has zero float: G and H are both critical.
For the third day, G can't go further. The cheapest remaining critical option is B for 1 day ($800). The finish moves to 37, and D-E's float drops from 2 to 1.

Total cost: $1,400 + $800 = $2,200 for 3 workdays. The owner can now decide whether opening 3 days earlier is worth $2,200. That's a business decision, not a scheduling one.

Recalculate after every step

Each time you shorten the critical path, another path may become critical, as H did here. If you crash blindly, you can pay for days that never show up in the finish date.

4. You need to save one day. Which option makes sense?

Pay $400 for faster delivery on a non-critical activity
Pay $700 to shorten a critical activity by one day
Pay $1,200 to shorten a critical activity by one day

Shortening a non-critical activity doesn't move the finish date at all. Between the two critical options, choose the cheapest one: $700.

Common mistakes

Assuming the critical path is the list of "most important" tasks.

It's the longest chain of dependencies, calculated, not voted on.

Ignoring near-critical paths.

Track any path within a few days of critical; it can take over quickly.

Giving every task owner "their" float.

Float belongs to the chain; manage it centrally.

Crashing whatever is cheapest.

Only critical activities, cheapest first, recalculating each time.

Practice exercise

A software team in Raleigh, North Carolina, plans a small release with five activities (durations in workdays):

Activity	Duration	Predecessors
P Requirements	4	–
Q Build the API	6	P
R Design the screens	3	P
S Build the screens	5	R
T Integration testing	2	Q, S

(a) Run the forward and backward passes. (b) What's the critical path and how long is it? (c) How much float does Q have? (d) If R takes one extra day, what's the new finish?

See solution

(a) Forward: P 0–4; Q 4–10; R 4–7; S 7–12; T starts at max(10, 12) = 12 and ends at 14. Backward: T LS 12; S LF 12, LS 7; R LF 7, LS 4; Q LF 12, LS 6; P LF = min(6, 4) = 4, LS 0.

(b) Paths: P-Q-T = 4 + 6 + 2 = 12; P-R-S-T = 4 + 3 + 5 + 2 = 14. The critical path is P-R-S-T, 14 workdays.

(c) Q: total float = LS − ES = 6 − 4 = 2 days. Its free float is also 2 (T's ES of 12 minus Q's EF of 10).

(d) R is critical, so one extra day moves the finish one day: 15 workdays.

Summary

The critical path is the longest chain of dependent activities; it sets the earliest finish.
Forward pass: ES = latest EF of predecessors. Backward pass: LF = earliest LS of successors.
Total float (LS − ES) protects the project date; free float protects the next activity.
Activities on the same chain share float. Watch near-critical paths.
Crash only critical activities, cheapest first, and recalculate after each step; fast-tracking trades risk for time.

Next lesson: Cost Estimating and the Project Budget. You'll estimate costs bottom-up, add contingency and management reserves, and build the cost baseline.