Module 6 · Networks and Critical Paths

Critical Path, Float & Scheduling

Identifying the critical path · Float calculation · Scheduling non-critical activities

MST-12-S2-07 MST-12-S2-07 Lesson 9 of 12
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Orient and prepare

Recall what matching EST and LST values reveal about a project.

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Think First

A project has 6 activities. After completing the forward and backward scans, you find that 3 nodes have EST = LST and the other 2 nodes have LST greater than EST.

Before reading on, how would you identify which activities are on the critical path? And what does it mean for a project manager if an activity has a lot of "float"?

Learning Intentions

  • Identify the critical path by locating nodes where EST = LST
  • Calculate float for any activity using the formula: float = LST(end node) − EST(start node) − duration
  • Interpret float in context: what it means for scheduling flexibility
  • Understand that delaying a critical activity always delays the project
  • Describe how non-critical activities can be shifted within their float without affecting the deadline

Key Terms

Critical path
The longest path through the network; the sequence of activities where every node has EST = LST and every activity has zero float.
Float (slack)
The amount of time an activity can be delayed without delaying the project. Float = LST(end node) − EST(start node) − duration.
Critical activity
An activity on the critical path with zero float. Any delay to a critical activity delays the whole project.
Non-critical activity
An activity with positive float. It has scheduling flexibility within its float window.
Scheduling
Planning when non-critical activities will actually start, within the window defined by their EST and LST.
Project duration
The minimum time to complete the project, equal to the length of the critical path (EST of the final node).
2

Identify the critical path

Use completed scans to locate the activities that control project duration.

1

Identifying the Critical Path

Once you have completed both forward and backward scans, identifying the critical path is straightforward:

Critical path rule: The critical path passes through every node where EST = LST. Activities connecting these nodes are critical activities with zero float.

The two-step identification method

  1. Mark critical nodes: Highlight every node where EST = LST on the network diagram.
  2. Trace the path: Follow connected critical nodes from start to finish. The path between them (using activities with zero float) is the critical path.
Exam tip: Always verify by summing the durations of activities on your critical path. The total must equal the EST of the final node (minimum project duration).
Worked Example 1, Identifying critical path from node boxes

Node box data (from Lesson 8 Worked Example 3):

Node EST LST EST = LST?
1 (Start)00Yes ✓
2 (after A)22Yes ✓
3 (after B)66Yes ✓
4 (after C)59No (float)
5 (End)1111Yes ✓

Critical nodes: 1, 2, 3, 5 (node 4 is non-critical).

Critical path: Start → A → B → D → End

Verification: A(2) + B(4) + D(5) = 11 = EST(End). ✓

Activity C (connecting node 2→4) has float and is NOT on the critical path.

Float (or slack) = LST − EST at a node, or for an activity: LST(end node) − EST(start node) − duration. An activity with zero float is on the critical path. Float shows how much delay is tolerable before the project is affected.

Pause, copy the activity float formula: float = LST(end node) − EST(start node) − duration, and the rule: float = 0 means the activity is on the critical path into your book.

A node has EST = 8 and LST = 8. This node is:
3

Calculate float

Measure how long a non-critical activity may be delayed.

2

Calculating Float

Float = LST − EST (or for an activity: LST of end node − EST of start node − duration) tells you how much an activity can be delayed without affecting the project. Activities with zero float are on the critical path. If you need to finish the project faster, you must shorten at least one critical-path activity, shortening a non-critical activity changes its float but has absolutely no effect on project duration.

Float (also called slack) is the maximum time an activity can be delayed without causing any delay to the project's minimum completion time.

Float formula: Float = LST(end node) − EST(start node) − duration of activity

This formula tells you: given that the activity can start as early as EST(start node) and the end node can be reached as late as LST(end node), how many days of spare time exist after accounting for the activity's own duration.

Special cases

  • Float = 0: Critical activity. No delay is possible.
  • Float > 0: Non-critical activity. Can be started later or extended (up to its float value).
  • Float never negative in a correctly solved network (if you get a negative, recheck your scans).
Worked Example 2, Computing float for every activity

Network: Start(0,0) → A(2) → Node2(2,2) → B(4) → Node3(6,6) → D(5) → End(11,11)

Also: Node2(2,2) → C(3) → Node4(5,9) → E(2) → End(11,11)

Activity Dur EST(start) LST(end) Float Critical?
A2020Yes
B4260Yes
C3294No
D56110Yes
E25114No

Interpretation: Activity C has 4 days float, it can start any time between day 2 (EST) and day 6 (= 9−3), as long as it finishes by day 9 (LST of end node). Activity E similarly has 4 days slack (noting C and E share the non-critical path, so using C's full float reduces E's available float).

Reducing project duration requires shortening at least one activity on the critical path (crashing). Shortening a non-critical activity has no effect on project completion time, only critical activities matter.

Pause, copy the crashing rule: reducing project duration requires shortening a critical-path activity; shortening any non-critical activity has no effect on the project completion time into your book.

Which statement does NOT correctly describe float?
4

Schedule with float

Use float to make practical scheduling and crashing decisions.

3

Scheduling with Float

Crashing, deliberately reducing a critical activity's duration, shortens the project, but only if that activity remains on the longest path after the reduction. Shortening one critical activity can make a previously non-critical path become the new longest path, changing which activities now have zero float. Always recalculate EST, LST, and float after every crashing decision.

Knowing float allows project managers to schedule non-critical activities flexibly for example, to spread worker load, reduce costs, or work around public holidays.

Scheduling window for a non-critical activity

A non-critical activity X with duration d, where the node before X has EST = e and the node after X has LST = L:

  • Earliest start: day e
  • Latest start: day L − d
  • Latest finish: day L
  • Float: L − e − d
Important constraint: Float is shared along a path. If activities C and E are both non-critical on the same path (C→E), using C's full float removes E's float. Always track float at the path level, not just per activity.
Worked Example 3, Scheduling a non-critical activity

Activity C (duration 3, float 4) runs between Node2 (EST=2) and Node4 (LST=9).

Scheduling window: Can start any day from day 2 to day 6 (= 9 − 3).

If workers are occupied on days 2–4 (critical activities A and B), the manager can schedule C to start on day 5 (within the window), finishing on day 8, which is before LST=9. Project is not delayed.

Decision: If C starts on day 5 it finishes on day 8, so E can no longer start before day 8. E's end node is End, with LST = 11, so E must still start by day 11 − 2 = 9. Remaining float for E = 9 − 8 = 1 day. (Measure E against the LST of the node it ends at, 11, not the LST of the node it starts at, 9.)

This illustrates that when C uses 3 of its 4 days float (starts day 5 instead of day 2), E has only 1 day remaining (not its original 4). Float is shared on a path.

After crashing a critical activity, the critical path may change, a previously non-critical path may become critical. Always recalculate EST/LST and identify the new critical path after any duration change.

Pause, copy the post-crash procedure: after shortening a critical activity, recalculate EST and LST for every node, re-identify the critical path (the new longest path), and check whether the crashed path is still critical into your book.

Activity M has duration 4 days. The node before M has EST = 3 and the node after M has LST = 11. The float of activity M is days. M can start as late as day .
5

Model the whole project

Bring the forward scan, backward scan and float together.

4

Modelling the whole project in a spreadsheet

Forward scan, backward scan and float are three passes over the same table, and each one is a formula a spreadsheet can hold. Build the columns once and the model recalculates every EST, LST and float the moment a duration changes, which is exactly what you need when a question asks what happens if an activity runs late.

By hand, changing one duration means redoing both scans. In a spreadsheet you change one cell. Start from the precedence table, which is how these projects are given to you in the first place:

ActivityImmediate predecessorsDuration (days)
A–4
B–3
CA5
DA, B2
EC3
FD5
GE, F2

The columns, and the one idea in each

Put the activities in rows 2 to 8, in an order where every activity appears below all of its predecessors. Then each column is a single formula filled down:

  • EFT in column E is =D2+C2 . An activity finishes when it starts plus how long it takes.
  • EST in column D is the latest of its predecessors' finish times, so =MAX(E6,E7) for activity G, and plain 0 for A and B, which wait for nobody. That single MAX is the entire forward scan.
  • LFT in column F is the earliest of its successors' latest starts, so =MIN(G4,G5) for activity A. The last activity is the exception: its LFT is the project duration, =MAX(E2:E8) .
  • LST in column G is =F2-C2 , and float in column H is =G2-D2 .
  • Critical? in column I is =IF(H2=0,"YES","") , which is the definition of a critical activity written as a formula.

Filled down, the model returns:

ActivityDurationESTEFTLSTLFTFloatCritical?
A404040YES
B303252
C549490YES
D246571
E39129120YES
F56117121
G2121412140YES

The project takes 14 days and the critical path is A → C → E → G . Activity B can start up to 2 days late without hurting anything, and D and F have 1 day each, which is the float shared along the path through them.

Three checks that catch a broken model.
  • No float may be negative. A negative float means an LFT was read from the wrong successor, or the project duration was typed in by hand instead of being computed from the EFT column.
  • The YES cells must join up. Critical activities form an unbroken chain from a starting activity to a finishing one. If the YES rows do not connect, a predecessor reference points at the wrong row.
  • Add the critical durations independently. Here 4 + 5 + 3 + 2 = 14, which must equal the largest EFT in the table. Those two numbers are produced by different columns, so agreement is real evidence rather than the same arithmetic twice.
Now solve with it. Change activity F's duration from 5 to 6 and watch the model rebuild itself: F has 1 day of float, so that one extra day is exactly absorbed. F's float falls to zero, the path through D and F becomes critical alongside A → C → E → G, and the project still takes 14 days. Push F to 7 and the project slips to 15 with a single new critical path through D and F. Answering "what if this activity runs late" by editing one cell is exactly what the syllabus means by solving related problems with a spreadsheet.

To model a project in a spreadsheet, list activities so each sits below its predecessors, then use one formula per column: EFT = EST + duration, EST = MAX of predecessors' EFTs, LFT = MIN of successors' LSTs (the last activity's LFT is the project duration), LST = LFT − duration, float = LST − EST, and critical when float = 0. Check that no float is negative and that the critical durations add to the project duration.

Pause, copy the six column formulas in order, and copy the three checks: no negative float, the critical activities form an unbroken chain, and the critical durations must add to the largest EFT.

Quick check: In the spreadsheet above, which formula computes the earliest start time of an activity that has two predecessors?
6

Show what you can do

Complete the fixed drills and short-answer questions, then check each worked answer.

Activities

Activity 1, Critical Path Identification

A project has five activities, each running between two of the events below:

  • P, duration 5: Start → “After P”
  • Q, duration 5: Start → “After Q”
  • R, duration 6: “After P” → “After R”
  • S, duration 3: “After R” → End
  • T, duration 6: “After Q” → End

The forward and backward scans give this node data:

Node EST LST
Start00
After P55
After Q58
After R (from P)1111
End1414
  1. Identify all critical nodes (EST = LST).
  2. State the critical path, naming the activities on it.
  3. Calculate the float for activity Q.
  4. Calculate the float for activity T. You should find that Q and T each have the same float, yet the path through them can absorb only that many days of delay in total, not twice as many. Explain why the two floats cannot be added.

Activity 2, Float and Scheduling

Activity W has duration 6 days. The node before W has EST = 4 and the node after W has LST = 15.

  1. Calculate the float for activity W.
  2. What is the earliest W can start? What is the latest it can start?
  3. If the project manager decides to start W on day 7, on what day will W finish? Is the project delayed?
  4. If there is another activity X (duration 3) after W, and the node after X has LST = 18, and W starts on day 7, what is the remaining float for X?

Multiple Choice

Q1. Activity T has duration 5 days. The node before T has EST = 3, and the node after T has LST = 12. The float of activity T is:

Q2. A critical activity is delayed by 2 days. The project completion time will:

Q3. All nodes on the critical path share which property?

Q4. An activity has EST(start node) = 6, LST(end node) = 14, and duration = 5. Its float is:

Q5. A non-critical activity has float = 6. If a project manager delays this activity by 6 days, the float for subsequent activities on the same path will be:

Short Answer

SAQ 1. A project has activities: A(3), B(5) after A, C(2) after A, D(4) after B and C. After completing forward and backward scans you find: Start(0,0), after A(3,3), after B(8,8), after C(5,8), End(12,12). Identify the critical path and calculate the float for activity C.

SAQ 2. Explain in practical terms what it means for a project if the critical path has no float, and describe one benefit of having non-critical activities with large float values.

Answers

Show MC Answers

Q1 → C (4) Float = LST(end) − EST(start) − duration = 12 − 3 − 5 = 4.

Q2 → C Delaying any critical activity delays the project by exactly that amount.

Q3 → C On the critical path, EST = LST at every node.

Q4 → C (3) Float = 14 − 6 − 5 = 3.

Q5 → B Float is shared along a path; using all 6 days leaves zero float for successors on the same path.

Show SAQ Model Answers

SAQ 1: Critical nodes: Start(0=0)✓, after A(3=3)✓, after B(8=8)✓, End(12=12)✓. After C has EST=5, LST=8, so NOT critical. Critical path: Start → A → B → D → End (3+5+4=12). Float for C: LST(after C) − EST(before C) − dur(C) = 8 − 3 − 2 = 3 days.

SAQ 2: No float on the critical path means any delay to a critical activity directly delays the project completion, there is no buffer. Benefit of large float: project managers can reschedule non-critical workers to other tasks, reduce peak resource demands, plan around external constraints (material delivery, inspections) without risking the overall deadline.

7

Retrieve, reflect and finish

Return to the opening problem, mark the lesson complete and choose the next destination.

Revisit, Think First Answer

The nodes where EST = LST are the critical nodes. The activities connecting them form the critical path. Any activity entering a node where LST > EST has float (LST − EST at that node, adjusted for duration).

For a project manager, large float on an activity is valuable flexibility: that activity can start later, be stretched, or have workers temporarily reassigned without risking the deadline.

Lesson Complete!

You can now identify critical paths, calculate float, and explain the scheduling implications of float to non-critical activities.