Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
Module 6 · Networks and Critical Paths

Critical Path, Float & Scheduling

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

MS12-7 MS-N3 Lesson 9 of 12

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).
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.

Book Notes

A node has EST = 8 and LST = 8. This node is:
2

Calculating Float

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).

Book Notes

Which statement does NOT correctly describe float?
3

Scheduling with Float

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, remaining float for E (which follows C) = 9 − 8 − 2 = −1. Wait — E must start after C finishes (day 8) but also LST of End = 11, dur(E)=2, so E must start by day 9 at latest. Starting E on day 8 is fine: 8+2=10 < 11. Float remaining for E if C starts day 5 = 1 day.

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.

Book Notes

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 _______.

Activities

Activity 1 — Critical Path Identification

A project network has the following node data after forward and backward scans:

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.
  3. Calculate the float for activity Q (duration 5, runs from Start to "After Q" node).

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 with LST(after X) = 15, 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(5,7), 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=7, 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) = 7 − 3 − 2 = 2 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.

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.