Module 6 · Networks and Critical Paths

Forward & Backward Scan

Earliest start times · Latest start times · Scanning a network

MST-12-S2-07 MST-12-S2-07 Lesson 8 of 12
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Orient and prepare

Recall the scan problem, then organise the essential EST and LST language.

Worksheets

Practise this lesson

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

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Think First

In a building project, Activity A (foundations, 4 days) must finish before Activities B (framing, 3 days) and C (plumbing, 5 days) can start. B and C both feed into D (roofing, 2 days).

Before reading on, what is the earliest day roofing (D) can start? And: what is the latest day framing (B) can start without delaying the final completion date?

Learning Intentions

  • Define Earliest Start Time (EST) and Latest Start Time (LST) for each activity
  • Apply the forward scan to find EST for every node, working left to right
  • Apply the backward scan to find LST for every node, working right to left
  • Record EST and LST in split node boxes on a network diagram
  • Interpret which activities are on the critical path using EST and LST values

Key Terms

EST (Earliest Start Time)
The earliest a node can be reached, the maximum of all incoming path totals at that node.
LST (Latest Start Time)
The latest a node can be reached without delaying the project, found by working backwards from the final node.
Forward scan
Working left to right through the network, calculating EST at each node by taking the maximum path total from the start.
Backward scan
Working right to left, calculating LST at each node by taking the minimum of (successor LST − activity duration).
Node box
A split box on each node showing EST (left) and LST (right). On the critical path, EST = LST.
Float
The slack in a non-critical activity: LST − EST. (Studied in detail in Lesson 9.)
2

Run a forward scan

Work left to right and take the maximum when paths meet.

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Earliest Start Time (EST), Forward Scan

The Earliest Start Time (EST) at any node is the earliest moment (in days, weeks, etc.) that the work after that node can begin. You find it by running a forward scan moving left to right through the network.

Forward scan rules

  1. Set EST = 0 at the start node.
  2. For each subsequent node, add each incoming activity's duration to the EST of the node it came from.
  3. If two or more paths arrive at the same node, take the maximum you must wait for all predecessors to finish.
  4. The EST of the final node is the minimum project duration (same as the critical path length found in Lesson 7).
Why maximum? All predecessors must be complete before the next activity can start. If path 1 arrives at day 5 and path 2 at day 8, you cannot start until day 8.

Worked example, forward scan

Network: activities sit on the arrows and the numbered nodes are the events. Node 1 → A(3) → node 2 → B(4) → node 3; node 1 → C(5) → node 3.

Worked Example 1, Four-node forward scan

Network: three nodes, three activities. Activity names sit on the arrows; the numbers 1, 2, 3 are the event nodes.

  • Path 1: node 1 —A(3)→ node 2 —B(4)→ node 3
  • Path 2: node 1 —C(5)→ node 3

Step 1, EST₁ = 0 (start node)

Step 2, EST₂ = EST₁ + dur(A) = 0 + 3 = 3

Step 3, EST₃: two arrows arrive. Via B: EST₂ + dur(B) = 3 + 4 = 7. Via C: EST₁ + dur(C) = 0 + 5 = 5.

Step 4, take the maximum: EST₃ = max(7, 5) = 7

Minimum project duration = 7 days.

EST (Earliest Start Time) at each node is found by the forward scan: set EST = 0 at start, then at each node take the maximum of (predecessor EST + activity duration). The final node's EST is the minimum project duration.

Pause, copy the forward scan rule: EST at each node = maximum of (predecessor EST + activity duration), and explain why maximum is used, you can only start when ALL predecessors have finished into your book.

When two paths arrive at the same node during a forward scan, you take the _____ of the incoming path totals.
3

Run a backward scan

Work right to left and take the minimum when paths split.

2

Latest Start Time (LST), Backward Scan

The forward scan sets EST = 0 at the start node and works left to right, taking the maximum of (predecessor EST + activity duration) at each merge node. This gives the earliest each event can occur. To find the latest each activity can start without pushing the project over time, the backward scan reverses the process: start at the final node with LST = EST, then subtract activity durations right to left, taking the minimum at each burst node.

The Latest Start Time (LST) at any node is the latest the work after that node can begin without pushing out the project completion date. You find it by running a backward scan moving right to left through the network.

Backward scan rules

  1. Set LST = EST at the final node (so the project is not delayed).
  2. For each node, subtract each outgoing activity's duration from the LST of the node it leads to.
  3. If two or more paths leave the same node, take the minimum you must start in time for the most urgent successor.
Why minimum? If one successor needs you by day 5 and another by day 8, you must start by day 5 to avoid delay.
Worked Example 2, Backward scan on same network

Network: the same three nodes as Worked Example 1, with EST₁ = 0, EST₂ = 3, EST₃ = 7.

Step 1, LST₃ = EST₃ = 7 (finish on time, so the final node's LST equals its EST)

Step 2, LST₂ = LST₃ − dur(B) = 7 − 4 = 3 (only one arrow leaves node 2)

Step 3, LST₁: two arrows leave node 1. Via A: LST₂ − dur(A) = 3 − 3 = 0. Via C: LST₃ − dur(C) = 7 − 5 = 2.

Step 4, take the minimum: LST₁ = min(0, 2) = 0

Node boxes: node 1 (0 | 0), node 2 (3 | 3), node 3 (7 | 7).

Watch the two arrows out of node 1. Both of node 1's end nodes have EST = LST, yet activity C still has slack: it could start at 0 and only has to finish by 7, but it takes just 5. Its float is LST₃ − EST₁ − dur(C) = 7 − 0 − 5 = 2 days. Float belongs to an activity (an arrow), so always measure it with that formula, not by subtracting EST from LST at a single node.

LST (Latest Start Time) is found by the backward scan: set LST = EST at the final node, then work right to left subtracting activity durations. When paths diverge backwards, take the minimum of outgoing LST − duration.

Pause, copy the backward scan rule: LST at each node = minimum of (successor LST − activity duration), and explain why minimum is used, you must not push any successor activity past its latest start time into your book.

Which statement does NOT describe the backward scan?
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Record node times

Combine EST and LST values in node boxes and interpret the result.

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Recording EST and LST in Node Boxes

The forward scan gives EST and the backward scan gives LST at every node. Recording both in split boxes (EST on the left, LST on the right) immediately shows which nodes are on the critical path: wherever EST = LST, there is zero float, that node must occur at exactly that time or the project is delayed. The difference LST − EST at any other node is the float available for those activities.

On network diagrams, each node is drawn as a split box:

EST LST 7 9 Node with float = 2
  • Left compartment → EST (found during forward scan)
  • Right compartment → LST (found during backward scan)
  • When EST = LST, the node is on the critical path
  • When LST > EST, the activity leading into this node has float (LST − EST days of slack)

Full worked example, 5 activities

Worked Example 3, Complete forward + backward scan

Precedence table:

Activity Duration (days) Depends on
A2
B4A
C3A
D5B
E2C

The network. Activities sit on the arrows, numbered nodes are the events:

  • node 1 —A(2)→ node 2
  • node 2 —B(4)→ node 3  and  node 2 —C(3)→ node 4
  • node 3 —D(5)→ node 5  and  node 4 —E(2)→ node 5

Forward scan (EST):

  • Node 1 (Start): EST = 0
  • Node 2 (end of A): EST = EST₁ + dur(A) = 0 + 2 = 2
  • Node 3 (end of B): EST = EST₂ + dur(B) = 2 + 4 = 6
  • Node 4 (end of C): EST = EST₂ + dur(C) = 2 + 3 = 5
  • Node 5 (End): two arrows arrive. Via D → EST₃ + dur(D) = 6 + 5 = 11. Via E → EST₄ + dur(E) = 5 + 2 = 7. Take the maximum: EST₅ = max(11, 7) = 11
  • EST(End) = 11 days

Backward scan (LST):

  • Node 5 (End): LST = EST₅ = 11
  • Node 3 (before D): LST = LST₅ − dur(D) = 11 − 5 = 6
  • Node 4 (before E): LST = LST₅ − dur(E) = 11 − 2 = 9
  • Node 2 (before B and C): min(LST₃ − dur(B), LST₄ − dur(C)) = min(6 − 4, 9 − 3) = min(2, 6) = 2
  • Node 1 (Start): LST = LST₂ − dur(A) = 2 − 2 = 0

Node boxes:

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

Critical path: node 1 → A → B → D → node 5 (total = 2 + 4 + 5 = 11 days, which matches EST₅). The other route, A → C → E, takes only 2 + 3 + 2 = 7 days, so activities C and E each carry 4 days of float — but they share it: the two of them together may absorb 4 days of delay, not 4 days each.

Each node box shows EST on the left and LST on the right. When EST = LST, the node is on the critical path (no float). A node with LST > EST has LST − EST days of slack. Activity float is measured on the arrow: float = LST(node the arrow ends at) − EST(node the arrow starts at) − duration, so an activity can have float even when both of its nodes are critical.

Pause, copy the split-box format (EST | LST), the critical-path condition EST = LST at a node, and the activity float formula: float = LST(end node) − EST(start node) − duration (zero on the critical path) into your book.

In a node box, when EST = LST, the node is on the . When LST > EST, the activity has .
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Apply both scans

Complete forward and backward scans on new project networks.

Activities

Activity 1, Forward Scan Practice

A project has the following activities:

Activity Duration Depends on
P3
Q6P
R4P
S2Q, R
  1. Perform a forward scan and find EST for each node.
  2. What is the minimum project duration?
  3. Which path is the critical path?

Activity 2, Forward and Backward Scan

Using the network from Activity 1 (P=3, Q=6, R=4, S=2), now perform a complete backward scan. Fill in the LST for each node and identify which activities have float.

  1. Complete the backward scan to find LST at every node.
  2. Draw split node boxes showing EST | LST for each node.
  3. Which node(s) have LST > EST? What does this mean?
  4. Calculate the float for activity R.
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Show what you can do

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

Multiple Choice

Q1. During a forward scan, when two paths arrive at the same node with path totals of 8 and 11, the EST of that node is:

Q2. The backward scan begins at the:

Q3. A node has LST = 7 and EST = 4. The float on the path through that node is:

Q4. Activity X has duration 5. The node before X has EST = 3 and the node after X has LST = 10. The float of activity X is:

Q5. Which statement about node boxes is correct?

Short Answer

SAQ 1. A project network has activities: A(5), B(3) after A, C(4) after A, D(2) after B and C. Perform a complete forward and backward scan. Show all EST and LST values in a table.

SAQ 2. Explain in your own words why we take the maximum in a forward scan but the minimum in a backward scan when multiple paths meet at a node.

Answers

Show MC Answers

Q1 → D (11) Forward scan takes the maximum of incoming paths.

Q2 → C Backward scan starts at the final node, setting LST equal to EST of that node.

Q3 → C (3) Float = LST − EST = 7 − 4 = 3.

Q4 → B (2) Float = LST(end node) − EST(start node) − duration = 10 − 3 − 5 = 2.

Q5 → D EST = LST means zero float, so the node is on the critical path.

Show SAQ Model Answers

SAQ 1: Forward: Start=0, A finishes at 5, B finishes at 8, C finishes at 9, and D finishes at max(8,9)+2=11. Backward: D has latest start 9. B has latest start 9−3=6 and C has latest start 9−4=5, so A must finish by min(6,5)=5 and start at 0. Activity table: A(EST 0, LST 0), B(EST 5, LST 6), C(EST 5, LST 5), D(EST 9, LST 9). B has 1 day of float. Critical path: Start→A→C→D→End = 5+4+2 = 11.

SAQ 2: Forward scan, at a merge node all predecessors must be complete before work can continue, so we wait for the last one: maximum ensures every required activity is done. Backward scan, at a fork node the activity must start early enough for the most urgent branch to meet its deadline: minimum ensures no deadline is missed.

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Retrieve, reflect and finish

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

Revisit, Think First Answer

Foundations (A=4 days) must finish before framing (B=3) and plumbing (C=5) start. Both B and C must finish before roofing (D=2) starts.

Forward scan: EST(start)=0, EST(after A)=4, EST(after B)=7, EST(after C)=9. EST(End) = max(7+2, 9+2) = max(9, 11) = 11. Roofing (D) can start on day 9 at earliest (after C finishes).

Backward scan: LST(End)=11. LST(before D)=9. LST(after A, B side)=9−3=6. LST(after A, C side)=9−5=4. LST(after A)=min(6,4)=4. LST(start)=4−4=0.

Framing (B) must start by day 6 (LST of node after A via B = 6). It can start as early as day 4 (EST), giving 2 days of float.

Lesson Complete!

You can now perform forward and backward scans on activity networks and record EST/LST values in node boxes.