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Module 6 · Networks and Critical Paths

Forward & Backward Scan

Earliest start times · Latest start times · Scanning a network

MS12-7 MS-N3 Lesson 8 of 12

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.)
1

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: Start → A(3) → B(4) → End; Start → C(5) → End (C also feeds into End).

Worked Example 1 — Four-node forward scan

Network: Two parallel paths from S to End.

  • Path 1: S → A (dur 3) → B (dur 4) → End
  • Path 2: S → C (dur 5) → End

Step 1 — EST(S) = 0

Step 2 — EST(A) = EST(S) + dur(SA) = 0 + 3 = 3

Step 3 — EST(C) = EST(S) + dur(SC) = 0 + 5 = 5

Step 4 — EST(B) = EST(A) + dur(AB) = 3 + 4 = 7

Step 5 — EST(End): Two paths arrive: via B → 7; via C → 5. Take max: EST(End) = 7

Minimum project duration = 7 days.

Book Notes

When two paths arrive at the same node during a forward scan, you take the _____ of the incoming path totals.
2

Latest Start Time (LST) — Backward Scan

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: EST values found above: S=0, A=3, C=5, B=7, End=7.

Step 1 — LST(End) = EST(End) = 7

Step 2 — LST(B) = LST(End) − dur(B→End) = 7 − 4 = 3

Step 3 — LST(C) = LST(End) − dur(C→End) = 7 − 5 = 2
Wait — but we calculated EST(C) = 5. Check: LST = 2, EST = 5? That cannot be right because LST < EST is impossible in a valid network. Let me re-examine: C has duration 5 and goes directly to End (LST 7). LST(C) = 7 − 5 = 2.

Note: In this network C is NOT on the critical path (it has float = 2 − 0 = 2 days at the start node side; the actual float equals LST(node after C) − EST(node before C) − dur(C) = 7 − 0 − 5 = 2). This is fine — float means slack.

Step 4 — LST(A) = LST(B) − dur(A→B) = 3 − 4 = −1? No — check again. B has duration 4 coming from A. LST(B) is the LST of node B. LST(A) = LST(B) − dur(AB) = 3 − 3 = 0.

Correction with proper labels: Let nodes be numbered 1(S), 2(A), 3(C merge), 4(End). Activity SA has dur 3 (1→2), AB has dur 4 (2→4), SC has dur 5 (1→4). Forward: EST₁=0, EST₂=3, EST₄=max(3+4, 0+5)=max(7,5)=7. Backward: LST₄=7, LST₂=7−4=3, LST₁=min(3−3, 7−5)=min(0,2)=0. Result: Node 1: EST=0,LST=0 ✓; Node 2: EST=3,LST=3 ✓ (critical); Node 4: EST=7,LST=7 ✓. Path S→C→End has float = 7−5−0 = 2 days.

Book Notes

Which statement does NOT describe the backward scan?
3

Recording EST and LST in Node Boxes

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, D

Forward scan (EST):

  • Node 1 (Start): EST = 0
  • Node 2 (after A): EST = 0 + 2 = 2
  • Node 3 (after B, from node 2): EST = 2 + 4 = 6
  • Node 4 (after C, from node 2): EST = 2 + 3 = 5
  • Node 5 (after D and E): paths arriving: via D → 6+5=11; via C → 5+2 is not right — E depends on both C and D. So node 5 gets: max(EST₃+dur(D)=6+5=11, EST₄+dur(C but C feeds E...).
  • Re-mapping: S→A(2)→node2. node2→B(4)→node3. node2→C(3)→node4. node3→D(5)→node5. node4→E but E also needs D — so node5 = max(node3+5, node4+2) = max(11, 7) = 11
  • EST(End) = 11 days

Backward scan (LST):

  • Node 5 (End): LST = 11
  • Node 3 (before D): LST = 11 − 5 = 6
  • Node 4 (before E): LST = 11 − 2 = 9
  • Node 2 (before B and C): min(LST₃−4, LST₄−3) = min(6−4, 9−3) = min(2, 6) = 2
  • Node 1 (Start): LST = 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: Start → A → B → D → End (total = 2+4+5 = 11 days). Activity C has 4 days float.

Book Notes

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

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.

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, after A=5, after B=8, after C=9, End=max(8+2, 9+2)=max(10,11)=11. Backward: End=11, before D (from B side)=11−2=9, before D (from C side)=11−2=9. Before B: 9−3=6. Before C: 9−4=5. Before A: min(6−5, 5−5)=min(1,0)=0. Node table: Start(0,0)✓, after A(5,5)✓, after B(8,6)float2, after C(9,9)✓, End(11,11)✓. 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.

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.