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Module 6 · Networks and Critical Paths

Complex Critical Path Problems

Multiple critical paths · Dummy activities · 7+ activity networks

MS12-7 MS-N3 Lesson 11 of 12

Think First

A project has two parallel paths from start to finish: Path 1 takes 10 days and Path 2 also takes 10 days. Both paths have zero float.

Before reading on — is this project easier or harder to manage than one with a single critical path? What happens if you try to crash one path by 2 days?

Learning Intentions

  • Recognise and handle networks with multiple critical paths
  • Understand dummy activities: what they are, when to use them, and how to draw them
  • Complete forward and backward scans on networks with 7 or more activities
  • Apply crashing logic when multiple critical paths exist
  • Interpret complex CPA results in practical scheduling contexts

Key Terms

Multiple critical paths
When two or more distinct paths through the network have equal length (equal to the project duration), all with zero float.
Dummy activity
An activity with duration zero, shown as a dashed arrow, used to represent a precedence dependency between activities that share some (but not all) predecessors.
Shared predecessor
When activity C depends on both A and B, but D depends only on A. A dummy from the "after A" node to the "after A,B" node handles this correctly.
Crashing (multiple paths)
To reduce project duration when multiple critical paths exist, you must crash at least one activity on EVERY critical path simultaneously.
7+ activity networks
Larger networks requiring systematic node labelling, careful scan progression, and complete float tables.
1

Multiple Critical Paths

In some networks, two or more paths have the same length — equal to the project duration. All such paths are critical, and all have zero float.

Implications of multiple critical paths

  • Any delay on any critical path delays the project.
  • To reduce project duration, you must crash activities on all critical paths simultaneously — crashing only one path does nothing if another remains at the same length.
  • More critical activities = less scheduling flexibility overall.
Identification: After scanning, look for multiple sequences of nodes with EST = LST all connected to the finish. If two such chains both sum to the project duration, both are critical paths.
Worked Example 1 — Network with two critical paths

Activities:

Activity Dur Depends on
A3—
B5A
C4A
D2B
E3C

Forward scan:

  • Start: 0. After A: 3. After B: 8. After C: 7. End: max(8+2, 7+3) = max(10, 10) = 10

Backward scan:

  • End: 10. Before D: 8. Before E: 7. After A: min(8−5, 7−4) = min(3, 3) = 3. Start: 3−3=0.

Node boxes: Start(0,0), After A(3,3), After B(8,8), After C(7,7), End(10,10) — all EST=LST!

Both paths are critical:

  • Path 1: Start → A → B → D → End (3+5+2 = 10)
  • Path 2: Start → A → C → E → End (3+4+3 = 10)

To reduce project to 9 days, you must crash one activity on Path 1 AND one on Path 2 (A is shared — crashing A by 1 day reduces both paths to 9).

Book Notes

A project has two critical paths both of length 12. To reduce the project to 11 days by crashing, you must:
2

Dummy Activities

A dummy activity is a special dashed arrow with duration zero. It represents a logical dependency, not real work. You need a dummy when:

When to use a dummy: Activity C depends on A and B, but activity D depends only on A. Without a dummy, you cannot correctly represent this in the network — a simple network would either make D depend on B (wrong) or lose the C's dependency on B (wrong).

How to draw a dummy

  1. Activity A finishes at node 2. Activity B finishes at node 3.
  2. Activity C (depends on A and B) starts at a new node 4.
  3. Draw: A → node 2 → C(real) → node 4 is wrong if B's node is 3.
  4. Correct: Draw dummy (dashed, dur=0) from node 3 to node 2 (or appropriate merge node), so node 2 now has both A and dummy (from B) as inputs, then C starts at node 2.
  5. D (depends only on A) comes directly from A's end node, before the dummy merges.
Forward/backward scan with dummies: Treat a dummy just like any other activity — duration = 0. EST passes through unchanged. LST passes through unchanged. Dummies can be on the critical path (if their connecting nodes both have EST = LST).
Worked Example 2 — Network with dummy activity

Situation: C requires both A and B. D requires only A.

Network structure:

  • Node 1 (Start) → A(4) → Node 2
  • Node 1 (Start) → B(3) → Node 3
  • Node 3 → dummy(0) → Node 2
  • Node 2 → C(5) → Node 4 (End)
  • Node 2 → D(2) → Node 4 (End)

Forward scan:

  • Node 1: EST = 0
  • Node 3: EST = 0 + 3 = 3
  • Node 2: max(0+4, 3+0) = max(4, 3) = 4
  • Node 4: max(4+5, 4+2) = max(9, 6) = 9

Backward scan:

  • Node 4: LST = 9
  • Node 2: min(9−5, 9−2) = min(4, 7) = 4
  • Node 3: LST = 4 − 0 = 4
  • Node 1: min(4−4, 4−3) = min(0, 1) = 0

Node boxes: Node 1(0,0)✓, Node 3(3,4) float 1, Node 2(4,4)✓, Node 4(9,9)✓

Critical path: Start → A → Node2 → C → End (4+5=9). Activity B has float 1. Activity D has float = 9−4−2=3.

Book Notes

Which statement does NOT correctly describe a dummy activity?
3

Larger Networks (7+ Activities)

HSC questions often involve networks with 7–10 activities. The method is identical to smaller networks; systematic organisation prevents errors.

Strategy for larger networks

  1. Number every node before scanning. Work left to right for node numbers.
  2. Forward scan in strict node order: never process a node until all predecessors are done.
  3. Backward scan in reverse node order: never process a node until all successors are done.
  4. Record EST/LST in a table as you go — don't rely on mental tracking for 8+ nodes.
  5. Float table — compute float for every activity to confirm the critical path.
Worked Example 3 — Full 7-activity network

Activities:

Activity Dur Depends on
A2—
B4—
C3A
D5A, B
E4B
F2C, D
G3E, F

Nodes: N1(Start), N2(after A), N3(after B), N4(after C), N5(after A+B → D), N6(after D+C → F), N7(after E), N8(End)

Note: D depends on A and B. We need the merge node for A and B before D. Let N5 = merge of N2 and N3 (via dummy from N2 if needed, or directly). Let's simplify: N1→A(2)→N2, N1→B(4)→N3, N2&N3→D(5)→N5, N2→C(3)→N4, N3→E(4)→N7, N4&N5→F(2)→N6, N6&N7→G(3)→N8(End).

Forward scan:

  • N1=0, N2=2, N3=4, N4=2+3=5, N5=max(2,4)=4 (merge of A and B paths to D), N5+D=4+5=9, N6=max(N4+0,N5)→need F node: after C(N4=5) and after D(N5D=9) both feed F. EST(F's start node)=max(5,9)=9. N6=9+2=11. N7=4+4=8. N8=max(11+3,8+3 — wait, G depends on E and F): EST(N8)=max(N6+3, N7+3... but G takes both): merge node before G = max(N6,N7)=max(11,8)=11. N8=11+3=14.

Backward scan:

  • N8=14. Before G merge node: 14−3=11. N6=11 (feeds G), N7: 11 (also feeds G). N7 backward: LST(N7)=11. N3 from E: 11−4=7. N6 from F: LST(N6)=11. F's start node LST: 11−2=9. N4: 9−0=9 (C feeds F: LST(N4)=9). N5: 9 (D feeds F: LST(N5)=9−5=4). N2: min(LST(N4)−3, LST(N5)−... D from N5: LST(N5)=4. C from N4: LST(N4)=9. N2 LST=min(4,9−3)=min(4,6)=4. N3: min(LST(N5)−from B contribution, LST(N7)−4)=min(4,11−4)=min(4,7)=4. N1: min(N2−2, N3−4)=min(4−2, 4−4)=min(2,0)=0.

Node summary: N1(0,0)✓, N2(2,4) float, N3(4,4)✓, N4(5,9) float, N5(4,4)✓, G-merge(11,11)✓, N6(11,11)✓, N7(8,11) float, N8(14,14)✓

Critical path: N1→B→N3→(merge with dummy)→D start→D→N5→F-start→F→N6→G-start→G→N8. In activity terms: B → D → F → G (4+5+2+3 = 14 days).

Float summary: A has float (arrives at N2 with LST=4, EST=2 → float via C: 9−2−3=4). C has float 4. E has float 3 (11−4−4=3).

Book Notes

A dummy activity has duration _______. It is used to represent a _______ dependency without adding extra time to the project.

Activities

Activity 1 — Two Critical Paths

A project has activities: A(4), B(3) after A, C(5) after A, D(2) after B, E(3) after C. Minimum project duration is 9 days.

  1. Perform forward and backward scans to find EST and LST for all nodes.
  2. Are there two critical paths? If so, state both.
  3. Activity B can be crashed by 1 day. Activity C can be crashed by 1 day. If the goal is to reduce the project to 8 days, describe the minimum crashing needed.

Activity 2 — Dummy Activity Recognition

Consider: Activity E depends on C and D. Activity F depends only on C.

  1. Explain why a dummy activity is needed in this network.
  2. Describe where the dummy arrow should be placed (from which node to which node).
  3. If C has duration 3 and D has duration 5, and the dummy has duration 0, what is the EST of the node where E can start (assuming both C and D start from a common start node at time 0)?

Multiple Choice

Q1. A project has two critical paths of equal length. Crashing only one activity on one path by 2 days will:

Q2. A dummy activity is always drawn as a:

Q3. Activities C and D both feed a dummy that leads to where E starts. C has duration 3 (starts from time 0), D has duration 5 (starts from time 0). The EST where E can start is:

Q4. In a larger network, the forward scan processes nodes in order from:

Q5. A project has two critical paths: A→B→D (total 10) and A→C→D (total 10). Activity A has duration 3 and can be crashed by 1. Crashing A by 1 will:

Short Answer

SAQ 1. A project has activities A(5), B(3) after A, C(4) after A, D(4) after B and C. Perform a complete forward and backward scan. State the critical path(s) and calculate float for all activities.

SAQ 2. Explain in your own words what a dummy activity is and describe a specific situation where one would be needed. Include a sketch of the network with and without the dummy.

Answers

Show MC Answers

Q1 → B — With two equal-length critical paths, crashing only one leaves the other at the original length, so the project duration does not change.

Q2 → B — Dummy activities are dashed arrows with duration zero.

Q3 → C (5) — E can only start when both C (finishes day 3) and D (finishes day 5) are done. EST = max(3, 5) = 5.

Q4 → C — Forward scan goes left to right (start to finish).

Q5 → B — A is shared by both paths. Crashing A by 1 reduces both paths from 10 to 9 simultaneously.

Show SAQ Model Answers

SAQ 1: Forward: Start=0, after A=5, after B=8, after C=9, End=max(8+4, 9+4)=max(12,13)=13. Backward: End=13, before D from B side=13−4=9, before D from C side=13−4=9. After A from B: 9−3=6. After A from C: 9−4=5. After A=min(6,5)=5. Start=5−5=0. Critical path: Start→A→C→D→End (5+4+4=13). Float for B=9−5−3=1. Float for C=0 (critical). A is critical (0).

SAQ 2: A dummy activity is a dashed arrow with duration zero, used to show a logical dependency without adding time. Example: Activity E depends on both C and D, but Activity F depends only on C. Without a dummy, drawing a single node after C and D means F also depends on D (incorrect). Solution: C→node X, D→node Y, dummy(0)→node X from Y (or vice versa), E starts from merged node, F starts from C's end node before the merge. Sketch: two separate nodes, dummy connecting them, E starting after the merge node.

Revisit — Think First Answer

Multiple critical paths are generally harder to manage — any delay anywhere on either path delays the project. For crashing: if the two paths share a common activity (like A in the example), crashing that shared activity reduces both paths simultaneously. If they share no activities, you must crash one activity on each path — adding resources to two separate areas, which is more expensive.

Lesson Complete!

You can now handle multiple critical paths, dummy activities, and systematic CPA for larger networks.