Module 6 · Networks and Critical Paths

Complex Critical Path Problems

Multiple critical paths · Dummy activities · 7+ activity networks

MST-12-S2-07 MST-12-S2-07 Lesson 11 of 12
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Orient and prepare

Recall how ties and dependencies make a project network more complex.

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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.
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Compare critical paths

Recognise and manage projects with more than one critical path.

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).

A complete CPA solution requires: (1) drawing the correctly-labelled network, (2) performing forward and backward scans to find EST and LST at every node, (3) identifying the critical path, and (4) stating the minimum project duration.

Pause, copy the four-step CPA solution: (1) draw the correctly labelled network from the precedence table; (2) forward scan to find EST at every node; (3) backward scan to find LST; (4) state the critical path and minimum project duration into your book.

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

Represent shared dependencies without adding time to the project.

2

Dummy Activities

The four-step solution, draw network, forward scan (EST), backward scan (LST), identify critical path and minimum duration, covers all marks. The most costly errors happen inside the scans: forgetting to take the maximum at a merge node in the forward scan, forgetting the minimum at a burst node in the backward scan, and omitting a dummy activity when shared predecessors are present.

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. Starting C at node 2 (where A ends) is wrong: C would then not require B at all.
  4. Correct: draw the dummy (dashed, dur = 0) from node 2 to node 3, so node 3 now has B and the dummy (carrying A) as inputs. C starts at node 3, and so requires both A and B.
  5. D depends only on A, so D leaves node 2 — upstream of the merge, which is exactly why the dummy has to point into node 3 rather than out of it.
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 2 → dummy(0) → Node 3
  • Node 3 → C(5) → Node 4 (End)
  • Node 2 → D(2) → Node 4 (End)

Forward scan:

  • Node 1: EST = 0
  • Node 2: EST = 0 + 4 = 4 (activity A)
  • Node 3: two arrows arrive — via B: 0 + 3 = 3; via the dummy: 4 + 0 = 4. Take the maximum: EST = max(3, 4) = 4
  • Node 4: via C: 4 + 5 = 9; via D: 4 + 2 = 6. Take the maximum: EST = max(9, 6) = 9

Backward scan:

  • Node 4: LST = EST = 9
  • Node 3: LST = 9 − 5 = 4 (only activity C leaves node 3)
  • Node 2: two arrows leave — via the dummy: 4 − 0 = 4; via D: 9 − 2 = 7. Take the minimum: LST = min(4, 7) = 4
  • Node 1: via A: 4 − 4 = 0; via B: 4 − 3 = 1. Take the minimum: LST = min(0, 1) = 0

Node boxes: Node 1(0,0)✓, Node 2(4,4)✓, Node 3(4,4)✓, Node 4(9,9)✓ — every node is critical.

Critical path: Start → A → node 2 → dummy → node 3 → C → End (4 + 0 + 5 = 9 days). The dummy lies on the critical path, which is normal: it costs nothing, but the dependency it carries is what holds C back. Activity B has float = LST(node 3) − EST(node 1) − 3 = 4 − 0 − 3 = 1 day. Activity D has float = LST(node 4) − EST(node 2) − 2 = 9 − 4 − 2 = 3 days.

Common CPA errors: forgetting to take the maximum at a merge node (forward scan), forgetting the minimum at a burst node (backward scan), and missing a dummy activity. Always check that EST ≤ LST at every node.

Pause, copy the three common CPA errors: (1) taking minimum instead of maximum at a forward-scan merge node; (2) taking maximum instead of minimum at a backward-scan burst node; (3) missing a dummy activity, plus the sanity check: EST ≤ LST at every node into your book.

Which statement does NOT correctly describe a dummy activity?
4

Solve larger networks

Apply the complete CPA workflow systematically to larger projects.

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Larger Networks (7+ Activities)

Avoiding the three common errors, wrong maximum at a merge, wrong minimum at a burst, missing a dummy, keeps your network valid. From a correctly drawn and fully scanned network, the HSC asks four question types: minimum project duration, identifying the critical path, activities to crash to meet a deadline, and the effect of a delay on float or completion time.

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)

With seven activities, three merge points and a dummy, node numbering gets in the way. Use the activity table from Lesson 9 instead: one row per activity, EST and EFT filled in top to bottom, then LFT and LST filled in bottom to top.

Forward scan:

  • A and B wait for nobody: EST = 0, so EFT(A) = 2 and EFT(B) = 4.
  • C follows A: EST = 2, EFT = 5. E follows B: EST = 4, EFT = 8.
  • D follows A and B: EST = max(2, 4) = 4, EFT = 9.
  • F follows C and D: EST = max(5, 9) = 9, EFT = 11.
  • G follows E and F: EST = max(8, 11) = 11, EFT = 14.
  • The project duration is the largest EFT: 14 days.

Backward scan:

  • G finishes the project: LFT = 14, so LST = 14 − 3 = 11.
  • E and F both feed G, so each has LFT = LST(G) = 11. LST(E) = 11 − 4 = 7; LST(F) = 11 − 2 = 9.
  • C and D both feed F, so each has LFT = LST(F) = 9. LST(C) = 9 − 3 = 6; LST(D) = 9 − 5 = 4.
  • B feeds D and E, so LFT(B) = min(LST(D), LST(E)) = min(4, 7) = 4, giving LST(B) = 4 − 4 = 0.
  • A feeds C and D, so LFT(A) = min(LST(C), LST(D)) = min(6, 4) = 4, giving LST(A) = 4 − 2 = 2.

Float = LST − EST for each activity: A = 2 − 0 = 2; B = 0 − 0 = 0; C = 6 − 2 = 4; D = 4 − 4 = 0; E = 7 − 4 = 3; F = 9 − 9 = 0; G = 11 − 11 = 0.

Critical path: the zero-float activities, in order — B → D → F → G (4 + 5 + 2 + 3 = 14 days). That total comes from a different column than the largest EFT, so the two agreeing is real evidence the scan is right.

Float summary: A has 2 days, C has 4 days and E has 3 days. A and C sit on the same path and share their float: together they can absorb 4 days of delay, not 6.

Mixed CPA exam questions may ask: minimum duration, which activities to crash, how much extra cost to meet a deadline, or what happens to float when one activity is delayed. Plan your response by drawing and annotating the network first.

Pause, copy the four HSC question types: minimum duration, critical path identification, crashing to meet a deadline, and float/delay effects, and the strategy: always draw and fully annotate the network before answering any sub-question into your book.

A dummy activity has duration . It is used to represent a dependency without adding extra time to the project.
5

Apply complex CPA

Test the methods on multiple-path and dummy-activity problems.

Activities

Activity 1, Two Critical Paths

A project has activities: A(4), B(3) after A, C(2) 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)?
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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. 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 ends at node X and D ends at node Y; draw the dummy (0) from X to Y, so node Y now carries both C and D. E starts from node Y, and F starts from node X, which carries C only. The dummy must point from X into Y: pointing it the other way would put D upstream of F and make F depend on D as well. Sketch: two separate nodes, dummy connecting them, E starting after the merge node.

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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

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.