Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
Multiple critical paths · Dummy activities · 7+ activity networks
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?
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.
Activities:
| Activity | Dur | Depends on |
|---|---|---|
| A | 3 | |
| B | 5 | A |
| C | 4 | A |
| D | 2 | B |
| E | 3 | C |
Forward scan:
Backward scan:
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:
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.
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:
Situation: C requires both A and B. D requires only A.
Network structure:
Forward scan:
Backward scan:
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.
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.
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.
Activities:
| Activity | Dur | Depends on |
|---|---|---|
| A | 2 | |
| B | 4 | |
| C | 3 | A |
| D | 5 | A, B |
| E | 4 | B |
| F | 2 | C, D |
| G | 3 | E, 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:
Backward scan:
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).
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 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.
Consider: Activity E depends on C and D. Activity F depends only on C.
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:
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.
Q1 → BWith two equal-length critical paths, crashing only one leaves the other at the original length, so the project duration does not change.
Q2 → BDummy 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 → CForward scan goes left to right (start to finish).
Q5 → BA is shared by both paths. Crashing A by 1 reduces both paths from 10 to 9 simultaneously.
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.
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.
You can now handle multiple critical paths, dummy activities, and systematic CPA for larger networks.
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