Mathematics Standard • Year 12 • Module 6 • Lesson 11
Mixed Project Problems — Skill Drill
Build fluency in the 6-step exam method: table → network → forward scan → backward scan → critical path → answer specific questions.
1. Quick recall — the 6-step method
Number the six steps in the correct order in the box below. 1 mark each
Q1.1 Backward scan (LS, LF) → step ____. Forward scan (ES, EF) → step ____. Build precedence table → step ____.
Q1.2 Identify critical path → step ____. Draw AOA network → step ____. Answer specific questions (float, crash, Gantt) → step ____.
Q1.3 Float formula: Float = LS − ES = ____ − EF.
2. Worked example — executing all six steps on one small network
Problem. A(3,−), B(4,A), C(2,A), D(5,B), E(3,C), F(2,D,E). Apply the 6-step method.
Step 1 — Table.
A(3, −); B(4, A); C(2, A); D(5, B); E(3, C); F(2, D and E).
Step 2 — AOA network sketch.
Events 1, 2, 3, 4 (with possible dummy). A:1→2. B:2→3. C:2→3' (or via dummy). D:3→4. E:3'→4. F:4→5.
Step 3 — Forward scan.
A: 0→3. B: 3→7. C: 3→5. D: 7→12. E: 5→8. F: max(12, 8) = 12 → 14.
Step 4 — Backward scan.
F: LF=14, LS=12. D: LF=12, LS=7. E: LF=12, LS=9. B: LF=7, LS=3. C: LF=9, LS=7. A: LF=min(3, 7)=3, LS=0.
Step 5 — Critical path.
Floats: A=0, B=0, C=4, D=0, E=4, F=0. Critical path: A → B → D → F (14 days).
Step 6 — Answer questions: Float on C = 4, float on E = 4. Project duration = 14.
3. Faded example — fill in the missing steps
Network: A(2,−), B(3,A), C(4,A), D(2,B), E(1,C), F(3,D,E). Fill in each blank line. 4 marks
Step 3 — Forward scan: A(0,____), B(____, ____), C(____, ____), D(____, ____), E(____, ____), F(max(____, ____) = ____, ____).
Step 4 — Backward scan: F: LF=____, LS=____. D: LF=____, LS=____. E: LF=____, LS=____. B: LF=____, LS=____. C: LF=____, LS=____. A: LF=min(____, ____)=____, LS=____.
Step 5 — Floats: A=____, B=____, C=____, D=____, E=____, F=____.
Step 6 — Critical path = ________________. Project duration = ____ days.
4. Graduated practice — Mixed exam-style skills
Foundation — single sub-step (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Forward scan: A(0,4), B(4,7). What is ES of B? | |
| 4.2 1 | Backward scan: F LF=15. F duration = 3. What is LS(F)? | |
| 4.3 1 | Activity X: ES=4, LS=4. What is the float? | |
| 4.4 1 | Common trap: should you crash an activity with float = 5? (Yes / No) |
Standard — small full networks (6 questions)
4.5 A(2,−), B(3,A), C(4,A), D(2,B), E(1,C), F(3,D,E). Forward scan and project duration. 2 marks
4.6 Same network as 4.5. Backward scan: LS and LF for all activities. 2 marks
4.7 Same network. State critical path and floats on non-critical activities. 2 marks
4.8 A(3,−), B(2,A), C(4,A), D(3,B), E(2,C), F(1,D,E). Find ES, EF, LS, LF and project duration. 3 marks
4.9 Using 4.8, state critical path and identify which activities have positive float. 2 marks
4.10 Using 4.8, if activity D can be crashed by 1 day at $400, and F by 1 day at $200, find the new project duration after both crashes and the total crash cost. 2 marks
Extension — full 6-step on a 7-activity network (2 questions)
4.11 A(3,−), B(4,A), C(2,A), D(5,B), E(3,C), F(2,D,E), G(4,F). Apply all 6 steps. State critical path, duration, and float of non-critical activities. 3 marks
4.12 Using 4.11, crash D by 1 day (cost $500) and F by 1 day (cost $300). Indirect cost $200/day. Find: new duration, total crash cost, and net change in total project cost. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Step order (backward, forward, table)
Backward scan = step 4. Forward scan = step 3. Build table = step 1.
Q1.2 — Step order (critical, network, answer)
Critical path = step 5. Draw AOA = step 2. Answer specific qs = step 6.
Q1.3 — Float formula
Float = LS − ES = LF − EF.
Q3 — Faded example
Step 3 (forward): A(0,2), B(2,5), C(2,6), D(5,7), E(6,7), F(max(7,7)=7, 10).
Step 4 (backward): F: LF=10, LS=7. D: LF=7, LS=5. E: LF=7, LS=6. B: LF=5, LS=2. C: LF=6, LS=2. A: LF=min(2,2)=2, LS=0.
Step 5 (floats): A=0, B=0, C=0, D=0, E=0, F=0.
Step 6: Critical paths: A → B → D → F and A → C → E → F (both 10 days). Project duration = 10 days.
Q4.1 — ES(B)
ES(B) = max(EF predecessors) = 4.
Q4.2 — LS(F)
LS = LF − duration = 15 − 3 = 12.
Q4.3 — Float of X
Float = LS − ES = 4 − 4 = 0 (critical).
Q4.4 — Crash with float = 5
No — float = 5 means activity is not critical. Crashing it doesn't reduce project duration.
Q4.5 — Forward scan and duration
A(0,2), B(2,5), C(2,6), D(5,7), E(6,7), F(max(7,7)=7, 10). Duration = 10 days.
Q4.6 — Backward scan
F: LF=10, LS=7. D: LF=7, LS=5. E: LF=7, LS=6. B: LF=5, LS=2. C: LF=6, LS=2. A: LF=min(2,2)=2, LS=0.
Q4.7 — Critical path and floats
Both A-B-D-F and A-C-E-F have all-zero float → two critical paths, each 10 days. No positive floats.
Q4.8 — A,B,C,D,E,F
Forward: A(0,3), B(3,5), C(3,7), D(5,8), E(7,9), F(max(8,9)=9, 10). Backward: F(10,9), D(9,6), E(9,7), B(6,4), C(7,3), A(min(4,3)=3, 0). Duration = 10 days.
Q4.9 — Critical path and float
Floats: A=0, B=1, C=0, D=1, E=0, F=0. Critical: A → C → E → F (10 days). Positive floats: B (1), D (1).
Q4.10 — Crashing D and F
D is not critical (float = 1). Crashing D does NOT reduce duration.
F is critical. Crash F by 1 day: new duration = 9. Cost = $200.
Crash D (waste): cost = $400 but no benefit.
If both crashes are applied: new duration = 9 days, total crash cost = $600. (Crashing D wastes $400 — only F's crash matters.)
Q4.11 — 7-activity network
Forward: A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14), G(14, 18). Backward: G(18,14), F(14,12), D(12,7), E(12,9), B(7,3), C(9,7), A(min(3,7)=3, 0). Floats: A=B=D=F=G=0; C=4, E=4. Critical path = A → B → D → F → G (18 days).
Q4.12 — Crash D + F
D and F both critical. Crashing each by 1 day cuts duration by 2: new = 16 days. Crash cost = $500 + $300 = $800. Indirect saving = 2 × $200 = $400. Net cost change = $800 − $400 = +$400 (project costs $400 more).