Mathematics Standard • Year 12 • Module 6 • Lesson 12

Module Review: Critical Path Analysis — Skill Drill

Rapid review of every Module 6 formula and concept: precedence tables, AOA, forward/backward scans, float, multiple critical paths, Gantt charts, lags, cost slope.

Build · Skill Drill

1. Quick recall — every formula

Answer each question in the space provided. 1 mark each

Q1.1 Forward scan: ES(successor) = ____(EF of predecessors). EF = ES + ____.

Q1.2 Backward scan: LF(predecessor) = ____(LS of successors). LS = LF − ____.

Q1.3 Float = ____ − ES = ____ − EF. Critical path has float = ____.

Stuck? Revisit lesson § Module 6 Summary panel — every formula listed in one place.

2. Worked example — module-wide single problem

Problem. Build a complete solution for: A(3,−), B(4,A), C(2,A), D(5,B), E(3,C), F(2,D and E), G(4,F). Find critical path, project duration and Gantt chart bars.

Step 1 — Precedence table.

A(3, −), B(4, A), C(2, A), D(5, B), E(3, C), F(2, D and E), G(4, F).

Step 2 — Forward scan.

A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14), G(14,18). Duration = 18 days.

Step 3 — Backward scan.

G(LF=18, LS=14), F(14,12), D(12,7), E(12,9), B(7,3), C(9,7), A(min(3,7)=3, 0).

Step 4 — Floats.

A=0, B=0, C=4, D=0, E=4, F=0, G=0.

Step 5 — Critical path.

A → B → D → F → G (zero-float chain). Duration 18 days.

Step 6 — Gantt bars. A solid 0–3, B solid 3–7, C solid 3–5 + extension 5–9, D solid 7–12, E solid 5–8 + extension 8–12, F solid 12–14, G solid 14–18.

3. Faded example — fill in the missing steps

Network: A(2,−), B(5,A), C(3,A), D(2,B), E(4,C), F(3,D and E). Fill in each blank line. 4 marks

Step 2 — Forward scan: A(0,____), B(____, ____), C(____, ____), D(____, ____), E(____, ____), F(max(____, ____)=____, ____).

Step 3 — Backward scan: F(LF=____, LS=____), D(____, ____), E(____, ____), B(____, ____), C(____, ____), A(min(____, ____)=____, ____).

Step 4 — Floats: A=____, B=____, C=____, D=____, E=____, F=____.

Step 5 — Critical path = ________________. Duration = ____ days.

Stuck? Revisit lesson § Worked Example.

4. Graduated practice — every Module 6 skill

Foundation — recall facts (4 questions)

Q	Problem	Answer
4.1 1	An AOA network uses arrows to represent ____ and circles to represent ____.
4.2 1	A zero-duration dashed arrow used to maintain unique events or show partial dependency is called a ____.
4.3 1	Cost slope = (Crash cost − Normal cost) ÷ (Normal time − ____ time).
4.4 1	True or False: A near-critical path has float < the critical path's float, which is zero, so near-critical means float < 0. (True / False)

Standard — apply each concept (6 questions)

4.5 A(3,−), B(4,A), C(5,A), D(2,B and C). Forward scan and project duration. 2 marks

4.6 Using 4.5, backward scan, critical path and floats. 2 marks

4.7 Activity X: normal 6 days $4,000; crash 4 days $5,200. Cost slope per day saved? 2 marks

4.8 A network has two critical paths X (10 days) and Y (10 days). Activity Z is on Path X only. Crashing Z by 1 day, what is the new project duration? 2 marks

4.9 Activity P: ES=4, EF=8, LF=10. Sketch the Gantt bar from P including its float extension. 2 marks

4.10 A project has lag 3 days between Pour and Frame. Pour finishes day 5. When does Frame start? 2 marks

Extension — full mini-project (2 questions)

4.11 A(2,−), B(4,A), C(3,A), D(5,B), E(4,C), F(2,D and E), G(3,F). Full 6-step solution. State critical path, duration, and floats. 3 marks

4.12 Using 4.11, with crash slopes D=$400/day max 2 days, F=$300/day max 1 day, G=$500/day max 1 day. Indirect $200/day. Find the optimal duration and total crash + indirect cost. 3 marks

Stuck on 4.12? Crash slopes < $200 are worth it; otherwise the crash adds to cost.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I said arrows = activities, circles = events.

For 4.2 I said dummy (zero-duration arrow, often dashed).

For 4.4 I said False — near-critical means small POSITIVE float, not negative.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Forward scan

ES = max(EF of predecessors). EF = ES + duration.

Q1.2 — Backward scan

LF = min(LS of successors). LS = LF − duration.

Q1.3 — Float and critical path

Float = LS − ES = LF − EF. Critical path has float = 0.

Q3 — Faded example

Forward: A(0,2), B(2,7), C(2,5), D(7,9), E(5,9), F(max(9,9)=9, 12).
Backward: F(12,9), D(9,7), E(9,5), B(7,2), C(5,2), A(min(2,2)=2, 0).
Floats: A=0, B=0, C=0, D=0, E=0, F=0. Two critical paths: A-B-D-F and A-C-E-F, each 12 days.

Q4.1 — AOA convention

Arrows = activities, circles = events.

Q4.2 — Dummy

Dummy activity (zero duration, dashed arrow).

Q4.3 — Cost slope denominator

... ÷ (Normal time − Crash time).

Q4.4 — Near-critical T/F

False. Near-critical means small POSITIVE float (close to zero, still not critical).

Q4.5 — Forward scan and duration

A(0,3), B(3,7), C(3,8), D(max(7,8)=8, 10). Duration = 10 days.

Q4.6 — Backward, critical, floats

D(10,8), C(8,3), B(8,4), A(min(4,3)=3, 0). Floats: A=0, B=1, C=0, D=0. Critical: A → C → D.

Q4.7 — Cost slope

($5,200 − $4,000) ÷ (6 − 4) = $1,200 ÷ 2 = $600/day.

Q4.8 — Crashing only Path X

Path X becomes 9, Path Y still 10. Project duration = 10 days (unchanged).

Q4.9 — Gantt bar for P

Solid bar from day 4 to day 8 (ES to EF), then lighter extension from day 8 to day 10 (EF to LF) — float = 2 days.

Q4.10 — Lag

Frame starts at day 5 + 3 = day 8.

Q4.11 — Full 7-activity solution

Forward: A(0,2), B(2,6), C(2,5), D(6,11), E(5,9), F(max(11,9)=11, 13), G(13,16). Backward: G(16,13), F(13,11), D(11,6), E(11,7), B(6,2), C(7,4), A(min(2,4)=2, 0). Floats: A=B=D=F=G=0; C=2, E=2. Critical: A → B → D → F → G (16 days).

Q4.12 — Optimal crash plan

Crash slopes: D $400, F $300, G $500 — all > indirect $200. Every crash NET LOSES money ($300 − $200 = $100 minimum loss per day). Don't crash. Optimal duration = 16 days. No crash cost. Total cost from indirect alone = 16 × $200 = $3,200.