Mathematics Standard • Year 12 • Module 6 • Lesson 12
Module Review: Critical Path Analysis — Past-Paper Style
Final HSC-style exam writing on Module 6 — every concept tested. Multi-mark short answers and one structured extended response.
1. Short-answer questions
1.1 A project has activities A(3,−), B(4,A), C(2,A), D(5,B), E(3,C), F(2,D and E). (a) Carry out a forward scan. (b) State the project duration. 3 marks Band 3
1.2 Using the network from 1.1: (a) Carry out a backward scan. (b) State the critical path and float of all non-critical activities. (c) Sketch a Gantt chart showing solid bars for ES → EF and lighter extensions for float. 4 marks Band 3-4
1.3 Activity X: normal 10 days $5,000; crash 7 days $6,500.
(a) Calculate the cost slope per day saved.
(b) If the indirect cost is $600/day, is it worth crashing X to its minimum 7 days? Show working.
(c) State the maximum number of days X should be crashed if indirect cost drops to $400/day. 4 marks Band 4
2. Extended response
2.1 A NSW agribusiness is upgrading a grain-handling facility.
Activity table (days):
A. Demolish old shed (3, −)
B. Pour slab (4, A)
Cure lag = 3 days after B
C. Frame structure (5, after cure)
D. Roofing (3, C)
E. Electrical (4, C)
F. Install grain auger (2, D and E)
G. Commissioning (1, F)
Direct cost $60,000. Indirect overheads $800/day. The facility must be operational within 22 days or a $1,200/day late penalty applies.
Crash options: C 5→3 at +$1,200 total (slope $600/day, max 2); E 4→3 at +$500 (slope $500/day, max 1); F 2→1 at +$400 (slope $400/day, max 1).
(a) Carry out forward and backward scans for all activities (treating the cure as a 3-day lag). State critical path and project duration. (b) Calculate total cost at the normal duration (direct + indirect + penalty). (c) Find the minimum-cost duration. Show your working in a table comparing the normal duration plus crash combinations. End with an explicit conclusion sentence naming the optimal duration and total cost. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correct forward scan including the 3-day cure lag between B and C.
• 1 mark — correct backward scan and critical path identified as A → B → cure → C → E → F → G.
• 1 mark — project duration stated as 22 days.
Part (b) — 2 marks
• 1 mark — indirect cost computed at 22 × $800 = $17,600; penalty = 0 (at exactly day 22, no overrun).
• 1 mark — total cost stated as $77,600 with all components shown.
Part (c) — 2 marks
• 1 mark — table of total costs at progressively crashed durations including 21, 20, 19, with cheapest crashes selected (F $400, E $500, C $600).
• 1 mark — explicit conclusion sentence naming the optimal duration (19 days) and total cost ($76,300).
Your response:
Stuck on (c)? At day 22, no penalty. So below day 22, only $800/day indirect saving applies. Compare each slope to $800.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Forward scan (3 marks)
Sample response. A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14). Duration = 14 days.
Marking notes. 1 mark — A, B, C. 1 mark — D, E (with max). 1 mark — F and duration stated.
1.2 — Backward, critical path, Gantt (4 marks)
(a) F(14,12), D(12,7), E(12,9), B(7,3), C(9,7), A(min(3,7)=3, 0).
(b) Floats: A=B=D=F=0, C=4, E=4. Critical path: A → B → D → F. C and E each have float 4.
(c) Gantt: A 0–3 solid, B 3–7 solid, C 3–5 solid + extension 5–9, D 7–12 solid, E 5–8 solid + extension 8–12, F 12–14 solid.
Marking notes. (a) 1 mark — backward scan correct. (b) 1 mark — critical path, 1 mark — floats. (c) 1 mark — Gantt with correct solid + extension bars.
1.3 — Cost slope and crash decision (4 marks)
(a) Slope = ($6,500 − $5,000) ÷ (10 − 7) = $1,500 ÷ 3 = $500/day.
(b) Slope $500 < indirect $600: net saving per day = $100. Crash all 3 days. Saving = 3 × $100 = $300. Yes — crash X to 7 days.
(c) Slope $500 > indirect $400: each crash day adds $100 net cost. Don't crash X at all — optimal duration is normal 10 days.
Marking notes. (a) 1 mark — slope. (b) 1 mark — comparison, 1 mark — final saving and decision. (c) 1 mark — correct "don't crash" answer with reason.
2.1 — Grain facility upgrade (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Scans + critical path.
Forward: A(0,3), B(3,7), cure lag 3 days, C(10,15), D(15,18), E(15,19), F(max(18,19)=19, 21), G(21,22). [1 mark — forward with lag]
Backward: G(22,21), F(21,19), D(19,16), E(19,15), C(15,10), B(10,7) — note B's LF is governed by start of cure + cure period = 10, so LF(B) = 10 − 3 (cure) = 7; LS(B) = 3. A(7,3) — actually A → B chain: LF(A) = LS(B) = 3, LS(A) = 0. [1 mark — backward and critical path: A → B → cure → C → E → F → G]
Floats: A=B=C=E=F=G=0, D=1.
Project duration = 22 days. [1 mark]
(b) Cost at normal duration.
Indirect = 22 × $800 = $17,600. Penalty = 0 (project finishes exactly at day 22). [1 mark — indirect and penalty]
Total cost = $60,000 + $17,600 + 0 = $77,600. [1 mark]
(c) Optimal duration.
Slopes ranked cheapest first: F $400, E $500, C $600/day (max 2).
Effective saving per day below day 22: only $800 indirect (no penalty). All slopes < $800 → all crashes save money.
Table:
Day 22: $60,000 + 0 + 22×$800 = $77,600.
Day 21 (crash F 1, $400): $60,000 + $400 + 21×$800 = $60,000 + $400 + $16,800 = $77,200.
Day 20 (F 1 + E 1, $400 + $500 = $900): $60,000 + $900 + 20×$800 = $60,000 + $900 + $16,000 = $76,900.
Day 19 (F 1 + E 1 + C 1, $900 + $600 = $1,500): $60,000 + $1,500 + 19×$800 = $60,000 + $1,500 + $15,200 = $76,700. Wait — recompute: $76,700. Actually: $60,000 + $1,500 + $15,200 = $76,700.
Day 18 (F 1 + E 1 + C 2, $1,500 + $600 = $2,100): $60,000 + $2,100 + 18×$800 = $60,000 + $2,100 + $14,400 = $76,500.
Cannot crash further (F, E, C all at max). [1 mark — table]
Conclusion: the optimal duration is 18 days at a total cost of $76,500. Every crash slope is cheaper than the $800/day indirect saving, so we crash all activities to their minima. The crash reduces total project cost by $1,100 from the normal 22-day schedule. [1 mark — explicit conclusion with duration and total]
Total: 7/7.
Band descriptors for marker.
Band 3: Forward scan attempted but cure lag forgotten; critical path mis-identified. ≈ 3 marks.
Band 4: Cure lag applied; scans correct; total at 22 days computed but no crashing attempted. ≈ 5 marks.
Band 5: Cost table started; identifies that all slopes < $800; conclusion sentence vague. ≈ 6 marks.
Band 6: Complete scans with cure, critical path stated, full crash table, explicit conclusion sentence naming 18 days AND $76,500 with reasoning that all crashes were worthwhile. 7/7.