Mathematics Standard • Year 12 • Module 6 • Lesson 11
Mixed Project Problems — Problem Set
Apply the full critical-path toolkit to multi-part real-world scenarios — construction, IT rollouts, manufacturing and events.
Problem 1 — Suburban single-storey build
A Sydney builder is scheduling a new house. Activities (days): Site prep SP(4,−), Foundation F(6,SP), Framing Fr(8,F), Roofing R(5,Fr), Plumbing P(4,Fr), Electrical E(3,Fr), Drywall D(6,P and E), Painting Pa(4,D and R).
Set up: What are we solving for?
(i) Carry out a forward scan for all 8 activities. State the project duration. 3 marks
(ii) Carry out a backward scan and find the float for each non-critical activity. State the critical path. 3 marks
(iii) The builder offers two crash options: Framing 8 → 6 days at $400/day; Plumbing 4 → 3 days at $300/day. Indirect cost $250/day. Reduce duration by 2 days at minimum cost. State the new duration and net change in total cost. 3 marks
Stuck on (iii)? Plumbing is on the path SP-F-Fr-P-D-Pa; check if it's critical before crashing.Problem 2 — Council IT rollout
A NSW council is rolling out a new payment system. Activities (days): Requirements R(3,−), Database design DB(4,R), API design API(3,R), Backend code BC(6,DB), Frontend code FC(5,API), Integration tests IT(4,BC and FC), User training UT(3,IT), Go-live GL(2,UT).
Set up: What are we solving for?
(i) Forward scan and project duration. 3 marks
(ii) Identify the critical path and the float on any non-critical activity. 2 marks
(iii) Council wants the system live 3 days earlier. Crash options: BC slope $700/day max 2 days; IT slope $500/day max 2 days. Show your minimum-cost plan and the total crash cost. 3 marks
Stuck on (iii)? Crash IT first (cheaper); then crash BC for the remaining days.Problem 3 — Boutique brewery launch
A regional NSW microbrewery is launching a new beer. Activities (weeks): Recipe R(2,−), Equipment hire EH(1,R), Mash and brew MB(2,EH), Fermentation F(4,MB), Bottling B(1,F), Marketing material MM(3,R), Label print LP(2,MM), Launch event LE(1,B and LP).
Set up: What are we solving for?
(i) Forward scan and project duration (in weeks). 2 marks
(ii) Backward scan and float for non-critical activities. State the critical path. 3 marks
(iii) Owner asks: "What's the maximum we can shorten Marketing material before it becomes critical?" Show your reasoning. 2 marks
Stuck on (iii)? Marketing material's float tells you how much shifting/shortening it can absorb before joining the critical path.Problem 4 — Manufacturing line upgrade
A NSW factory is installing a new conveyor line. Activities (days): Power-off PO(1,−), Remove old line RO(2,PO), Install frame IF(3,RO), Install conveyor IC(4,IF), Install electrical IE(3,IF), Calibration C(2,IC and IE), Commission C2(1,C).
Set up: What are we solving for?
(i) Forward scan, backward scan and critical path. State project duration. 4 marks
(ii) The factory loses $4,000 per day of production while the line is offline. Direct project cost is $30,000. Calculate total cost at normal duration. 2 marks
(iii) Two crash options: IC 4 → 2 days at $1,500/day; IE 3 → 2 days at $2,500. Find the optimal duration and total cost. 3 marks
Stuck on (iii)? Compare slopes ($1,500 and $2,500) to indirect saving ($4,000). Both worth crashing if they shorten duration.Problem 5 — Music festival main-stage build
A festival is building the main stage. Activities (hours): Site survey S(2,−), Truss assembly T(4,S), Lighting rig LR(5,T), Sound PA(3,T), Backdrop B(4,T), Plug-in tests PT(2,LR and PA and B), Soundcheck SC(2,PT).
Set up: What are we solving for?
(i) Forward scan, backward scan, critical path. 3 marks
(ii) A storm warning is forecast in 17 hours from project start. Will the stage be ready in time? 2 marks
(iii) The promoter offers $300/hour bonus to crash any one activity by up to 1 hour. Which single activity should be crashed to give the biggest impact on finishing earlier? Justify in one sentence. 3 marks
Stuck on (iii)? Only crashing a CRITICAL activity reduces duration. Pick the one with highest float? No — pick the critical one with most flexibility.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Suburban build
Set up. We run forward and backward scans on 8 activities, identify the critical path, then choose a 2-day crash plan.
(i) SP(0,4), F(4,10), Fr(10,18), R(18,23), P(18,22), E(18,21), D(max(22,21)=22, 28), Pa(max(28,23)=28, 32). Duration = 32 days.
(ii) Backward: Pa(32,28), D(28,22), R(28,23), P(22,18), E(22,19), Fr(18,10), F(10,4), SP(4,0). Floats: SP=0, F=0, Fr=0, R=5, P=0, E=1, D=0, Pa=0. Critical path: SP → F → Fr → P → D → Pa. R has float 5; E has float 1.
(iii) Both Framing ($400) and Plumbing ($300) are critical. Cheaper: Plumbing first (max 1 day @ $300). Then Framing 1 day @ $400. Total crash $700. New duration 30 days. Indirect saving 2×$250 = $500. Net cost change = $700 − $500 = +$200. (Crashing increases cost slightly here.)
Problem 2 — Council IT rollout
Set up. We do scans, identify the critical path, then crash 3 days cheaply.
(i) R(0,3), DB(3,7), API(3,6), BC(7,13), FC(6,11), IT(max(13,11)=13, 17), UT(17,20), GL(20,22). Duration = 22 days.
(ii) Backward: GL(22,20), UT(20,17), IT(17,13), BC(13,7), FC(13,8), DB(7,3), API(8,5), R(min(3,5)=3, 0). Floats: R=0, DB=0, API=2, BC=0, FC=2, IT=0, UT=0, GL=0. Critical: R → DB → BC → IT → UT → GL. API float 2, FC float 2.
(iii) Need 3 days. IT cheaper ($500). Crash IT 2 days max = $1,000. Then BC 1 day = $700. Total = $1,700. New duration = 19 days. Minimum-cost plan: IT 2 days + BC 1 day, total crash cost $1,700.
Problem 3 — Brewery launch
Set up. We compute scans, find float on Marketing material, then determine how much Marketing can shrink before becoming critical.
(i) R(0,2), EH(2,3), MB(3,5), F(5,9), B(9,10), MM(2,5), LP(5,7), LE(max(10,7)=10, 11). Duration = 11 weeks.
(ii) Backward: LE(11,10), B(10,9), F(9,5), MB(5,3), EH(3,2), R(min(2,?)=2,0). For MM, LP path: LP(LF=10, LS=8); MM(LF=8, LS=5). MM float = 5−2 = 3. LP float = 8−5 = 3. Critical path: R → EH → MB → F → B → LE. MM and LP each have float = 3.
(iii) MM currently 3 weeks. Float of MM is 3 weeks — Marketing path total = 2+3+2+1 = 8, critical = 11, gap = 3. Marketing material can shrink by up to 3 weeks (down to 0 weeks) without making its path critical. Beyond 3 weeks of saving on MM, it wouldn't matter — the critical path is still 11.
Problem 4 — Conveyor line upgrade
Set up. We do full scans, total cost at normal duration, then evaluate crashing.
(i) PO(0,1), RO(1,3), IF(3,6), IC(6,10), IE(6,9), C(max(10,9)=10, 12), C2(12,13). Backward: C2(13,12), C(12,10), IC(10,6), IE(10,7), IF(6,3), RO(3,1), PO(1,0). Floats: all 0 except IE = 1. Critical: PO → RO → IF → IC → C → C2 (13 days).
(ii) Total at 13 days = direct $30,000 + indirect 13 × $4,000 = $30,000 + $52,000 = $82,000.
(iii) IC slope $1,500 < $4,000 (worth crashing). IE slope $2,500 < $4,000 (worth crashing) — but IE has float 1, so first day of IE crash doesn't shorten duration. Crash IC 2 days ($3,000): new duration 11. IE now becomes critical because IF→IE→C path: 1+2+3+3+2+1 = wait recompute. Actually re-scan: after IC 2-day crash, IC duration=2. PO(0,1), RO(1,3), IF(3,6), IC(6,8), IE(6,9), C(max(8,9)=9, 11), C2(11,12). Duration = 12. So IC crash 2 days saves only 1 day (because IE then constrains). Crashing IE 1 day too: IE=2, EF(IE)=8. C starts at max(8,8)=8, ends 10. C2 ends 11. With IC 2 and IE 1: duration = 11. Crash cost = $3,000 + $2,500 = $5,500. Total = $30,000 + $5,500 + 11×$4,000 = $30,000 + $5,500 + $44,000 = $79,500. Versus normal $82,000. Saving $2,500. Optimal duration = 11 days, total cost = $79,500.
Problem 5 — Festival stage
Set up. We compute scans, check storm timing, then pick best single-activity crash.
(i) S(0,2), T(2,6), LR(6,11), PA(6,9), B(6,10), PT(max(11,9,10)=11, 13), SC(13,15). Backward: SC(15,13), PT(13,11), LR(11,6), PA(11,8), B(11,7), T(6,2), S(2,0). Floats: S=T=LR=PT=SC=0, PA=2, B=1. Critical: S → T → LR → PT → SC. Duration = 15 hours.
(ii) 15 hours < 17-hour storm warning. Yes, ready 2 hours before the storm.
(iii) Crash a critical activity. Among S(2), T(4), LR(5), PT(2), SC(2), the one with most flexibility (longest current duration) and biggest gain is LR (currently 5 hours). Crash LR by 1 hour for $300 → duration 14 hours. Crash LR — it's critical and has the most room to shrink among the critical activities.