Full module assessment covering all twelve lessons: project networks, precedence tables, AOA diagrams, forward and backward scans, critical paths, float, Gantt charts, multiple critical paths, constraints, crashing, and cost-time trade-offs.
Multiple Choice
15 questions drawn from the full module bank, feedback shown immediately
Short Answer
1. A construction project has: Site prep(4,-), Foundation(6,Site prep), Framing(8,Foundation), Roofing(5,Framing), Plumbing(4,Framing), Electrical(3,Framing), Drywall(6,Plumbing+Electrical), Painting(4,Drywall+Roofing). (a) Build the precedence table. (b) Draw the AOA network. (c) Perform forward and backward scans. (d) Find the critical path and project duration. (e) Calculate float for all non-critical activities. (f) Draw a Gantt chart. 8 MARKS
2. Using the project from Q1: (a) Framing can be crashed from 8 to 6 days at $\$500$/day. Drywall from 6 to 4 at $\$400$/day. Reduce duration by 2 days at minimum cost. Show working. (2 marks) (b) A resource constraint means only 2 trades can work at a time on Framing, Plumbing and Electrical. Plumbing and Electrical both follow Framing. How should these activities be scheduled, and what is the effect on the finish date? (2 marks) (c) Evaluate the claim: “Reducing Framing by 2 days might not reduce the project duration by 2 days because another route could become critical.” Explain using the project network. (2 marks) 6 MARKS
3. (a) Prove that the critical path is the longest path through the network. (3 marks) (b) A project has normal cost $\$20,000$ and duration 25 days. Indirect costs are $\$600$/day. The project has one critical path containing both activities X and Y; every other path remains shorter throughout the crashes described below. Each day of crashing reduces project duration by one day, and crash cost increases linearly: X can be shortened from 10 to 7 days for an additional $\$1500$ total, and Y from 8 to 6 days for an additional $\$1200$ total. Find the optimal duration and minimum total cost. Show all working. (4 marks) (c) Evaluate the statement: "Crashing is always worth it if the indirect cost per day exceeds the crash cost per day." (3 marks) 10 MARKS
(a) Table correct [1].
(b) Network with dummies if needed [1.5].
(c) Forward and backward scans [2].
(d) Critical path and duration = 32 [1].
(e) Floats: Roofing=5, Electrical=1 [1.5].
(f) Gantt chart [1].
Q1 worked model: A precedence table lists each activity with its duration and immediate predecessor(s): Site prep 4 (none); Foundation 6 (Site prep); Framing 8 (Foundation); Roofing 5, Plumbing 4 and Electrical 3 (each after Framing); Drywall 6 (after both Plumbing and Electrical); Painting 4 (after both Drywall and Roofing). One valid AOA node/arc listing is: Site prep 1→2, Foundation 2→3, Framing 3→4, Plumbing 4→5, Electrical 4→6, dummy 6→5, Drywall 5→7, Roofing 4→7, Painting 7→8. The dummy joins Plumbing and Electrical before Drywall; node 7 joins Drywall and Roofing before Painting. The forward scan (ES–EF) is Site prep 0–4, Foundation 4–10, Framing 10–18, Roofing 18–23, Plumbing 18–22, Electrical 18–21, Drywall 22–28, Painting 28–32. For the backward scan, start Painting at LF 32/LS 28. Then Drywall has LF 28/LS 22 and Roofing LF 28/LS 23; Plumbing has LF 22/LS 18 and Electrical LF 22/LS 19. Framing has LF 18/LS 10, Foundation LF 10/LS 4, and Site prep LF 4/LS 0. The critical path is Site prep–Foundation–Framing–Plumbing–Drywall–Painting, with duration 32 days. Roofing float is 23−18=5 days; Electrical float is 19−18=1 day. A valid Gantt schedule uses the ES–EF intervals above; Roofing ends at 23 and Electrical at 21, while Drywall waits for both Plumbing and Electrical and starts at 22.
(a) Crash plan and cost [2].
(b) Schedule analysis [2].
(c) Explanation [2].
Q2 worked model: (a) The critical route runs through Framing and Drywall. Crashing Framing by 2 days costs 2×$500=$1000. Crashing Drywall by 2 costs 2×$400=$800, so choose Drywall; the finish moves from day 32 to day 30, while the Roofing route finishes at day 27. (b) Framing runs from day 10 to 18. Plumbing and Electrical then start together at day 18, finishing at 22 and 21 respectively, so the two-trade cap is met without extra delay. Drywall runs 22–28, Roofing 18–23, and Painting 28–32; the project still finishes on day 32. (c) The claim is false for this network: Framing is a common predecessor on every route, so reducing it by 2 days shifts every route's finish 2 days earlier and the project finish also moves from 32 to 30. A route may become critical after other reductions, but that does not undo the two-day shift caused by shortening a shared activity. Drywall is not on the route through Roofing, so shortening Drywall can stop reducing the project once that other route controls.
(a) Proof [3].
(b) Optimal duration and cost [4].
(c) Evaluation [3].
Q3 worked model: (a) At each event, the earliest event time is the maximum of the earliest finish times of all incoming activities. Repeatedly applying that rule takes the longest accumulated duration among all routes reaching each event. At the final event this maximum is the project duration, so the critical path is a longest path through the network. (b) X costs $1500÷3=$500 per day saved, below the $600/day indirect cost, so crash all 3 available days. Y costs $1200÷2=$600 per day, exactly equal to the indirect saving, so crashing Y by 0, 1 or 2 days leaves total cost unchanged. The direct cost after crashing X is $21,500; indirect cost at 22 days is $13,200, total $34,700. At 21 days, direct cost is $22,100 and indirect cost is $12,600; at 20 days, direct cost is $22,700 and indirect cost is $12,000. Both also total $34,700. Thus the minimum is $34,700, at durations 20, 21 or 22 days. (c) The claim is too broad. Crashing is worthwhile only while the marginal crash cost is below the indirect saving and the crash reduces the actual controlling route; other critical paths, crash limits, and direct-cost effects can change the decision. At equal rates it breaks even, as Y does here.
Tick when you have finished all questions and checked your answers.