Mathematics Standard • Year 12 • Module 6 • Lesson 4
Forward and Backward Scan — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on forward/backward scans: clean tables, max/min reasoning, and one structured analysis of slack.
1. Short-answer questions
1.1 A project has activities A(3, —), B(4, A), C(2, A), D(5, B and C). Perform the forward scan and state the project duration. 3 marks Band 3
1.2 A student writes "At a merge point in the forward scan, take the minimum EF of the predecessors." State the correct rule, and explain in one sentence why the rule is the way it is. Then apply your rule to ES(X) when X's predecessors have EFs of 8 and 12. 3 marks Band 3-4
1.3 A project has the precedence table A(2, —), B(4, A), C(3, A), D(5, B), E(2, C), F(3, D and E).
(a) Perform both scans and present the results in a six-column table (Activity | Duration | ES | EF | LS | LF).
(b) State the project duration and the slack on activity C. 4 marks Band 4
2. Extended response
2.1 A small NSW road-bridge upgrade is described by the precedence table below (days):
Permits PM(3, —)
Demolition DM(5, PM)
Steel order SO(7, PM)
Site prep SP(4, DM)
Steel erect SE(6, SP and SO)
Concrete CO(4, SE)
Inspect IN(2, CO)
(a) Perform both scans and present the results in a six-column table (Activity | Duration | ES | EF | LS | LF).
(b) State the project duration and identify the activities with zero slack (LS = ES).
(c) The project manager asks: "If we can shorten DM by 2 days, will the bridge upgrade finish 2 days earlier?" Use your scan results to give a clear yes/no answer with reasoning. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correct ES, EF for every activity (forward scan).
• 1 mark — correct LS, LF for every activity (backward scan).
• 1 mark — clean six-column table with units consistent across all rows.
Part (b) — 2 marks
• 1 mark — correct project duration.
• 1 mark — all activities with LS = ES correctly listed.
Part (c) — 2 marks
• 1 mark — applies the 2-day change to DM and recomputes the relevant path-sum (or recognises DM is/is not on the controlling path).
• 1 mark — clear yes/no conclusion sentence with reasoning that references the scan results.
Your response:
Stuck on (c)? Check the path DM is on. If DM has slack (LS > ES), shortening DM does not shorten the project.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Forward scan ABCD (3 marks)
Sample response.
A: ES = 0, EF = 0 + 3 = 3.
B: ES = 3, EF = 3 + 4 = 7.
C: ES = 3, EF = 3 + 2 = 5.
D: ES = max(EF(B), EF(C)) = max(7, 5) = 7, EF = 7 + 5 = 12.
Project duration = 12 days.
Marking notes. 1 mark — A and B (or A and C) correct. 1 mark — merge at D using MAX. 1 mark — project duration stated with units. A bare "12 days" with no scan loses 2 marks.
1.2 — Merge-rule correction (3 marks)
Sample response. The correct rule is "At a merge point, take the maximum EF of the predecessors." The rule is MAX because the activity cannot start until every predecessor has finished — so the slowest predecessor (largest EF) controls the start time. Applying this: ES(X) = max(8, 12) = 12.
Marking notes. 1 mark — names MAX as the correct rule. 1 mark — explains the "wait for all predecessors" reason. 1 mark — correct numerical answer 12 (not 8). Answering only "max" with no reason scores 1/3.
1.3 — Full scan table for A-F (4 marks)
(a) Sample response.
Forward: A(0,2), B(2,6), C(2,5), D(6,11), E(5,7), F(max(11,7)=11, 14).
Backward (F LF=14): F LS=11. D LF=11, LS=6. E LF=11, LS=9. B LF=6, LS=2. C LF=9, LS=6. A LF=min(2,6)=2, LS=0.
Table:
A | 2 | 0 | 2 | 0 | 2
B | 4 | 2 | 6 | 2 | 6
C | 3 | 2 | 5 | 6 | 9
D | 5 | 6 | 11 | 6 | 11
E | 2 | 5 | 7 | 9 | 11
F | 3 | 11 | 14 | 11 | 14
(b) Sample response. Project duration = 14 days. Slack on C = LS − ES = 6 − 2 = 4 days.
Marking notes. 1 mark — forward scan correct. 1 mark — backward scan correct (including MIN at A). 1 mark — clean table. 1 mark — duration and C's slack both correct.
2.1 — Bridge upgrade (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Scans.
Forward:
PM(0, 3)
DM(3, 8)
SO(3, 10)
SP(8, 12)
SE(max(12, 10) = 12, 18)
CO(18, 22)
IN(22, 24)
Project duration = EF(IN) = 24 days. [1 mark — forward scan correct.]
Backward (IN LF = 24):
IN: LF = 24, LS = 22
CO: LF = 22, LS = 18
SE: LF = 18, LS = 12
SP: LF = 12, LS = 8
SO: LF = 12, LS = 5
DM: LF = 8, LS = 3
PM: LF = min(LS(DM), LS(SO)) = min(3, 5) = 3, LS = 0
[1 mark — backward scan correct including MIN at PM.]
Six-column table:
Act | Dur | ES | EF | LS | LF
PM | 3 | 0 | 3 | 0 | 3
DM | 5 | 3 | 8 | 3 | 8
SO | 7 | 3 | 10 | 5 | 12
SP | 4 | 8 | 12 | 8 | 12
SE | 6 | 12 | 18 | 12 | 18
CO | 4 | 18 | 22 | 18 | 22
IN | 2 | 22 | 24 | 22 | 24
[1 mark — clean table with consistent units.]
(b) Duration and zero-slack activities.
Project duration = 24 days. [1 mark.]
Activities with LS = ES (zero slack): PM, DM, SP, SE, CO, IN. SO is the only activity with slack (LS − ES = 5 − 3 = 2 days). [1 mark.]
(c) Shortening DM by 2 days.
DM has zero slack (LS = ES = 3), so DM lies on the controlling path PM → DM → SP → SE → CO → IN of length 3 + 5 + 4 + 6 + 4 + 2 = 24. [1 mark — applies change and identifies DM is on the controlling path.]
Shortening DM by 2 days reduces this path-sum to 22 days; the other path through SO is now PM(3) + SO(7) + SE(6) + CO(4) + IN(2) = 22 days. Both paths give 22 days. Conclusion: yes — the bridge upgrade finishes 2 days earlier, taking 22 days, but both paths are now critical, so any further reduction must happen on activities common to both. [1 mark — clear conclusion with reasoning that references the scans.]
Total: 7/7.
Band descriptors for marker.
Band 3: Forward scan correct, project duration stated, but no backward scan or no table. ≈ 3 marks.
Band 4: Both scans complete and table present, but does not list zero-slack activities or does not analyse DM. ≈ 5 marks.
Band 5: Full table, duration and zero-slack activities correct; (c) gives "yes, 22 days" but does not note that both paths now control the project. ≈ 6 marks.
Band 6: Complete table, full backward scan including MIN at PM, all zero-slack activities listed, AND (c) notes that the SO path now also equals 22 so both paths become critical. 7/7.