Mathematics Standard • Year 12 • Module 6 • Lesson 4

Forward and Backward Scan — Skill Drill

Build fluency in the forward and backward scan: earliest start (ES), earliest finish (EF), latest start (LS) and latest finish (LF) — with the right rule at every merge and burst.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the formulas.

EF = ES + ____ LS = LF − ____.

Q1.2 At a merge point in the forward scan, take the ____________ (max/min) of the predecessors' EFs.

At a burst point in the backward scan, take the ____________ (max/min) of the successors' LSs.

Q1.3 For a start activity (no predecessor), ES = ____.

For a finish activity (no successor), LF = ____________ (give the value in terms of the project total).

Stuck? Revisit lesson § Key Ideas — Forward scan, Backward scan, merge vs burst.

2. Worked example — full forward and backward scan

Follow each line of working. Every step has a reason on the right.

Problem. A project has A(3, —), B(4, A), C(2, A), D(5, B and C). Perform the forward and backward scans and state the project duration.

Step 1 — Forward scan: ES and EF for every activity, starting at A.

A: ES = 0, EF = 0 + 3 = 3
B: ES = EF(A) = 3, EF = 3 + 4 = 7
C: ES = EF(A) = 3, EF = 3 + 2 = 5
D: ES = max(EF(B), EF(C)) = max(7, 5) = 7, EF = 7 + 5 = 12

Reason: at the merge into D, take the MAX so D waits for the slowest predecessor.

Step 2 — Project duration = EF of the (final) finish activity.

Project duration = EF(D) = 12 days

Step 3 — Backward scan: LF and LS for every activity, starting at the finish.

D: LF = 12 (project duration), LS = 12 − 5 = 7
B: LF = LS(D) = 7, LS = 7 − 4 = 3
C: LF = LS(D) = 7, LS = 7 − 2 = 5
A: LF = min(LS(B), LS(C)) = min(3, 5) = 3, LS = 3 − 3 = 0

Reason: at the burst out of A, take the MIN so A finishes early enough for its most restrictive successor.

Step 4 — Tabulate.

Act | Dur | ES | EF | LS | LF
A | 3 | 0 | 3 | 0 | 3
B | 4 | 3 | 7 | 3 | 7
C | 2 | 3 | 5 | 5 | 7
D | 5 | 7 | 12 | 7 | 12

Conclusion. Project duration = 12 days. Notice C has LS ≠ ES, so it has some "slack"; A, B and D have ES = LS (zero slack).

3. Faded example — fill in the missing scan values

Activities P(2, —), Q(3, P), R(4, P), S(2, Q), T(3, R), U(1, S and T). Complete the scans. 4 marks

Forward scan:

P: ES = 0, EF = 0 + 2 = ____
Q: ES = ____, EF = ____ + 3 = ____
R: ES = ____, EF = ____ + 4 = ____
S: ES = EF(Q) = ____, EF = ____ + 2 = ____
T: ES = EF(R) = ____, EF = ____ + 3 = ____
U: ES = max(EF(S), EF(T)) = max(____, ____) = ____, EF = ____ + 1 = ____

Project duration = ____ days.

Backward scan:

U: LF = ____, LS = ____ − 1 = ____
S: LF = LS(U) = ____, LS = ____ − 2 = ____
T: LF = LS(U) = ____, LS = ____ − 3 = ____
Q: LF = LS(S) = ____, LS = ____ − 3 = ____
R: LF = LS(T) = ____, LS = ____ − 4 = ____
P: LF = min(LS(Q), LS(R)) = min(____, ____) = ____, LS = ____ − 2 = ____

Stuck? Revisit lesson § Worked Example — forward uses MAX at merges; backward uses MIN at bursts.

4. Graduated practice — scans and small networks

Show your working in the space below each part. For full scans, write the formula line then the substitution.

Foundation — single-activity calculations (4 questions)

Q	Problem	Answer
4.1 1	Activity X has ES = 4 and duration 6. Find EF.
4.2 1	Activity Y has LF = 18 and duration 5. Find LS.
4.3 1	Activity Z has two predecessors with EFs 7 and 10. Find ES(Z).
4.4 1	Activity W has two successors with LSs 12 and 9. Find LF(W).

Standard — typical HSC difficulty (6 questions)

For each scan question, show at least the formula line then the substituted values.

4.5 Perform the forward scan only for A(4, —), B(3, A), C(2, A), D(5, B), E(1, C and D). Project duration? 2 marks

4.6 Perform the backward scan only for the network in Q4.5 (using your duration from 4.5). Report LS and LF for every activity. 2 marks

4.7 Perform both scans for A(3, —), B(4, A), C(2, A), D(5, B), E(3, C), F(2, D and E). Project duration? 2 marks

4.8 A student computes ES(D) using min(EF of predecessors) at a merge point. State the correct rule and give the corrected answer if predecessors have EFs 5 and 8. 2 marks

4.9 A student computes LF(A) using max(LS of successors) at a burst point. State the correct rule and give the corrected answer if successors have LSs 6 and 11. 2 marks

4.10 Perform both scans for A(2, —), B(3, —), C(4, A), D(3, A and B), E(2, C and D). Project duration? 2 marks

Extension — full table (2 questions)

4.11 Activities: A(4, —), B(3, A), C(2, A), D(5, B and C), E(1, D). Build a six-column table (Act | Dur | ES | EF | LS | LF). 3 marks

4.12 Activities: A(3, —), B(2, A), C(4, A), D(2, B), E(3, C), F(1, D and E), G(2, E). Build a six-column table. Project duration? 3 marks

Stuck on 4.12? F and G are both finish activities — set both LFs equal to the project duration (the max EF).

5. Self-check the easy 3

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

For Q4.1 (ES = 4, dur = 6) I used EF = ES + duration = 4 + 6 = 10.

For Q4.3 (two predecessors with EFs 7, 10) I used MAX, not MIN, to get ES = 10.

For Q4.4 (two successors with LSs 12, 9) I used MIN, not MAX, to get LF = 9.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formulas

EF = ES + duration. LS = LF − duration.

Q1.2 — Merge vs burst rules

Forward at a merge: take the maximum EF of the predecessors. Backward at a burst: take the minimum LS of the successors.

Q1.3 — Start and finish defaults

For a start activity, ES = 0. For a finish activity, LF = project duration (the maximum EF in the project).

Q3 — Faded scans (P, Q, R, S, T, U)

Forward. P: ES=0, EF=2. Q: ES=2, EF=5. R: ES=2, EF=6. S: ES=5, EF=7. T: ES=6, EF=9. U: ES=max(7,9)=9, EF=10.
Project duration = 10 days.

Backward. U: LF=10, LS=9. S: LF=9, LS=7. T: LF=9, LS=6. Q: LF=7, LS=4. R: LF=6, LS=2. P: LF=min(4,2)=2, LS=0.

Q4.1 — EF given ES and duration

EF = 4 + 6 = 10.

Q4.2 — LS given LF and duration

LS = 18 − 5 = 13.

Q4.3 — ES at a merge

ES(Z) = max(7, 10) = 10.

Q4.4 — LF at a burst

LF(W) = min(12, 9) = 9.

Q4.5 — Forward scan

A: ES=0, EF=4. B: ES=4, EF=7. C: ES=4, EF=6. D: ES=7, EF=12. E: ES=max(6,12)=12, EF=13. Project duration = 13.

Q4.6 — Backward scan (project duration 13)

E: LF=13, LS=12. D: LF=12, LS=7. C: LF=12, LS=10. B: LF=7, LS=3. A: LF=min(3,10)=3, LS=0.

Q4.7 — Both scans on A-F

Forward: A(0,3), B(3,7), C(3,5), D(7,12), E(5,8), F(max(12,8)=12, 14). Duration = 14.
Backward: F LF=14, LS=12. D LF=12, LS=7. E LF=12, LS=9. B LF=7, LS=3. C LF=9, LS=7. A LF=min(3,7)=3, LS=0.

Q4.8 — Merge rule fix

The correct rule is MAX at a merge. ES(D) = max(5, 8) = 8. (Reason: D cannot start until the slowest predecessor finishes.)

Q4.9 — Burst rule fix

The correct rule is MIN at a burst. LF(A) = min(6, 11) = 6. (Reason: A must finish early enough for the most restrictive successor.)

Q4.10 — Two starts, one merge into E

Forward: A(0,2), B(0,3), C(2,6), D(max(2,3)=3, 6), E(max(6,6)=6, 8). Duration = 8.
Backward: E LF=8, LS=6. C LF=6, LS=2. D LF=6, LS=3. A LF=min(2,3)=2, LS=0. B LF=3, LS=0.

Q4.11 — Six-column table for A-E

A | 4 | 0 | 4 | 0 | 4
B | 3 | 4 | 7 | 4 | 7
C | 2 | 4 | 6 | 5 | 7
D | 5 | 7 | 12 | 7 | 12
E | 1 | 12 | 13 | 12 | 13
Project duration = 13.

Q4.12 — Table with two finish activities

Forward: A(0,3), B(3,5), C(3,7), D(5,7), E(7,10), F(max(7,10)=10, 11), G(10, 12). Duration = max(11, 12) = 12.
Backward: F LF=12, LS=11. G LF=12, LS=10. D LF=11, LS=9. E LF=min(11,10)=10, LS=7. B LF=9, LS=7. C LF=7, LS=3. A LF=min(7,3)=3, LS=0.

Table:
A | 3 | 0 | 3 | 0 | 3
B | 2 | 3 | 5 | 7 | 9
C | 4 | 3 | 7 | 3 | 7
D | 2 | 5 | 7 | 9 | 11
E | 3 | 7 | 10 | 7 | 10
F | 1 | 10 | 11 | 11 | 12
G | 2 | 10 | 12 | 10 | 12