Mathematics Standard • Year 12 • Module 6 • Lesson 9

Scheduling with Constraints — Skill Drill

Build fluency in applying lags and leads to ES/EF calculations, and in adjusting schedules when resources are limited.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Lag formula: ES(successor) = EF(predecessor) + ____________.

Q1.2 Lead formula: ES(successor) = EF(predecessor) − ____________.

Q1.3 Resource smoothing uses an activity's ____________ to shift its start time so that resource demand does not exceed the limit.

Stuck? Revisit lesson § Lags and Leads and § Resource Constraints.

2. Worked example — applying a lag and a lead

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

Problem. A(5,−), B(3, A with lag 1), C(4, A with lead 2), D(2, B and C). Find ES, EF and project duration.

Step 1 — A has no predecessor.

A: ES = 0, EF = 0 + 5 = 5.

Reason: starting activity always begins at time 0.

Step 2 — B follows A with a LAG of 1 day (mandatory wait).

B: ES = EF(A) + 1 = 5 + 1 = 6. EF = 6 + 3 = 9.

Reason: lag adds a delay between predecessor finish and successor start.

Step 3 — C follows A with a LEAD of 2 days (overlap).

C: ES = EF(A) − 2 = 5 − 2 = 3. EF = 3 + 4 = 7.

Reason: lead allows C to start before A finishes (an overlap window).

Step 4 — D needs B AND C complete (merge).

D: ES = max(9, 7) = 9. EF = 9 + 2 = 11.

Reason: at a merge, ES = max(EF of predecessors).

Conclusion. Project duration = 11 days. A and C overlap on days 3 to 5 (the lead window).

3. Faded example — fill in the missing steps

Network: A(4,−), B(3, A lag 2), C(2, A lead 1), D(5, B), E(3, C), F(2, D and E). Fill in each blank line. 4 marks

Step 1 — A: ES = 0, EF = ____.

Step 2 — B with lag 2: ES = EF(A) + 2 = ____ + 2 = ____. EF = ____ + 3 = ____.

Step 3 — C with lead 1: ES = EF(A) − 1 = ____ − 1 = ____. EF = ____ + 2 = ____.

Step 4 — D: ES = EF(B) = ____. EF = ____ + 5 = ____.

Step 5 — E: ES = EF(C) = ____. EF = ____ + 3 = ____.

Step 6 — F merge: ES = max(EF(D), EF(E)) = max(____, ____) = ____. EF = ____ + 2 = ____.

Conclusion. Project duration = ____ days. Activities that overlap: ____ and ____ on days ____ to ____.

Stuck? Revisit lesson § Worked Example — lag adds, lead subtracts.

4. Graduated practice — Lag, lead and resource skills

Foundation — apply one lag/lead each (4 questions)

Q	Problem	Answer
4.1 1	A finishes on day 6. B follows A with lag 3. What is ES(B)?
4.2 1	A finishes on day 8. C follows A with lead 3. What is ES(C)?
4.3 1	Concrete cures for 3 days. This is an example of a (lag / lead / resource constraint).
4.4 1	Painters can start hallway 1 day before the previous trade finishes. This is a (lag / lead / resource constraint).

Standard — small networks with constraints (6 questions)

4.5 A(3,−), B(2, A lag 1), C(4, A). Find ES and EF for all three. 2 marks

4.6 A(5,−), B(3, A), C(4, A lead 2). Find ES and EF for all three. 2 marks

4.7 A(2,−), B(4, A lag 1), C(3, A), D(2, B and C). Find ES and EF for all four and state the project duration. 2 marks

4.8 Three activities P (5 days), Q (5 days), R (5 days) can all start on day 0. Each needs 2 workers. The site has 4 workers. Can all three run in parallel? Explain. 2 marks

4.9 Two activities X (3 days, no float) and Y (4 days, float = 2) can run in parallel but share one truck. How should they be scheduled? 2 marks

4.10 A project has 4 critical activities running in parallel on day 5, each needing 1 specialist crane operator. Only 2 operators are available. What is the impact on project duration if no float exists on any of these activities? 2 marks

Extension — combine constraints (2 questions)

4.11 A(4,−), B(3, A lag 1), C(5, A lead 2), D(2, B and C). Find ES, EF and project duration. Identify any overlap and its length. 3 marks

4.12 A small renovation has Paint (3 days) and Hang fixtures (1 day) with a mandatory 2-day drying lag between them. Plus a 1-day inspection that can start 1 day before Hang fixtures finishes (lead 1). Calculate total time from Paint start to Inspection finish. 3 marks

Stuck on 4.12? Paint 0–3, then 2-day lag (no work, just drying), then Hang fixtures 5–6, then Inspection starts 1 day early at 5, ends 6.

5. Self-check the easy 3

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

For 4.1 I added the lag (3) to A's EF (6), getting ES(B) = 9.

For 4.2 I subtracted the lead (3) from A's EF (8), getting ES(C) = 5.

For 4.3 I said lag — concrete curing is a mandatory wait between activities.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Lag formula

ES(successor) = EF(predecessor) + lag.

Q1.2 — Lead formula

ES(successor) = EF(predecessor) − lead.

Q1.3 — Resource smoothing

Uses an activity's float to shift its start time.

Q3 — Faded example

Step 1: A EF = 4.
Step 2: B: ES = 4 + 2 = 6, EF = 6 + 3 = 9.
Step 3: C: ES = 4 − 1 = 3, EF = 3 + 2 = 5.
Step 4: D: ES = 9, EF = 9 + 5 = 14.
Step 5: E: ES = 5, EF = 5 + 3 = 8.
Step 6: F: ES = max(14, 8) = 14, EF = 14 + 2 = 16.
Conclusion: Duration = 16 days. Overlap: A and C on days 3 to 4 (C started 1 day early during A's final day).

Q4.1 — Lag

ES(B) = 6 + 3 = 9.

Q4.2 — Lead

ES(C) = 8 − 3 = 5.

Q4.3 — Concrete curing

Lag — a mandatory wait between activities.

Q4.4 — Painters starting early

Lead — overlap allowing successor to start before predecessor finishes.

Q4.5 — A, B (lag 1), C

A: 0→3. B: ES = 3 + 1 = 4, EF = 4 + 2 = 6. C: ES = 3, EF = 3 + 4 = 7.

Q4.6 — A, B, C (lead 2)

A: 0→5. B: 5→8. C: ES = 5 − 2 = 3, EF = 3 + 4 = 7.

Q4.7 — A, B (lag 1), C, D merge

A: 0→2. B: ES = 2 + 1 = 3, EF = 7. C: 2→5. D: ES = max(7, 5) = 7, EF = 9. Duration = 9 days.

Q4.8 — Three activities, 2 workers each, 4 workers available

Total demand if all parallel = 3 × 2 = 6 workers. Available = 4. Cannot run all three in parallel. Must shift one activity (using float) or extend duration.

Q4.9 — X and Y share a truck

X has no float (must start on time). Y has 2 days float. Run X first (3 days), then Y (4 days) starting 3 days later. Y's 2-day float partially absorbs the delay, but Y will finish at day 7 instead of day 4 — check whether Y can still meet its LF deadline.

Q4.10 — 4 critical activities, 2 operators

With no float and only 2 operators, 2 of the 4 activities must wait. Each wait day delays the critical path. Project duration increases by the total wait time across the activities — at minimum, by the duration of the activities that must run sequentially instead of in parallel.

Q4.11 — Lag + lead + merge

A: 0→4. B (lag 1): ES = 5, EF = 8. C (lead 2): ES = 2, EF = 7. D merge: ES = max(8, 7) = 8, EF = 10. Duration = 10 days. Overlap: A (0–4) and C (2–7) overlap on days 2 to 4 (length = 2 days).

Q4.12 — Renovation with lag and lead

Paint: 0 → 3. Drying lag: must wait until day 5 before Hang fixtures starts. Hang fixtures: 5 → 6. Inspection (lead 1): ES = EF(Hang) − 1 = 6 − 1 = 5, EF = 5 + 1 = 6. Total time = 6 days from Paint start to Inspection finish.