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.