Mathematics Standard • Year 12 • Module 6 • Lesson 8

Multiple Critical Paths — Skill Drill

Build fluency in finding ALL critical paths in a network, spotting shared activities, and planning efficient crashing when more than one path is critical.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 A network has paths of length 11, 11 and 9 days. The project duration is ____ days, and ____ paths are critical.

Q1.2 An activity that lies on more than one critical path is called a ____________ activity.

Q1.3 To reduce the project duration when there are two critical paths sharing no activities, you must crash at least one activity on ____________ path.

Stuck? Revisit lesson § Identifying Multiple Paths and § Shared Activities.

2. Worked example — finding all critical paths

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

Problem. A network has: A(3,−), B(4,A), C(2,A), D(3,B), E(5,C), F(2,D,E). Find ALL critical paths.

Step 1 — Forward scan (ES and EF).

A: ES=0, EF=3. B: ES=3, EF=7. C: ES=3, EF=5. D: ES=7, EF=10. E: ES=5, EF=10. F: ES=max(10,10)=10, EF=12.

Reason: at burst points, ES = max(EF of predecessors).

Step 2 — List every path and its length.

Path 1: A → B → D → F = 3 + 4 + 3 + 2 = 12.
Path 2: A → C → E → F = 3 + 2 + 5 + 2 = 12.

Reason: project duration is the longest path. Any path equal to that length is critical.

Step 3 — Identify shared activities.

Both paths contain A (at the start) and F (at the end). B, D are unique to Path 1; C, E are unique to Path 2.

Step 4 — State the answer clearly.

Project duration = 12 days. Critical paths: A → B → D → F and A → C → E → F. Shared critical activities: A, F.

3. Faded example — fill in the missing steps

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

Step 1 — Forward scan:

A: ES=0, EF=____. B: ES=____, EF=____. C: ES=____, EF=____. D: ES=____, EF=____. E: ES=____, EF=____. F: ES=max(____,____)=____, EF=____.

Step 2 — Length of each path:

Path A-B-D-F = 2 + ____ + ____ + ____ = ____.
Path A-C-E-F = 2 + ____ + ____ + ____ = ____.

Step 3 — Critical paths:

Project duration = ____ days. Critical path(s) = ________________.

Step 4 — Shared activities: ____________ (list those on every critical path).

Stuck? Revisit lesson § Worked Example — list every path, compute its length, compare to project duration.

4. Graduated practice — Multiple critical path skills

Show working: list each path and its length, then state which are critical.

Foundation — read off given path lengths (4 questions)

Q	Problem	Answer
4.1 1	Path lengths: 10, 8, 7. Which path(s) are critical and what is the project duration?
4.2 1	Path lengths: 14, 14, 13. How many critical paths?
4.3 1	Path lengths: 20, 20, 20. How many critical paths?
4.4 1	True or False: A near-critical path has float greater than zero but small.

Standard — find critical paths from a small network (6 questions)

Show working: list every path and its length, then identify the critical set.

4.5 A(2,−), B(3,A), C(4,A), D(2,B), E(1,C), F(3,D,E). Find all critical path(s). 2 marks

4.6 A(3,−), B(2,A), C(2,A), D(4,B), E(4,C), F(1,D,E). Find critical path(s) and shared activities. 2 marks

4.7 A(4,−), B(3,A), C(3,A), D(5,B), E(5,C), F(2,D,E). What is the project duration and how many critical paths are there? 2 marks

4.8 Project has two critical paths: A → B → D → F and A → C → E → F (each 12 days). Activity B can be crashed by 1 day. Will the project duration shorten? Explain in one sentence. 2 marks

4.9 Same project as 4.8. If activity A (shared) is crashed by 1 day, what is the new project duration? 2 marks

4.10 A project's critical path is 18 days. A near-critical path is 17 days. If a delay adds 1 day to the critical path, what now happens? 2 marks

Extension — crash strategy with multiple paths (2 questions)

4.11 A project has THREE critical paths sharing no common activities. You need to cut 1 day. Explain the minimum number of activity-crashes needed and why. 3 marks

4.12 A project has critical paths X (A → B → D → F) and Y (A → C → E → F). Crashing A costs $300/day; B costs $200/day; C costs $400/day. Cut the duration by 1 day at minimum cost. Show your choice and the cost. 3 marks

Stuck on 4.12? Crashing the shared activity A cuts BOTH paths for one cost; crashing B alone leaves Path Y at 12.

5. Self-check the easy 3

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

For 4.1 I picked 10 days as the project duration (longest path) and said only the 10-day path is critical.

For 4.2 I counted exactly 2 critical paths (both 14-day paths) — the 13-day path is not critical.

For 4.4 I said False — near-critical means float close to zero but greater than zero, with the activity still NOT on a critical path.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Two paths of equal max length

Project duration = 11 days. 2 paths are critical.

Q1.2 — Shared activity

A shared critical activity.

Q1.3 — Crashing two disjoint critical paths

You must crash at least one activity on each (every) critical path.

Q3 — Faded example

Step 1: A(0,2), B(2,5), C(2,4), D(5,9), E(4,9), F(9,10).
Step 2: A-B-D-F = 2 + 3 + 4 + 1 = 10. A-C-E-F = 2 + 2 + 5 + 1 = 10.
Step 3: Project duration = 10 days. Critical paths: A → B → D → F and A → C → E → F.
Step 4: Shared = A and F.

Q4.1 — Path lengths 10, 8, 7

Project duration = 10 days. Only the 10-day path is critical.

Q4.2 — Path lengths 14, 14, 13

2 critical paths (both 14-day paths). The 13-day path has 1-day float.

Q4.3 — Path lengths 20, 20, 20

3 critical paths — all share the same maximum length.

Q4.4 — Near-critical definition

True — near-critical paths have small but positive float; they are not currently critical but could become so if the critical path is delayed.

Q4.5 — Network A,B,C,D,E,F

Forward: A(0,2), B(2,5), C(2,6), D(5,7), E(6,7), F(7,10). Paths: A-B-D-F = 2+3+2+3 = 10. A-C-E-F = 2+4+1+3 = 10. Both critical (10 days). Shared: A, F.

Q4.6 — Network A,B,C,D,E,F (alt)

A(0,3), B(3,5), C(3,5), D(5,9), E(5,9), F(9,10). Paths: A-B-D-F = 3+2+4+1 = 10. A-C-E-F = 3+2+4+1 = 10. Both critical (10 days). Shared: A and F.

Q4.7 — Symmetric network

A(0,4), B(4,7), C(4,7), D(7,12), E(7,12), F(12,14). Paths: A-B-D-F = 4+3+5+2 = 14. A-C-E-F = 4+3+5+2 = 14. Project duration = 14 days. 2 critical paths.

Q4.8 — Crashing only one path

No. Crashing B shortens A-B-D-F to 11 days, but A-C-E-F is unchanged at 12. Project duration stays at 12 (controlled by the unshortened critical path).

Q4.9 — Crashing a shared activity

A is on BOTH critical paths. Crashing A by 1 day shortens both: A-B-D-F = 11, A-C-E-F = 11. New duration = 11 days. Shared-activity crashes give double value.

Q4.10 — Near-critical promotion

Critical becomes 19 days; near-critical (still 17 + 0 = 17? but if the delay is on critical, near-critical is still 17). Actually: critical now 19, near-critical still 17 — gap is 2, so near-critical is no longer "near". Corrected: If the delay applies generally — both paths slip by 1 — critical is 19 and near-critical is 18. Difference shrinks but near-critical still NOT critical. If the delay applies only to the critical, near-critical stays 17 and the gap grows.

Q4.11 — Three disjoint critical paths, cut 1 day

You must crash one activity on each of the three paths = minimum 3 crashes. Reason: with no shared activities, each path is reduced independently. Shortening only one or two leaves the others at the original duration, so the project duration is unchanged.

Q4.12 — Choose cheapest 1-day cut

Option 1 — crash A (shared) by 1 day = $300, both paths drop by 1 ⇒ new duration = original − 1.
Option 2 — crash B (Path X only) by 1 day = $200, but Path Y unchanged ⇒ project unchanged.
Option 3 — crash B + C (one per path) = $200 + $400 = $600, both paths drop ⇒ new duration = original − 1.
Cheapest = $300 (crash A by 1 day). Insight: a shared-activity crash beats two separate crashes whenever cost(shared) < sum of separates.