GATE CE 2024 Set 2 — Question 52

NAT+2 / -0MediumCritical PathConstruction ManagementStructural Engineering

Structural Engineering → Construction Management → Critical Path

Last updated 5 September 2026

Question

The table shows the activities and their durations and dependencies in a project. Activity Duration (Days) Depends on A 8 - B 4 A C 4 B D 4 C, L F 4 A G 4 F H 6 G, L K 10 A L 6 F, K The total duration (in days) of the project is___________(in integer).

Correct answer 30 to 30 Solution We calculate the Earliest Finish (EF) for each activity based on dependencies: A : EF = 8 B : EF = EF(A) + 4 = 8 + 4 = 12 C : EF = EF(B) + 4 = 12 + 4 = 16 F : EF = EF(A) + 4 = 8 + 4 = 12 G : EF = EF(F) + 4 = 12 + 4 = 16 K : EF = EF(A) + 10 = 8 + 10 = 18 L : EF = max(EF(F), EF(K)) + 6 = max(12, 18) + 6 = 18 + 6 = 24 D : EF = max(EF(C), EF(L)) + 4 = max(16, 24) + 4 = 24 + 4 = 28 H : EF = max(EF(G), EF(L)) + 6 = max(16, 24) + 6 = 24 + 6 = 30 The project duration is the maximum of the EF of all terminal activities (D and H): Duration = max(EF(D), EF(H)) = max(28, 30) = 30 days. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer 30 to 30 Solution We calculate the Earliest Finish (EF) for each activity based on dependencies: A : EF = 8 B : EF = EF(A) + 4 = 8 + 4 = 12 C : EF = EF(B) + 4 = 12 + 4 = 16 F : EF = EF(A) + 4 = 8 + 4 = 12 G : EF = EF(F) + 4 = 12 + 4 = 16 K : EF = EF(A) + 10 = 8 + 10 = 18 L : EF = max(EF(F), EF(K)) + 6 = max(12, 18) + 6 = 18 + 6 = 24 D : EF = max(EF(C), EF(L)) + 4 = max(16, 24) + 4 = 24 + 4 = 28 H : EF = max(EF(G), EF(L)) + 6 = max(16, 24) + 6 = 24 + 6 = 30 The project duration is the maximum of the EF of all terminal activities (D and H): Duration = max(EF(D), EF(H)) = max(28, 30) = 30 days.

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