CPM network times: Which statements are correct regarding earliest and latest times for activities and events in CPM?

Civil Engineering Construction Management Difficulty: Easy
Choose an option
  • A
    Earliest occurrence time of the event from which an activity originates is the earliest start time (ES) of that activity
  • B
    Earliest occurrence time of the event from which the activity originates plus the activity duration is the earliest finish time (EF)
  • C
    Latest occurrence time of the node where the activity terminates minus the activity duration is the latest start time (LS)
  • D
    Latest occurrence time of the node at which the activity terminates is the latest finish time (LF)
  • E
    All the above.

Answer

Correct Answer: All the above.

Explanation

Introduction / Context:Forward and backward pass calculations in CPM determine schedule flexibility (float) and the critical path. Understanding the definitions of ES, EF, LS, and LF is essential for correct computations.

Given Data / Assumptions:

  • Activity-on-Arrow or Activity-on-Node representation with consistent logic.
  • Deterministic durations are given.
  • No resource constraints are applied during time analysis.

Concept / Approach:Forward pass yields earliest event times and thus ES/EF; backward pass yields latest allowable times without delaying project finish, thus LS/LF. Float metrics derive from these times.

Step-by-Step Solution:ES of an activity equals the earliest time its start node occurs.EF = ES + duration.LF equals the latest allowable time of its finish node.LS = LF - duration.Zero float means EF = LF and ES = LS on critical activities.

Verification / Alternative check:Check consistency: EF of predecessors must be less than or equal to ES of successors. Any violation indicates a logic or calculation error.

Why Other Options Are Wrong:Each statement accurately defines schedule times; hence 'All the above' is correct.

Common Pitfalls:Mixing event times with activity times; neglecting multiple predecessors by not taking the maximum EF when computing successor ES.

Final Answer:All the above.

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