Difficulty: Medium
Correct Answer: Incorrect
Explanation:
Introduction / Context:Dummy activities in Activity-on-Arrow (AOA) networks have zero duration and exist only to preserve logical relationships. Understanding when any activity (including a dummy) is “critical” requires correct use of early/late times and float definitions.
Given Data / Assumptions:
Concept / Approach:A dummy has d = 0, hence EF = ES and LS = LF. It lies on the critical path only if ES = LS (equivalently EF = LF). The statement compares ES to LF (earliest start to latest finish), which is not the criterion for criticality. ES equal to LF is neither necessary nor sufficient to establish TF = 0.
Step-by-Step Solution:
Write EF = ES + 0 and LS = LF - 0 for a dummy.Compute TF = LS - ES = LF - EF.Critical if and only if ES = LS (or EF = LF).Note that ES = LF is unrelated to TF unless additional equalities hold.Verification / Alternative check:Construct a small network where a dummy connects two critical events; you will find ES = LS and EF = LF, not ES = LF.
Why Other Options Are Wrong:Marking the statement as “Correct” misapplies the definition of criticality.
Common Pitfalls:Mixing early and late times across different ends of an activity; forgetting that zero duration does not change the logic for float conditions.
Final Answer:Incorrect
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