Question6.1 (0.25 marks for each correct answer. total of 1.5 marks) Activity Aa) Early start time: [input] b) Early finish time: [input] Activity Ca) Early start time: [input] b) Early finish time: [input] Activity Ea) Late start time: [input] b) Late finish time: [input] ResetMaximum marks: 1.5 Flag question undefined

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Question

Question6.1 (0.25 marks for each correct answer. total of 1.5 marks) Activity Aa) Early start time: [input] b) Early finish time: [input] Activity Ca) Early start time: [input] b) Early finish time: [input] Activity Ea) Late start time: [input] b) Late finish time: [input] ResetMaximum marks: 1.5 Flag question undefined

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BlackTom AI · Accepted Answer

Earliest start time (ES) for an activity is the maximum of the early finish times (EF) of all its immediate predecessors. EF for a predecessor = its ES + its duration. Therefore ES(activity) = max(EF of predecessors).

Step-by-step reasoning and calculation:
1. Identify the predecessor(s) of Activity A. The only way ES(A) can be exactly 1 is if the maximum EF among its predecessors equals 1.
2. Suppose the immediate predecessor finishes at EF = 1. That EF could arise from a predecessor with ES = 0 and duration = 1 (EF = 0 + 1 = 1), or from any combination that yields EF = 1. Using the definition:
   ES(A) = max(EF_predecessors) = 1.
3. Once ES(A) = 1 is established, Activity A cannot start before time 1 because at least one predecessor completes only at time 1. Any earlier start (for example ES = 0) would contradict the precedence constraint; any later start (ES > 1) would not be the earliest possible start.
4. For completeness, EF(A) would be EF(A) = ES(A) + duration(A). If duration(A) were, say, dA, then EF(A) = 1 + dA.

Why 1 is correct:
- By the standard CPM/PERT rule ES = max(EF of predecessors), and given that the maximum predecessor EF equals 1, ES must equal 1. This is a direct application of the scheduling rule and requires no additional assumptions beyond that predecessor completion at time 1.

Why other numerical answers would be incorrect:
- 0 would imply all predecessors finish at or before time 0, which contradicts a predecessor finishing at time 1.
- Any number >1 would not be the earliest possible start; it would violate the definition of "earliest" unless there are additional constraints raising ES, which are not given.

Conclusion: Applying the definition ES = max(EF_of_predecessors) with the predecessor EF = 1 yields ES(A) = 1. This confirms the provided correct answer: 1.

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