Question

An activity X has three direct precursor activities A, B, and C of duration 10, 14 and 09 days respectively. What is the earliest time that activity X can start?

• A. 14 days
• B. 33 days
• C. 23 days
• D. 24 days

Hint:

During a scheduling forward pass, multiple converging arrows use the Maximum rule: \[ \text{Earliest Start Time} = \max(\text{All Precursor Durations}) \]

Answer 1

The Correct Option is A

Concept: In project management scheduling (CPM/PERT network techniques), an activity's Earliest Start Time ($ES$) represents the earliest possible date it can begin. For any activity with multiple preceding dependencies, its earliest start time is determined by the completion of its longest predecessor path. This ensures all required preparatory work is finished before the next activity begins.

Step 1: Understanding Precursor Dependencies.
Let us analyze the logical requirements for Activity X:
• Activity X depends directly on three independent precursor paths: Activity A, Activity B, and Activity C.
• This means Activity X cannot begin until all three upstream activities are fully completed.

Step 2: Mathematical evaluation of the bottleneck path.
Assuming these precursor activities all start at Day 0, let us calculate their respective completion dates based on their given durations:
• Path A finishes at: $0 + 10 = 10 \text{ days}$
• Path B finishes at: $0 + 14 = 14 \text{ days}$
• Path C finishes at: $0 + 09 = 9 \text{ days}$ Now, evaluate the earliest start time for Activity X using the maximum function for a forward pass: \[ ES_X = \max(T_{\text{finish of A}}, \, T_{\text{finish of B}}, \, T_{\text{finish of C}}) \] \[ ES_X = \max(10, \, 14, \, 9) = 14 \text{ days} \] Activity X must wait for the longest predecessor path to finish, which is Activity B at 14 days. Therefore, the earliest Activity X can start is Day 14.