Question

The activities of W, X, Y and Z are the direct precursors of A. What is the earliest starting time for A, if the activities of earliest finishing times are 12, 15, 10 and 6?

• A. 6
• B. 10
• C. 15
• D. 12

Hint:

Remember these path routing rules for network diagrams: - Forward Pass (Earliest Times): Use the MAXIMUM value of preceding finish times when paths merge. - Backward Pass (Latest Times): Use the MINIMUM value of succeeding start times when paths split.

Answer 1

The Correct Option is C

Concept: In project scheduling frameworks using the Critical Path Method (CPM), the timing of network tasks is calculated via a forward-pass analysis. For any given activity $A$, its Earliest Start Time ($ES_A$) represents the earliest possible time step at which the activity can begin. If an activity has multiple immediate preceding tasks (precursors), it cannot start until *every single one* of those prerequisite tasks has finished. Therefore, during the forward pass, the Earliest Start time of a dependent activity is determined by selecting the absolute maximum value among the Earliest Finish times ($EF$) of all its direct preceding tasks: \[ ES_A = \max \{ EF_{\text{precursor 1}}, \, EF_{\text{precursor 2}}, \, \dots, \, EF_{\text{precursor } n} \} \]

Step 1: Isolate the precursor data.
We are given that activity $A$ depends on four independent parallel precursors: $W, X, Y,$ and $Z$. Their respective earliest finishing times are: * $EF_W = 12$ * $EF_X = 15$ * $EF_Y = 10$ * $EF_Z = 6$

Step 2: Apply the forward pass maximum selection rule.
Because activity $A$ cannot begin if any of its prerequisites remain incomplete, it must wait for the longest task to finish. Let us calculate the maximum value from the pool of finishing times: \[ ES_A = \max \{ EF_W, \, EF_X, \, EF_Y, \, EF_Z \} \] \[ ES_A = \max \{ 12, \, 15, \, 10, \, 6 \} \] \[ ES_A = 15 \] Thus, the earliest starting time for activity $A$ is $15$, which corresponds to Option (3).