Given the following information, draw the project network

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Given the following information, draw the project network. Compute the early, late, and slack times for the project network. Which activities on the critical path have only the start or finish of the activity on the critical path? Finish-to-Start Finish-to-StartAdditional Lag ID Duration Predecessor Lag Relationships Lag None None Finish-finish C to F 7 None None Finish-finish E to G 9 18 10 None Start-start G to H None Finish-finish ItoJ None 10 None 5 14 15 G, H

The Correct Answer and Explanation is :

o analyze the project network, we’ll construct an Activity-on-Node (AON) diagram, compute the early start (ES), early finish (EF), late start (LS), late finish (LF), and slack times for each activity, and identify which activities have only their start or finish on the critical path. 1. Constructing the Project Network Diagram:

ased on the provided information, the activities (A to J), their durations, and dependencies (including finish-to-finish and start-to-start relationships) are as follows:

  • A: Duration 18 days; no predecessor.
  • B: Duration 10 days; no predecessor.
  • C: Duration 5 days; predecessor A.
  • D: Duration 14 days; predecessor A.
  • E: Duration 15 days; predecessor B.
  • F: Duration 7 days; finish-to-finish with C (F can finish 7 days after C finishes).
  • G: Duration 9 days; finish-to-finish with E (G can finish 9 days after E finishes).
  • H: Duration 10 days; predecessor D; start-to-start with G (H can start when G starts).
  • I: Duration 5 days; predecessor F.
  • J: Duration 14 days; predecessor I; finish-to-finish with H (J can finish when H finishes).

2. Calculating Early Start (ES), Early Finish (EF), Late Start (LS), Late Finish (LF), and Slack Times:

e’ll perform a forward pass to determine ES and EF, followed by a backward pass to determine LS and LF. Slack is calculated as LS – ES or LF – EF. Forward Pass:

  • A: ES = 0; EF = ES + Duration = 0 + 18 = 18
  • B: ES = 0; EF = 10
  • C: ES = EF of A = 18; EF = 18 + 5 = 23
  • D: ES = EF of A = 18; EF = 18 + 14 = 32
  • E: ES = EF of B = 10; EF = 10 + 15 = 25
  • F: EF = EF of C + 7 (lag) = 23 + 7 = 30; ES = EF – Duration = 30 – 7 = 23
  • G: EF = EF of E + 9 (lag) = 25 + 9 = 34; ES = EF – Duration = 34 – 9 = 25
  • H: ES = ES of G = 25; EF = 25 + 10 = 35
  • I: ES = EF of F = 30; EF = 30 + 5 = 35
  • J: EF = EF of H = 35; ES = EF – Duration = 35 – 14 = 21

Backward Pass:

  • J: LF = EF = 35; LS = LF – Duration = 35 – 14 = 21
  • I: LF = LS of J = 21; LS = LF – Duration = 21 – 5 = 16
  • H: LF = LF of J = 35; LS = LF – Duration = 35 – 10 = 25
  • G: LF = LS of H = 25; LS = LF – Duration = 25 – 9 = 16
  • F: LF = LS of I = 16; LS = LF – Duration = 16 – 7 = 9
  • E: LF = LF of G – 9 (lag) = 16 – 9 = 7; LS = LF – Duration = 7 – 15 = -8 (indicating a scheduling conflict)
  • D: LF = LS of H = 25; LS = LF – Duration = 25 – 14 = 11
  • C: LF = LS of F – 7 (lag) = 9 – 7 = 2; LS = LF – Duration = 2 – 5 = -3 (indicating a scheduling conflict)
  • B: LF = LS of E = -8; LS = LF – Duration = -8 – 10 = -18 (indicating a scheduling conflict)
  • A: LF = min(LS of C, LS of D) = min(-3, 11) = -3; LS = LF – Duration = -3 – 18 = -21 (indicating a scheduling conflict)

Slack Calculation:

  • A: Slack = LS – ES = -21 – 0 = -21
  • B: Slack = -18
  • C: Slack = -3
  • D: Slack = LS – ES = 11 – 18 = -7
  • E: Slack = -8
  • F: Slack = LS – ES = 9 – 23 = -14
  • G: Slack = LS – ES = 16 – 25 = -9
  • H: Slack = LS – ES = 25 – 25 = 0
  • I: Slack = LS – ES = 16 – 30 = -14
  • J: Slack = LS – ES = 21 – 21 = 0

3. Identifying Activities on the Critical Path:

he critical path is the sequence of activities with zero slack. From the calculations, activities H and J have zero slack. Activities with Only Start or Finish on the Critical Path:

  • H: Both start and finish are on the critical path.
  • J: Both start and finish are on the critical path.

herefore, in this project network, activities H and J are on the critical path, with both their start and finish times critical. Explanation:

he project network is analyzed using the Precedence Diagramming Method (PDM), which allows for various types of dependencies, including finish-to-finish (FF) and start-to-start (SS) relationships.n the forward pass, we calculate the earliest possible start and finish times for each activity, considering their durations and dependencies.he backward pass then determines the latest possible start and finish times without delaying the project.lack time, or float, indicates the flexibility available for each activity; activities with zero slack are critical and cannot be delayed without affecting the project’s completion time.n this analysis, activities H and J are identified as critical, meaning any delay in these activities will directly impact the project’s finish date. t’s important to note that negative slack values indicate scheduling conflicts, suggesting that the planned schedule is not feasible as currently structured.his

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