A small project has 12 activities – N, P, Q, R, S, T, U, V, W, X, Y, and Z. The relationship among these activities and the duration of these activities are given in the Table.Activity Duration(in weeks) Depends upon N 2 - P 5 N Q 3 N R 4 P S 5 Q T 8 R U 7 R, S V 2 U W 3 U X 5 T, V Y 1 W Z 3 X, Y
The total float of the activity "V" (in weeks, in integer) is ________
| Activity | Duration(in weeks) | Depends upon |
|---|---|---|
| N | 2 | - |
| P | 5 | N |
| Q | 3 | N |
| R | 4 | P |
| S | 5 | Q |
| T | 8 | R |
| U | 7 | R, S |
| V | 2 | U |
| W | 3 | U |
| X | 5 | T, V |
| Y | 1 | W |
| Z | 3 | X, Y |
Correct Answer
Solution & Step-by-step Explanation
Activity Network and Calculations
To determine the total float for activity "V", we perform a Critical Path Method (CPM) analysis. This involves calculating the Earliest Start (ES), Earliest Finish (EF), Latest Start (LS), and Latest Finish (LF) for each activity.
Forward Pass (ES and EF)
The forward pass calculates the earliest possible times for starting and finishing activities.
- N: ES=0, EF = ES + Duration = 0 + 2 = 2
- P: ES=EF(N)=2, EF = 2 + 5 = 7
- Q: ES=EF(N)=2, EF = 2 + 3 = 5
- R: ES=EF(N)=2, EF = 2 + 4 = 6
- S: ES=EF(P)=7, EF = 7 + 5 = 12
- T: ES=EF(Q)=5, EF = 5 + 8 = 13
- U: ES=EF(R)=6, EF = 6 + 3 = 9
- V: ES=max(EF(R), EF(S)) = max(6, 12) = 12, EF = 12 + 2 = 14
- W: ES=EF(U)=9, EF = 9 + 3 = 12
- X: ES=EF(U)=9, EF = 9 + 5 = 14
- Y: ES=EF(T)=13, EF = 13 + 1 = 14
- Z: ES=max(EF(X), EF(Y)) = max(14, 14) = 14, EF = 14 + 3 = 17
The minimum project completion time is the maximum EF, which is 17 weeks.
Backward Pass (LF and LS)
The backward pass calculates the latest possible times activities can start and finish without delaying the project.
- Z: LF=17 (Project Duration), LS = LF - Duration = 17 - 3 = 14
- Y: LF=LS(Z)=14, LS = 14 - 1 = 13
- X: LF=LS(Z)=14, LS = 14 - 5 = 9
- W: LF=EF(W)=12 (Assuming no successors delay its LF), LS = 12 - 3 = 9
- V: LF=EF(V)=14 (Assuming no successors delay its LF), LS = 14 - 2 = 12
- T: LF=LS(Y)=13, LS = 13 - 8 = 5
- U: LF=min(LS(W), LS(X)) = min(9, 9) = 9, LS = 9 - 3 = 6
- S: LF=LS(V)=12, LS = 12 - 5 = 7
- R: LF=min(LS(U), LS(V)) = min(6, 12) = 6, LS = 6 - 4 = 2
- Q: LF=LS(T)=5, LS = 5 - 3 = 2
- P: LF=LS(S)=7, LS = 7 - 5 = 2
- N: LF=min(LS(P), LS(Q), LS(R)) = min(2, 2, 2) = 2, LS = 2 - 2 = 0
Total Float Calculation
Total Float (TF) is the amount of time an activity can be delayed without affecting the project's earliest finish time. It's calculated using the formula: or .
| Activity | EF (Weeks) | LF (Weeks) | TF = LF - EF (Weeks) |
|---|---|---|---|
| N | 2 | 2 | 0 |
| P | 7 | 7 | 0 |
| Q | 5 | 5 | 0 |
| R | 6 | 6 | 0 |
| S | 12 | 12 | 0 |
| T | 13 | 13 | 0 |
| U | 9 | 9 | 0 |
| V | 14 | 14 | 0 |
| W | 12 | 12 | 0 |
| X | 14 | 14 | 0 |
| Y | 14 | 14 | 0 |
| Z | 17 | 17 | 0 |
- Earliest Finish (EF) = 14 weeks
- Latest Finish (LF) = 14 weeks
- Total Float (TF) = LF - EF = 14 - 14 = 0 weeks.
Alternatively, using LS and ES:
- Earliest Start (ES) = 12 weeks
- Latest Start (LS) = 12 weeks
- Total Float (TF) = LS - ES = 12 - 12 = 0 weeks.
The total float for activity "V" is 0 weeks.