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mediumSHORT_ANSWERGATE CE 2021 Question Paper (06-Feb-2021) (Shift-1)CE
2 marks

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.
ActivityDuration(in weeks)Depends upon
N2-
P5N
Q3N
R4P
S5Q
T8R
U7R, S
V2U
W3U
X5T, V
Y1W
Z3X, Y
The total float of the activity "V" (in weeks, in integer) is ________

Correct Answer

Solution & Step-by-step Explanation

In the range: 0 - 0

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 .
ActivityEF (Weeks)LF (Weeks)TF = LF - EF (Weeks)
N220
P770
Q550
R660
S12120
T13130
U990
V14140
W12120
X14140
Y14140
Z17170
For activity "V":

- 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.

Practice this question

Try it yourself before checking the explanation above.

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.
ActivityDuration(in weeks)Depends upon
N2-
P5N
Q3N
R4P
S5Q
T8R
U7R, S
V2U
W3U
X5T, V
Y1W
Z3X, Y
The total float of the activity "V" (in weeks, in integer) is ________

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