Question

9) Use the network diagram and activity lengths in the nodes to determine the early start time for activity D. ( F 3 B 8 А G 6 5 с 11 H 6 4 9 A) 13 B) 21 C) 8 D) 14 10) What is the early start time for activity C if it has the start to start relationship shown? ( 20 A 27 25 B 29 ?? C 30 30 D 42 4 ?? 12 WA A) day 20 B) day 29 C) day 25 D) day 27 11) An activity has an optimistic time estimate of 15 days, a most likely estimate of 24 days, and a pessimistic estimate of 40 days. What is the variance of the activity? ( A) 17.36 B) 4.17 C) 10.33 D) 25.17

Hilf mir.

Answer 1

Answer 9: 14

Early Start time of Activity D = Sum of Duration of predecessor activities (A &B) = 6 + 8 = 14

Answer 10: 25

Start to Start Relationship: Successor Activity Can't start until Predecessor activities have started.

Early Start time of Activity C = Max (Start time of A, Start time of B) = Max (20,25) = 25

Answer 11: 17.36

a = optimistic time = 15

m = most likely time = 24

b = pessimistic time = 40

Variance of Activity (VAR) = [(b-a)/6]^2 = [(40-15)/6]^2 = (25/6)^2 = 17.36

Answer 12: day 23

Activity A is first activity of the project

ES of A = 0

EF of A = 8

A has two successor activities B & C

For C LS = 8 which is equal to EF of A so Activity C is a critical activity

C has two successor activities D & E

But there is a slack (LS - ES or LF-EF) for activity D so Activity D can't be a critical activity.

so Activity E is the critical activity.

E has one successor activity F

so Critical path for the project is A-C-E-F

Now, Late start of Activity E = Early Start of Activity E = Early Finish of Activity C = 23

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