Question

# Activity Immediate Predecessor Optimistic Time Most-likely time Pessimistic Time A None 3 6 9 B A...

 Activity Immediate Predecessor Optimistic Time Most-likely time Pessimistic Time A None 3 6 9 B A 3 5 7 C A 4 7 12 D B 4 8 10 E C 5 10 16 F D, E 3 4 5 G D, E 3 6 8 H F 5 6 10 I G 5 8 11 J H, I 3 3 3

(a). What is the expected time for each activity? (2 points)

(b.) What is the variance for each activity? (2 points)

Use both the longest path method and ES/EF; LS/LF method for the following questions

(c) What is the expected project completion time? (2 point)

(d) What are the critical activities? (1 points)

(e) What are the slack values for non-critical activities? (4 points)

(f) What is the probability that the project will be completed in 38 weeks? (2 points)

(g) What is the probability that the project will be completed in 42 weeks? (2 points)

Expected Duration = (Optimistic time + 4* Most likely time + Pessimistic time) / 6

Variance= ((Pessimistic time - Optimistic time)/ 6 )2

Standard Deviation = square root of Variance

For Activity A,

Expected duration = (3+4*6+9) / 6 = 6 weeks

Variance = ((9-3)/6)2 = 1

Stadndard Deviation = square root (1) = 1

Remaining calculation are as below

 Activity Immediate Predecessor Optimistic Time Most-likely time Pessimistic Time Expected mean time Variance Std Dev ES EF LS LF Slack Critical (1) (2) (3) (4) (5) (6) = [(3)+4(4)+(5)]/6 (7)=[[(5)-(3)]/6]^2 (8)=(7)^0.5 (9) (10)=(9)+(6) (11)=(12)-(6) (12) A None 3 6 9 6 1 1 0 6 0 6 0 Yes B A 3 5 7 5 0.444444444 0.666666667 6 11 10.83333333 15.83333 4.833333 No C A 4 7 12 7.333333333 1.777777778 1.333333333 6 13.33333333 6 13.33333 0 Yes D B 4 8 10 7.666666667 1 1 11 18.66666667 15.83333333 23.5 4.833333 No E C 5 10 16 10.16666667 3.361111111 1.833333333 13.33333 23.5 13.33333333 23.5 0 Yes F D, E 3 4 5 4 0.111111111 0.333333333 23.5 27.5 26.83333333 30.83333 3.333333 No G D, E 3 6 8 5.833333333 0.694444444 0.833333333 23.5 29.33333333 23.5 29.33333 0 Yes H F 5 6 10 6.5 0.694444444 0.833333333 27.5 34 30.83333333 37.33333 3.333333 No I G 5 8 11 8 1 1 29.33333 37.33333333 29.33333333 37.33333 0 Yes J H, I 3 3 3 3 0 0 37.33333 40.33333333 37.33333333 40.33333 0 Yes Critical Path A-C-E-G-I-J Critical duration 40.33 weeks

f) Probability that project will be completed in 38 weeks

Variance along critical path = 1+1.7778+3.3611+0.69444+1+0 = 7.8333

Std Dev along critical path = 7.8333^0.5 = 2.79888

Z = (x- ) / = (38-40.33) / 2.7988 = -0.8325

Prob = Norm.s.dist(-0.8325,1) = 0.2056 = 20.56%

g) Probability that project will be completed in 42 weeks

Variance along critical path = 1+1.7778+3.3611+0.69444+1+0 = 7.8333

Std Dev along critical path = 7.8333^0.5 = 2.79888

Z = (x- ) / = (42-40.33) / 2.7988 = 0.59669

Prob = Norm.s.dist(0.59669,1) = 0.72464 = 72.46%

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