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The following table shows the activities their optimistic, most likely and pessi

ID: 1844176 • Letter: T

Question

The following table shows the activities their optimistic, most likely and pessimistic durations and their immediate predecessors

Activity a   m b Immediate Predecessors

A 4 8 12 ---

B 4 10 13 A

C 7 14 18 B

D 9 16 20 B

E 6 9 12 B

F 2 4 6 D,E

G 4 7 13 C,F

H 3 5 7 G

I 2 3 4 G,H

A- Determine the expected times and variances for each activity

B- Construct a project network for this problem

C- What is the probability that the project will be finished in less than 57 days?

D- What is the probability that the project will need at least 50 days?

Explanation / Answer

Expected time and varience for activiy A

a = 4 m =8 b =12

Expected time = (a+4m+b)/6 = (4+4*8+12)/6 = 8

Varience = (b-a)/6 = (12-4)/6 = 1.33

Total project duration,

0.33333

Taking all this values in network diagram,

critical path will be,

A-B-C-G-H-I

Project Duration = 8+9.5+13.5+7.5+5+3 = 46.5

Varience of project = sqrt(Va^2 + Vb^2 + Vc^2 + Vg^2 + Vh^2 + Vi^2)

V = sqrt(10.19)

V = 3.19

P(x<57) = P(z< (57-46.5)/3.19 )

P(x<57) = P(z< 3.29)

P(x<57) = 0.9995

P(x>50) = 1 - P(x<50)

P(x>50) = 1 - P(z<(50-46.5)/3.19)

P(x>50) = 1 - P(z<1.09)

P(x>50) = 1 - 0.8621

P(x>50) = 0.1379

a m b E V A 4 8 12 8 1.33333 B 4 10 13 9.5 1.5 C 7 14 18 13.5 1.83333 D 9 16 12 14.16667 0.5 E 6 9 12 9 1 F 2 4 6 4 0.66667 G 4 7 13 7.5 1.5 H 3 5 7 5 0.66667 I 2 3 4 3

0.33333

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