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2. Consider the network described in the table below. Immediate Activity Predecessor(s)Pessimistic Probable Optimistic 15 10
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Answer #1

ANSWER:

a)

consider:

a - optimistic

b - pessimistic

m - probable

Te - Expected time completion

Activity a b m Te (Var)^0.5 Var
J 8 15 10 10.5 1.166666667 1.361111111
K 7 9 8 8 0.333333333 0.111111111
L 5 10 6 6.5 0.833333333 0.694444444
M 3 3 3 3 0 0
N 1 9 5 5 1.333333333 1.777777778
O 4 10 7 7 1 1
P 3 10 8 7.5 1.166666667 1.361111111

b)

ES - Early start

EF - Early finish

LS - Latest start

LF - Latest finish

Expected time completion:

\small =\frac{a+4m+b}{6}

variance = standard deviation2

standard deviation:

\small = \frac{b-a}{6}

Activity ES EF LS LF SLACK
J 0 10.5 0 10.5 0
K 0 8 5.5 13.5 5.5
L 10.5 17 12 18.5 1.5
M 10.5 13.5 10.5 13.5 0
N 13.5 18.5 13.5 18.5 0
O 13.5 20.5 19 26 5.5
P 18.5 26 18.5 26 0

Critical path is JMNP with duration 26

Critical path = J-M-N-P

Expected duration of the critical path = 10.5 + 3 + 5 + 7.5

Expected duration of the critical path = 26 weeks

Variance of the critical path = 1.36 + 0 + 1.78 + 1.36

Variance of the critical path = 4.5

c)

If the project completion time is considered to be normal, the z score for the project to be completed in 28 weeks is found as

\small Z = \frac{\overline{X}-mean}{standard deviation}

\small =\frac{28-26}{2.1213}

Z = 0.94

From the Z score table, the probability the project will be completed in 28 weeks = 0.8264

The probability of completing the project in more than 28 weeks time = 1 -  0.8264

Probability of completing the project in more than 28 weeks time = 0.174 or ( 17.4%).

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