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:
variance = standard deviation2
standard deviation:
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
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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