ANSWER:
given that :
Doug Casey is in charge of planning and coordinating next springs sales management training program for his company.
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Cheerslll
Answer(a):
Answer(b):
Activity | Optimistic (O) |
Most probable (M) |
Pessimistic (P) |
Expected Time (O+4M + P)/6 |
Variance [(P -O)/6]2 |
A | 1.5 | 2 | 2.5 | 2 | 0.03 |
B | 2 | 2.5 | 6 | 3 | 0.44 |
C | 1 | 2 | 3 | 2 | 0.11 |
D | 1.5 | 2 | 2.5 | 2 | 0.03 |
E | 0.5 | 1 | 1.5 | 1 | 0.03 |
F | 1 | 2 | 3 | 2 | 0.11 |
G | 3 | 3.5 | 7 | 4 | 0.44 |
H | 3 | 4 | 5 | 4 | 0.11 |
I | 1.5 | 2 | 2.5 | 2 | 0.03 |
Answer(c):
Activity | Earliest start | Latest start | Earliest Finish | Latest Finish | Slack | Critical Activity |
A | 0 | 0 | 2 | 2 | 0 | Yes |
B | 2 | 2 | 5 | 5 | 0 | Yes |
C | 0 | 1 | 2 | 3 | 1 | |
D | 2 | 3 | 4 | 5 | 1 | |
E | 5 | 10 | 6 | 11 | 5 | |
F | 6 | 11 | 8 | 13 | 5 | |
G | 5 | 5 | 9 | 9 | 0 | Yes |
H | 9 | 9 | 13 | 13 | 0 | Yes |
I | 13 | 13 | 15 | 15 | 0 | Yes |
The critical path followed -
End
Estimated completion time = 2+3+4=4+2 = 15 Weeks
Answer(d):
We calculate the variance sum of all critical path activities
0.03 (A) +0.44(B)+0.44(G) + 0.11 (H) + 0.03 (I) = 1.05
For probability of 0.99 , from the distribution table We find that Z = 2.56
Z = ( T .15)/1.050.5 = 2.56
T = 15 + 2.56 * 1.050.5 = 17.62 Weeks.
** As per HOMEWORKLIB RULES we should solve only the first
question. So I have done it.
For the other question please post differently mentioning your
requirement
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