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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