In: Operations Management
Use the following activities and their three duration estimates (in days) to do the following
Path |
Activity |
Optimistic |
Most Likely |
Pessimistic |
a-b-c |
A |
4 |
5 |
6 |
B |
7 |
8 |
10 |
|
C |
3 |
5 |
9 |
|
d-e-f |
D |
7 |
8 |
11 |
E |
2 |
3 |
4 |
|
f |
1 |
4 |
6 |
Expected Duration of Activity = (Optimistic Time + 4*Most Likely Time + Pessimistic Time)/6
Expected Duration of Activity A = (4+4*5+6)/6 = 5 days
Expected Duration of Activity B = (7+4*8+10)/6 = 8.17 days
Expected Duration of Activity C = (3+4*5+9)/6 = 5.33 days
Expected Duration of Path A-B-C = 5+8.17+5.33 = 18.50 days
Expected Duration of Activity D = (7+4*8+11)/6 = 8.33 days
Expected Duration of Activity E = (2+4*3+4)/6 = 3 days
Expected Duration of Activity F = (1+4*4+6)/6 = 3.83 days
Expected Duration of Path D-E-F = 8.33+3+3.83 = 15.16 days
From above,
Critical Path = A-B-C
Variance = ((Pessimistic Time - Optimistic Time)/6)^2
Varince of A = ((6-4)/6)^2 = 0.11
Varince of B = ((10-7)/6)^2 = 0.25
Varince of C = ((9-3)/6)^2 = 1.00
Total Variance = 1.36
SD = Variance^(1/2)
SD = (1.36)^(1/2) = 1.166 = 1.17
Expected Duration = 18.50
Z = (X-Mean)/SD
Z = (19-18.50)/1.17
Z = 0.4274
Probability = NORM.S.DIST(0.4274,1)
Probability = 0.6655 = 66.55%