The following estimates of activity times (in days) are available for a small pr
ID: 3111372 • Letter: T
Question
The following estimates of activity times (in days) are available for a small project:
Compute the expected activity completion times and the variance for each activity. If necessary, round your answer to two decimal places.
Most
Expected
Activity
Optimistic
Probable
Pessimistic
Times
Variance
A
4
5
6
B
8
9
10
C
7
7.5
11
D
7
9
10
E
6
7
9
F
5
6
7
An analyst determined that the critical path consists of activities B–D–F. Compute the expected project completion time and the variance. If necessary, round your answer to two decimal places.
Expected project completion time: .
Variance of projection completion time: .
Most
Expected
Activity
Optimistic
Probable
Pessimistic
Times
Variance
A
4
5
6
B
8
9
10
C
7
7.5
11
D
7
9
10
E
6
7
9
F
5
6
7
Explanation / Answer
Expected time (t)=1/6 (optimistic time + 4×most probable time+pessimistic time)
variance=((pessimistic time-optimistic time)/6)²
For activity A , Expected time=1/6 (4+4×5+6)=30/6=5
variance = ((6-4)/6)²=(1/3)² =0.11
For activity B , Expected time = 1/6 (8+4×9+10)=54/6=9
Variance = ((10-8)/6)²=(1/3)²=0.11
For activity C, Expected time = 48/6 =8
Variance = (2/3)²= 0.44
For activity D , Expected time= 53/6=8.83
Variance= (3/6)²=0.25
for activity E, Expected time = 43/6=7.17
Variance = (3/6)²=0.25
for activity F, Expected time= 36/6=6
Variance = (2/6)²=0.11
Critical path consists B-D- F. So,it has maximum project completion time and variance.
Expected project completion time = 9+8.83+6=23.83
Variance = 0.11+0.25+0.11=0.47
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