Score: 0 of 5 pts 10 of 10 (9 complete) HW Score: 66.67%, 10 of 15 Instructor-cr
ID: 345007 • Letter: S
Question
Score: 0 of 5 pts 10 of 10 (9 complete) HW Score: 66.67%, 10 of 15 Instructor-created question Question Help Time (weeks) Immediate Time (weeks) Immediate Activity m b Predecessors Activity a 23 10 bPredecessor(s) C, E 5 4 6 D, G, H 20 C, E a) The expected (estimated) time for activity C is 13.5 weeks. (Round your response to two decimal places.) b) The variance for activity C is 1.36 weeks. (Round your response to two decimal places.) c) Based on the calculation of the estimated times, the critical path is A-C F-H-J - K d) The estimated time for the critical path is weeks. (Round your response to two decimal places.) e) The activity variance along the critical path is weeks. (Round your response to two decimal places.)Explanation / Answer
Below table calculates expected duration as well as variance of each activity on the critical path :
Activity
Optimistic
Most likely
Pessimistic
Expected duration
Variance
A
6
9
12
9.00
1.00
C
9
14
16
13.50
1.36
F
5
9
20
10.17
6.25
H
2
2
2
2.00
0.00
J
6
7
12
7.67
1.00
K
1
1
4
1.50
0.25
SUM =
43.83
9.86
It is to be noted :
Expected duration of an activity = ( Optimistic duration + 4 x Most likely duration + Pessimistic duration ) / 6
Variance of an activity = ( Pessimistic duration – Optimistic duration)^2/36
Thus sum of expected duration of all activities on critical path = 43.83 weeks
Sum of variances of all activities on critical path = 9.86 weeks
THE ESTIMATED TIME FOR THE CRITICAL PATH IS = 43.83 WEEKS
THE ACTIVITY VARIANCE ALONG THE CRITICAL PATH IS = 9.86 DAYS
Activity
Optimistic
Most likely
Pessimistic
Expected duration
Variance
A
6
9
12
9.00
1.00
C
9
14
16
13.50
1.36
F
5
9
20
10.17
6.25
H
2
2
2
2.00
0.00
J
6
7
12
7.67
1.00
K
1
1
4
1.50
0.25
SUM =
43.83
9.86
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