Following table calculates and presents expected completion time and standard de
ID: 373414 • Letter: F
Question
Following table calculates and presents expected completion time and standard deviation of each activity :
Activity
Optimistic
Most likely
Pessimistic
Expected time
Standard deviation
A
1.00
4.00
7.00
4.00
1.00
B
3.00
5.00
10.00
5.50
1.17
C
2.00
5.00
8.00
5.00
1.00
D
4.00
6.00
11.00
6.50
1.17
E
1.00
2.00
3.00
2.00
0.33
F
3.00
5.00
7.00
5.00
0.67
G
1.00
2.00
6.00
2.50
0.83
H
2.00
5.00
6.00
4.67
0.67
It is to be noted that :
Expected time = ( Optimistic time + 4 x Most likely time + Pessimistic time)/6
Standard deviation of an activity = ( Pessimistic time – Optimistic time ) / 6
A
B
C
D
E
F
G
H
The parallel paths and their cumulative expected durations as follows :
A-C-F-G = 4 + 5 +5 + 2.5 = 16.5
A-D-F-H = 4 + 6.5 + 5 + 4.67 = 20.17
A-D-G-H = 4 + 6.5 + 2.5 + 4.67 = 17.67
B-E-G-H = 5.5 + 2 + 2.5 + 4.67 = 14.67
Out of above, A-D-F-H has the longest duration and hence forms the critical path Duration of the critical path is the expected project completion time. Therefore, expected project completion time will be 20.17 days
CRITICAL PATH = A-D-F-H
EXPECTED PROJECT COMPLETION TIME = 20.17 DAYS
Let Z value corresponding to the probability of completing the project within 22 days = Z1
Variance of duration of critical path
= Sum of variances of A,D,F, and H
= 1^2 + 1.17^2 + 0.67^2 + 0.67^2
= 1 + 1.3689 + 0.4489 + 0.4489
= 3.2667
Standard deviation of critical path
= Square root ( Variance )
= Square root ( 3.2667)
= 1.807
Thus,
Expected duration of critical path + Zvalue x Standard deviation of critical path = 22 days
Or, 20.17 + 1.807.Z1 = 22
Or, 1.807.Z1 = 1.83
Or, Z1 = 1.83/1.807 = 1.0127
Corresponding probability for Z1 =1.0127 ( 1.01 rounded to 2 decimal places) as derived from standard normal distribution table is 0.8437
PROBABILITY OF COMPLETING THE PROJECT WITHIN 22 DAYS = 0.8437
Activity
Optimistic
Most likely
Pessimistic
Expected time
Standard deviation
A
1.00
4.00
7.00
4.00
1.00
B
3.00
5.00
10.00
5.50
1.17
C
2.00
5.00
8.00
5.00
1.00
D
4.00
6.00
11.00
6.50
1.17
E
1.00
2.00
3.00
2.00
0.33
F
3.00
5.00
7.00
5.00
0.67
G
1.00
2.00
6.00
2.50
0.83
H
2.00
5.00
6.00
4.67
0.67
Explanation / Answer
The following represents a project that should be scheduled using CPM IMMEDIATE TIMES (DAYS) ACTIVITY PREDECESSORS a mb 3 5 10 4 6 11 C,D D,E F,G 6 6 b. What is the critical path? A-C-F-H O B-E-G-HH AD-G-H A-D-F-H c. What is the expected project completion time? (Round your answer to 3 decimal places.) Project completion time days d. What is the probability of completing this project within 22 days? (Do not round intermediate calculations. Round your answer to 4 decimal places.) Probability