2. In the following table, task durations are given in weeks. The estimates were
ID: 345566 • Letter: 2
Question
2. In the following table, task durations are given in weeks. The estimates were made at the 95 percent level. Answer the following questions Activity Predecessor Optimistic Normal Pessimistic 10 E BC 4 5 a. Draw the AOA network diagram (5 pts) b. Find the expected time and variance for each task (15 pts) c. Find the critical path and critical time (5 pts) d. Find the probability that the critical path will be completed in 22 weeks (15 pts) e. What is the probability that the entire network/project will be completed in 22 weeks? (10 pts)Explanation / Answer
The expected durations and variances of each activity are presented in tabular form :
Activity
Optimistic
Most likely
Pessimistic
Expected duration
Variance
A
2
4
6
4.00
0.44
B
3
5
9
5.33
1.00
C
4
5
7
5.17
0.25
D
5
6
10
6.50
0.69
E
4
5
7
5.17
0.25
F
3
4
8
4.50
0.69
G
3
6
8
5.83
0.69
Following may be noted :
Expected duration = ( Optimistic + 4 x Most Likely + Pessimistic )/6
Variance of each activity = ( Pessimistic duration – Optimistic Duration)^2 / 36
The precedence diagram of activities as follows :
A
B
D
C
F
E
G
The alternative paths and their expected durations as follows :
A-D-F = 4 + 6.5 + 4.5 = 15 weeks
A-C-E-G = 4 + 5.17 + 5.17 + 5.83 = 20.17 weeks
B-E-G = 5.33 + 5.17 + 5.83 = 16.33 weeks
Since A-C-E-G has the longest duration, it forms the critical path .
The expected duration of the critical path = 20.17 weeks
CRITICAL PATH = A-C-E-G
CRITICAL TIME = 20.17 WEEKS
Variance of the critical path
= Sum of variances of A, C, E and G
= 0.44 + 0.25 + 0.25 + 0.69
= 1.63
Standard deviation of the critical path = Square root ( Variance ) = Square root ( 1.63 ) = 1.276
Let,
Z value corresponding to the probability that the project will be completed in 22 weeks = Z1
Therefore ,
Expected duration of critical path + Z1 x Standard deviation of critical path = 22
Or, 20.17 + 1.63.Z1 = 22
Or, 1.63.Z1 = 22 – 20.17
Or, 1.63.Z1 = 1.83
Or, Z1 = 1.83/1.63
Or, Z1 = 1.12
Probability corresponding to Z = 1.12 as derived from standard normal table will be 0.86864
Therefore, probability that the critical path will be completed in 22 weeks = 0.86864
PROBABILITY THAT THE CRITICAL PATH WILL BE COMPLETED IN 22 WEEKS = 0.86864
The variance of the duration of path A-D-F
= Variance of A + Variance of D + Variance of F
= 0.44 + 0.69 + 0.69
= 1.82
Standard deviation of duration of A-D-F = Square root ( 1.82 ) = 1.25
Expected duration of A-D-F = 15
Let Z value that of the probability that this path will be completed in 22 days = Z1
Therefore .
15 + 1.25,Z1 = 22
Or, 1.25.Z1 = 7
Or, Z1 = 7/1.25 = 5.6
Corresponding probability for Z = 5.6 is approx. 1
The variance of duration of path B-E-G is 16.33 weeks
Variance of the path B-E-G = Sum of variances of B , E and G = 1 + 0.25 + 0.69 = 1.94
Standard deviation of duration of critical path = Square root ( 1.94) = 1.39
Let z value for probability that project will be completed in 22 weeks is Z2
Therefore,
Expected duration of B-E-G + Z2 x Standard deviation of duration of B-E-G = 22
Or , 16.33 + 1.39.Z2 = 22
Or, 1.39.Z2 = 5.67
Or, Z2 = 5.67/1.39
Or, Z2 = 4.08
Corresponding value of probability for Z = 4.08 is approx. 1
To summarise,
The probability values of three parallel paths of the project are
1 ( path : A-D-F ), 0.86864 ( path A-C-E-G ) and 1 ( path : B-E-G )
Since the least probability is 0.86864, we can conclude that probability that entire project will be completed in 22 weeks is 0.86864
Activity
Optimistic
Most likely
Pessimistic
Expected duration
Variance
A
2
4
6
4.00
0.44
B
3
5
9
5.33
1.00
C
4
5
7
5.17
0.25
D
5
6
10
6.50
0.69
E
4
5
7
5.17
0.25
F
3
4
8
4.50
0.69
G
3
6
8
5.83
0.69
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