The project manager of Good public relations gathered the data shown in Table 2.11 for a new advertising campaign.
1. How long is the project likely to take?
2. What is the probability that the project will take more than 38 weeks?
3. Consider the path A-E-G-H-J. What is the probability that this path will exceed the expected project duration?
Answers
Good public relations Gathered Data information
Step-by-step explanation:
a) The critical path is A-E-F-H-J.
The expected time for each activity is:
A: (8 + 4*10 + 12) / 6 = 10
E: (8 + 4*10 + 12) / 6 = 10
F: (5 + 4*6 + 7) / 6 = 6
H: (2 + 4*5 + 8) / 6 = 5
J: (4 + 4*5 + 8) / 6 = 5.3333
The expected time to completion is then 10 + 10 + 6 + 5 + 5.3333 = 36.3333 weeks
b) The variance for the tasks on the critical path are:
A: [(12 - 8)/6]² = 0.4444
E: [(12 - 8)/6]² = 0.4444
F: [(7 - 5)/6]² = 0.1111
H: [(8 - 2)/6]² = 1.0000
J: [(8 - 4)/6]² = 0.4444
The total variance for the critical path is 0.4444 + 0.4444 + 0.1111 + 1.0000 + 0.4444 = 2.4444
The standard variation is √2.4444 = 1.5635
The z-score corresponding to a completion time of 38 weeks is:
z = (38 - 36.3333) / 1.5635 = 1.0660
P(z > 1.0660) = 0.1432
The probability that the project will take more than 38 weeks is 0.1432.
c) The expected time for activity G is:
G: (1 + 4*3 + 5)/6 = 3
The expected time for path A-E-G-H-J is 10 + 10 + 3 + 5 + 5.3333 = 33.3333
The variance for task G is:
G: [(5 - 1)/6]² = 0.4444
The total variance for the path A-E-G-H-J is 0.4444 + 0.4444 + 0.4444 + 1.0000 + 0.4444 = 2.7776
The standard deviation for the path is √2.7776 = 1.6666
The z-score corresponding to a completion time of 38 weeks is:
z = (38 - 33.3333) / 1.6666 = 2.8001
P(z > 2.8001) = 0.0026
The probability that path A-E-G-H-J will take more than 38 weeks to complete is 0.0026