In: Statistics and Probability
QUESTION 7 [18 MARKS)
In the following table, optimistic, most likely and pessimistic time estimates have been given for each activity for a project.
Task |
Immediate Predecessor(s) |
Time (days) |
||
Optimistic (a) |
Most Likely (m) |
Pessimistic (b) |
||
A |
None |
5 |
7 |
12 |
B |
None |
8 |
8 |
8 |
C |
A |
2 |
6 |
10 |
D |
A,B |
12 |
14 |
19 |
E |
C,D |
6 |
6 |
12 |
F |
B,E |
3 |
12 |
18 |
G |
F |
6 |
8 |
10 |
Solution:
Expected time for each activity can be calculated as
Expected time = (O+4*M+P)/6
Variance = ((P-O)^2)/36
Task | O | M | P | Expected Time | Variance |
A | 5 | 7 | 12 | 7.5 | 1.361111111 |
B | 8 | 8 | 8 | 8 | 0 |
C | 2 | 6 | 10 | 6 | 1.777777778 |
D | 12 | 14 | 19 | 14.5 | 1.361111111 |
E | 6 | 6 | 12 | 7 | 1 |
F | 3 | 12 | 18 | 11.5 | 6.25 |
G | 6 | 8 | 10 | 8 | 0.444444444 |
Paths through which this task can be completed as
A-C-E-F-G with the completion time (7.5+6+7+11.5+8) = 40
A-D-E-F-G with the completion time (7.5+14.5+7+11.5+8) = 48.5
B-D-E-F-G with the completion time (8+14.5+7+11.5+8) = 49.5
B-F-G with the completion time (8+11.5+8) = 27.5
So critical path is B-D-E-F-G with a critical time 49.5 days and
Variance = (0+1.36+1+6.25+0.44) = 9.06
Standard deviation = sqrt(9.06) = 3.01
Solution(b)
We need to calculate the probability that the project is due to
complete in 47 days
P(X<47) = ?
Here we will use the standard normal distribution, First, we will
calculate Z-score which can be calculated as
Z= (47-49.5)/3.01 = -0.83
So from Z table, we found a p-value
P(X<47) = 0.2033
So there is a 20.33% probability that the project is due to be
completed in 47 days.
Solution(c)
Given that P-value = 0.95, So Z score from Z table is 1.645
So Due date can be calculated as
Due date = 49.5 + 1.645*3.01 = 49.5 + 5 = 54.5 days
So the shortest project time due date that will satisfy the manager
is 54.5 days.