In: Operations Management
The following table contains information related to the major activities of a research project. Use the information to do the following:
a. Compute the expected duration, variance, and standard deviation for each activity
b. What is the probability that the project finishes in less than 60 days?
Activity |
Predecessor |
To |
Tm |
Tp |
A |
- |
3 |
7 |
12 |
B |
A |
8 |
11 |
17 |
C |
B |
7 |
13 |
16 |
D |
B |
7 |
10 |
13 |
E |
C |
14 |
17 |
23 |
F |
D |
12 |
16 |
19 |
G |
E, F |
5 |
7 |
9 |
Expected time = (To + 4*Tm + Tp)/6
Variance = (Tp-To)^2/36
Standard deviation = sqrt(Variance)
Activity | To | Tm | Tp | Expected Time | Variance | Standard deviation |
A | 3 | 7 | 12 | 7.17 | 2.25 | 1.50 |
B | 8 | 11 | 17 | 11.50 | 2.25 | 1.50 |
C | 7 | 13 | 16 | 12.50 | 2.25 | 1.50 |
D | 7 | 10 | 13 | 10.00 | 1.00 | 1.00 |
E | 14 | 17 | 23 | 17.50 | 2.25 | 1.50 |
F | 12 | 16 | 19 | 15.83 | 1.36 | 1.17 |
G | 5 | 7 | 9 | 7.00 | 0.44 | 0.67 |
Probability to complete in less than 60 days will be product of probability of two path to complete the project in 60 days
z value = (Estimated time - Expected time)/Standard deviation
Estimated time = 60
z value for path ABCEG = (60 - 55.67)/3.07 = 1.41
Probability for z value 1.41 is 0.9207
z value for path ABDFG = (60 - 51.50)/2.70 = 3.14
Probability for z value 3.14 is 0.9992
Probability for project = 0.9207*0.9992 = 0.9200