Assignment: # 01 (Fall-2012)
Mth601 (Operations Research)
Lecture: 04– 10
Question-01
An industrial project has the following data:
| Activity | Immediate Predecessor(s) | Time(Days) |
|
A |
– |
3 |
|
B |
– |
4 |
|
C |
A,B |
5 |
|
D |
B |
6 |
|
E |
D |
7 |
|
F |
C,E |
8 |
|
G |
D |
9 |
i) Draw the network flow diagram.
ii) Find the critical path.
iii) What is the project completion time?
iv) Compute the total floats (slacks) and free floats for the activities.
Solution:
i) Network flow diagram:
The longest path is so our critical path is.
iii) Project completion time:
Expected duration of the project is 25 which is of course the length of the critical path.
iv )
|
Activity |
Early start time |
Early Finish time |
Late Start time |
Late Finish time |
Total Float |
Free Float |
|
A |
0 |
3 |
9 |
12 |
9 |
1 |
|
B |
0 |
4 |
0 |
4 |
0 |
0 |
|
C |
4 |
9 |
12 |
17 |
8 |
8 |
|
D |
4 |
10 |
4 |
10 |
0 |
0 |
|
E |
10 |
17 |
10 |
17 |
0 |
0 |
|
F |
17 |
25 |
17 |
25 |
0 |
0 |
|
G |
10 |
19 |
16 |
25 |
6 |
6 |
Total float = (Latest finish time – Early start time) – duration of the activity
Free float = total float – (Latest finish time – Earliest start time for tail event)
Question-02
The following table shows the jobs of a network along with their time estimates. The time estimates are in weeks:
| Job | 1-2 | 2-3 | 2-4 | 3-5 | 4-6 | 4-5 | 5-7 | 6-7 |
| Optimistic Time | 2 | 4 | 4 | 6 | 1 | 3 | 4 | 6 |
| Most likely time | 5 | 7 | 9 | 10 | 3 | 6 | 5 | 8 |
| Pessimistic Time | 8 | 10 | 14 | 20 | 5 | 9 | 12 | 10 |
i) Find the Critical Path.
ii) Compute the probability of completing the project in 36 weeks.
Solution:
i) Critical Path
First calculate the expected time, standard deviation and variance.
The PERT diagram is as follows:
The possible paths with their expected time of completion are as follows:
Possible Paths Time
1-2-3-5-7 5+7+11+6 = 29* Critical Path
1-2-4-5-7 5+9+6+6 = 26
1-2-4-6-7 5+9+3+8 = 25
Expected project length is 29 weeks.
ii) Probability of completing project:
Probability that the project completion time T£ 36
Due date=36 weeks (T)
Expected duration=29 weeks (Te)
Variance of Critical Path =1 + 1 + 5.44 + 1.78
= 9.22
S.D of Critical Path =
= 3.04
Z =
=
= 2.30
The area under the normal curve z = 2.30 is 0.4896
P (T£ 36 = P (z£2.30) =0.5+0.4896= 0.9896
_______________________________________