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MTH601 Operations Research Assignment 1 Solution Fall 2012

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 competition time?

iv)                Compute the total floats (slacks) and free floats for the activities.

 Solution:

i)

B-D-G =          4+6+9  =19

There is no other higher value path is available

 

ii)                   What is the project competition time?

B-D-G =          4+6+9  =19

There is no other longest path is available

 

iii)                 Compute the total floats (slacks) and free floats for the activities.

Activity

Difference

Free Float

Total Float

A

3

0

3

3

6

0

3

B

4

0

4

4

8

0

4

C

5

4

9

9

14

0

5

D

6

4

10

10

16

0

6

E

7

10

17

17

24

0

7

F

8

17

25

25

33

0

8

G

9

10

19

19

28

0

9

 

 

Question-02

The following table shows the jobs of a network along with their time estimates. The time estimates are in days:

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.

 

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

S.D

5

7

9

11

3

6

6

8

 

The network flow diagram

i)                    Find the Critical Path.

1-2-4-7=26

1-2-3-5-7=29

1-2-4-6-8=25

The longest path / Critical Path is                      1-2, 2-3, 3-5, 5-7=29

ii)                   Compute the probability of completing the project in 36 weeks.           

Expected length of the critical path is 29 days. The variance for 1-2, 2-3, 3-5 and 5-7 are 1, 1, 5 and 2 respectively and variance of the projection duration is 9 and hence

Standard deviation of the project duration = 9 = 3 days.

 

(c)       Due date = 36 days (T)

Expected duration = 29 days (Te) and S.D = 3 days (o)

The area under the normal curve for Z = 2.33 is 0.3849.

Therefore, the probability of completing the project in 36 days

 

= 0.3000 + 0.9893

= 1.2893

= 12.893%

 

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