# 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%