Page 207 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes
Table 10.15: Job 4 Sequenced
J1 J4

Now select Job 5 and sequence it as shown in Table 6.16.

Table 10.16: Job 5 Sequenced

J1 J5 J4

The reduced table is shown in Table 10.17

Table 10.17: Reduced Table, Job 1, 4 & 5 Deleted

Job J2 J3 J6
Machine I 3 8 3
Machine II 6 3 10

Now we have Job 2, Job 3 and Job 6 having the least time which is 3. Compare these times with
the adjacent machine time and select the least time. Here we have Job 2 with least adjacent time
and hence sequence is as shown below in Table 10.18.
Table 10.18: Job 2 Sequenced

J1 J2 J5 J4

Now select Job 3 and sequence it, as shown in Table 6.19.

Table 10.19: Job 3 Sequenced

J1 J2 J3 J5 J4

Finally, select Job 6 and sequence it, as shown in Table 6.20.

Table 10.20: Job 6 Sequenced

J1 J2 J6 J3 J5 J4

The optimal sequence thus obtained is,

J  J  J  J  J  J
1 2 6 3 5 4
The total elapsed time and idle time for Machine I and Machine II is calculated as shown in Table
10.21 below:

202 LOVELY PROFESSIONAL UNIVERSITY