Page 206 - DCOM303_DMGT504_OPERATION_RESEARCH

Unit 10: Sequencing Problems and Replacement Theory

Notes
Table 10.11: Final Optimal Sequence Table
Job Machine A Machine B Idle Time Idle Time
Sequence A B
Time In Time Out Time In Time Out
2 0 1 1 7 0 1
4 1 4 7 15 0 0
3 4 13 15 22 0 0
5 13 23 23 27 0 1
1 23 28 28 30 0 1
30-28=2
2 3

In Table 10.11, the minimum elapsed time, (from Job 2 to Job 1) is 30 hours. The idle time on
machine A is 2 hours and on machine B is 3 hours.

Example: Six jobs go first over machine I and then over machine II. The order of the
completion of jobs has no significance. The table 10.12 shows the machine times in hours for six
jobs and the two machines.

Table 10.12: Sequence Problem

Job J J J J J 5 J
3
1
2
5
4
Machine I 2 4 9 6 7 4
Machine II 6 7 4 3 3 11

Find the sequence of the jobs that minimizes the total elapsed time to complete the jobs. Also
find the idle time for Machine I and Machine II.
Solution:
Establish a sequence table containing six job cells. Find the least time available for both Machine
I and Machine II. Job 1 has the least processing time, i.e., 1. Place the sequence from left to right
(or the first cell) as shown in Table 10.13, since it occurs Machine I.
Table 10.13: Job 1 has Least Processing Time

J1

Deleting job 1, we get the reduced table as shown in Table 10.14

Table 10.14: Reduced Table, Job 1 Deleted

Job J2 J3 J4 J5 J6
Machine I 3 8 5 6 3
Machine II 6 3 2 2 10

The least time available in the reduced table is 2, which is on Job 4 and Job 5 for Machine II. Now,
compare the adjacent time available for Machine I. Here, Job 4 time is less than that for Job 5.
Select Job 4 first and sequence it as shown in Table 10.15.

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