Sequencing Problems - Processing 2 Jobs Through m Machines Problem calculator

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Book	1	2	3	4	5	6	7
Printing time (hours)	20	90	80	20	120	15	65
Binding time (hours)	25	60	75	30	90	35	50

Decide the optimum sequence of processing of books in order to minimize the total time required to bring out all the books.

2. A manufacturing company processes 6 different jobs on two mahcines A and B. Number of units of each job and its processing times on A and B are given below. Find the optimal sequence, the tatal minimum elapsed time and idle time for either machine

Job	Number of Units of each job	Processing time Machine A	Processing time Machine B
1	3	5	8
2	4	16	7
3	2	6	11
4	5	3	5
5	2	9	7.5
6	3	6	14

3. There are seven jobs, each of which has to go through the machines A and B in the order AB. Processing times in hours are as follows.

Job	1	2	3	4	5	6	7
Machine A	3	12	15	6	10	11	9
Machine B	8	10	10	6	12	1	3

Decide a sequence of these jobs that will minimize the total elapsed time T. Also find T and idle time for machines A and B.

4. Find the sequence that minimizes the total time required in performing the following job on three machines in the order ABC. Processing times (in hours) are given in the following table.

Job	1	2	3	4	5
Machine A	8	10	6	7	11
Machine B	5	6	2	3	4
Machine C	4	9	8	6	5

Example

1. A book binder has one printing press, one binding machine and manuscripts of 7 different books. The times required for performing printing and binding operations for different books are shown below.
Book 1 2 3 4 5 6 7

Printing time (hours) 20 90 80 20 120 15 65

Binding time (hours) 25 60 75 30 90 35 50

Decide the optimum sequence of processing of books in order to minimize the total time required to bring out all the books.

Solution:

Job	1	2	3	4	5	6	7
Machine `M_1`	20	90	80	20	120	15	65
Machine `M_2`	25	60	75	30	90	35	50

1. The smallest processing time is 15 hour for job 6 on Machine-1. So job 6 will be processed first.

2. The next smallest processing time is 20 hour for job 1,4 on Machine-1 and for this jobs 30 is largest on Machine-2. So job 4 will be processed after job 6.

3. The next smallest processing time is 20 hour for job 1 on Machine-1. So job 1 will be processed after job 4.

4. The next smallest processing time is 50 hour for job 7 on Machine-2. So job 7 will be processed last.

5. The next smallest processing time is 60 hour for job 2 on Machine-2. So job 2 will be processed before job 7.

6. The next smallest processing time is 75 hour for job 3 on Machine-2. So job 3 will be processed before job 2.

7. The next smallest processing time is 90 hour for job 5 on Machine-2. So job 5 will be processed before job 3.

According to Johanson's algorithm, the optimal sequence is as below

Job	`M_1` In time	`M_1` Out time	`M_2` In time	`M_2` Out time	Idle time `M_2`
6	0	0 + 15 = 15	15	15 + 35 = 50	15
4	15	15 + 20 = 35	50	50 + 30 = 80	-
1	35	35 + 20 = 55	80	80 + 25 = 105	-
5	55	55 + 120 = 175	175	175 + 90 = 265	70
3	175	175 + 80 = 255	265	265 + 75 = 340	-
2	255	255 + 90 = 345	345	345 + 60 = 405	5
7	345	345 + 65 = 410	410	410 + 50 = 460	5

The total minimum elapsed time = 460

Idle time for Machine-1
`=460 - 410`

`=50`

Idle time for Machine-2
`=(15)+(175-105)+(345-340)+(410-405)+(460-460)`

`=15+70+5+5+0`

`=95`