Sequencing problems & mathematical Solution
Download as PPTX, PDF1 like210 views
The document describes a job sequencing problem to minimize idle time across two machines. It provides processing times for 7 jobs on each of the two machines. The optimal job sequence is determined to be A, D, E, C, B, G, F, with a minimum total time of 134 hours and idle times of 2 hours for machine 1 and 34 hours for machine 2.
Sequencing problems & mathematical Solution
- 1. Md. Tarek Aziz Id: M21221111049 Batch:28th
- 3. It is the selection of an appropriate order in which a number of jobs (Operations) can be assigned to a finite number of service facilities (Machines or equipment's) so as to optimize the outputs in terms of time, cost or profit. Sequencing problems
- 4. Problem No 11.1. Seven jobs go first over Machine 1 and then over Machine 2. Processing times in Hours are given as: Job A B C D E F G Machine 1 6 24 30 12 20 22 18 Machine 1 16 20 20 12 24 2 6 Find the optimum sequence in which jobs should be processed.
- 5. Solution: We find the Smallest processing time between two machine is 2 hours for job F on machine 2. Thus we shall schedule job F in the last as shown below: F The Reduced set of processing times becomes: Job A B C D E G Machine 1 6 24 30 12 20 18 Machine 1 16 20 20 12 24 6
- 6. We find the Smallest processing time between two machine is 6 hours for job A in machine 1 & job G in machine 2. we shall schedule job A in the First and job G in the 2nd last as shown below: A G F The Reduced set of processing times becomes: Job B C D E Machine 1 24 30 12 20 Machine 1 20 20 12 24
- 7. We find the Smallest processing time between two machine is 12 hours for job D in machine 1 and machine 2. we shall schedule job D in the Next to Job A or previous to Job G. Here, we sequences Job D in next to Job A. shown below: A D G F The Reduced set of processing times becomes: Job B C E Machine 1 24 30 20 Machine 1 20 20 24
- 8. Here, We find the Smallest processing time between two machine is 20 hours for job E in machine 1 and Job B & C in machine 2. we shall schedule job E in the Next to Job D and Job B in previous to Job G again job C in previous to Job B. Here, we sequences Job B,C And job E shown below: A D E C B G F
- 9. The minimum elapsed time can now be computed cumulatively As flow: Job Time in Time out Time in Time out Idle time for machine 2 A 0 6 6 22 6 D 6 18 22 34 0 E 18 38 38 62 4 C 38 68 68 88 6 B 68 92 92 112 4 G 92 110 112 118 0 F 110 132 132 134 14 Machine 1 Machine 2 Idle time for machine 2 is 34 hours
- 10. From the above table, it is apparent that the minimum total time to finish all the 7 jobs is 134 hours and idle time for machine 1 is 2 hours (132-134) and for machine 2 is 34 hours.