Two jobs, A and B, have to be done on three machines, M1, M2, and M3 in the operation sequence given below. The operation times and the due dates are also given. Draw Gantt chart showing the scheduling of these jobs based on the Least Slack first rule.
(2) Two jobs, A and B, have to be done on three machines, M1, M2, and M3 in the operation sequence given below. The operation times and the due dates are also given. Draw Gantt chart showing the scheduling of these jobs based on the Least Slack first rule. Machines/Hours Operations Due 1 | 2 | 3 | date Jobs | Solution: The information given in the table is that job A needs to work on Mi first for 2 hours, then M3 for 4 hours, then on to M2 for 3 hours, and finally on to M4 for 6 hours. Similarly, job B needs to work on Mi first, then M2, and finally on M3. Same way, jobs C and D are interpreted. Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours A typical Gantt chart is shown in the above figure. This chart has a horizontal line for each machine to load jobs. Assume that all machines are free from time 0 to do the jobs. From the first operation column, only job C requires M2, and only job D requires M4 (at time 0), we can go ahead and schedule them on the chart for the time they need these machines as shown below Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 TTTTTT 24 26 28 30 32 34 36 38 40 hours 113 It makes sense to apply any scheduling rule only when there are two or more jobs waiting for the same machine. If there is only one job waiting for a machine that becomes free, common sense will indicate to load that job on the machine right away. Jobs A and B require MI. We need to apply the least slack rule to determine which job to schedule on Mi. At time 0, available time of A is 18 hours, and available time of B is 12 hours. Remaining operation time on A is all the work left to be done on A. Remaining operation time on B is all the work left to be done on B. Therefore, Slack of A = 18 - (2+4+3+6)= 3 Slack of B = 12-(4+3+4)=1 Job B has the least slack. B is loaded on Ml as shown in the chart below. A waits for Mi. Now that we have scheduled jobs on machines at time 0, the next time to schedule anything else on a machine is when a job gets done. This is at time 2 when D is done on M4. D goes from M4 to M1 according to the operation sequence of D. But M1 is busy with B at time 2. So, D has to wait for M1 along with A. Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours Next activity happens at time 4. In fact, number of things happen at time 4. B and C get off from Ml and M2 respectively. Next operation of B is M2, and next operation of C is M3. M2 has only B to do, and M3 has only C to do. There is no need to apply any rule for them. Schedule B on M2 between time 4-7, and C on M3 from time 4-9. 114 Gantt Chart B NO 0 TTTTTTTTTTTTTTTT 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours Since MI gets free at time 4, we have to decide between waiting jobs A and D for Ml. As there are two jobs waiting for MI, we need to apply the least slack rule to A and D at time 4. At time 4, available time of A is 14 = 18 - 4) hours. Available time of Dis 10 = 14 - 4) hours. Remaining operation time on A is all the work left to be done on A. Remaining operation time on D is all the work left to be done on D. Therefore, Slack of A = 14 - (2+4+3+6)=-1 Slack of D = 10 - (3+5) = 2 Job A has the least slack. A is loaded next on M1 at time 4. D continues to wait for MI. Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours 115 Next time to stop is 6 when A gets of Mi. A has to go to M3, but M3 is busy with C, so A has to wait for M3. D has been waiting for M1 and since it is the only job for M1, we can schedule it on M1 as shown in the figure. Gantt Chart TTTTTTTTTTTTTTT 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 0 34 36 38 40 hours Next event happens at time 7 when B gets off M2 and goes to M3. But again, M3 is busy, so B has wait along with A. At time 9, again many things are happening. D gets off M1 and goes to M3 and joins A and B. M3 gets done with C that heads to M4. C is the only job for M4, so no need for any rule. Gantt Chart CA 9 12 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours Apply the least slack rule for A, B, and D for M3 at time 9. 116 (2) Two jobs, A and B, have to be done on three machines, M1, M2, and M3 in the operation sequence given below. The operation times and the due dates are also given. Draw Gantt chart showing the scheduling of these jobs based on the Least Slack first rule. Machines/Hours Operations Due 1 | 2 | 3 | date Jobs | (2) Two jobs, A and B, have to be done on three machines, M1, M2, and M3 in the operation sequence given below. The operation times and the due dates are also given. Draw Gantt chart showing the scheduling of these jobs based on the Least Slack first rule. Machines/Hours Operations Due 1 | 2 | 3 | date Jobs | Solution: The information given in the table is that job A needs to work on Mi first for 2 hours, then M3 for 4 hours, then on to M2 for 3 hours, and finally on to M4 for 6 hours. Similarly, job B needs to work on Mi first, then M2, and finally on M3. Same way, jobs C and D are interpreted. Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours A typical Gantt chart is shown in the above figure. This chart has a horizontal line for each machine to load jobs. Assume that all machines are free from time 0 to do the jobs. From the first operation column, only job C requires M2, and only job D requires M4 (at time 0), we can go ahead and schedule them on the chart for the time they need these machines as shown below Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 TTTTTT 24 26 28 30 32 34 36 38 40 hours 113 It makes sense to apply any scheduling rule only when there are two or more jobs waiting for the same machine. If there is only one job waiting for a machine that becomes free, common sense will indicate to load that job on the machine right away. Jobs A and B require MI. We need to apply the least slack rule to determine which job to schedule on Mi. At time 0, available time of A is 18 hours, and available time of B is 12 hours. Remaining operation time on A is all the work left to be done on A. Remaining operation time on B is all the work left to be done on B. Therefore, Slack of A = 18 - (2+4+3+6)= 3 Slack of B = 12-(4+3+4)=1 Job B has the least slack. B is loaded on Ml as shown in the chart below. A waits for Mi. Now that we have scheduled jobs on machines at time 0, the next time to schedule anything else on a machine is when a job gets done. This is at time 2 when D is done on M4. D goes from M4 to M1 according to the operation sequence of D. But M1 is busy with B at time 2. So, D has to wait for M1 along with A. Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours Next activity happens at time 4. In fact, number of things happen at time 4. B and C get off from Ml and M2 respectively. Next operation of B is M2, and next operation of C is M3. M2 has only B to do, and M3 has only C to do. There is no need to apply any rule for them. Schedule B on M2 between time 4-7, and C on M3 from time 4-9. 114 Gantt Chart B NO 0 TTTTTTTTTTTTTTTT 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours Since MI gets free at time 4, we have to decide between waiting jobs A and D for Ml. As there are two jobs waiting for MI, we need to apply the least slack rule to A and D at time 4. At time 4, available time of A is 14 = 18 - 4) hours. Available time of Dis 10 = 14 - 4) hours. Remaining operation time on A is all the work left to be done on A. Remaining operation time on D is all the work left to be done on D. Therefore, Slack of A = 14 - (2+4+3+6)=-1 Slack of D = 10 - (3+5) = 2 Job A has the least slack. A is loaded next on M1 at time 4. D continues to wait for MI. Gantt Chart 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours 115 Next time to stop is 6 when A gets of Mi. A has to go to M3, but M3 is busy with C, so A has to wait for M3. D has been waiting for M1 and since it is the only job for M1, we can schedule it on M1 as shown in the figure. Gantt Chart TTTTTTTTTTTTTTT 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 0 34 36 38 40 hours Next event happens at time 7 when B gets off M2 and goes to M3. But again, M3 is busy, so B has wait along with A. At time 9, again many things are happening. D gets off M1 and goes to M3 and joins A and B. M3 gets done with C that heads to M4. C is the only job for M4, so no need for any rule. Gantt Chart CA 9 12 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 hours Apply the least slack rule for A, B, and D for M3 at time 9. 116 (2) Two jobs, A and B, have to be done on three machines, M1, M2, and M3 in the operation sequence given below. The operation times and the due dates are also given. Draw Gantt chart showing the scheduling of these jobs based on the Least Slack first rule. Machines/Hours Operations Due 1 | 2 | 3 | date Jobs |
