a.
To complete: The table given below.
Given information:
It is given that C machine center schedules for every 2 weeks. Today is 241st working day, and they are all set to schedule the jobs for next 2 weeks.
Job | Date job received | Production days needed | Date job due |
B | 228 | 15 | 300 |
C | 225 | 25 | 270 |
D | 230 | 35 | 320 |
R | 235 | 40 | 360 |
S | 231 | 30 | 310 |
Explanation:
Early due date:
The job with the earliest due date is selected first and the rest of the jobs are taken in an ascending order.
The order of selection of jobs using early due date method is:
Job sequence | Date job due |
C | 270 |
B | 300 |
S | 310 |
D | 320 |
R | 360 |
Table no: 1
Note: Flow time is the cumulative addition of production days needed / processing time (column 3).
Note: In column 6, the date job due (from column 5) is reduced by 240 since the job is started from day 241.
Note: In column 7, when subtracting flow time (column 4) and job starting date (column 6), assign 0 instead of negative numbers.
It is calculated that the total late days are 50. Refer to Table no: 1.
It is given that the total number of jobs is 5.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 1.
It is calculated that the sum of total flow time is 385 days. Refer to Table no: 1.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 1.
It is calculated that the sum of total flow time is385 days. Refer to Table no: 1.
Shortest processing time:
The job with shortest processing time is taken first and the rest of the jobs are taken in an ascending order.
The order of selection of jobs using shortest processing time method is:
Job sequence | Production days needed |
B | 15 |
C | 25 |
S | 30 |
D | 35 |
R | 40 |
Table no: 2
Note: Flow time is the cumulative addition of production days needed / processing time (column 3).
Note: In column 6, the date job due (from column 5) is reduced by 240 since the job is started from day 241.
Note: In column 7, when subtracting flow time (column 4) and job starting date (column 6), assign 0 instead of negative numbers.
It is calculated that the total late days are 60. Refer to Table no: 2.
It is given that the total number of jobs is 5.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 2.
It is calculated that the sum of total flow time is 375 days. Refer to Table no: 2.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 2.
It is calculated that the sum of total flow time is 375 days. Refer to Table no: 2.
Longest processing time:
The job with the longest processing time is taken first and the rest of the jobs are taken in a descending order.
The order of selection of jobs using longest processing time method is:
Job sequence | Production days needed |
R | 40 |
D | 35 |
S | 30 |
C | 25 |
B | 15 |
Table no: 3
Note: Flow time is the cumulative addition of production days needed / processing time (column 3).
Note: In column 6, the date job due (from column 5) is reduced by 240 since the job is started from day 241.
Note: In column 7, when subtracting flow time (column 4) and job starting date (column 6), assign 0 instead of negative numbers.
It is calculated that the total late days are 220. Refer to Table no: 3.
It is given that the total number of jobs is 5.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 3.
It is calculated that the sum of total flow time is 495 days. Refer to Table no: 3.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 3.
It is calculated that the sum of total flow time is 495 days. Refer to Table no: 3.
First come first serve method:
The processes for the job are done according to the order they arrive.
The order of selection of jobs using first come first serve method is:
Job sequence | Date job received |
B | 228 |
C | 225 |
D | 230 |
R | 235 |
S | 231 |
Table no: 4
Note: Flow time is the cumulative addition of production days needed / processing time (column 3).
Note: In column 6, the date job due (from column 5) is reduced by 240 since the job is started from day 241.
Note: In column 7, when subtracting flow time (column 4) and job starting date (column 6), assign 0 instead of negative numbers.
It is calculated that the total late days are 85. Refer to Table no: 3.
It is given that the total number of jobs is 5.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 4.
It is calculated that the sum of total flow time is 390 days. Refer to Table no: 4.
It is calculated that the total production days needed (processing time) is 145 days. Refer to Table no: 4.
It is calculated that the sum of total flow time is 390 days. Refer to Table no: 4.
Therefore, the completed table is:
b.
To determine: The dispatching rule that has the best score for flow time.
Given information:
It is given that C machine center schedules for every 2 weeks. Today is 241st working day, and they are all set to schedule the jobs for next 2 weeks.
Job | Date job received | Production days needed | Date job due |
B | 228 | 15 | 300 |
C | 225 | 25 | 270 |
D | 230 | 35 | 320 |
R | 235 | 40 | 360 |
S | 231 | 30 | 310 |
Explanation:
Now, find the dispatching rule that has best score for flow time.
Flow time shows the time taken for all the processes in the job. Least value denotes quick processing and completion of jobs, which is the best option. SPT has the least score for flow time compared to all other methods.
c.
To determine: The dispatching rule that has the best score for utilization metric.
Given information:
It is given that C machine center schedules for every 2 weeks. Today is 241st working day, and they are all set to schedule the jobs for next 2 weeks.
Job | Date job received | Production days needed | Date job due |
B | 228 | 15 | 300 |
C | 225 | 25 | 270 |
D | 230 | 35 | 320 |
R | 235 | 40 | 360 |
S | 231 | 30 | 310 |
Explanation:
Now, find the dispatching rule that has best score for utilization metric.
Utilization metric shows how effectively the jobs are done. Higher the score higher is the effectiveness. SPT has the highest score; therefore, it must be chosen.
d.
To determine: The dispatching rule that has the best score for lateness.
Given information:
It is given that C machine center schedules for every 2 weeks. Today is 241st working day, and they are all set to schedule the jobs for next 2 weeks.
Job | Date job received | Production days needed | Date job due |
B | 228 | 15 | 300 |
C | 225 | 25 | 270 |
D | 230 | 35 | 320 |
R | 235 | 40 | 360 |
S | 231 | 30 | 310 |
Explanation:
Now, find the dispatching rule that has best score for lateness.
Average job lateness shows the delay in the processing of the job. Lower the score quicker the jobs are done. EDD has the least score for average job lateness.
e.
To determine: The best dispatching rule.
Given information:
It is given that C machine center schedules for every 2 weeks. Today is 241st working day, and they are all set to schedule the jobs for next 2 weeks.
Job | Date job received | Production days needed | Date job due |
B | 228 | 15 | 300 |
C | 225 | 25 | 270 |
D | 230 | 35 | 320 |
R | 235 | 40 | 360 |
S | 231 | 30 | 310 |
Explanation:
Now, find the best dispatching rule.
Either SPT or EDD can be chosen, since SPT has the best score for flow time and utilization rate and EDD has the best score for job lateness.
The following jobs are waiting to be processed at Rick Carlson’s machine center.
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