Question Number: 557

Engineering homework help problemsUnit 8 ProblemsNameInstitutional AffiliationUnit 8 ProblemsProblem 6A. Sequence the jobs using (1) FCFS, (2) SPT, (3) EDD, and (4) CR. Assume the list is by order of arrival. First Come, First Served (FCFS) SequenceJob TimeFlow TimeDue DateLateness A1414200 B1024168 C7311516 D6371720 TOTALS3710644The sequencing of the jobs based on the First Come, First Served (FCFS) basis is A – B – C – D, and the chart shows that shows that jobs B, C, and D will be late by a total of 44 days. Shortest Possible Time (SPT) SequenceJob TimeFlow Time Due DateLateness D66170 C713150 B1023167 A14372017 TOTALS377924The sequencing on the jobs based on the Shortest Possible Time is D – C – B – A, and the chart shows that jobs B and A will be late by a total of 24 days. Earliest Due Date (EDD) SequenceJob TimeFlow TimeDue DateDays Tardy C77150 B1017161 D623176 A14372017 TOTALS378424The sequencing of the jobs based on the Earliest Due Date (EDD) is given as C – B – D – A, and the average completion time is 42 days. Critical Ratio SequenceJob TimeFlow TimeDue DateTardiness A14372017 C7441529 D6501733 B10601644 TOTALS37191123All the critical ratios are greater than 1, and this indicates that all the jobs will be completed in time if scheduled as A – C – D – B. B. For each of the methods in part a, determine (1) the average job flow time, (2) the average tardiness, and (3) the average number of jobs at the work center. Average time flow a) FCFS = average completion time/number of jobs = 106/4 = 26.5 days. b) SPT = average completion time/number of jobs = 79/4 = 19.75 daysc) EDD = average completion time/number of jobs = 84/4 = 21 days d) CR = average completion time/number of jobs = 191/4 = 47.75 days Average tardiness a) FCFS = 44/4 = 11 daysb) SPT = 24/4 = 6 days c) EDD =24/4 = 6 days d) CR = 123/4 = 30.75 days Average number of jobs a) FCFS = 106/37 = 2.865b) SPT = 79/37 = 2.135c) EDD = 84/37 = 2.270d) CR = 191/37 = 5.162C. Is one method superior to the others? Explain. None of the methods is superior than the others because they each have their inherent weaknesses in addition to their strengths. For example, using the FCFS and SPT priority rules results in the increased delays in jobs that require longer times to complete while the EDD approach tends to ignore the processing times during the scheduling. However, the CR method has the most weaknesses in scheduling of the jobs. Problem 11Given the operation times provided: a. Develop a job sequence that minimizes idle time at the two work centers. The determination of the job sequence that would reduce the idle time at the two work centers is completed using the Johnson’s as shown in the table below: Developing the Job Sequence IterationJob SequenceComments ADThe shortest job time is 12 minutes for D at Center 2, and thus D is scheduled last. BBDD is then removed from the table of eliminated times and the next least job time is observed as 16 minutes for B at Center 1. Thus, B is scheduled first. CBADB is then eliminated from the table and the next least job time is 20 minutes for A at Center 1 and is there scheduled next. DBAFDA is then eliminated from the table and the next least time is 24 minutes for F at Center 2 and this is scheduled next. EBAEFDF is then eliminated from the table and the next least time is E at Center 2 and this is scheduled next. FBACEFDE is then eliminated from the table and the last least time is 43 minutes at Center 1 and it is scheduled the earliest. Thus the job sequence that reduces the idle time at Centers 1 and 2 is given as B – A – C- E – F – Db. Construct a chart of the activities at the two centers, and determine each one’s idle time, assuming no other activities are involved. The chart of activities for the two centers is shown below: 0 16 0 36 79 114 156 216 BACEFD BACEFD 16 46 73 79 130 158 218 216 228 From the chart, the idle time at Center is given as 0 minutes while the total idle time at Center 2 is given as ((16 – 0) + (79 – 73) + (216 – 182)) = 16 + 6 + 34 = 56 minutes.