Unit 8 Problems Name Institutional Affiliation Unit 8 Problems 14 14 20 0 10 24 16 8 7 31 15 16 6 37 17 20 37 106 44 The 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. Sequence D 6 6 17 0 C 7 13 15 0 B 10 23 16 7 A 14 37 20 17 37 79 24 The 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. C 7 7 15 0 B 10 17 16 1 D 6 23 17 6 A 14 37 20 17 37 84 24 The 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. 14 37 20 17 7 44 15 29 6 50 17 33 10 60 16 44 37 191 123 All 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. 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 days c) EDD = average completion time/number of jobs = 84/4 = 21 days d) CR = average completion time/number of jobs = 191/4 = 47.75 days a) FCFS = 44/4 = 11 days b) SPT = 24/4 = 6 days c) EDD =24/4 = 6 days d) CR = 123/4 = 30.75 days a) FCFS = 106/37 = 2.865 b) SPT = 79/37 = 2.135 c) EDD = 84/37 = 2.270 d) CR = 191/37 = 5.162 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. 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: A D The shortest job time is 12 minutes for D at Center 2, and thus D is scheduled last. B B D D 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. C B A D B 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. D B A F D A is then eliminated from the table and the next least time is 24 minutes for F at Center 2 and this is scheduled next. E B A E F D F is then eliminated from the table and the next least time is E at Center 2 and this is scheduled next. F B A C E F D E 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 – D The chart of activities for the two centers is shown below: 0 16 0 36 79 114 156 216 B A C E F D B A C E F D 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.