Scheduling Seminar exercises 2010.10.14 Process and Production Management

Scheduling Estabilishing the timing of the use of equipment, facilities and human activities in an organization. High volume systems (characterized by standardized equipment and activities that provide identical or highly similar operations on products – autos, personnal computers, radios, TVs) Intermediate – volume systems (ouput fall between the standardized type of output of the high volume systems and made-to-order output of job shops. Work centers periodically shift from one product to another, but lot size relatively high. Low-volume systems (products are made to order, and orders differ in terms of processing requirements, materials needed, processing time.)

Priority rules Select the order in which job will be processed First come first served (FCFS) – jobs are processed in the order which they arrive at a machine or work center Shortest processing time (SPD) – according to processing time, shortest job first Earliest due date (EDD) – according to due date, earliest due date first. Critical ratio (CR) – smallest ratio of time remaining until due date to processing time remaining.

Assumptions The set of jobs is known Setup time is independent of processing sequence Setup time is deterministic Process time are deterministic There will be no interruptions in processing such as machine breakdowns, accidents, or workes illness.

Processing time (days) Example 1 Determine the sequence of jobs/products Determine average flow time Determine average tardiness Determine average number of jobs at the workcenter Job/ product Processing time (days) Due date (days) A 2 7 B 8 16 C 4 D 10 17 E 5 15 F 12 18

Job flow time – the length of time a job is at a particular workstation or work center. (it includes not only procesing time, but waiting time, transportation time) Average flow time = total flow time /number of jobs Job lateness – length of time the job completion date is exceed the date the job was due to promised to the customer Average tardiness=total lateness/number of products Makespan – is the total time needed to complete a group of jobs (time between start the first and completion of the last one) Average number of jobs=total flow time/makespan (it reflects the average work-in-process inventory if the jobs represent equal amount of inventory)

FCFS ABCDEF Aft=120/6=20 days At=54/6=9 days Job/ product Processin time (days) Due date (days) (2) Flow time (3) Days tardy (0 if negative) (3)-(2) A 2 7 2-7=-50 B 8 16 2+8=10 10-16=-60 C 4 10+4=14 14-4=10 D 10 17 14+10=24 25-17=7 E 5 15 24+5=29 29-15=14 F 12 18 29+12=41 41-18=23 41 120 54 Aft=120/6=20 days At=54/6=9 days M=41  Aj=120/41=2,93 pieces

 ACEBDF SPT Aft=108/6=18 days At=40/6=6,67 days Aj=108/41=2,63 pieces Job/ product Processin time (days) Due date (days) (2) Flow time (3) Days tardy (0 if negative) (3)-(2) A 2 7 2-7=-50 C 4 2+4=6 6-4=2 E 5 15 5+6=11 11-15=-40 B 8 16 11+8=19 19-16=3 D 10 17 19+10=29 29-17=12 F 12 18 29+12=41 41-18=23 41 108 40 Aft=108/6=18 days At=40/6=6,67 days Aj=108/41=2,63 pieces

EDD CAEBDF Aft=110/6=18,33 days At=38/6=6,33 days Job/ product Processin time (days) Due date (days) (2) Flow time (3) Days tardy (0 if negative) (3)-(2) C 4 4-4=0 A 2 7 4+2=6 6-7=-10 E 5 15 5+6=11 11-15=-40 B 8 16 11+8=19 19-16=3 D 10 17 19+10=29 29-17=12 F 12 18 29+12=41 41-18=23 41 110 38 Aft=110/6=18,33 days At=38/6=6,33 days Aj=110/41=2,68 pieces

CR ??? Job/ product Processin time (days) Due date (days) (2) CR0 A 2 Remaining time untill due to dat (on the 0th day) Job/ product Processin time (days) Due date (days) (2) CR0 A 2 7 (7-0)/2=3,5 B 8 16 (16-0)/8=2 C 4 (4-0)/4=1 D 10 17 (17-0)10=1,7 E 5 15 (15-0)/5=3 F 12 18 (18-0)/12=1,5 41 CR4 (7-4)/2=1,5 (16-4)/8=1,5 ------- (17-4)/10=1,3 (15-4)/5=2,2 (18-4)/12=1,17 1. 2.

Job/ product CR16 A (7-16)/2=-4,5 B (16-16)/8=0 C ------- D (17-16)/10=0,1 E (15-16)/5=-0,2 F ----- CR23 -------- (16-23)/8=-0,875 ------- (17-23)/10=-0,6 --------- CR18 -------- (16-18)/8=-0,25 ------- (17-18)/10=-0,1 (15-18)/5=-0,6 --------- 3. 4. 1. 5. 4. 2.

CRCFAEBD Aft=133/6=22,16 days At=58/6=9,66 days Aj=133/41=3,24 pieces Job/ product Processin time (days) Due date (days) (2) C 4 F 12 18 A 2 7 E 5 15 B 8 16 D 10 17 41 Flow time (3) 4 4+12=16 16+2=18 18+5=23 23+8=31 31+10=41 133 Days tardy (0 if negative) (3)-(2) 4-4=0 16-18=-20 18-7=11 23-15=8 31-16=15 41-17=24 58 Aft=133/6=22,16 days At=58/6=9,66 days Aj=133/41=3,24 pieces

Sequencing Jobs through two Work Center Job time is known and constant Job time is independent of job sequence All job must follow the same two-step job sequence Job priorities cannot be used All units must be completed at the first work center before moving on to the second work center

Example 2 E C F A B D Select the job with the shortest time. If the shortest time at first wc, schedule that job first. If the time at the 2nd wc, schedule the work last. Eliminate the job and its time from further consideration Repeat steps Job/ product #1 #2 A 5 B 4 3 C 8 9 D 2 7 E 6 F 12 15 E C F A B D

Chart 2 8 16 28 33 37 #1 D E C F A B #2 D E C F A B 2 9 17 26 28 43 48 51 Idle time Flow time: 51 hours

Thank you for your attention!