1. Scheduling Seminar exercises
Process and Production Management

2. 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.)

3. 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.

4. 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.

5. 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

6. 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)

7. 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

8.  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

9. 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

10. 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.

11. 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.

12. 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

13. 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

14. 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

15. 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

16. Thank you for your attention!