Scheduling 2012.10.16 Process and Production Management

Scheduling

2012.10.16

Process and Production Management

Scheduling

• Establishing 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, personal computers, radios, TVs)

– Intermediate – volume systems (output 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.)

High volume systems

• Standardized equipments and parts• Standardized process• Smooth flow of work flow-shop scheduling• Main problem: line balancing • Success of the system:

– Process and product design– Preventive maintenance– Rapid repair– Optimal product mix– Minimization of quality problems– Reliability and timing of suppliers

Intermediate – volume systems

• Standardized output, volume is not so large, produce intermittently

• Main problem: run size of jobs (economic run size), timing of jobs, sequence

up

p

H

DSQopt

2

• D-demand• S-setup cost• H-production cost• P- production rate• U-usage rate

Low-volume systems

• Products are made to order

• Different jobs,

• Main problem: job-shop scheduling– Loading (assigning jobs to processing

centers)– Sequencing (determine the order of jobs at

work centers)

• Gantt chart: visualize the problem

• Loading:– Infinite loading (no regards to the capacity of

workstations)– Finite loading (capability is taken into account)

• Scheduling:– Forward scheduling: scheduling ahead from a point of

time– Backward Scheduling: scheduling backward from a

due date

Loading

• Hungarian rule:– Subtract the smallest number in each row from, every number in

the row– Subtract the smallest number in each column from, every number

in the column.– Test weather an optimum assignment can be made (determining

the minimum number of lines needed to cross out all zeros, if it is equals the number of rows optimum assignment) Go to step 6

– Modify the table• Subtract the smallest uncovered number from every uncovered

number• Add the smallest uncovered number to the numbers at intersections

of cross-out lines.

– Repeat step 3 and 4 until obtain optimal solution– Make assignments

Exercise

Worker

A B C D

Job 1 8 6 2 4

2 6 7 11 10

3 3 5 7 6

4 5 10 12 9

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.

Example 1Job/

productProcessing time (days)

Due date

(days)

A 2 7

B 8 16

C 4 4

D 10 17

E 5 15

F 12 18

• Determine the sequence of jobs/products

• Determine average flow time

• Determine average tardiness

• Determine average number of jobs at the workcenter

• Job flow time – the length of time a job is at a particular workstation or work center. (it includes not only processing 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)

FCFSJob/

productProcessin time (days)

Due date (days)

(2)

Flow time

(3)

Days tardy (0 if negative)

(3)-(2)

A 2 7 2 2-7=-50

B 8 16 2+8=10 10-16=-60

C 4 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

ABCDEF

SPTJob/

productProcessin time (days)

Due date (days)

(2)

Flow time

(3)

Days tardy (0 if negative)

(3)-(2)

A 2 7 2 2-7=-50

C 4 4 2+4=6 6-4=2

E 5 15 5+6=11 11-15=-40

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

ACEBDF

EDD

Job/ product

Processin time (days)

Due date (days)

(2)

Flow time

(3)

Days tardy (0 if negative)

(3)-(2)

C 4 4 4 4-4=0

A 2 7 4+2=6 6-7=-10

E 5 15 5+6=11 11-15=-40

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

CAEBDF

CR

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 (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

Remaining time untill due to dat (on the 0th day)

1.

CR4

(7-4)/2=1,5

(16-4)/8=1,5

-------

(17-4)/10=1,3

(15-4)/5=2,2

(18-4)/12=1,172.

???

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 -----

CR18

--------

(16-18)/8=-0,25

-------

(17-18)/10=-0,1

(15-18)/5=-0,6

---------

1.

2.

3.

4.

CR23

--------

(16-23)/8=-0,875

-------

(17-23)/10=-0,6

-------

---------

4.

5.

CRCFAEBD

Job/ product

Processin time

(days)

Due date (days)

(2)

C 4 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

• Aft=133/6=22,16 days• At=58/6=9,66 days• Aj=133/41=3,24 pieces

Days tardy (0 if negative)

(3)-(2)

4-4=0

16-18=-20

18-7=11

23-15=8

31-16=15

41-17=24

58

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

Job/ product

#1 #2

A 5 5

B 4 3

C 8 9

D 2 7

E 6 8

F 12 15

• 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

D BAE C F

Chart

D E C F A B#1

D E C F A B#2

2 8

1512F

86E

72D

98C

34B

55A

#2#1Job/ product

1512F

86E

72D

98C

34B

55A

#2#1Job/ product

16 28 33 37

2 9 17 26

Idle time

28 43 48 51

Flow time: 51 hours

Thank you for your attention!