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SCHEDULING TECHNIQUE

Scheduling Rules

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Page 1: Scheduling Rules

SCHEDULING TECHNIQUE

Page 2: Scheduling Rules

• BASICS OF SCHEDULING

• In case of job-shop, job needs to be routed between functionally organized work centres to complete the work

• Scheduling, involves determining the order for running jobs and also assigning a machine

• Proper scheduling enables firms allocate their production capacity to meet their customer requirements on time. Scheduling types:

• Forward scheduling,

• Backward scheduling

• Combination of both.

Page 3: Scheduling Rules

In Forward Scheduling system takes order and then schedules each operation that must be completed forward in time. System can tell (gives) the earliest date that an order can be completed.

Backward Scheduling starts from some date in future (possibly a due date) and schedules required operations in reverse sequence.

Backward schedule tells when an order must be started in order to be done by a specific date.

Page 4: Scheduling Rules

Important elements of scheduling :1.Routing decisions the sequence of operations and the work centers that perform the work. 2.Loading assigns jobs to various work centers.3.Dispatching is the release of an order to start the production operation. The different priority rules in dispatching are: 1.earliest due date, 2.longest processing time, 3.shortest processing time,4. first in first serve, 5.critical ratio method and 6.slack time remaining.

Page 5: Scheduling Rules

Priority Rules for job sequencing:

1. FCFS (first come, first serve). Orders are run in the order they arrive in the dept.

2. SOT (Shortest Operating Time) / SPT {Shortest Processing Time). : Run the job with shortest completion time first, next shortest second & so on. (Identical to

3. Due Date: Earliest due date first; run the job with earliest due date first. D Date – when referring to the entire job; OPNDD – when referring to next operation.

Page 6: Scheduling Rules

4.. STR (slack time remaining :Orders with shortest STR are run first.

5. LCFS (Last Come First Served): works on default. Orders as they arrive placed on top of stack. Operator usually pricks up order on top to run first.

6. Random order or whim : Supervisors or operators often pick up / select which ever job they feel like running.

7. LOT (Longest Operating Time) / LPT (Longest Processing Time) : Run the job with longest completion time first, next longest second & so on.

Page 7: Scheduling Rules

7.Critical ratio method

• The critical ratio method is a job sequencing technique that an operations manager can use to verify whether a job is being operated on schedule. In this method, the operations manager calculates the critical ratio of a job, which is the ratio of the actual time remaining to complete the job and the scheduled time remaining to complete the job.

• Critical Ratio = Actual time remaining

• Scheduled time remaining

Page 8: Scheduling Rules

If the critical ratio of an operation is less than one, then it can be said that the operation is behind schedule. If the critical ratio is more than unity, the operation is being processed ahead of schedule. The method helps the operations manager revise the status of the jobs from time to time and re-prioritize them.

Page 9: Scheduling Rules

Schedule n jobs on 1 machine (n/1)Mike Morales is supervisor of legal copy-express, which provides copy services for dun time Los Angeles law firms. Five customers submitted the orders at beginning of week. Specific scheduling dates are as follows:

Job Process Due Date Time (Days) (Days Hence) A 3 5 B 4 6 C 2 7 D 6 9 E 1 2

Page 10: Scheduling Rules

All orders require use of only color copy machine available. Morales must decide on processing sequence for five orders. Evaluation criterion is minimum flow time. Suppose that Morales decides use the FCFS rule in an attempt to make Legal Copy Express appear fair to its customers.

Page 11: Scheduling Rules

1.FCFS Schedule

Job Sequence

Process Time (Days)

Due Date (Days Hence)

Flow Time(Days)

Lateness

A 3 5 0 + 3 = 3 0

B 4 6 3 + 4 = 7 1

C 2 7 7 + 2 = 9 2

D 6 9 9 + 6 = 15 6

E 1 2 15 + 1 = 16 14

Total Flow Time = 3 + 7 + 9 + 15 + 16 = 50 days. Mean Flow time = 50/5 = 10 days. Only job A will be on time. Job B, C, D & E will be late by 1,2,6 and 14 days respectively. On an average job will be late by 0+1+2+6+14 = 4 . 6 days

Page 12: Scheduling Rules

2.SOT ( shortest Operation time) RuleSOT Schedule

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

Flow Time(Days)

Lateness

E 1 2 0 + 1 = 1 0

C 2 7 1 + 2 = 3 0

A 3 5 3 + 3 = 6 1

B 4 6 6 + 4 = 10 4

D 6 9 10 + 6 = 16 7

Total Flow Time = 1+3+6+10+16 = 36 daysMean Flow time = 36 = 7.2 daysAverage delay 0+0+1+4+7 = 2.4 days. SOT results in lower average flow time than FCFS rule. Jobs E & C will be ready before due date, Job A is late by only one day, on an average job will be late by 0+0+1+4+7 = 2.4 days.

Page 13: Scheduling Rules

SOT results in lower average flow time than FCFS rule. Jobs E & C will be ready before due date, Job A is late by only one day, on an average job will be late by 0+0+1+4+7 = 2.4 days. 5

Page 14: Scheduling Rules

3.Due Date Rule : Resulting Schedule is

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

Flow Time(Days)

Lateness

E 1 2 0 + 1 = 1 0

A 3 5 1 + 3 = 4 0

B 4 6 4 + 4 = 8 2

C 2 7 8 + 2 = 10 3

D 6 9 10 + 6 = 16 7

Total completion time = 1+4+8+10+16=39 days.Mean Flow Time = 39/5 = 7.8 days.Jobs BCD will be late. On an average jobs will be late by 0+0+2+3+7 = 2.4 days.

Page 15: Scheduling Rules

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

Flow Time(Days)

Lateness

E 1 2 0 + 1 = 1 0

D 6 9 1 + 6 = 7 0

C 2 7 7 + 2 = 9 2

B 4 6 9+ 4 = 13 7

A 3 5 13 + 3 = 16 11

Total Flow Time = 46 days, Mean Flow Time = 9.2 days 0+0+2+7+11 20Average Lateness = -------------------- = ------- = 4 days 5 5

4.LCFS

Page 16: Scheduling Rules

5.Random Schedule

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

Flow Time(Days)

Lateness

D 6 9 0 + 6 = 6

C 2 7 6 + 2 = 8 1

A 3 5 8 + 3 = 11

6

E 1 2 11 + 1= 12

10

B 4 6 12 + 4 = 16

10

Total Flow Time = 53 days. Mean Flow Time = 10.6 days 0+1+6+10+10Average Lateness =----------------------------- = 5.4 days. 5

Page 17: Scheduling Rules

6.STR ( slack time remaining )Schedule

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

Slack Time

Flow Time(Days)

Lateness

E 1 2 2-1=1 0 + 1 = 1

0

A 3 5 5-3 =2 1 + 3 = 4

0

B 4 6 6-4 = 2 4 + 4 = 8

2

D 6 9 9-6 = 3 8 + 6 = 14

5

C 2 7 7-2 = 5 14 + 2 = 16

9

Total Flow time = 43 days. Mean Flow Time = 8.6 days 0+0+2+5+9Average Lateness =----------------------------- = 3.2 days.

5

Page 18: Scheduling Rules

7.CR ( Critical Ratio )Schedule

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

 Critical Ratio

Flow Time(Days)

Lateness

D 6 9  1.5 0 + 6 = 6 0

B 4 6  1.5 6 + 4 = 10

4

A 3 5  1.67 10 + 3 = 13

8

E 1 2  2 13 + 1 = 14

12

C 2 7  3.5 14+2 = 16

9

Total Flow time = 59 days. Mean Flow Time = 11.8 daysAverage Lateness 33 / 5 = 6.6 days.

Page 19: Scheduling Rules

8.LOT ( Longest Operating Time First )Schedule

Job Sequence

Processing Time (Days)

Due Date (Days Hence)

  Flow Time(Days)

D 6 9   0 + 6 = 6

B 4 6   6 + 4 = 10

A 3 5   10 + 3 = 13

C 2 7   13 + 2 = 15

E 1 2   15+ 1 = 16

Total Flow time = 60 days. Mean Flow Time = 12 daysAverage Lateness 34 / 5 = 6.8 days.

Page 20: Scheduling Rules

Comparison of Priority Rules

RuleTotal Completion Time (Days)

Average Completion Time (Days)

Average Lateness (Days)

FCFS 50 10 4.6

SOT 36 7.2 2.4

D Date 39 7.8 2.4

LCFS 46 9.2 4.0

Random 53 10.6 5.4

STR 43 8.6 3.2

CR 59 11.8 6.6

LOT 60 12 6.8

Obviously SOT is better than the Rest. It is always the case SOT is the most important concept in entire subject of sequencing. It is optimum solution n/1 case in other criteria mean waiting time & mean completion time.

Page 21: Scheduling Rules

Johnson’s Job Sequencing Rules:(For 2 stage production)Scheduling n jobs to 2 machines (n/2)2 or more jobs to be processed on 2 machine

common sequencing.Objective of this approach, called Johnson’s

Rule or method is to minimize flow time from beginning of first job until the finish of the last.

Page 22: Scheduling Rules

Rule consists of following steps.1. List operation time for each job

on both machines.2. Select the shortest operation time.3. If shortest time is for first

machine, do the job first, if it is for the second do the job last.

4. Repeat steps 2 & 3 for each remaining job until schedules is complete.

Page 23: Scheduling Rules

• Johnson’s Rule for 2 stage Production:- n jobs on 2 m/c A & B in that order.

• (A printing / B )• Expected processing the A1 – A2 – An, B,

B2 - - Bn • Sequence Johnson & Billnon……• Plenty least processing time, if it is least on

first m/c place in bringing, if it is least on B then place at end.

Page 24: Scheduling Rules

If tie them in A, the any job in beginning, tie at B, then any at end.Next processing time and repeat.Calculated total elapsed time = Time shifting from job & but job in sequence.Idle time on m/c AIdle time on m/c B

Page 25: Scheduling Rules

Example n jobs on 2 machines, in example 4 jobs 2 machinesStep 1 : List Operation times

JobOperation Time on machine 1

Operation Time on machine 2

A 3 2

B 6 8

C 5 6

D 7 4

Page 26: Scheduling Rules

Step 2 & 3: Select shortest operation time and assign. Job A is shortest on machine 2 and is assigned first and performed last (once assigned job A is no longer available to be scheduled).Step 4: Repeat step 2 and 3 until completion of schedule.

Page 27: Scheduling Rules

Select the shortest operation time amongst remaining jobs. Job D is second shortest on machine 2 so to be performed second last. Job C – shortest on machine 1 so performed first. Now only B is left with shortest operation time on machine 1 so first among remaining or second overall (after C ). Solution sequence is C →B→D→A flow time is 25 days which is minimum. Also are total idle time and mean idle time

Page 28: Scheduling Rules

Fan Blade Type

1 2 3 4 5 6

m/c A 30 100 50 20 90 100

m/c B 70 95 90 60 30 15

41 3 2 5 6

Then least 20 hrs for type 4 in ANext 30 Hrs. for type 1 m/c A, 30 Hrs type 5 – m/c BNext 50 hrs type 3 on m/c, & remaining 2 in last slot.

Least type 6 (15) at B

Fan Blade six types on m/c A & B

Page 29: Scheduling Rules

Job Sequence

Machine - A Machine - B

Time In

Processing Time

Time Out

Time In

Processing Time

Time Out

4 0 20 20 20 60 80

1 20 30 50 80 70 150

3 50 50 100 150 90 240

2 100 100 200 240 95 335

5 200 90 290 335 30 365

6 290 100 390 390 15 405

So final results Schedule of 6 jobs on 2 – m/c .

Total elapsed time 405 hrsIdle time A is 15 hrs., for 390 to 405 hrs.Idle time for m/c B is 20 + 25 = 45 hrs. (0-20) & (365 – 390)hrs.

Page 30: Scheduling Rules

Gantt ChartsSmaller job shops and individual departments of large ones employ the venerable Gantt Chart to help plan & track jobs.

Gantt Chart is a type of bar chart that plots tasks against time.

Note: Whether job is ahead of schedule or behind schedule is based on where it stands compared to where we are now.

Page 31: Scheduling Rules

Gantt Work Load Chart Date

JOB

Week Number

1 2 3 4 5 6 7 8

A

B

C

D

Scheduled Activity Time Time for Non Productive Activity.

Actual Preference - Review Period(point in time where project is now)

closing date. -starting date