sums print operationsscheduling

7/31/2019 Sums Print OperationsScheduling

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Sums: Operations Scheduling

Type 1 - Scheduling n jobs on one machine

Priority Rules for Job sequencing:

1. FCFS (First-come, first-served) orders are run in the order they arrive in the department.2. SOT (Shortest operating time.)Run the job with the shortest completion time first, next shortest second and

so on. It is also referred to as SPT (Shortest processing time).

3. EDD (earliest Due date first) Run the job with earliest due date first.4. STR (Slack time remaining) Orders with shortest slack time remaining (STR) are run first.

STR = Time remaining before due date Remaining processing time.

5. STR/OP (slack time remaining per operation). Orders with the shortest slack time per number of operationsare run first.

STR/OP = STR/ Number of remaining operations

6. CR (Critical ratio) this is calculated as the difference between the due date & the current date divided by thenumber of work days remaining. Orders with smallest CR are run first.

7. LCFS (Last-come, first-served). As orders arrive they are placed on top of the stack the operator usuallypicks up the order on top to run first.

8. Random order or Whim. The supervisor or operators usually select whichever job they feel like running.Sum: 1

Job ( in order of

Arrival)

Processing Time

(Days)

Due Date

A 3 5

B 4 6

C 2 7

D 6 9

E 1 2

Solution:

Method :1 First Come First Serve ( FCFS )

Job Seq.

Processing

Time

(Days)

Due Date

(Days

Hence)

Flow Time

Late by

(Comparing

Due date with

Flow time )

A 3 5 0 + 3 = 3 0


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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 /

Late by

50 23

Average Flow

Time

/ Late by

10 divide 5 23 divide 5

10 4.6

Method :2 Shortest Operating Time ( SOT )

Job Seq.

Processing

Time

(Days)

Due Date

(Days

Hence)

Flow Time

Late by

(Comparing

Due date with

Flow time )

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 /

Late by

36 12

Average FlowTime

/ Late by

36 divide 5 12 divide 57.2 2.4

Method :3 Earliest Due Date ( EDD )

Job Seq.

Processing

Time

(Days)

Due Date

(Days

Hence)

Flow Time

Late by

(Comparing

Due date with

Flow time )

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


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Total

Flow Time /

Late by

39 12

Average Flow

Time

/ Late by


7.8 2.4

Method :4 Last Come First Serve ( LCFS )

Job Seq.

Processing

Time

(Days)

Due Date

(Days

Hence)

Flow Time

Late by

(Comparing

Due date with

Flow time )

E 1 2 0+1 = 1 0

D 6 9 6+1 = 7 0

C 2 7 7 + 2 = 9 2

B 4 6 9 + 4 = 13 7A 3 5 13+ 3 = 16 11

Total

Flow Time /

Late by

46 20

Average Flow

Time

/ Late by


9.2 4

Method :5 Slack Time Remaining Per Operation ( STR )

Calculation of Slack

Job ( in order of

Arrival)

Processing

Time

(Days)

Due Date

STR

= DD - PT

A 3 5 2

B 4 6 2

C 2 7 5D 6 9 3

E 1 2 1

Note : Orders with shortest slack time run first


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Job Seq.

Processing Time

(Days)

Due Date

(Days Hence)

STR

= DD - PT

Flow

Time

Late by

(Comparing Du

date with Flow

time )

E 1 2 1 0 + 1 = 1 0

A 3 5 2 1 + 3 = 4 0

B 4 6 2 4 + 4 = 8 2

D 6 9 3 8 + 6 = 14 5

C 2 7 5

14 + 2 =

16 9

Total

Flow Time /

Late by

43 16

Average Flow

Time

/ Late by

43 divide

5 16 divide 5

8.6 3.2


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Type 2 Scheduling n jobs on two machines (Johnsons rule)

Johnsons Rule is:

1. From the list of unscheduled jobs, select the one with the shortest processing time in either work center.2. If the shortest time is at the first work center, do the job first in the schedule otherwise do the job last in the

schedule.

3. Remove the job assigned in Step 2 from the list of unscheduled jobs.4. Repeat steps 1,2 and 3 filling in the schedule from the front and the back until all jobs have been scheduled.

Sum: 1

Job Machine A

( Times in hours)

Machine B

( Times in hours)

J1 8 4

J2 6 7

J3 5 5

J4 7 4

J5 3 6

J6 9 8

Solution:

Optimum

Sequence

1 2 3 4 5 6

J5 J3 J2 J6 J4 J1

Choose the smallest time 3 hrs J5 ( on 1st Machine) so placed this job in the beginning; Next lowest time J1 or J4 4 hrs. Since tie choose J1 (earlier in numbering of jobs.) ( on 2 nd machine )

So place J1 end of sequence. And place J4 at the end (unfilled space)

Next lowest J3 5 hrs (same time on both machine so tie) choose 1st machine so place J3 frombeginning after J5.

J2 6 hrs (1st Machine) Place this job from the beginning of sequence. J6 in vacant place.Note:

When there is a tie in two stages for two different jobs. In this case, place the job with the smallest time inthe first stage as early as possible in the unfilled job sequence.


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Calculation: Cycle time / idle time Machine A & B

JobMachine A

( Times in hours)

Machine B

( Times in hours)

Max Time

Required

J1 8 4 8

J2 6 7 7

J3 5 5 5J4 7 4 7

J5 3 6 6

J6 9 8 9

Operation time

Machine A /B38 34

42

Idle time

machine A/B4 8

Determination of Cycle time & idle time of the two machines: Two methods that can be used:

1. Graphical method or Gantt chart method2. Tabular column method

Tabular Column method:

Job

Sequence

Machine A Machine B Idle time

OperationTime

OT

Time In

TI

Time out

TO

OperationTime

OT

TimeIn

TI

Time out

TOMachine A Machine B

J5 3 0 3 6 3 6 + 3 = 9 - 3

J3 5 3 5 + 3 = 8 5 9 9 + 5 = 14 - -

J2 6 8 14 7 14 21 - 2

J6 9 14 23 8 23 31 -

J4 7 23 30 4 31 34 - 3

J1 8 30 38 4 38 42 4 -

4 8


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Graphical method or Gantt chart method: - last page.


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Type 3 Scheduling a set of number of jobs on the same number of machines (n things

n destination) The Assignment method.

The assignment method is appropriate method in solving problems that have the following characteristics:

1.There are n things to be distributed to n destination2.Each thing must be assigned to one & only one destination.3.Only one criterion can be used (minimum cost, maximum profit, or minimum completion time etc...)

Steps:

Step : 1 Subtract the smallest number in each row from itself & all other numbers in that row. (There will

be at least 1 zero in each row.)

Step : 2 Subtract the smallest number in each column from itself & all other numbers in that column

(There will be at least 1 zero in each column.)

Step : 3 Determine if the minimum number of lines required to cover each zero is equal to n.

If so optimal solution has been found, because job machine assignments must be made at the zero

entries, & this test proves that this is possible.

If minimum number of lines required is less than n, go to step 4.

Step : 4 Draw the least possible number of lines through all the zeros.

Subtract the smallest number not covered by lines from itself & all other uncovered numbers &

add it to the number at each intersection of lines Repeat step 3.


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Sum: 17

Machine

1 2 3 4 5

Job

A 6 11 12 3 10

B 5 12 10 7 9

C 7 14 13 8 12

D 4 15 16 7 9

E 5 13 17 11 12

Solution:

Step: 1 Row reduction

Machine

1 2 3 4 5

JobA 3 8 9 0 7

B 0 7 5 2 4

C 0 7 6 1 5

D 0 11 12 3 5

E 0 8 12 6 7


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Step: 2 Column reduction

Machine

1 2 3 4 5

JobA 3 1 4 0 3

B 0 0 0 2 0

C 0 0 1 1 1

D 0 4 7 3 1

E 0 1 7 6 3

Step: 3 Minimum number of lines

Lines = 4 not equal to machines = 5


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Step: 4

Smallest uncovered number was 1.

So subtracting 1 from uncovered numbers and adding in to the number at each intersection of lines.

Optimum Solution:

Note:

Start assigning with line with 1 zero.

Delete row & column & carry out the same process for all jobs.


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Assign Cost

1st

- A 4 Rs3

4th - B 3 10

5th - C 2 14

3rd

- D 5 9

2nd

- E 1 5

Total Rs41

Type 4 - Consecutive days-off (Sums 4, 8, 9, 16)

Objective:Finding the schedule that minimizes the number of five-day workers with two consecutive days off.

Steps:

Step: 1 Assign the first worker to all of the days that require staffing.

Simply recopy the total requirement for each day.

Circle the two consecutive days with the lowest numbers, These will be considered for days off.

In case of ties:

Choose the days-off pair with the lowest requirement on an adjacent day.

If tie still remains choose the first of the available tied pairs.

Step: 2 For worker 2,Subtract 1 from each of the days not circled with positive number & enter this in the worker 2

row.

This indicates that 1 less worker is required on these days because the first worker has just been

assigned to them.

The 2 steps are repeated for the second worker third so forth until no more workers are required

to satisfy the schedule.


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Sum: 4

Jumbos restaurant is trying to create a consecutive-days-off schedule that uses the fewest workers. Use thefollowing information to create a 5 days on 2 days off schedule.

Day M Tu W Th F S Su

Requirements 2 2 1 3 3 4 2

Solution:


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Type 5 First-hour principle

The first-hour rule says that for the first hour we assign a number of workers equal to the requirement in

that period.

For each subsequent period assign the exact number of additional workers to meet the requirements.

When in a period one or more workers come to the end of their shifts add more workers only if they are

needed to meet the requirement.

Sum: 2

A hotel has to schedule its receptionist according to hourly loads. Mgt has identified the number of receptionist

needed to meet the hourly requirement which changes from day to day, Assume each receptionist works a 4

hour shift. Given the following requirement in a certain day use first-hour principle to find the personnel

schedule:

Period

8AM 9AM 10AM 11AM Noon 1PM 2PM 3PM 4PM 5PM 6PM 7PM

Requirement 2 3 5 8 8 6 5 8 8 6 4 3

Solution:

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