Session – XII

Advanced Schedulingi.e. Multiple Machine Scheduling

Beyond Classical Job Shop n = 1 (done)

Rule was : SOT gives best results and apply this to when n = 2

Ex.-1

4 Jobs A B C & D are performed on machine I and machine II in series and time taken is given as follows :

Jobs Time Taken onMachine – I Machine – II

A 3 2B 6 8C 5 6D 7 4

Find minimum time to be taken to perform all four jobs on both machines and also find idle time on each machine.

SOLUTION :

Rule is called Johnson’s rule which states that if SOT is on machine–I, give 1st priority and if on machine– II, give last priority.

SOT is 2 on machine – II for job A last priority (A is over) then 4 on II for D, 2nd last, 5 on I, 1st for C and 6 on I for B, 2nd priority.

Sequencing is C B D A

Jobs In I Out In II OutC 0 5 5 5 6 11B 5 6 11 11 8 19D 11 7 18 19 4 23A 18 3 21 23 2 25

Total time taken is 25

-2-

Gantt Chart and idle time calculation

5 11 18 21 25I C B D A Idle

5 6 7 3 4

5 11 19 23 25II Idle C B D A

5 6 8 4 2

Idle time on I is 4 & II is 5._____________________________________________________________________

Cross Check by F C F S

Jobs In I Out In II OutA 0 3 3 3 2 5 xB 3 6 9 9 8 17C 9 5 14 17 6 23D 14 7 21 23 4 27

Rule : For II machine, higher out time is to be considered

Total time is 27 (High)

3 9 14 21 27I A B C D Idle

3 6 5 7 6

3 5 9 17 23 27II Idle A Idl

eB C D

3 2 4 8 6 4

Idle time on I is 6 & II is 7 (both high), so not suitable

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Ex-2 : DATA

Job I IIA 9 6 3rd lastB 8 5 2nd lastC 7 4th 7D 6 3 lastE 1 1st 2F 2 2nd 6G 4 3rd 7

Sequencing is E F G C A B D

Jobs In I Out In II Out

E 0 1 1 1 2 3

F 1 2 3 3 6 9

G 3 4 7 9 7 16

C 7 7 14 16 7 23

A 14 9 23 23 6 29 x

B 23 8 31 31 5 36 x

D 31 6 37 37 3 40

TOTAL 37 36

Total time is 40

Idle on 1st is 40 – 37 = 3Idle on 2nd is 40 – 36 = 4

Gantt Charts are for Senior Executives

1 3 7 14 23 31 37 40I E F G C A B D Idle

1 2 4 7 9 8 6 3

1 3 9 16 23 29 31 36 37 40II Idle E F G C A Idle B Idle D

1 2 6 7 7 6 2 5 1 3

-4-

Ex - 3When n = 3

Rules : Convert 3 machines to 2 machines and apply Johnsons rule

I II III

A 18 10 8

B 19 12 18

C 12 5 16

D 16 6 14

E 21 9 10

Time on I + II becomes I and III + II becomes II

I + II I III + II IIA 28 18 LastB 31 30 4th lastC 17 1st 21D 22 20 3rd lastE 30 19 2nd last

Sequence is C B D E A

Time calculation of all three machines

Jobs In I Out In II Out In III OutC 0 12 12 12 5 17 x 17 16 33 xB 12 19 31 31 12 43 x 43 18 61 D 31 16 47 47 6 53 x 61 14 75 xE 47 21 68 68 9 77 x 77 10 87 xA 68 18 86 86 10 96 96 8 104

TOTAL 86 42 66

Time of completion is 104

Idle time on I is 104 – 86 = 18 II is 104 – 42 = 62III is 104 – 66 = 38

-5-

Gantt Chart

12 31 47 68 86 104I C B D E A Idl

e12 19 16 21 18 18

12 17 31 43 47 53 68 77 86 96 104II Idle C Idle B Idle D Idle E Idle A Idle

12 5 14 12 4 6 15 9 9 10 8

17 33 43 61 75 77 87 96 104

III Idle C Idle B D Idle

E Idle A

17 16 10 18 14 2 10 9 8

Idle time on I is =18 II 12 + 14 + 4 + 15 + 9 + 8 = 62 III 17 + 10 + 2 + 9 = 38

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Ex-4

n = 4

Jobs DataI II III IV

A 10 7 5 9B 9 6 4 7C 8 4 2 6D 12 8 3 9

Total 39 25 14 31

Rule I + II + III = I For sequencing IV + III + II = II

JobsI II

I + II + III

IV + III + II

A 22 1st 21 4th last orB 19 3rd 17 2nd last orC 14 4th 12 Last orD 23 2nd 20 3rd last or

Sequence is A D B C -6-

Flow time Calculation

I out II out III out IV outA 10 10 7 17 x 5 22 x 9 31 xD 12 22 8 30 x 3 33 x 9 42 B 9 31 6 37 x 4 41 x 7 49 C 8 39 4 43 2 45 x 6 55

39 25 14 31

Time of Completion is = 55

Idle time I = 55 – 39 = 16II = 55 – 25 = 30III = 55 – 14 = 41IV = 55 – 31 = 24

Grantt Chart for n = 4

10 22 31 39 55I A D B C Idle

10 12 9 8 16

Idle Time = 55 – 39 = 16

10 17 22 30 31 37 39 43 55II Idle A Idl

eD Idle B Idle C Idle

10 7 5 8 1 6 2 4 12

Idle Time = 10 + 5 + 1 + 2 + 12 = 30

17 22 30 33 37 41 43 45 55III Idle A Idle D Idle B Idle C Idle

17 5 8 3 4 4 2 2 10

Idle Time = 17 + 8 + 4 + 2 + 10 = 41

22 31 33 42 49 55IV Idle A Idle D B C

22 9 2 9 7 6

Idle Time = 22 + 2 = 24-7-

Assignment – II : A garment manufacturer has to process 7 orders A to G through two stages of cutting and sewing. The estimated time of 7 orders is.

Order No.

A B C D E F G

Stage – I 5 7 3 4 6 7 12Stage –II 2 6 7 5 9 5 8

Then third stage of packing is added and time taken is

10 12 11 13 12 10 11

Find sequence of Stage-I & II and I, II & III separately and total time of completion. Also find idle time on Stage- I, II and III.

-8-

Scheduling for moving Assembly line called Line Balancing.

It is to balance the time of some activities (or task) in such a way, so as to increase production and minimize idle time on assembly line.

Steps for line balancing.

1. All activities are planned and time is set.

2. Production per Shift (of 8 hrs) is decided with lunch break and rest time.

3. Cycle time Ct = Time available . desired production

4. No of work N wt = Total task timeStations (theoretical) Ct

5. Efficiency of line = Total task time Ct x Nwt

6. Actual work stations Nwa = to work out ?

Example 1 :

Production desired is 500 nos. from 8 hr shift with 30 min lunch and 2 rest limes of 15 min each. There are 11 activities to be performed to produce one product as follows.

Sr. No.

Activity Activity time (in sec)

Precedence

1 A 45 Nil2 B 11 A3 C 9 B4 E 50 Nil5 E 15 D6 F 12 C7 G 12 C8 H 12 E9 I 12 E10 J 8 F G H & I

11 K 9 J195 Sec

-9-

Calculate the actual efficiency and improve the same by balancing the line after splitting the task time.

Solution :

Total time available is 480 – 30 – 30 = 420Production desired is 500

Cycle time for one no production

Ct = 420 = 0.84 minutes 500

= 50.4 Seconds

as task time is given in seconds

Nwt = Task Time = 195 = 3.87Ct 50.4

Approx = 4 Nos.

Theoratical efficiency = 195 = 96.7% 50.4 x 4

Actual work stations ?

Network diagram

A B

11450

start

D

50

startE

15 I

12

F

12 J

8

K

9

G

12H

12

C

9 finish

-10-

Each work station has to be within the cycle time of 50.4 sec.

A will be 1st

D will be 2nd B, C, E & F can be added up for 3rd G, H, I & J can be added up for 4th K will be 5th (more idle time)

Actually the line has 5 work stations and not 4 as theoratically calculated

Actual efficiency is 195 = 77.3%50.4 x 5

Now the line has to be balanced for increasing the efficiency to 96.7% by splitting the task time and convert 5 work station to 4.

If time of G, H & I is reduced from 12 sec each to 11 sec by further training the workers and give them better tools, then G, H, I, J & K can be added up to one work station.

It is balancing the assembly line by time splitting and increase the efficiency of line to 96.7%.

Example : 2

Production desired is 750 nos.Time available is 480 – 60 = 420 min.8 tasks are performed as follows :

A 20 NilB 7 AC 20 BD 22 BE 15 CF 10 DG 16 E & FH 8 G

118 Sec

Balance the line to increase efficiency. -11-

SOLUTION :

Network diagram

Actual work stations ? It is 5

Ct = 420 x 60 = 33.6 Sec 750

N wt = 118 = 3.51 approx = 4 33.6

Efficiency = 118 = 87.79% 33.6 x 4

Actual eff is 70.20%, since Nwa is 5

Split the task time of C from 20 to 19 & E to 14, so as to have actual work stations as 4, and increase efficiency to 87.79%.

Assignment – III : Time available is 440 min and production is 400 nos. There are 8 tasks as follows and time is given in sec.

Task A B C D E F G HTime 54 24 36 12 18 24 42 66Precedence

Nil A B C C D&E F G

It is possible to achieve 100% efficiency by time splitting. Do some research work on task A & B and find answer.

A B

C E

D F

G H20 15

22 10

16 8720start

finish

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