Kuliah-responsi Versi B TATA LETAK

8/8/2019 Kuliah-responsi Versi B TATA LETAK

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To AccompanyKrajewski & Ritzman

Implementchanges

6

Redesign Process(Revise layout)

5

Redesigning a Process

Through Layout Revision

Documentprocess

3

Definescope

2

Evaluateperformance

4

Figure 7.1Figure 7.1

Identifyopportunity

1


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Absolute Locations

(a) Original layout(a) Original layout

Frozen

foodsDry

groceries

Bread Vegetables

Meats



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Absolute Locations

(b) Revised layout(b) Revised layout

Frozen

foodsDry

groceries

BreadVegetables

Meats



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Layout Types

FoundryMilling

machines

LathesGrinding

Painting Drills

Office

Welding

Forging

(a) Layout of a job shop(a) Layout of a job shop



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Layout Types

(b) Layout of a production line(b) Layout of a production line


Station 1 Station 2 Station 3 Station 4


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Group Technology

Machine

1

Machine2

Machine

3

Machine

4Machine

5

Materials in

Finished

goods out


One Worker, Multiple MachinesOne Worker, Multiple Machines


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Group Technology


Drilling

D D

D D

Grinding

G G

G G

G G

Milling

M M

M M

M M

Assembly

A A

A A

Lathing

Receiving and

shipping

L

L L

L L

L L

L

(a) Jumbled flows in a job shop without GT cells(a) Jumbled flows in a job shop without GT cells


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Group Technology

(b) Line flows in a job shop with three GT cells(b) Line flows in a job shop with three GT cells

Cell 3

L M G G

Cell 1 Cell 2

Assembly

area

A A

L M DL

L MShipping

D

Receiving

G



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Department Area Needed (ft2)

1. Administration 3,500

2. Social services 2,600

3. Institutions 2,400

4. Accounting 1,600

5. Education 1,500

6. Internal audit 3,400

Total 15,000

100'

150'

Office ofBudget

Management

3 6 4

1 2 5



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Office ofBudget

ManagementTrips between Departments

Department 1 2 3 4 5 6

1. Administration

2. Social services

3. Institutions

4. Accounting

5. Education

6. Internal audit

3 6 5 6 10

8 1 1

3 9

2

1

100'

150'

3 6 4

1 2 5


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Office ofBudgetManagement

y Departments 1 and 6 close together




100'

150'

1 5

6

100'

150'

3 6 4

1 2 5


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100'

150'

1 5

6 3

100'

150'

3 6 4

1 2 5


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100'

150'

1 5

6 32

4

100'

150'

3 6 4

1 2 5


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100'

150'

3 6 4

1 2 5





100'

150'

1 5

6 32

4


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Out-and-Back Warehouse


Dock Aisle

Storage area

Storage area

3 5 5 6 4 2 7

1 5 5 4 4 2 7


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Zone Systems


Click to add title

Zones Zones Control

station Shipping

doors

Tractor

trailer

Tractor

trailer

Overflow

Feeder

linesFeeder

lines


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17

Design Product Layouts: Line

BalancingLine balancing is the process of assigning tasks to workstations

in such a way that the workstations have approximately the same

processing time requirements. This results in the minimized idle time

along the line and high utilization of labor and equipment.

Cycle time is the maximum time allowed at each workstation

to complete its set of tasks on a single unit

What is the cycle time for the system above?

Worker 1 Worker 2

4 tasks 2 tasks

Each task takes 1 minutes, how to balance?


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18

Parallel Workstations

1 min.2 min.1 min.1 min.30/hr. 30/hr. 30/hr. 30/hr.

1 min.

2 min.

1 min.1 min.60/hr.

30/hr. 30/hr.

60/hr.

2 min.

30/hr.30/hr.

Bottleneck

Parallel Workstations


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19

The obstacle

The difficulty to forming task bundles that have thesame duration.

The difference among the elemental task lengths

can not be overcome by grouping task. Ex: Can you split the tasks with task times {1,2,3,4} intotwo groups such that total task time in each group is thesame?

Ex: Try the above question with {1,2,2,4}

A required technological sequence prohibit thedesirable task combinations Ex: Let the task times be {1,2,3,4} but suppose that the

task with time 1 can only done after the task with time 4 iscompleted. Moreover task with time 3 can only done after

the task with time 2 is completed. How to group?


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20

Cycle time is the maximum time allowed at each

workstation to complete its tasks on a unit.

Cycle Time

The major determinant: cycle time

Minimum cycle time: longest task time by

assigning each task to a workstation

Maximum cycle time: sum of the task time by

assigning all tasks to a workstation


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21

Determine Maximum Output

Cycle Time: Time to process 1 unit

Example: If a student can answer a multiple choice question in 2 minutes

but gets a test with 30 questions and is given only 30 minutes then

OT=30 minutes; D=30

Desired cycle time=1 minute < 2 minutes = Cycle time from the processcapability

OT

D

OT

D

CT

OT

DCT

OT

D

CT

:

:

OperatingTimePerDay

DesiredOutputRate

DesiredCycleTime

CycleTime FromProcessDesign

Can produce at the desired level, design is feasible

Cannot produce at the desired level, design is infeasible

!

! n

u


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22

Determine the Minimum Number

of Workstations Required: Efficiency

stask timeofsum=t

CTt

productafortimeAvailabaleproductafortask timeTotal

OT/DtN

OT

t)(D)(

dayaintimeAvailabale

dayainproducedproductsallfortask timeTotal=N

min

min

!!!

!

Example: Students can answer a multiple choice question in 2 minutes but

given a test with 30 questions and is given only 30 minutes. What is the

minimum number of students to collaborate to answer all the questions in

the exam?

Total operation (task) time = 60 minutes = 30 x 2 minutes

Operating time=30 minutes60/3=2 students must collaborate. This Nmin below.


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24

Example 1: Precedence Diagram

Precedence diagram: Tool used in line balancing

to display elemental tasks and sequence

requirements

a b

c d e

0.1 min.

0.7 min.

1.0 min.

0.5 min. 0.2 min.


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25

Example 1: Assembly Line

Balancing

Arrange tasks shown in the previous slideinto workstations.

Use a cycle time of 1.0 minute Every 1 minute, 1 unit must be completed

Rule: Assign tasks in order of the most numberof followers

If you are to choose between a and c, choose a

If you are to choose between b and d, choose b

Number of followers: a:3, b:2, c:2, d:1, e:0

Eligible task fits into the remaining time and allof its predecessors are assigned.


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26

o u on o xamp e .

Assigning operations by the number

of followersWork-Station

TimeRemaining Eligible

AssignTask

StationIdle Time

1 1.0.9

.2

a,cc

none

ac

- .22 1.0

0b

noneb- 0

3 1.0.5

.3

de

-

de

- .3.5

- Eligible operation fits into the remaining time and its predecessors are already

assigned.- What is the minimum cycle time possible for this example?


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28

Line Balancing Heuristic Rules

Assign tasks in order of most following

tasks.

Assign task in the order of the greatest tasktime.

Assign tasks in order of greatest positional

weight.

Positional weight is the sum of each tasks

time and the times of all following tasks.


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29

Solution to Example 1. Assigning

operations using their task times.Wor -

t tion

i

ining ligi l

ign t tion

I l i1 1.

.9

.2

a,ca

none

ca

- .22 1.

none -1...

e-

e- .

.

Eligible operation fits into the remaining time and its predecessors are already

assigned.


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30

Positional Weights

Assign tasks in order of greatest positional weight.

Positional weight is the sum of each tasks time

and the times of all following tasks.

a:1.8 mins; b: 1.7 mins; c:1.4 mins; d: 0.7 mins;

e:0.2 mins.


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31

Solution to Example 1. Assigning

operations using their task times.Wor -

t tion

i

ining ligi l

ign t tion

I l i1 1.

.

.

a,cc

none

ac

- .1.

none -3 1.

.

.3e-

e- .3

.

Eligible operation fits into the remaining time and its predecessors are already

assigned.


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32

c d

a b e

f g h

0.2 0.2 0.3

0.8 0.6

1.0 0.4 0.3

Example 2


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33

Station 1 Station 2 Station 3 Station 4

a b e

f

d

g h

c

Solution to Example 2


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Line Balancing

Green Grass, Inc.Green Grass, Inc.

Big BroadcasterBig Broadcaster


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Line Balancing Big BroadcasterBig Broadcaster


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AA Bolt leg frame to hopper Bolt leg frame to hopper 4040 NoneNone

BB Insert impeller shaftInsert impeller shaft 3030 AACC Attach axleAttach axle 5050 AA

DD Attach agitatorAttach agitator 4040 BB

EE Attach drive wheelAttach drive wheel 66 BB

FF Attach free wheelAttach free wheel 2525 CC

GG Mount lower postMount lower post 1515 CC

HH Attach controlsAttach controls 2020 D, ED, E

II Mount nameplateMount nameplate 1818 F, GF, G

TotalTotal 244244

WorkWork TimeTime ImmediateImmediate

ElementElement DescriptionDescription (sec)(sec) Predecessor(s)Predecessor(s)

Example 7.3Example 7.3


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BB Insert impeller shaftInsert impeller shaft 3030 AACC Attach axleAttach axle 5050 AADD Attach agitatorAttach agitator 4040 BBEE Attach drive wheelAttach drive wheel 66 BBFF Attach free wheelAttach free wheel 2525 CCGG Mount lower postMount lower post 1515 CCHH Attach controlsAttach controls 2020 D, ED, EII Mount nameplateMount nameplate 1818 F, GF, G

TotalTotal 244244



4040

3030

B

A



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TotalTotal 244244



4040

5050

B

C

A

3030



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TotalTotal 244244



4040

5050

3030

4040

D

B

C

A



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BB Insert impeller shaftInsert impeller shaft 3030 AACC Attach axleAttach axle 5050 AADD Attach agitatorAttach agitator 4040 BBEE Attach drive wheelAttach drive wheel 66 BBFF Attach free wheelAttach free wheel 2525 CCGG Mount lower postMount lower post 1515 CCHH Attach controlsAttach controls 2020 D, ED, E


TotalTotal 244244



4040

66

5050

E3030

4040

D

B

C

A



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TotalTotal 244244



4040

66

5050

E3030

2525

4040

D

B

F

C

A



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TotalTotal 244244



4040

66

5050

1515

E3030

2525

4040

D

B

F

C

A

G



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TotalTotal 244244



4040

66

2020

5050

1515

E3030

2525

4040H

D

B

F

C

A

G



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TotalTotal 244244



4040

66

2020

5050

1515

1818

E3030

2525

4040H

I

D

B

F

C

A

G



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4040

66

2020

5050

1515

1818

E3030

2525

4040H

I

D

B

F

C

A

G


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Desired output rate, rmust be matched to the

staffing or production plan.

Cycle time, cis the maximum time allowed for work on a unit at

each station:1

rc=


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Desired output rate, r = 2400 unit/week

Plant operates 40 hours/week

r=2400/40 = 60 units/hour

Cycle time, c =1/60

=1 minute/unit

= 60 seconds/unit

1

r


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Theoretical minimum (TM ) - sum of all work-element

standard times divided by the cycle time.

TM =244 seconds/60 seconds =4.067

It must be rounded up to 5stations

Cycle time: c=1/60=1 minute/unit = 60 seconds/unit

Efficiency (%) - ratio of productive time to total time.

Efficiency= [244/5(60)]100= 81.3%

Balance Delay - amount by which efficiency falls short of 100%.

(100 81.3) =18.7%


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c = 60 seconds/unitc = 60 seconds/unitTM = 5 stationsTM = 5 stations

Efficiency = 81.3%Efficiency = 81.3%

S1

S2

S3

S5S4


40

6

20

50

15

18

E30

25

40H

I

D

B

F

C

A

G

S1 A A 40 20

S2 B,C C 50 10

S3 B,F,G B 30 30

E,F,G F 55 5

Cumm Idle

Station Candidate Choice Time Time


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From-To Chart

Processes:

X, Y, Z, Q, and R

Parts that are manufactured through theseprocesses:

Part Nos.: 1, 2, & 3


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From-To ChartTo

From

X

Y

Z

Q

R

X Y Z Q R Total

Total


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From-To ChartTo

From

X

Y

Z

Q

R

X Y Z Q R Total

Total

Process Sequence: Part No. 1= X to Y to Z to R to Q

1

1

1

1


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From-To ChartTo

From

X

Y

Z

Q

R

X Y Z Q R Total

Total

1

1

1

1

Process Sequence: Part No. 3= X to Y to Z to Q to R

2

2

2

3

3

3

3


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From-To ChartTo

From

X

Y

Z

Q

R

X Y Z Q R Total

Total

1

1

1

1

2

2

2

3

3

3

3

3

2

2

2

3

3 33

1

1111


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Soal 1

Tata letak sebuah rumah sakit ditunjukkan sebagai berikut

Proses

Awal

(1)

RuangRecovery

(7)

Cast-SettingRoom

(8)

X-Ray

(4)

RuangOperasi

(6)

Exam

Room 2

(3)

Lab EKG

(5)

Exam

Room-1

(2)

40

10

10

Adapun pergerakan jumlah pasien antar diperlihatkan

sebagai berikut :


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(1) (2) (3) (4) (5) (6) (7) (8)

(1) 100 100

(2) 50 20

(3) 30 30

(4) 20 20

(5) 20 10(6) 30

(7)

(8)

Tentukan tata letak yang optimal berdasarkan

orientasi proses


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(1) (2) (3) (4) (5)

(1)

(5)

(4)

(3)(2)

(5)(1)

(2) (4)(3)

4 4 4 4

4 4

4

3

14

12

10 8

8 13

11 11

4 4

5 58 8

(1) Refrigerator

(2) Counter

(3) Sink

(4) Storage

(5) Stove


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Soal 4

Sebuah industri memiliki target produksi sebanyak 500

unit perminggu dengan waktu efektif kerja sejumlah 4000

menit per minggu. Pengerjaan produk tersebut dilakukanmelalui proses (tugas) sebagai berikut

Tugas waktu (menit) tugas sebelumnya

A 5 -

B 3 A

C 4 B

D 3 B

E 6 C

F 1 C

G 4 D,E,F

H 2 G

Tentukan waktu siklus, jumlah stasiun kerja, dan efisiensi.

Gambarkan diagram precedence kegiatan tsb


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Soal 5

Sebuah industri perakitan perahu menyediakan waktu

sebanyak 200 menit per harinya untuk kegiatan perakitan

perahu. Permintaan setiap harinya adalah 60 unit perahu.Adapun kegiatan perakitan dapat diuraikan dengan tugas-

tugas sebagai berikut :


A 1 -

B 1 A

C 2 A

D 1 C

E 3 C

F 1 C

G 1 D,E,F

H 2 B

I 1 G,H

Tentukan kondisi keseimbangan lini yang diinginkan

sehingga kegiatan dapat optimal


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Soal 6

Sebuah perusahaan minuman memiliki waktu kerja 5 jam per hari dan

5 hari seminggu, mempunyai target permintaan sebesar 3000 botol

per minggu Pengerjaan produk tersebut dilakukan melalui proses(tugas) sebagai berikut


1 0,1 -

2 0,1 13 0,1 2

4 0,2 2

5 0,1 2

6 0,2 3,4,5

7 0,1 1

8 0,1 79 0,2 7,8

10 0,1 9

11 0,2 6

12 0,2 10,11

13 0,1 12

Kuliah-responsi Versi B TATA LETAK

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