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Departemen Teknik Industri FTI-ITB

TI-3003 Perencanaan dan Pengendalian Produksi

Keseimbangan Lintasan Perakitan(Assembly Line Balancing – Chapter 8.10)

Laboratorium Sistem Produksiwww.lspitb.org

©2013


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Hasil Pembelajaran

• Umum

� Mahasiswa mampu menerapkan model matematik, heuristik dan teknik statistik untuk menganalisis dan merancang suatu sistem perencanaan dan pengendalian produksi

• Khusus

� Memahami konsep keseimbangan lintasan


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Various Objects

• Part: is composed of a single material and is an individual part of the product

• Component: ranges from a part to a combination of parts which are included in the product

• Building block: a composite part of the product which because of assembly requirements represents a sub-assembly

• Base component: a (larger) component onto which others are assembled

• Formless material: e.g. viscose components such as glue, paint, liquids

• Sub-assembly: one component is assembled with another component, a base component or building block

• Final assembly: describes the construction of a building block or the finished product


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Assembly Example


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Exploded Assembly


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The assembly of a joint


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The concept of sub and final assembly


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Step by step assembly


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An integrated approach


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Fabrication/Assembly Line


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Fabrication/Assembly Line

• A production/fabrication line builds components on a series of machines

• An assembly line puts the fabricated parts together at a series of workstations

• Both are repetitive processes and in both cases, the line must be balanced

• Fabrication lines tend to be machine-paced and require mechanical and engineering changes to facilitate balancing

• Assembly lines tend to be paced by work tasks assigned to individuals or workstations

• Assembly lines therefore can be balanced by moving task form one individual or workstation to another. In this manner, the amount of time required by each individual or workstation is equalized


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Assembly Line Balancing

• Assembly line is a production line in which material moves continuously at a uniform average rate through a sequence of workstations where assembly work is performed

• Assembly accounts for between 40-60% of the total production time

• Down time of an assembly line costs a major car manufacturer $98,000 per minute

• Line balancing is usually done to minimize imbalance between individual/machine/workstation while meeting a required output from the line

• Problems in assembly lines: balancing the workstations and keeping the assembly line in continuous production


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Types of assembly line balancing problems

• Simple assembly line balancing type I (SALB-I): is to determine the minimal number of workstations necessary to maintain the production rate (1/CT) while observing the precedence constraints

• Simple assembly line balancing type II (SALB-II): is to assign task to a fixed number of workstations to maximize the production rate while observing the precedence constraints

• SALB-I is a more common problem than SALB-II


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Definitions(1)

• Assembled product: the product that passes through a sequence of workstations where tasks are performed on the product until it is completed at the final workstation. The throughput of the assembly line is measured by the number of assembled products per unit time

• Work element: a part of the total work content in an assembly process.

• N: The total number of work elements required to complete the assembly

• i: the work element number in the process (1≤i≤ N)

• Workstation (WS): a location on the assembly line where a work element or elements are performed on the product.

• K denotes the minimum number of workstations, K≥1


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Definitions(2)

• Cycle time (CT): the time between the completion of 2 successive assemblies, assumed constant for all assemblies for a given conveyor speed. Conveyors are the key material movers in most assembly lines: belt, chain, overhead, pneumatic, and screw conveyors

T= production time available per dayd= demand per day or production per day

• Station time (ST): the sum of the times of work elements that are performed at the same workstation. ST≤CT

• Delay/idle time of a workstation: the difference between the cycle time (CT) and the station time (ST). D=CT-ST

d

TCT =


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Definitions(3)

• The number of work statations

• Precedence diagram: a diagram that describes the ordering in which work elements should be performed. It shows that some jobs cannot be performed unless their predecessors are completed. The layout of workstations along the assembly line depends on the precedence diagram

CT

tK

m

ii∑

== 1m= the number of elementsti= the time for element i


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Definitions(4)

• Perfect balance means to combine the elements of work to be done in such a manner that at each station the sum of the elemental times just equals the cycle time (D=CT-ST=0)

• Line efficiency (LE): the ratio of total station time to the cycle time multiplied by the number of workstations

STi =the time for station i

• Smoothness index (SI): an index to indicate the relative smoothness of a given assembly line balance. A smoothness index of zero indicates a perfect balance

%100))((

1 xCTK

STLE

K

ii∑

==


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Definitions(5)

• SI is expressed as

• Restrictions in designing an assembly line:

� Precedence relationship

� The number of workstations cannot be greater than the number of work elements (operations). The minimum number of workstations is 1. (1≤i ≤ N)

� The cycle time is greater than or equal to the maximum time of any station time and of the time of any work element ti. The station time should not be exceed the cycle time. ti≤STi≤CT

2

1max )(∑

=

−=K

iiSTSTSI


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Contoh 1

• Diketahui precedence diagram berikut:

1

4 5

2 3

6 9

7 8

12

1110

5 1

2

63

4

5

3

44

6

7


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Kilbridge-Weston Heuristic(1)

1. Gambarkan precedence diagram. Bagi elemen-elemen kerja dalam diagram tersebut ke dalam kolom-kolom. Kolom I adalah elemen-elemen kerja yang tidak memiliki elemen kerja pendahulu (predecessor). Kolom II adalah elemen-elemen kerja dengan elemen kerja pendahulu di Kolom I. Kolom III adalah elemen-elemen kerja dengan elemen kerja pendahulu di Kolom II, dan seterusnya.


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1

4 5

2 3

6 9

7 8

12

1110

5 1

2

63

4

5

3

44

6

7

I II III IV V VI VII


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2. Tentukan waktu siklus (CT) dari bilangan prima waktu total elemen kerja , dan tentukan jumlah stasiun kerja

= 50

Bilangan prima untuk 50 adalah 2 x 5 x 5, sehingga alternatif waktu siklus adalah: 2, 5, 10, 25 dan 50.

Alternatif waktu siklus mana yang tidak mungkin?

∑=

m

iit

1

CT

tK

m

ii∑

== 1 ∑=

m

iit

1

CT=2 dan CT=5

Karena syarat: 7≤CT ≤ 50

Kenapa?

Bila dipilih CT=10, maka jumlah stasiun kerja minimum adalah 50/10 = 5


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3. Tempatkan elemen-elemen kerja ke stasiun kerja sedemikian sehingga total waktu elemen kerja tidak melebihi waktu siklus. Hapus elemen kerja yang sudah ditempatkan dari daftar elemen kerja

4. Bila penempatan suatu elemen kerja mengakibatkan total waktu elemen kerja melebihi waktu siklus maka elemen kerja tersebut ditempatkan di stasiun kerja berikutnya

5. Ulangi Langkah 3 dan 4 sampai seluruh elemen kerja ditempatkan


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Salah satu solusi feasible untuk Contoh 1 adalah:

Work Station Elemen Waktu stasiun (ST)

I 1 5

II 2 dan 4 6

III 3 dan 5 10

IV 6 5

V 7, 9, dan 10 7

VI 8 dan 11 10

VII 12 7


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Untuk menjalankan Langkah 3 pada Metoda Kilbridge-Weston, hitung jumlah elemen kerja pendahulu untuk setiap elemen. Elemen dengan jumlah pendahulu terkecil ditempatkan terlebih dahulu.

Elemen Jumlah Elemen Jumlah

1 0 7 6

2 1 8 7

3 2 9 6

4 1 10 6

5 2 11 7

6 5 12 11

Tempatkan Elemen 1 di Stasiun 1, kemudian Elemen 2 atau 4. Bila dipilih Elemen 2 maka jumlah total waktu elemen adalah 8. Elemen 4 tidak bisa ditempatkan ke Statsiun 1 karena akan menyebabkan total waktu elemen 11, yang melebihi CT


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Elemen 4 kemudian ditempatkan di Stasiun 2. Lanjutkan langkah ini untuk elemen dengan jumlah pendahulu terkecil berikutnya sampai seluruh elemen ditempatkan.

Stasiun Elemen ST CT-ST

I 1 dan 2 8 2

II 4 dan 5 9 1

III 3 dan 6 9 1

IV 7, 9 dan 10 7 3

V 8 dan 11 10 0

VI 12 7 3

LE= 50/(6x10)

= 83,3%

SI=

=4,89

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Dengan coba-coba dapat diperoleh alternatif berikut:


I 1 dan 2 8 1

II 4 dan 5 9 0

III 3 dan 6 9 0

IV 7 dan 8 8 1

V 10 dan 11 8 1

VI 9 dan 12 8 1

LE= 50/(6x9) = 92,6%

SI = =24


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Metoda Helgeson-Birnie(1)

• Metoda ini disebut juga Teknik Bobot Posisi.

• Gambarkan precedence diagram.

1. Hitung bobot posisi setiap elemen kerja. Bobot posisi suatu elemen adalah jumlah waktu elemen-elemen pada rantai terpanjang mulai elemen tersebut sampai elemen terakhir

2. Urut elemen-elemen menurut bobot posisi dari besar ke kecil

3. Tempatkan elemen kerja dengan bobot terbesar pada stasiun kerja sepanjang tidak melanggar hubungan precedence dan waktu stasiun tidak melebihi waktu siklus

4. Ulangi Langkah 3 sampai seluruh elemen ditempatkan


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• Contoh: perhitungan bobot Elemen 6 adalah:

Max{(5+2+6+7), (5+1+7), (5+4+4+7)}= 20


1 34 7 15

2 27 8 13

3 24 9 8

4 29 10 15

5 25 11 11

6 20 12 7


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• Hasil pengurutan menurut bobot posisi dari besar ke kecil


1 34 7 15

4 29 10 15

2 27 8 13

5 25 11 11

3 24 9 8

6 20 12 7


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Solusi dari metoda ini


I 1 dan 4 8 2

II 2 dan 5 9 1

III 3 dan 6 9 1

IV 7 dan 10 6 4

V 8 dan 11 10 0

VI 9 dan 12 8 2

LE= 50/(6x10) = 83,3%

SI = =5,0926


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Mathematical model(1)


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