slide 0 of 96 manufacturing facility layout. slide 1 of 96 basic layout forms process product...

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Manufacturing Facility Layout

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Basic Layout Forms

• Process

• Product

• Cellular

• Fixed position

• Hybrid

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Sistem produksi IntermittenLay out by process

Saw

Mill

Grind

Lathe

Paint

Drill

Ass

emb

ly

War

ehou

se

Sto

res

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Tata Letak Proses (Manufacturing Process Layout)

L

L

L

L

L

L

L

L

L

L M

M

M

M

D

D

D

D

D

D

D

D

G

G

G

G

G

G

P

P

A A AReceiving andShipping

Assembly

Painting Department

Lathe Department

Milling Department

Drilling Department

Grinding Department

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Sistem produksi ContinousLay out by product

Saw

Saw

Grind

Weld

Lathe

Mill

Mill

Grind

Mill

Drill

Drill

Lathe

Drill

Paint

Paint

Drill

Sto

res

Ass

e mb

ly

War

ehou

s e

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Process (Job Shop) Layouts

• Equipment that perform similar processes are grouped together

• Used when the operations system must handle a wide variety of products in relatively small volumes (i.e., flexibility is necessary)

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Characteristics of Process Layouts

• General-purpose equipment is used

• Changeover is rapid

• Material flow is intermittent

• Material handling equipment is flexible

• Operators are highly skilled

• . . . more

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Characteristics of Process Layouts

• Technical supervision is required

• Planning, scheduling and controlling functions are challenging

• Production time is relatively long

• In-process inventory is relatively high

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Product (Assembly Line) Layouts

• Operations are arranged in the sequence required to make the product

• Used when the operations system must handle a narrow variety of products in relatively high volumes

• Operations and personnel are dedicated to producing one or a small number of products

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Characteristics of Product Layouts

• Special-purpose equipment are used

• Changeover is expensive and lengthy

• Material flow approaches continuous

• Material handling equipment is fixed

• Operators need not be as skilled

• . . . more

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Characteristics of Product Layouts

• Little direct supervision is required

• Planning, scheduling and controlling functions are relatively straight-forward

• Production time for a unit is relatively short

• In-process inventory is relatively low

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Product Layout-Advantages/Disadvanta

gesAdvantages:Low cost variable cost

per unitLower material

handling costs reduction in work in-

process inventorieseasier training and

supervision

Disadvantages:High volume required

because of large initial investment

Lack of flexibility in handling variety of products or production rates

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Tata Letak Proses (Manufacturing Process Layout)

L

L

L

L

L

L

L

L

L

L M

M

M

M

D

D

D

D

D

D

D

D

G

G

G

G

G

G

P

P

A A AReceiving andShipping

Assembly

Painting Department

Lathe Department

Milling Department

Drilling Department

Grinding Department

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Part Routing MatrixReordered To Highlight Cells

MACHINESPARTS 1 2 4 8 10 3 6 9 5 7 11 12 A x x x x x D x x x x x F x x x C x x x G x x x x B x x x H x x x E x x x

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Cellular Layout Solution

12

12 3

4

5

6

7

8 910

11

A BCRaw materials

Cell1 Cell 2 Cell 3

Assembly

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Designing and Analyzing a Product Layout

• Line Balancing

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Designing and Analyzing a Product Layout

• Characteristics

• Inputs

• Design Procedure

• How Good Is The Layout?

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Line Balancing Problem• Work stations are arranged so that the output of one

is an input to the next, i.e., a series connection

• Layout design involves assigning one or more of the tasks required to make a product to work stations

• . . . more

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Line Balancing Problem

• The objective is to assign tasks to minimize the workers’ idle time, therefore idle time costs, and meet the required production rate for the line

• In a perfectly balanced line, all workers would complete their assigned tasks at the same time (assuming they start their work simultaneously)

• This would result in no idle time• . . . more

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Inputs

• The production rate required from the product layout or the cycle time.– The cycle time is the reciprocal of the

production rate and visa versa

• All of the tasks required to make the product– It is assumed that these tasks can not be

divided further

• . . . more

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Inputs

• The estimated time to do each task

• The precedence relationships between the tasks– These relationships are determined by the

technical constraints imposed by the product– These relationships are displayed as a

network known as a precedence diagram

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Design Procedure

1. If not provided, find the cycle time for the line. Remember the cycle time is the reciprocal of the production rate. Make sure the cycle time is expressed in the same time units as the estimated task times.

2. Select the line-balancing heuristic that may be used to help with the assignments. (Two heuristics are described at the end of this procedure.)

. . . more

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Design Procedure

3. Open a new work station with the full cycle time remaining.

4. Determine which tasks are feasible, i.e., can be assigned to this work station at this time. For a task to be feasible, two conditions must be met:– All tasks that precede that task must have already

been assigned– The estimated task time must be less than or equal to

the remaining cycle time for that work station.

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Line-Balancing HeuristicsLine-Balancing Heuristics

• Heuristic methods, based on simple rules, have been used to develop very good, not optimal, solutions to line balancing problems.

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How Good Is the Design?

• Utilization is one way of objectively determining how near perfectly balanced an assignment scheme is.

• Utilization is the percentage of time that a production line is working.

• Utilization is calculated as:

or

100 stations) work ofnumber (Actual Time) (Cycle

s task timeall of Sumx

x

100ons workstatiofnumber Actual

ons workstatiofnumber Minimumx

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Station 1

Min/Unit 6

Station 2

7

Station 3

3

Why is Balancing the Line Important?

What’s Going to Happen?

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Example 1: The ALB Problem• You’ve just been assigned the job a setting up an

electric fan assembly line with the following tasks:

Task Time (Mins) Description PredecessorsA 2 Assemble frame NoneB 1 Mount switch AC 3.25 Assemble motor housing NoneD 1.2 Mount motor housing in frame A, CE 0.5 Attach blade DF 1 Assemble and attach safety grill EG 1 Attach cord BH 1.4 Test F, G

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Example 1: The ALB Problem The Precedence Diagram

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

• Which process step defines the maximum rate of production?

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Example 1: The ALB Problem We want to assemble 100 fans per day

Required Cycle Time, C = Production time per period

Required output per period

C = 420 mins / day

100 units / day= 4.2 mins / unit

What do these numbers this represent?

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Example 1: The ALB Problem We want to assemble 100 fans per day

Theoretical Min. Number of Workstations, N

N = Sum of task times (T)

Cycle time (C)

t

t

N = 11.35 mins / unit

4.2 mins / unit= 2.702, or 3t

Why should we always round up?Why should we always round up?

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Example 1: The ALB ProblemSelected Task Selection Rules

• Primary: Assign tasks in order the the largest number of following tasks.

• Secondary (tie-breaking): Assign tasks in order of the longest operating time

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Example 1: The ALB ProblemSelected Task Selection Rules

Precedence Diagram

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

Station 1 Station 2 Station 3

Task Followers Time (Min)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4

Slide 34 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)

Slide 35 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)

Slide 36 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2

Slide 37 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2

C (4.2-3.25)=.95

Idle = .95

Slide 38 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2)=3

Slide 39 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2)=3E (3-.5)=2.5

Slide 40 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2

C (4.2-3.25)=.95

Idle = .95

D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5

Slide 41 of 96

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1



A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

C (4.2-3.25)=.95 D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5H (1.5-1.4)=.1

Idle=.2 Idle=.95 Idle=.1

Slide 42 of 96

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Example 1: The ALB Problem

• Which station is the bottleneck?

• What is the effective cycle time?

Efficiency =Sum of task times (T)

Actual number of workstations (Na) x Cycle time (C)

Efficiency =11.35 mins / unit

(3)(4.2mins / unit)=.901

slide 0 of 96 manufacturing facility layout. slide 1 of 96 basic layout forms process product...

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