Manufacturing Facility Layout
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Basic Layout Forms
• Process
• Product
• Cellular
• Fixed position
• Hybrid
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Sistem produksi Intermitten
Lay out by process
Saw Grind Paint
Mill
Lathe
Drill
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L
L
L
Tata Letak Proses
(Manufacturing Process Layout)
L
L
L
L L
L L
Lathe Department
M M
M M
Milling
Department
G G
G G
Grinding
Department
Receiving and
Shipping
G
G
D
D
D D D
D D D
Drilling Department
P
P
Painting Department
Assembly
A A A
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Sistem produksi Continous
Lay out by product
Saw
Saw
Grind
Weld
Lathe
Mill
Mill
Grind
Mill
Drill
Drill
Lathe
Drill
Paint
Paint
Drill
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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/Disadvantages
Advantages:
Disadvantages:
 Low cost variable cost
 High volume required per unit because of large
 Lower material initial investment handling costs
 Lack of flexibility in
 reduction in work inhandling variety of process inventories products or
 easier training and production rates supervision
Slide 11 of 96

L
L
L
Tata Letak Proses
(Manufacturing Process Layout)
L
L
L
L L
L L
Lathe Department
M M
M M
Milling
Department
G G
G G
Grinding
Department
Receiving and
Shipping
G
G
D
D
D D D
D D D
Drilling Department
P
P
Painting Department
Assembly
A A A
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Part Routing Matrix
Reordered To Highlight Cells
PARTS
A
D
F
C
G
B
H
E
MACHINES
1 2 4 8 10 3 6 9 5 7 11 12 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
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Cellular Layout Solution
4 Cell1
2
8 10
Assembly
9
Raw materials
1
Cell 2
A
6
3
C B
12
Cell 3
11
7
5
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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 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 ons x 100 ons
Sum of all task time s x 100
(Cycle Time) x (Actual number of work stations) or
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Why is Balancing the Line
Important?
Min/
Unit
Station 1
6
Station 2
7
What’s Going to Happen?
Station 3
3
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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
A 2 Assemble frame
B
C
1
3.25
Mount switch
Assemble motor housing
D
E
F
G
H
1.2
0.5
1
1
1.4
Mount motor housing in frame
Attach blade
Assemble and attach safety grill
Attach cord
Test
Predecessors
None
A
None
A, C
D
E
B
F, G
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Example 1: The ALB Problem
The Precedence Diagram
• Which process step defines the maximum rate of production?
2
A
1
B
1
G
1.4
H
C
3.25
D
1.2
E
.5
F
1
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Example 1: The ALB Problem
We want to assemble 100 fans per day
Production time per period
Required Cycle Time, C =
Required output per period
420 mins / day
C =
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 t
N = t
Sum of task times (T)
Cycle time (C)
N = t
11.35 mins / unit
4.2 mins / unit
= 2.702, or 3
Why should we always round up?
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Example 1: The ALB Problem
Selected 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 Problem
Selected Task Selection Rules
Precedence Diagram
2
A
1
B
1
G
1.4
H
C
3.25
D
1.2
E
.5
F
1
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
Station 1
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3 Station 1
A (4.2-2=2.2)
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3 Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3 Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
C (4.2-3.25)=.95
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3
Idle = .95
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
C (4.2-3.25)=.95
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
Idle = .95
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
C (4.2-3.25)=.95
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5
Idle = .95
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
C (4.2-3.25)=.95
Followers Time (Min)
6
4
2
3.25
3
2
2
1
1
0
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5
F (2.5-1)=1.5
Idle = .95
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2
A
C
3.25
1
B
1
G
D
1.2
E
.5
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle=.2
F
1
1.4
H
Station 2
Task
A
C
D
B
E
F
G
H
C (4.2-3.25)=.95
Followers Time (Min)
6 2
4
3
2
3.25
1.2
1
2
1
1
0
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5
F (2.5-1)=1.5
H (1.5-1.4)=.1
Idle=.95
Idle=.1
Slide 41 of 96

Example 1: The ALB Problem
Which station is the bottleneck?
What is the effective cycle time?
Sum of task times (T)
Efficiency =
Actual number of workstations (Na) x Cycle time (C)
11.35 mins / unit
Efficiency =
(3)(4.2mins / unit)
=.
901
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