Manufacturing Facility Layout
Slide 0 of 96
Basic Layout Forms
•
•
•
•
•
Process
Product
Cellular
Fixed position
Hybrid
Slide 1 of 96
Sistem produksi Intermitten
Lay out by process
Grind
Paint
Stores
Assembly
Lathe
Mill
Warehouse
Saw
Drill
Slide 2 of 96
Tata Letak Proses
(Manufacturing Process Layout)
L
L
M
M
D
D
D
D
L
L
M
M
D
D
D
D
Milling
Department
Drilling Department
L
L
G
G
G
P
L
L
G
G
G
P
L
L
Lathe Department
Grinding
Department
Receiving and
Shipping
Painting Department
Assembly
A
A
A
Slide 3 of 96
Sistem produksi Continous
Lay out by product
Saw
Lathe
Mill
Drill
Saw
Mill
Drill
Paint
Grind
Mill
Drill
Paint
Weld
Grind
Lathe
Drill
Slide 4 of 96
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)
Slide 5 of 96
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
Slide 6 of 96
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
Slide 7 of 96
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
Slide 8 of 96
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
Slide 9 of 96
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
Slide 10 of 96
Product LayoutAdvantages/Disadvantages
Advantages:
 Low cost variable cost
per unit
 Lower material
handling costs
 reduction in work inprocess inventories
 easier training and
supervision
Disadvantages:
 High volume required
because of large
initial investment
 Lack of flexibility in
handling variety of
products or
production rates
Slide 11 of 96
Tata Letak Proses
(Manufacturing Process Layout)
L
L
M
M
D
D
D
D
L
L
M
M
D
D
D
D
Milling
Department
Drilling Department
L
L
G
G
G
P
L
L
G
G
G
P
L
L
Lathe Department
Grinding
Department
Receiving and
Shipping
Painting Department
Assembly
A
A
A
Slide 12 of 96
Part Routing Matrix
Reordered To Highlight Cells
PARTS
A
D
F
C
G
B
H
E
1
x
x
x
2
x
x
4
x
x
x
8
x
x
x
MACHINES
10 3 6 9
x
x
x
x
x
x
5
11 12
x
x
x
x
7
x
x
x
x
x
x
x
x
Slide 13 of 96
Cellular Layout Solution
Assembly
8
10
9
12
11
4
Cell1
Cell 2
6
Cell 3
7
2
Raw materials
1
3
A
C
5
B
Slide 14 of 96
Designing and Analyzing a
Product Layout
• Line Balancing
Slide 15 of 96
Designing and Analyzing a
Product Layout
•
•
•
•
Characteristics
Inputs
Design Procedure
How Good Is The Layout?
Slide 16 of 96
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
Slide 17 of 96
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
Slide 18 of 96
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
Slide 19 of 96
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
Slide 20 of 96
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
Slide 21 of 96
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.
Slide 22 of 96
Line-Balancing Heuristics
• Heuristic methods, based on simple rules,
have been used to develop very good, not
optimal, solutions to line balancing
problems.
Slide 23 of 96
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
linenumber
is working.
Minimum
of workstati ons
x100
Actual
number of workstati
• Utilization
is calculated
as: ons
Sum of all task time s
x 100
(Cycle Time) x (Actual number of work stations)
or
Slide 24 of 96
Why is Balancing the Line
Important?
Min/
Unit
Station 1
Station 2
Station 3
6
7
3
What’s Going to Happen?
Slide 25 of 96
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
A
B
C
D
E
F
G
H
Time (Mins)
2
1
3.25
1.2
0.5
1
1
1.4
Description
Assemble frame
Mount switch
Assemble motor housing
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
Slide 26 of 96
Example 1: The ALB Problem
The Precedence Diagram
• Which process step defines the maximum
rate of production?
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
1.4
H
Slide 27 of 96
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=
= 4.2 mins / unit
100 units / day
What do these numbers this
represent?
Slide 29 of 96
Example 1: The ALB Problem
We want to assemble 100 fans per day
Theoretical Min. Number of Workstations, N t
Sum of task times (T)
Nt =
Cycle time (C)
11.35 mins / unit
Nt =
= 2.702, or 3
4.2 mins / unit
Why should we always round up?
Slide 30 of 96
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
Slide 31 of 96
Example 1: The ALB Problem
Selected Task Selection Rules
Precedence Diagram
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
1.4
H
Slide 32 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
Slide 33 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
A (4.2-2=2.2)
Slide 34 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
A (4.2-2=2.2)
B (2.2-1=1.2)
Slide 35 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
Slide 36 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
Slide 37 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
Slide 38 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5
Slide 39 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
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
Slide 40 of 96
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Min)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
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)=3
E (3-.5)=2.5
F (2.5-1)=1.5
H (1.5-1.4)=.1
Idle=.2
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 =
=.901
(3)(4.2mins / unit)
Slide 42 of 96