Transcript Step 4 - Industrial Engineering 2011

Facility Design-Week6
Group Technology and Facility
Layout
Anastasia L. Maukar
1
Introduction
2

Group technology was introduced by Frederick Taylor in
1919 as a way to improve productivity.

One of long term benefits of group technology is it helps
implement a manufacturing strategy aimed at greater
automation.
WHAT IS GROUP TECHNOLOGY?
Group technology (GT) is a philosophy that implies the
notion of recognizing and exploiting similarities in three
different ways:
1. By performing like activities together
2. By standardizing similar tasks
3. By efficiently storing and retrieving information about
recurring problems
3
What is Group Technology

4
Group Technology examines products,
parts and assemblies. It then groups
similar items to simplify design,
manufacturing, purchasing and other
business processes.
Benefits of GT and Cellular
Manufacturing (CM)
REDUCTIONS
 Setup time
 Inventory
 Material handling cost
 Direct and indirect labor
cost
5
IMPROVEMENTS
 Quality
 Material Flow
 Machine and operator
Utilization
 Space Utilization
 Employee Morale
Group technology emphasis on part
families based on similarities in
design attributes and manufacturing,
therefore GT contributes to the
integration of CAD and CAM.
6
The Basic Key Features for a Successful Group
Technology Applications:
•Group Layout
•Short Cycle Flow Control
•A Planned Machine Loading
7
Process layout
DM
TM
TM
TM
TM
VMM
8
DM
DM
VMM
BM
BM
Group technology layout
TM
DM
BM
TM
VMM
9
VMM
TM
DM
TM
DM
BM
Sample part-machine processing
indicator matrix
P1
[aij] =
P
P2
a
P3
r
P4
t
P5
P6
10
M
a
c
h
i
n
e
M1
M2
M3
M4
M5
M6
M7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Rearranged part-machine processing
indicator matrix
[aij] =
11
P
a
r
t
P1
P3
P2
P4
P5
P6
M
M1
1
a
M4
1
1
c
M6
1
1
h
M2
i
M3
n
M5
1
1
1
1
1
1
1
1
e
M7
1
1
Rearranged part-machine processing
indicator matrix
P1
[aij] =
P
P3
a
P2
r
P4
t
P5
P6
12
M1
M4
M6
1
1
1
1
1
1
M2
M3
M5
1
1
1
1
1
1
M7
1
1
1
1
Rearranged part-machine processing
indicator matrix
[aij] =
13
P
a
r
t
P1
P3
P2
P4
P5
P6
M
M1
2
3
a
M4
3
1
c
M6
1
2
h
M2
i
M3
n
M5
1
2
4
1
2
1
1
2
e
M7
2
3
Classification and Coding Schemes



14
Hierarchical
Non-hierarchical
Hybrid
Classification and Coding Schemes
1
2
3
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n-1
n
15
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.
.
.
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Classification and Coding Schemes
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Name of
system
TOYODA
MICLASS
TEKLA
BRISCH
Country
Developed
Japan
The Netherlands
Norway
United Kingdom
DCLASS
USA
NITMASH
USSR
OPITZ
West Germany
Characteristics
Ten digt code
Thirty digit code
Twelve digit code
Based on four to six digit primary
code and a number of secondary
digits
Software-based system without
any fixed code structure
A hierarchical code of ten to
fifteen digits and a serial number
Based on a five digit primary code
with a four digit secondary code
Advantages of Classification and
Coding Systems



17
Maximize design efficiency
Maximize process planning efficiency
Simplify scheduling
Clustering Approach





18
Rank order clustering
Bond energy
Row and column masking
Similarity coefficient
Mathematical Programming
Rank Order Clustering Algorithm
Step 1: Assign binary weight BWj = 2m-j to each column j of the
part-machine processing indicator matrix.
Step 2: Determine the decimal equivalent DE of the binary value
of each row i using the formula
m
DEi   2m j aij
j 1
Step 3: Rank the rows in decreasing order of their DE values.
Break ties arbitrarily. Rearrange the rows based on this ranking.
If no rearrangement is necessary, stop; otherwise go to step 4.
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Rank Order Clustering Algorithm
Step 4: For each rearranged row of the matrix, assign binary
weight BWi = 2n-i.
Step 5: Determine the decimal equivalent of the binary value of
each column j using the formula
m
DE j   2n i aij
i 1
Step 6: Rank the columns in decreasing order of their DE
values. Break ties arbitrarily. Rearrange the columns based on
this ranking. If no rearrangement is necessary, stop; otherwise
go to step 1.
20
Rank Order Clustering – Example 1
Binary weight
[aij] =
21
P1
P2
P3
P4
P5
P6
M1
64
1
M2
32
M3
16
1
1
M4
8
1
M5
4
1
M7
1
1
1
1
M6
2
1
1
1
1
1
1
1
Binary value
74
52
10
48
17
37
Rank Order Clustering – Example 1
Binary value
[aij] =
22
P1
P2
P4
P6
P5
P3
M1
32
1
M2
28
M3
26
1
1
1
1
1
M4
33
1
M5
20
M6
33
1
M7
6
1
1
1
1
1
1
1
Binary weight
32
16
8
4
2
1
Rank Order Clustering – Example 1
Binary weight
[aij] =
23
P1
P2
P4
P6
P5
P3
M4
64
1
M6
32
1
M1
16
1
M2
8
M3
4
M5
2
1
1
1
1
1
1
1
1
1
1
M7
1
1
1
Binary value
112
14
12
11
5
96
Rank Order Clustering – Example 1
Binary value
[aij] =
24
P1
P3
P2
P4
P6
P5
M4
48
1
1
M6
48
1
1
M1
32
1
M2
14
M3
12
M5
10
1
1
1
1
1
1
1
M7
3
1
1
Binary weight
32
16
8
4
2
1
ROC Algorithm Solution – Example 1
Binary value
[aij] =
25
P1
P3
P2
P4
P6
P5
M4
48
1
1
M6
48
1
1
M1
32
1
M2
14
M3
12
M5
10
1
1
1
1
1
1
1
M7
3
1
1
Binary weight
32
16
8
4
2
1
Bond Energy Algorithm
Step 1: Set i=1. Arbitrarily select any row and place it.
Step 2: Place each of the remaining n-i rows in each of the i+1 positions
(i.e. above and below the previously placed i rows) and determine the
row bond energy for each placement using the formula
i 1 m
 a  a
ij
i 1, j
 ai 1 , j 
i 1 j 1
Select the row that increases
the bond energy the most and place it in
the corresponding position.
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Bond Energy Algorithm
Step 3: Set i=i+1. If i < n, go to step 2; otherwise go to step 4.
Step 4: Set j=1. Arbitrarily select any column and place it.
Step 5: Place each of the remaining m-j rows in each of the j+1
positions (i.e. to the left and right of the previously placed j columns) and
determine the column bond energy for each placement using the
formula
n
j 1
 a
i 1 j 1
ij
(ai , j 1  ai , j 1 )
Step 6: Set j=j+1. If j < m, go to step 5; otherwise stop.
27
BEA – Example 2
Row
1
2
3
4
28
Column 1
2
3
4
1
0
0
1
0
1
1
0
1
0
0
1
0
1
1
0
BEA – Example 2
29
Row Selected
Where Placed
1
Above Row 2
1
Below Row 2
3
Above Row 2
3
Below Row 2
4
Above Row 2
4
Below Row 2
Row
Arrangement
1010
0101
0101
1010
0101
0101
0101
0101
1010
0101
0101
1010
Row Bond
Energy
0
Maximize
Energy
No
0
No
4
Yes
4
Yes
0
No
0
No
BEA – Example 2
1
1
0
0
30
0
0
1
1
1
1
0
0
0
0
1
1
BEA – Example 2
31
Column
Selected
2
Where Placed
2
Right of Column 1
3
Left of Column 1
3
Right of Column 1
4
Left of Column 1
4
Right of Column 1
Left of Column 1
Column
Arrangement
01
01
10
10
10
10
01
01
11
11
00
00
11
11
00
00
01
01
10
10
10
10
01
01
Column Bond
Energy
0
Maximize
Energy
No
0
No
4
Yes
4
Yes
0
No
0
No
BEA Solution – Example 2
1
1
0
0
32
1
1
0
0
0
0
1
1
0
0
1
1
Row and Column Masking Algorithm
Step 1: Draw a horizontal line through the first row. Select any 1 entry in
the matrix through which there is only one line.
Step 2: If the entry has a horizontal line, go to step 2a. If the entry has a
vertical line, go to step 2b.
Step 2a: Draw a vertical line through the column in which this 1 entry
appears. Go to step 3.
Step 2b: Draw a horizontal line through the row in which this 1 entry
appears. Go to step 3.
Step 3:If there are any 1 entries with only one line through them, select
any one and go to step 2. Repeat until there are no such entries left.
Identify the corresponding machine cell and part family. Go to step 4.
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Step 4: Select any row through which there is no line. If there are no such
rows, STOP. Otherwise draw a horizontal line through this row, select any
1 entry in the matrix through which there is only one line and go to Step 2.
R&CM Algorithm – Example 3
P1
[aij] =
P
P2
a
P3
r
P4
t
P5
P6
34
M
a
c
h
i
n
e
M1
M2
M3
M4
M5
M6
M7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
R&CM Algorithm – Example 3
P1
[aij] =
35
P
P2
a
P3
r
P4
t
P5
P6
M
a
c
h
i
n
e
M1
M2
M3
M4
M5
M6
M7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
R&CM Algorithm - Solution
[aij] =
36
P
a
r
t
P1
P3
P2
P4
P5
P6
M
M1
1
a
M4
1
1
c
M6
1
1
h
M2
i
M3
n
M5
1
1
1
1
1
1
1
1
e
M7
1
1
Similarity Coefficient (SC) Algorithm
n
sij 
a
k 1
a
n
k 1
37
ki
a
ki kj
 akj  aki akj 
,
1 if part k requires machine i
where aki  
0 otherwise
SC Algorithm – Example 4
P1
[aij] =
P
P2
a
P3
r
P4
t
P5
P6
38
M
a
c
h
i
n
e
M1
M2
M3
M4
M5
M6
M7
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
SC Algorithm – Example 4
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Machine Pair
{1,2}
{1,3}
{1,4}
{1,5}
{1,6}
{1,7}
{2,3}
{2,4}
{2,5}
{2,6}
{2,7}
{3,4}
{3,5}
{3,6}
{3,7}
{4,5}
{4,6}
{4,7}
{5,6}
{5,7}
{6,7}
SC Value
0/4=0
0/4=0
1/2
0/3=0
1/2
0/3=0
1/2
0/5=0
2/3
0/5=0
1/4
0/5=0
1/4
0/5=0
1/4
0/4=0
2/2=1
0/4=0
0/4=0
1/3
0/4=0
Combine into one cell?
No
No
No
No
No
No
No
No
Yes
No
No
No
No
No
No
No
Yes
No
No
No
No
SC Algorithm – Example 4
Machine/Cell Pair
{1, (2,5)}
{1, (4,6)}
{1,3}
{1,7}
{(2,5), (4,6)}
{(2,5), 3}
{(2,5), 7}
{(4,6), 3}
{(4,6), 7}
{3,7}
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SC Value
0
1/2
0
0
0
1/2
1/3
0
0
1/4
Combine into one cell?
No
Yes
No
No
No
Yes
No
No
No
No
SC Algorithm – Example 4
Machine/Cell Pair
{(1,4,6) (2,3,5)}
{(1,4,6), 7}
{(2,3,5), 7}
Machine/Cell Pair
{(1,4,6) (2,3,5,7)}
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SC Value
0
0
1/3
SC Value
0
Combine into one cell?
No
No
Yes
Combine into one cell?
No
SC Algorithm Solution – Example 4
42