Group Technology

GT

Job shop production System

Batch production System

Mass production System

Group Technology

Group technology (GT) is a manufacturing philosophy that seeks to improve productivity by grouping parts and products with similar characteristics into families and forming production cells with a group of dissimilar machines and processes.

The group of similar parts is known as part family and the group of machineries used to process an individual part family is known as machine cell. It is not necessary for each part of a part family to be processed by every machine of corresponding machine cell.

Group technology begun by grouping parts into families, based on their attributes (Geometry, manufacturing process ).

Geometric classification of families is normally based on size and shape, while production process classification is based on the type, sequence, and number of operations. The type of operation is determined by such things as the method of processing, the method of holding the part, the tooling.

There are three methods that can be used to form part families:–Manual visual inspection–Production flow analysis–Classification and coding

Manual visual inspection involves arranging a set of parts into groups by visually inspecting the physical characteristics of the parts.

Group Technology

Manual visual inspection

Part Family 1

Part Family 2

Production flow analysis: Parts that go through common operations are grouped into part families.

The machines used to perform these common operations may be grouped as a cell, consequently this technique can be used in facility layout (factory layout)

Production flow analysis

Rank Order Clustering Algorithm:

Rank Order Clustering Algorithm is a simple algorithm used to form machine-part groups.

Step 1: Assign binary weight and calculate a decimal weight for each row.

Step 2: Rank the rows in order of decreasing decimal weight values.

Step 3: Repeat steps 1 and 2 for each column.

Step 4: Continue preceding steps until there is no change in the position of each element in the row and the column.

Part ‘Number’

Machine ID

1 2 3 4 5 6

A 1 1

B 1 1

C 1 1

D 1 1 1

E 1 1 1

Example #1

Consider a problem of 5 machines and 6 parts. Try to group them by using Rank Order Clustering Algorithm.

Step 1:

Part Numbers Decimalequivalent

Rank

Machine ID

1 2 3 4 5 6

B. Wt:

25 24 23 22 21 20

A 1 1 23+21 = 10 5

B 1 1 24+23 = 24 4

C 1 1 25+22=36 2

D 1 1 124+23+21 =

263

E 1 1 1 25+22+20=37 1

Step 2: Must Reorder!

Step 2:

Part Number

1 2 3 4 5 6

Ma

chine ID

E 1 1 1

C 1 1

D 1 1 1

B 1 1

A 1 1

Step 3:

Part Number

B. WT.

1 2 3 4 5 6

Ma

chine ID

E 24 1 1 1

C 23 1 1

D 22 1 1 1

B 21 1 1

A 20 1 1

Decimalequivalent

24+23 = 24

22+21= 6

22+21+20=7

24+23=24

22+20=5

24=16

Rank 1 5 4 2 6 3

Step 4: Must Reorder

Back at Step 1:

Part Number D. Eqv Rank

1 4 6 3 2 5

B Wt: 25 24 23 22 21 20

E 1 1 1 25+24+ 23=56 1

C 1 1 25+24= 48 2

D 1 1 1 22+21+ 20 = 7 3

B 1 1 22+21=6 4

A 1 1 22+20=5 5

Order stays the same: STOP!

Example #2

Part Number

Ma

chine ID

1 2 3 4 5 6 7

A 1 1 1

B 1 1

C 1 1 1 1

D 1 1 1

E 1 1 1 1

Example #2

Part Number

1 2 3 4 5 6 7 Equivalent decimal value

Rank

Ma

chine ID

Binary wt. 26 25 24 23 22 21 20

A 1 1 1 41 3

B 1 1 20 5

C 1 1 1 1 105 1

D 1 1 1 82 2

E 1 1 1 1 40 4

Step 1: Assign binary weight and calculate a decimal weight for each row

Example #2

Part Number

Ma

chine ID

1 2 3 4 5 6 7

C 1 1 1 1

D 1 1 1

A 1 1 1

E 1 1 1 1

B 1 1

Step 3: Reorder the matrix according to rank

Example #2

Part Number

Ma

chine ID

Binary wt. 1 2 3 4 5 6 7

C 16 1 1 1 1

D 8 1 1 1

A 4 1 1 1

E 2 1 1 1 1

B 1 1 1Equ. Decimal Value

24 20 11 22 3 10 20

Rank 1 3 5 2 7 6 4

Step 4: Assign binary weight and calculate a decimal weight for each Column

Example #2

Part Number

Ma

chine ID

1 4 2 7 3 6 5

C 1 1 1 1

D 1 1 1

A 1 1 1

E 1 1 1 1

B 1 1

Step 5: Reorder the matrix according to rank

Example #2

Part Number

1 4 2 7 3 6 5 Equivalent decimal value

Rank

Ma

chine ID

Binary wt. 26 25 24 23 22 21 20

C 1 1 1 1 120 1

D 1 1 1 70 2

A 1 1 1 56 3

E 1 1 1 1 39 4

B 1 1 5 5

Repeat Step 1&2: Assign binary weight and calculate a decimal weight for each row

Order stays the same:

Example #2

Part Number

Ma

chine ID

Binary wt. 1 4 2 7 3 6 5

C 16 1 1 1 1

D 8 1 1 1

A 4 1 1 1

E 2 1 1 1 1

B 1 1 1Equ. Decimal Value

24 22 20 20 11 10 3

Rank 1 2 3 4 5 6 7

Repeat Step 4 & 5

Order stays the same: STOP!

Example #2

Part Number

Ma

chine ID

1 4 2 7 3 6 5

C 1 1 1 1

D 1 1 1

A 1 1 1

E 1 1 1 1

B 1 1

Part family 1: Part Nos. 1, 4, 2 & 7

Machine Cell 1: C, D & A

Part family 2: Part Nos. 3, 5, and 5Machine Cell 2: E & B

Voids Exceptional parts

No. of exceptional Parts: 3

No. of Voids: 5

No. of bottleneck machines: 2(Machines D & E)

Solutions for overcoming this problem?

• Duplicate machines• Alternate process plans• Subcontract these operations

Duplicate machines

Part Number

Ma

chine ID

1 4 2 7 3 6 5

C 1 1 1 1

D 1 1 1

A 1 1 1

E 1 1 1 1

B 1 1

Part Number

Ma

chine ID

1 4 2 7 3 6 5

C 1 1 1 1

D 1

A 1 1 1

E 1

E 1 1 1

B 1 1

D 1 1

No. of exceptional Parts: 0

No. of Voids: 9

No. of bottleneck machines: 0

No. of duplicate machine: 2(Machines D & E

No. of exceptional Parts: 3

No. of Voids: 5

No. of bottleneck machines: 2(Machines D & E)

Alternate process plans

Part Number

Ma

chine ID

1 4 2 7 3 6 5

C 1 1 1 1

D 1 1 1

A 1 1 1

E 1 1 1 1

B 1 1

Part Number

Ma

chine ID

1 4 2 7 3 6 5

C 1 1 1 1

A 1 1 1

E 1 1 1 1

B 1 1

D 1 1 1

No. of exceptional Parts: 2

No. of Voids: 3

No. of bottleneck machines: 2(Machines D & E

No. of exceptional Parts: 3

No. of Voids: 5

No. of bottleneck machines: 2(Machines D & E)

Step 1:

Part Numbers Decimalequivalent

Rank

Machine ID

1 2 3 4 5 6

B. Wt:

25 24 23 22 21 20

A 1 1 23+21 = 10 5

B 1 1 24+23 = 24 4

C 1 1 25+22=36 2

D 1 1 124+23+21 =

263

E 1 1 1 25+22+20=37 1

Step 1:

Part Numbers Decimalequivalent

Rank

Machine ID

6 2 3 4 5 1

B. Wt:

25 24 23 22 21 20

A 1 1 23+21 = 10 4

B 1 1 24+23 = 24 3

C 1 1 22 +20=5 5

D 1 1 1 24+23+21 = 26 2

E 1 1 1 25+ 22 +20=37 1

Step 2: Must Reorder!

Step 3:

Part Number

B. WT. 6 2 3 4 5 1

Ma

chine ID

E 24 1 1 1 1

D 23 1 1

B 22 1 1

A 21 1 1

C 20 1 1

Decimalequivalent

24 = 1623+22= 12

23+22+21=14

24+20= 17

24+21

=1824+20

=17

Rank 1 5 4 2 6 3

Step 4: Must Reorder

Back at Step 1:

Part Number D. Eqv Rank

6 4 1 3 2 5

B. Wt:

25 24 23 22 21 20

Ma

chine ID

E 1 1 1 25+24+ 23=56 1

D 1 1 1 22+21+ 20 = 7 3

B 1 1 22+21= 6 4

A 1 1 22+ 20 = 5 5

C 1 1 24+23= 24 2

Step 2: Must Reorder

Back at Step 1:

Part Number

B. WT.

6 4 1 3 2 5

Ma

chine ID

E 24 1 1 1

C 23 1 1

D 22 1 1 1

B 21 1 1

A 20 1 1

Decimalequivalent

24 = 16

24+23= 24

24+23= 24

22+21+20= 7

22+21

=622+20

=5

Rank 3 1 2 4 5 6

Step 2: Must Reorder

Part Number D. Eqv Rank

4 1 6 3 2 5

B. Wt:

25 24 23 22 21 20

Ma

chine ID

E 1 1 1 25+24+23=56 1

C 1 1 25+24= 48 2

D 1 1 1 22+21+20 = 7 3

B 1 1 22+21 = 6 4

A 1 1 22+20= 5 5

Order stays the same: STOP!

Clustering Methods

• Using Process Similarity methods:– Create Machine – Part Matrices– Compute machine ‘pair-wise’ similarity Coefficient

comparisons:

Part ‘Number’

Machine ID

1 2 3 4 5 6

A 1 1

B 1 1

C 1 1

D 1 1 1

E 1 1 1

Example:

:

is # of parts (in matrix) visiting

both machines of the pair

is # of parts visiting one but not both machines

ij

jjx

ijij

ij jj

here

x

xS

x x

Computing Similarity Coefficients:

• Total number of coefficient is: • [(N-1)N]/2 = [(5-1)5]/2 = 10

• For 25 machines (typical number in a small Job Shop): 300 S ij’s

Part ‘Number’

Machine ID

1 2 3 4 5 6

A 1 1

B 1 1

C 1 1

D 1 1 1

E 1 1 1

jjij

ijij xx

xS

1.33

1 20

00 42

.672 1

AB

AC

AD

S

S

S

Continuing:

Part ‘Number’

Machine ID

1 2 3 4 5 6

A 1 1

B 1 1

C 1 1

D 1 1 1

E 1 1 1

jjij

ijij xx

xS

Grouping of parts using Clustering Methods

Similarity coefficients of machine pairs

Machinepairs

AB AC AD AE BC BD BE CD CE DE

SC .33 0 0.67 0 0 0.67 0 0 .67 0

A D C E

0.67

0.33

0Dendrogram

B

• Coding refers to the process of assigning symbols to the parts

• The symbols represent design attributes of parts or manufacturing features of part families

• Although well over 100 classification and coding systems have been developed for group technology applications, all of them can be grouped into three basic types:

– Monocode or hierarchical code– Polycode or attribute– Hybrid or mixed code

Classification and Coding

MONOCODE (HIERARCHICAL CODE)

• This coding system was originally developed for biological classification in 18th century.

• In this type of code, the meaning of each character is dependent on the meaning of the previous character; that is, each character amplifies the information of the previous character. Such a coding system can be depicted using a tree structure

Monocode of Fig. 1: A-1-1-B-2Fig. 1 Spur gear

hierarchical code for the spur gear (Fig. 1)

POLYCODE (ATTRIBUTE CODE):

• The code symbols are independent of each other

• Each digit in specific location of the code describes a unique property of the workpiece– it is easy to learn and useful in manufacturing

situations where the manufacturing process have to be described

– the length of a polycode may become excessive because of its unlimited combinational features

POLYCODE (ATTRIBUTE CODE):

Polycode for the spur gear (Fig. 1): 22213

MIXED CODE (HYBRID CODE):

• In reality, most coding systems use a hybrid (mixed) code so that

the advantages of each type of system can be utilized. The first digit

for example, might be used to denote the type of part, such as gear.

The next five position might be reserved for a short attribute code

that would describe the attribute of the gear. The next digit (7th digit)

might be used to designate another subgroup, such as material,

followed by another attribute code that would describe the attributes.