# Group Technology GT Job shop production System Batch production System

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Group Technology GT Job shop production System Batch production System
Mass production System GT

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 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.

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.

Example #1 Part ‘Number’ Machine ID 1 2 3 4 5 6 A B C D E
Consider a problem of 5 machines and 6 parts. Try to group them by using Rank Order Clustering Algorithm. Part ‘Number’ Machine ID 1 2 3 4 5 6 A B C D E

Step 1: Part Numbers Machine ID 1 2 3 4 5 6 B. Wt: 25 24 23 22 21 20 A
Decimal equivalent Rank Machine ID 1 2 3 4 5 6 B. Wt: 25 24 23 22 21 20 A 23+21 = 10 B 24+23 = 24 C 25+22=36 D = 26 E =37 Step 2: Must Reorder!

Step 2: Part Number 1 2 3 4 5 6 Machine ID E C D B A

Step 3: Part Number B. WT. 1 2 3 4 5 6 Machine ID E 24 C 23 D 22 B 21
20 Decimal equivalent 24+23 = 24 22+21= 6 =7 24+23=24 22+20=5 24=16 Rank Step 4: Must Reorder

Order stays the same: STOP!
Back at Step 1: Part Number D. Eqv Rank 1 4 6 3 2 5 B Wt: 25 24 23 22 21 20 E =56 C 25+24= 48 D = 7 B 22+21=6 A 22+20=5 Order stays the same: STOP!

Example #2 Part Number Machine ID 1 2 3 4 5 6 7 A B C D E

Example #2 Part Number 1 2 3 4 5 6 7 A 41 B C 105 D 82 E 40
Step 1: Assign binary weight and calculate a decimal weight for each row Part Number 1 2 3 4 5 6 7 Equivalent decimal value Rank Machine ID Binary wt. 26 25 24 23 22 21 20 A 41 B C 105 D 82 E 40

Example #2 Part Number 1 2 3 4 5 6 7 C D A E B
Step 3: Reorder the matrix according to rank Part Number Machine ID 1 2 3 4 5 6 7 C D A E B

Example #2 Part Number 1 2 3 4 5 6 7 C 16 D 8 A E B 24 20 11 22 10
Step 4: Assign binary weight and calculate a decimal weight for each Column Part Number Machine ID Binary wt. 1 2 3 4 5 6 7 C 16 D 8 A E B Equ. Decimal Value 24 20 11 22 10 Rank

Example #2 Part Number 1 4 2 7 3 6 5 C D A E B
Step 5: Reorder the matrix according to rank Part Number Machine ID 1 4 2 7 3 6 5 C D A E B

Example #2 Part Number 1 4 2 7 3 6 5 C 120 D 70 A 56 E 39 B
Repeat Step 1&2: Assign binary weight and calculate a decimal weight for each row Part Number 1 4 2 7 3 6 5 Equivalent decimal value Rank Machine ID Binary wt. 26 25 24 23 22 21 20 C 120 D 70 A 56 E 39 B Order stays the same:

Order stays the same: STOP!
Example #2 Repeat Step 4 & 5 Part Number Machine ID Binary wt. 1 4 2 7 3 6 5 C 16 D 8 A E B Equ. Decimal Value 24 22 20 11 10 Rank Order stays the same: STOP!

Example #2 Part Number Machine ID 1 4 2 7 3 6 5 C D A E B Voids
Exceptional parts Part Number Machine ID 1 4 2 7 3 6 5 C D A E B Part family 1: Part Nos. 1, 4, 2 & 7 Machine Cell 1: C, D & A Part family 2: Part Nos. 3, 5, and 5 Machine Cell 2: E & B 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 Machine ID 1 4 2 7 3 6 5 C D A E B
No. of exceptional Parts: 3 No. of Voids: 5 No. of bottleneck machines: 2(Machines D & E) No. of exceptional Parts: 0 No. of Voids: 9 No. of bottleneck machines: 0 No. of duplicate machine: 2(Machines D & E

Alternate process plans
Part Number Machine ID 1 4 2 7 3 6 5 C A E B D Part Number Machine ID 1 4 2 7 3 6 5 C D A E B No. of exceptional Parts: 3 No. of Voids: 5 No. of bottleneck machines: 2(Machines D & E) No. of exceptional Parts: 2 No. of Voids: 3 No. of bottleneck machines: 2(Machines D & E

Step 1: Part Numbers Machine ID 1 2 3 4 5 6 B. Wt: 25 24 23 22 21 20 A
Decimal equivalent Rank Machine ID 1 2 3 4 5 6 B. Wt: 25 24 23 22 21 20 A 23+21 = 10 B 24+23 = 24 C 25+22=36 D = 26 E =37

Step 1: Part Numbers Machine ID 6 2 3 4 5 1 B. Wt: 25 24 23 22 21 20 A
Decimal equivalent Rank Machine ID 6 2 3 4 5 1 B. Wt: 25 24 23 22 21 20 A 23+21 = 10 B 24+23 = 24 C 22 +20=5 D = 26 E =37 Step 2: Must Reorder!

Step 3: Part Number B. WT. 6 2 3 4 5 1 Machine ID E 24 D 23 B 22 A 21
20 Decimal equivalent 24 = 16 23+22= 12 =14 24+20= 17 24+21=18 24+20=17 Rank 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 Machine ID E =56 D = 7 B 22+21= 6 A = 5 C 24+23= 24 Step 2: Must Reorder

Back at Step 1: Part Number B. WT. 6 4 1 3 2 5 Machine ID E 24 C 23 D
22 B 21 A 20 Decimal equivalent 24 = 16 24+23= 24 = 7 22+21=6 22+20=5 Rank Step 2: Must Reorder

Order stays the same: STOP!
Part Number D. Eqv Rank 4 1 6 3 2 5 B. Wt: 25 24 23 22 21 20 Machine ID E =56 C 25+24= 48 D = 7 B 22+21 = 6 A 22+20= 5 Order stays the same: STOP!

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

Example: Part ‘Number’ Machine ID 1 2 3 4 5 6 A B C D E

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 Sij’s Part ‘Number’ Machine ID 1 2 3 4 5 6 A B C D E

Continuing: Part ‘Number’ Machine ID 1 2 3 4 5 6 A B C D E

Grouping of parts using Clustering Methods
Similarity coefficients of machine pairs Machine pairs AB AC AD AE BC BD BE CD CE DE SC .33 0.67 .67 A D B C E 0.67 0.33 Dendrogram

Classification and Coding
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

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-2 Fig. 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):

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.

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