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

2. 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.

3. 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.

4. Manual visual inspection
Part Family 1
Part Family 2

5. 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)

6. Production flow analysis

7. 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.

8. 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

9. 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!

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

11. Step 3: Part Number B. WT. 1 2 3 4 5 6 Machine ID E 24 C 23 D 22 B 2120
Decimal
equivalent
24+23 = 24
22+21= 6 =7
24+23=24
22+20=5
24=16
Rank
Step 4: Must Reorder

12. 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!

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

14. 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

15. 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

16. 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

17. 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

18. 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:

19. 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!

20. 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

21. 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

22. 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

24. 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

25. 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!

26. Step 3: Part Number B. WT. 6 2 3 4 5 1 Machine ID E 24 D 23 B 22 A 2120
Decimal
equivalent
24 = 16
23+22= 12
=14
24+20= 17
24+21=18
24+20=17
Rank
Step 4: Must Reorder

27. Back at Step 1: Part Number D. Eqv Rank 6 4 1 3 2 5 B. Wt: 25 24 23 2221 20
Machine ID E
=56 D
= 7 B
22+21= 6 A
= 5 C
24+23= 24
Step 2: Must Reorder

28. Back at Step 1: Part Number B. WT. 6 4 1 3 2 5 Machine ID E 24 C 23 D22 B 21 A 20
Decimal
equivalent
24 = 16
24+23= 24
= 7
22+21=6
22+20=5
Rank
Step 2: Must Reorder

29. 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!

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

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

32. 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

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

34. 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

35. 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

36. 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

37. hierarchical code for the spur gear (Fig. 1)

38. 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

39. POLYCODE (ATTRIBUTE CODE):
Polycode for the spur gear (Fig. 1):

40. 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.