Taguchi's Designs Ver1.12 Print -...

Part 2

Static response designsThe Process

X1 Xn

(Reduce variance in controllable factors)

Control parameters (Xs)

N1 Nn

(Minimise effect of variance of noise factors)

Noise factors

Y

Response

Example [Taguchi’s Design]

To understand the step by step approach to create the design using Linear Graph, let us take an example:

Case I [Factors with 2 levels]:

o Silicon microprocessors are made on a die and cut as per specifications. Lets say, we have to improve the micro-processor yield [Output response].

o In manufacturing, the critical control factors include the temperature (A), the deposition rate (B) of silicon layers, pressure (C) and doping amount (D)

o We have to arrive at a robust design for this manufacturing process.

o For this design, the main effects of the factors A, B, C, D and the interaction effects A x B and B x C need to be studied. Assume that all the factors are taken at 2 levels.

o The Factor Levels are:

o This is an example of Static Response design as the output response has to be optimised at a fixed target

©Harish Haridasan6

Factors Level 1 Level 2

Temperature [A] 1200 C 2500 C

Deposition rate of silicon layers [B] 0.1 mg/s 0.2 mg/s

Pressure [C] 0.15 bar 0.45 bar

Doping amount [D] 5% 10%


Systems Design

Step 1:

o First objective is to determine the number of experimental runs required

o Taguchi’s designs use the Degrees of Freedom [DoF] method to determine the number of experimental runs

o DoF refers to the number of ways a system can independently vary when a constraint is imposed

o While conducting an experiment, every Factor is at specific Levels. Lets say a Factor has 2 Levels. These levels are chosen as per the process tolerance. Hence, the freedom of choice, as per this tolerance, will be limited to only 1 as the 2nd level must match the tolerance

o Hence, the Degrees of Freedom for a Factor is given by: [No. of Levels – 1]. Total DoF is summation of DoF for all Factors & Interactions

o In this case, the DoF would be:

©Harish Haridasan7

Factors Levels (S) Degrees of Freedom (Df = S – 1)

Temperature [A] 2 1

Deposition rate of silicon layers [B] 2 1

Pressure [C] 2 1

Doping amount [D] 2 1

A x B [Temperature – Deposition rate] (2-1) x (2-1) = 1 x 1 = 1

B x C [Deposition rate - Pressure] (2-1) x (2-1) = 1 x 1 = 1

Total Degrees of Freedom ∑ [𝑆$−1]$)*+ 6


Step 2:

o Total number of Experimental runs, TExperiments = 1 + ∑ [𝑆$−1]$)*+ , where k= No. of Factors, S represents Levels

= 1 + (6) = 7

Step 3:

o We select the suitable Orthogonal Array

o It should be ≥Number of Experiments [ 7 ]

o Referring to the Standard OA Table, the nearest OA is L.

©Harish Haridasan8

Array Experimental runs

Max. # of Factors

Max. # of Factors that can be considered at various Factor Levels

Level 2 Level 3 Level 4 Level 5

L/ 4 3 3

L. 8 7 7

L0 9 4 4

L+1 12 11 11


Step 4:

o We visualise the “Required Linear Graph” (RLG)

o Since, we have to study the main effects for factors A, B, C, D and the interaction effects A x B and B x C, the RLG will look as,

©Harish Haridasan9

A

B

A x B

CB x C D

Nodes: A,B,C,D || Edges: AB,BC


Step 5:

o The “Standard Linear Graph” (SLG) to be used in this case, for OA

L. is as shown,

©Harish Haridasan10

1

2

3

46

5

7

Step 6:

o We will modify this “Standard Linear Graph” (SLG) to match it

with the “Required Linear Graph”

1

2

3

46

5

7 5

A

B CB x C D

A x B


Step 7:

o In preparation to determine the experimental layout, we need to allocate the Factors to the Columns of OA

o This can be done as shown:


Factors To be assigned to Column/ Node

A 1

B 2

C 4

D 7

A x B 3

B X C 6

1/ A

2/ B

3/ A x B

4/ C6/ B x C 7/ D 5


Step 8:

o We choose the experimental layout referring to

the default OA design layout

The default layout can be depicted as à

o Based on our factor allotment of Aà1, Bà2,

Cà4, Dà7, we choose the 1st, 2nd, 4th and 7th

columns from the default array

o We leave out Column 3 & 6 as they represent

the confounded interaction of Col-1,2 and Col-

2,4 respectively [Only then can we study the

interactions independently]

o Hence, the final design layout shall be:


Runs Column-1 Column-2 Column-3 Column-4 Column-5 Column-6 Column-7

1 1 1 1 1 1 1 1

2 1 1 1 2 2 2 2

3 1 2 2 1 1 2 2

4 1 2 2 2 2 1 1

5 2 1 2 1 2 1 2

6 2 1 2 2 1 2 1

7 2 2 1 1 2 2 1

8 2 2 1 2 1 1 2

Runs 1/A 2/B 4/C 7/D

1 1 1 1 1

2 1 1 2 2

3 1 2 1 2

4 1 2 2 1

5 2 1 1 2

6 2 1 2 1

7 2 2 1 1

8 2 2 2 2


Case II [Factors with 3 levels]:

o Lets consider Factors A, B, C, D, E, F

o All the factors are taken at 3 levels

o For this design, the main effects of all the factors and the interaction effects A x B and B x C need to be studied

Step 1: Determine the Degrees of Freedom-



[A] 3 2

[B] 3 2

[C] 3 2

[D] 3 2

[E] 3 2

[F] 3 2

A x B (3-1) x (3-1) = 2 x 2 = 4

B x C (3-1) x (3-1) = 2 x 2 = 4



Max. # of Factors



L12 25 6 6

L13 27 13 1 13

L41 32 31 31

L41 32 10 1 9


Step 2:

o Total number of Experimental runs, TExperiments = 1 + ∑ [𝑆$−1]$)*+ , would be 1 + (20) = 21

Step 3:

o We select the suitable Orthogonal Array

o It should be ≥Number of Experiments [ 21 ]

o Referring to the Standard OA Table, the nearest OA is L13



Step 4:


o Since, we have to study the main factors A, B, C, D, E, F and the interaction effects A x B and B x C, the RLG will look as,


A

B

A x B

CB x C D

Nodes: A,B,C,D,E,F || Edges: AB,BC

E F


Step 5:


L13 is as shown,


Step 6:



1

2

3,4

5

8,11

6,

9 6

1

2

3,4

5

8,11

6,7

9 10 12 13 710 12 13

7


Step 7:





A 1

B 2

C 5

D 9

E 10

F 12

A x B 3,4

B X C 8,11

1/ A

2/ B

3,4/ A x B

5/ C8,11/ B x C

9/ D

6 710/ E

12/ F

13

RunsColumns

1 2 3 4 5 6 7 8 9 10 11 12 131 1 1 1 1 1 1 1 1 1 1 1 1 12 1 1 1 1 2 2 2 2 2 2 2 2 23 1 1 1 1 3 3 3 3 3 3 3 3 34 1 2 2 2 1 1 1 2 2 2 3 3 35 1 2 2 2 2 2 2 3 3 3 1 1 16 1 2 2 2 3 3 3 1 1 1 2 2 27 1 3 3 3 1 1 1 3 3 3 2 2 28 1 3 3 3 2 2 2 1 1 1 3 3 39 1 3 3 3 3 3 3 2 2 2 1 1 110 2 1 2 3 1 2 3 1 2 3 1 2 311 2 1 2 3 2 3 1 2 3 1 2 3 112 2 1 2 3 3 1 2 3 1 2 3 1 213 2 2 3 1 1 2 3 2 3 1 3 1 214 2 2 3 1 2 3 1 3 1 2 1 2 315 2 2 3 1 3 1 2 1 2 3 2 3 116 2 3 1 2 1 2 3 3 1 2 2 3 117 2 3 1 2 2 3 1 1 2 3 3 1 218 2 3 1 2 3 1 2 2 3 1 1 2 319 3 1 3 2 1 3 2 1 3 2 1 3 220 3 1 3 2 2 1 3 2 1 3 2 1 321 3 1 3 2 3 2 1 3 2 1 3 2 122 3 2 1 3 1 3 2 2 1 3 3 2 123 3 2 1 3 2 1 3 3 2 1 1 3 224 3 2 1 3 3 2 1 1 3 2 2 1 325 3 3 2 1 1 3 2 3 2 1 2 1 326 3 3 2 1 2 1 3 1 3 2 3 2 127 3 3 2 1 3 2 1 2 1 3 1 3 2



Step 8:


the default OA design layout


o Based on our factor allotment of Aà1, Bà2,

Cà5, Dà9, Eà10, Fà12, we choose the 1st,

2nd, 5th, 9th , 10th and 12th columns from the

default array

o We leave out Column 3,4 & 8,11 as they

represent the confounded interaction of Col-1,2

and Col-2,5 respectively [Only then can we

study the interactions independently]

A B D E FC

Mixed Designs

o When the Factors have a linear relationship, 2-Level designs are used

o When the Factors follow a non-linear relation, designs with more than 2-Levels would be required for apt representation & analysis

o Taguchi designs are used when we need to experiment with many factors (like, >5). Hence, chances are that some factors would have 2-levels and

few may have more than 2-levels

o Such a scenario calls for a Mixed design where following approaches may be applied:

§ “Dummy Level Technique” or “Pseudo Level Technique”: Lower level Factors can be used in the design layout for Higher level Factors

§ “Combination Design”: Lower level Factors [which do not interact] are to be used in the design layout for Higher level Factors

§ “Collapsing Technique”: Higher level Factors can be used in the design layout for lower level Factors

o Such a modification will still keep the arrays orthogonal but the design will seize to be balanced due to the level modification



Case III [Factors with mixed levels]:

o Lets consider 4 Factors A, B, C, D where A, B has 2 levels and C, D has 3 levels

o For this design, the main effects of all the factors and the interaction effects A x B need to be studied

o In such a case, we can use the “Dummy Level technique”

Step 1: Determine the Degrees of Freedom-



[A] 2 1

[B] 2 1

[C] 3 2

[D] 3 2

A x B (2-1) x (2-1) = 1 x 1 = 1



Max. # of Factors



L/ 4 3 3

L. 8 7 7

L0 9 4 4

L+1 12 11 11

L+5 16 15 15

L+5 16 5 5

L+. 18 8 1 7

L12 25 6 6

L13 27 13 1 13

L41 32 31 31

L41 32 10 1 9

L45 36 23 11 12

L45 36 16 3 13

L26 50 12 1 11

L2/ 54 26 1 25

L5/ 64 63 63

L5/ 64 21 21

L.+ 81 40 40


Step 2:

o Total number of Experimental runs, TExperiments = 1 + ∑ [𝑆$−1]$)*+ ,

would be 1 + (7) = 8

Step 3:

o The suitable Orthogonal Array should be ≥Number of

Experiments [ 8 ]

o Referring to the Standard OA Table,

§ the relevant OA which can be used to study Dummy level technique

will follow 3n series

§ L0 can handle 4 factors but in our case we wish to study an interaction beyond these 4 factors

§ Hence ,the nearest 3n series OA would be L13



Step 4:


o Since, we have to study the main factors A, B, C, D and the interaction effects A x B, the RLG will look as,


A

B

A x B

C D

Nodes: A,B,C,D || Edges: AB


Step 5:


L13 is as shown,


Step 6:



6

1

2

3,4

5

8,11

6,7

9 10 12 13 7 8 11

1

2

3,4

58,

6,

9 10 12 13

7

11


Step 7:





A 1

B 2

C 5

D 6

A x B 3,4

6/D 7 8 11

1/A

2/B

3,4/A x B

5/C 9 10 12 13

RunsColumns

1 2 3 4 5 6 7 8 9 10 11 12 131 1 1 1 1 1 1 1 1 1 1 1 1 12 1 1 1 1 2 2 2 2 2 2 2 2 23 1 1 1 1 3 3 3 3 3 3 3 3 34 1 2 2 2 1 1 1 2 2 2 3 3 35 1 2 2 2 2 2 2 3 3 3 1 1 16 1 2 2 2 3 3 3 1 1 1 2 2 27 1 3 3 3 1 1 1 3 3 3 2 2 28 1 3 3 3 2 2 2 1 1 1 3 3 39 1 3 3 3 3 3 3 2 2 2 1 1 110 2 1 2 3 1 2 3 1 2 3 1 2 311 2 1 2 3 2 3 1 2 3 1 2 3 112 2 1 2 3 3 1 2 3 1 2 3 1 213 2 2 3 1 1 2 3 2 3 1 3 1 214 2 2 3 1 2 3 1 3 1 2 1 2 315 2 2 3 1 3 1 2 1 2 3 2 3 116 2 3 1 2 1 2 3 3 1 2 2 3 117 2 3 1 2 2 3 1 1 2 3 3 1 218 2 3 1 2 3 1 2 2 3 1 1 2 319 3 1 3 2 1 3 2 1 3 2 1 3 220 3 1 3 2 2 1 3 2 1 3 2 1 321 3 1 3 2 3 2 1 3 2 1 3 2 122 3 2 1 3 1 3 2 2 1 3 3 2 123 3 2 1 3 2 1 3 3 2 1 1 3 224 3 2 1 3 3 2 1 1 3 2 2 1 325 3 3 2 1 1 3 2 3 2 1 2 1 326 3 3 2 1 2 1 3 1 3 2 3 2 127 3 3 2 1 3 2 1 2 1 3 1 3 2



Step 8:

o We choose the experimental layout referring to the default OA design layout


o Based on our factor allotment of Aà1, Bà2, Cà5, Dà6, we choose the 1st, 2nd, 5th, 6th columns from the default array

o We leave out Column 3,4 as they represent the confounded interaction of Col-1,2 [Only then can we study the interactions independently]

o Since Factor A,B have only 2 levels, we introduce a dummy level to replace Level 3 values in the design

o Dummy level, if we use 1’, it means, we repeat the experiment at Level 1 for the Factor, Level 1 being more representative

A B DC

⑊1’⑊1’⑊1’⑊1’⑊1’⑊1’⑊1’⑊1’⑊1’

⑊2’⑊2’⑊2’

⑊2’⑊2’⑊2’

⑊2’⑊2’⑊2’


Case IV [Factors with mixed levels]:

o Lets consider 4 Factors A, B, C, D where A, B has 2 levels and C, D has 3 levels

o For this design, only the main effects of all the factors need to be studied [No interaction exists]

o In such a case, it’s easier to design with the “Combination design technique or Compounding Factors technique”

o In this technique, we combine the 2 level Factors [A, B] to form a new 3 level Factor -


Original Factor [A] Original Factor [B] New Factor [AB]

A1 B1 A1B1 = X1

A2 B2 A2B1 = X2

A1B2 = X3




[A] 2 1

[B] 2 1

[C] 3 2

[D] 3 2

Total Degrees of Freedom ∑ [𝑆$−1]$

)*+6

Step 1: Determine the Degrees of Freedom

Step 2:

o Total number of Experimental runs, TExperiments = 1 + ∑ [𝑆$−1]$)*+ ,

would be 1 + (6) = 7

Step 3:


Experiments [ 7 ]

o Referring to the Standard OA Table,

§ Since we have Factors C, D with 3 levels and we are combining 2

level Factors into a 3 level Factor, it will be useful to select an OA in

3n series

§ Hence, L0 is the nearest 3n series OA


Max. # of Factors



L/ 4 3 3

L. 8 7 7

L0 9 4 4

L+1 12 11 11


Step 4:


o Since, we have to study only the main factors A, B, C, D and no interaction effects, the RLG will look as,


A

Nodes: A,B,C,D

B C D


Step 5:


L0 is as shown,


Step 6:



1 2

3,4

3 41 2

3, 4


Step 7:





X=A x B 1

C 2

D 33/D 41/A x B 2/C

RunsColumns

1 2 3 41 1 1 1 12 1 2 2 23 1 3 3 34 2 1 2 35 2 2 3 16 2 3 1 27 3 1 3 28 3 2 1 39 3 3 2 1



Step 8:


the default OA design layout for L0


o Based on our factor allotment of X = A x Bà1,

Cà2, Dà3, we choose the 1st, 2nd, 3rd columns

from the default array

X C DA x B


Case V [Factors with mixed levels]:

o Lets consider 4 Factors A, B, C, D where A has 4 levels and B, C, D has 2 levels

o For this design, only the main effects of all the factors need to be studied [No interaction exists]

o In such a case, we may use “Collapsing technique”

o This technique finds its application in cases where we have to use a a 4 or 8 level factor in a 2 level OA or a 9 or 27 level factor in a 3 level OA

o The Factor Levels may look as -


Factors Levels

A 4

B 2

C 2

D 2


Max. # of Factors



L/ 4 3 3

L. 8 7 7

L0 9 4 4

L+1 12 11 11




[A] 4 3

[B] 2 1

[C] 2 1

[D] 2 1

Total Degrees of Freedom ∑ [𝑆$−1]$

)*+6

Step 1: Determine the Degrees of Freedom

Step 2:

o Total number of Experimental runs, TExperiments = 1 +

∑ [𝑆$−1]$)*+ , would be 1 + (6) = 7

Step 3:


Experiments [ 7 ]

o Referring to the Standard OA Table, the nearest 2 level

OA is L. [2**4 4**1]


Step 4:


o Since, we have to study only the main factors A, B, C, D and no interaction effects, the RLG will look as,


A B C D

Note:

o A Factor with 4 levels (in this case, A), is represented by 2 nodes and an edge joining them.

o Hence, 3 Columns in the array are used to represent a 4-level Factor


Step 5:


L. is as shown,


Step 6:



1

2

3

46

5

5

1

2

3

46

5

7 67


Step 7:




Factors Levels To be assigned to Column/ Node

A 4 1, 2, 3

B 2 4

C 2 5

D 2 6

1

2

3

4 5 6 7

A B C D



Step 8:


the default OA design layout for L.

o In this case, Factor A has 4 levels & hence we

represent it with 3 Columns. So, A à Column

1,2,3

o Similar factor levels for Column 1,2,3 are

grouped as shown into the 4 levels of Factor A

o Other Factors: Bà4, Cà5, Dà6

Columns

Runs 1 2 3 4 5 6 7

1 1 1 1 1 1 1 1 1

2 1 1 1 1 2 2 2 2

3 1 2 2 2 1 1 2 2

4 1 2 2 2 2 2 1 1

5 2 1 2 3 1 2 1 2

6 2 1 2 3 2 1 2 1

7 2 2 1 4 1 2 2 1

8 2 2 1 4 2 1 1 2

Factors allocated A Levels for

A B C D

Taguchi’s design: Primary objective

o As per Taguchi’s (Quality) Loss Function, the loss to customer depends on the deviation of the process mean from the

target and the variance

o Hence, Taguchi’s design aim at minimising the variance, while keeping the Mean on target

o The difficult task of variance reduction is achieved by optimising the Signal to Noise ratio [S/N], where

§ Signal refers to the primary factor whose relationship with the output response quality characteristic is critical to find in the experiments. (Ex: The Accelerator pedal position is a

Signal while improving the Response – Car speed)

§ All experimental runs are carried out at each level of the signal factor & response is measured


(S/N)1 (S/N)2 (S/N)3> >

Taguchi’s static design: Signal to Noise ratio

Three common optimisation scenarios:


o Nominal – The Best: When a specified value is MOST desired (Neither a smaller nor a larger value is desirable)

§ Signal to noise ratio is given by: [S/N]n =10.log+6μFσFwhere 𝛍 is the signal mean and 𝛔 is the standard deviation

o Larger - The Better: Signal to noise ratio is given by: [S / N]L = -10 . log+6[MSD] = -10 . log+6∑ KμF��

Mwhere MSD is

the mean standard deviation, 𝛍 is the signal mean and n is the no. of trials/ sample size

o Smaller – The Better: Signal to noise ratio is given by: [S/N]S =-10.log+6[𝑀𝑆𝐷] =-10.log+6∑ μF��Qwhere MSD

is the mean standard deviation, 𝛍 is the signal mean and nis the no. of trials/ sample size

o Looked at the Static Response design

o As a part of the Systems design, we looked at how thefactors & levels are chosen & represented

o Using Degrees of Freedom, how to decide on thenumber of experimental runs

o Choosing the relevant Orthogonal Array

o Using the Linear Graphs

o Allocating Factors to OA Columns

o Finalising the Experimental layout


Thank you !!!

Harish Haridasan || [email protected] || +91- 94492 40463 || Bangalore, India

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