PARAMETER DESIGN


This section provides an example of how the DOE technique is used to determine design factor target levels. This approach is an upstream attempt to develop a robust product that will avoid problems later in production. The emphasis at this stage is on using wide tolerance levels to provide a product that is easy to manufacture and still meets all requirements.

DISCUSSION

After the basic design of a product is determined, the next step is to determine to what levels the components of that product should be set to ensure that the target will be met. The experience of the designer or design team is useful in establishing the starting values for the investigation. This investigation begins by determining what the component target values should be using wide component tolerances. This is called parameter design. If the resultant variability around the product response target is too great, the next step is to determine which tolerances should be tightened. This approach, Tolerance Design, will be discussed in Section 9.

Example

A particular product has been designed with five components (A through E). The target response for the product is 59.0 units. Field experience has indicated that when the response differs from the target by five units, the average customer would complain and the repair cost would be $150. From this information, the k value in the loss function can be calculated.

k

=

$150/5 2

 

=

$6.00 per unit 2

A brainstorming group that consisted of the designer and other experts in this area determined that the response is linear over the component range of interest and that the components should be evaluated at the levels shown in Table 9.54.

Table 9.54: Components and Their Levels

Component (Factor)

Levels (Units are those appropriate to each component)

Low

High

A

1000

1500

B

400

700

C

50

70

D

1300

2200

E

1200

1600

Note  

Factor A is more expensive to manufacture at the higher level than at the lower level.

These factors will be assigned to an L8 inner array. The two unassigned columns will be used for an estimate of the experimental error.

An L8 outer array contains the low cost tolerances as follows :

 

Low Cost Tolerance

(Factor)

Low

High

A

-50

+50

B

-15

+15

C

-5

+5

D

-200

+200

E

-100

+100

The tolerance amounts are added to/subtracted from the control level as indicated by the outer array. The brainstorming group suspects that two other noise factors are significant, namely, the temperature (T) and humidity (H) of the assembly environment. The noise and tolerance factors are combined into an L8 outer array. The testing setup and test results are shown in Table 9.55.

Table 9.55: L8 Inner OA with L8 Outer OA and Test Results
 

Outer Array

T

1

2

2

1

2

1

1

2

 

H

1

2

2

1

1

2

2

1

 

E

1

2

1

2

2

1

2

1

 

D

1

2

1

2

1

2

1

2

 

C

1

1

2

2

2

2

1

1

 

L8 Inner Array

B

1

1

2

2

1

1

2

2

NTB S/N

A

B

C

D

E

X

Y

A

1

1

1

1

2

2

2

2

1

1

1

1

1

1

1

 

51.4

49.5

48.9

56.6

52.7

50.8

46.0

51.9

24.34

1

1

1

2

2

2

2

 

58.6

56.7

56.8

60.3

58.2

56.3

52.8

56.7

28.36

1

2

2

1

1

2

2

 

52.1

59.1

58.9

62.7

61.5

51.1

47.0

50.9

19.58

1

2

2

2

2

1

1

 

62.6

60.0

61.6

67.2

62.5

59.4

56.0

62.7

25.59

2

1

2

1

2

1

2

 

47.2

45.3

45.2

51.3

47.3

45.4

41.3

47.4

24.28

2

1

2

2

1

2

1

 

50.1

48.8

48.4

54.6

51.0

49.1

44.5

50.4

24.86

2

2

1

1

2

2

1

 

40.1

38.6

37.7

44.6

40.7

38.7

34.8

39.7

22.99

2

2

1

2

1

1

2

 

46.6

43.2

42.6

49.4

46.6

39.2

39.2

45.4

23.12

Note: X and Y are the unassigned columns that will be used to estimate error.

An understanding of the way the testing matrix is interpreted can be reached by considering the factor A. When the inner array column associated with factor A has a value of 1, the value of A is 1000. When there is a 2 in that column, the value of A is 1500. The actual test values of A are also determined by the tolerance value of A in the outer array. If the outer array value of A is 1, then 50 is subtracted from the value of A determined in the inner array. If the outer array value is 2, then 50 is added to it. This can be summarized as follows:

 
Actual Test Values of A

Inner Array Value of A

Outer Array Tolerance Value of A

1

2

1

950

1050

2

1450

1550

The ANOVA table and level averages for the most significant factors for the S/N and raw data are shown in Table 9.56.

Table 9.56: ANOVA Table (NTB) and Level Averages for the Most Significant Factors

Source

df

SS

MS

F Ratio

S'

%

A

1*

0.860

0.860

     

B

1

13.965

13.965

21.992

13.330

30.50

C

1

2.533

2.533

3.989

1.898

4.34

D

1

14.455

14.455

22.764

13.820

31.62

E

1

10.845

10.845

17.079

10.210

23.36

X

1*

2.477

2.477

     

Y

1*

10.424

10.424

     

Error

         

(pooled error)

3

1.906

0.635

 

4.446

10.17

Total

7

43.704

6.243

     
 

S/N Level Averages

Factor

Level 1

Level 2

D

22.80

25.48

B

25.46

22.82

E

22.98

25.30

 

Raw Data ANOVA Table

Source

df

SS

MS

F Ratio

S'

%

A

1

2032.883

2032.883

680.577

2029.896

56.85

B

1

9.533

9.533

3.192

6.546

0.18

C

1

435.244

435.244

145.713

432.257

12.11

D

1

427.973

427.973

143.279

424.986

11.90

E

1

13.231

13.231

4.430

10.244

0.29

X

1*

3.658 3.563

3.658 3.563

     

Y

1*

88.125

88.125

29.503

85.138

2.38

A-Tol.

1

1.995

1.995

     

B-Tol.

1*

113.156

113.156

37.883

110.169

3.09

C-Tol.

1

49.879

49.879

16.699

46.892

1.31

D-Tol.

1

3.754

3.754

     

E-Tol.

1*

239.089

239.089

80.043

236.102

6.61

H

1

30.754

30.754

     

T

1*

140.969

140.969

     

Error

49*

161.297

161.297

 

188.180

5.27

(pooled error)

54

         

Total

63

3570.410

56.673

     
 

Level Averages

Factor

Level 1

Level 2

A

56.23

44.96

C

47.99

53.21

D

48.01

53.18

Average of all data = 24.14

Average of all data = 50.60

From the S/N level averages, D 2 B 1 E 2 is clearly the best setting for S/N. The estimated S/N at that setting is:

click to expand

Since A 1 is preferred from a cost standpoint and D 2 is preferred from the S/N analysis, the next step is to determine if the value of C can be adjusted to attain the target of 59. The average response at A 1 D 2 is:

  • Average Response = 50.6 + (56.23 - 50.6) + (53.18 - 50.6) = 58.8

To reach a target of 59, the value of C that is included in the average response calculation must have a level average of 50.8 since:

Target = Average Response at A 1 D 2 + Effect due to C

59.0 = 58.8 + (50.8 - 50.6)

The target value for C can be interpreted from the tested levels and the level averages as follows:

Note  

Value of C

Response

50

47.99

Target

50.8

70

53.21

This value, 60.77, is at the same percentile between 50 and 70 as 50.8 is at between 47.99 and 53.21.

In summary, the recommended target values are:

Factor

Target Value

A

1000

B

400

C

60.77

D

2200

E

1600

The estimated average is 59.0 and the estimated S/N is 27.96. The 90% confidence on the average is:

59.0 ±

or,

59.0 ±1.50

A set of verification runs is not made using the recommended factor target values given previously. The tolerance levels from the outer array are used to define an L8 verification run experiment as shown in Table 9.57.

Table 9.57: Variation Runs Using Recommended Factor Target Values

A-Tol

B-Tol

C-Tol

D-Tol

E-Tol

H

T

Test Result

1

1

1

1

1

1

1

60.2

1

1

1

2

2

2

2

57.9

1

2

2

1

1

2

2

59.5

1

2

2

2

2

1

1

64.8

2

1

2

1

2

1

2

59.4

2

1

2

2

1

2

1

58.6

2

2

1

1

2

2

1

55.7

2

2

1

2

1

1

2

59.9

The average response is 59.5 and the S/N is 27.3. Since the average response and the S/N are close to the predicted values, the verification runs confirm the prediction. If the average response and S/N did not confirm the predictions , the data could be analyzed to determine which factors have response characteristics different from those predicted.

The information from the verification runs cannot be used directly in the loss function, since the observed variability may be affected by testing only at the tolerance limits. The center portions of the factor distributions are not represented in these tests. For the situation where the change in response is assumed to be a linear increase or decrease across the tolerance levels, the loss function can be easily calculated as follows:

  1. If it can be assumed that the C pk in production will be 1.0 or greater for all specified tolerances, then the difference between the tolerance limits will be equal to or greater than six times the production standard deviation for each component parameter.

  2. The difference between the response level averages for the two tolerance limits will equal six times the production response standard deviation since the product response is linearly related to the component parameter level.

  3. The response variance due to each tolerance is additive since the response effect of each component tolerance is additive. (Variance = Std. Dev. 2 )

  4. The effect of noise factors can be treated in a similar manner.

In this example, the levels of humidity were set at the average humidity ± 2 times the humidity standard deviation. The change in response is assumed to be linear across the change in humidity. The difference in response between the two levels represents four times the response standard deviation. The response variance can be calculated as shown in Table 9.58.

Table 9.58: Calculated Response Variance

Tolerance for Factor

Response Difference Between Tol Limits

Response Production Std. Dev.

Variance

A

2.2

0.37

0.1334

B

1.0

0.16

0.0278

C

2.2

0.37

0.1344

D

1.6

0.27

0.0711

E

0.1

0.02

0.0003

     

0.3680

Humidity (H)

3.2

0.80

0.6400

Error Variance

   

2.9890

     

3.9970

The response variance will be 3.9970. The loss function can be calculated from the equation:

For a production run of 50,000 pieces, the total loss would be $1,274,100.

In the situation where the change in response is non-linear across the tolerance levels or noise factor levels, a computer simulation can be used to determine the distribution of product response for each component taken singly and for the total assembly. This situation occurs when the highest ( lowest ) response occurs at the component nominal and the response decreases (increases) as the distance from the component nominal increases . The purpose of these calculations is to estimate the response variance for the total assembly population so that the loss function can be calculated. Once the value of the loss function has been calculated, it can be compared to the cost of tightening the tolerances so as to determine the optimal tolerance limits. This technique will be discussed in the following section.




Six Sigma and Beyond. Design for Six Sigma (Vol. 6)
Six Sigma and Beyond: Design for Six Sigma, Volume VI
ISBN: 1574443151
EAN: 2147483647
Year: 2003
Pages: 235

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