Design of Experimentansytech.com/doe-day7.pdf ·  · 2008-02-08§2k Factorial with Center Point §32 Factorial Design §33 Factorial Design §Confounding ... as some blocking effects.

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Design of Experiment

ANSY Tech. Inc.Muhammad Sohail Ahmed, Ph.D.Mubashir Siddiqui

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Day 7

Factorial Design – 3 Levels

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Factorial Designs (Day 7)

§ 2k Factorial with Center Point§ 32 Factorial Design§ 33 Factorial Design§ Confounding

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Methods to Obtain Curvature Estimate

§ 2k design with center points

§ 3-Level Designs

§ Response Surface Method

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2k Factorial with Center Pointεβββ +++= ∑∑∑

<= jijiij

k

jjj xxx

10 y First Order Model

To model CURVATURE in Response Function

εββββ ++++= ∑∑∑∑<== ji

jiij

k

jjjj

k

jjj xxxx

1

2

10 y

where ßjj: pure second order or quadratic effects

(1) a

b ab

A-B- A+B-

A+B+A-B+

Low High

Low

HighB

A

Let nc be the observations at center point (0,0)Let yF be the observations at corner points (0,0).

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Problem

Response: Yield of a chemical processFactors: Reaction Time and

Reaction Temperature

(a) Find significant factors, their effects.(b) Is there any evidence of curvature?

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Entering Data in Minitab

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Solution

Stat à DOE à Define Custom Factorial Design

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Solution (cont’d)

Stat à DOE à Analyze Factorial Design

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Solution (cont’d)

No evidence of second order curvature for the region tested

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32 Design

Low level Intermediate Level High Level

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32 Design Model

Low level Intermediate Level High Level

Notice the quadratic nature of the model

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ProblemAn experiment is run in a chemical process using a 32 factorial design. The design factors

are temperature and pressure, and the response variable is yield. Data is shown in the table.

(a) Using ANOVA, which factor has significant effects?

(b) Analyze the residuals.

(c) Verify that the second order model could be written, in natural variables, as

y = -1335.63 + 18.56T + 8.59P – 0.072T2 – 0.0196P2 – 0.0384TP(d) Verify that if we let low, medium, and high levels be represented by -1, 0, and

+1, then the second order model for yield isy = 86.81 + 10.4x1 + 8.42x2 – 7.17x1

2 – 7.86x22 – 7.69x1x2

76.58, 83.0496.57, 88.7271.18, 92.77100

93.95, 88.5488.47, 84.2351.86, 82.4390

80.92, 72.6064.97, 69.2247.58, 48.7780

140120100

Pressure (psig)Temperature

0C

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Solution

83.04100140

88.5490140

72.680140

88.72100120

84.2390120

69.2280120

92.77100100

82.4390100

48.7780100

76.58100140

93.9590140

80.9280140

96.57100120

88.4790120

64.9780120

71.18100100

51.8690100

47.5880100

YTP

Entering Data in Minitab

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Solution

Solution (a) Using ANOVA, which factors are significant?Stat à ANOVA à Two-WaySelect Y as response and Pressure and Temperature as Factors.Click on Graphs, Select Four in One for Residual Plots.

Pressure and Temperature are significant. Their interaction is not.

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Solution (cont’d)Solution (b) Analyze residuals.We will get residuals plot by selecting Four in One for Residual Plots.

20100-10-20

99

90

50

10

1

Residual

Per

cent

9080706050

10

0

-10

Fitted Value

Res

idua

l

151050-5-10-15

6.0

4.5

3.0

1.5

0.0

Residual

Freq

uenc

y

18161412108642

10

0

-10

Observation Order

Res

idua

l

Normal Probability Plot Versus Fits

Histogram Versus Order

Residual Plots for Y

Residuals seem to be normally distributed in Normal Probability Plot.But the plot of residuals vs fits shows a converging nozzle. Hence there is non-constant variance.

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Residuals VS Fitted Values

n As the magnitude of observation increases, the variance increases.

n Error in measuring instrument.n Data follows a non-normal or skewed

distribution.n F-test is only slightly affected in balanced

designs.n In worse cases, data transformation is required.

§

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Solution (cont’d)Solution (c)Verify that the second order model could be written, in natural variables, as

y = -1335.63 + 18.56T + 8.59P – 0.072T2 – 0.0196P2 – 0.0384TP

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Solution (cont’d)

Stat à Regression à RegressionSelect Response and Predictors.

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Solution (cont’d)Solution (d) Verify that if we let low, medium, and high levels be represented by -1, 0, and +1, then the second order model for yield is

y = 86.81 + 10.4x1 + 8.42x2 – 7.17x12 – 7.86x2

2 – 7.69x1x2

Enter data in Minitab using Coded Variablesrepresenting low, medium , and high levels as -1, 0, and +1 respectively.Then get square- and product-columns as shown in C24, C25, and C26.

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Solution (cont’d)Stat à Regression à RegressionSelect Response and Predictors.

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3-D view Of Response y=f(T,P)

h t tp : / /www. l i vephys ics .com/p too ls /on l i ne - 3 d-f u n c t i o n - grapher.php?ymin=- 1 & x m i n = - 1 & z m i n = A u t o & y m a x = 1 & x m a x = 1 & z m a x =Auto&f =86.81%2B%2810.4*x%29%2B%288.42*y%29 -% 2 8 7 . 1 * x % 5 E 2 % 2 9 -%287.86*y%5E2%29 -% 2 8 7 . 6 9 * x * y % 2 9

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Same Problem Using Factorial Design

Stat à DOE à Define Custom Factorial Design

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Same Problem Using Factorial DesignStat à DOE à Analyze Factorial DesignSelect Response

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33 Design

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33 DesignThree factors are being studied viz bottle type, shelf type, and workers to find theireffects on the time it takes to stock 10 cases on shelves. Time data is as shown here.(a) Which of the factors are significant?(b) Specify appropriate levels of factors..

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Entering Data in Minitab

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SolutionSolution (a) – Significant FactorsStat à DOE à Define Custom Factorial DesignSelect Factors.Check ‘General Factorial Design’

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Solution (cont’d)Stat à DOE à Analyze Factorial DesignSelect Response.Click Graphs. Select Four In One under Residuals Plot.

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Solution (cont’d)

0.500.250.00-0.25-0.50

99

90

50

10

1

Residual

Pe

rce

nt

6543

0.4

0.2

0.0

-0.2

-0.4

Fitted Value

Re

sid

ua

l

0.320.160.00-0.16-0.32

8

6

4

2

0

Residual

Fre

qu

ency

50454035302520151051

0.4

0.2

0.0

-0.2

-0.4

Observation Order

Re

sid

ua

l

Normal Probability Plot Versus Fits

Histogram Versus Order

Residual Plots for Time

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Solution (cont’d)

321

5.6

5.2

4.8

4.4

4.0321

321

5.6

5.2

4.8

4.4

4.0

Worker

Me

an

Bottle Type

Shelf Type

Main Effects Plot for TimeData Means

Solution (b)Stat à DOE à Factorial Plots

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Solution (cont’d)

321 321

6

5

4

6

5

4

Wor ker

Bottle T ype

Shelf T ype

123

Worker

123

TypeBottle

Interaction Plot for TimeData Means

Shortest Time:Worker: 1 (or 3)Shelf Type:1

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Confounding

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Confounding

n Confounding is a design technique for arranging a complete factorial experiment in blocks, where the block size is smaller than the number of treatment combinations in one replicate.

n A confounding design is one where some treatment effects (main or interactions) are estimated by the same linear combination, of the experimental observations, as some blocking effects.

n In this case, the treatment effect and the blocking effect are said to be confounded.

n Thus confounding causes some information, usually about higher order interactions to be indistinguishable from blocks.

Block 1(1)ab

Block 2ab

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Confounding

ab++4

b+-3

a-+2

(1)--1

CombinationBARuns

Block 1(1)ab

Block 2ab

Assignment of four runs of a 22 Design in Two BLOCKS

])1([

])1([

])1([

21

21

21

baabABEffectnInteractio

abbaBBofeffectMain

abbaAAofeffectMain

n

n

n

−−+==

++−−==

+−+−==

Which effects are confounded?See the SIGNS in Main and Interaction Effects. Here AB effect is confounded.

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ProblemFour factors are studied to find their effects on Filtration Rate. They are:Temperature(A), Pressure(B), Concentration of formaldehyde(C), and Stirring Rate(D).All combinations cannot be run using one batch of raw materials.So there will be two blocks.

Block 1(1)=25ab=45ac=40bc=60ad=80bd=25cd=55

abcd=76

Block 2a=71b=48c=68d=43

abc=65bcd=70acd=86

abd=104

1

1

1

1

1

1

1

1

-1

-1

-1

-1

-1

-1

-1

-1

D

abcd

bcd

acd

cd

abd

bd

ad

d

abc

bc

ac

c

ab

b

a

(1)

Comb

111

11-1

1-11

1-1-1

-111

-11-1

-1-11

-1-1-1

111

11-1

1-11

1-1-1

-111

-11-1

-1-11

-1-1-1

CBA

ABCD = + ABCD = -

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Entering Data in Minitab

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SolutionStat à DOE à Define Factorial DesignSelect Low/High and select Worksheet Data as Coded.

Click on Designs, and specify the block column.

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Solution (cont’d)

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Solution (cont’d)

20100-10-20

99

95

90

80

7060504030

20

10

5

1

Effect

Per

cen

t

A TempB PressC ConcD Rate

Factor Name

Not SignificantSignificant

Effect Type

AD

AC

DC

A

Normal Plot of the Effects(response is Y, Alpha = .05)

Lenth's PSE = 3.1875

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End of Day 7