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Arizona State University *
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Course
572
Subject
Mathematics
Date
Feb 20, 2024
Type
Pages
10
Uploaded by BailiffLlama2777
Characterization
of
the
Response
Surface
*
Find
out
where
our
stationary
point
is
»
Find
what
type
of
surface
we
have
—
Graphical
Analysis
—
Canonical
Analysis
«
Determine
the
sensitivity
of
the
response
variable
to
the
optimum
value
—
Canonical
Analysis
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
21
Montgomery
Finding
the
Stationary
Point
»
After
fitting
a
second
order
model
take
the
partial
derivatives
with
respect
to
the
x;'s
and
set
to
zero
—Qy/dx;=...=
0y/dx,=0
Stationary
point
represents...
—
Maximum
Point
—
Minimum
Point
—
Saddle
Point
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
Montgomery
Stationary
Point
x,=
—B7'b
X
Bl
Bll'BL!’z
X
>
x=|"]1
b=
B.'
and
B
=
Xi
B
sym.
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
Montgomery
B\\.-
.
v
vee
s
Bul2
B/2
B
kk
23
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Canonical
Analysis
+
Used
for
sensitivity
analysis
and
stationary
point
identification
»
Based
on
the
analysis
of
a
transformed
model
called:
canonical
form
of
the
model
«
Canonical Model
form:
Y
=
Ys
AW+
AW
+
L+
A
w2
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
Montgomery
24
x,
u
FIGURE
11.9
Canonical
form
of
\
the
second-order
model
*1
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
25
Montgomery
Eigenvalues
»
The
nature
of
the
response
can
be
determined
by
the
signs
and
magnitudes
of
the
eigenvalues
—
{e}
all
positive:
a
minimum
is
found
—
{e}
all
negative:
a
maximum
is
found
—
{e}
mixed:
a
saddle
point
is
found
»
Eigenvalues
can
be
used
to
determine
the
sensitivity
of
the
response
with
respect
to
the
design
factors
»
The
response
surface
is
steepest
in
the
direction
(canonical)
corresponding
to
the
largest
absolute
eigenvalue
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
26
Montgomery
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Chapter
11
exampLE
11.2
We
will
continue
the
analysis
of
the
chemical
process
in
Example
11.1.
A
second-order
model
in
the
vanables
x,
and
v;
cannot
be
fit
using
the
design
in
Table
11.4.
The
experi-
menter
decides
to
augment
this
design
with
enough
points
to
fit
a
second-order
model.'
She
obtains
four
observations
at
(x,
=0,5
=
*1414)
and
(x,
=
£
1414,
x;
=
0).
The
complete
experiment
is
shown
in
Table
11.6,
and
the
design
is
dis-
played
in
Figure
11.10.
This
design
is
called
a
central
com-
posite
design
(or
a
CCD)
and
will
be
discussed
in
more
detail
in
Section
11.4.2.
In
this
second
phase
of
the
study,
two
=
TABLE
11.6
Central
Composite
Design
for
Example
11.2
Natural
Vartables
Coded
Varfables
Responses
&
&
x,
X
¥y
(yhedd)
¥y
A
viscosity
)
¥y
(mwbecular
weighty
%0
170
1
I
76.5
62
2040
0
150
1
!
7.0
0
470
o«
170
I
-1
50
o
w0
180
I
1
0.5
»
85
175
0
0
99
”
~
175
0
)
0.3
0
85
175
0
0
0.0
o8
'
175
0
0
9.7
70
290
5
175
0
0
9%
1
1500
920
175
1414
0
54
o8
1360
7798
175
1414
0
756
T
20
85
182.0
0
1414
8.5
8
1630
85
167.93
0
~1.414
77.0
57
1150
Design
&
Analysis
of
Experiments
10E
2020
Montgomery
27
X2
+2
0,
1.4149)
=11
(LN
|
]
-2
(-1.414,0)
(0,0
(1.414,0)
+2
=1,-1)
(1,-1
0.-1.413)
2
s
FIGURE
11.10
Central
composite
design
for
Example
11.2
additional
responses
were
of
interest:
the
viscosity
and
the
molecular
weight
of
the
product.
The
responses
are
also
shown
in
Table
11.6.
We
will
focus
on
fitting
a
quadratic
model
to
the
yield
response
v,
(the
other
responses
will
be
discussed
in
Section
11.3.4).
We
generally
use
computer
software
to
fit
a
response
surface
and
to
construct
the
contour
plots.
Table
11.7
contains
the
output
from
Design-Expert.
From
exam-
ining
this
table,
we
notice
that
this
software
package
first
computes
the
“sequential
or
extra
sums
of
squares™
for
the
Chapter
11
Design
&
Analysis
of
Experiments
10E
2020
Montgomery
28
TABLE
11.7
Y
ST
Ty
S
S
T
P
I
7
E
T
S
T
S
—
e
——
Computer
Output
from
Design-Expert
for
Fitting
a
Model
to
the
Yield
Response
in
Exam,
Response:
yield
***WARNING:
The
Cubic
Model
is
Aliased!***
Sequential
Model
Sum
of
Squares
Source
Sum
of
Squares
DF
Mean
Square
FValue
Prob
>
F
Mean
80062.16
1
80062.16
Linear
10.04
2
5.02
2.69
0.1166
2FI
0.25
1
0.25
0.12
0.7350
Quadratic
17.95
2
8.98
126.88
<0.001
Suggested
Cubic
2.042E-003
2
1.021E-003
0.010
0.9897
Aliased
Residual
0.49
5
0.099
Total
80090.90
13
6160.84
“Sequential
Model
Sum
of
Squares”:
Select
the
highest
order
polynomial
where
the
additional
terms
are
significant.
Lack-of-Fit
Tests
Source
Sum
of
Squares
DF
Mean
Square
FValue
Prob
>
F
Linear
18.49
6
3.08
58.14
0.0008
2FI
18.24
5
3.65
68.82
0.0006
Quadratic
0.28
3
0.094
1.78
0.2897
Suggested
Cubic
0.28
1
0.28
5.31
0.0826
Aliased
Pure
Error
0.21
4
0.053
“Lack-of-Fit
Tests”:
Want
the
selected
model
to
have
insignificant
lack-of-fit.
Model
summary
Statistics
Source
Std.
Dev.
R-Squared
Adjusted
R-Squared
Predicted
R-Squared
PRESS
Linear
137
0.3494
0.2193
-0.0435
29.99
2FI
1.43
0.3581
0.1441
—-0.2730
36.59
Quadratic
0.27
0.9828
0.9705
0.9184
2.35
Suggested
Cubic
0.31
0.9828
0.9588
0.3622
18.33
Aliased
“Model
Summary
Statistics”:
Focus
on
the
model
minimizing
the
“PRESS,”
or
equivalently
maximizing
the
“PRED
R-SQR.”
Response:
yield
ANOVA
for
Response
Surface
Quadratic
Model
Analysis
of
variance
table
[Partial
sum
of
squares]
Source
Sum
of
Squares
DF
Mean
Square
FValue
Prob
>
F
Model
28.25
5
5.65
79.85
<0.0001
A
7.92
1
7.92
111.93
<0.0001
B
2.12
1
2.12
30.01
0.0009
A2
13.18
1
13.18
186.22
<0.0001
B?
6.97
1
6.97
98.56
<0.0001
AB
0.25
1
0.25
3.53
0.1022
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Response:
yiel&
]
S
ANOVA
for
Response
Surface
Quadratic
Model
]
Analysis
of
variance
table
[Partial
sum
of
squares]
Source
Sum
of
Squares
)
DF
M;an
Square
F
Vaiue
Prob
>
F
Model
S
28.25
5
o
5.65
79.85
<0.0001
A
7.92
7.92
111.93
<0.0001
B
2.12
1
2.12
30.01
0.0009
A2
13.18
1
13.18
186.22
<0.0001
B?
6.97
1
6.97
98.56
<0.0001
AB
0.25
1
0.25
3.53
0.1022
Residual
0.50
7
0.071
Lack
of
Fit
0.28
3
0.094
1.78
0.2897
Pure
Error
0.21
4
0.053
Cor
Total
28.74
12
Std.
Dev.
0.27
R-Squared
0.9828
Mean
78.48
Adj
R-Squared
0.9705
C.V.
0.34
Pred
R-Squared
0.9184
PRESS
2.35
Adeq
Precision
23.018
F;lctor
DF
Standard
Error
95%
CI
Low
95%
CI
High
I;xtercep‘t
‘
79.94
N
o
0.12
79.66
80.22
A-time
0.99
1
0.094
0.77
1.22
li;ten;l.).
0.52
1
0.094
0.29
0.74
“A{
-1.38
1
0.10
-1.61
-1.14
F
-1.00
1
0.10
-1.24
-0.76
AB
0.25
1
0.13
—0.064
0.56
Final
Equation
in
Terms
of
Coded
i‘:actors:
[
yield
=
+79.94
+0.99
+0.52
—1.38
—1.00
+0.25
e
VWM
_at
s
A
AT