DESIGN OF EXPERIMENT

1. ANIS ATIKAH AHMAD
DESIGN OF EXPERIMENTS
ERT 427
ponse Surface Methodology (RSM)

2. OUTLINE
Definition of Response Surface Methodology
Method of Steepest Ascent
Second-Order Response Surface
Design for Fitting First Order and Second Order
Model
Blocking in Response Surface Designs

3. RESPONSE SURFACE METHODS
AND DESIGNS
If we denote the expected response by
y = f (x1, x2 )+e
  

 )
,
( 2
1 x
x
f
y
E
oPrimary focus of previous chapters is factor screening.
Two-level factorials, fractional factorials are widely used.
oResponse Surface Methodology (RSM) is useful for the modeling and
analysis of programs in which a response of interest is influenced by
several variables.
oObjective of RSM is optimization
For example: Find the levels of temperature (x1) and pressure (x2) to
maximize the
yield (y) of a process. The process yield:
Then the surface is represented by )
,
( 2
1 x
x
f



4. EXAMPLE OF RESPONSE SURFACE
GRAPH

5. (CONTINUED)
RSM is a sequential procedure.
When we are at a point on the response
surface that is remote from the
optimum, such as the current operating
conditions in Figure 11.3, there is little
curvature in the system and the first-
order model will be appropriate.
Once the region of the optimum has
been found, a more elaborate model,
such as the second-order model, may
be employed, and an analysis may be
performed to locate the optimum.
The eventual objective of RSM is to
determine the optimum operating
conditions for the system or to
determine a region of the factor space in
which operating requirements are
satisfied.
*the analysis of response surface can be
thought as ‘’climbing a hill’’, where the
top of the hill represents the point of
maximum response.

6. METHOD OF STEEPEST
ASCENT
The first-order model is an adequate
approximation to the true surface in a
small region of the x’s WHEN the initial
estimate of the optimum conditions is
FAR from the actual optimum.
The method of steepest ascent is a
procedure for moving sequentially from
an initial “guess” towards to region of
the optimum.
Experiments are conducted along the
path of steepest ascent until no further
increase in response is observed.
Then a new first-order model may be
fit, a new path of steepest ascent
determined, and the procedure
continued. Eventually, the experimenter
will arrive in the vicinity of the
optimum. (This is usually indicated by
lack of fit of a first-order model.)
At that time, additional experiments are
Figure 11.4: First-order response
surface and path of steepest ascent

7. EXAMPLE 11.1:
A chemical engineer is interested in determining the
operating conditions that maximize the yield of a process.
Two controllable variables: reaction time (ξ1) and reaction
temperature (ξ2). The engineer is currently operating the
process with reaction time of 35 minutes and temperature
of 155 °F, which result in yield around 40 percent. Because
it is unlikely that this region contains the optimum, she fits
a first-order model and applies the method of steepest
ascent.

8. STEP 1: FIT TO FIRST-ORDER
MODEL
The design employed was a 22 factorial design with five centerpoints
5
35
1
1



x
5
155
2
2



x

9. (CONTINUED)
A first-order model may be fit to these data by
least squares.

10. STEP 2: METHOD OF
STEEPEST ASCENT
To move away from the design center,
the point (x1=0,x2=0) along the path of
steepest ascent, we would move 0.775
units in the x1 direction for every 0.325
units in the x2 direction. If we decide to
use 5 minutes of reaction time as the
basic step size, then.
Δx1 =1.0000
Δx2= 0.325/0.775 =0.4194
The engineer computes points along
this path and observes yields at these
points until a decrease in response is
noted. The results are shown in Table1-
3.The steps are shown in both coded
and natural variables.

11. 11
The step size is 5 minutes of reaction time and 2 degrees F
We can see from the table1-3, increases in response are
observed through the tenth step. However, the eleventh step
produces a decrease in yield. Therefore, another first-order
model must be fit in the general vicinity of the point( ξ1 = 85,
ξ2 =175).

12. STEP 3: FIT TO NEW FIRST-
ORDER MODEL
The first-order model fit to the coded data in Table 1-4 is
y=78.97+1.00X1 +0.50X2
The region of exploration: ξ1 is [80,90] AND ξ2 is [170-180].

13. ANALYSIS OF DATA FOR SECOND
FIRST-MODEL
The interaction and pure quadratic checks imply that the first-order
model is not an adequate approximation. This curvature in the true
surface may indicate that we are near the optimum. At this point,
additional analysis must be done to locate the optimum more precisely.

14. SUMMARY OF STEEPEST
ASCENT
 Points on the path of steepest ascent are proportional to the
magnitudes of the model regression coefficients.
 The direction depends on the sign of the regression
coefficient.
 Step-by-step procedure:

15. THE SECOND-ORDER MODEL
When the experimenter is relative closed to the optimum, the
second-order model is used to approximate the response.
It is used to find the stationary point.
Determine whether the stationary point is a point of
maximum or minimum response or a saddle point.
There is a lot of empirical evidence that they work very well.
This model is used widely in practice.

16. RESPONSE SURFACE WITH
MAXIMUM POINT

17. RESPONSE SURFACE WITH
MINIMUM POINT

18. RESPONSE SURFACE WITH
SADDLE POINT

19. SECOND ORDER MODEL: HOW
TO FIND STATIONARY POINT?
Writing the second-order model in matrix notation, we
have:
b
x
y
b
B
x
B
b
x
Bx
x
b
x
s
1
s
'
0
2
22
1
12
11
2
1
2
1
0
2
1
ˆ
ˆ
2
1
ˆ
2
/
ˆ
ˆ
2
/
ˆ
2
/
ˆ
ˆ
and
ˆ
ˆ
ˆ
,
,
'
'
ˆ
ˆ
s
kk
k
k
k
k
x
x
x
y



































































 Stationary point
 The response at the
stationary point

20. SECOND ORDER MODEL:
CHARACTERISTIC OF
STATIONARY POINT
In derivatives form:
Stationary point represents:
 Maximum Point
 Minimum Point
 Saddle Point (minimax)
How to know?

21. CANONICAL ANALYSIS
The canonical form:
the {λi} are the
eigenvalues or
characteristic roots of
the matrix B
to transform the model into a
new coordinate system with the
origin at the stationary point xs
and then to rotate the axes of
this system until they are parallel
to the principal axes of the fitted
response surface.

22. CANONICAL ANALYSIS
The nature of the response surface can be determined from
the stationary point & the signs and magnitudes of the {λi }.
 all positive: a minimum is found
 all negative: a maximum is found
 mixed: a saddle point is found

23. QUIZ 3
https://forms.gle/m4wTVsaae3s1tvKk6

24. ATTENDANCE

25. EXAMPLE 11.2
A second-order model in the variables x1 and x2 cannot be fit
using the design in Table 1-4. The experimenter decides to
augment this design with enough points to fit a second-order
model. She obtains four observations at (x1±0, x2±1.414) and
(x1±1.414, x2±0). The complete experiment is shown in Table
1-5

26. TABLE 1-5

27. CENTRAL COMPOSITE
DESIGN

28. SEQUENTIAL MODEL SUM OF
SQUARES
Source
Sum of
Squares
df
Mean
Square
F-value p-value
Mean
80062.1
6
1 80062.16
Linear 10.04 2 5.02 2.69 0.1166
2FI 0.2500 1 0.2500 0.1220 0.7350
Quadratic 17.95 2 8.98 126.88
<
0.0001
Suggested
Cubic 0.0020 2 0.0010 0.0103 0.9897 Aliased
Residual 0.4933 5 0.0987
Total
80090.9
0
13 6160.84
Select the highest order polynomial where additional terms
are SIGNIFICANT

29. LACK-OF-FIT-TESTS
Source
Sum of
Squares
df
Mean
Square
F-value p-value
Linear 18.49 6 3.08 58.14 0.0008
2FI 18.24 5 3.65 68.82 0.0006
Quadrati
c
0.2833 3 0.0944 1.78 0.2897
Suggeste
d
Cubic 0.2813 1 0.2813 5.31 0.0826 Aliased
Pure
Error
0.2120 4 0.0530
Select the model with INSIGNIFICANT lack-of-fit

30. MODEL SUMMARY STATISTICS
Source
Std.
Dev.
R²
Adjusted
R²
Predicte
d R²
PRESS
Linear 1.37 0.3494 0.2193 -0.0435 29.99
2FI 1.43 0.3581 0.1441 -0.2730 36.59
Quadrati
c
0.266
0
0.9828 0.9705 0.9184 2.35
Suggeste
d
Cubic
0.314
1
0.9828 0.9588 0.3622 18.33 Aliased
Focus on the model that MINIMIZE the “PRESS” or
equivalently maximize the “PRED R-SQR”

31. ANOVA FOR RESPONSE SURFACE
QUADRATIC MODEL
Source
Sum of
Squares
df
Mean
Square
F-value p-value
Model 28.25 5 5.65 79.85 < 0.0001
significan
t
A-A 7.92 1 7.92 111.93 < 0.0001
B-B 2.12 1 2.12 30.01 0.0009
AB 0.2500 1 0.2500 3.53 0.1022
A² 13.18 1 13.18 186.22 < 0.0001
B² 6.97 1 6.97 98.56 < 0.0001
Residual 0.4953 7 0.0708
Lack of
Fit
0.2833 3 0.0944 1.78 0.2897
not
significan
t
Pure Error 0.2120 4 0.0530
Cor Total 28.74 12

32. COEFFICIENTS IN TERMS OF
CODED FACTORS
Factor
Coeffici
ent
Estimat
e
df
Standar
d Error
95% CI
Low
95% CI
High
VIF
Interce
pt
79.94 1 0.1190 79.66 80.22
A-A 0.9950 1 0.0940 0.7726 1.22 1.0000
B-B 0.5152 1 0.0940 0.2928 0.7375 1.0000
AB 0.2500 1 0.1330
-
0.0645
0.5645 1.0000
A² -1.38 1 0.1009 -1.61 -1.14 1.02
B² -1.00 1 0.1009 -1.24
-
0.7628
1.02

33. FINAL EQUATION IN TERMS
OF CODED FACTORS
yield =
+79.94
+0.9950 *A
+0.5152 *B
+0.2500 *A*B
-1.38 *A²
-1.00 *B²

34. FINAL EQUATION IN TERMS
OF ACTUAL FACTORS
yield =
-1430.52285
+7.80749 *time
+13.27053 *temperature
+0.010000 *time * temperature
-0.055050 *time²
-0.040050 *temperature²

35. CONTOUR PLOT

36. RESPONSE SURFACE PLOT

37. LOCATION OF STATIONARY
POINT
















22
21
12
11
2
1
ˆ
2
ˆ
2
ˆ
ˆ
and
ˆ
ˆ






B
b
b
B
x 1
s



2
1

























306
.
0
389
.
0
515
.
0
995
.
0
0096
.
1
0917
.
0
0917
.
0
7345
.
0
2
1
s
x
















001
.
1
1250
.
0
1250
.
0
376
.
1
and
515
.
0
995
.
0
B
b
389
.
0
1 
s
x 306
.
0
2 
s
x Coded values

38. In terms of natural variables (time and
temperature);
LOCATION OF STATIONARY
POINT (cont.)
5
85
1
1



x
5
175
2
2



x
5
85
389
.
0 1 


5
175
306
.
0 2 


87
95
.
86
1 

 minut
es
F


 5
.
176
53
.
176
2

Actual values
CAN YOU CALCULATE ŷs (response at the stationary po

39. CHARACTERISTIC OF
RESPONSE SURFACE
Using canonical analysis, find the eigenvalues
1 and 2.:
which reduces to
0

 I
B 
0
001
.
1
1250
.
0
1250
.
0
376
.
1







0
3639
.
1
3788
.
2
2


 
 9634
.
0
1 


4141
.
1
2 


Thus, the canonical form of the fitted model is
2
2
2
1 4141
.
1
9634
.
0
21
.
80
ˆ w
w
y 


the {λi} are the
characteristic
roots of the
matrix B.
Matrix B minus
eigenvalues for
diagonal column

40. ATTENDANCE

41. DESIGN FOR FITTING THE
FIRST ORDER MODEL
Orthogonal first order
A first-order design is orthogonal if the cross products of the columns
of the X matrix equal to zero.
A design is orthogonal if the effects of any factor balance out (sum to
zero) across the effects of the other factors. This guarantees that the
effect of one factor or interaction can be estimated separately from the
effect of any other factor or interaction in the model.
Consist of 2k factorial and fractions of the 2k series in which main effects
are not aliased with each other. In using these designs, we assume that
the low and high levels of the k factors are coded to the usual ±1 levels.
The addition of center points to the 2k design can be used to estimate
pure error
 Another design: Simplex

42. DESIGN FOR FITTING THE SECOND
ORDER MODEL
1. Central Composite Design
Five level designs (-1, 1, 0, α , -α)
 Consists of
 2k factorial designs (or 2k- p fractional
factorial designs with resolution V)
with nF runs
 2k axial runs
 nc center runs
 Two parameters  and nc must be specified
where  is the distance of the axial runs
from the design center
 Generally, 3~5 central runs (nc ) are
recommended.
 Choose  which makes CCD “rotatable”

43. ROTATABILITY
 Rotatable means that the variance of the
predicted response at any point, x is the
same at all points that are the same distance
from the design center.
 The variance of the predicted response at
x:
 A design with this property will leave the
variance of the predicted response
unchanged when the design is rotated about
the center
 A CCD is rotatable when =(nf) ¼
*nf is the number of factorial runs
Concentric
circles
Contours of constant
standard deviation of
predicted response for

44. ROTATABILITY (cont.)
Determining for rotatability =(nf) 1/4

45. DESIGN FOR FITTING THE SECOND
ORDER MODEL
2. Box-Behnken Design
Three level designs (-1, 1, 0)
No corner points, and no star points (α)
Less number of runs compared to CCD

46. BOX-BEHNKEN DESIGN (cont.)
For three factors, the
CCD contains 14+nc
while the BBD contains
12+nc
BBD is used when one
is not interested in
predicting response at
the extremes.

47. BLOCKING IN RESPONSE
SURFACE DESIGN
For second-order design to block orthogonally, TWO
conditions must be satisfied:
1. Each block must be a first-order order orthogonal design:
Where 𝑥𝑖𝑢 and 𝑥𝑗𝑢 are the levels of ith and jth variables in the
uth run of the experiment with 𝑥𝑜𝑢 = 1 for all u.
𝑢=1
𝑛𝑏
𝑥𝑖𝑢𝑥𝑗𝑢 = 0 𝑖 ≠ 𝑗 = 0,1, … 𝑘 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑏

48. BLOCKING IN RESPONSE
SURFACE DESIGN
2. The fraction of the total sum of squares for each variable
contributed by every block must be equal to the fraction of the
total observations that occur in the block;
i= 1,2,…,k
Where N is the number of runs in the design
𝑢=1
𝑛𝑏
𝑥𝑖𝑢
2
𝑢−1
𝑁
𝑥𝑖𝑢
2 =
𝑛𝑏
𝑁

49. Example: Consider a rotatable CCD with k = 2 variables with N =
12 runs
BLOCKING IN RESPONSE
SURFACE DESIGN
Notice that the design has been
arranged in TWO blocks, with the
first block consisting of the
factorial portion of the design plus
two center points and the second
block consisting of the axial points
plus two additional center points.

50. BLOCKING IN RESPONSE
SURFACE DESIGN
Condition 1:
Block 1
Block 2
𝑢=1
𝑛𝑏
𝑥𝑖𝑢𝑥𝑗𝑢 = 0 𝑖 ≠ 𝑗 = 0,1, … 𝑘 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑏
−1 × −1 + 1 × −1 + −1 × 1 + 1 × 1 =0
1.414 × 0 + −1.414 × 0 + 0 × 1.414 + 0 × −1.414 =0
Condition 1 is met!

51. BLOCKING IN RESPONSE
SURFACE DESIGN
Condition 2:
𝑢=1
𝑛𝑏
𝑥𝑖𝑢
2
𝑢−1
𝑁
𝑥𝑖𝑢
2 =
𝑛𝑏
𝑁
Block 1
Therefore;
𝑢−1
𝑁
𝑥1𝑢
2
𝑢=1
𝑛1
𝑥1𝑢
2
𝑢=1
𝑛1
𝑥2𝑢
2
= −12 + 12 + −12 + 12 = 4 =
= −12 + 12 + −12 + 12 + 1.4142 + −1.4142 =8 =
𝑢=1
𝑁
𝑥2𝑢
2
4
8
=
6
12
Condition 2 is satisfied in
Block 1!

52. BLOCKING IN RESPONSE
SURFACE DESIGN
Condition 2:
𝑢=1
𝑛𝑏
𝑥𝑖𝑢
2
𝑢−1
𝑁
𝑥𝑖𝑢
2 =
𝑛𝑏
𝑁
Block 2
Therefore;
𝑢−1
𝑁
𝑥1𝑢
2
𝑢=1
𝑛2
𝑥1𝑢
2
𝑢=1
𝑛2
𝑥2𝑢
2
= 1.4142
+ −1.4142
=4 =
= −12 + 12 + −12 + 12 + 1.4142 + −1.4142 =8 =
𝑢=1
𝑁
𝑥2𝑢
2
4
8
=
6
12
Condition 2 is also satisfied in
Block 2. Thus, this design blocks
orthogonally.

54. ATTENDANCE
https://forms.gle/ux21ec2qz3sNmNL6
6