1. RESPONSE SURFACE METHODOLOGY (R S M)
Par
Mariam MAHFOUZ

2. Remember that: General Planning
Part I
A - Introduction to the RSM method
B - Techniques of the RSM method
C - Terminology
D - A review of the method of least squares
Part II
A - Procedure to determine optimum
conditions – Steps of the RSM method
B – Illustration of the method with an example

3. Part II

4. A - Procedure to determine optimum conditions steps of the method
This method permits to find the settings of the input variables which produce the most desirable response values.
The set of values of the input variables which result in the most desirable response values is called the set of optimum conditions.

5. Steps of the method
The strategy in developing an empirical model through a sequential program of experimentation is as follows:
The simplest polynomial model is fitted to a set of data collected at the points of a first-order design.
If the fitted first-order model is adequate, the information provided by the fitted model is used to locate areas in the experimental region, or outside the experimental region, but within the boundaries of the operability region, where more desirable values of the response are suspected to be.

6. In the new region, the cycle is repeated in that the first-order model is fitted and testing for adequacy of fit.
If nonlinearity in the surface shape is detected through the test for lack of fit of the first-order model, the model is upgraded by adding cross-product terms and / or pure quadratic terms to it. The first-order design is likewise augmented with points to support the fitting of the upgraded model.

7. If curvature of the surface is detected and a fitted second-order model is found to be appropriate, the second-order model is used to map or describe the shape of the surface, through a contour plot, in the experimental region.
If the optimal or most desirable response values are found to be within the boundaries of the experimental region, then locating the best values as well as the settings of the input variables that produce the best response values.

8. 7. Finally, in the region where the most desirable response values are suspected to be found, additional experiments are performed to verify that this is so.

9. B- Illustration of the method with an example
For simplicity of presentation we shall assume that there is only one response variable to be studied although in practice there can be several response variables that are under investigation simultaneously.

10. Two controlled Factors
Experience
Two controlled Factors
Chemical reaction
One response
Temperature (X1)
percent yield
Time (X2)
An experimenter, interested in determining if an increase in the percent yield is possible by varying the levels of the two factors.

11. And two repetitions at each point
Two levels of temperature: 70° and 90°.
Two levels of time: 30 sec and 90 sec.
Four different design points
Four temperature-time settings
(factorial combinations)
And two repetitions at each point
The total number of observations is N = 8

12. Detail
The response of interest is the percent yield, which is a measure of the purity of the end product.
The process currently operates in a range of percent purity between 55 % and 75 %, but it is felt that a higher percent yield is possible.

13. x1 and x2 are the coded variables which are defined as:
Design 1
Original variables
Coded variables
Percent yield
Temperature
X1 (C°)
Time
X2 (sec.) x1 x2 Y 70 30 -1
49.8
48.1 90 1
57.3
52.3
65.7
69.4
73.1
77.8
x1 and x2 are the coded variables which are defined as:

14. Representation of the first design

15. The remaining term, , represents random error in the yield values.
First-order model
Expressed in terms of the coded variables, the observed percent yield values are modeled as:
The remaining term, , represents random error in the yield values.
The eight observed percent yield values, when expressed as function of the levels of the coded variables, in matrix notation, are:
Y = X  + 

16. Vector of response values
Matrix form
Vector of error terms
Vector of response values
Matrix of the design =
Vector of unknown parameters

17. Estimations
The estimates of the coefficients in the first-order model are found by solving the normal equations:
The estimates are:
The fitted first-order model in the coded variables is:

18. ANOVA table – design 1 Source Degrees of freedom d.f. Sum of squaresSS
Mean square F
Model 2
63.71
Residual 5
6.7873
Lack of fit 1
2.1013
0.264
Pure error 4
7.9588
Test of
adequacy

19. Individual tests of parameters
To do that the Student-test is used.
For the test of: we have
And for we have
Each of the null hypotheses is rejected at the  = 0.05 level of significance owing to the calculated values, 3.73 and 10.65, being greater in absolute value than the tabled value,
T5;0.025 =2.571.

20. Conclusion of the first analysis
The first – order model is adequate.
That both temperature and time have an effect on percent yield.
Since both b1 and b2 are positive, the effects are positive.
Thus, by raising either the temperature or time of reaction, this produced a significant increase in percent yield.

21. Second stage of the sequential program
At this point, the experimenter quite naturally might ask:
If additional experiments can be performed
At what settings of temperature and time should the additional experiments be run?”
To answer this question, we enter the second stage of our sequential program of experimentation.

22. Contour plots The fitted model:
can now be used to map values of the estimated response surface over the experimental region.
This response surface is a hyper-plane; their contour plots are lines in the experimental region.
The contour lines are drawn by connecting two points (coordinate settings of x1 and x2) in the experimental region that produce the same value of

23. In the figure above are shown the contour lines of the estimated planar surface for percent yield corresponding to values of = 55, 60, 65 and 70 %.

24. Performing experiments along the path of steepest ascent
To describe the method of steepest ascent mathematically, we begin by assuming the true response surface can be approximated locally with an equation of a hyper-plane
Data are collected from the points of a first-order design and the data are used to calculate the coefficient estimates to obtain the fitted first-order model
Estimated response function

25. subject to the constraint
The next step is to move away from the center of the design, a distance of r units, say, in the direction of the maximum increase in the response.
By choosing the center of the design in the coded variable to be denoted by O(0, 0, …, 0), then movement away from the center r units is equivalent to find the values of which maximize
subject to the constraint
Maximization of the response function is performed by using Lagrange multipliers. Let
where  is the Lagrange multiplier.

27. To maximize subject to the above-mentioned constraint, first we set equal to zero the partial derivatives
i=1,…,k and
Setting the partial derivatives equal to zero produces:
i =1,…,k, and
The solutions are the values of xi satisfying
or i = 1,…,k, where the value of  is yet to be determined. Thus the proposed next value of xi is directly proportional to the value of bi.

28. Let us the change in Xi be noted by i , and the change in xi be noted by i. The coded variables is obtained by these formulas where
(respectively si) is the mean (respectively the standard deviation) of the two levels of Xi .
Thus , then or

29. Let us illustrate the procedure with the fitted first-order model:
that was fitted early to the percent yield values in our example.
To the change in X2, 2=45 sec. corresponds the change in x2, 2=45/30=1.5 units.
In the relation , we can substitute i to xi:
, thus and 1 = 0.526, so
1=0.526*10=5.3°C .

30. Observed percent yield
Points along the path of steepest ascent and observed percent yield values at the points
Temperature
X1 (°C)
Time
X2 (sec.)
Observed percent yield
Base
80.0 60 i
5.3 45
Base + i
85.3
105
74.3
Base i
87.95
127.5
78.6
Base + 2 I
90.6
150
83.2
Base + 3 I
95.9
195
84.7
Base + 4 i
101.2
240
80.1

32. Design two: For this design the coded variables are defined as:
Sequence of experimental trials performed in moving to a region of high percent yield values
Design two: For this design the coded variables are defined as: x1 x2 X1 X2
% yield -1
85.9
165
82.9; 81.4 +1
105.9
87.4; 89.5
225
74.6; 77.0
84.5; 83.1
95.9
195
84.7; 81.9

34. The model is jugged adequate
The fitted model corresponding to the group of experiments of design two is:
The corresponding analysis of variance is:
Source
d.f. SS MS F
Model 2
81.372
42.34
Residual 7
13.455
1.922
Lack of fit
2.345
1.173
0.53
Pure error 5
11.110
2.222
Total (variations) 9
176.2
The model is jugged adequate

35. sequence of experimental trials that were performed in the direction two:
Steps x1 x2 X1 X2
% yield 1
Center + i +1
- 0.77
105.9
171.9
89.0 2
Center + 2 i +2
- 1.54
115.9
148.8
90.2 3
Center + 3 i +3
- 2.31
125.9
125.7
87.4 4
Center + 4 i +4
- 3.08
135.9
102.6
82.6

37. Retreat to center + 2 i and proceed in direction three
Steps x1 x2 X1 X2
% yield 5
Replicated 2 +2
- 1.54
115.9
148.8
91.0 6 +3
- 0.77
125.9
171.9
93.6 7 +4
135.9
195
96.2 8 +5
0.77
145.9
218.1
92.9

38. Set up design three using points of steps 6, 7, and 8 along with the following two points:x1 x2 X1 X2
% yield 9 +3
0.77
125.9
218.1
91.7 10 +5
- 0.77
145.9
171.9
92.5 11
Replicated 7 +4
135.9
195
97.0

40. Design three was set up using the point at step 7 as its center
Design three was set up using the point at step 7 as its center. It includes steps 6 – 11. If we redefine the coded variables:
and
then the fitted first-order model is:
The corresponding analysis of variance table is:

41. ANOVA table source d.f. SS MS F Model 2 0.5650 0.2825 0.04 Residual 3
7.3944
total 5
It is obvious from the ANOVA table that the least model does not explain a significant amount of the overall variation in the percent yield values, and it is necessary to fit a curved surface.

42. Fitting a second-order model
A second-order model in k variables is of the form:
The number of terms in the model above is p=(k+1)(k+2)/2; for example, when k=2 then p=6.
Let us return to the chemical reaction example of the previous section. To fit a second-order model (k=2), we must perform some additional experiments.

43. Central composite rotatable design
Suppose that four additional experiments are performed, one at each of the axial settings (x1,x2):
These four design settings along with the four factorial settings: (-1,-1); (-1,1); (1,-1); (1,1) and center point comprise a central composite rotatable design.
The percent yield values and the corresponding nine design settings are listed in the table below:

44. Central composite rotatable design

45. Percent yield values at the nine points of a central composite rotatable design

46. The fitted second-order model, in the coded variables, is:
The analysis is detailed in this table, using the RSREG procedure in the SAS software:

47. SAS output 1

48. SAS output 2

49. Response surface and the contour plot

50. More explanations
The contours of the response surface, showing above, represent predicted yield values of 95.0 to 96.5 percent in steps of 0.5 percent.
The contours are elliptical and centered at the point
(x1; x2)=( ; )
or (X1; X2)=(135.85°C; sec).
The coordinates of the centroid point are called the coordinates of the stationary point.
From the contour plot we see that as one moves away from the stationary point, by increasing or decreasing the values of either temperature or time, the predicted percent yield (response) value decreases.

52. Determining the coordinates of the stationary point
A near stationary region is defined as a region where the surface slopes (or gradients along the variables axes) are small compared to the estimate of experimental error.
The stationary point of a near stationary region is the point at which the slope of the response surface is zero when taken in all direction.
The coordinates of the stationary point
are calculated by differentiating the estimated response equation with respect to each xi, equating these derivatives to zero, and solving the resulting k equations simultaneously.

53. Remember that the fitted second-order model in k variables is:
To obtain the coordinates of the stationary point, let us write the above model using matrix notation, as:

54. where
and

55. Some details
The partial derivatives of with respect to x1, x2, …, xk are :

56. More details
Setting each of the k derivatives equal to zero and solving for the values of the xi, we find that the coordinate of the stationary point are the values of the elements of the kx1 vector x0 given by:
At the stationary point, the predicted response value, denoted by , is obtained by substituting x0 for x:

57. Return to our example
The fitted second-order model was:
so the stationary point is:
In the original variables, temperature and time of the chemical reaction example, the setting at the stationary point are: temperature=135.85°C and time= sec.
And the predicted percent yield at the stationary point is:

58. Moore details
Note that the elements of the vector x0 do not tell us anything about the nature of the surface at the stationary point.
This nature can be a minimum, a maximum or a mini_max point.
For each of these cases, we are assuming that the stationary point is located inside the experimental region.
When, on the other hand, the coordinates of the stationary point are outside the experimental region, then we might have encountered a rising ridge system or a falling ridge system, or possibly a stationary ridge.

60. Nest Step
The next step is to turn our attention to expressing the response system in canonical form so as to be able to describe in greater detail the nature of the response system in the neighborhood of the stationary point.

61. The canonical Equation of a Second-Order Response System
The first step in developing the canonical equation for a k-variable system is to translate the origine of the system from the center of the design to the stationary point, that is, to move from (x1,x2,…,xk)=(0,0,…,0) to x0.
This is done by defining the intermediate variables (z1,z2,…,zk)=(x1-x10,x2-x20,…,xk-xk0) or z=x-x0.
Then the second-order response equation is expressed in terms of the values of zi as:

62. This transformation is a rotation of the zi axes to form the wi axes.
Now, to obtain the canonical form of the predicted response equation, let us define a set of variables w1,w2,…,wk such that W’=(w1,w2,…,wk) is given by
where M is a kxk orthogonal matrix whose columns are eigenvectors of the matrix B.
The matrix M has the effect of diagonalyzing B, that is, where 1,2,…,k are the corresponding eigenvalues of B.
The axes associated with the variables w1,w2,…,wk are called the principal axes of the response system.
This transformation is a rotation of the zi axes to form the wi axes.

64. It is easy to see that if 1,2,…,k are:
So we obtain the canonical equation:
The eigenvalues i are real-valued (since the matrix B is a real-valued, symmetric matrix) and represent the coefficients of the terms in the canonical equation.
It is easy to see that if 1,2,…,k are:
1) All negative, then at x0 the surface is a maximum.
2) All positive, then at x0 the surface is a minimum.
3) Of mixed signs, that is, some are positive and the
others are negative, then x0 is a saddle point of
the fitted surface.
The canonical equation for the percent yield surface is:

65. Moore details
The magnitude of the individual values of the i tell how quickly the surface height changes along the Wi axes as one moves away from x0.
Today there are computer software packages available that perform the steps of locating the coordinates of the stationary point, predict the response at the stationary point, and compute the eigenvalues and the eigenvectors.

66. For example, the solution for optimum response generated from PROC RSREG of the Statistical Analysis System (SAS) for the chemical reaction data, is in following table:

67. Contours and optimal direction New series of experiments
Recapitulate
Process to
optimize
Contours and optimal direction
Input and output
variables
Experiments in the
Optimal direction
Experimental and
Operational regions
Locate a new
Experimental region
Series of experiments
New series of experiments
Yes
Fitting
First-order
model
Fitting
First-order
model
Model
Adequate ?
Model
Adequate ?
Yes
Fitting a
Second-order
model No No

68. Doing this: Coordinates of the stationary point.
Description of the shape of the response surface near the stationary point by contour plots.
Canonical analysis.
If needed, Ridge analysis (not detailed here).

69. Field of use of the method
In agriculture
In food industry
In pharmaceutical industry
In all kinds of the light and heavy industries
In medical domain
Etc….

70. Bibliography
André KHURI and John CORNELL: “Response Surfaces – Designs and Analyses” , Dekker, Inc., ASQC Quality Press, New York.
Irwin GUTTMAN: “Linear Models: An Introduction”, John Wiley & Sons, New York.
George BOX, William HUNTER & J. Stuart HUNTER: “Statistics for experimenters: An Introduction to Design, Data Analysis, and Model Building” , John Wiley & Sons, New York.
George BOX & Norman DRAPPER: “Empirical Model-Building and Response Surfaces” , John Wiley & Sons, New York.

71. Thank you
Questions?