RESPONSE SURFACE METHODOLOGY

DESIGN OF EXPERIMENT

1. ANIS ATIKAH AHMAD DESIGN OF EXPERIMENTS ERT 427 Response 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 Primary focus of previous chapters is factor screening. Two-level factorials, fractional factorials are widely used. Response Surface Methodology (RSM) is useful for the modeling and analysis of programs in which a response of interest is influenced by several variables. Objective of RSM is optimization For example: Find the levels of temperature (x 1 ) and pressure (x 2 ) to maximize the yield (y) of a process. The process yield: Then the surface is represented by

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 conducted to obtain a more precise estimate of the optimum. 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 2 2 factorial design with five centerpoints

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 (x 1 =0,x 2 =0) along the path of steepest ascent, we would move 0.775 units in the x 1 direction for every 0.325 units in the x 2 direction. If we decide to use 5 minutes of reaction time as the basic step size, then. Δx 1 = 1.0000 Δx 2 = 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( ξ 1 = 85, ξ 2 =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+1.00X 1 +0.50X 2 The region of exploration: ξ 1 is [80,90] AND ξ 2 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: 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 { λ i } 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 x s 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 x 1 and x 2 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 (x 1 ±0, x 2 ±1.414) and (x 1 ±1.414, x 2 ±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 80062.16 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 80090.90 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 Quadratic 0.2833 3 0.0944 1.78 0.2897 Suggested 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 Predicted R² PRESS Linear 1.37 0.3494 0.2193 -0.0435 29.99 2FI 1.43 0.3581 0.1441 -0.2730 36.59 Quadratic 0.2660 0.9828 0.9705 0.9184 2.35 Suggested Cubic 0.3141 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 significant 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 significant Pure Error 0.2120 4 0.0530 Cor Total 28.74 12

32. Coefficients in Terms of Coded Factors Factor Coefficient Estimate df Standard Error 95% CI Low 95% CI High VIF Intercept 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 *t ime +13.27053 *t emperature +0.010000 *time * temperature -0.055050 *time² -0.040050 *temperature²

35. Contour plot

36. Response surface plot

37. LOCATION OF STATIONARY POINT Coded values

38. In terms of natural variables (time and temperature); LOCATION OF STATIONARY POINT (cont.) minutes Actual values CAN YOU CALCULATE ŷ s ( response at the stationary point) ?

39. Characteristic of response surface Using canonical analysis, find the eigenvalues  1 and  2 .: which reduces to Thus, the canonical form of the fitted model is the { λ i } 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 2 k factorial and fractions of the 2 k 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 2 k 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 2 k factorial designs (or 2 k- p fractional factorial designs with resolution V) with n F runs 2 k axial runs n c center runs Two parameters  and n c must be specified where  is the distance of the axial runs from the design center Generally, 3~5 central runs ( n c ) 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  =( n f ) ¼ * n f is the number of factorial runs Concentric circles Contours of constant standard deviation of predicted response for the rotatable CCD

44. ROTATABILITY ( cont.) Determining for rotatability  =( n f ) 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 and are the levels of i th and j th variables in the u th run of the experiment with = 1 for all u.    

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  

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   + + =0   + + =0   Condition 1 is met!

51. BLOCKING IN RESPONSE SURFACE DESIGN Condition 2:   Block 1 Therefore;       +   + =8 =       Condition 2 is satisfied in Block 1!

52. BLOCKING IN RESPONSE SURFACE DESIGN Condition 2:   Block 2 Therefore;       4 =   + =8 =       Condition 2 is also satisfied in Block 2. Thus, this design blocks orthogonally .

53. ATTENDANCE https://forms.gle/ux21ec2qz3sNmNL66