5. Process Improvement
5.5.3. How do you optimize a process?
5.5.3.1. Single response case

## Single response: Choosing the step length

A procedure for choosing how far along the direction of steepest ascent to go for the next trial run Once the search direction is determined, the second decision needed in Phase I relates to how far in that direction the process should be "moved". The most common procedure for selecting a step length is based on choosing a step size in one factor and then computing step lengths in the other factors proportional to their parameter estimates. This provides a point on the direction of maximum improvement. The procedure is given below. A similar approach is obtained by choosing increasing values of $$\rho$$ in
$$x_{i}^{*} = \rho \frac{b_{i}}{\sqrt{\sum_{j=1}^{k}{b_{j}^{2}}}} \hspace{.3in} i = 1, 2, \ldots , k$$
However, the procedure below considers the original units of measurement which are easier to deal with than the coded "distance" $$\rho$$.
Procedure: selection of step length
Procedure for selecting the step length The following is the procedure for selecting the step length.
1. Choose a step length $$\Delta X_{j}$$ (in natural units of measurement) for some factor j. Usually, factor j is chosen to be the one engineers feel more comfortable varying, or the one with the largest |bj|. The value of $$\Delta X_{j}$$ can be based on the width of the confidence cone around the steepest ascent/descent direction. Very wide cones indicate that the estimated steepest ascent/descent direction is not reliable, and thus $$\Delta X_{j}$$ should be small. This usually occurs when the R2 value is low. In such a case, additional experiments can be conducted in the current experimental region to obtain a better model fit and a better search direction.
2. Transform to coded units:

$$\Delta x_{j} = \frac{\Delta X_{j}}{s_{j}}$$

with sj denoting the scale factor used for factor j (e.g., sj = rangej/2).

3. Set $$\Delta x_{i} = \frac{b_{i}}{b_{j}} \Delta x_{j}$$ for all other factors i.
4. Transform all the $$\Delta x_{i}$$'s to natural units:
$$\Delta X_{i} = (\Delta x_{i})(s_{i})$$.
Example: Step Length Selection.
An example of step length selection The following is an example of the step length selection procedure.
• For the chemical process experiment described previously, the process engineer selected $$\Delta X_{2} = 50$$ minutes. This was based on process engineering considerations. It was also felt that $$\Delta X_{2} = 50$$ does not move the process too far away from the current region of experimentation. This was desired since the R2 value of 0.6580 for the fitted model is quite low, providing a not very reliable steepest ascent direction (and a wide confidence cone, see Technical Appendix 5B).
• $$\Delta x_{2} = \frac{50}{50} = 1.0$$
• $$\Delta x_{1} = \frac{-1.2925}{11.14} = -0.1160$$
• $$\Delta X_{2} = (-0.1160)(30) = -3.48^{\circ}C$$
Thus the step size is $$\Delta X$$' = (-3.48oC, 50 minutes).
Procedure: Conducting Experiments Along the Direction of Maximum Improvement
Procedure for conducting experiments along the direction of maximum improvement The following is the procedure for conducting experiments along the direction of maximum improvement.
1. Given current operating conditions $$X_{0}^{'} = (X_{1}, X_{2}, \dots , X_{k})$$ and a step size $$\Delta X' = (\Delta X_{1}, \Delta X_{2}, \dots , \Delta X_{k})$$, perform experiments at factor levels $$X_{0} + \Delta X, X_{0} + 2 \Delta X, X_{0} + 3 \Delta X, \dots$$ as long as improvement in the response Y (decrease or increase, as desired) is observed.
2. Once a point has been reached where there is no further improvement, a new first-order experiment (e.g., a 2k-p fractional factorial) should be performed with repeated center runs to assess lack of fit. If there is no significant evidence of lack of fit, the new first-order model will provide a new search direction, and another iteration is performed as indicated in Figure 5.3. Otherwise (there is evidence of lack of fit), the experimental design is augmented and a second-order model should be fitted. That is, the experimenter should proceed to "Phase II".
Example: Experimenting Along the Direction of Maximum Improvement
Step 1: increase factor levels by $$\delta$$ Step 1:

Given X0 = (200oC, 200 minutes) and $$\Delta X$$ = (-3.48oC, 50 minutes), the next experiments were performed as follows (the step size in temperature was rounded to -3.5oC for practical reasons):

X1 X2 x1 x2 Y (= yield)
X0 200 200 0 0
X0 + $$\Delta X$$ 196.5 250 -0.1160 1 56.2
X0 + $$2 \Delta X$$ 193.0 300 -0.2320 2 71.49
X0 + $$3 \Delta X$$ 189.5 350 -0.3480 3 75.63
X0 + $$4 \Delta X$$ 186.0 400 -0.4640 4 72.31
X0 + $$5 \Delta X$$ 182.5 450 -0.5800 5 72.10
Since the goal is to maximize Y, the point of maximum observed response is X1 = 189.5oC, X2 = 350 minutes. Notice that the search was stopped after 2 consecutive drops in response, to assure that we have passed by the "peak" of the "hill".
Step 2: new factorial experiment Step 2:

A new 22 factorial experiment is performed with X' = (189.5, 350) as the origin. Using the same scaling factors as before, the new scaled controllable factors are:

$$x_{1} = \frac{X_{1} - 189.5} {30} \hspace{.5in}$$ and $$\hspace{.5in} x_{2} = \frac{X_{2} - 350} {50}$$
Five center runs (at X1 = 189.5, X2 = 350) were repeated to assess lack of fit. The experimental results were:
x1 x2 X1 X2 Y (= yield)
-1 -1 159.5 300 64.33
+1 -1 219.5 300 51.78
-1 +1 159.5 400 77.30
+1 +1 219.5 400 45.37
0 0 189.5 350 62.08
0 0 189.5 350 79.36
0 0 189.5 350 75.29
0 0 189.5 350 73.81
0 0 189.5 350 69.45

The corresponding ANOVA table for a linear model is

                SUM OF            MEAN     F
SOURCE         SQUARES    DF     SQUARE  VALUE  PROB > F

MODEL          505.376     2    252.688  4.731   0.0703
CURVATURE      336.364     1    336.364  6.297   0.0539
RESIDUAL       267.075     5     53.415
LACK OF FIT   93.896     1     93.896  2.168   0.2149
PURE ERROR   173.179     4     43.295

COR TOTAL     1108.815     8

From the table, the linear effects (model) are significant and there is no evidence of lack of fit. However, there is a significant curvature effect (at the 5.4 % significance level), which implies that the optimization should proceed with Phase II; that is, the fit and optimization of a second-order model.