5.
Process Improvement
5.5. Advanced topics 5.5.5. How do you optimize a process? 5.5.5.1. Single response case
|
|||
Starting
at the current operating conditions, fit a linear model, determine the
directions of steepest ascent and continue experimenting until no further
improvement occurs - then iterate the process
Flow chart of iterative search process
|
If experimentation is initially
performed in a new, poorly understood production process, chances are that
the initial operating conditions X1, X2, ...,Xk
are located far from the region where the factors achieve a maximum or
minimum for the response of interest
Y. A first order model will
serve as a good local approximation in a small region close to the initial
operating conditions and far from where the process exhibits curvature.
Therefore, it makes sense to fit a simple first order (or linear polynomial)
model of the form:
Experimental strategies for fitting this type of models were discussed earlier. Usually, a 2k-p fractional factorial experiment is conducted with repeated runs at the current operating conditions (which serve as the origin of coordinates in orthogonally coded factors). The idea behind "Phase I'' is to keep experimenting along the direction of steepest ascent (or descent, as required) until there is no further improvement in the response. At that point, a new fractional factorial experiment with center runs is conducted to determine a new search direction. This process is repeated until at some point significant curvature in is detected. This implies that the operating conditions X1, X2, ...,Xk are close to where the maximum (or minimum, as required) of Y occurs. When significant curvature, or lack of fit, is detected, the experimenter should proceed with "Phase II". Figure 5.2 illustrates a sequence of line searches when seeking a region where curvature exists in a problem with 2 factors (i.e., k=2). FIGURE 5.2: A Sequence of Line Searches for a 2 Factor Optimization Problem
|
||
The direction of steepest ascent is determined by the gradient of the fitted model and depends on the scaling convention - equal variance scaling is recommended | Procedure for Finding the Direction
of Maximum Improvement.
Suppose a first order model like above has been fitted and provides a useful approximation. As long as lack of fit (due to pure quadratic curvature and interactions) is small compared to the main effects, steepest ascent can be attempted. To determine the direction of maximum improvement we use
This coding convention is recommended since it provides better parameter estimates, and therefore, a more reliable search direction. The coordinates of the factor settings on the direction of steepest ascent separated a distance from the origin are given by:
This problem can be solved with the aid of an optimization solver (e.g. like the solver option of a spreadsheet). However, in this case this is not really needed, as the solution is a simple equation which yields the coordinates An engineer can compute this equation for different increasing values of and get different factor settings all on the steepest ascent direction. To see the details of why this equation is true, see Technical Appendix 5A.
|
||
Optimization by search example | Example.
Optimization of a Chemical Process.
It has been concluded (perhaps after a factor screening experiment) that the yield (Y, in %) of a chemical process is mainly affected by the temperature (, in C ) and by the reaction time (, in minutes). Due to safety reasons, the region of operation is limited to
The process is currently run at a temperature of 200 C and a reaction time of 200 minutes. A process engineer decides to run a full factorial experiment with factor levels at
Five repeated runs at the center levels are conducted to assess lack of fit. The orthogonally coded factors are
The experimental results were:
The corresponding ANOVA table for a first order polynomial model, obtained using the DESIGN EASE statistical software, is SUM OF MEAN F SOURCE SQUARES DF SQUARE VALUE PROB>F MODEL 503.3035 2 251.6517 4.810 0.0684 CURVATURE 8.1536 1 8.1536 0.1558 0.7093 RESIDUAL 261.5935 5 52.3187 LACK OF FIT 37.6382 1 37.6382 0.6722 0.4583 PURE ERROR 223.9553 4 55.9888 COR TOTAL 773.0506 8It can be seen from the ANOVA table that there is no significant lack of linear fit due to an interaction term, and there is no evidence of curvature. Furthermore, there is evidence that the first order model is significant. Using the DESIGN EXPERT statistical software we obtain the resulting model (in the coded variables) as
Usual diagnostic checks show conformance to the regression assumptions, although the R2 value is not very high: R2 = 0.6580. To maximize, we use the direction of steepest ascent. The engineer selects , since a point on the steepest ascent direction distanced one unit (in the coded units) from the origin is desired. Then from the equation above for the predicted Y response, the coordinates of the factor levels for the next run are given by:
and
This means that to improve the process, for every C
that temperature is varied (decreased), the reaction time should be varied
by minutes.
|
||
Details of how to determine the path of steepest ascent |
Technical Appendix 5A: finding the factor settings on the steepest ascent direction a distance from originThe problem of finding the factor settings on the steepest ascent/descent direction that are located at a distance from origin is given by the optimization problem,
To solve it, use a Lagrange multiplier. First, add a penalty for solutions not satisfying the constraint (since we want a direction of steepest ascent, we maximize, and therefore the penalty is negative. For steepest descent we minimize and the penalty term is added instead):
Compute the partials and equate them to zero
These two equations have two unknowns (the vector and the scalar) and thus can be solved yielding the desired solution:
or, in non-vector notation:
From this equation, we can see that any multiple of the direction of the gradient (given by ) will lead to points on the steepest ascent direction. For steepest descent, use instead in the numerator of the equation above. |