5.
Process Improvement
5.5. Advanced topics 5.5.3. How do you optimize a process? 5.5.3.2. Multiple response case


The mathematical programming approach maximizes or minimizes a primary response, subject to appropriate constraints on all other responses 
The analysis of multiple response systems usually involves some type
of optimization problem. When one response can be chosen as the
"primary", or most important response, and bounds or targets can be
defined on all other responses, a mathematical programming approach
can be taken. If this is not possible, the desirability approach
should be used instead.
In the mathematical programming approach, the primary response is maximized or minimized, as desired, subject to appropriate constraints on all other responses. The case of two responses ("dual" responses) has been studied in detail by some authors and is presented first. Then, the case of more than 2 responses is illustrated. 

Dual response systems  
Optimization of dual response systems 
The optimization of dual response systems (DRS) consists of finding
operating conditions x that
\( \begin{array}{ll} \mbox{subject to:} \hspace{.2in} & \hat{Y}_{s}(x) = T \\ & x'x \le \rho^{2} \end{array} \) 

Nonlinear programming software required for DRS  In a DRS, the response models \( \hat{Y}_{p} \) and \( \hat{Y}_{s} \) can be linear, quadratic or even cubic polynomials. A nonlinear programming algorithm has to be used for the optimization of a DRS. For the particular case of quadratic responses, an equality constraint for the secondary response, and a spherical region of experimentation, specialized optimization algorithms exist that guarantee global optimal solutions. In such a case, the algorithm DRSALG can be used (download from http://www.stat.cmu.edu/jqt/293), but a Fortran compiler is necessary.  
More general case  In the more general case of inequality constraints or a cubical region of experimentation, a general purpose nonlinear solver must be used and several starting points should be tried to avoid local optima. This is illustrated in the next section.  
Example for more than 2 responses  
Example: problem setup 
The values of three components
(x_{1}, x_{2}, x_{3})
of a propellant need to be selected to maximize a primary response,
burning rate (Y_{1}), subject to satisfactory
levels of two secondary reponses; namely, the variance of the burning
rate (Y_{2}) and the cost
(Y_{3}). The three components must add to 100%
of the mixture. The fitted models are:
\( \begin{array}{lcl} \hat{Y}_{2} & = & 3.88 x_{1} + 9.03 x_{2} + 13.63 x_{3}  0.1904 x_{1} x_{2} \\ & &  16.61 x_{1} x_{3}  27.67 x_{2} x_{3} \end{array} \) \( \hat{Y}_{3} = 23.13 x_{1} + 19.73 x_{2} + 14.73 x_{3} \) 

The optimization problem 
The optimization problem is therefore:


Solution 
The solution is
(x^{*})' = (0.212, 0.343, 0.443) which provides
\( \hat{Y}_{1} = 106.62 \),
\( \hat{Y}_{2} = 4.17 \), and
\( \hat{Y}_{3} = 18.23 \).
Therefore, both secondary responses are below the specified upper bounds.
The optimization should be implemented using a variety of starting points
to avoid local optima. Once again, confirmatory experiments should be
conducted at the estimated optimal operating conditions.
The solution to the optimization problem can be obtained using R code. 