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

## Multiple responses: The mathematical programming approach

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
$$\mbox{optimize:} \hspace{.5in} \hat{Y}_{p}(x)$$

$$\begin{array}{ll} \mbox{subject to:} \hspace{.2in} & \hat{Y}_{s}(x) = T \\ & x'x \le \rho^{2} \end{array}$$

with T denoting the target value for the secondary response, p the number of primary responses (i.e., responses to be optimized), s the number of secondary responses (i.e., responses to be constrained), and ρ is the radius of a spherical constraint that limits the region in the controllable factor space where the search should be undertaken. The value of ρ should be chosen with the purpose of avoiding solutions that extrapolate too far outside the region where the experimental data were obtained. For example, if the experimental design is a central composite design, choosing ρ = α (axial distance) is a logical choice. Bounds of the form LxiU can be used instead if a cubical experimental region were used (e.g., when using a factorial experiment). Note that a Ridge Analysis problem is related to a DRS problem when the secondary constraint is absent. Thus, any algorithm or solver for DRS's will also work for the Ridge Analysis of single response systems.
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/29-3), 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 (x1, x2, x3) of a propellant need to be selected to maximize a primary response, burning rate (Y1), subject to satisfactory levels of two secondary reponses; namely, the variance of the burning rate (Y2) and the cost (Y3). The three components must add to 100% of the mixture. The fitted models are:
$$\begin{array}{lcl} \hat{Y}_{1} & = & 35.4 x_{1} + 42.77 x_{2} + 70.36 x_{3} + 16.02 x_{1} x_{2} \\ & & + 36.33 x_{1} x_{3} + 136.8 x_{2} x_{3} + 854.9 x_{1} x_{2} x_{3} \end{array}$$

$$\begin{array}{lcl} \hat{Y}_{2} & = & 3.88 x_{1} + 9.03 x_{2} + 13.63 x_{3} - 0.1904 x_{1} x_{2} \\ & & - 16.61 x_{1} x_{3} - 27.67 x_{2} x_{3} \end{array}$$

$$\hat{Y}_{3} = 23.13 x_{1} + 19.73 x_{2} + 14.73 x_{3}$$

The optimization problem The optimization problem is therefore:

 maximize $$\hat{Y}_{1}(x)$$ subject to: $$\begin{array}{c} \hat{Y}_{2}(x) \le 4.5 \\ \hat{Y}_{3}(x) \le 20 \\ x_{1} + x_{2} + x_{3} = 1 \\ 0 \le x_{1} \le 1 \\ 0 \le x_{2} \le 1 \\ 0 \le x_{3} \le 1 \end{array}$$

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.