DOW Experimental Design

DOW Experimental Design
State dimension:	1
Differential states:	11
Discrete control functions:	2
Path constraints:	4
Interior point equalities:	11

The DOW Experimental Design problem models the OED problem for the parameter estimation problem formulated by the DOW Chemical Co. in 1981. The following formulation is taken from "Nonlinear Parameter Estimation: a Case Study Comparison" by L. T. Biegler and J. J. Damiano.

The chemical species are disguised for proprietary reasons and the desired reaction is given by $HA+2BM\to AB+HBMH$, where AB is the desired product. The reactions are described as follows:

Slow Kinetic Reactions:

$\begin{array}{c}{M}^{-}+BM\underset{k_{-1}}{\leftarrow }\stackrel{k_1}{\to }MB{M}^{-}\\ A^{-}+BM\stackrel{k_2}{\to }AB{M}^{-}\\ M^{-}+AB\underset{k_{-3}}{\leftarrow }\stackrel{k_3}{\to }AB{M}^{-}\end{array}$

Acid-Base Reactions:

$\begin{array}{c}MBMH\leftarrow K_1\to MB{M}^{-}+H^{+}\\ HA\leftarrow K_2\to A^{-}+H^{+}\\ HABM\leftarrow K_2\to AB{M}^{-}+H^{+}\end{array}$

In order to devise a model to account for these reactions, it is first necessary to distinguish between the overall concentration of a species and the concentration of its neutral form. Overall con- centrations are defined for three components based on neutral and ionic species

$\begin{array}{c}\left[HBMH\right]=[(MBMH{)}_N]+\left[MB{M}^{-}\right]\\ \left[HA\right]=[(HA{)}_N]+\left[A^{-}\right]\\ \left[HABM\right]=[(HABM{)}_N]+\left[AB{M}^{-}\right]\end{array}$

Here $[]$ denotes the concentration of the species in $gmol/kg$. By assuming the rapid acid-base reactions are at equilibrium, the equilibrium constants $K_1,K_2,K_3$ can be defined as

$\begin{array}{c}{K}_1=\frac{[MB{M}^{-}][H^{+}]}{[(MBMH{)}_N]}\\ K_1=\frac{[A^{-}][H^{+}]}{[(HA{)}_N]}\\ K_1=\frac{[AB{M}^{-}][H^{+}]}{[(HABM{)}_N]}\end{array}$

The anionic species may then be represented by

$\begin{array}{cr}\left[MB{M}^{-}\right]=\frac{K_1[MBMH]}{K_1+[H^{+}]}& (a)\\ \left[A^{-}\right]=\frac{K_2[HA]}{K_2+[H^{+}]}& (b)\\ \left[AB{M}^{-}\right]=\frac{K_3[HABM]}{K_3+[H^{+}]}& (c)\end{array}$

Material balance equations for the three reactants in the slow kinetic reactions yield:

$\begin{array}{cr}\frac{d[M^{-}]}{dt}=-k_1\left[M^{-}\right]\left[BM\right]+k_{-1}\left[MB{M}^{-}\right]-k_3\left[M^{-}\right]\left[AB\right]+k_{-1}\left[AB{M}^{-}\right]& (d)\\ \frac{d[BM]}{dt}=-k_1\left[M^{-}\right]\left[BM\right]+k_{-1}\left[MB{M}^{-}\right]-k_2\left[A^{-}\right]\left[BM\right]& (e)\\ \frac{d[AB]}{dt}=-k_3\left[M^{-}\right]\left[AB\right]+k_{-3}\left[AB{M}^{-}\right]& (f)\end{array}$

From stoichiometry, rate expressions can also be written for the total species:

$\begin{array}{cr}\frac{d[MBMH]}{dt}=k_1\left[M^{-}\right]\left[BM\right]+k_{-1}\left[MB{M}^{-}\right]& (g)\\ \frac{d[HA]}{dt}=k_2\left[A^{-}\right]\left[BM\right]& (h)\\ \frac{d[HABM]}{dt}=k_2\left[A^{-}\right]\left[BM\right]+k_3\left[M^{-}\right]\left[AB\right]-k_{-3}\left[AB{M}^{-}\right]& (i)\end{array}$

An electroneutrality constraint gives the hydrogen ion con- centration $\left[H^{+}\right]$ as

$\left[H^{+}\right]+\left[Q^{+}\right]=\left[M^{-}\right]+\left[MB{M}^{-}\right]+\left[A^{-}\right]+\left[AB{M}^{-}\right](j)$

Based on similarities of reacting species, we assume

$k_3=k_1,k_{-3}=\frac{1}{2}{k}_{-1}$

With these assumptions, the three rate constants $k_1,k_2$ and $k_3$ must be estimated. Each of these can be fitted with two adjustable model parameters, assuming an Arrhenius temperature dependence. That is

$\begin{array}{c}{k}_i={\alpha }_i\exp (-E_i/(RT)),i\in \{-1,1,2\}\end{array}$

Here $R$ is the gas constant, $T$ is reaction temperature in Kelvins and the parameters ${\alpha }_i,E_i$ represent the pre-exponential factor and activation energy, respectively, for the appropriate rate constant.

Mathematical formulation

The chemical processes $(a)-(j)$ can be expressed mathematically as six differential equations and four algebraic equations:

$\begin{array}{lr}\frac{d{y}_1}{dt}=-k_2{y}_8{y}_2& (1),(h)\\ \frac{d{y}_2}{dt}=-k_1{y}_6{y}_2+k_{-1}{y}_{10}-k_2{y}_8{y}_2& (2),(e)\\ \frac{d{y}_3}{dt}=-k_2{y}_8{y}_2+k_1{y}_6{y}_4-\frac{1}{2}{k}_{-1}{y}_9& (3),(i)\\ \frac{d{y}_4}{dt}=-k_1{y}_6{y}_4+\frac{1}{2}{k}_{-1}{y}_9& (4),(f)\\ \frac{d{y}_5}{dt}=-k_1{y}_6{y}_2+k_{-1}{y}_{10}& (5),(g)\\ \frac{d{y}_6}{dt}=-k_1(y_6{y}_2+y_6{y}_4)+k_{-1}(y_{10}+\frac{1}{2}{y}_9)& (6),(d)\\ y_7=-\left[Q^{+}\right]+y_6+y_8+y_9+y_{10}& (7),(j)\\ y_8=\frac{{\theta }_8{y}_1}{{\theta }_8+y_7}& (8),(b)\\ y_9=\frac{{\theta }_9{y}_3}{{\theta }_9+y_7}& (9),(c)\\ y_{10}=\frac{{\theta }_7{y}_5}{{\theta }_7+y_7}& (10),(a)\end{array}$

Here the letter stands for the corresponding chemical process. The nine parameters form the vector

$\theta =({\alpha }_1,E_1,{\alpha }_2,E_2,{\alpha }_{-1},E_{-1},K_1,K_2,K_3)$

The predicted concentrations form the vector

$y=(HA,BM,HABM,AB,MBMH,M^{-},H^{+},A^{-},AB{M}^{-},MB{M}^{-})$

Let $f_k(\cdot )$ denote the right hand side of equation $(k)$ for $k=1,\dots ,6$. Moreover, let

$\begin{array}{l}{f}_{ymn}(\cdot )=\frac{\partial f_m(\cdot )}{\partial y_n},m,n\in \{1,\dots ,6\}\\ f_{\theta mn}(\cdot )=\frac{\partial f_m(\cdot )}{\partial y_n}\end{array}$

The non-zero derivatives are given by

$\begin{array}{l}{f}_{y12}=-k_2{y}_8\\ f_{y18}=-k_2{y}_2\\ f_{y22}=-k_1{y}_6-k_2{y}_8\\ f_{y26}=-k_1{y}_2\\ f_{y28}=-k_2{y}_2\\ f_{y32}=-k_2{y}_8\\ f_{y34}=k_1{y}_6\\ f_{y36}=k_1{y}_4\\ f_{y44}=-k_1{y}_6\\ f_{y46}=-k_1{y}_4\\ f_{52}=-k_1{y}_6\\ f_{y56}=-k_1{y}_2\\ f_{y62}=-k_1{y}_6\\ f_{y64}=-k_1{y}_6\\ f_{y66}=-k_1(y_2+y_4)\end{array}$

We are interested in when to measure, with an upper bound $M$ on the measuring time. We can measure the states directly, $h^1(x(t))=x_1(t)$ and $h^2(x(t))=x_2(t)$. We use two different sampling functions, $w^1(\cdot )$ and $w^2(\cdot )$ in the same experimental setting. This can be seen either as a two-dimensional measurement function $h(x(t))$, or as a special case of a multiple experiment, in which $u(\cdot ),x(\cdot )$, and $G(\cdot )$ are identical. The experimental design problem then reads

Parameters

We use $t_f=12$, $p_1=p_2=p_3=p_4=1$, and $p_5=0.4$, $p_6=0.2$. The upper bound on the measurement time intervals is chosen as $M=4$.

Miscellaneous and Further Reading

The Lotka Volterra fishing problem was introduced by Sebastian Sager in a proceedings paper [Sager2006]Address: Heidelberg
Author: S. Sager; H.G. Bock; M. Diehl; G. Reinelt; J.P. Schl\"oder
Booktitle: Recent Advances in Optimization
Editor: A. Seeger
Note: ISBN 978-3-5402-8257-0
Pages: 269--289
Publisher: Springer
Series: Lectures Notes in Economics and Mathematical Systems
Title: Numerical methods for optimal control with binary control functions applied to a Lotka-Volterra type fishing problem
Volume: 563
Year: 2009
and revisited in his PhD thesis [Sager2005]Address: Tönning, Lübeck, Marburg
Author: S. Sager
Editor: ISBN 3-89959-416-9
Publisher: Der andere Verlag
Title: Numerical methods for mixed--integer optimal control problems
Url: http://mathopt.de/PUBLICATIONS/Sager2005.pdf
Year: 2005
. These are also the references to look for more details. The experimental design problem has been described in the habilitation thesis of Sager, [Sager2011d]Author: S. Sager
How published: University of Heidelberg
Month: August
Note: Habilitation
Title: On the Integration of Optimization Approaches for Mixed-Integer Nonlinear Optimal Control
Url: http://mathopt.de/PUBLICATIONS/Sager2011d.pdf
Year: 2011
.

References

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