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Subsections


14 Catalyst Mixing

Determine the optimal mixing policy of two catalysts along the length of a tubular plug flow reactor involving several reactions.

Formulation

The nonlinear model [23] that describes the reactions is

$\displaystyle x_1'(t)$ $\textstyle =$ $\displaystyle u(t) (10 x_2(t)-x_1(t))$ (11)
$\displaystyle x_2'(t)$ $\textstyle =$ $\displaystyle u(t) (x_1(t)-10 x_2(t))-(1-u(t)) x_2(t).$  

Initial conditions for (14.1) are $ x_1 (0) = 1 $ and $ x_2 (0) = 0 $. The control variable $u$ represents the mixing ratio of the catalysts and must satisfy the bounds

\begin{displaymath}
0 \leq u(t) \leq 1.
\end{displaymath}

The problem is to minimize
\begin{displaymath}
-1+x_1(t_f)+x_2 (t_f),
\end{displaymath} (12)

where the final time is fixed at $t_f=1$.

We discretize the control and state variables along a uniform mesh with $n_h$ intervals and with the standard trapezoidal rule. Data for this problem appears in Table 14.1.



Table 14.1: Catalyst mixing data
Variables $3(n_h+1)$
Constraints $ 2 n_h $
Bounds $n_h+1$
Linear equality constraints 0
Linear inequality constraints 0
Nonlinear equality constraints $2 n_h$
Nonlinear inequality constraints 0
Nonzeros in $ \nabla ^2 f(x) $ 0
Nonzeros in $ c'(x) $ $12 n_h$

Performance

Results for the AMPL implementation are shown in Table 14.2. For starting points we use $ u = 0 $, $ x_1 = 1 $, and $ x_2 = 0 $ evaluated at the grid points.

The catalyst mixing problem is a typical bang-singular-bang problem. The singularity leads to nonunique values of the control in the singular region, and thus it is possible to obtain different values for the control. Figure 14.1 shows the controls obtained by two different solvers.



Table 14.2: Performance on catalyst mixing problem
Solver $n_h=100$ $n_h=200$ $n_h=400$ $n_h=800$
LANCELOT 7.71 s 31.74 s 87.26 s 424.61 s
$f$ -4.77480e-02 -4.80163e-02 -4.78620e-02 -4.71856e-02
$c$ violation 9.31890e-06 1.03790e-06 2.09770e-06 3.95740e-06
iterations 75 97 100 104
LOQO 0.66 s 1.37 s 3.1 s 8.25 s
$f$ -4.80694e-02 -4.80591e-02 -4.80565e-02 -4.80559e-02
$c$ violation 7.0e-08 6.4e-08 1.2e-08 1.2e-08
iterations 24 24 24 25
MINOS 2.39 s 5.54 s 5.34 s 17.88 s
$f$ -4.80605e-02 -4.80302e-02 -4.79881e-02 -4.74787e-02
$c$ violation 2.2e-16 2.2e-16 1.1e-16 1.1e-16
iterations 11 11 7 8
SNOPT 3.99 s 17.32 s 78.72 s 181.24 s
$f$ -4.80579e-02 -4.80471e-02 -4.80429e-02 -4.80451e-02
$c$ violation 1.8e-08 1.8e-07 3.5e-08 2.0e-05
iterations 22 22 22 6
$\dagger$ Errors or warnings. $\ddagger$ Timed out.

The results in Table 14.2 show that all the solvers are successful for $ n_h \ge 100 $ but that the objective function value fluctuates somewhat. This is probably due to the bang-singular-bang nature of the problem. The most common approach to dealing with singular control problems is to add a penalty to the objective function that leads to a smooth control, for example,

\begin{displaymath}
\alpha \int_{0}^{1} u'(t)^2 \, dt
\end{displaymath}

for some positive value of $ \alpha $. Values of $ \alpha \ge 1 $ seems to work well for this problem, but an appropriate value is difficult to find.

Figure 14.1: Controls obtained by two different solvers for the catalyst mixing problem
\includegraphics[width=2.5in]{ps/catmix-snopt_u.eps} \includegraphics[width=2.5in]{ps/catmix-loqo_u.eps}


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Liz Dolan
2001-01-02