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Subsections


5 Isometrization of $\alpha$-pinene

Determine the reaction coefficients in the thermal isometrization of $\alpha$-pinene. The linear kinetic model [6] is

$\displaystyle y_1'$ $\textstyle =$ $\displaystyle -(\theta_1 + \theta_2) y_1$  
$\displaystyle y_2'$ $\textstyle =$ $\displaystyle \theta_1 y_1$  
$\displaystyle y_3'$ $\textstyle =$ $\displaystyle \theta_2 y_1 - (\theta_3 + \theta_4 )y_3 + \theta_5 y_5$ (1)
$\displaystyle y_4'$ $\textstyle =$ $\displaystyle \theta_3 y_3$  
$\displaystyle y_5'$ $\textstyle =$ $\displaystyle \theta_4 y_3 - \theta_5 y_5,$  

where $ \theta_i \ge 0 $ are the reaction coefficients. Initial conditions for are known. The problem is to minimize
\begin{displaymath}
\sum _ {j=1}^8 \Vert y ( \tau_j ; \theta ) - z_j \Vert ^ 2 ,
\end{displaymath} (2)

where $z_j$ are concentration measurements for $y$ at time points $\tau_1, \ldots ,\tau_8$.

Formulation

Our formulation of the $\alpha$-pinene problem as an optimization problem follows [21,3]. We use a $k$-stage collocation method, a uniform partition with $n_h$ subintervals of $[0,\tau_8]$, and the standard [2, pages 247-249] basis representation,

\begin{displaymath}
\ v_{i} + \sum_{j=1}^{k}
\frac{(t - t_i)^j}{j!\ h^{j-1}} w_{ij} ,
\qquad
t \in [t_i,t_{i+1}] ,
\end{displaymath}

for the components of the solution $ y $ of . The constraints in the optimization problem are the initial conditions in , the continuity conditions, and the collocation equations. The continuity equations at each interior grid point are a set of $ 5(n_h -1 ) $ linear equations. The collocation equations are a set of $5 k n_h$ nonlinear equations obtained by requiring that the collocation approximation satisfy at the collocation points. Data for this problem appears in Table 5.1.



Table 5.1: Isometrization of $\alpha$-pinene data
Variables $ 5(k+1) n_h + 5$
Constraints $ 5(k+1) n_h$
Bounds 5
Linear equality constraints $5 n_h$
Linear inequality constraints 0
Nonlinear equality constraints $5 k n_h$
Nonlinear inequality constraints 0
Nonzeros in $ \nabla ^2 f(x) $ $ 40 (k+1)^2 $
Nonzeros in $ c'(x) $ $10 k(k+1) n_h$

Performance

We provide results for the AMPL formulation with $k=3$ in Table 5.2. The initial values for the $\theta$ parameters are $ \theta_i = 0.0 $. The initial basis parameters are chosen so that the collocation approximation is piecewise constant and interpolates the data. The solution and data are shown in Figure 5.1.



Table 5.2: Performance on isometrization problem
Solver $n_h=25$ $n_h=50$ $n_h=100$ $n_h=200$
LANCELOT 1426.01 s 2720.49 s $\ddagger$ $\ddagger$
$f$ 1.96766e+01$\dagger$ 1.93937e+01$\dagger$ $\ddagger$ $\ddagger$
$c$ violation 1.87900e-06$\dagger$ 6.09920e-06$\dagger$ $\ddagger$ $\ddagger$
iterations 305 179 $\ddagger$ $\ddagger$
LOQO 28.85 s 6.15 s 6.77 s 16.87 s
$f$ 1.98715e+01 1.98721e+01 1.98721e+01 1.98721e+01
$c$ violation 1.3e-11 2.2e-13 7.6e-13 8.4e-13
iterations 389 32 23 21
MINOS 1.98 s 6.74 s 21.66 s 194.84 s
$f$ 1.98715e+01 1.98721e+01 1.98721e+01 0.00000e+00$\dagger$
$c$ violation 4.2e-13 4.4e-13 2.3e-12 1.7e+04$\dagger$
iterations 7 8 7 49
SNOPT 3.74 s 13.1 s 48.91 s 235.44 s
$f$ 1.98715e+01 1.98721e+01$\dagger$ 1.98721e+01 1.98721e+01$\dagger$
$c$ violation 3.9e-13 4.2e-13$\dagger$ 6.7e-13 5.1e-13$\dagger$
iterations 13 18 21 37
$\dagger$ Errors or warnings. $\ddagger$ Timed out.

LANCELOT stops with the message step got too small, near the solution for $n_h \leq 50$. MINOS fails completely on $n_h = 200$ with unbounded (or badly scaled) problem, while SNOPT manages a [p]rimal feasible solution, which could not satisfy dual feasibility for both $n_h = 50,\ 200$.

Figure 5.1: Solution and data for the $\alpha$-pinene problem
\includegraphics[width=3.in]{ps/pinene.eps}


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Liz Dolan
2001-01-02