Double Tank multimode problem

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Double Tank multimode problem
State dimension: 1
Differential states: 3
Discrete control functions: 3
Interior point equalities: 2

This site describes a Double tank problem variant with three binary controls instead of only one control.

Mathematical formulation

The mixed-integer optimal control problem is given by


\begin{array}{llll}
 \displaystyle \min_{x,w} &  \displaystyle \int_{0}^{T} & k_1(x_2-k_2)^2  \; \text{d}t\\[1.5ex]
 \mbox{s.t.} &  \dot{x}_1 & = \sum\limits_{i=1}^{3} c_{i}\; w_i,-\sqrt{x_1}, \\[1.5ex]
 &  \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[1.5ex]
 &  x(0) & = (2,2)^T, \\[1.5ex]
& 1 & = \sum\limits_{i=1}^{3}w_i(t), \\
 & w_i(t) &\in  \{0, 1\}, \quad i=1\ldots 3.
\end{array}


Here the differential states (x_0, x_1) describe the biomasses of prey and predator, respectively. The third differential state is used here to transform the objective, an integrated deviation, into the Mayer formulation \min \; x_2(t_f). This problem variant allows to choose between three different fishing options.

Parameters

These fixed values are used within the model.


\begin{array}{rcl}
[t_0, t_f] &=& [0, 12],\\
(c_{0,1}, c_{1,1}) &=& (0.2, 0.1),\\
(c_{0,2}, c_{1,2}) &=& (0.4, 0.2),\\
(c_{0,3}, c_{1,3}) &=& (0.01, 0.1).
\end{array}

Reference Solutions

If the problem is relaxed, i.e., we demand that w(t) be in the continuous interval [0, 1] instead of the binary choice \{0,1\}, the optimal solution can be determined by means of direct optimal control.

The optimal objective value of the relaxed problem with  n_t=12000, \, n_u=400  is x_2(t_f) =1.82875272. The objective value of the binary controls obtained by Combinatorial Integral Approimation (CIA) is x_2(t_f) =1.82878681.