Double Tank multimode problem

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

$\begin{array}{lll}\underset{x,w}{\min }& \underset{0}{\overset{T}{\int }}& k_1(x_2-k_2{)}^2\text{d}t\\ \text{s.t.}& {\dot{x}}_1& =\sum _{i=1}^3{c}_i{w}_i,-\sqrt{x_1},\\ & {\dot{x}}_2& =\sqrt{x_1}-\sqrt{x_2},\\ & x(0)& =(2,2{)}^T,\\ & 1& =\sum _{i=1}^3{w}_i(t),\\ & w_i(t)& \in \{0,1\},i=1\dots 3.\end{array}$

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 $x_2(t_f)=1.82878681$.

• Reference solution plots
• 
  Optimal relaxed controls and states determined by an direct approach with ampl_mintoc (Radau collocation) and $n_t=12000,n_u=400$.
• 
  Optimal binary controls and states determined by an direct approach (Radau collocation) with ampl_mintoc and $n_t=12000,n_u=400$. The relaxed controls were approximated by Combinatorial Integral Approximation.