Three Tank multimode problem

Three 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 and three tanks, i.e., three differential states representing different compartments.

Mathematical formulation

The mixed-integer optimal control problem is given by

$\begin{array}{llll} \min_{x, w} & \int_{0}^{T} & k_{1} (x_{2} - k_{2})^{2} + k_{3} (x_{3} - k_{4})^{2} d t \\ s.t. & {\dot{x}}_{1} & = - \sqrt{x_{1}} + c_{1} w_{1} + c_{2} w_{2} - w_{3} \sqrt{c_{3} x_{3}}, \\ {\dot{x}}_{2} & = \sqrt{x_{1}} - \sqrt{x_{2}}, \\ {\dot{x}}_{3} & = \sqrt{x_{2}} - \sqrt{x_{3}} + w_{3} \sqrt{c_{3} x_{3}}, \\ x (0) & = (2, 2, 2)^{T}, \\ 1 & = \sum_{i = 1}^{3} w_{i} (t), \\ w_{i} (t) & \in {0, 1}, i = 1 \dots 3 . \end{array}$

Parameters

These fixed values are used within the model.

$T = 12, c_{1} = 1, c_{2} = 2, c_{3} = 0.8, k_{1} = 2, k_{2} = 3, k_{3} = 1, k_{4} = 3 .$

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 $n_{t} = 12000, n_{u} = 100$ is $2.59106823$ . The objective value of the binary controls obtained by Combinatorial Integral Approimation (CIA) is $2.59121008$ .

Reference solution plots
Optimal relaxed controls and states determined by an direct approach with ampl_mintoc (Radau collocation) and $n_{t} = 12000, n_{u} = 100$ .
Optimal binary controls and states determined by an direct approach (Radau collocation) with ampl_mintoc and $n_{t} = 12000, n_{u} = 100$ . The relaxed controls were approximated by Combinatorial Integral Approximation.