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

$\begin{array}{llll} \displaystyle \min_{x,w} & \displaystyle \int_{0}^{T} & k_1(x_2-k_2)^2 + k_3(x_3-k_4)^2 \; \text{d}t\\[1.5ex] \mbox{s.t.} & \dot{x}_1 & = -\sqrt{x_1}+c_1 w_1 + c_2 w_2 - w_3 \sqrt{c_3 x_3}, \\[1.5ex] & \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[1.5ex] & \dot{x}_3 & = \sqrt{x_2}-\sqrt{x_3}+w_3 \sqrt{c_3 x_3}, \\[1.5ex] & x(0) & = (2,2,2)^T, \\[1.5ex] & 1 & = \sum\limits_{i=1}^{3}w_i(t), \\ & w_i(t) &\in \{0, 1\}, \quad i=1\ldots 3. \end{array}$

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 and the CIA decomposition. We denote the relaxed control values with $a(t)\in[0, 1]$.

The optimal objective value of the relaxed problem with $n_t=100, \, n_u=100$ is $8.775979$. The objective value of the binary controls obtained by Combinatorial Integral Approimation (CIA) is $8.789487$.

• Reference solution plots
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  Optimal relaxed controls and states determined by an direct approach with python 3.6 and CasADi, applied Multiple Shooting, 4th order Runge Kutta scheme and $n_u=100$.

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  According optimal binary controls and states determined by the direct approach. The relaxed controls were approximated by Combinatorial Integral Approximation.

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  Binary and relaxed control values as part of the Combinatorial Integral Approximation problem