Three Tank multimode problem - Revision history

RobertLampel: /* Mathematical formulation */

2025-10-14T07:16:17Z

Mathematical formulation

← Older revision		Revision as of 07:16, 14 October 2025
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	<math>		<math>
	\begin{array}{llll}		\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 \; \~~mathrm~~{d}t\\[1.5ex]		\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_1}, \\[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_1}, \\[1.5ex]
	& \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[1.5ex]		& \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[1.5ex]

RobertLampel: /* Mathematical formulation */

2025-10-14T07:16:04Z

Mathematical formulation

← Older revision		Revision as of 07:16, 14 October 2025
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	<math>		<math>
	\begin{array}{llll}		\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]		\displaystyle \min_{x,w} & \displaystyle \int_{0}^{T} & k_1(x_2-k_2)^2 + k_3(x_3-k_4)^2 \; \mathrm{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_1}, \\[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_1}, \\[1.5ex]
	& \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[1.5ex]		& \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[1.5ex]

RobertLampel: /* Mathematical formulation */

2025-10-14T07:15:22Z

Mathematical formulation

← Older revision		Revision as of 07:15, 14 October 2025
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	\begin{array}{llll}		\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]		\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]		\mbox{s.t.} & \dot{x}_1 & = -\sqrt{x_1}+c_1 w_1 + c_2 w_2 - w_3 \sqrt{c_3 x_1}, \\[1.5ex]
	& \dot{x}_2 & = \sqrt{x_1}-\sqrt{x_2}, \\[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]		& \dot{x}_3 & = \sqrt{x_2}-\sqrt{x_3}+w_3 \sqrt{c_3 x_1}, \\[1.5ex]
	& x(0) & = (2,2,2)^T, \\[1.5ex]		& x(0) & = (2,2,2)^T, \\[1.5ex]
	& 1 & = \sum\limits_{i=1}^{3}w_i(t), \\		& 1 & = \sum\limits_{i=1}^{3}w_i(t), \\
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	\end{array}		\end{array}
	</math>		</math>


	== Parameters ==		== Parameters ==

ClemensZeile at 09:18, 14 March 2020

2020-03-14T09:18:16Z

← Older revision		Revision as of 09:18, 14 March 2020
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	</gallery>		</gallery>

			== Source Code ==

			Model description is available in (using pycombina for solving the (CIA) rounding problem step):
			* [[:Category:Casadi \| Casadi code]] at [[Three Tank multimode problem (python/casadi)]]

ClemensZeile: /* Reference Solutions */

2020-03-14T09:13:31Z

Reference Solutions

← Older revision		Revision as of 09:13, 14 March 2020
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	<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">		<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">
	Image:three_tank_relaxed_solution.png\| 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 ~~<math>n_u=~~100~~</math>~~.		Image:three_tank_relaxed_solution.png\| 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 100 discretization intervals.
	Image:three_tank_binary_solution.png\| According optimal binary controls and states determined by the direct approach. The relaxed controls were approximated by Combinatorial Integral Approximation.		Image:three_tank_binary_solution.png\| According optimal binary controls and states determined by the direct approach. The relaxed controls were approximated by Combinatorial Integral Approximation.
	Image:three_tank_rounding_solution.png\| Binary and relaxed control values as part of the Combinatorial Integral Approximation problem		Image:three_tank_rounding_solution.png\| Binary and relaxed control values as part of the Combinatorial Integral Approximation problem

ClemensZeile: /* Reference Solutions */

2020-03-14T09:12:39Z

Reference Solutions

← Older revision		Revision as of 09:12, 14 March 2020
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	== Reference Solutions ==		== Reference Solutions ==

	If the problem is relaxed, i.e., we demand that <math>w(t)</math> be in the continuous interval <math>[0, 1]</math> instead of the binary choice <math>\{0,1\}</math>, the optimal solution can be determined by means of direct optimal control.		If the problem is relaxed, i.e., we demand that <math>w(t)</math> be in the continuous interval <math>[0, 1]</math> instead of the binary choice <math>\{0,1\}</math>, the optimal solution can be determined by means of direct optimal control and the CIA decomposition. We denote the relaxed control values with <math>a(t)\in[0, 1]</math>.

	The optimal objective value of the relaxed problem with <math> n_t=~~12000~~, \, n_u=100 </math> is <math>2.~~59106823~~</math>. The objective value of the binary controls obtained by Combinatorial Integral Approimation (CIA) is <math>2.~~59121008~~</math>.		The optimal objective value of the relaxed problem with <math> n_t=100, \, n_u=100 </math> is <math>8.775979</math>. The objective value of the binary controls obtained by Combinatorial Integral Approimation (CIA) is <math>8.789487</math>.

	<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">		<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">
	Image:~~MmdoubletankRelaxed 12000 120 1~~.png\| Optimal relaxed controls and states determined by an direct approach with ~~ampl_mintoc (Radau collocation)~~ and <math>~~n_t=12000, \,~~ n_u=100</math>.		Image:three_tank_relaxed_solution.png\| 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 <math>n_u=100</math>.
	Image:~~MmdoubletankCIA_12000_120_1~~.png\| ~~Optimal~~ binary controls and states determined by an direct approach ~~(Radau collocation) with ampl_mintoc and <math>n_t=12000, \, n_u=100</math>~~. The relaxed controls were approximated by Combinatorial Integral Approximation.		Image:three_tank_binary_solution.png\| According optimal binary controls and states determined by the direct approach. The relaxed controls were approximated by Combinatorial Integral Approximation.
			Image:three_tank_rounding_solution.png\| Binary and relaxed control values as part of the Combinatorial Integral Approximation problem
	</gallery>		</gallery>

ClemensZeile: Created page with "{{Dimensions |nd = 1 |nx = 3 |nw = 3 |nre = 2 }}This site describ..."

2020-03-14T07:48:51Z

Created page with "{{Dimensions |nd = 1 |nx = 3 |nw = 3 |nre = 2 }}This site describ..."

New page

{{Dimensions
|nd = 1
|nx = 3
|nw = 3
|nre = 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

<math>
\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}
</math>

== Parameters ==

These fixed values are used within the model.

<math> T=12, c_1=1, c_2=2, c_3=0.8, k_1=2, k_2=3, k_3 = 1, k_4 = 3. </math>

== Reference Solutions ==

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

The optimal objective value of the relaxed problem with <math> n_t=12000, \, n_u=100 </math> is <math>2.59106823</math>. The objective value of the binary controls obtained by Combinatorial Integral Approimation (CIA) is <math>2.59121008</math>.

<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">
Image:MmdoubletankRelaxed 12000 120 1.png| Optimal relaxed controls and states determined by an direct approach with ampl_mintoc (Radau collocation) and <math>n_t=12000, \, n_u=100</math>.
Image:MmdoubletankCIA_12000_120_1.png| Optimal binary controls and states determined by an direct approach (Radau collocation) with ampl_mintoc and <math>n_t=12000, \, n_u=100</math>. The relaxed controls were approximated by Combinatorial Integral Approximation.
</gallery>


[[Category:MIOCP]]
[[Category:ODE model]]
[[Category:Tracking objective]]
[[Category:Chattering]]
[[Category:Sensitivity-seeking arcs]]