Electric Car - Revision history

RobertLampel: /* Reference Solutions */

2026-01-29T15:01:37Z

Reference Solutions

← Older revision		Revision as of 15:01, 29 January 2026
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	== Reference Solutions ==		== Reference Solutions ==
	We look at the particular instance of our problem with <math> t_f = 10s </math>and target set <math> \mathcal{T} = \mathbb{R} \times \mathbb{R} \times \{100\} \times \mathbb{R} </math>, in which the car needs to cover 100m in 10s.		We look at the particular instance of our problem with <math> t_f = 10s </math> and target set <math> \mathcal{T} = \mathbb{R} \times \mathbb{R} \times \{100\} \times \mathbb{R} </math>, in which the car needs to cover 100m in 10s.
	Figure 1 shows a plot of the differential states of the optimal trajectory of the relaxed problem (i.e. <math> u \in [-1,1] </math>instead of <math> u \in \{-1,1\} </math>) for <math> N = 1000, N </math>being the number of time discretization points. The current <math> x_0 </math>increases to its maximal value of 150A, stays there for a certain time, decreases on its minimal value of -150A, stays on this value and eventually increases slightly. This behavior corresponds to the different arcs bang, path-constrained, singular, path-constrained, bang and can be observed also in Figure 2. It shows the corresponding switching function and the optimal control. Note that the plots show data from the solution with the indirect approach.		Figure 1 shows a plot of the differential states of the optimal trajectory of the relaxed problem (i.e. <math> u \in [-1,1] </math>instead of <math> u \in \{-1,1\} </math>) for <math> N = 1000, N </math>being the number of time discretization points. The current <math> x_0 </math>increases to its maximal value of 150A, stays there for a certain time, decreases on its minimal value of -150A, stays on this value and eventually increases slightly. This behavior corresponds to the different arcs bang, path-constrained, singular, path-constrained, bang and can be observed also in Figure 2. It shows the corresponding switching function and the optimal control. Note that the plots show data from the solution with the indirect approach.

RobertLampel: /* Parameters */

2026-01-29T14:49:57Z

Parameters

← Older revision		Revision as of 14:49, 29 January 2026
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	\|-		\|-
	\|Battery voltage		\|Battery voltage
	\|<math>V_{alim}</math>		\|<math>V_\mathrm{alim}</math>
	\|150		\|150
	\|<math> V </math>		\|<math> V </math>
	\|-		\|-
	\|Resistance of the battery		\|Resistance of the battery
	\|<math>R_{bat}</math>		\|<math>R_\mathrm{bat}</math>
	\|0.05		\|0.05
	\|<math> Ohms </math>		\|<math> Ohms </math>

RobertLampel: /* Mathematical formulation */

2026-01-29T14:49:30Z

Mathematical formulation

← Older revision		Revision as of 14:49, 29 January 2026
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	\displaystyle \min_{x, u} & x_3(t_f) \\[1.5ex]		\displaystyle \min_{x, u} & x_3(t_f) \\[1.5ex]
	\mbox{s.t.}		\mbox{s.t.}
	& \dot{x}_0 & = & (V_{alim} u - R_m x_0 - K_m x_1) / L_m, \\		& \dot{x}_0 & = & (V_{\mathrm{alim}} u - R_m x_0 - K_m x_1) / L_m, \\
	& \dot{x}_1 & = & \frac{K_r^2}{Mr^2} (K_m x_0 - \frac{r}{K_r} ( M g K_f + \frac{1}{2} \rho S C_x \frac{r^2}{K_r^2} x_1^2)), \\		& \dot{x}_1 & = & \frac{K_r^2}{Mr^2} (K_m x_0 - \frac{r}{K_r} ( M g K_f + \frac{1}{2} \rho S C_x \frac{r^2}{K_r^2} x_1^2)), \\
	& \dot{x}_2 & = & \frac{r}{K_r} x_1, \\		& \dot{x}_2 & = & \frac{r}{K_r} x_1, \\
	& \dot{x}_3 & = & V_{alim} u x_0 + R_{bat} x_0^2, \\[1.5ex]		& \dot{x}_3 & = & V_{\mathrm{alim}} u x_0 + R_{bat} x_0^2, \\[1.5ex]
	& x(t_0) &=& (0,0,0,0)^T, \\		& x(t_0) &=& (0,0,0,0)^T, \\
	& x(t_f) & \in & \mathcal{T} \subseteq \mathbb{R}^4,\\		& x(t_f) & \in & \mathcal{T} \subseteq \mathbb{R}^4,\\

RobertLampel: /* Mathematical formulation */

2026-01-29T14:39:34Z

Mathematical formulation

← Older revision		Revision as of 14:39, 29 January 2026
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	& x(t_0) &=& (0,0,0,0)^T, \\		& x(t_0) &=& (0,0,0,0)^T, \\
	& x(t_f) & \in & \mathcal{T} \subseteq \mathbb{R}^4,\\		& x(t_f) & \in & \mathcal{T} \subseteq \mathbb{R}^4,\\
	& x_0(t) & \in & [-i_{max}, i_{max}], \\		& x_0(t) & \in & [-i_{\mathrm{max}}, i_{\mathrm{max}}], \\
	& u(t) &\in& \{-1, 1\}.		& u(t) &\in& \{-1, 1\}.
	\end{array}		\end{array}

FelixMueller at 08:29, 27 July 2016

2016-07-27T08:29:02Z

← Older revision		Revision as of 08:29, 27 July 2016
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	\|nx = 4		\|nx = 4
	\|nw = 1		\|nw = 1
			\|nc = 2
	\|nri = 2		\|nri = 2
	\|nre = 5		\|nre = 5

FelixMueller: /* Reference Solutions */

2016-07-26T15:33:40Z

Reference Solutions

← Older revision		Revision as of 15:33, 26 July 2016
Line 111:		Line 111:

	== Reference Solutions ==		== Reference Solutions ==
	We look at the particular instance of our problem with <math> t_f = 10s </math>and target set <math> \mathcal{T} = = \mathbb{R} \times \mathbb{R} \times \{100\} \times \mathbb{R} </math>, in which the car needs to cover 100m in 10s.		We look at the particular instance of our problem with <math> t_f = 10s </math>and target set <math> \mathcal{T} = \mathbb{R} \times \mathbb{R} \times \{100\} \times \mathbb{R} </math>, in which the car needs to cover 100m in 10s.
	Figure 1 shows a plot of the differential states of the optimal trajectory of the relaxed problem (i.e. <math> u \in [-1,1] </math>instead of <math> u \in \{-1,1\} </math>) for <math> N = 1000, N </math>being the number of time discretization points. The current <math> x_0 </math>increases to its maximal value of 150A, stays there for a certain time, decreases on its minimal value of -150A, stays on this value and eventually increases slightly. This behavior corresponds to the different arcs bang, path-constrained, singular, path-constrained, bang and can be observed also in Figure 2. It shows the corresponding switching function and the optimal control. Note that the plots show data from the solution with the indirect approach.		Figure 1 shows a plot of the differential states of the optimal trajectory of the relaxed problem (i.e. <math> u \in [-1,1] </math>instead of <math> u \in \{-1,1\} </math>) for <math> N = 1000, N </math>being the number of time discretization points. The current <math> x_0 </math>increases to its maximal value of 150A, stays there for a certain time, decreases on its minimal value of -150A, stays on this value and eventually increases slightly. This behavior corresponds to the different arcs bang, path-constrained, singular, path-constrained, bang and can be observed also in Figure 2. It shows the corresponding switching function and the optimal control. Note that the plots show data from the solution with the indirect approach.

FelixMueller: /* Reference Solutions */

2016-07-04T10:05:40Z

Reference Solutions

← Older revision		Revision as of 10:05, 4 July 2016
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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="3">
	Image:ElectricCar PrimalStates Plot.PNG\| Primal states of an optimal trajectory for the relaxed problem on a control discretization grid with <math> N = 1,000 </math>.		Image:ElectricCar PrimalStates Plot.PNG\| Primal states of an optimal trajectory for the relaxed problem on a control discretization grid with <math> N = 1,000 </math>.
	Image:ElectricCar Control PathConstraints SwitchingFunction Plot.PNG\| The optimal control and switching function. The dotted vertical lines show the switching times <math> \tau_i </math> where transitions between different kinds of arcs occur.		Image:ElectricCar Control PathConstraints SwitchingFunction Plot.PNG\| The optimal control and switching function. The dotted vertical lines show the switching times <math> \tau_i </math> where transitions between different kinds of arcs occur.

FelixMueller: /* Reference Solutions */

2016-07-04T10:05:18Z

Reference Solutions

← Older revision		Revision as of 10:05, 4 July 2016
Line 118:		Line 118:

	The direct and indirect approaches are local optimization techniques and only provide upper bounds for the relaxed problem and hence for the original problem. Here the indirect solution of the relaxed problem gives us a bound of <math> x_3(t_f) = 22777.2 </math>.		The direct and indirect approaches are local optimization techniques and only provide upper bounds for the relaxed problem and hence for the original problem. Here the indirect solution of the relaxed problem gives us a bound of <math> x_3(t_f) = 22777.2 </math>.


			<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">
			Image:ElectricCar PrimalStates Plot.PNG\| Primal states of an optimal trajectory for the relaxed problem on a control discretization grid with <math> N = 1,000 </math>.
			Image:ElectricCar Control PathConstraints SwitchingFunction Plot.PNG\| The optimal control and switching function. The dotted vertical lines show the switching times <math> \tau_i </math> where transitions between different kinds of arcs occur.
			Image:ElectricCar PrimalStates SumUpRounding.PNG\| Primal states with Sum up rounding on grid with <math> \Delta t = 10^{-2} </math>.
			</gallery>

	== Source Code ==		== Source Code ==

FelixMueller: /* Mathematical formulation */

2016-07-04T09:52:56Z

Mathematical formulation

← Older revision		Revision as of 09:52, 4 July 2016
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	\displaystyle \min_{x, u} & x_3(t_f) \\[1.5ex]		\displaystyle \min_{x, u} & x_3(t_f) \\[1.5ex]
	\mbox{s.t.}		\mbox{s.t.}
	& \dot{x}_0 & = & (R_m x_0 -K_m x_1) / L_m, \\		& \dot{x}_0 & = & (V_{alim} u - R_m x_0 - K_m x_1) / L_m, \\
	& \dot{x}_1 & = & \frac{K_r^2}{Mr^2} (K_m x_0 - \frac{r}{K_r} ( M g K_f + \frac{1}{2} \rho S C_x \frac{r^2}{K_r^2} x_1^2)), \\		& \dot{x}_1 & = & \frac{K_r^2}{Mr^2} (K_m x_0 - \frac{r}{K_r} ( M g K_f + \frac{1}{2} \rho S C_x \frac{r^2}{K_r^2} x_1^2)), \\
	& \dot{x}_2 & = & \frac{r}{K_r} x_1, \\		& \dot{x}_2 & = & \frac{r}{K_r} x_1, \\

FelixMueller: /* Mathematical formulation */

2016-07-04T09:51:39Z

Mathematical formulation

← Older revision		Revision as of 09:51, 4 July 2016
Line 20:		Line 20:
	\displaystyle \min_{x, u} & x_3(t_f) \\[1.5ex]		\displaystyle \min_{x, u} & x_3(t_f) \\[1.5ex]
	\mbox{s.t.}		\mbox{s.t.}
	& \dot{x}_0 & = & (K_m x_1) / L_m, \\		& \dot{x}_0 & = & (R_m x_0 -K_m x_1) / L_m, \\
	& \dot{x}_1 & = & \frac{K_r^2}{Mr^2} (K_m x_0 - \frac{r}{K_r} ( M g K_f + \frac{1}{2} \rho S C_x \frac{r^2}{K_r^2} x_1^2)), \\		& \dot{x}_1 & = & \frac{K_r^2}{Mr^2} (K_m x_0 - \frac{r}{K_r} ( M g K_f + \frac{1}{2} \rho S C_x \frac{r^2}{K_r^2} x_1^2)), \\
	& \dot{x}_2 & = & \frac{r}{K_r} x_1, \\		& \dot{x}_2 & = & \frac{r}{K_r} x_1, \\