Optimierung - Optimierung - Versionsgeschichte

JUNGBAUER Christoph am 1. Februar 2024 um 07:25 Uhr

2024-02-01T07:25:07Z

← Nächstältere Version		Version vom 1. Februar 2024, 07:25 Uhr
Zeile 894:		Zeile 894:


	\nabla f\left(\vec{x}_2\right)=\left(\begin{array}{l}		\begin{aligned}
	~~3227~~ \\		\nabla f\left(\vec{x}_2\right)= & \left(\begin{array}{l}
	~~3227~~		32 / 27 \\
	\end{array}\right), H_f\left(\vec{x}_1\right)=\left(\begin{array}{cc}		32 / 27
	~~163~~ & 0 \\		\end{array}\right), \quad H_f\left(\vec{x}_1\right)=\left(\begin{array}{cc}
	0 & ~~163~~		16 / 3 & 0 \\
	\end{array}\right) \vec{x}_3=\left(\begin{array}{l}		0 & 16 / 3
	23 \\		\end{array}\right) \\
	23		& \vec{x}_3=\left(\begin{array}{l}
			2 / 3 \\
			2 / 3
	\end{array}\right)-\left(\begin{array}{cc}		\end{array}\right)-\left(\begin{array}{cc}
	~~316~~ & 0 \\		3 / 16 & 0 \\
	0 & ~~316~~		0 & 3 / 16
	\end{array}\right)\left(\begin{array}{l}		\end{array}\right)\left(\begin{array}{l}
	~~3227~~ \\		32 / 27 \\
	~~3227~~		32 / 27
	\end{array}\right)=\left(\begin{array}{l}		\end{array}\right)=\left(\begin{array}{l}
	49 \\		4 / 9 \\
	49		4 / 9
	\end{array}\right)		\end{array}\right)
			\end{aligned}
			</math>
			<br>

			<br>
			3. Schritt:
			<math display="block">
			\begin{aligned}
			\nabla f\left(\vec{x}_3\right)= & \left(\begin{array}{l}
			4^4 / 9^3 \\
			4^4 / 9^3
			\end{array}\right), \quad H_f\left(\vec{x}_1\right)=\left(\begin{array}{cc}
			4^3 / 3^3 & 0 \\
			0 & 4^3 / 3^3
			\end{array}\right) \\
			& \vec{x}_4=\left(\begin{array}{l}
			4 / 9 \\
			4 / 9
			\end{array}\right)-\left(\begin{array}{cc}
			3^3 / 4^3 & 0 \\
			0 & 3^3 / 4^3
			\end{array}\right)\left(\begin{array}{l}
			4^4 / 9^3 \\
			4^4 / 9^3
			\end{array}\right)=\left(\begin{array}{l}
			8 / 27 \\
			8 / 27
			\end{array}\right)
			\end{aligned}
	</math>		</math>
	<br>		<br>
	~~<br>~~
	Aufgabe 2.4:<br>		Aufgabe 2.4:<br>
	<math display="block">		<math display="block">

JUNGBAUER Christoph am 1. Februar 2024 um 07:23 Uhr

2024-02-01T07:23:22Z

← Nächstältere Version		Version vom 1. Februar 2024, 07:23 Uhr
Zeile 848:		Zeile 848:
	<math display="block">		<math display="block">

	\nabla ~~\mathrm{~~f}=\left(\begin{array}{c}		\nabla f=\left(\begin{array}{l}
	4 x^3 \\		4 x^3 \\
	4 y^3		4 y^3
	\end{array}\right), H_f=\left(\begin{array}{cc}		\end{array}\right), \quad H_f=\left(\begin{array}{cc}
	12 x^2 & 0 \\		12 x^2 & 0 \\
	0 & 12 y^2		0 & 12 y^2
	\end{array}\right), \vec{x}_1=\left(\begin{array}{l}		\end{array}\right) \quad, \vec{x}_1=\left(\begin{array}{l}
	1 \\		1 \\
	1		1
	\end{array}\right)		\end{array}\right)
	</math>		</math>

	<br>		<br>
	1. Schritt:		1. Schritt:
	<math display="block">		<math display="block">

			\begin{aligned}
	\nabla f\left(\vec{x}_1\right)=\left(\begin{array}{l}		\nabla f\left(\vec{x}_1\right)=\left(\begin{array}{l}
	4 \\		4 \\
	4		4
	\end{array}\right), H_f\left(\vec{x}_1\right)=\left(\begin{array}{cc}		\end{array}\right), \quad H_f\left(\vec{x}_1\right)=\left(\begin{array}{cc}
	12 & 0 \\		12 & 0 \\
	0 & 12		0 & 12
	\end{array}\right) \vec{x}_2=\left(\begin{array}{l}		\end{array}\right) \\
			\vec{x}_2=\left(\begin{array}{l}
	1 \\		1 \\
	1		1
	\end{array}\right)-\left(\begin{array}{cc}		\end{array}\right)-\left(\begin{array}{cc}
	~~112~~ & 0 \\		1 / 12 & 0 \\
	0 & ~~112~~		0 & 1 / 12
	\end{array}\right)\left(\begin{array}{l}		\end{array}\right)\left(\begin{array}{l}
	4 \\		4 \\
	4		4
	\end{array}\right)=\left(\begin{array}{l}		\end{array}\right)=\left(\begin{array}{l}
	23 \\		2 / 3 \\
	23		2 / 3
	\end{array}\right)		\end{array}\right)
			\end{aligned}
	</math>		</math>


	<br>		<br>
	2. Schritt:		2. Schritt:

JUNGBAUER Christoph am 1. Februar 2024 um 07:18 Uhr

2024-02-01T07:18:51Z

← Nächstältere Version		Version vom 1. Februar 2024, 07:18 Uhr
Zeile 397:		Zeile 397:
	\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\		\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\
	\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\		\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\
	g(x, y) \leq c \; \; \; \lambda & \leq 0 \\		g(x, y) \leq c \; \; \; \; \; \; \; \lambda & \leq 0 \\
	\lambda(g(x, y)-c) &=0		\lambda(g(x, y)-c) &=0
	\end{aligned}		\end{aligned}

JUNGBAUER Christoph am 1. Februar 2024 um 07:18 Uhr

2024-02-01T07:18:25Z

← Nächstältere Version		Version vom 1. Februar 2024, 07:18 Uhr
Zeile 397:		Zeile 397:
	\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\		\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\
	\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\		\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\
	g(x, y) \leq c \lambda & \leq 0 \\		g(x, y) \leq c \; \; \; \lambda & \leq 0 \\
	\lambda(g(x, y)-c) &=0		\lambda(g(x, y)-c) &=0
	\end{aligned}		\end{aligned}

JUNGBAUER Christoph am 1. Februar 2024 um 07:15 Uhr

2024-02-01T07:15:25Z

← Nächstältere Version		Version vom 1. Februar 2024, 07:15 Uhr
Zeile 397:		Zeile 397:
	\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\		\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\
	\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\		\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\
	g(x, y) \leq c		g(x, y) \leq c \lambda & \leq 0 \\
	\lambda & \leq 0 \\
	\lambda(g(x, y)-c) &=0		\lambda(g(x, y)-c) &=0
	\end{aligned}		\end{aligned}

JUNGBAUER Christoph am 1. Februar 2024 um 07:14 Uhr

2024-02-01T07:14:18Z

← Nächstältere Version		Version vom 1. Februar 2024, 07:14 Uhr
Zeile 397:		Zeile 397:
	\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\		\frac{\partial}{\partial x} L(x, y, \lambda)=\frac{\partial}{\partial x} f(x, y)+\lambda \frac{\partial}{\partial x} g(x, y) &=0 \\
	\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\		\frac{\partial}{\partial y} L(x, y, \lambda)=\frac{\partial}{\partial y} f(x, y)+\lambda \frac{\partial}{\partial y} g(x, y) &=0 \\
	g(x, y) \leq c \lambda & \leq 0 \\		g(x, y) \leq c
			\lambda & \leq 0 \\
	\lambda(g(x, y)-c) &=0		\lambda(g(x, y)-c) &=0
	\end{aligned}		\end{aligned}

JUNGBAUER Christoph am 18. Oktober 2022 um 15:48 Uhr

2022-10-18T15:48:57Z

← Nächstältere Version		Version vom 18. Oktober 2022, 15:48 Uhr
Zeile 761:		Zeile 761:
	\vec{d}_1=-\left(\begin{array}{l}		\vec{d}_1=-\left(\begin{array}{l}
	2 \\		2 \\
	2\end{array}\right)		2
			\end{array}\right)
	</math>		</math>
	<br>		<br>
Zeile 768:		Zeile 769:
	1 \\		1 \\
	1		1
			\end{array}\right)
	-h\left(\begin{array}{l}		-h\left(\begin{array}{l}
	2 \\		2 \\

JUNGBAUER Christoph am 18. Oktober 2022 um 15:46 Uhr

2022-10-18T15:46:40Z

← Nächstältere Version		Version vom 18. Oktober 2022, 15:46 Uhr
Zeile 765:		Zeile 765:
	<br>		<br>
	<math display="block">		<math display="block">
	g(h)=f~~\left(~~\left(\begin{array}{l}		g(h)=f\left(\begin{array}{l}
	1 \\		1 \\
	1		1
	~~\end{array}\right)~~
	~~</math>~~
	~~<br>~~
	~~<math display="block">~~
	-h\left(\begin{array}{l}		-h\left(\begin{array}{l}
	2 \\		2 \\

JUNGBAUER Christoph am 18. Oktober 2022 um 15:44 Uhr

2022-10-18T15:44:49Z

← Nächstältere Version		Version vom 18. Oktober 2022, 15:44 Uhr
Zeile 759:		Zeile 759:
	Abstiegsrichtung		Abstiegsrichtung
	<math display="block">		<math display="block">
	\vec{d}_1=-\left(\begin{array}{l}2 \\ 2\end{array}\right)		\vec{d}_1=-\left(\begin{array}{l}
			2 \\
			2\end{array}\right)
	</math>		</math>
	<br>		<br>
Zeile 767:		Zeile 769:
	1		1
	\end{array}\right)		\end{array}\right)
			</math>
			<br>
			<math display="block">
	-h\left(\begin{array}{l}		-h\left(\begin{array}{l}
	2 \\		2 \\

JUNGBAUER Christoph am 18. Oktober 2022 um 15:41 Uhr

2022-10-18T15:41:30Z

← Nächstältere Version		Version vom 18. Oktober 2022, 15:41 Uhr
Zeile 766:		Zeile 766:
	1 \\		1 \\
	1		1
	\end{array}\right)-h\left(\begin{array}{l}		\end{array}\right)
			-h\left(\begin{array}{l}
	2 \\		2 \\
	2		2
			\end{array}\right)
	</math>		</math>
	<br>		<br>
	<math display="block">		<math display="block">
	~~\end{array}\right)\right)~~=2(1-2 h)^2 g^{\prime}(h)=-8(1-2 h)=0 h=12, \vec{x}_2=\left(\begin{array}{l}		=2(1-2 h)^2
			</math>
			<br>
			<math display="block">
			g^{\prime}(h)=-8(1-2 h)=0
			</math>
			<br>
			<math display="block">
			h=12
			</math>
			<br>
			<math display="block">
			\vec{x}_2=\left(\begin{array}{l}
	1 \\		1 \\
	1		1