Optimizasyon Problemleri

Question

2. ve 3. soruları cevaplarsanız sevinirim.

Comments

Sorunuz bircok kurala uymuyor. Lutfen daha sonraki sorularinizi kurallar dahilinde sorunuz. Kurallar soruyu sorarken beliriyor. Ordan okuyabilirsiniz.

Answer 1

1)  a)

$\dfrac{\partial f}{\partial x}=2(x-2)+2(x-2y^2)=0\qquad (1)$

$\dfrac{\partial f}{\partial y}=2(x-2y^2)(-4y)=0\qquad (2)$

$(2)\Longrightarrow y=0$ veya $x=2y^2$

$(1)\Longrightarrow y=0$ ise  $2(x-2)+2(x)=0 \Longrightarrow x=1$

$(1)\Longrightarrow x=y^2$ icin cozum yok.  O zaman kritik noktalarimiz

$(2)\, x=1 \Longrightarrow 2(1-2y^2)(-4y)=0\Longrightarrow y=0$ veya $y\mp\dfrac{\sqrt{2}}{2}$

$(1)\, x=1 \Longrightarrow 2(1-2)+2(1-2y^2)=0\Longrightarrow y=0$

$(1,0),(1,\dfrac{\sqrt{2}}{2}),(1,-\dfrac{\sqrt{2}}{2})$ olur.

2)  $V=\pi r^2 h=20\pi$. Maliyet $M$ olsun. O zaman

$M=(2\pi r^2)10+(2\pi r h)8$ olur. Mailyeti minimize etmemiz lazim. Oncelikle yukaridaki iliskiyi kullanarak bir $M$'yi bir degiskene dusurelim.

$\pi r^2 h=20\pi \Longrightarrow h=\dfrac{20}{r^2}$, $M$'de yerine koyalaim.

$M=(2\pi r^2)10+(2\pi r\dfrac{20}{r^2})8=20\pi r^2+ \dfrac{320\pi}{r}$

$\dfrac{dM}{dr}=40\pi r-\dfrac{320\pi}{r^2}=0$

$\dfrac{40\pi r^3-320\pi}{r^2}=0\Longrightarrow 40\pi r^3-320\pi=0\Longrightarrow  r^3=8\Longrightarrow  r=2$. $r=2$'nin gercekten  maliyeti minimize eden deger oldugunu gostermek icin $M''(2)$ isaretine bakilabilir veya birinci turevde tablo yapilabilir ($r=0$'i da goz onune alarak). Bunu size birakiyorum.

$h=\dfrac{20}{2^2}=5$

3) Once  Newton metodunu 1 degiskenli fonksiyonlar icin hatirlayalim. Newton methodu, $x_0$ noktasini baslangic olarak alan ve itarasyon ile $f(x)=0$ fonsiyonun kokunu bulan bir metod.

$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\qquad n=0,1,\dots$

$n=0:\qquad x_{1}=x_0-\dfrac{f(x_0)}{f'(x_0)}$

Cok degiskenlilerde ise formul su haldedir. $J=\left[ \begin{array}{ccc}  \dfrac{\partial f_1}{\partial x_1} & \dots & \dfrac{\partial f_1}{\partial x_N} \\  \vdots & \ddots & \vdots \\   \dfrac{\partial f_N}{\partial x_1} & \dots & \dfrac{\partial f_N}{\partial x_N}\\ \end{array} \right]$ Jacobian matrix olmak uzere

$\mathbf{x}_{n+1}=\mathbf{x}_{n}-J(\mathbf{x}_{n})^{-1}f(\mathbf{x}_{n})$ dir.

2 degiskenliler icin yazalim.

$\left[ \begin{array}{c}  x_{n+1} \\  y_{n+1} \\ \end{array} \right]=\left[ \begin{array}{c}  x_{n} \\  y_{n} \\ \end{array} \right]-\left[ \begin{array}{cc}  \dfrac{\partial f_1}{\partial x}( x_{n}) & \dfrac{\partial f_1}{\partial y}( x_{n}) \\   \dfrac{\partial f_2}{\partial x}(y_{n}) & \dfrac{\partial f_2}{\partial y}( y_{n}) \\ \end{array} \right]^{-1}\left[ \begin{array}{c} f_1( x_{n}) \\  f_2(y_{n}) \\ \end{array} \right]$

Sizin sorunuza gelince, $(x_0,y_0)=(1,1)$

$F(x,y)=\left[ \begin{array}{c}  f_1(x,y) \\ f_2(x,y)  \\ \end{array} \right]=\left[ \begin{array}{c} x^2+y^2+3 \\  -2x^2-\dfrac{1}{2}y^2 +2\\ \end{array} \right]$

$J=\left[ \begin{array}{cc}  \dfrac{\partial f_1}{\partial x} & \dfrac{\partial f_1}{\partial y} \\   \dfrac{\partial f_2}{\partial x} & \dfrac{\partial f_2}{\partial y} \\ \end{array} \right]=\left[ \begin{array}{cc} 2x &2y \\   -4x & -y \\ \end{array} \right]$

$J^{-1}=\left[ \begin{array}{cc}  -\dfrac{1}{6 x} & -\dfrac{1}{3 x} \\  \dfrac{2}{3 y} & \dfrac{1}{3 y} \\ \end{array} \right]$

$\left[ \begin{array}{c}  x_{n+1} \\  y_{n+1} \\ \end{array} \right]=\left[ \begin{array}{c}  x_{n} \\  y_{n} \\ \end{array} \right]-\left[ \begin{array}{cc}  -\dfrac{1}{6 x}(x_n) & -\dfrac{1}{3 x} (x_n)\\  \dfrac{2}{3 y}(y_n) & \dfrac{1}{3 y}(y_n) \\ \end{array} \right]\left[ \begin{array}{c} f_1( x_{n}) \\  f_2(y_{n}) \\ \end{array} \right]$

n=0: $(x_0,y_0)=(1,1)$

$\left[ \begin{array}{c}  x_{1} \\  y_{1} \\ \end{array} \right]=\left[ \begin{array}{c}  x_0 \\  y_0 \\ \end{array} \right]-\left[ \begin{array}{cc}  -\dfrac{1}{6 x}(x_0,y_0) & -\dfrac{1}{3 x} (x_0,y_0)\\  \dfrac{2}{3 y}(x_0,y_0) & \dfrac{1}{3 y}(x_0,y_0) \\ \end{array} \right]\left[ \begin{array}{c} f_1( x_0,y_0) \\  f_2(x_0,y_0) \\ \end{array} \right]$

$\left[ \begin{array}{c}  x_{1} \\  y_{1} \\ \end{array} \right]=\left[ \begin{array}{c}  1 \\  1 \\ \end{array} \right]-\left[ \begin{array}{cc}  -\dfrac{1}{6 x}(1,1) & -\dfrac{1}{3 x} (1,1)\\  \dfrac{2}{3 y}(1,1) & \dfrac{1}{3 y}(1,1) \\ \end{array} \right]\left[ \begin{array}{c} f_1( 1,1) \\  f_2(1,1) \\ \end{array} \right]$

$\left[ \begin{array}{c}  x_{1} \\  y_{1} \\ \end{array} \right]=\left[ \begin{array}{c}  1 \\  1 \\ \end{array} \right]-\left[ \begin{array}{cc}  -\dfrac{1}{6} & -\dfrac{1}{3} \\  \dfrac{2}{3} & \dfrac{1}{3} \\ \end{array} \right]\left[ \begin{array}{c} -1 \\ -\dfrac{1}{2} \\ \end{array} \right]$

$\left[ \begin{array}{c}  x_{1} \\  y_{1} \\ \end{array} \right]=\left[ \begin{array}{c}  1 \\  1 \\ \end{array} \right]-\left[ \begin{array}{c} \dfrac{1}{3} \\ -\dfrac{5}{6}\\ \end{array} \right]$

$\left[ \begin{array}{c}  x_{1} \\  y_{1} \\ \end{array} \right]=\left[ \begin{array}{c} \dfrac{2}{3} \\ \dfrac{11}{6}\\ \end{array} \right]$

Siyah noktada baslayip ilk iterasyonla magenta noktasina geldik, bence bir iterasyon icin cok iyi. Amacimiz en yakin kirmizi koktaya ulasmak.

Mathematica programi:

ClearAll["Global`*"]
val = Values@NSolve[{x^2 + y^2 - 3, -2 x^2 - y^2/2 + 2}, {x, y}];

F = {x^2 + y^2 - 3, -2 x^2 - y^2/2 + 2};
b = {x, y};
MatrixForm[J = Grad[F, b]];
MatrixForm[Jinv = Inverse[J]];
{x[0], y[0]} = {2, 1};
val2 = Prepend[Table[

{x[n + 1],
y[n + 1]} = {x[n],
y[n]} - (Jinv /. {x -> x[n], y -> y[n]}).(F /. {x -> x[n],
y -> y[n]}), {n, 0, 2}], {x[0], y[0]}] // N


{{2., 1.}, {1.08333, 1.83333}, {0.695513, 1.64394}, {0.587388,   1.63303}}

Comments

Epey uzun cevap yazmissin. Ellerine saglik :) 
Soran pek kurallar dahilinde sormamis. 

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Kapatmaya yetişemedim.