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Thread: How to get the solution of this simple optimization problem?

  1. #1
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    Question How to get the solution of this simple optimization problem?

    How to get the solution of this simple optimization problem?

    $\displaystyle
    minimize~~\Phi = x_1^2 + x_2^2
    $
    $\displaystyle
    subject~to~~x_1 +2x_2 = 5
    $

    I konw we may use Moore-Penrose inverse like,

    $\displaystyle
    Ax = b => x = (A^TA)^{-1}A^Tb
    $

    but what are $\displaystyle A$ and $\displaystyle x$ now?

    if $\displaystyle A = (1~~2) $ and $\displaystyle A^T = (1~~2)^T $ then there is no inverse of $\displaystyle A^TA$. I am quite confused how to formulate these $\displaystyle A$ and $\displaystyle x$ to do the matrix multiplication.


    Thanks a lot
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  2. #2
    MHF Contributor

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    Quote Originally Posted by ggyyree View Post
    How to get the solution of this simple optimization problem?

    $\displaystyle
    minimize~~\Phi = x_1^2 + x_2^2
    $
    $\displaystyle
    subject~to~~x_1 +2x_2 = 5
    $

    I konw we may use Moore-Penrose inverse like,

    $\displaystyle
    Ax = b => x = (A^TA)^{-1}A^Tb
    $

    but what are $\displaystyle A$ and $\displaystyle x$ now?

    if $\displaystyle A = (1~~2) $ and $\displaystyle A^T = (1~~2)^T $ then there is no inverse of $\displaystyle A^TA$. I am quite confused how to formulate these $\displaystyle A$ and $\displaystyle x$ to do the matrix multiplication.


    Thanks a lot
    "Moore-Penrose" is much too advanced for me ("Moore" isn't so bad but I cringe when I hear "Penrose") so I would do it with Lagrange multipliers (I much better with "Lagrange").

    You want to minimize $\displaystyle F(x,y)= x^2+ y^2$ subject to the constraint G(x,y)= x+ 2y= 5. The gradients are $\displaystyle \nabla F= 2x\vec{i}+ 2y\vec{j}$ and $\displaystyle \nabla G= \vec{i}+ 2\vec{j}$. Now we need to find (x,y) so that $\displaystyle 2x\vec{i}+ 2y\vec{j}= \lambda(\vec{i}+ 2\vec{j})$. That means we must have $\displaystyle 2x= \lambda$ and $\displaystyle 2y= 2\lambda$. Dividing the second equation by the first give $\displaystyle \frac{y}{x}= 2$ or y= 2x. Putting that into the constraint x+ 2y= 5, x+ 2(2y)= x+ 4x= 5x= 5 or x= 1, y= 2.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ggyyree View Post
    How to get the solution of this simple optimization problem?

    $\displaystyle
    minimize~~\Phi = x_1^2 + x_2^2
    $
    $\displaystyle
    subject~to~~x_1 +2x_2 = 5
    $
    Why not solve the constraint for $\displaystyle x_1$ in terms of $\displaystyle x_2$ and substitute into the objective to reduce this to a 1-D optimisation??

    $\displaystyle x_1=5-2x_2$, then:

    $\displaystyle
    \Phi=5x_2^2-20x_2+25
    $

    CB
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