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Math Help - Penalty function

  1. #1
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    Question Penalty function

    How can i minimize
    f(x,y) = 1/2 (x+1) + y
    with Penalty function method

    with constraints of
    1-x \leq 0
    2-y \leq 0
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  2. #2
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    First i set Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2

    than take the derivatives

    \frac{\partial Q}{\partial x_1}=\frac1{2} +2 \mu (1-x_1) + 2 \mu (2-x_1)

    \frac{\partial Q}{\partial x_2}=1 +2 \mu (1-x_1) + 2 \mu (2-x_1)

    Then?

    Quote Originally Posted by trante View Post
    How can i minimize
    f(x,y) = \frac1{2} (x_1+1) + x_2
    with Penalty function method

    with constraints of
    1-x_1 \leq 0
    2-x_2 \leq 0
    Last edited by trante; January 2nd 2009 at 11:52 AM.
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  3. #3
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    Question

    The function will be
    as
    <br /> <br />
a) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2<br />
    or
    b) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1)^2 + (2-x_1)^2]<br />
    or
    c) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1 + 2-x_1)^2]<br />

    ???


    Quote Originally Posted by trante View Post
    First i set Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2

    than take the derivatives

    \frac{\partial Q}{\partial x_1}=\frac1{2} +2 \mu (1-x_1) + 2 \mu (2-x_1)

    \frac{\partial Q}{\partial x_2}=1 +2 \mu (1-x_1) + 2 \mu (2-x_1)

    Then?
    Last edited by trante; January 3rd 2009 at 03:54 AM.
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  4. #4
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    Jan 2009
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    Question

    f(x,y) = \frac1{2} (x_1+1) + x_2
    with constraints of
    1-x_1 \leq 0
    2-x_2 \leq 0

    i set  Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1)^2 + (2-x_2)^2]

    \frac{\partial Q}{\partial x_1}=\frac1{2} +2 \mu (1-x_1)=0

    \frac{\partial Q}{\partial x_2}=1 +2 \mu (2-x_2)=0

    then \frac1{2(1-x_1)}=\frac1{2-x_2}
    then 2x_2=x_1

    2 \mu (2-x_2)=4\mu-2\mu x_2=-1
    x_2=\frac{4\mu+1}{2\mu}

    x_2=2
    x_1=1

    Is it the solution?
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