Results 1 to 4 of 4

Thread: Penalty function

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    9

    Question Penalty function

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

    with constraints of
    $\displaystyle 1-x \leq 0$
    $\displaystyle 2-y \leq 0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2009
    Posts
    9
    First i set $\displaystyle Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2$

    than take the derivatives

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

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

    Then?

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

    with constraints of
    $\displaystyle 1-x_1 \leq 0$
    $\displaystyle 2-x_2 \leq 0$
    Last edited by trante; Jan 2nd 2009 at 11:52 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    9

    Question

    The function will be
    as
    $\displaystyle

    a) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2
    $
    or
    $\displaystyle b) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1)^2 + (2-x_1)^2]
    $
    or
    $\displaystyle c) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1 + 2-x_1)^2]
    $

    ???


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

    than take the derivatives

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

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

    Then?
    Last edited by trante; Jan 3rd 2009 at 03:54 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2009
    Posts
    9

    Question

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

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

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

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

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

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

    $\displaystyle x_2=2$
    $\displaystyle x_1=1$

    Is it the solution?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: Nov 27th 2012, 05:28 AM
  2. Replies: 4
    Last Post: Oct 27th 2010, 05:41 AM
  3. Replies: 3
    Last Post: Sep 14th 2010, 02:46 PM
  4. Penalty reward contrast analysis for identifying Kano model
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Jun 4th 2009, 06:38 PM
  5. Replies: 5
    Last Post: May 3rd 2009, 03:16 PM

Search Tags


/mathhelpforum @mathhelpforum