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Re: FindMinimum within a specific interval
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32651] Re: [mg32600] FindMinimum within a specific interval
*From*: Brett Champion <brettc at wolfram.com>
*Date*: Fri, 1 Feb 2002 16:10:53 -0500 (EST)
*References*: <200201310645.BAA04291@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
At 1:45 AM -0500 1/31/02, Wassim T. Jalbout wrote:
>Dear all,
>
>I have the following 3 simultaneous equations
>
>eq1 x + y + z - 1=0
>eq2 -65.96x - 55.87y + 132.19z=0
>eq3 174.23x + 148.29y - 380.11z=0
>
>and I want to find the triplet (x,y,z) that best satisfies them within the
>interval 0>x>1,
>0>y>1 and 0>z>1.
>
>I realize, by using Solve, that the exact solution to simultaneous eqs 1 to
>3 above, namely
>(x=-3.6,y=4.5,z=0.1), lies outside the [0,1] interval for x and y and is
>thus uninteresting to my work.
>
>So I thought of minimizing the following expression:
>eq1^2+eq2^2+eq3^2
>within the interval 0 to 1 for (x,y,z)
>
>to do so I used FindMinimum as follows:
>
>FindMinimum [ (x + y + z - 1)^2 + (-65.96x - 55.87y + 132.19z)^2 +
>
> (174.23x + 148.29y - 380.11z)^2, {x, 0.1}, {y, 0.6},
>
> {z,0.3}]
>
>where I can only specify starting points rather than fixed intervals for x,
>y and z.(unfortunately).
>
>The result I get is:
>{1.0107607*^-28, {x->-3.6, y->4.5, z->0.10}}
>with x,y,z values still outside the 0,1 interval.
>
>
>How to to find the x,y,z value between 0 and 1 that results in the smallest
>value to my second degree expression written above? Is my approach wrong
>alltogether?
>
Here is one way. We'll use the Boole function in Calculus, which
returns 1 only when its argument is True, to create a hefty penalty
outside the box.
In[5]:= <<Calculus`
In[6]:= FindMinimum[(x + y + z - 1)^2 + (-65.96x - 55.87y + 132.19z)^2 +
(174.23x + 148.29y - 380.11z)^2 + 100*(1-Boole[0<=x<=1]) +
100*(1-Boole[0<=y<=1]) + 100*(1-Boole[0<=z<=1]),
{x, .5, .6}, {y, .5, .6}, {z, .5, .6},
MaxIterations->100]
-12
Out[6]= {0.894632, {x -> 9.25568 10 , y -> 0.0756025, z -> 0.0297652}}
Brett
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