[R] Optimization

[Moshe Olshansky]

You are right!!!

For some strange reason I substituted ^
(exponentiation) for *, so the problem became

Max x1^0.021986+x2^0.000964+x3^0.02913

with these conditions:

x1+x2+x3=1;

sqrt((x1^0.114434)^2+(x2^0.043966)^2+(x3^0.100031)^2)=0.04;

which is clearly non-linear.

--- Berwin A Turlach <berwin at maths.uwa.edu.au> wrote:

> G'day Moshe,
> 
> On Tue, 17 Jul 2007 17:32:52 -0700 (PDT)
> Moshe Olshansky <m_olshansky at yahoo.com> wrote:
> 
> > This is partially true since both the function to
> be
> > maximized and the constraint are non-linear. 
> 
> I am not sure what your definition of non-linear is,
> but in my book,
> and I believe by most mathematical/statistical
> definitions, the
> objective function is linear.
> 
> The only non-linearity comes in through the second
> constraint.
> 
> > One may substitute 1-x1-x2 for x3 and use (let
> say) Lagrange
> > multipliers to get two non-linear equations with 2
> > unknowns for which there should be a function
> solving
> > them. 
> 
> Why would you want to use Lagrange multipliers? 
> Isn't that a bit of an
> overkill?  Once you substitute 1-x1-x2 for x3 in the
> second constraint,
> you have a quadratic equations in x1 and x2.  So for
> any given value of
> x1 you can solve for x2 (or for any given value of
> x2 you can solve for
> x1).  They still teach how to solve quadratic
> equations at school,
> don't they? ;-)
> 
> > Then you must find the points where the
> > constraint function intersects with the triangle
> > {x1>=0,x2>=0,x1+x2<=1}, which is easier (for each
> of
> > the 3 edges you get a non-linear equation in one
> > variable).
> 
> Even easier.  Take an x1 between 0 and 1.  If for
> that x1 the quadratic
> equation in x2 has no real solution, then x1 is not
> feasible.
> Otherwise find the values of x2 that solve the
> equation.  Use each of
> these values together with x1 to calculate
> corresponding values of x3.
> Then check these tuples for feasibility.  If they
> are feasible,
> evaluate the objective function and return the tuple
> with the larger
> function value.
> 
> All the calculations outlined in the paragraph above
> are easily
> implemented in R, e.g. the function polyroot()
> returns the roots of a
> polynomial.
> 
> Cheers,
> 
> 	Berwin
> 
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