[R] Quadratic Constraints

[Hans W. Borchers]

The package lpSolve (that I have recommended before) supports so-called
'semi-continuous variables', that is

    "Semi-continuous variables are variables that must take a value between
    their their minimum and maximum or zero. So these variables are treated
    the same as regular variables, except that a value of zero is also
    accepted, even if a minimum bigger than zero is set on the variable."

which exactly how you want to handle your variable x3.
For an example, see the documentation at <lpsolve.sourceforge.net/5.5/>.

By the way, the minimum of your problem is 44.64 (manual calculation).



vikrant S wrote:
> 
> HI All,
> I am unable to solve a optimization Problem Please Help Me out of this to
> solve. The Optimization problem is as follows :- 
> My objective function is linear and one of the constraint is quadratic. 
> 
> Min z = 5 * X1 + 9* X2  + 7.15 *X3 + 2 * X4
> subject to
>  X1 + X2 + X3 +X4  = 9
>  X1  + X4 < = 6.55
>  X3(X3 - 3.5) >=0
>  X1,X2,X3,X4 >=0
>  Now the problem is how to solve this kind of problem. Which package
> should be used to handle such problems. Please explain with an example. 
> Another problem is that I have to cases to  be solve in this problem.
> case 1:-) If X3 = 0
> case 2 :-) If X3 > 0 then X3 > 3.6
> I want to handle both this case in one problem so the quadratic
> constraints is written
> The thing is that I want to evaluate my objective function for both cases
> and which ever is optimum that solution i need,
> Here I don't want to use the If Else condition and repeat the program. IS
> there any other better way in which i could solve this problem?
> If not please try to provide me the solution for my original problem
> having a quadratic constraint.
> 

-- 
View this message in context: http://www.nabble.com/Quadratic-Constraints-tp25528480p25529374.html
Sent from the R help mailing list archive at Nabble.com.