Optimization with constraint.

Andreas Klein <klein82517 <at> yahoo.de> writes:

Hello.

I have some problems, when I try to model an
optimization problem with some constraints.

The original problem cannot be solved analytically, so
I have to use routines like "Simulated Annealing" or
"Sequential Quadric Programming".

But to see how all this works in R, I would like to
start with some simple problem to get to know the
basics:

The Problem:
min f(x1,x2)= (x1)^2 + (x2)^2
s.t. x1 + x2 = 1

The analytical solution:
x1 = 0.5
x2 = 0.5

Does someone have some suggestions how to model it in
R with the given functions optim or constrOptim with
respect to the routines "SANN" or "SQP" to obtain the
analytical solutions numerically?

In optimization problems, very often you have to replace an equality by two
inequalities, that is you replace  x1 + x2 = 1  with

    min f(x1,x2)= (x1)^2 + (x2)^2
    s.t.  x1 + x2 >= 1
          x1 + x2 <= 1

The problem with your example is that there is no 'interior' starting point for
this formulation while the documentation for constrOptim requests:

    The starting value must be in the interior of the feasible region,
    but the minimum may be on the boundary.

You can 'relax', e.g., the second inequality with  x1 + x2 <= 1.0001 and use
(1.00005, 0.0) as starting point, and you will get a solution:

A <- matrix(c(1, 1, -1, -1), 2)
b <- c(1, -1.0001)

fr <- function (x) { x1 <- x[1]; x2 <- x[2]; x1^2 + x2^2 }

constrOptim(c(1.00005, 0.0), fr, NULL, ui=t(A), ci=b)

$par
    [1] 0.5000232 0.4999768
    $value
    [1] 0.5
    [...]
    $barrier.value
    [1] 9.21047e-08

where the accuracy of the solution is certainly not excellent, but the solution
is correctly fulfilling  x1 + x2 = 1.