
The penalty constraint worked great with "optim". Thanks Andy.


--- On Wed, 11/4/09, apjaworski@mmm.com <apjaworski@mmm.com> wrote:

From: apjaworski@mmm.com <apjaworski@mmm.com>
Subject: Re: [R] Constrained Optimization
To: "s t" <thampy77@yahoo.com>
Cc: r-help@r-project.org, r-help-bounces@r-project.org
Date: Wednesday, November 4, 2009, 5:54 PM



Hi,



This probably does not answer your question,
which I presume is about the workings of constrOptim function, but I have
a couple of comments and different solutions of your problem.



1.  In general, in the problem
of this type, one can incorporate the equality constraint(s) into the objective
function by adding a penalty term for each constraint.  A pretty crude
solution would be to redefine your objective function as, say,

function(x){

  x1 = x[1]

  x2 = x[2]

  x3 = x[3]

  -(x1*(1-x1) + x2*(1-x2) + x3*(1-x3))
+ 1000*(x1+x2+x3-1)^2

}

and solve a minimization problem with
simple 0/1 box constraints.  Notice the minus in front of your objective
function to turn the problem into the minimization one.



2.  Second possibility would be
to formulate the Lagrangian using just the equality constraint.  Then
we can analytically differentiate the Lagrangian WRT x1, x2, x3, and the
multiplier and equate the derivatives to zero.  This will form a set
of, generally nonlinear, equations to solve.  Again, one can use a
minimization with box constraints to minimize the sum of squared left-hand-sides
of the equations.  This is probably not the most efficient method
of solving a system of equations but it often works and is easy to program.
 Here is an example of the objective function in this case

function(x){

  x1 = x[1]

  x2 = x[2]

  x3 = x[3]

  x4 = x[4]

  (1-2*x1+x4)^2 + (1-2*x2+x4)^2
+ (1-2*x3+x4)^2 + (x1+x2+x3-1)^2

}



3.  This is just a follow-up to
2.  In your case, the partial derivatives of the Lagrangian and the
equality constraint are linear.  So if this was not a toy problem
but a real one, one can just solve a system of linear equations

and check that the solution is inside
the box constraints.  In your case

> aa

     [,1] [,2] [,3] [,4]

[1,]   -2    0  
 0    1

[2,]    0   -2  
 0    1

[3,]    0    0  
-2    1

[4,]    1    1  
 1    0

> bb

[1] -1 -1 -1  1

> solve(aa, bb)

[1]  0.3333333  0.3333333
 0.3333333 -0.3333333



Hope this helps,



Andy

__________________________________

Andy Jaworski

518-1-01

Process Laboratory

3M Corporate Research Laboratory

-----

E-mail: apjaworski@mmm.com

Tel:  (651) 733-6092

Fax:  (651) 736-3122








From:


To:
r-help@r-project.org

Date:
11/04/2009 04:48 PM

Subject:
[R] Constrained Optimization

Sent by:
<r-help-bounces@r-project.org>








Hi All,

I'm trying to do the following constrained optimization example.

Maximize x1*(1-x1) + x2*(1-x2) + x3*(1-x3)

s.t. x1 + x2 + x3 = 1

x1 >= 0 and x1 <= 1

x2 >= 0 and x2 <= 1

x3 >= 0 and x3 <= 1

which are the constraints.

I'm expecting the answer x1=x2=x3 = 1/3.



I tried the "constrOptim" function in R and I'm running into
some issues.

I first start off by setting my equalities to inequalities

x1+x2+x3 >= 1

x1+x2+x3 <= 1.001



However, I get a 

"Error in constrOptim(c(1.00004, 0, 0), fr, grr, ui = t(A), ci = b)
: 

  initial value not feasible

Execution halted" error.

The values 1.00004,0,0 are well within the constraints domain.



fr = function(x){

  x1 = x[1]

  x2 = x[2]

  x3 = x[3]

  x1*(1-x1) + x2*(1-x2) + x3*(1-x3)

}



grr = function(x){

  x1 = x[1]

  x2 = x[2]

  x3 = x[3]

  c(1-2*x1,1-2*x2,1-2*x3)

}



A = matrix(c(1,1,1,-1,-1,-1,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1),3,8,byrow=T)

b = c(1,-1.001,0,0,0,1,1,1)



y = constrOptim(c(1.00004,0,0),fr,grr,ui = t(A),ci = b)



Any help/pointers greatly appreciated.



Thanks

















      

                
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