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matlaboptimizationminimization
Frey
Frey

Reputation: 325

Solve optimization using fmincon MATLAB when objective function is in constraints

I want to solve: enter image description here

I use following MATLAB code, but it does not work.

Can someone please guide me? 

function f=objfun

f=-f;

function [c1,c2,c3]=constraint(x)
a1=1.1; a2=1.1; a3=1.1;
c1=f-log(a1)-log(x(1)/(x(1)+1)); 
c2=f-log(a2)-log(x(2)/(x(2)+1))-log(1-x(1)); 
c3=f-log(a3)-log(1-x(1))-log(1-x(2));


x0=[0.01;0.01]; 
[x,fval]=fmincon('objfun',x0,[],[],[],[],[0;0],[1;1],'constraint')

Upvotes: 1

Views: 1034

Answers (2)

Matthew Gunn
Matthew Gunn

Reputation: 4519

Can also solve this in the convex optimization package CVX with the following MATLAB code:

cvx_begin
 variables T(1);
 variables x1(1);
 variables x2(1);

 maximize(T)
 subject to:
  log(a1) + x1 - log_sum_exp([0, x1]) >= T;
  log(a2) + x2 - log_sum_exp([0, x2]) + log(1 - exp(x1)) >= T;
  log(a3) +  log(1 - exp(x1)) +  log(1 - exp(x2)) >= T;
  x1 <= 0;
  x2 <= 0;
cvx_end
l1 = exp(x1); l2 = exp(x2);

To use CVX, each constraint and the objective function has to be written in a way that is proveably convex using CVX's ruleset. Making the substitution x1 = log(l1) and x2 = log(l2) allows one to do that. Note that: log_sum_exp([0,x1]) = log(exp(0) + exp(x1)) = log(1 + l1)

This also returns the answers: l1 = .3383, l2 = .6180, T = -1.2800

Upvotes: 0

David
David

Reputation: 8459

You need to flip the problem around a bit. You are trying to find the point x (which is (l_1,l_2)) that makes the minimum of the 3 LHS functions the largest. So, you can rewrite your problem as, in pseudocode,

maximise, by varying x in [0,1] X [0,1]
       min([log(a1)+log(x(1)/(x(1)+1)) ...
            log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
            log(a3)+log(1-x(1))+log(1-x(2))])

Since Matlab has fmincon, rewrite this as a minimisation problem,

minimise, by varying x in [0,1] X [0,1]
       max(-[log(a1)+log(x(1)/(x(1)+1)) ...
             log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
             log(a3)+log(1-x(1))+log(1-x(2))])

So the actual code is

F=@(x) max(-[log(a1)+log(x(1)/(x(1)+1)) ...
             log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
             log(a3)+log(1-x(1))+log(1-x(2))])
[L,fval]=fmincon(F,[0.5 0.5])

which returns

L =
    0.3383    0.6180
fval =
    1.2800

Upvotes: 3

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