Writing nonlinear constraint in fmincon

Hello,

I minimize a function with 4 parameters on fmincon. The vector of parameters is b=[a1 a2 b1 b2 ]. How can I add a nonlinear constraint such that 5/b1<b2?

[Matt J]

Note that it can be critically misleading to people to say you want 5/b1<b2 if you really mean 5/b1<=b2. Theoretically, for example, the following minimization problem no solution:

min. x, 
s.t. x>0

but the solution to,

min. x
s.t. x>=0

is x=0.

[george pepper]

Thanks a lot! This is great.

Answer 1

[Walter Roberson]

5/b1 < b2 implies 5 < b2*b1 implies 0 < b2*b1 - 5 implies b2*b1 - 5 < 0 implies b2*b1 - 5 + delta = 0 for some positive delta.

This leads to the constraint

delta = eps(realmin);
b(3)*b(4) - 5 + delta  %<= 0 implied

However I would suggest you think more about your boundary constraint. Is 5/b1 == b2 an actual problem for your situation? If it is then you run serious risks that due to round-off issues, that whatever calculation fails with 5/b1 == b2, will not round in a "fortunate" way.

I personally would probably not use eps(realmin) for the delta: I would be more likely to use 5*(1-eps) instead of 5+delta

[Matt J]

5/b1 < b2 implies 5 < b2*b1

That is only true under the assumption that b1>0

[Walter Roberson]

True, I forgot about the case of negatives.

You could always code

5/b(3) - b(4)

and make the appropriate alteration for the border equality... provided that you know that b(3) is never 0.

[george pepper]

Thank you very much!