Writing nonlinear constraint in fmincon

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george pepper 2020-10-23

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评论： george pepper 2020-10-27

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?

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Matt J 2020-10-24

编辑：Matt J 2020-10-24

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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 2020-10-27

Thanks a lot! This is great.

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Answer 1

Walter Roberson 2020-10-24

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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