I am optimizing a complicated function. That function depends on 2 variables t1 and t2. I can derive first order conditions that result in a system of 2 equations. I can also derive conditions for when t1 < t2, t1 > t2, and t1 == t2. Obviously, only one of those conditions will contain the optimum. …

I replaced fullsol! with this one: function fullsol2!(t; x=10.0) sol12 = [NaN, NaN] sol21 = [NaN, NaN] try sol12 = foc12!(t, x=x) catch sol12 = [NaN, NaN] end try sol21 = foc21!(t, x=x) catch sol21 = [NaN, NaN] end true…

Let me try to clarify: You have a parameter x, a function f1(y; x) and another function f2(y; x) (the FOCs). You know from theory that there is exactly one y* for which either f1(y*; x) == 0 or f2(y*; x) == 0 (but not both) (over the interval y* in [x, 30]). You are trying to find this unique y*.…

@hendri54 thanks for taking the time to read this convoluted writeup. Let me try to clarify the problem. I have a function N() to maximize that depends on t1 and t2, and some parameters x (you can think of these as covariates). I can try to optimize N directly, but there is a theoretical case when …

OK - then let’s go back to my proposals. If the FOCs are monotone: easy. If they are not: are you sure there is at most one root for each case? If not: find_zeros would be needed instead of find_zero. If yes: I would use a form of grid search to find one interval where the sign changes. How to i…

Well, Bisection (and other bracketing methods) throws the error when you don’t have a bracketing interval. You can check this before the call (avoiding try-catch) with 2 function evaluations (ie. is f(a)*f(b) < 0, which if costly to compute, the values may be reused in the bisection call). You might…

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amrods June 30, 2021, 12:36am 6

BTW, I should have linked this writeup of the problem from another post: Status of integration, optimization, and nl solvers - #14 by amrods

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