Instead of solving max f(x) subject to sum(x) = 1 g(x) = 0 h(x) <= 0
where x is in R^n solve max f(x(y)) subject to g(x(y)) = 0 h(x(y)) <= 0 where y is in R^(n-1) and x(y) = [1-sum(y) y(1) y(2) … y(n-1)] This way sum(x(y)) = 1 for all y. For a simple example max u(c1,c2) c1 + c2 = 1 c1>= 0 c2 >= 0 becomes max u(1-c2, c2) 1-c2 >= 0 c2 >= 0 -Grey > On Apr 27, 2016, at 11:11 AM, David Morris <otha...@othalan.net> wrote: > > On Wed, Apr 27, 2016 at 9:44 PM, Grey Gordon <greygor...@gmail.com> wrote: > > Instead of using sum(x) = 1 as an equality constraint, perhaps you can > directly take x(1) = 1 - sum(x(2:end)) and substitute it into the problem > directly. > > Grey, can you expand on this a bit? Do you mean use that as a penalty in the > optimization function, or something else? > > David _______________________________________________ NLopt-discuss mailing list NLopt-discuss@ab-initio.mit.edu http://ab-initio.mit.edu/cgi-bin/mailman/listinfo/nlopt-discuss
