Finite-budget Nelder–Mead always returns a status

Each continuing iteration consumes at least one call for reflection. Every further expansion, contraction, or shrink evaluation also consumes a call, and the counter is bounded by $B$; before any call that cannot fit in the remaining budget, the procedure returns budget-exhausted. It therefore cannot continue indefinitely and returns a listed status after at most $B$ calls.

The only tolerance-met branch is the explicit test $D\le {\epsilon }_x$ and $S\le {\epsilon }_f$ on a committed simplex, so that status implies both inequalities. None of the algorithm's finite statuses asserts a minimizer without an additional theorem.