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Nelder–Mead can stop on a nonminimizing flat plateau

Example

Define the continuous objective on R by f(x)=0 for x≤2, f(x)=2−x for 2<x<3, and f(x)=−1 for x≥3. Run Finite-budget Nelder–Mead simplex procedure from the one-dimensional simplex (0,1) with diameter tolerance 0.1, spread tolerance 0.01, and call budget 14. It returns tolerance-met after four shrink updates, with simplex (0,1/16), returned best point 0, and objective spread 0, although the global minimum is −1 on [3,∞).

Facts & Assumptions

Given: Exact branch comparisons, tie rules, shrink rule, tolerance test, and call accounting Finite-budget Nelder–Mead simplex procedure.

[A1]

The stated piecewise formula defines f on all of R; its pieces agree at 2 and 3.

[F1]

The algorithm keeps the lower-slot vertex on objective ties, uses strict inside-contraction acceptance, and shrinks toward the best vertex (Finite-budget Nelder–Mead simplex procedure).

Verification

1.1givenA1F1algebra

The two initial values at 0,1 are both 0; the slot tie rule makes 0 best and 1 worst. At a simplex (0,h) with 0<h≤1, reflection is −h, with value 0=f(h). Since the reflected value ties the worst value, the inside-contraction branch applies; its point is h/2 and also has value 0, so strict inside-contraction acceptance fails and shrink produces (0,h/2).

2.1givenstep 1.1algebra∎

Starting at h=1, four repetitions give h=1/16=0.0625<0.1 and the spread remains 0<0.01. Initial evaluations use two calls and each iteration uses reflection, inside contraction, and a fresh shrink evaluation, so the total is 2+4⋅3=14 calls. All evaluated points lie in [−1,1]⊂(−∞,2], where the objective is flat; meanwhile f(x)≥−1 everywhere and equals −1 for x≥3. Thus the prescribed finite tolerance is met at a non-global minimizer.

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