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Subsections

# Termination Test. In other words, if is small, then the reduction in that occurs at the point is close to the greatest reduction that is allowed by the trust region constraint.
Proof
for any , we have the identity:  ( using : )  ( using : )    (4.23)

If we choose such that , we have:  (4.24)

From 4.27, using the 2 hypothesis:    ( Using Equation 4.25: )    (4.25)

Combining 4.28 and 4.29, we obtain finally 4.26.

## is near the boundary of the trust region: normal case

Lemma From the hypothesis:      (4.26)

Combining 4.31 and 4.27 when reveals that:       (4.27)

The required inequality 4.30 is immediate from 4.28 and 4.32.
We will use this lemma with .

## is inside the trust region: hard case

We will choose as (see paragraph containing Equation 4.15 for the meaning of and ): (4.28)

Thus, the condition for ending the trust region calculation simplifies to the inequality: (4.29)

We will choose .    Next: An estimation of the Up: The Trust-Region subproblem Previous: The Rayleigh quotient trick   Contents
Frank Vanden Berghen 2004-04-19