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ID: past-exam-of-the-continuous-optimisation-course-of-the-university-of-oxford/2011/1/iv/solution

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Past exam of the continuous optimisation course of the University of Oxford / 2011 / 1 / iv / Solution by

Codex  0 2026-10-31

Convert to minimization with

f = - F

and retain

g \geq 0

. A continuously differentiable inequality augmented Lagrangian method uses the merit function

Φ (x, λ, μ) = f (x) + \frac{1}{2 μ} i = 1 \sum 3 (min {g_{i} (x) - μ λ_{i}, 0}^{2} - μ^{2} λ_{i}^{2}), μ > 0, λ \geq 0.

For

g_{i} \leq μ λ_{i}

its contribution is

- λ_{i} g_{i} + g_{i}^{2} / (2 μ)

; otherwise it is the constant

- μ λ_{i}^{2} /2

. This shifted truncation permits multipliers to decrease as well as increase while remaining nonnegative.

A useful initial estimate comes from the preceding pure quadratic penalty method at

μ_{- 1}

:

λ_{i}^{[0]} = max {0, - g_{i} (x^{[0]}) / μ_{- 1}} .

Here

g (- 1, - 1) = (3, 18, - 2)

, so with

μ_{- 1} = 2

choose

λ^{[0]} = (0, 0, 1) .

Then use this estimate and

μ_{0} = 1

in the first augmented subproblem. Initial estimates are a choice, rather than uniquely specified data; zero multipliers would also define an initialization.

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