Convex Optimization. Mikhail Moklyachuk
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Название: Convex Optimization

Автор: Mikhail Moklyachuk

Издательство: John Wiley & Sons Limited

Жанр: Математика

Серия:

isbn: 9781119804086

isbn:

СКАЧАТЬ alt="image"/> ∈ absmin, Smin = 3.

      EXAMPLE 1.5.– An example of the irregular problem. Consider the constrained optimization problem

image image

      The vector image cannot be represented as a linear combination of vectors image and image. The relation

image

      at the point image can only be performed with

image

      Gradients image and image in this case are linearly dependent.

      Answer. image ∈ absmin, Smin = 0.

      EXAMPLE 1.6.– Solve the convex constrained optimization problem

image

      Solution. Slater’s condition is fulfilled. Therefore, we write the regular Lagrange function:

image

      The system for finding stationary points in this case (s = 0, n = 2, k = m = 3) can be written in the form

image

      At point image, the first and third restrictions are active, and the second is passive. Therefore, λ2 = 0. As a result, we obtain a system for determining λ1 and λ3:

image

      Answer. image ∈ absmin, image.

      EXAMPLE 1.7.– Let the numbers a > 0, b > 0, and let a < b. Find points of the local minimum and the local maximum of the function

image

      on the set of solutions of the system

image

      Solution. We denote this set by X. Let us write the Lagrange function

image Schematic illustration of a solution with no stationary points.

      The system for determining stationary points has the form

      [1.4]image

      [1.5]image

      [1.6]image

      [1.7]image

      [1.8]image

      Let xi = 0. Then it follows from the system that x2 ≥ 1, image. Hence either x2 = 1 or x2 ≥ –1. In other case, λ1 = 0. If x2 < –1, then λ2 = 0. But then λ1 = 0, which contradicts the conditions of the problem. Now we easily get the first two groups of system solutions.

      1 1) x1 = 0, x2 = 1, bλ0 + 3λ1 − 2λ2 = 0, λ1 ≥ 0, λ2 ≥ 0, (λ1, λ2) ≠ 0;

      2 2) x1 = 0, x2 = −1, bλ0 − 2λ2 = 0, λ1 = 0, λ2 > 0.

      Similarly, assuming that x2 = 0, we obtain two other groups of solutions:

      1 3) x1 = 1, x2 = 0, aλ0 + 3λ1 − 2λ2 = 0, λ1 ≥ 0, λ2 ≥ 0, (λ1, λ2) ≠ 0;

      2 4) x1 = −1, x2 = 0, aλ0 − 2λ2 = 0, λ1 = 0, λ2 > 0.

      Assume that x1 ≠ 0, x2 ≠ 0. Then equations of the system can be presented in the form

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