Mikhail Moklyachuk - Convex Optimization

4 4) The function as ∥x∥ → ∞. By the corollary of the Weierstrass theorem, a solution of the problem exists. Since the critical point is unique, the solution of the problem can be only this point.

Answer . ∈ absmin, S min= 3.

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

Figure 1.1. Example 1.5

Solution . Figure 1.1depicts the admissible set of the problem and the level line of the objective function. The solution of the problem is the point . Active at this point are the first and the second restrictions. Wherein

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

at the point can only be performed with

Gradients and in this case are linearly dependent.

Answer . ∈ absmin, S min= 0.

EXAMPLE 1.6.– Solve the convex constrained optimization problem

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

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

At point , the first and third restrictions are active, and the second is passive. Therefore, λ 2= 0. As a result, we obtain a system for determining λ 1and λ 3:

System solution , λ 3= 1/2. The point is a solution of the problem. Make sure that there are no other stationary points and, therefore, no solution to the problem (see Figure 1.2).

Answer . ∈ absmin, .

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

on the set of solutions of the system

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

Figure 1.2. Example 1.6

The system for determining stationary points has the form

[1.4]

[1.5]

[1.6]

[1.7]

[1.8]

Let x i= 0. Then it follows from the system that x 2≥ 1, . Hence either x 2= 1 or x 2≥ –1. In other case, λ 1= 0. If x 2< –1, then λ 2= 0. But then λ 1= 0, which contradicts the conditions of the problem. Now we easily get the first two groups of system solutions.