5. Models of bilevel disjunctive programming problems

The manifold M is understood from the context. The connection Γ<sup>h</sup> ij can be realized in each case, imposing convexity conditions.

Example 5.1 Let us solve the problem (cite [7], p. 7; [9]):

$$\min\_{(\mathbf{x}\_1, \mathbf{x}\_2)} F(\mathbf{x}\_1, \mathbf{x}\_2, y) = (\mathbf{x}\_1 - y, \mathbf{x}\_2)$$

subject to

Theorem 4.2 Suppose M<sup>1</sup> is a compact manifold. If for each x ∈ M1, at least one partial function

Proof. In our hypothesis, the set ψð Þx is nonvoid, for any x, and the compacity assures the

In the next Theorem, we shall use the Value Function Method or Utility Function Method. □

y ! L xð Þ¼ ; y u Fð Þ <sup>1</sup>ð Þ x; y ;…; Fkð Þ x; y

min<sup>x</sup> ½ � f xð Þ ; <sup>y</sup> : <sup>y</sup> <sup>∈</sup>ψð Þ<sup>x</sup>

Proof. Let minyL xð Þ¼ ; <sup>y</sup> L x; <sup>y</sup><sup>∗</sup> ð Þ. Suppose that for each <sup>i</sup> <sup>¼</sup> <sup>1</sup>, …, k, minyFið Þ <sup>x</sup>; <sup>y</sup> <sup>&</sup>lt; Fi <sup>x</sup>; <sup>y</sup><sup>∗</sup> ð Þ. Then <sup>y</sup><sup>∗</sup> would not be minimum point for the partial function <sup>y</sup> ! L xð Þ ; <sup>y</sup> . Hence, there exists

Boundedness of f implies that the bilevel problem has solution once it is well-posed, but the

An important concept for making wise tradeoffs among competing objectives is bilevel dis-

We present an exact algorithm for obtaining the bilevel disjunctive solutions to the multi-

min<sup>x</sup> ½ � f xð Þ ; <sup>y</sup> ; <sup>y</sup>∈ψð Þ<sup>x</sup> :

From numerical point of view, we can use the Newton algorithm for optimization on affine

junctive programming optimality, on affine manifolds, introduced in this chapter.

ð Þ¼ x Argmin<sup>y</sup><sup>∈</sup> <sup>M</sup><sup>2</sup>

fact that the problem is well-posed is shown in the first part of the proof.

ð Þ<sup>x</sup> . □

Fið Þ x; y , i ¼ 1, …, m:

ð Þx be a subset in M<sup>2</sup> representing the mapping of optimal solutions for the

ð Þx 6¼ ∅. Moreover, if f xð Þ ; y is bounded, then

y ! Fið Þ x; y is affine convex and has a critical point, then the problem OBDP ð Þ has a solution.

existence of minxf xð Þ ; ψð Þx .

128 Optimization Algorithms - Examples

the bilevel problem

an index i such that y<sup>∗</sup> ∈ψ<sup>i</sup>

has solution.

Step 1: Solve

Let <sup>ψ</sup>ð Þ¼ <sup>x</sup> <sup>∪</sup><sup>r</sup>

<sup>i</sup>¼<sup>1</sup>ψ<sup>i</sup>

follower multi-objective function.

manifolds, which is given in [19].

Step 2: Build the mapping f x, ð ψð Þx .

Step 3: Solve the leader's following program

Theorem 4.3 If a C<sup>1</sup> increasing scalarization partial function

has a minimum, then there exists an index i such that ψ<sup>i</sup>

4.1. Bilevel disjunctive programming algorithm

objective optimization in the following section.

ψi

$$(\mathbf{x}\_1, \mathbf{x}\_2) \in A \operatorname{argmin}\_{(\mathbf{x}\_1, \mathbf{x}\_2)} \left\{ (\mathbf{x}\_1, \mathbf{x}\_2) \, | \, y^2 - \mathbf{x}\_1^2 - \mathbf{x}\_2^2 \ge \mathbf{0} \right\}.$$

$$1 + \mathbf{x}\_1 + \mathbf{x}\_2 \ge \mathbf{0}, \, -1 \le \mathbf{x}\_1, \mathbf{x}\_2 \le 1, \, 0 \le y \le 1.$$

Both the lower and the upper level optimization tasks have two objectives each. For a fixed y value, the feasible region of the lower-level problem is the area inside a circle with center at origin xð Þ <sup>1</sup> ¼ x<sup>2</sup> ¼ 0 and radius equal to y. The Pareto-optimal set for the lower-level optimization task, preserving a fixed y, is the bottom-left quarter of the circle,

$$\left\{ (\mathfrak{x}\_1, \mathfrak{x}\_2) \in \mathbb{R}^2 \, | \, \mathfrak{x}\_1^2 + \mathfrak{x}\_2^2 = y^2, \, \mathfrak{x}\_1 \le 0, \, \mathfrak{x}\_2 \le 0 \right\}.$$

The linear constraint in the upper level optimization task does not allow the entire quarter circle to be feasible for some y. Thus, at most a couple of points from the quarter circle belongs to the Pareto-optimal set of the overall problem. Eichfelder [8] reported the following Pareto-optimal set of solutions

$$A = \left\{ (\mathbf{x}\_1, \mathbf{x}\_2, y) \in \mathbb{R}^3 \, | \, \mathbf{x}\_1 = -1 - \mathbf{x}\_2, \mathbf{x}\_2 = -\frac{1}{2} \pm \frac{1}{2}\sqrt{2y^2 - 1}, y \in \left[\frac{1}{\sqrt{2}}, 1\right] \right\}.$$

The Pareto-optimal front in F<sup>1</sup> � F<sup>2</sup> space can be written in parametric form

$$\left\{ (F\_1, F\_2) \in \mathbb{R}^2 \, | \, F\_1 = -1 - F\_2 - t, F\_2 = -\frac{1}{2} \pm \frac{1}{2}\sqrt{2t^2 - 1}, t \in \left[\frac{1}{\sqrt{2}}, 1\right] \right\}.$$

Example 5.2 Consider the bilevel programming problem

$$\min\_{\mathbf{x}} \left[ (\mathbf{x} - \mathbf{y})^2 + \mathbf{x}^2 : -2\mathbf{0} \le \mathbf{x} \le 2\mathbf{0}, \mathbf{y} \in \psi(\mathbf{x}) \right]\_{\mathbf{x}}$$

where the set-valued function is

$$
\psi(\mathbf{x}) = \operatorname{Argmin}\_{y} [\mathbf{x}y : -\mathbf{x} - \mathbf{1} \lhd y \lhd -\mathbf{x} + 1].
$$

Explicitly,

$$\psi(\mathbf{x}) = \begin{cases} [-1, 1] & \text{if } \mathbf{x} = \mathbf{0} \\ -\mathbf{x} - 1 & \text{if } \mathbf{x} > \mathbf{0} \\ -\mathbf{x} + 1 & \text{if } \mathbf{x} < \mathbf{0} \end{cases}$$

Since F xð Þ¼ ; y ð Þ x � y <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>, we get

$$F(\mathbf{x}, \psi(\mathbf{x})) = \begin{cases} [0, 1] & \text{if } \mathbf{x} = \mathbf{0} \\ (-2\mathbf{x} - 1)^2 + \mathbf{x}^2 & \text{if } \mathbf{x} > \mathbf{0} \\ (-2\mathbf{x} + 1)^2 + \mathbf{x}^2 & \text{if } \mathbf{x} < \mathbf{0} \end{cases}$$

The objective f xð Þ¼ ; y ð Þ x � y

f xð Þ¼ ;ψð Þx

In context, we find the inferior envelope

active, then the optimal solution is ð Þ 0; 1 .

function with two vector variables

which is called minimum function.

ties related to increase and convexity ideas.

6. Properties of minimum functions

<sup>1</sup>Γ � �, the leader decision affine manifold, and M2,

and taking the infimum after one variable, let say y, we build a function

f xð Þ¼ inf y

and then

Let M1,

8 >>>>>><

>>>>>>:

f xð Þ¼ ; y xð Þ

y xð Þ¼

8 ><

>:

Since ð Þ <sup>2</sup><sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>&</sup>gt; <sup>0</sup>, the unique optimal solution is x<sup>∘</sup> ; <sup>y</sup><sup>∘</sup> ð Þ¼ ð Þ <sup>0</sup>; <sup>0</sup> .

8 >>><

>>>:

<sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> and the multi-function <sup>ψ</sup>ð Þ<sup>x</sup> produce a multi-function

n o if <sup>0</sup> <sup>&</sup>lt; <sup>x</sup> <sup>≤</sup> <sup>1</sup>

0 if x ¼ 0 �x þ 1 if 0 < x ≤ 1 0 if x > 1 �x þ 1 if x < 0

0 if x ¼ 0 ð Þ <sup>2</sup><sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> if x<sup>∈</sup> ð Þ �∞; <sup>0</sup> <sup>∪</sup>ð � <sup>0</sup>; <sup>1</sup> 2x<sup>2</sup> if x > 1:

If we consider only ψ1ð Þx as active, then the unique optimal solution ð Þ 0; 0 is maintained. If ψ2ð Þx is

be two connected affine manifolds of dimension n<sup>1</sup> and n2, respectively. Starting from a

φ : M<sup>1</sup> � M<sup>2</sup> ! R, xð Þ! ; y φð Þ x; y ,

A minimum function is usually specified by a pointwise mapping a of the manifold M<sup>1</sup> in the subsets of a manifold M<sup>2</sup> and by a functional φð Þ x; y on M<sup>1</sup> � M2. In this context, some differential properties of such functions were previously examined in [4]. Now we add new proper-

First we give a new proof to Brian White Theorem (see Mean Curvature Flow, p. 7, Internet 2017).

f g φð Þ x; y : y∈ a xð Þ ,

n o if x <sup>&</sup>gt; <sup>1</sup>

ð Þ <sup>2</sup><sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> if x <sup>&</sup>lt; <sup>0</sup>:

<sup>2</sup>Γ � �, the follower decision affine manifold,

ð Þ <sup>2</sup><sup>x</sup> <sup>þ</sup> <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>;ð Þ <sup>2</sup><sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>

ð Þ <sup>2</sup><sup>x</sup> <sup>þ</sup> <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>; <sup>2</sup>x<sup>2</sup>

½ � 0; 1 if x ¼ 0

Bilevel Disjunctive Optimization on Affine Manifolds http://dx.doi.org/10.5772/intechopen.75643 131

on the regions where the functions are defined.

Taking into account ð Þ �2<sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>&</sup>gt; <sup>0</sup> and ð Þ �2<sup>x</sup> <sup>þ</sup> <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>&</sup>gt; <sup>0</sup>, it follows that x<sup>∘</sup> ; <sup>y</sup><sup>∘</sup> ð Þ¼ ð Þ <sup>0</sup>; <sup>0</sup> is the unique optimistic optimal solution of the problem. Now, if the leader is not exactly enough in choosing his solution, then the real outcome of the problem has an objective function value above 1 which is far away from the optimistic optimal value zero.

Example 5.3 Let F xð Þ¼ ; y ð Þ F1ð Þ x; y ; F2ð Þ x; y and a Pareto disjunctive problem

$$
\psi(\mathbf{x}) = \operatorname{Argmin}\_{y} F(\mathbf{x}, y) = \operatorname{Argmin}\_{y} F\_1(\mathbf{x}, y) \cup \operatorname{Argmin}\_{y} F\_2(\mathbf{x}, y).
$$

Then it appears a bilevel disjunctive programming problem of the form

$$\min\_{\mathbf{x}} \ [f(\mathbf{x}, y), y \in \psi(\mathbf{x})].$$

This problem is interesting excepting the case ψð Þ¼ x Ø, ∀x. If y ! F1ð Þ x; y and y ! F2ð Þ x; y are convex functions, then ψð Þx 6¼ Ø.

To write an example, we use

$$F\_1(\mathbf{x}, y) = \left[ \mathbf{x}y : -\mathbf{x} - 1 \lhd y \lessdot -\mathbf{x} + 1 \right] \\ F\_2(\mathbf{x}, y) = \left[ \mathbf{x}^2 + y^2 : y \rhd -\mathbf{x} + 1 \right] \mathbf{x}$$

and we consider a bilevel disjunctive programming problem of the form

$$\min\_{\mathbf{x}} \left[ (\mathbf{x} - \mathbf{y})^2 + \mathbf{x}^2 \; :\; -20 \le \mathbf{x} \le 20, y \in \psi(\mathbf{x}) \right]\_{\mathbf{y}}$$

with

$$
\psi(\mathbf{x}) = \psi\_1(\mathbf{x}) \cup \psi\_2(\mathbf{x}) \,.
$$

where

$$\psi\_1(\mathbf{x}) = \operatorname{Argmin}\_y \left[ \mathbf{x} y : -\mathbf{x} - 1 \lessgtr y \lessgtr -\mathbf{x} + 1 \right] = \begin{cases} [-1, 1] & \text{if} \quad \mathbf{x} = \mathbf{0} \\ -\mathbf{x} - 1 & \text{if} \quad \mathbf{x} > 0 \\ -\mathbf{x} + 1 & \text{if} \quad \mathbf{x} < 0 \end{cases}$$

$$\psi\_2(\mathbf{x}) = \operatorname{Argmin}\_y \left[ \mathbf{x}^2 + y^2 : y \succ -\mathbf{x} + 1 \right] = \begin{cases} -\mathbf{x} + 1 & \text{if} \quad \mathbf{x} \le 1 \\ 0 & \text{if} \quad \mathbf{x} > 1 \end{cases}$$

$$\psi(\mathbf{x}) = \begin{cases} [-1, 1] & \text{if} \quad \mathbf{x} = 0 \\ \{-\mathbf{x} - 1, -\mathbf{x} + 1\} & \text{if} \quad 0 < \mathbf{x} \le 1 \\ \{-\mathbf{x} - 1, 0\} & \text{if} \quad \mathbf{x} > 1 \\ -\mathbf{x} + 1 & \text{if} \quad \mathbf{x} < 0. \end{cases}$$

The objective f xð Þ¼ ; y ð Þ x � y <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> and the multi-function <sup>ψ</sup>ð Þ<sup>x</sup> produce a multi-function

$$f(\mathbf{x}, \psi(\mathbf{x})) = \begin{cases} \begin{bmatrix} 0, 1 \end{bmatrix} & \text{if} \quad \mathbf{x} = \mathbf{0} \\\\ \begin{Bmatrix} \left(2\mathbf{x} + 1\right)^2 + \mathbf{x}^2, \left(2\mathbf{x} - 1\right)^2 + \mathbf{x}^2 \end{Bmatrix} & \text{if} \quad \mathbf{0} < \mathbf{x} \le \mathbf{1}\_2 \\\\ \begin{Bmatrix} \left(2\mathbf{x} + 1\right)^2 + \mathbf{x}^2, 2\mathbf{x}^2 \end{Bmatrix} & \text{if} \quad \mathbf{x} > \mathbf{1} \\\\ \begin{Bmatrix} \left(2\mathbf{x} - 1\right)^2 + \mathbf{x}^2 & \text{if} \quad \mathbf{x} < \mathbf{0} . \end{Bmatrix} & \text{if} \quad \mathbf{x} < \mathbf{0} . \end{cases}$$

In context, we find the inferior envelope

$$y(\mathbf{x}) = \begin{cases} 0 & \text{if} \quad \mathbf{x} = \mathbf{0} \\ -\mathbf{x} + \mathbf{1} & \text{if} \quad \mathbf{0} < \mathbf{x} \le \mathbf{1} \\ 0 & \text{if} \quad \mathbf{x} > \mathbf{1} \\ -\mathbf{x} + \mathbf{1} & \text{if} \quad \mathbf{x} < \mathbf{0} \end{cases}$$

and then

Since F xð Þ¼ ; y ð Þ x � y

130 Optimization Algorithms - Examples

<sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup>, we get

on the regions where the functions are defined.

convex functions, then ψð Þx 6¼ Ø.

To write an example, we use

with

where

which is far away from the optimistic optimal value zero.

F xð Þ¼ ; ψð Þx

8 ><

>:

Example 5.3 Let F xð Þ¼ ; y ð Þ F1ð Þ x; y ; F2ð Þ x; y and a Pareto disjunctive problem

Then it appears a bilevel disjunctive programming problem of the form

and we consider a bilevel disjunctive programming problem of the form

min<sup>x</sup> ð Þ <sup>x</sup> � <sup>y</sup>

ψ1ð Þ¼ x Argminy½xy : �x � 1 ⩽y ⩽ � x þ 1� ¼

8 >>><

>>>:

ψð Þ¼ x

Taking into account ð Þ �2<sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>&</sup>gt; <sup>0</sup> and ð Þ �2<sup>x</sup> <sup>þ</sup> <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>&</sup>gt; <sup>0</sup>, it follows that x<sup>∘</sup> ; <sup>y</sup><sup>∘</sup> ð Þ¼ ð Þ <sup>0</sup>; <sup>0</sup> is the unique optimistic optimal solution of the problem. Now, if the leader is not exactly enough in choosing his solution, then the real outcome of the problem has an objective function value above 1

ψð Þ¼ x Argminy F xð Þ¼ ; y Argminy F1ð Þ x; y ∪ Argminy F2ð Þ x; y :

min<sup>x</sup> ½ � f xð Þ ; <sup>y</sup> ; <sup>y</sup>∈ψð Þ<sup>x</sup> :

This problem is interesting excepting the case ψð Þ¼ x Ø, ∀x. If y ! F1ð Þ x; y and y ! F2ð Þ x; y are

<sup>F</sup>1ð Þ¼ <sup>x</sup>; <sup>y</sup> ½ � xy : �<sup>x</sup> � <sup>1</sup> <sup>⩽</sup><sup>y</sup> <sup>⩽</sup> � <sup>x</sup> <sup>þ</sup> <sup>1</sup> , F2ð Þ¼ <sup>x</sup>; <sup>y</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> : <sup>y</sup> <sup>⩾</sup> � <sup>x</sup> <sup>þ</sup> <sup>1</sup> � �

ψð Þ¼ x ψ1ð Þx ∪ψ2ð Þx ,

<sup>ψ</sup>2ð Þ¼ <sup>x</sup> Argminy <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> : <sup>y</sup><sup>⩾</sup> � <sup>x</sup> <sup>þ</sup> <sup>1</sup> � � <sup>¼</sup> �<sup>x</sup> <sup>þ</sup> <sup>1</sup> if x <sup>≤</sup> <sup>1</sup>

<sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> : �<sup>20</sup> <sup>≤</sup> <sup>x</sup> <sup>≤</sup> <sup>20</sup>; <sup>y</sup>∈ψð Þ<sup>x</sup> h i

> 8 ><

> >:

�

½ � �1; 1 if x ¼ 0 f g �x � 1; �x þ 1 if 0 < x ≤ 1 f g �x � 1; 0 if x > 1 �x þ 1 if x < 0:

,

½ � �1; 1 if x ¼ 0 �x � 1 if x > 0 �x þ 1 if x < 0,

0 if x > 1,

½ � 0; 1 if x ¼ 0 ð Þ �2<sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> if x <sup>&</sup>gt; <sup>0</sup> ð Þ �2<sup>x</sup> <sup>þ</sup> <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> if x <sup>&</sup>lt; <sup>0</sup>:

$$f(\mathbf{x}, y(\mathbf{x})) = \begin{cases} 0 & \text{if} \quad \mathbf{x} = \mathbf{0} \\ \left(2\mathbf{x} - \mathbf{1}\right)^2 + \mathbf{x}^2 & \text{if} \quad \mathbf{x} \in (-\infty, 0) \cup (0, 1] \\ 2\mathbf{x}^2 & \text{if} \quad \mathbf{x} > 1. \end{cases}$$

Since ð Þ <sup>2</sup><sup>x</sup> � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>x</sup><sup>2</sup> <sup>&</sup>gt; <sup>0</sup>, the unique optimal solution is x<sup>∘</sup> ; <sup>y</sup><sup>∘</sup> ð Þ¼ ð Þ <sup>0</sup>; <sup>0</sup> .

If we consider only ψ1ð Þx as active, then the unique optimal solution ð Þ 0; 0 is maintained. If ψ2ð Þx is active, then the optimal solution is ð Þ 0; 1 .
