Bilevel optimization with a multiobjective problem in the lower level

Abstract

Bilevel problems model instances with a hierarchical structure. Aiming at an efficient solution of a constrained multiobjective problem according with some pre-defined criterion, we reformulate this semivectorial bilevel optimization problem as a classic bilevel one. This reformulation intents to encompass all the objectives, so that the properly efficient solution set is recovered by means of a convenient weighted-sum scalarization approach. Inexact restoration strategies potentially take advantage of the structure of the problem under consideration, being employed as an alternative to the Karush-Kuhn-Tucker reformulation of the bilevel problem. Genuine multiobjective problems possess inequality constraints in their modeling, and these constraints generate theoretical and practical difficulties to our lower level problem. We handle these difficulties by means of a perturbation strategy, providing the convergence analysis, together with enlightening examples and illustrative numerical tests.

Appendix: Test Problems

Problem 1.

This is the problem (10)–(11) of example 1 on page 8.

Problem 2.

We define problem 2 as the minimization of $F(x_{1},x_{2})={x_{1}^{2}}+{x_{2}^{2}}$ over the set $\mathscr X^{*}$ of efficient solutions of the multiobjective constrained problem [28]

$$\begin{array}{ll} \min & \{ x_{1}, : x_{2} \}\\ \text{s.t.} & 1 + 0.1\cos\left( 16\arctan\frac{x_{2}}{x_{1}}\right) - {x_{1}^{2}} - {x_{2}^{2}}\leq 0 \\ &\left( x_{1}-\frac{1}{2}\right)^{2} + \left( x_{2}-\frac{1}{2}\right)^{2} \leq \frac{1}{2}\\ &0\leq x_{1},x_{2}\leq \pi. \end{array} $$

$\mathscr X^{*}$ is discontinuous and nonconvex. There are four optimal solutions with functional value F^∗≈ 0.95 (see Fig. 3).

Fig. 3

Geometry of problem 2. The feasible set $\mathscr X^{*}$ is discontinuous and nonconvex. There are four solutions, with optimal value F^∗≈ 0.95

Problem 3.

Problem 3 consists in minimizing $F(x_{1},x_{2})={x_{1}^{2}}+{x_{2}^{2}}$ over the set of efficient solutions of

$$\begin{array}{cc} \min & \{ -x_{1},: x_{1}+{x_{2}^{2}} \}\\ \text{s.t.} & {x_{1}^{2}}-x_{2}\leq 0 \\ &x_{1} + 2x_{2}\leq 3\\ &x_{1},x_{2}\geq 0. \end{array} $$

The feasible set is $\mathscr X^{*}=\{ (x_{1},x_{2})\in \mathbb{R} ^{2};: x_{2}={x_{1}^{2}}, : 0\leq x_{1}\leq 2 \}$ and (0, 0)^T is the optimal solution.

Problem 4.

The lower level multiobjective problem is exactly the same of the problem 3, and thus the feasible set $\mathscr X^{*}$ is the same. The upper level objective function of problem 4 is F(x₁, x₂) = (x₁ − 1)² + (x₂ − 1)², and (1, 1)^T is the optimal solution.

Problem 5.

This is a slight modification of problem 4. It consists in minimizing F(x₁, x₂) = (x₁ − 1)² + (x₂ − 1)² over the set of efficient solutions of

$$\begin{array}{cc} \min & \{ x_{1},: -x_{1}+{x_{2}^{2}} \}\\ \text{s.t.} & {x_{1}^{2}}-x_{2}\leq 0 \\ &-x_{1} + 2x_{2}\leq 3\\ &x_{1},x_{2}\geq 0. \end{array} $$

The feasible set is $\mathscr X^{*}=\{ (x_{1},x_{2})\in \mathbb{R} ^{2}; x_{2}={x_{1}^{2}}, 0\leq x_{1}\leq 2^{-2/3} \}$ and the optimal solution is (2^− 2/3, 2^− 4/3)^T ≈ (0.63, 0.40)^T.

Problem 6.

The lower level problem is

$$\begin{array}{cc} \min & \left\{ {x_{1}^{2}} + \frac{1}{4}\left( x_{2}-\frac{9}{2}\right)^{2}, {x_{2}^{2}} + \frac{1}{4}\left( x_{1}-\frac{9}{2}\right)^{2} \right\}\\ \text{s.t.} & -x_{1}-x_{2}+ 2\leq 0 \\ &x_{1},x_{2}\geq 0. \end{array} $$

Geometrically, the set $\mathscr X^{*}$ of efficient solutions of this multiobjective problem is composed by the simultaneously tangential points of the level sets of both objective functions, when it is contained in the semispace defined by x₁ + x₂ ≥ 2, and by the line x₁ + x₂ = 2 otherwise. More specifically, $\mathscr X^{*}$ is the union of the curves

$$\begin{array}{@{}rcl@{}} &C_{1}: \left\{ \left( \frac{9t}{2t-8},\frac{9}{2-8t} \right); -\frac{1}{2}\leq t\leq 0 \right\},\\ &C_{2}: \left\{ \left( t,2-t \right); \frac{1}{2}\leq t\leq \frac{3}{2} \right\},\\ &C_{3}: \left\{ \left( \frac{9}{2-8t},\frac{9t}{2t-8} \right); -\frac{1}{2}\leq t\leq 0 \right\}. \end{array} $$

The upper level objective function is $F(x_{1},x_{2}) = (x_{1}-9/2)^{2} + {x_{2}^{2}}$, for which (9/2, 0)^T is the optimal solution. Figure 4 illustrates its geometry.

Fig. 4

Geometry of problem 6. The feasible set is the union of curves C₁, C₂, and C₃. The optimal solution is x^∗ = (9/2,0)^T

Problem 7.

This is the problem (12)–(13) of example 2 on page 9.

Problems 8 to 11 (ZDT family instances).

Multiobjective problems of [29] (here named ZDT problems) are constructed with three functions f₁, g, h in the following way:

$$\begin{array}{cc} \min & \{ f_{1}(x_{1}), f_{2}(x_{1},\ldots,x_{n}) \} \end{array} $$

where

$$f_{2}(x_{1},\ldots,x_{n}) = g(x_{2},\ldots,x_{n}) h(f_{1}(x_{1}),g(x_{2},\ldots,x_{n})) $$

(see also [9]). We were inspired in four combinations suggested in Definition 4 of [29]. For each combination, we minimize the following:

$$F(x_{1},x_{2},\ldots,x_{n}) = (x_{1}-1)^{2} + \sum\limits_{i = 2}^{n} {x_{i}^{2}} $$

over the set of efficient solutions defined by the minimization of functions f₁ and f₂ on certain box constraints. We summarized the properties of these test problems in Table 5. For these problems, the set of efficient solutions is split into discontinuous components, one for each (local) minimum of g, formed with all possible values of f₁ (see Section 5.1 of [9]). Thus, problems 8 to 10 have a unique set of efficient solutions component, for which x₂ = ⋯ = x_n = 0 (i.e. g(x₂,…,x_n) = 1), while Problem 11 has several components. For problems 8, 9, and 11, the possible values of x₁ = f₁(x₁) are the whole interval [0, 1]. For problem 10, the possible values of x₁ are the intervals for which h decreases (see Fig. 5). Hence, (1, 0,…, 0) is the optimal solution of problems 8, 9, and 11, whereas (0.954, 0…, 0)^T is the (approximated) optimal solution of problem 10. Problem 11 has 21^n− 1 sets of efficient solutions components, one for each combination of the local minimizers of ${x_{i}^{2}}-2\cos (4\pi x_{i})$, i = 2,…,n (see Fig. 6).

Table 5 ZDT family instances.

Fig. 5

Geometry of problem 10. In the set of efficient solutions, $g(x_{2}^{*},\ldots ,x_{n}^{*})= 1$ and x₁ lies on the intervals for which $\tilde f_{2}(x_{1})=f_{2}(x_{1},x_{2}^{*},\ldots ,x_{n}^{*}) = 1-x_{1}(1-\sin (10\pi x_{1}))$ decreases (huge lines). These are the non-dominated values corresponding to the two objectives f₁(x₁) = x₁ and $\tilde f_{2}(x_{1})$

Fig. 6

Geometry of problem 11. There is one set of efficient solutions component for each combination of local minimizers of $\tilde g_{i}(x_{i})={x_{i}^{2}}-2\cos (4\pi x_{i})$, i = 2,…,n

Problems 12 and 13 (WFG family instances).

Based on the Walking Fish Group (WFG) Toolkit [19], we consider two instances, for which particular choices of the parameters were made to come up with smooth functions:

• Problem 12, that consists in minimizing $F(x) = {\sum }_{i = 1}^{n} {x_{i}^{2}}$, over the set of efficient solutions defined by the minimization of

  $$\begin{array}{@{}rcl@{}} \displaystyle f_{1}(x) &=& \frac{x_{n}}{2n} + 2\overset{n-1}{\underset{i = 1}{\prod}}\left[1 - \cos\left( \frac{\pi x_{i}}{4i}\right)\right],\\ \displaystyle f_{j}(x) &=& \frac{x_{n}}{2n} + 2j\left[1-\sin\left( \frac{\pi x_{n-j + 1}}{4(n-j + 1)}\right)\right]\overset{n-j}{\underset{i = 1}{\prod}}\left[1 - \cos\left( \frac{\pi x_{i}}{4i}\right)\right],\ \\j&=&2,\ldots,n-1,\\ \displaystyle f_{n}(x) &=& \frac{x_{n}}{2n} + 2n\left[1-\frac{x_{1}}{2} - \frac{\cos(5\pi x_{1} + \pi/2)}{10\pi} \right] \end{array} $$

  on [0, 1]ⁿ. The optimal solution is the origin.

• Problem 13, that consists in minimizing $F(x) = {\sum }_{i = 1}^{n} {x_{i}^{2}}$, over the set of efficient solutions defined by the minimization of

  $$\begin{array}{@{}rcl@{}} \displaystyle f_{1}(x) &=& \frac{x_{n}}{2n} + 2\overset{n-1}{\underset{i = 1}{\prod}}\left[1 - \cos\left( \frac{\pi x_{i}}{4i}\right)\right],\\ \displaystyle f_{j}(x) &=& \frac{x_{n}}{2n} + 2j\left[1-\sin\left( \frac{\pi x_{n-j + 1}}{4(n-j + 1)}\right)\right]\overset{n-j}{\underset{i = 1}{\prod}}\left[1 - \cos\left( \frac{\pi x_{i}}{4i}\right)\right],\\ j&=&2,\ldots,n-1,\\ \displaystyle f_{n}(x) &=& \frac{x_{n}}{2n} + 2n\left[ 1-\frac{x_{1}}{2}\cos^{2}\left( \frac{5\pi x_{1}}{2}\right) \right] \end{array} $$

  on [0, 1]ⁿ. Again, the optimal solution is the origin.