Decision uncertainty in multiobjective optimization

Abstract

In many real-world optimization problems, a solution cannot be realized in practice exactly as computed, e.g., it may be impossible to produce a board of exactly 3.546 mm width. Whenever computed solutions are not realized exactly but in a perturbed way, we speak of decision uncertainty. We study decision uncertainty in multiobjective optimization problems and we propose the concept of decision robust efficiency for evaluating the robustness of a solution in this case. This solution concept is defined by using the framework of set-valued maps. We prove that convexity and continuity are preserved by the resulting set-valued maps. Moreover, we obtain specific results for particular classes of objective functions that are relevant for solving the set-valued problem. We furthermore prove that decision robust efficient solutions can be found by solving a deterministic problem in case of linear objective functions. We also investigate the relationship of the proposed concept to other concepts in the literature.

Appendix: An example for monotonic objective functions

This example illustrates Theorem 27 and shows that, after having excluded a specific part of the decision robust feasible set, all decision robust efficient solutions can be determined by Theorem 27.

Example 4

Let the feasible set, the perturbation set and the objective function be

$$\begin{aligned}&\varOmega = conv \left\{ \left( {\begin{matrix} 0 \\ 0 \end{matrix}}\right) , \left( {\begin{matrix} 1.5 \\ 0 \end{matrix}}\right) , \left( {\begin{matrix} 1.5 \\ 1 \end{matrix}}\right) , \left( {\begin{matrix} 1 \\ 1 \end{matrix}}\right) \right\} , \quad Z= [0,0.5]\times \{0\} \\ \text {and}\qquad&f:\mathbb {R}^2\rightarrow \mathbb {R}^2,\quad x\mapsto \left( {\begin{matrix} -(x_1+x_2) \\ x_1^2+x_2^2 \end{matrix}}\right) . \end{aligned}$$

Then $f_1$ is strongly decreasing and $f_2$ is strongly increasing on $\mathbb {R}^2_+$ and therefore also on the decision robust feasible subset X of $\varOmega$, which is illustrated in Fig. 4.

We show next that the set of decision robust [$\cdot$/strictly] efficient solutions is

$$\begin{aligned} Y=conv \left\{ \left( {\begin{matrix} 0 \\ 0 \end{matrix}}\right) , \left( {\begin{matrix} 1 \\ 1 \end{matrix}}\right) \right\} . \end{aligned}$$

We first show that the elements of $X\backslash Y$ can not be decision robust efficient.

For all $s\in (0,2)$ we define the set $L_s=\{x\in X\ | \ x_1+x_2=s\}$ and $y^s=\frac{1}{2}\cdot \left( {\begin{matrix} s \\ s \end{matrix}}\right) \in L_s$, which is illustrated in Fig. 5.

Let $s\in (0,2)$ and let $x\in L_s\backslash \{y^s\}$. Then $x_1>x_2$ and there exists $0<t\le \frac{s}{2}$ such that $x_1=\frac{s}{2}+t$ and $x_2=\frac{s}{2}-t$. For each $z\in Z$ it holds

$$\begin{aligned} f_1(x+z) = -(x_1+x_2+z_1) = -(s + z_1) = -\left( y^s_1 + y^s_2 + z_1\right) = f_1(y^s + z) \end{aligned}$$

as well as

$$\begin{aligned} f_2(y^s +z) = \frac{s^2}{2} + s\cdot z_1 + z_1^2 \end{aligned}$$

and

$$\begin{aligned} f_2(x+z) = \underbrace{\frac{s^2}{2} + s\cdot z_1 + z_1^2}_{=f_2(y^s +z)} + \underbrace{2t^2}_{>0} + \underbrace{2tz_1}_{\ge 0} > f_2(y^s +z). \end{aligned}$$

Consequently, $f(y^s+ z) \le f(x + z)$ for all $z\in Z$. Since $x\in L_s$ was arbitrarily chosen, we have for all $x\in L_s\backslash \{y^s\}$

$$\begin{aligned} f_Z(y^s) \subseteq f_Z(x)-\mathbb {R}^2_+\backslash \{0\}, \end{aligned}$$

which is visualized in Fig. 5. Hence, the set of decision robust [$\cdot$/strictly] efficient solutions is a subset of Y. Furthermore, because of the transitivity of $\leqq$, every element in Y that is not decision robust [$\cdot$/strictly] efficient is dominated by another element of Y. Since Y and Z are totally ordered with respect to $\leqq$, $f_1$ is strongly de- and $f_2$ is strongly increasing, Theorem 27 can be applied, leading to the conclusion that all $y\in Y$ are decision robust [$\cdot$/strictly] efficient.

Fig. 4

The feasible set $\varOmega$ and the decision robust feasible set X in Example 4

Fig. 5

Illustration of Example 4. The sets Y and $L_s$ for $s=1$ are illustrated on the left hand side. On the right hand side, inclusions of the sets $f_Z$ for different elements of $L_1$ are displayed