Concepts of efficiency for uncertain multi-objective optimization problems based on set order relations

Abstract

In this paper we present new concepts of efficiency for uncertain multi-objective optimization problems. We analyze the connection between the concept of minmax robust efficiency presented by Ehrgott et al. (Eur J Oper Res, 2014, doi:10.1016/j.ejor.2014.03.013) and the upper set less order relation $\preceq _s^u$ introduced by Kuroiwa (1998, 1999). From this connection we derive new concepts of efficiency for uncertain multi-objective optimization problems by replacing the set ordering with other set orderings. Those are namely the lower set less ordering (see Kuroiwa 1998, 1999), the set less ordering (see Nishnianidze in Soobshch Akad Nauk Gruzin SSR 114(3):489–491, 1984; Young in Math Ann 104(1):260–290, 1931, doi:10.1007/BF01457934; Eichfelder and Jahn in Vector Optimization. Springer, Berlin, 2012), the certainly less ordering (see Eichfelder and Jahn in Vector Optimization. Springer, Berlin, 2012), and the alternative set less ordering (see Ide et al. in Fixed Point Theory Appl, 2014, doi:10.1186/1687-1812-2014-83; Köbis 2014). We analyze the resulting concepts of efficiency and present numerical results on the occurrence of the various concepts. We conclude the paper with a short comparison between the concepts, and an outlook to further work.

Appendix

Proof of Proposition 12

Suppose that $x^0$ is not lower set less ordered (strictly/weakly) efficient. Then there exists an ${\overline{x}}\in {\mathcal {X}}{\setminus }\{x^0\}$ s.t.

$$\begin{aligned} {f_{{\mathcal {U}}}}({\overline{x}}) + {{\mathbb {R}}^k_{\left[ \geqq />\right] }}\supseteq {f_{{\mathcal {U}}}}(x^0). \end{aligned}$$

This is equivalent to

$$\begin{aligned}&\forall \xi \in {\mathcal {U}}~ \exists \eta \in {\mathcal {U}}:~ f({\overline{x}}, \eta ) + {{\mathbb {R}}^k_{\left[ \geqq />\right] }}\ni f(x^0, \xi ) \nonumber \nonumber \\&\quad \Longleftrightarrow \forall \xi \in {\mathcal {U}}~ \exists \eta \in {\mathcal {U}}:~ f({\overline{x}}, \eta ) \in f(x^0, \xi ) - {{\mathbb {R}}^k_{\left[ \geqq />\right] }} \end{aligned}$$

(5)

Now choose $\uplambda ^* \in {{\mathbb {R}}^k_{>}}$ arbitrary, but fixed. Hence, we obtain from (5)

$$\begin{aligned}&\forall \xi \in {\mathcal {U}}~ \exists \eta \in {\mathcal {U}}:~ {\uplambda }_{i}^* f_i({\overline{x}}, \eta ) ~[\leqq /<]~ {\uplambda }_{i}^{*} f_i(x^0, \xi ) ,~i=1, \ldots ,k,\\&\quad \Longrightarrow \left[ \inf _{\eta \in {\mathcal {U}}}/\min _{\eta \in {\mathcal {U}}}\right] {\uplambda }_{i}^{*} f_i({\overline{x}}, \eta )~[\leqq /<]~ \left[ \inf _{\xi \in {\mathcal {U}}}/\min _{\xi \in {\mathcal {U}}}\right] {\uplambda }_i^* f_i(x^0, \xi ), \\&\quad i=1, \ldots ,k. \end{aligned}$$

Since this holds for all $i=1,\ldots ,k$, it is a contradiction to the assumption. $\square$

Proof of Theorem 13

Suppose that $x^0$ is not lower set less ordered robust efficient. Then there exists ${\overline{x}}\in {\mathcal {X}}{\setminus } \{x^0\}$ s.t. ${f_{{\mathcal {U}}}}({\overline{x}})+{{\mathbb {R}}^k_\ge }\supseteq {f_{{\mathcal {U}}}}(x^0)$, i.e.,

$$\begin{aligned} \forall \xi \in {\mathcal {U}}~ \exists \eta \in {\mathcal {U}}: ~ f({\overline{x}},\eta ) \le f(x^0,\xi ). \end{aligned}$$

This implies on the one hand

$$\begin{aligned} \displaystyle \min _{\xi \in {\mathcal {U}}} f_i({\overline{x}},\xi )\leqq \displaystyle \min _{\xi \in {\mathcal {U}}} f_i(x^0,\xi ) \end{aligned}$$

for all $i\in \{1,\ldots ,k\}$. On the other hand

$$\begin{aligned} \forall \xi \in {\mathcal {U}}~ \exists \eta \in {\mathcal {U}}: ~ \sum _{i=1}^kf_i({\overline{x}},\eta ) < \sum _{i=1}^kf_i(x^0,\xi ). \end{aligned}$$

Since $\uplambda \in {{\mathbb {R}}^k_\geqq }$ and $\rho >0$,

$$\begin{aligned}&\displaystyle \max _{i=1, \ldots ,k} \min _{\xi \in {\mathcal {U}}}{\uplambda }_i f_i({\overline{x}},\xi ) + \displaystyle \min _{\xi \in {\mathcal {U}}}~\rho \sum _{i=1}^kf_i({\overline{x}},\xi ) \\&\quad <\max _{i=1, \ldots ,k} \min _{\xi \in {\mathcal {U}}} {\uplambda }_i f_i(x^0,\xi ) + \displaystyle \min _{\xi \in {\mathcal {U}}}~\rho \sum _{i=1}^kf_i(x^0,\xi ), \end{aligned}$$

in contradiction to the optimality of $x^0$ for ${{\mathcal {P}}({\mathcal {U}})}_{\uplambda ^*,\rho }^{max}$. $\square$

Proof of Proposition 16

Suppose that $x^0$ is not lower set less ordered (strictly/weakly) efficient for $P(\mathcal{{U}})$. Thus there is an element ${\overline{x}}\in {\mathcal {X}}{\setminus } \{ x^0 \}$ with

$$\begin{aligned} {f_{{\mathcal {U}}}}({\overline{x}}) + \mathbb {R}^k_{[\geqq />]} \supseteq {f_{{\mathcal {U}}}}(x^0). \end{aligned}$$

In other words,

$$\begin{aligned}&~ \forall \ \xi \in \mathcal{{U}} \; \exists \ \eta \in \mathcal{{U}}:~ f({\overline{x}}, \eta ) \in f(x^0,\xi ) - \mathbb {R}^k_{[\geqq />]} \\&\quad \Longleftrightarrow ~ \forall \ \xi \in \mathcal{{U}}~ \exists \ \eta \in \mathcal{{U}}:~ f({\overline{x}}, \eta ) ~[\leqq /<] ~ f(x^0,\xi ) \\&\quad \Longleftrightarrow ~\forall \ \xi \in \mathcal{{U}}~ \exists \ \eta \in \mathcal{{U}}: ~ f_j({\overline{x}}, \eta ) ~[\leqq /<] ~ f_j(x^0,\xi ),~j=1, \ldots ,k \\&\quad \Longleftrightarrow ~ \exists \ \eta \in \mathcal{{U}}: f_j({\overline{x}}, \eta )~ [\leqq /<] ~ \inf _{\xi \in {\mathcal {U}}}f_j(x^0,\xi ),~j=1, \ldots ,k \end{aligned}$$

and this yields

$$\begin{aligned} \inf _{\xi \in {\mathcal {U}}}f_j({\overline{x}}, \xi ) ~[\leqq /<]~ \inf _{\xi \in {\mathcal {U}}}f_j(x^0,\xi ),~j=1, \ldots ,k. \end{aligned}$$

(6)

Because $\inf _{\xi \in {\mathcal {U}}}f_j(x^0,\xi ) \leqq \epsilon _j$ for every $j=1, \ldots ,k,~j \ne i$, ${\overline{x}}$ is also feasible and has an equal or better objective function value (a better objective function value, respectively) than $x^0$, contradicting the assumption that $x^0$ is the unique optimal (an optimal, respectively) solution of ${{\mathcal {P}}({\mathcal {U}})}_{\epsilon , i}$. Note that the strict inequality in (6) holds because the existence of $\min _{\xi \in \mathcal{{U}}} f_j(x, \xi )$ for all $x \in {\mathcal {X}}$ and $j=1, \ldots ,k$ is presumed. $\square$

Proof of Proposition 18

Suppose that $x^0$ is not lower set less ordered (strictly/weakly) efficient. Then there exists ${\overline{x}}\in {\mathcal {X}}{\setminus } \{x^0\}$ s.t.

$$\begin{aligned}&{f_{{\mathcal {U}}}}({\overline{x}}) + \mathbb {R}^k_{[\geqq />]} \supseteq {f_{{\mathcal {U}}}}(x^0) \\&\Longleftrightarrow ~ \forall \ \xi \in {\mathcal {U}}~ \exists \ \eta \in {\mathcal {U}}:~ f({\overline{x}}, \eta ) + \mathbb {R}^k_{[\geqq />]} \ni f(x^0, \xi ) \\&\Longleftrightarrow ~ \forall \ \xi \in {\mathcal {U}}~ \exists \ \eta \in {\mathcal {U}}:~ f_i({\overline{x}}, \eta )~ [\leqq /<]~ f_i(x^0, \xi ), ~ i=1, \ldots ,k,\\&\Longrightarrow ~ \exists \ \eta \in {\mathcal {U}}:~ f_i({\overline{x}}, \eta )~ [\leqq /<]~ \inf _{\xi \in {\mathcal {U}}} f_i(x^0, \xi ) ,~ i=1, \ldots ,k,\\&\Longrightarrow ~ \inf _{\xi \in {\mathcal {U}}} f_i({\overline{x}}, \xi )~ [\leqq /<]~ \inf _{\xi \in {\mathcal {U}}} f_i(x^0, \xi ), ~ i=1, \ldots ,k, \end{aligned}$$

in contradiction to the assumption of $x^0$ being (strictly/weakly) efficient for $({\mathcal {O}}{\mathcal {W}}{\mathcal {I}}_{{\mathcal {P}}({\mathcal {U}})})$. $\square$

Proof of Theorem 23

Let $x^0$ be (strictly/weakly/weakly) efficient for $({{\mathcal {P}}({\mathcal {U}})}^{biobj}_{\uplambda ^*})$ with some $\uplambda ^* \in {{{{\mathbb {R}}^k}_{[\ge />/\ge ]}}}$, i.e., there is no ${\overline{x}}\in {\mathcal {X}}{\setminus }\{x^0\}$ such that

$$\begin{aligned}&\inf _{\xi \in {\mathcal {U}}} \sum _{i=1}^k {\uplambda }_{i}^*f_i({\overline{x}}, \xi ) ~[\leqq ,<,<]~\inf \limits _{\xi \in {\mathcal {U}}} \sum \limits _{i=1}^k {\uplambda }_{i}^*f_i(x^0, \xi ) \\&\quad \text {and~} \sup _{\xi \in {\mathcal {U}}} \sum \limits _{i=1}^k {\uplambda }_{i}^*f_i({\overline{x}}, \xi ) ~[\leqq ,<,<]~ \sup \limits _{\xi \in {\mathcal {U}}} \sum \limits _{i=1}^k {\uplambda }_{i}^*f_i(x^0, \xi ). \end{aligned}$$

Now assume that $x^0$ is not set less ordered (strictly/$\cdot$/weakly) efficient. Then there exists an element ${\overline{x}}\in {\mathcal {X}}{\setminus }\{x^0\}$ such that

$$\begin{aligned} {f_{{\mathcal {U}}}}({\overline{x}}) + {{\mathbb {R}}^k_{\left[ \geqq /\ge />\right] }}\supseteq {f_{{\mathcal {U}}}}(x^0) \text {~and~} {f_{{\mathcal {U}}}}({\overline{x}}) \subseteq {f_{{\mathcal {U}}}}(x^0) - {{\mathbb {R}}^k_{\left[ \geqq /\ge />\right] }}. \end{aligned}$$

This is equivalent to

$$\begin{aligned}&\exists ~{\overline{x}}\in {\mathcal {X}}{\setminus }\{x^0\}:~ \forall ~ \xi _1, \xi _2 \in {\mathcal {U}}~ \exists ~\eta _1, \eta _2 \in {\mathcal {U}}:~ f({\overline{x}}, \eta _1) + {{\mathbb {R}}^k_{\left[ \geqq /\ge />\right] }}\ni f(x^0, \xi _1) \nonumber \\&\quad \text {and~} f({\overline{x}}, \xi _2) \in f(x^0, \eta _2) - {{\mathbb {R}}^k_{\left[ \geqq /\ge />\right] }}. \end{aligned}$$

(7)

Now choose $\uplambda ^* \in {{\mathbb {R}}^k}_{[\ge />/\ge ]}$ as in problem ${{\mathcal {P}}({\mathcal {U}})}^{biobj}_{\uplambda ^*}$. We obtain from (7)

$$\begin{aligned}&\exists ~{\overline{x}}\in {\mathcal {X}}{\setminus }\{x^0\}:~ \forall ~\xi _1, \xi _2 \in {\mathcal {U}}~ \exists ~\eta _1, \eta _2 \in {\mathcal {U}}:\\&~~~~~~~\sum \limits _{i=1}^k {\uplambda }_{i}^* f_i({\overline{x}}, \eta _1) ~[\leqq ,<,<]~ \sum \limits _{i=1}^k {\uplambda }_{i}^* f_i(x^0, \xi _1) \\&\text {~and~} \sum \limits _{i=1}^k {\uplambda }_{i}^* f_i({\overline{x}}, \xi _2) ~[\leqq ,<,<]~\sum \limits _{i=1}^k {\uplambda }_{i}^* f_i(x^0, \eta _2)\\&\Rightarrow ~\left[ \inf _{\xi \in {\mathcal {U}}}/\min _{\xi \in {\mathcal {U}}}/\min _{\xi \in {\mathcal {U}}}\right] \sum _{i=1}^k {\uplambda }_{i}^* f_i({\overline{x}}, \xi )\\&[\leqq ,<,<]~ \left[ \inf _{\xi \in {\mathcal {U}}}/\min _{\xi \in {\mathcal {U}}}/\min _{\xi \in {\mathcal {U}}}\right] \sum _{i=1}^k {\uplambda }_{i}^* f_i(x^0, \xi ) \\&\text {~and~} \left[ \sup _{\xi \in {\mathcal {U}}}/\max _{\xi \in {\mathcal {U}}}/\max _{\xi \in {\mathcal {U}}}\right] \sum _{i=1}^k {\uplambda }_{i}^* f_i({\overline{x}}, \xi ) \\&[\leqq ,<,<]~ \left[ \sup _{\xi \in {\mathcal {U}}}/\max _{\xi \in {\mathcal {U}}}/\max _{\xi \in {\mathcal {U}}}\right] \sum _{i=1}^k {\uplambda }_{i}^* f_i(x^0, \xi ). \end{aligned}$$

The last two strict inequalities hold because the minimum and maximum exist. But this is a contradiction to the assumption. $\square$