Multi-criteria decision making in Pythagorean fuzzy environment

Abstract

Pythagorean fuzzy set, initially extended by Yager from intuitionistic fuzzy set, is capable of modeling information with more uncertainties in the process of multi-criteria decision making (MCDM), thus can be used on wider range of conditions. The fuzzy decision analysis of this paper is mainly built upon two expressions in Pythagorean fuzzy environment, named Pythagorean fuzzy number (PFN) and interval-valued Pythagorean fuzzy number (IVPFN), respectively. We initiate a novel axiomatic definition of Pythagorean fuzzy distance measurement, including PFNs and IVPFNs. After that, corresponding theorems are put forward and then proved. Based on the defined distance measurements, the closeness indexes are developed for both expressions, inspired by the idea of technique for order preference by similarity to ideal solution (TOPSIS) approach. After these basic definitions have been established, the hierarchical decision approach is presented to handle MCDM problems under Pythagorean fuzzy environment. To address hierarchical decision issues, the closeness index-based score function is defined to calculate the score of each permutation for the optimal alternative. To determine criterion weights, a new method based on the proposed similarity measure and aggregation operator of PFNs and IVPFNs is presented according to Pythagorean fuzzy information from decision matrix, rather than being provided in advance by decision makers, which can effectively reduce human subjectivity. An experimental case is then conducted to demonstrate the applicability and flexibility of the proposed decision approach. Finally, extension forms of Pythagorean fuzzy decision approach for heterogeneous information are briefly introduced to show its potentials on further applications in other processing fields with information uncertainties.

Appendices

Appendix A: Proof of Theorem 1

Proof

Let

$$ d_{Temp}(\beta_{1}, \beta_{2}) = \sqrt{(\mu_{\beta_{1}} - \mu_{\beta_{2}})^{2} + (\nu_{\beta_{1}} - \nu_{\beta_{2}})^{2}} $$

(36)

$$ \begin{array}{ll} s.t. &\left\{ \begin{array}{lll} &{0 \leq (\mu_{\beta_{1}})^{2} + (\nu_{\beta_{1}})^{2}} \leq 1 \\ &{0 \leq (\mu_{\beta_{2}})^{2} + (\nu_{\beta_{2}})^{2}} \leq 1 \\ &{0 \leq \mu_{\beta_{1}}, \nu_{\beta_{1}}, \mu_{\beta_{2}}, \nu_{\beta_{2}}} \leq 1 \end{array} \right. \end{array} $$

A coordinate system is established and shown in Fig. 8, which takes μ as the vertical axis with ν as the horizontal axis. According to the constraints, we know that $P(\mu _{\beta _{1}} ,\nu _{\beta _{1}})$ and $P(\mu _{\beta _{2}} ,\nu _{\beta _{2}})$ are the two points of the shaded part in the coordinate system. Based on (41), d_Temp(β₁, β₂) is the distance between point β₁ and β₂. The maximum and minimum values between them are 0 and $\sqrt {2}$, respectively. That is, $d_{Temp}(A, B) \in [0, \sqrt {2}]$. Based on (2), we have 0 ≤ d(β₁, β₂) ≤ 1. □

Fig. 8

The coordinate system for μ and ν

Appendix B: Proof of Theorem 4

Proof

As β₁ ≤ β₂ ≤ β₃, we have $\mu _{\beta _{1}} \leq \mu _{\beta _{2}} \leq \mu _{\beta _{3}}$ and $\nu _{\beta _{1}} \geq \nu _{\beta _{2}} \geq \nu _{\beta _{3}}$ based on the Definition in [62]. As shown in Fig. 8, we take arbitrary values $\mu _{\beta _{1}}$, $\mu _{\beta _{2}}$, $\mu _{\beta _{3}}$ and $\nu _{\beta _{1}}$, $\nu _{\beta _{2}}$, $\nu _{\beta _{3}}$ on the coordinate axis. Their intersection points β₁, β₂ and β₃ form a triangle. Obviously, ∠β₁β₂β₃ is always an obtuse angle, so β₁β₂ < β₁β₃ and β₂β₃ < β₁β₃. That is, d_Temp(β₁, β₂) ≤ d_Temp(β₁, β₃) and d_Temp(β₂, β₃) ≤ d_Temp(β₁, β₃). So d(β₁, β₂) ≤ d(β₁, β₃) and d(β₂, β₃) ≤ d(β₁, β₃). □

Appendix C: Proof of Theorem 6

Proof

According to Definition 4, we have

$$ \begin{array}{@{}rcl@{}} && {\boldsymbol{\Re}}(\beta_{1}) -{\boldsymbol{\Re}}(\beta_{2})\\ \!&=&\!\frac{\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}}-1)^{2}}}{\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}} + \sqrt{(\mu_{\beta_{1}} - 1)^{2} + \nu_{\beta_{1}}^{2}}} \\ &&- \frac{\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}}-1)^{2}}}{\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}} + \sqrt{(\mu_{\beta_{2}} - 1)^{2} + \nu_{\beta_{2}}^{2}}} \\ \!&=&\! \frac{\begin{array}{lll} \left( ({\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}})(\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}} + \sqrt{(\mu_{\beta_{2}} - 1)^{2} + \nu_{\beta_{2}}^{2}}}) \right. \\ \left. {-(\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}})(\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}} + \sqrt{(\mu_{\beta_{1}} - 1)^{2} + \nu_{\beta_{1}}^{2}})} \right) \end{array}}{\begin{array}{lll} \left( {(\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}} + \sqrt{(\mu_{\beta_{1}} - 1)^{2} + \nu_{\beta_{1}}^{2}})} \right. \\ \left. { (\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}} + \sqrt{(\mu_{\beta_{2}} - 1)^{2} + \nu_{\beta_{2}}^{2}})} \right) \end{array}} \\ \!&=&\! \frac{\begin{array}{lll} \left( {(\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}})(\sqrt{(\mu_{\beta_{2}} - 1)^{2}+\nu_{\beta_{2}}^{2}})} \right. \\ \left. { \!-(\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}})(\sqrt{(\mu_{\beta_{1}} - 1)^{2} + \nu_{\beta_{1}}^{2}})} \right) \end{array}}{\begin{array}{lll} \left( {(\sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}} + \sqrt{(\mu_{\beta_{1}} - 1)^{2} + \nu_{\beta_{1}}^{2}})} \right. \\ \left. { (\sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}} + \sqrt{(\mu_{\beta_{2}} - 1)^{2} + \nu_{\beta_{2}}^{2}})} \right) \end{array}} \end{array} $$

(37)

Since

$$ \begin{array}{@{}rcl@{}} && (\mu_{\beta_{1}}^{2}+(\nu_{\beta_{1}}-1)^{2})-(\mu_{\beta_{2}}^{2}+(\nu_{\beta_{2}}-1)^{2})\\ &=& \mu_{\beta_{1}}^{2}-\mu_{\beta_{2}}^{2}+((\nu_{\beta_{1}}-1)^{2})-\left.\left.(\nu_{\beta_{2}}-1)^{2}\right)\right)\\ &=& (\mu_{\beta_{1}}+\mu_{\beta_{2}})(\mu_{\beta_{1}}-\mu_{\beta_{2}})+(\nu_{\beta_{1}}+\nu_{\beta_{2}}-2)(\nu_{\beta_{1}}-\nu_{\beta_{2}})\\ \end{array} $$

(38)

Since β₁ ≤ β₂, we have $0\leq \mu _{\beta _{1}}\leq \mu _{\beta _{2}} \leq 1$ and $1 \geq \nu _{\beta _{1}} \geq \nu _{\beta _{2}} \geq 0$ according to Definition in [62]. So, $\mu _{\beta _{1}}+\mu _{\beta _{2}} \geq 0$, $\mu _{\beta _{1}}-\mu _{\beta _{2}} \leq 0$, $\nu _{\beta _{1}}+\nu _{\beta _{2}}-2 \leq 0$, $\nu _{\beta _{1}}-\nu _{\beta _{2}} \geq 0$; that is, $(\mu _{\beta _{1}}+\mu _{\beta _{2}})(\mu _{\beta _{1}}-\mu _{\beta _{2}}) \leq 0$ and $(\nu _{\beta _{1}}+\nu _{\beta _{2}}-2)(\nu _{\beta _{1}}-\nu _{\beta _{2}}) \leq 0$, so $(\mu _{\beta _{1}}^{2}+(\nu _{\beta _{1}}-1)^{2})-(\mu _{\beta _{2}}^{2}+(\nu _{\beta _{2}}-1)^{2}) \leq 0$, then $(\mu _{\beta _{1}}^{2}+(\nu _{\beta _{1}}-1)^{2}) \leq (\mu _{\beta _{2}}^{2}+(\nu _{\beta _{2}}-1)^{2})$, then $\sqrt {\mu _{\beta _{1}}^{2}+(\nu _{\beta _{1}}-1)^{2}} \leq \sqrt {\mu _{\beta _{2}}^{2}+(\nu _{\beta _{2}}-1)^{2}}$. Similarly, we have $\sqrt {(\mu _{\beta _{2}}-1)^{2}+\nu _{\beta _{2}}^{2}} \leq \sqrt {(\mu _{\beta _{1}}-1)^{2}+\nu _{\beta _{1}}^{2}}$. So, in (37), we have

$$ \begin{array}{@{}rcl@{}} &&\left( \sqrt{\mu_{\beta_{1}}^{2} + (\nu_{\beta_{1}} - 1)^{2}}\right)\left( \sqrt{(\mu_{\beta_{2}} - 1)^{2} + \nu_{\beta_{2}}^{2}}\right)\\ &&-\left( \sqrt{\mu_{\beta_{2}}^{2} + (\nu_{\beta_{2}} - 1)^{2}}\right)\left( \sqrt{(\mu_{\beta_{1}} - 1)^{2} + \nu_{\beta_{1}}^{2}}\right) \!\leq\! 0 \end{array} $$

(39)

and

$$ \begin{array}{@{}rcl@{}} &&\left( \sqrt{\mu_{\beta_{1}}^{2}+(\nu_{\beta_{1}}-1)^{2}}+\sqrt{(\mu_{\beta_{1}}-1)^{2}+\nu_{\beta_{1}}^{2}}\right) \\ &&\left( \sqrt{\mu_{\beta_{2}}^{2}+(\nu_{\beta_{2}}-1)^{2}}+\sqrt{(\mu_{\beta_{2}}-1)^{2}+\nu_{\beta_{2}}^{2}} \right) \geq 0 \end{array} $$

(40)

So, R(β₁) −R(β₂) ≤ 0, that is, R(β₁) ≤R(β₂). □

Appendix D: Proof of Theorem 7

Proof

Let

$$ \begin{array}{@{}rcl@{}} d_{Temp}^{L}(\beta_{1}, \beta_{2}) = \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}} - \widetilde{\mu}^{L}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} - \widetilde{\nu}^{L}_{\widetilde{\beta}_{2}})^{2} } \end{array} $$

(41)

$$ \begin{array}{@{}rcl@{}} d_{Temp}^{U}(\beta_{1}, \beta_{2}) = \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}} - \widetilde{\mu}^{U}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} - \widetilde{\nu}^{U}_{\widetilde{\beta}_{2}})^{2}} \end{array} $$

(42)

$$ \begin{array}{lll} s.t. &\left\{ \begin{array}{lll} &{0 \leq (\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}})^{2}} \leq 1 \\ &{0 \leq (\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}})^{2}} \leq 1 \\ &{0 \leq \widetilde{\mu}^{U}_{\widetilde{\beta}_{1}}, \widetilde{\nu}^{U}_{\widetilde{\beta}_{1}}, \widetilde{\mu}^{U}_{\widetilde{\beta}_{2}}, \widetilde{\nu}^{U}_{\widetilde{\beta}_{2}}} \leq 1 \end{array} \right. \end{array} $$

According to the proof of Theorem 1, it is easy to have $d_{Temp}^{L}(\beta _{1}, \beta _{2}) \in [0, \sqrt 2]$ and $d_{Temp}^{U}(\beta _{1}, \beta _{2}) \in [0, \sqrt 2]$, that is, $0 \leq d(\widetilde {\beta }_{1}, \widetilde {\beta }_{2}) \leq 1$ based on (11). □

Appendix E: Proof of Theorem 12

Proof

$$ \begin{array}{@{}rcl@{}} && \Re(\widetilde{\beta}_1) - \Re(\widetilde{\beta}_2) \\ \! &=&\! \frac{d(\widetilde{\beta}_1,\widetilde{\xi}^-)}{d(\widetilde{\beta}_1,\widetilde{\xi}^-)+d(\widetilde{\beta}_1,\widetilde{\xi}^+)} - \frac{d(\widetilde{\beta}_2,\widetilde{\xi}^-)}{d(\widetilde{\beta}_2,\widetilde{\xi}^-)+d(\widetilde{\beta}_2,\widetilde{\xi}^+)} \\ \!&=&\! \frac{(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} - 1)^2})}{\begin{array}{llllllllll} \left( {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} - 1)^2})} \right. +\\ \left. {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1} - 1)^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} )^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1}-1)^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} )^2})} \right) \end{array}} \end{array} $$

$$ \begin{array}{@{}rcl@{}} &&- \frac{(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} - 1)^2})}{\begin{array}{llllllllll} \left( {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} - 1)^2})} \right. +\\ \left. { (\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2} - 1)^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} )^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2}-1)^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} )^2})} \right) \end{array}} \\ \!&= &\! \frac{\begin{array}{llllllllll} \left( {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} - 1)^2})} \right. \\ \left. {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2} - 1)^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} )^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2}-1)^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} )^2})} \right. -\\ \left. {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1} - 1)^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} )^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1}-1)^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} )^2})} \right. \\ \left. { (\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} - 1)^2})} \right) \end{array}} {\begin{array}{llllllllll} \left( {((\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} - 1)^2})} \right.+ \\ \left. {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_1} - 1)^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_1} )^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_1}-1)^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_1} )^2}))} \right. \\ \left. {((\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} - 1)^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2})^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} - 1)^2})} \right. +\\ \left. {(\sqrt{(\widetilde{\mu}^L_{\widetilde{\beta}_2} - 1)^2 + (\widetilde{\nu}^L_{\widetilde{\beta}_2} )^2 } + \sqrt{(\widetilde{\mu}^U_{\widetilde{\beta}_2}-1)^2 + (\widetilde{\nu}^U_{\widetilde{\beta}_2} )^2}))} \right) \end{array}}\\\end{array} $$

(43)

According to (39) and corresponding definition in [72], we have

$$ \begin{array}{@{}rcl@{}} &&\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} - 1)^{2}}\right) \\ &\leq& \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} - 1)^{2}}\right),\\ &&\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} )^{2}}\right) \\ &\leq& \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} )^{2}}\right) \end{array} $$

(44)

so,

$$ \begin{array}{@{}rcl@{}} &\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} - 1)^{2}}\right) \\ & \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} )^{2}}\right) \\ & \leq \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} )^{2}}\right) \\ & \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} - 1)^{2}}\right) \end{array} $$

(45)

so,

$$ \begin{array}{@{}rcl@{}} && \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} - 1)^{2}} \right)\\ && \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} )^{2}}\right) \\ &&-\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} )^{2}}\right) \\ & &\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} - 1)^{2}}\right) \leq 0\\ \end{array} $$

(46)

and,

$$ \begin{array}{@{}rcl@{}} &&\left( \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} - 1)^{2}}\right) \right.\\ &&+\left.\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{1}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{1}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{1}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{1}} )^{2}}\right)\right)\\ && \left( \left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}})^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} - 1)^{2}}\right)\right.\\ && +\left.\left( \sqrt{(\widetilde{\mu}^{L}_{\widetilde{\beta}_{2}} - 1)^{2} + (\widetilde{\nu}^{L}_{\widetilde{\beta}_{2}} )^{2} } + \sqrt{(\widetilde{\mu}^{U}_{\widetilde{\beta}_{2}}-1)^{2} + (\widetilde{\nu}^{U}_{\widetilde{\beta}_{2}} )^{2}}\right)\right)\\ &\geq& 0 \end{array} $$

(47)

so, ${\boldsymbol {\Re }}(\widetilde {\beta }_{1}) - {\boldsymbol {\Re }}(\widetilde {\beta }_{2}) \leq 0$, that is ${\boldsymbol {\Re }}(\widetilde {\beta }_{1}) \leq {\boldsymbol {\Re }}(\widetilde {\beta }_{2})$. □