基于双向投影法的概率语言多属性群决策

doi:10.3969/j.issn.1001-506X.2020.09.21

Abstract

Abstract:

Aiming at the multi-attribute group decision-making problem with the probabilistic linguistic information, a new operation of the probabilistic linguistic term sets (PLTS) is defined based on the adjustment rules and linguistic scale function of the PLTS, which is proved to satisfy the basic properties of algebraic operations. Then, to solve the inconsistency between the calculation results of the existing vector module formula in subscript symmetric language and the actual situation, an improved vector module formula in the probabilistic linguistic information is proposed. On this basis, the vector expressions formed by the scheme, positive and negative ideal solutions are defined, and the bidirectional projection method for the probabilistic linguistic information is established, thus a new closeness degree formula is presented. Finally, the effectiveness and feasibility of the proposed method are verified by the comparative analysis of the application examples. This method fully considers the data characteristics of the PLTS, and avoids the information loss while simplifying the calculation, which provides a new method and idea to solve the problem of small numerical differences in the probabilistic linguistic information.

Key words: multi-attribute group decision-making, probabilistic linguistic term sets (PLTS), bidirectional projection method

CLC Number:

C934

Yue HU, Dengying JIANG, He LI. Probabilistic linguistic multi-attribute group decision-making based on bidirectional projection method[J]. Systems Engineering and Electronics, 2020, 42(9): 2052-2059.

Figures/Tables 11

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References 22

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文献	公式
文献[3]	$L_{1}(p) \oplus L_{2}(p)=\bigcup\limits_{L_{1}^{(k)} \in L_{1}(p), L_{2}^{(k)} \in L_{2}(p)}\left\{p_{1}^{(k)} L_{1}^{(k)} \oplus p_{2}^{(k)} L_{2}^{(k)}\right\}\left(k=1, 2, \cdots, \max \left(\# L_{1}(p), \# L_{2}(p)\right)\right) $ $\lambda L(p)=\bigcup\limits_{L^{(k)} \in L(p)} \lambda p^{(k)} L^{(k)}(k=1, 2, \cdots, \# L(p))$
文献[4]	$ L_{1}(p) \oplus L_{2}(p)=g^{-1}\left(\bigcup\limits_{\eta_{1}^{(i)} \in g\left(L_{1}\right), \eta_{2}^{(j)} \in g\left(L_{2}\right)}\left\{\left(\eta_{1}^{(i)}+\eta_{2}^{(j)}-\eta_{1}^{(i)} \eta_{2}^{(j)}\right)\left(p_{1}^{(i)} p_{2}^{(j)}\right)\right\}\right)\left(i=1, 2, \cdots, \# L_{1}(p) ; j=1, 2, \cdots, \# L_{2}(p)\right)$ $\lambda L(p)=g^{-1}(\bigcup\limits_{\eta^{(i)} \in g(L)}\left\{\left(1-\left(1-\eta^{(i)}\right)^{\lambda}\right)\left(p^{(i)}\right)\right\})(i=1, 2, \cdots, \# L(p)) $
文献[5]	$ L_{1}(p) \oplus L_{2}(p)=\bigcup\limits_{L_{1}^{(k)} \in L_{1}(p), L_{2}^{(k)} \in L_{2}(p)}\left\{L_{3}^{(k)}\left(p_{3}^{(k)}\right)\right\}\left(L_{3}^{(k)}=L_{1}^{(k)} \oplus L_{2}^{(k)}, p_{3}^{(k)}=p_{1}^{(k)} \cdot p_{2}^{(k)}, k=1, 2, \cdots, \max \left(\# L_{1}(p), \# L_{2}(p)\right)\right)$ $\lambda L(p)=\bigcup\limits_{L^{(k)} \in L(p)}\left\{\lambda L^{(k)}\left(p^{(k)}\right)\right\}(k=1, 2, \cdots, \# L(p)) $
文献[19]	$L_{1}(p) \oplus L_{2}(p)=\bigcup\limits_{\eta \in\left(L_{1}(p) L_{2}(p)\right)}{\cup}\left\{\eta^{(w)}\left(p_{1}^{(w)}+p_{2}^{(w)}\right)\right\}\left(w=\# L_{1}(p)+\# L_{2}(p)-\#\left(L_{1}(p) \cap L_{2}(p)\right)\right) $ $ \lambda L(p)=\bigcup\limits_{\eta^{(k)} \in L(p)}\left\{\eta^{(k)}\left(\lambda p^{(k)}\right)\right\}(k=1, 2, \cdots, \# L(p))$

文献	L₁(p)⊕L₂(p)
文献[3]	L₁(p)⊕L₂(p)={s_-0.4, s_0.3, s_1.3}
文献[4]	L₁(p)⊕L₂(p)={s_0.5(0.08), s₁(0.06), s_1.5(0.15), s_1.25(0.12), s_1.75(0.09), s₂(0.5)}
文献[5]	L₁(p)⊕L₂(p)={s_-1(0.08), s₀(0.18), s₁(0.35), s₂(0.24), s₃(0.15)}
文献[19]	L₁(p)⊕L₂(p)={s_-1(0.4), s₀(0.5), s₁(0.3), s₂(0.5)}
本文	L₁(p)⊕L₂(p)={s_-0.6(0.2), s_-0.28(0.2), s_0.44(0.1), s_0.84(0.2), s_1.48(0.3)}

文献	λ₁L₁(p)⊕λ₂L₂(p), λ₁=0.3, λ₂=0.7
文献[3]	λ₁L₁(p)⊕λ₂L₂(p)={s_-0.28, s_0.09, s_0.51}
文献[4]	λ₁L₁(p)⊕λ₂L₂(p)={s_-0.66(0.08), s_-0.16(0.12), s₀(0.16), s_0.38(0.09), s_0.77(0.06), s₁(0.09), s₂(0.5)}
文献[5]	λ₁L₁(p)⊕λ₂L₂(p)={s_-0.7(0.08), s_-0.4(0.12), s_-0.1(0.2), s₀(0.06), s_0.3(0.09), s_0.6(0.15), s_0.7(0.06), s₁(0.09), s_1.3(0.15)}
文献[19]	λ₁L₁(p)⊕λ₂L₂(p)={s_-1(0.28), s₀(0.27), s₁(0.3), s₂(0.15)}
本文	λ₁L₁(p)⊕λ₂L₂(p)={s_-0.6(0.2), s_-0.28(0.2), s_0.44(0.1), s_0.84(0.2), s_1.48(0.3)}

模公式	贴近度结果	排序结果
文献[16]	C₁=1＞C₂=0.933 9＞C₄=0.932 1＞C₃=0.877 4＞C₅=0.868 5	A₁?A₂?A₄?A₃?A₅
本文	C₁=1＞C₂=0.896 8＞C₄=0.865 7＞C₃=0.791 3＞C₅=0.773 3	A₁?A₂?A₄?A₃?A₅

	G₁	G₂	G₃	G₄	G₅
A₁	{s₂(0.4), s₁(0.4), s₀(0.2)}	{s₂(0.6), s₁(0.4)}	{s₂(0.6)}	{s₁(0.4)}	{s₂(0.2), s₁(0.4), s₀(0.2)}
A₂	{s_-2(0.4), s₀(0.2), s₁(0.2)}	{s₂(0.6), s₁(0.4)}	{s₂(0.4), s₁(0.6)}	{s₁(0.4), s₀(0.4)}	{s_-1(0.6), s₀(0.2), s₂(0.2)}
A₃	{s₁(0.6), s₀(0.4)}	{s₁(0.2), s₀(0.4), s_-1(0.2)}	{s_-2(0.6), s₁(0.2), s₀(0.2)}	{s₁(0.4), s₀(0.2), s_-2(0.2)}	{s₂(0.2), s₁(0.4), s₀(0.4)}
A₄	{s_-2(0.4), s₀(0.4), s_-1(0.2)}	{s₀(0.6), s₁(0.2)}	{s₀(0.2), s₁(0.2), s₂(0.4)}	{s₂(0.4), s₁(0.4), s₀(0.2)}	{s₁(0.2), s₂(0.4), s_-2(0.2)}
A₅	{s₀(0.4), s₁(0.6)}	{s₁(0.2), s₀(0.2), s_-2(0.2)}	{s_-1(0.4), s_-2(0.4), s₀(0.2)}	{s₂(0.8)}	{s_-2(0.4), s₀(0.6)}

Probabilistic linguistic multi-attribute group decision-making based on bidirectional projection method

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行业领域	属性
行业领域	G₁	G₂	G₃	G₄	G₅
A₁	{s₂(0.4), s₁(0.4), s₀(0.2)}	{s₂(0.6), s₁(0.4)}	{s₂(1)}	{s₁(1)}	{s₁(0.5), s₂(0.25), s₀(0.25)}
A₂	{s₁(0.25), s₀(0.25), s_-2(0.5)}	{s₂(0.6), s₁(0.4)}	{s₁(0.6), s₂(0.4)}	{s₁(0.5), s₀(0.5)}	{s₂(0.2), s_-1(0.6), s₀(0.2)}
A₃	{s₁(0.6), s₀(0.4)}	{s₀(0.5), s₁(0.25), s_-1(0.25)}	{s₁(0.2), s₀(0.2), s_-2(0.6)}	{s₁(0.5), s₀(0.25), s_-2(0.25)}	{s₁(0.4), s₂(0.2), s₀(0.4)}
A₄	{s₀(0.4), s_-1(0.2), s_-2(0.4)}	{s₀(0.75), s₁(0.25)}	{s₂(0.5), s₁(0.25), s₀(0.25)}	{s₂(0.4), s₁(0.4), s₀(0.2)}	{s₂(0.5), s₁(0.25), s_-2(0.25)}
A₅	{s₁(0.6), s₀(0.4)}	{s₁(0.33), s₀(0.33), s_-2(0.34)}	{s₀(0.2), s_-1(0.4), s_-2(0.4)}	{s₂(1)}	{s₀(0.6), s_-2(0.4)}

	A^-A⁺		A^-A⁺
A^-A₁	Prj_A^-A⁺(A^-A₁)=0.959 5	A₁A⁺	Prj_A₁A⁺(A^-A⁺)=1.164 0
A^-A₂	Prj_A^-A⁺(A^-A₂)=0.783 6	A₂A⁺	Prj_A₂A⁺(A^-A⁺)=1.303 2
A^-A₃	Prj_A^-A⁺(A^-A₃)=0.610 7	A₃A⁺	Prj_A₃A⁺(A^-A⁺)=1.145 1
A^-A₄	Prj_A^-A⁺(A^-A₄)=0.734 1	A₄A⁺	Prj_A₄A⁺(A^-A⁺)=1.307 0
A^-A₅	Prj_A^-A⁺(A^-A₅)=0.329 3	A₅A⁺	Prj_A₅A⁺(A^-A⁺)=1.033 5

方法	排序	最优解
PLWA算子^[3]	A₁?A₂?A₄?A₃?A₅	A₁
扩展TOPSIS法^[3]	A₁?A₂?A₄?A₃?A₅	A₁
投影法^[16]	A₁?A₂?A₄?A₃?A₅	A₁
本文方法	A₁?A₂?A₄?A₃?A₅	A₁

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