Dealing with interaction between bipolar multiple criteria preferences in PROMETHEE methods

Abstract

In this paper we extend the PROMETHEE methods to the case of interacting criteria on a bipolar scale, introducing the bipolar PROMETHEE method based on the bipolar Choquet integral. In order to elicit parameters compatible with preference information provided by the Decision Maker (DM), we propose to apply the Robust Ordinal Regression (ROR). ROR takes into account simultaneously all the sets of parameters compatible with the preference information provided by the DM considering a necessary and a possible preference relation.

Appendix

Proof of Proposition 3.2

Let us prove that if $\hat{\mu}(C,D)= -\hat{\mu}(D,C)$ for each ${(C, D) \in P({\mathcal{J}})},$ then $Ch^B(P^B(a,b), \hat{\mu})= - Ch^B(P^B(b,a), \hat{\mu}).$ As noticed, P ^B _j (a, b) =  − P ^B _j (b, a) for all $j \in \mathcal{J},$ and consequently $\vert P_{(j)}^{B}(a,b) \vert = \vert -P_{(j)}^{B}(b,a) \vert = \vert P_{(j)}^{B}(b,a) \vert$ for all ${j\in{\mathcal J}}.$

By this, it follows that:

$$ \begin{aligned} (\alpha)\quad\quad C_{(j)}(a,b) &= \{ i \in {{\mathcal{J}}}^{>} \, : \, P_{i}^{B}(a,b) \geq \vert P_{(j)}^{B}(a,b) \vert \} = \{ i \in {{\mathcal{J}}}^{>} \, : \, -P_{i}^{B}(b,a) \geq \vert P_{(j)}^{B}(b,a) \vert \}\\ &= D_{(j)}(b,a);\\ (\beta) \quad\quad D_{(j)}(a,b) &= \{ i \in {{\mathcal{J}}}^{>} \, : \, - P_{i}^{B}(a,b) \geq \vert P_{(j)}^{B}(a,b) \vert \} = \{ i \in {{\mathcal{J}}}^{>} \, : \, P_{i}^{B}(b,a) \geq \vert P_{(j)}^{B}(b,a) \vert \}\\ &= C_{(j)}(b,a). \end{aligned} $$

By (α) and (β) we have that

$$ \begin{aligned} (\gamma)\quad\quad { Ch^B(P^B(a,b), \hat{\mu})}&={ \sum_{j \in {{\mathcal{J}}}^{>}} \vert P_{(j)}^{B}(a,b) \vert \Big[\hat{\mu}(C_{(j)}(a,b), D_{(j)}(a,b)) - \hat{\mu}(C_{(j+1)}(a,b), D_{(j+1)}(a,b))\Big]}\\ &= \sum_{j \in {{\mathcal{J}}}^{>}} \vert P_{(j)}^{B}(b,a) \vert \Big[\hat{\mu}(D_{(j)}(b,a), C_{(j)}(b,a)) - \hat{\mu}(D_{(j+1)}(b,a), C_{(j+1)}(b,a)) \Big]. \end{aligned} $$

Since ${\hat{\mu}(C,D) =- \hat{\mu}(D,C), \forall (C,D)\in P({\mathcal J})},$ by (γ) we have that,

$$ \begin{aligned} (\delta)\quad\quad { Ch^B(P^B(b,a), \hat{\mu})} &= \sum_{j \in {{\mathcal{J}}}^{>}} \vert P_{(j)}^{B}(b,a) \vert \Big[\hat{\mu}(C_{(j)}(b,a), D_{(j)}(b,a)) - \hat{\mu}(C_{(j+1)}(b,a), D_{(j+1)}(b,a))\Big] \\ &= \sum_{j \in {{\mathcal{J}}}^{>}} \vert P_{(j)}^{B}(b,a) \vert \Big[ - \hat{\mu}(D_{(j)}(b,a), C_{(j)}(b,a)) + \hat{\mu}(D_{(j+1)}(b,a), C_{(j+1)}(b,a)) \Big] \\ &= -Ch^B(P^B(a, b), \hat{\mu}). \end{aligned} $$

Let us now prove that if $Ch^B(P^B(a,b), \hat{\mu})= - Ch^B(P^B(b,a), \hat{\mu}),$ then $\hat{\mu}(C,D) = - \hat{\mu}(D,C).$ Let us consider the pair (a, b) such that,

$$ P_{j}^{B}(a,b)= \left\{ \begin{array}{lll} 1& \hbox{if} & j\in C\\ -1 & \hbox{if} & j\in D\\ 0 & & \hbox{otherwise} \\ \end{array} \right. $$

(19)

In this case we have that $Ch^B(P^B(a,b), \hat{\mu})= \hat{\mu}(C,D)$ and $Ch^B(P^B(b,a), \hat{\mu})= \hat{\mu}(D,C).$ Thus if $Ch^B(P^B(a,b), \hat{\mu})=-Ch^B(P^B(b,a), \hat{\mu}),$ by (iv) we obtain that $\hat{\mu}(C,D) = - \hat{\mu}(D,C)$ and the proof is concluded. □

Proof of Corollary 3.1

This can be seen as a Corollary both of Propositions 3.2 and 3.3. In fact,

• μ⁺(C, D) =  μ⁻(D, C) for each ${(C, D) \in P({\mathcal{J}})}$ implies that $\hat{\mu}(C,D)=-\hat{\mu}(D,C)$ for each ${(C, D) \in P({\mathcal{J}})},$ and by Proposition 3.2, it follows the thesis.

• μ⁺(C, D) =  μ⁻(D, C) for each ${(C, D) \in P({\mathcal{J}})}$ implies that $Ch^{B+}(P^{B}(a, b),\hat{\mu}) = Ch^{B-}(P^{B}(b, a),\hat{\mu})$ (by Proposition 3.3) and from this it follows obviously the thesis by Eq. (10). □

Proof of Proposition 3.4

We shall prove only part 1 Proof of part 2 can be obtained analogously.

If the bicapacity $\hat{\mu}$ is 2-additive decomposable, then

$$ \begin{aligned} { Ch^{B+}(x, \hat{\mu})} &= \sum_{j \in {{\mathcal{J}}}^>}\vert x_{(j)}\vert \big[{\mu}^{+}(C_{(j)}, D_{(j)}) - {\mu}^{+}(C_{(j+1)}, D_{(j+1)}) \Big] \\ & = \sum_{j \in {{\mathcal{J}}}^>}\vert x_{(j)} \vert \Big[\Big( \sum_{k \in {{\mathcal{J}}}^>, x_k \geq \vert x_{(j)} \vert} a^{+}_{k} - \sum_{k \in {{\mathcal{J}}}^>, x_k \geq \vert x_{(j+1)} \vert} a^{+}_{k}\Big) \\ & \quad+ \Big(\sum_{h, k \in {{\mathcal{J}}}^>, h \neq k, x_h, x_k \geq \vert x_{(j)} \vert} a^{+}_{hk} - \sum_{h, k \in {{\mathcal{J}}}^>, h \neq k, x_h, x_k \geq \vert x_{(j+1)} \vert} a^{+}_{hk}\Big)\\ & { \quad+\Big(\sum_{h, k \in {{\mathcal{J}}}^>, h\neq k, x_h, -x_k \geq \vert x_{(j)} \vert} a^{+}_{h \vert k} - \sum_{h, k \in {{\mathcal{J}}}^>, h\neq k, x_h, -x_k \geq \vert x_{(j+1)} \vert} a^{+}_{h \vert k} \Big)\Big] } \\ \end{aligned} $$

Let us remark that,

$$ \begin{aligned} &a) \quad\quad \Big( \sum_{k \in {{\mathcal{J}}}^>, x_k \geq \vert x_{(j)} \vert} a^{+}_{k} - \sum_{k \in {{\mathcal{J}}}^>, x_k \geq \vert x_{(j+1)} \vert} a^{+}_{k}\Big) = \left\{ {\begin{array}{ll} \sum\limits_{k\in{\mathcal J}^{>}, x_{k}=|x_{(j)}|} a^{+}_{k} & \hbox{if}\,\,|x_{(j)}| < |x_{(j+1)}| \\ 0 & \hbox{otherwise} \end{array}} \right. \\ &b) \quad\quad \Big( \sum\limits_{\substack{h,k \in {{\mathcal{J}^{>}}}, {h\neq k}, \\ {x_{h}},{x_{k}} \geq {\vert {x_{(j)}} \vert} }} a^{+}_{hk} - \sum\limits_{\substack{h,k \in {{\mathcal{J}^{>}}}, {{h}\neq {k}}, \\ {x_{h}},{x_{k}} \geq {\vert {x_{(j+1)}} \vert}}} a^{+}_{hk} \Big) = \left\{ {\begin{array}{ll} {\sum\limits_{\substack{{h,k} \in {{\mathcal{J}^{>}}}, {{h}\neq {k}}, \\ \min\{{x_{h}},{x_{k}}\} = {\vert {x_{(j)}} \vert} }}} a^{+}_{hk}&\hbox{if} \,\, |x_{(j)}| < |x_{(j+1)}| \\ 0 & \hbox{otherwise} \end{array}} \right. \\ &c) \quad\quad \Big( \sum\limits_{\substack{{h,k} \in {{\mathcal{J}^{>}}}, {{h}\neq {k}}, \\ {x_{h}},{{-x}_{k}} \geq {\vert {x_{(j)}} \vert} }} {a^{+}_{{h} \vert {k}}} - \sum\limits_{\substack{{h,k} \in {{\mathcal{J}^{>}}}, {{h}\neq {k}}, \\ {x_{h}},-{x_{k}} \geq {\vert {x_{(j+1)}} \vert} }} a^{+}_{{h} \vert {k}}\Big) = \left\{{\begin{array}{ll} { \sum\limits_{\substack{{h,k} \in {{\mathcal{J}^{>}}}, {{h}\neq {k}}, \\ \min\{{x_{h}},-{x_{k}}\} = {\vert {x_{(j)}} \vert} }}} a^{+}_{{h} \vert {k}} &\hbox{if} \,\, |x_{(j)}| < |x_{(j+1)}| \\ 0& \hbox{otherwise} \end{array}} \right. \end{aligned} $$

Considering a) − c) we get that:

$$ \chi) = \sum_{\substack{j \in {{\mathcal{J}}}^>,\\ |x_{(j)}|<|x_{(j+1)}|}}\vert x_{(j)} \vert \Big[ \sum_{k\in{\mathcal J}^{>}, x_{k}=|x_{(j)}|} a^{+}_{k} + \sum_{\substack{h,k \in {{\mathcal{J}}}^>, h\neq k, \\ \min\{x_h,x_k\} = \vert x_{(j)} \vert }} a^{+}_{hk} + \sum_{\substack{h,k \in {{\mathcal{J}}}^>, h\neq k, \\ \min\{x_h,-x_k\} = \vert x_{(j)} \vert }} a^{+}_{h \vert k} \Big]\\ $$

and from this it follows the thesis. □

Proof of Proposition 3.5

First, let us prove that

$$ (a) \quad \hat{\mu}(C,D) = - \hat{\mu}(D,C) $$

implies 1, 2 and 3 For each $j \in \mathcal{J},$

$$ (b)\quad \hat{\mu}(\{j\}, \emptyset)= a^{+}_{j}\,\,\hbox{and}\,\, \hat{\mu}(\emptyset,\{j\})= -a^{-}_{j}. $$

From (a) and (b) we have,

$$ a^{+}_{j}= \hat{\mu}(\{j\}, \emptyset )= - \hat{\mu}(\emptyset ,\{j\})=a^{-}_{j} $$

which is 1.

For each $\{ j,k \} \subseteq \mathcal{J}$ we have that,

$$ (c)\quad \hat{\mu}(\{j,k\}, \emptyset) = a^{+}_{j} + a^{+}_{k} + a^{+}_{jk}\,\hbox{and}\,\hat{\mu}(\emptyset, \{j,k\}) = -a^{-}_{j} - a^{-}_{k} - a^{-}_{jk} $$

Being $\hat{\mu}(\{j,k\}, \emptyset)=-\hat{\mu}(\emptyset,\{j,k\}),$ and being a ⁺ _j  = a ⁻ _j and a ⁺ _k  = a ⁻ _k by 1, we have that for each $\{j,k\} \subseteq \mathcal{J},$ a ⁺ _jk  = a ⁻ _jk , i.e. 2.

For all ${j,k\in {\mathcal J}}$ with j ≠ k, we have:

$$ \begin{aligned} \hat{\mu}(\{j\}, \{k\})&=a_{j}^{+}-a_{k}^{-}+a^{+}_{j \vert k}-a^{-}_{j \vert k}\\ \hat{\mu}(\{k\}, \{j\})&=a_{k}^{+}-a_{j}^{-}+a^{+}_{k \vert j}-a^{-}_{k \vert j} \end{aligned} $$

Being $\hat{\mu}(\{j\}, \{k\})=-\hat{\mu}(\{k\}, \{j\})$ and having proved that a ⁺ _j  = a ⁻ _j , ∀ j, we obtain that $a^{+}_{j \vert k}-a^{-}_{j \vert k}=-a^{+}_{k \vert j}+a^{-}_{k \vert j}$ i.e. 3.

It is straightforward to prove that 1, 2, and 3 imply $\hat{\mu}(C,D)=-\hat{\mu}(D,C).$ □