Multiple criteria hierarchy process for sorting problems based on ordinal regression with additive value functions

Abstract

A hierarchical decomposition is a common approach for coping with complex decision problems involving multiple dimensions. Recently, the multiple criteria hierarchy process (MCHP) has been introduced as a new general framework for dealing with multiple criteria decision aiding in case of a hierarchical structure of the family of evaluation criteria. This study applies the MCHP framework to multiple criteria sorting problems and extends existing disaggregation and robust ordinal regression techniques that induce decision models from data. The new methodology allows the handling of preference information and the formulation of recommendations at the comprehensive level, as well as at all intermediate levels of the hierarchy of criteria. A case study on bank performance rating is used to illustrate the proposed methodology.

Appendices

Appendix

Proof of Proposition 3.1

$(1)\Rightarrow (2)$ Let $a\xrightarrow [{(\mathbf {r},j)}]{}\left[ C_{h_j},C_{k_j}\right]$ for all $j=1,\ldots ,n(\mathbf {r})$. This means that $b_{h_j-1}^{(\mathbf {r},j)}\le U_{(\mathbf {r},j)}(a)<b_{k_j}^{(\mathbf {r},j)}$ for all $j=1,\ldots ,n(\mathbf {r})$. Let us consider $h=\min _{j=1,\ldots ,n(\mathbf {r})}h_j$ and $k=\max _{j=1,\ldots ,n(\mathbf {r})}k_j$. For the monotonicity of the thresholds, we shall have for all $j=1,\ldots ,n(\mathbf {r})$ that $b_{h-1}^{(\mathbf {r},j)}\le b_{h_j-1}^{(\mathbf {r},j)}\le U_{(\mathbf {r},j)}(a)<b_{k_j}^{(\mathbf {r},j)}\le b_{k}^{(\mathbf {r},j)}$ for all $j=1,\ldots ,n(\mathbf {r})$ and, consequently, $b_{h-1}^{(\mathbf {r},j)}\le U_{(\mathbf {r},j)}(a)< b_{k}^{(\mathbf {r},j)}$. Adding up with respect to j, we get

$$\begin{aligned} b^{\mathbf {r}}_{h-1}=\sum _{j=1}^{n(\mathbf {r})}b_{h-1}^{(\mathbf {r},j)}\le \sum _{j=1}^{n(\mathbf {r})}U_{(\mathbf {r},j)}(a)<\sum _{j=1}^{n(\mathbf {r})}b_{k}^{(\mathbf {r},j)}=b_{k}^{\mathbf {r}}. \end{aligned}$$

From Eq. (1), it follows that $b^{\mathbf {r}}_{h-1}\le U_{\mathbf {r}}(a)<b_{k}^{\mathbf {r}}$ and, consequently, $a\xrightarrow [\mathbf {r}]{}\left[ C_{h},C_k\right]$.

$(2)\Rightarrow (3)$ follows directly by setting $h_{j}=k_{j}=h$ for all $j=1,\ldots ,n(\mathbf {r})$.

$(3)\Rightarrow (1)$ follows by contradiction, when we suppose that $b_h^{\mathbf {r}}\ne \sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}$ for some h. This implies that $b_h^{\mathbf {r}}>\sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}$ or $b_h^{\mathbf {r}}<\sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}$.

Let $b_h^{\mathbf {r}}>\sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}$ and $a\in A$ an alternative, such that

$$\begin{aligned} U_{(\mathbf {r},j)}(a)=b_{h}^{(\mathbf {r},j)} \;\;\text{ for } \text{ all }\;\; j=1,\ldots ,n(\mathbf {r}). \end{aligned}$$

(11)

Obviously, this implies that $a\xrightarrow [(\mathbf {r},j)]{}C_{h+1}$ for all $j=1,\ldots ,n(\mathbf {r})$. Adding up with respect to j in the two members of Eq. (11), we get $U_{\mathbf {r}}(a)=\sum _{j=1}^{n(\mathbf {r})}U_{(\mathbf {r},j)}(a)=\sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}<b_{h}^{\mathbf {r}}$ and, consequently, $a\xrightarrow [\mathbf {r}]{}C_{\le h}$, being in contradiction with the hypothesis.

Now, let $b_h^{\mathbf {r}}<\sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}$ and $a\in A$ an alternative, such that

$$\begin{aligned} U_{(\mathbf {r},j)}(a)=b_{h}^{(\mathbf {r},j)}-\frac{\varepsilon }{n(\mathbf {r})} \;\;\text{ for } \text{ all }\;\; j=1,\ldots ,n(\mathbf {r}) \end{aligned}$$

(12)

where $\varepsilon >0$. This choice implies that $a\xrightarrow [\mathbf {(r,j)}]{}C_{\le h}$, for all $j=1,\ldots ,n(\mathbf {r})$. Now, adding up with respect to j in the two members of Eq. (12), we get $U_{\mathbf {r}}(a)=\sum _{j=1}^{n(\mathbf {r})}U_{(\mathbf {r},j)}(a)=\sum _{j=1}^{n(\mathbf {r})}\left[ b^{(\mathbf {r},j)}_{h}-\frac{\varepsilon }{n(\mathbf {r})}\right] =\sum _{j=1}^{n(\mathbf {r})}b^{(\mathbf {r},j)}_{h}-\varepsilon .$ If we choose $\varepsilon$ such that

$$\begin{aligned} 0<\varepsilon \le \min \left\{ \min \left\{ n(\mathbf {r})\cdot \left[ b_{h}^{(\mathbf {r},j)}-b_{h-1}^{(\mathbf {r},j)}\right] ,\; {j=1,\ldots ,n(\mathbf {r})}\right\} ,\sum _{j=1}^{n(\mathbf {r})}b_{h}^{(\mathbf {r},j)}-b_{h}^{\mathbf {r}}\right\} \end{aligned}$$

we obtain that $b_{h-1}^{(\mathbf {r},j)}\le U_{(\mathbf {r},j)}(a)<b_{h}^{(\mathbf {r},j)}$ for all $j=1,\ldots ,n(\mathbf {r})$ ² and $U_{\mathbf {r}}(a)>b_{h}^{\mathbf {r}}$ ³ implying that $a\xrightarrow [(\mathbf {r},j)]{}C_{h}$ for all $j=1,\ldots ,n(\mathbf {r})$ and $a\xrightarrow [\mathbf {r}]{}C_{\ge h+1}$, thus leading to a contradiction. $\square$

Proof of Proposition 4.1

• 1. Let $a\in A$, $\mathbf {r}\in \mathcal{I}_\mathcal{G}{\setminus } EL$ and $h=2,\ldots ,p$ such that not$\left( a\xrightarrow [\mathbf {r}]{N}C_{\ge h}\right)$. This means that there exists at least one (U, b) such that $U_{\mathbf {r}}(a)<b_{h-1}^{\mathbf {r}}$. Therefore $a\xrightarrow [\mathbf {r}]{P}C_{\le h-1}$. Let us observe that $a\xrightarrow [\mathbf {r}]{N}C_{\ge h}$ and $a\xrightarrow [\mathbf {r}]{P}C_{\le h-1}$ do not hold simultaneously because, otherwise, a couple $(\overline{U},\overline{b})$ should exist, such that $\overline{U}_{\mathbf {r}}(a)\ge \overline{b}_{h-1}^{\mathbf {r}}$ and $\overline{U}_{\mathbf {r}}(a)<\overline{b}_{h-1}^{\mathbf {r}}$, which is impossible.

• 2. Let $a\in A$, $\mathbf {r}\in \mathcal{I}_\mathcal{G}{\setminus } EL$ and $k=1,\ldots ,p-1$ such that not$\left( a\xrightarrow [\mathbf {r}]{N}C_{\le k}\right)$. This means that there exists at least one (U, b) such that $U_{\mathbf {r}}(a)\ge b_{k}^{\mathbf {r}}$. Therefore $a\xrightarrow [\mathbf {r}]{P}C_{\ge k+1}$. Let us observe that $a\xrightarrow [\mathbf {r}]{N}C_{\le k}$ and $a\xrightarrow [\mathbf {r}]{P}C_{\ge k+1}$ do not hold simultaneously because, otherwise, a couple $(\overline{U},\overline{b})$ should exist, such that $\overline{U}_{\mathbf {r}}(a)<\overline{b}_{k}^{\mathbf {r}}$ and $\overline{U}_{\mathbf {r}}(a)\ge \overline{b}_{k}^{\mathbf {r}}$, which is impossible.

• 3. $a\xrightarrow [(\mathbf {r},j)]{N}C_{\ge h_{j}}$ for all $j=1,\ldots ,n(\mathbf {r})$ implies that $U_{(\mathbf {r},j)}(a)\ge b_{h_j-1}^{(\mathbf {r},j)}$ for all (U, b) and for all $j=1,\ldots ,n(\mathbf {r})$. Considering $h=\min _{j=1,\ldots ,n(\mathbf {r})}h_j$, for the monotonicity of the thresholds we have that $U_{(\mathbf {r},j)}(a)\ge b_{h-1}^{(\mathbf {r},j)}$ for all (U, b) and for all j. As a consequence, adding up with respect to j, we get $U_{\mathbf {r}}(a)=\sum _{j=1}^{n(\mathbf {r})}U_{(\mathbf {r},j)}(a)\ge \sum _{j=1}^{n(\mathbf {r})}b^{(\mathbf {r},j)}_{h-1}=b_{h-1}^{\mathbf {r}}$ for all (U, b), which proves point 2.

• 4. $a\xrightarrow [(\mathbf {r},j)]{N}C_{\le k_{j}}$ for all $j=1,\ldots ,n(\mathbf {r})$ implies that $U_{(\mathbf {r},j)}(a)<b_{k_j}^{(\mathbf {r},j)}$ for all (U, b) and for all $j=1,\ldots ,n(\mathbf {r})$. Considering $k=\max _{j=1,\ldots ,n(\mathbf {r})}k_j$, for the monotonicity of the thresholds we have that $U_{(\mathbf {r},j)}(a)<b_{k}^{(\mathbf {r},j)}$ for all (U, b) and for all j. As a consequence, adding up with respect to j, we get $U_{\mathbf {r}}(a)=\sum _{j=1}^{n(\mathbf {r})}U_{(\mathbf {r},j)}(a)<\sum _{j=1}^{n(\mathbf {r})}b^{(\mathbf {r},j)}_{k}=b_{k}^{\mathbf {r}}$ for all (U, b), which implies point 3.

• 5. $a\xrightarrow [(\mathbf {r},j)]{N}C_{\ge h_{j}}$, for all $j\in \left\{ 1,\ldots ,n(\mathbf {r})\right\} {\setminus } \left\{ \overline{j}\right\}$ implies that for all (U, b), $U_{(\mathbf {r},j)}(a)\ge b^{(\mathbf {r},j)}_{h_j-1}$ for all $j\in \left\{ 1,\ldots ,n(\mathbf {r})\right\} {\setminus } \left\{ \overline{j}\right\}$. Analogously, $a\xrightarrow [(\mathbf {r},\overline{j})]{P}C_{\ge h_{\overline{j}}}$ implies that there exists at least one $(\overline{U},\overline{b})$ such that $\overline{U}_{(\mathbf {r},\overline{j})}(a)\ge \overline{b}^{(\mathbf {r},\overline{j})}_{h_{\overline{j}}-1}$. Considering $h=\min _{j=1,\ldots ,n(\mathbf {r})} h_j$, for $(\overline{U},\overline{b})$ and for the monotonicity of the thresholds we have that $\overline{U}_{(\mathbf {r},j)}(a)\ge \overline{b}^{(\mathbf {r},j)}_{h-1}$ for all $j=1,\ldots ,n(\mathbf {r})$. Adding up with respect to j we get $\overline{U}_{\mathbf {r}}(a)=\sum _{j=1}^{n(\mathbf {r})}\overline{U}_{(\mathbf {r},j)}(a)\ge \sum _{j=1}^{n(\mathbf {r})}\overline{b}^{(\mathbf {r},j)}_{h-1}=\overline{b}^{\mathbf {r}}_{h-1}$, which proves point 4.

• 6. $a\xrightarrow [(\mathbf {r},j)]{N}C_{\le k_{j}}$, for all $j\in \left\{ 1,\ldots ,n(\mathbf {r})\right\} {\setminus } \left\{ \overline{j}\right\}$ implies that for all (U, b), $U_{(\mathbf {r},j)}(a)< b^{(\mathbf {r},j)}_{k_j}$ for all $j\in \left\{ 1,\ldots ,n(\mathbf {r})\right\} {\setminus } \left\{ \overline{j}\right\}$. Analogously, $a\xrightarrow [(\mathbf {r},\overline{j})]{P}C_{\le k_{\overline{j}}}$ implies that there exists at least one $(\overline{U},\overline{b})$ such that $\overline{U}_{(\mathbf {r},\overline{j})}(a)< \overline{b}^{(\mathbf {r},\overline{j})}_{k_{\overline{j}}}$. Considering $k=\max _{j=1,\ldots ,n(\mathbf {r})} k_j$, for $(\overline{U},\overline{b})$ and for the monotonicity of the thresholds we have that $\overline{U}_{(\mathbf {r},j)}(a)< \overline{b}^{(\mathbf {r},j)}_{k}$ for all $j=1,\ldots ,n(\mathbf {r})$. Adding up with respect to j we get $\overline{U}_{\mathbf {r}}(a)=\sum _{j=1}^{n(\mathbf {r})}\overline{U}_{(\mathbf {r},j)}(a)< \sum _{j=1}^{n(\mathbf {r})}\overline{b}^{(\mathbf {r},j)}_{k}=\overline{b}^{\mathbf {r}}_{k}$, which proves point 5. $\square$

Proof of Proposition 4.2

• 1. Let $L_{(\mathbf {r},j)}^{\mathcal{U}, P}(a)=h_j$ for all $j=1,\ldots ,n(\mathbf {r})$. This means that $a\xrightarrow [_{(\mathbf {r},j)}]{N} C_{\ge h_j}$ and not$\left( a\xrightarrow [_{(\mathbf {r},j)}]{N} C_{\ge l}\right)$ with $l>h_j$ for all $j=1,\ldots ,n(\mathbf {r})$. By Proposition 4.1 we get $a\xrightarrow [_{\mathbf {r}}]{N} C_{\ge h}$ with $h=\min _{j=1,\ldots ,n(\mathbf {r})}h_j$. As a consequence we get the thesis.

• 2. Let $R_{(\mathbf {r},j)}^{\mathcal{U}, P}(a)=k_j$ for all $j=1,\ldots ,n(\mathbf {r})$. This means that $a\xrightarrow [_{(\mathbf {r},j)}]{N} C_{\le k_j}$ and not$\left( a\xrightarrow [_{(\mathbf {r},j)}]{N} C_{\ge l}\right)$ with $l>k_j$ for all $j=1,\ldots ,n(\mathbf {r})$. By Proposition 4.1 we get $a\xrightarrow [_{\mathbf {r}}]{N} C_{\ge k}$ with $k=\max _{j=1,\ldots ,n(\mathbf {r})}k_j$. As a consequence we get the thesis. $\square$