A VIKOR-based method for hesitant fuzzy multi-criteria decision making

Abstract

Since it was firstly introduced by Torra and Narukawa (The 18th IEEE International Conference on Fuzzy Systems, Jeju Island, Korea, 2009, pp. 1378–1382), the hesitant fuzzy set has attracted more and more attention due to its powerfulness and efficiency in representing uncertainty and vagueness. This paper extends the classical VIKOR (vlsekriterijumska optimizacija i kompromisno resenje in serbian) method to accommodate hesitant fuzzy circumstances. Motivated by the hesitant normalized Manhattan distance, we develop the hesitant normalized Manhattan $L_p$—metric, the hesitant fuzzy group utility measure, the hesitant fuzzy individual regret measure, and the hesitant fuzzy compromise measure. Based on these new measures, we propose a hesitant fuzzy VIKOR method, and a practical example is provided to show that our method is very effective in solving multi-criteria decision making problems with hesitant preference information.

Appendices

Appendix 1

1.1 The computational procedures of score values

Here we take the first column as an example:

$$\begin{aligned} s(z_{11})&= \frac{0.6+0.7+0.9}{3}=0.7333, \quad s(z_{21})=\frac{0.7+0.8+0.9}{3}=0.8000\\ s(z_{31})&= \frac{0.5+0.6+0.8}{3}=0.6333, \quad s(z_{41})=\frac{0.6+0.9}{2}=0.7500 \end{aligned}$$

Since $s(z_{21})>s(z_{41})>s(z_{11})>s(z_{31})$, according to the scheme in Sect. 2.1, there is no need to calculate the variance values and we can derive that $z_{21} \succ z_{41} \succ z_{11} \succ z_{31}$. Since $\varsigma _1$ is a benefit-type criterion, then we obtain $h_1^*=z_{21} = \left\{ {0.7,0.8,0.9}\right\}$ and $h_1^- =z_{31} = \left\{ {0.5,0.6,0.8}\right\}$.

1.2 The computational procedures of variance values

Here we take the last column as an example:

$$\begin{aligned} s(z_{14})&= \frac{0.4+0.5+0.9}{3}=0.6000, \quad s(z_{24})=\frac{0.5+0.6+0.7}{3}=0.6000\\ s(z_{34})&= \frac{0.5+00.7}{2}=0.6000, \quad s(z_{44})=\frac{0.4+0.5}{2}=0.4500 \end{aligned}$$

Since $s(z_{14})=s(z_{24})=s(z_{34})>s(z_{44})$, according to the scheme in Sect. 2.1, we need to calculate the variance values of $z_{14}, z_{24}$ and $z_{34}$:

$$\begin{aligned} v(z_{14})&= \frac{\sqrt{0.1^{2}+0.4^{2}+0.5^{2}}}{3}=0.6481\\ v(z_{24})&= \frac{\sqrt{0.1^{2}+0.1^{2}+0.2^{2}}}{3}=0.2449\\ v(z_{34})&= \frac{\sqrt{0.2^{2}}}{2}=0.2000 \end{aligned}$$

Because $v(z_{14})>v(z_{24})>v(z_{34})$, thus, $z_{34} \succ z_{24} \succ z_{14} \succ z_{44}$. Since $\varsigma _4$ is a benefit-type criterion, then we obtain $h_4^*=z_{34} = \left\{ {0.5,0.7}\right\}$ and $h_4^- =z_{44} =\left\{ {0.4,0.5}\right\}$.

Appendix 2

Here we take the fourth alternative as an example:

According to Example 4, we have $d\left( {h_1^*,h_1^-}\right) = \frac{1}{3}\left( \left| {0.7-0.5}\right| +\left| {0.8-0.6}\right| \right. \left. +\left| {0.9-0.8}\right| \right) =0.1667$. Similarly, we can obtain $d\left( {h_2^*,h_2^-}\right) =0.1, d\left( {h_3^*, h_3^-}\right) =0.1333, d\left( {h_4^*,h_4^-}\right) =0.15, d\left( {h_1^*, h_{41}}\right) =0.1333, d\left( {h_2^*, h_{42}} \right) =0,d\left( {h_3^*, h_{43}}\right) =0.1333$ and $d\left( {h_4^*, h_{44}}\right) =0.15$.

Hence, $\tilde{S}_4 =\sum _{j=1}^4 {\omega _j \frac{d\left( {h_j^*, h_{4j}}\right) }{d\left( {h_j^*, h_j^-}\right) }} =0.1\times \frac{0.1333}{0.1667}+0.1\times \frac{0}{0.1}+0.1\times \frac{0.1333}{0.1333}+0.1\times \frac{0.15}{0.15}=0.778$ and $\tilde{R}_4 =\mathop {\max }\limits _j \left( {\omega _j\frac{d\left( {h_j^*, h_{4j}}\right) }{d\left( {h_j^*, h_j^-}\right) }} \right) =0.4$. Similarly, we can get $\tilde{S}_1 =0.704, \tilde{S}_2 =0.1334, \tilde{S}_3 =0.6, \tilde{R}_1 =0.3, \tilde{R}_2 =0.0667$ and $\tilde{R}_3 =0.3$. So, $\tilde{S}^{*}=\mathop {\min }\nolimits _i \tilde{S}_i =0.1334, \tilde{S}^{-}=\mathop {\max }\nolimits _i \tilde{S}_i =0.778, \tilde{R}^{*}=\mathop {\min }\nolimits _i \tilde{R}_i =0.0667, \tilde{R}^{-}=\mathop {\max }\nolimits _i \tilde{R}_i =0.4$.

Then, $\tilde{Q}_4 =\upsilon \frac{\tilde{S}_4 -\tilde{S}^{*}}{\tilde{S}^{-}-\tilde{S}^{*}}+\left( {1-\upsilon }\right) \frac{\tilde{R}_4 -\tilde{R}^{*}}{\tilde{R}^{-}-\tilde{R}^{*}}=0.5\times \frac{0.778-0.1334}{0.778-0.1334}+0.5\times \frac{0.4-0.0667}{0.4-0.0667}=0.5$. Similarly, $\tilde{Q}_1 =0.7926, \tilde{Q}_2 =0$ and $\tilde{Q}_3 =0.7119$.

Therefore, $\tilde{Q}_2 <\tilde{Q}_4 <\tilde{Q}_3 <\tilde{Q}_1, \tilde{S}_2 <\tilde{S}_3 <\tilde{S}_1 <\tilde{S}_4$, and $\tilde{R}_2 <\tilde{R}_3 =\tilde{R}_1 <\tilde{R}_4$.