Type-2 fuzzy numbers made simple in decision making

Abstract

For the decision-making problems based on decision makers’ judgments in terms of linguistic terms, we propose type-2 fuzzy numbers (T2FNs) that allow decision makers better formalize their judgments. A T2FN has two components: a primary membership and a secondary membership. Compared with T1FSs and interval type-2 fuzzy sets, T2FNs consider an additional dimension by introducing the secondary membership. The primary membership indicates the truth degree of judgment, and the secondary membership further indicates the reliability degree of the truth. We define simple operation rules on T2FNs such that they can be easily used to deal with decision-making problems, such as multi-criteria decision making and multi-stages decision making. Compared with existing related approaches, we verify our approach with several numerical examples.

Appendix

Proof of Theorem 1

We use mathematical induction to prove the aggregation result of T2WA.

• (1) When $n = 2$, we can calculate T2WA as

  $$\begin{aligned} {\text{T2WA}}_{\omega } (t_{1} ,t_{2} ) & = \omega _{1} t_{1} \oplus \omega _{2} t_{2} \\ & = (1 - (1 - u_{1} )^{{\omega _{1} }} ,1 - (1 - \mu _{1} (u))^{{\omega _{1} }} ) \oplus (1 - (1 - u_{2} )^{{\omega _{2} }} ,1 - (1 - \mu _{2} (u))^{{\omega _{2} }} ) \\ & = (1 - (1 - u_{1} )^{{\omega _{1} }} (1 - u_{2} )^{{\omega _{2} }} ,1 - (1 - \mu _{1} (u)^{{\omega _{1} }} )(1 - \mu _{2} (u)^{{\omega _{2} }} )). \\ \end{aligned}$$

• (2) Suppose that when $n = k$,

  $${\text{T2WA}}_{\omega }^{{}} (t_{1}^{{}} ,t_{2}^{{}} , \ldots ,t_{k}^{{}} ) = \omega _{1}^{{}} t_{1}^{{}} \oplus \omega _{2}^{{}} t_{2}^{{}} \oplus \ldots \oplus \omega _{k}^{{}} t_{k}^{{}} = \left( {1 - \prod\limits_{{j = 1}}^{k} {(1 - u_{j} )^{{\omega _{j} }} } ,} \right.\left. {1 - \prod\limits_{{j = 1}}^{k} {(1 - \mu _{j} (u))^{{\omega _{j} }} } } \right)$$

holds. When $n = k + 1$, we have.

$$\begin{aligned} {\text{T2WA}}_{\omega }^{{}} (t_{1}^{{}} ,t_{2}^{{}} , \ldots ,t_{{k + 1}}^{{}} ) & = \omega _{1}^{{}} t_{1}^{{}} \oplus \omega _{2}^{{}} t_{2}^{{}} \oplus \cdots \oplus \omega _{{k + 1}}^{{}} t_{{k + 1}}^{{}} \\ & = \left( {1 - \prod\limits_{{j = 1}}^{k} {(1 - u_{j} )^{{\omega _{j} }} } ,} \right.\left. {\left. {1 - \prod\limits_{{j = 1}}^{k} {(1 - \mu _{j} (u))^{{\omega _{j} }} } } \right)} \right) \oplus (1 - (1 - u_{{k + 1}} )_{{}}^{{\omega _{{k + 1}} }} ,1 - (1 - \mu _{{k + 1}} (u))_{{}}^{{\omega _{{k + 1}} }} ) \\ & = \left( {1 - \prod\limits_{{j = 1}}^{{k + 1}} {(1 - u_{j} )^{{\omega _{j} }} } ,} \right.\left. {1 - \prod\limits_{{j = 1}}^{{k + 1}} {(1 - \mu _{j} (u))^{{\omega _{j} }} } } \right). \\ \end{aligned}$$

Thus, when $n = k + 1$, the expression in Theorem 1 holds.

Therefore, ${\text{T2WA}}_{\omega }^{{}} (t_{1}^{{}} ,t_{2}^{{}} , \ldots ,t_{n}^{{}} ) = \left( {1 - \prod\limits_{{j = 1}}^{n} {(1 - u_{j} )^{{\omega _{j} }} } ,} \right.\left. {1 - \prod\limits_{{j = 1}}^{n} {(1 - \mu _{j} (u))^{{\omega _{j} }} } } \right)$ holds for all $n$, which completes the proof. □

Proof of Theorem 2

We use mathematical induction to prove this theorem.

• (1) When $n = 2$, we can obtain

  $$\begin{aligned} {\text{T2WG}}_{\omega } (t_{1}^{{}} ,t_{2}^{{}} ) & = t_{1}^{{\omega _{1}^{{}} }} \otimes t_{2}^{{\omega _{2}^{{}} }} \\ & = (u_{1}^{{\omega _{1} }} ,\mu _{1} (u)^{{\omega _{1} }} ) \otimes (u_{2}^{{\omega _{2} }} ,\mu _{2} (u)^{{\omega _{2} }} ) \\ & = (u_{1}^{{\omega _{1} }} u_{2}^{{\omega _{2} }} ,\mu _{1} (u)^{{\omega _{1} }} \mu _{2} (u)^{{\omega _{2} }} ). \\ \end{aligned}$$

• (2) Suppose that when $n = k$,

${\text{T2WG}}_{\omega }^{{}} (t_{1}^{{}} ,t_{2}^{{}} , \ldots ,t_{k}^{{}} ) = t_{1}^{{\omega _{1}^{{}} }} \otimes t_{2}^{{\omega _{2}^{{}} }} \otimes \ldots \otimes t_{k}^{{\omega _{k}^{{}} }} = \left( {\prod\limits_{{j = 1}}^{k} {\mu _{j} ^{{\omega _{j} }} } ,} \right.\left. {\prod\limits_{{j = 1}}^{k} {\mu _{j} (u)^{{\omega _{j} }} } } \right)$ holds. When $n = k + 1$, we get

$$\begin{aligned} {\text{T2WG}}_{\omega } (t_{1} ,t_{2} , \ldots ,t_{{k + 1}} ) & = t_{1}^{{\omega _{1}^{{}} }} \otimes t_{2}^{{\omega _{2}^{{}} }} \otimes \cdots \otimes t_{{k + 1}}^{{\omega _{{k + 1}}^{{}} }} \\ & = \left( {\prod\limits_{{j = 1}}^{k} {\mu _{j}^{{\omega _{j} }} } ,} \right.\left. {\prod\limits_{{j = 1}}^{k} {\mu _{j} (u)^{{\omega _{j} }} } } \right) \otimes (u_{{k + 1}}^{{\omega _{{k + 1}} }} ,\mu _{{k + 1}} (u)^{{\omega _{{k + 1}} }} ) \\ & = \left( {\prod\limits_{{j = 1}}^{{k + 1}} {\mu _{j}^{{\omega _{j} }} } ,\;\;\prod\limits_{{j = 1}}^{{k + 1}} {\mu _{j} (u)^{{\omega _{j} }} } } \right). \\ \end{aligned}$$

Thus, when $n = k + 1$, the expression in Theorem 2 holds.

Thus, ${\text{T2WG}}_{\omega }^{{}} (t_{1}^{{}} ,t_{2}^{{}} , \ldots ,t_{n}^{{}} ) = \left( {\prod\limits_{{j = 1}}^{n} {\mu _{j} ^{{\omega _{j} }} } ,\;\;\prod\limits_{{j = 1}}^{n} {\mu _{j} (u)^{{\omega _{j} }} } } \right)$ holds for all $n$, which completes the proof. □

Proof of Proposition 1

Suppose that there are $n$ T2FNs $t = (u,\mu (u))$ have been added to an existing T2FN $t_{1} = (u_{1}^{{}} ,\mu _{1}^{{}} (u))$. According to T2WA and T2WG, the aggregation results are calculated as

$${\text{T2WA}} = \left( {1 - (1 - u_{1} )^{{\frac{1}{{n + 1}}}} (1 - u)^{{\frac{n}{{n + 1}}}} ,1 - (1 - \mu _{1} (u))^{{\frac{1}{{n + 1}}}} (1 - \mu (u))^{{\frac{n}{{n + 1}}}} } \right),$$

and

$${\text{T2WG}} = \left( {u_{1} ^{{\frac{1}{{n + 1}}}} u^{{\frac{n}{{n + 1}}}} ,\mu _{1} (u)^{{\frac{1}{{n + 1}}}} \mu (u)^{{\frac{n}{{n + 1}}}} } \right).$$

Since

$$\mathop {\lim }\limits_{{n \to \infty }} \left[ {1 - (1 - u_{1} )^{{\frac{1}{{n + 1}}}} (1 - u)^{{\frac{n}{{n + 1}}}} } \right] = u,\,{\text{and}}\,\mathop {\lim }\limits_{{n \to \infty }} \left[ {1 - (1 - \mu _{1} (u))^{{\frac{1}{{n + 1}}}} (1 - \mu (u))^{{\frac{n}{{n + 1}}}} } \right] = \mu (u),$$

the result of T2WA approximates the $t = (u,\mu (u))$ as $n$ increases. Similarly, the result of T2WG approximates the $t = (u,\mu (u))$ as $n$ increases. The proof is completed. □

Proof of Proposition 2

Suppose that there exist $n\;(n \ge 2)$ T2FNs $t_{i} \;(i = 1,2,...,n)$. According to T2A and T2G, we get

$$T2A = \left( {1 - \prod\limits_{{i = 1}}^{n} {(1 - u_{i} )} ,1 - \prod\limits_{{i = 1}}^{n} {(1 - \mu _{i} (u))} } \right),$$

and

$$T2G = \left( {\prod\limits_{{i = 1}}^{n} {u_{i} } ,\;\prod\limits_{{i = 1}}^{n} {\mu _{i} (u)} } \right).$$

Since

$$\mathop {\lim }\limits_{{n \to \infty }} \left[ {1 - \prod\limits_{{i = 1}}^{n} {(1 - u_{i} )} } \right] = 1 - 0 = 1,\,{\text{and}}\,\mathop {\lim }\limits_{{n \to \infty }} \left[ {1 - \prod\limits_{{i = 1}}^{n} {(1 - \mu _{i} (u))} } \right] = 1 - 0 = 1,$$

the result of T2A approximates $(1,1)$ as $n$ increases. Since

$$\mathop {\lim }\limits_{{n \to \infty }} \prod\limits_{{i = 1}}^{n} {u_{i} } = 0,\,{\text{and}}\,\mathop {\lim }\limits_{{n \to \infty }} \prod\limits_{{i = 1}}^{n} {\mu _{i} (u)} = 0,$$

the result of T2G approximates $(0,0)$ as $n$ increases. The proof is completed. □