An approximation method of intuitionistic fuzzy numbers

Abstract

A method of approximating intuitionistic fuzzy numbers by trapezoidal intuitionistic fuzzy numbers with respect to a well-known metric is studied in this paper. Firstly, the representations of extended trapezoidal intuitionistic approximations of intuitionistic fuzzy numbers are discussed. Then the representations of trapezoidal intuitionistic and triangular intuitionistic approximations of intuitionistic fuzzy numbers are provided. Finally, the algorithms and examples of trapezoidal intuitionistic and triangular intuitionistic approximations are given, and the results of approximations are applied on ranking of intuitionistic fuzzy numbers.

Keywords

Fuzzy number intuitionistic fuzzy number trapezoidal intuitionistic approximation

1 Introduction

Fuzzy sets are able to express and convey uncertain information effectively. The importance of the application of fuzzy set theory has been shown in many papers. The readers who want to learn fuzzy sets refer to [1 –4]. The numbers acts as the main carriers for conveying information, and thus fuzzy numbers play an important role among fuzzy sets. The shapes of membership functions of fuzzy numbers have an obvious impact on the result of calculations which are made by fuzzy numbers. The complicated membership functions will produce complex calculation, and it is much more direct and comprehensible for fuzzy numbers with simple membership functions. That results in approximations of fuzzy numbers. Currently, lots of literatures have investigated approximation methods of fuzzy numbers (see [5 , 22, 23]).

The intuitionistic fuzzy set is the generalization of the fuzzy set, and the intuitionistic fuzzy number is the particular case of intuitionistic fuzzy set in this paper. There are a large number of literatures on intuitionistic fuzzy sets and intuitionistic fuzzy numbers and their applications (see [24 –34]). In these literatures of applications, trapezoidal and triangular intuitionistic fuzzy numbers are used. Especially in intuitionistic fuzzy decision making, ranking of trapezoidal intuitionistic fuzzy numbers plays an important role. Different ranking methods for trapezoidal and triangular intuitionistic fuzzy numbers have been studied in [18 –21]. It is important that intuitionistic fuzzy numbers are approximated by trapezoidal intuitionistic fuzzy numbers. The results of approximations can be applied on ranking of intuitionistic fuzzy numbers and intuitionistic fuzzy decision making. Some scholars have studied the approximations of intuitionistic fuzzy numbers by fuzzy numbers. A.I. Ban investigated the approaches of approximating intuitionistic fuzzy numbers by trapezoidal fuzzy numbers both with condition of preserving the expected interval and without one (see [9]). A.I. Ban and L.C. Coroianu achieved an approach of approximating intuitionistic fuzzy numbers by trapezoidal fuzzy numbers with condition of preserving the expected interval from trapezoidal approximations of fuzzy numbers (see [10]). Besides A.I.Ban and L.C. Coroianu provided an application of intuitionistic fuzzy numbers in intuitionistic fuzzy linear systems (see [11]).

In this paper, the representations and algorithms of approximating intuitionistic fuzzy numbers by trapezoidal intuitionistic fuzzy numbers are mainly discussed. Section 2 reviews some basic concepts and propositions needed in the latter parts. In Section 3, the representations of extended trapezoidal intuitionistic approximations of intuitionistic fuzzy numbers are proposed. In Section 4, the representations of trapezoidal and triangular intuitionistic approximations of intuitionistic fuzzy numbers are offered. In Section 5, the algorithms of trapezoidal and triangular intuitionistic approximations of intuitionistic fuzzy numbers are discussed, and the examples of calculating trapezoidal and triangular intuitionistic approximations by applying above algorithms are demonstrated. Ranking of intuitionistic fuzzy numbers according to ranking of their approximations is given. This paper concludes in Section 6.

2 Preliminaries

For two arbitrary fuzzy numbers μ and ν with α-cuts [μ_L (α) , μ_U (α)] and [ν_L (α) , ν_U (α)], respectively, α ∈ [0, 1], the quantity $d (μ, ν) = \sqrt{\int_{0}^{1} {(μ_{L} (α) - ν_{L} (α))}^{2} d α + \int_{0}^{1} (μ_{U} (α) - ν_{U} (α))^{2} d α} .$ (1) is the distance between μ and ν (see [6]).

A fuzzy number μ is a trapezoidal fuzzy number iff (if and only if) its α-cuts are of the form [t₁ + (t₂ - t₁) α, t₄ - (t₄ - t₃) α], where t₁, t₂, t₃ and t₄ are real numbers (denote by R) with the constraint t₁ ≤ t₂ ≤ t₃ ≤ t₄. Since the trapezoidal fuzzy number is completely characterized by four real numbers t₁, t₂, t₃ and t₄, it is often denoted in brief as μ (t₁, t₂, t₃, t₄). A trapezoidal fuzzy number is called a triangular fuzzy number iff t₂ = t₃ holds. Let the symbol μ = μ (l, u ; x, y) denote an extended trapezoidal fuzzy number (see [7]) with the following α-cuts $μ_{α} = [l + x (α - \frac{1}{2}), u - y (α - \frac{1}{2})] .$ (2)

Proposition 2.1. [7] Let μ = μ (l, u ; x, y) be an extended trapezoidal fuzzy number. Then

μ is a trapezoidal fuzzy number iff x, y ≥ 0, x + y ≤ 2 (u - l).

μ is a triangular fuzzy number iff x, y ≥ 0, x + y = 2 (u - l).

Proposition 2.2. [7] Let μ = (l₁, u₁ ; x₁, y₁) and ν = (l₂, u₂ ; x₂, y₂) be two extended trapezoidal fuzzy numbers. Then $\begin{matrix} d^{2} (μ, ν) & = & (l_{1} - l_{2})^{2} + (u_{1} - u_{2})^{2} \\ + \frac{1}{12} [(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}] . \end{matrix}$

The definitions of an intuitionistic fuzzy set and an intuitionistic fuzzy number are given in [12 –14].

Definition 2.1. [12, 13] Let X be a given nonempty set. An intuitionistic fuzzy set in X is an object A given by A = {〈x, μ_A (x) , ν_A (x) 〉 : x ∈ X} , where μ_A : X → [0, 1] and ν_A : X → [0, 1] satisfy the condition 0 ≤ μ_A (x) + ν_A (x) ≤1, for every x ∈ X.

Definition 2.2. [14] An intuitionistic fuzzy set A = {〈x, μ_A (x) , ν_A (x) 〉 : x ∈ R} such that μ_A and 1 - ν_A, where (1 - ν_A) (x) =1 - ν_A (x) , ∀ x ∈ R, are fuzzy numbers is called an intuitionistic fuzzy number.

The set of all intuitionistic fuzzy numbers is denoted by IF (R). An intuitionistic fuzzy number A = 〈μ_A, ν_A〉, where μ_A and 1 - ν_A are trapezoidal fuzzy numbers, is called a trapezoidal intuitionistic fuzzy number. The set of all trapezoidal intuitionistic fuzzy numbers is denoted by IF^T (R). An intuitionistic fuzzy number A = 〈μ_A, ν_A〉, where μ_A and 1 - ν_A are triangualr fuzzy numbers, is called a triangular intuitionistic fuzzy number. An intuitionistic fuzzy number A = 〈μ_A, ν_A〉, where μ_A and 1 - ν_A are extended trapezoidal fuzzy numbers, is called an extended trapezoidal intuitionistic fuzzy number. The set of all extended trapezoidal intuitionistic fuzzy numbers is denoted by $\tilde{{IF}^{T}} (R)$ . Then from Proposition 2.1, the follow proposition can be obtained immediately.

Proposition 2.3. Let A = 〈μ_A, ν_A〉 ∈ IF (R), where μ_A = μ_A (l, u ; x, y) and 1 - ν_A = (1 - ν_A) (l′, u′ ; x′, y′) . Then

A = 〈μ_A, ν_A〉 is a trapezoidal intuitionistic fuzzy number iff x, y, x′, y′ ≥ 0, x + y ≤ 2 (u - l) and x′ + y′ ≤ 2 (u′ - l′).

A = 〈μ_A, ν_A〉 is a triangular intuitionistic fuzzy number iff x, y, x′, y′ ≥ 0, x + y = 2 (u - l) and x′ + y′ = 2 (u′ - l′).

The quantity $d (A, B) = \sqrt{\frac{1}{2} d^{2} (μ_{A}, μ_{B}) + \frac{1}{2} d^{2} (1 - ν_{A}, 1 - ν_{B})} .$ (3) is the distance between two arbitrary intuitionistic fuzzy numbers A = 〈μ_A, ν_A〉 and B = 〈μ_B, ν_B〉 (see [16]).

By Proposition 2.2 and Equation (3), the following proposition can be obtained.

Proposition 2.4. Let $A = 〈 μ_{A}, ν_{A} 〉 \in \tilde{{IF}^{T}} (R)$ and $B = 〈 μ_{B}, ν_{B} 〉 \in \tilde{{IF}^{T}} (R)$ , where μ_A = μ_A (l₁, u₁ ; x₁, y₁) , $1 - ν_{A} = (1 - ν_{A}) (l_{1}^{'}, u_{1}^{'}; x_{1}^{'}, y_{1}^{'}),$ μ_B = μ_B (l₂, u₂ ; x₂, y₂) and $1 - ν_{B} = (1 - ν_{B}) (l_{2}^{'}, u_{2}^{'};$ $x_{2}^{'}, y_{2}^{'}) .$ Then $\begin{matrix} d^{2} (A, B) & = & \frac{1}{2} (l_{1} - l_{2})^{2} + \frac{1}{2} (u_{1} - u_{2})^{2} \\ + \frac{1}{24} [(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}] \\ + \frac{1}{2} (l_{1}^{'} - l_{2}^{'})^{2} + \frac{1}{2} (u_{1}^{'} - u_{2}^{'})^{2} \\ + \frac{1}{24} [(x_{1}^{'} - x_{2}^{'})^{2} + (y_{1}^{'} - y_{2}^{'})^{2}] . \end{matrix}$

Definition 2.3. [21] Let A = 〈μ_A, ν_A〉 ∈ IF^T (R), where μ_A = μ_A (t₁, t₂, t₃, t₄) and $1 - ν_{A} = (1 - ν_{A}) (t_{1}^{'}, t_{2}^{'}, t_{3}^{'}, t_{4}^{'}) .$ Then the membership score function L of A is defined as $\begin{matrix} L (A) & = & \frac{2 (t_{1} + t_{2} + t_{3} + t_{4}) - 2 (t_{1}^{'} + t_{2}^{'} + t_{3}^{'} + t_{4}^{'})}{8}, \\ + \frac{(t_{1} + t_{2}) (t_{1}^{'} + t_{2}^{'}) + (t_{3} + t_{4}) (t_{3}^{'} + t_{4}^{'})}{8} . \end{matrix}$

Definition 2.4. [21] Let A, B ∈ IF^T (R). A relation < on IF^T (R) is defined by A < B if L (A) < L (B).

3 Extended trapezoidal intuitionistic approximations

In this section, the representations of the extended trapezoidal intuitionistic approximations of intuitionistic fuzzy numbers are discussed. The extended trapezoidal intuitionistic fuzzy number T_e (A), which minimizes the distance d (A, X) (with respect to Equation (3)), is called the extended trapezoidal intuitionistic approximation of the intuitionistic fuzzy number A, where $X \in \tilde{{IF}^{T}} (R)$ .

Let $T_{e} (A) = 〈 μ_{T_{e}}, ν_{T_{e}} 〉 \in \tilde{{IF}^{T}} (R)$ , where μ_{T
_e} = μ_{T
_e} (t₁, t₂, t₃, t₄) = μ_{T
_e} (l_e, u_e ; x_e, y_e) and $1 - ν_{T_{e}} = (1 - ν_{T_{e}}) (t_{1}^{'}, t_{2}^{'}, t_{3}^{'}, t_{4}^{'})$ $= (1 - ν_{T_{e}}) (l_{e}^{'}, u_{e}^{'}; x_{e}^{'}, y_{e}^{'}) .$ Let (see [15]): $t_{1} = - 6 \int_{0}^{1} α (μ_{A})_{L} (α) d α + 4 \int_{0}^{1} (μ_{A})_{L} (α) d α,$ (4) $t_{2} = 6 \int_{0}^{1} α (μ_{A})_{L} (α) d α - 2 \int_{0}^{1} (μ_{A})_{L} (α) d α,$ (5) $t_{3} = 6 \int_{0}^{1} α (μ_{A})_{U} (α) d α - 2 \int_{0}^{1} (μ_{A})_{U} (α) d α,$ (6) $t_{4} = - 6 \int_{0}^{1} α (μ_{A})_{U} (α) d α + 4 \int_{0}^{1} (μ_{A})_{U} (α) d α,$ (7) $\begin{matrix} t_{1}^{'} = - 6 \int_{0}^{1} α (1 - ν_{A})_{L} (α) d α + 4 \int_{0}^{1} (1 - ν_{A})_{L} (α) d α, \end{matrix}$ (8) $\begin{matrix} t_{2}^{'} = 6 \int_{0}^{1} α (1 - ν_{A})_{L} (α) d α - 2 \int_{0}^{1} (1 - ν_{A})_{L} (α) d α, \end{matrix}$ (9) $\begin{matrix} t_{3}^{'} = 6 \int_{0}^{1} α (1 - ν_{A})_{U} (α) d α - 2 \int_{0}^{1} (1 - ν_{A})_{U} (α) d α, \end{matrix}$ (10) $\begin{matrix} t_{4}^{'} = - 6 \int_{0}^{1} α (1 - ν_{A})_{U} (α) d α + 4 \int_{0}^{1} (1 - ν_{A})_{U} (α) d α . \end{matrix}$ (11)

By Equation (2), we have $l_{e} = \frac{t_{1} + t_{2}}{2},$ $u_{e} = \frac{t_{3} + t_{4}}{2},$ x_e = t₂ - t₁, y_e = t₄ - t₃, $l_{e}^{'} = \frac{t_{1}^{'} + t_{2}^{'}}{2},$ $u_{e}^{'} = \frac{t_{3}^{'} + t_{4}^{'}}{2},$ $x_{e}^{'} = t_{2}^{'} - t_{1}^{'},$ $y_{e}^{'} = t_{4}^{'} - t_{3}^{'} .$ So by Equations (4)–(11), we may get the following formulas (see [8]): $l_{e} = \int_{0}^{1} (μ_{A})_{L} (α) d α,$ (12) $u_{e} = \int_{0}^{1} (μ_{A})_{U} (α) d α,$ (13) $x_{e} = 12 \int_{0}^{1} (α - \frac{1}{2}) (μ_{A})_{L} (α) d α,$ (14) $y_{e} = - 12 \int_{0}^{1} (α - \frac{1}{2}) (μ_{A})_{U} (α) d α,$ (15) $l_{e}^{'} = \int_{0}^{1} (1 - ν_{A})_{L} (α) d α,$ (16) $u_{e}^{'} = \int_{0}^{1} (1 - ν_{A})_{U} (α) d α,$ (17) $x_{e}^{'} = 12 \int_{0}^{1} (α - \frac{1}{2}) (1 - ν_{A})_{L} (α) d α,$ (18) $y_{e}^{'} = - 12 \int_{0}^{1} (α - \frac{1}{2}) (1 - ν_{A})_{U} (α) d α .$ (19) By Definition 2.1 and Lemma 2.1 in [17], we have $u_{e} - l_{e} \geq 0, u_{e}^{'} - l_{e}^{'} \geq 0,$ (20) $x_{e} \geq 0, y_{e} \geq 0, x_{e}^{'} \geq 0, y_{e}^{'} \geq 0 .$ (21)

Similar to proofs of Proposition 3.2 and Proposition 3.3 in [8], we may obtain the following Proposition 3.1 and Proposition 3.2.

Proposition 3.1. Let A = 〈μ_A, ν_A〉 ∈ IF (R) and $T_{e} (A) = 〈 μ_{T_{e}}, ν_{T_{e}} 〉 \in \tilde{{IF}^{T}} (R)$ , where μ_{T
_e} = μ_{T
_e} (l_e, u_e ; x_e, y_e) and $1 - ν_{T_{e}} = (1 - ν_{T_{e}}) (l_{e}^{'}, u_{e}^{'}; x_{e}^{'}, y_{e}^{'})$ are defined as Equations (12)–(19). Then

$d^{2} (A, X) = d^{2} (A, T_{e} (A)) + d^{2} (T_{e} (A), X)$ holds for every $X = 〈 μ_{X}, ν_{X} 〉 \in \tilde{{IF}^{T}} (R)$ .

Proposition 3.2. Let A = 〈μ_A, ν_A〉 ∈ IF (R) and $T_{e} (A) = 〈 μ_{T_{e}}, ν_{T_{e}} 〉 \in \tilde{{IF}^{T}} (R)$ , where μ_{T
_e} = μ_{T
_e} (l_e, u_e ; x_e, y_e) and $1 - ν_{T_{e}} = (1 - ν_{T_{e}}) (l_{e}^{'}, u_{e}^{'}; x_{e}^{'}, y_{e}^{'})$ are defined as Equations (12)–(19). Then T_e (A) = 〈μ_{T
_e}, ν_{T
_e}〉 equals the extended trapezoidal intuitionistic approximation of A = 〈μ_A, ν_A〉.

Because the extended trapezoidal intuitionistic approximation T_e (A) of the intuitionistic fuzzy number A can be calculated by Equations (12)–(19), T_e (A) exists and is unique.

4 Trapezoidal intuitionistic approximations

If T_n (A) = 〈μ_{T
_n}, ν_{T
_n}〉 is the nearest trapezoidal intuitionistic fuzzy number (with respect to Equation (3)) to a given intuitionistic fuzzy number A = 〈μ_A, ν_A〉, then T_n (A) = 〈μ_{T
_n}, ν_{T
_n}〉 is called trapezoidal intuitionistic approximation of A = 〈μ_A, ν_A〉. we may Similarly define the concept of triangular intuitionistic approximation.

Let B = 〈μ_B, ν_B〉 ∈ IF^T (R) and T_e (A) = 〈μ_{T
_e}, ν_{T
_e}〉 be extended trapezoidal intuitionistic approximation of A, where μ_B = μ_B (l, u ; x, y) and 1 - ν_B = (1 - ν_B) (l′, u′ ; x′, y′) . Since d² (A, T_e (A)) is constant, by Propositions 2.4 and 3.1, to minimize d (A, B) is as to minimizing the following function

$\begin{matrix} f (l, u, x, y, l^{'}, u^{'}, x^{'}, y^{'}) \\ = d^{2} (T_{e} (A), B) \\ = \frac{1}{2} (l - l_{e})^{2} + \frac{1}{2} (u - u_{e})^{2} \\ + \frac{1}{24} [(x - x_{e})^{2} + (y - y_{e})^{2}] \\ + \frac{1}{2} (l^{'} - l_{e}^{'})^{2} + \frac{1}{2} (u^{'} - u_{e}^{'})^{2} \\ + \frac{1}{24} [(x^{'} - x_{e}^{'})^{2} + (y^{'} - y_{e}^{'})^{2}] . \end{matrix}$ (22) Taking into account Proposition 2.3, to find the minimum of Equation (22) can be reduced to the following problem.

Problem 4.1. Find (l, u, x, y, l′, u′, x′, y′) ∈ Ω, which minimizes Equation (22), where

$\begin{matrix} Ω & = & {(l, u, x, y, l^{'}, u^{'}, x^{'}, y^{'}) | x, y, x^{'}, y^{'} \geq 0, \\ x + y \leq 2 (u - l), x^{'} + y^{'} \leq 2 (u^{'} - l^{'})} . \end{matrix}$ (23)

In order to solve Problem 4.1, we introduce two lemmas as follows.

Lemma 4.1. Every solution (l, u, x, y, l′, u′, x′, y′) of Problem 4.1 satisfies

$u_{e} - l_{e} \leq u - l, u_{e}^{'} - l_{e}^{'} \leq u^{'} - l^{'}$ .

$x_{e} + y_{e} \geq x + y, x_{e}^{'} + y_{e}^{'} \geq x^{'} + y^{'}$ .

Let $δ = x + y - 2 (u - l),$ (24) $δ^{'} = x^{'} + y^{'} - 2 (u^{'} - l^{'}),$ (25) $δ_{e} = x_{e} + y_{e} - 2 (u_{e} - l_{e}),$ (26) $δ_{e}^{'} = x_{e}^{'} + y_{e}^{'} - 2 (u_{e}^{'} - l_{e}^{'}) .$ (27) Then $δ \leq δ_{e}, δ^{'} \leq δ_{e}^{'} .$

Proof. (1) Applying proof of Lemma 4.2 in [17], we may certify case (1) in the same vein.

(2) Assume that (l, u, x, y, l′, u′, x′, y′) is a solution of Problem 4.1, satifying x_e + y_e < x + y or $x_{e}^{'} + y_{e}^{'} < x^{'} + y^{'}$ . When x_e + y_e < x + y, we may get a contradiction in the following. Since x_e, y_e ≥ 0, x_e + y_e < x + y ≤ 2 (u - l), then (l, u, x_e, y_e, l′, u′, x′, y′) ∈ Ω. While $\begin{matrix} f (l, u, x, y, l^{'}, u^{'}, x^{'}, y^{'}) \\ > \frac{1}{2} (l - l_{e})^{2} + \frac{1}{2} (u - u_{e})^{2} \\ + \frac{1}{2} (l^{'} - l_{e}^{'})^{2} + \frac{1}{2} (u^{'} - u_{e}^{'})^{2} \\ + \frac{1}{24} [(x^{'} - x_{e}^{'})^{2} + (y^{'} - y_{e}^{'})^{2}] \\ = f (l, u, x_{e}, y_{e}, l^{'}, u^{'}, x^{'}, y^{'}) . \end{matrix}$ Hence, f (l, u, x, y, l′, u′, x′, y′) is not minimal, which is a contradiction.

In the same vein,we may also get a contradiction when $x_{e}^{'} + y_{e}^{'} < x^{'} + y^{'}$ .

(3) Because x + y ≤ x_e + y_e, u_e - l_e ≤ u - l, we have $x + y - 2 (u - l) \leq x_{e} + y_{e} - 2 (u_{e} - l_{e}),$ that is δ ≤ δ_e . We can certify $δ^{'} \leq δ_{e}^{'}$ in the same vein. □

Lemma 4.2. Let (l, u, x, y, l′, u′, x′, y′) be a solution of Problem 4.1, then the following items hold:

If $x_{e} + y_{e} \geq 2 (u_{e} - l_{e}) (x_{e}^{'} + y_{e}^{'} \geq 2 (u_{e}^{'} - l_{e}^{'}))$ ,then x + y = 2 (u - l) (x′ + y′ = 2 (u′ - l′)).

If $x_{e} + y_{e} < 2 (u_{e} - l_{e}) (x_{e}^{'} + y_{e}^{'} < 2 (u_{e}^{'} - l_{e}^{'}))$ , then x + y < 2 (u - l) (x′ + y′ < 2 (u′ - l′)).

Proof. (1) Applying proof of Lemma 4.3 in [17], we can prove case (1) in the same vein.

(2) If x_e + y_e < 2 (u_e - l_e), then by Equations (24), (26) and the case (3) of Lemma 4.1, we have $\begin{matrix} x + y - 2 (u - l) \\ = δ \leq δ_{e} = x_{e} + y_{e} - 2 (u_{e} - l_{e}) < 0 . \end{matrix}$ Therefore x + y < 2 (u - l). A similar reasoning can be done for another case. So we complete the proof. □

In the following the representations of trapezoidal and triangular intuitionistic approximations of intuitionistic fuzzy numbers are provided.

Theorem 4.1. Let A = 〈μ_A, ν_A〉 ∈ IF (R) and T_e (A) = 〈μ_{T
_e}, ν_{T
_e}〉 be extended trapezoidal intuitionistic approximation of A, and T_n (A) = 〈μ_{T
_n}, ν_{T
_n}〉 be trapezoidal intuitionistic approximation of A, where μ_{T
_n} = μ_{T
_n} (l_n, u_n ; x_n, y_n) and $1 - ν_{T_{n}} = (1 - ν_{T_{n}}) (l_{n}^{'}, u_{n}^{'}; x_{n}^{'}, y_{n}^{'}) .$ δ_e and $δ_{e}^{'}$ are given by Equations (26) and (27), respectively. Then T_n (A) can be calculated in the following cases:

If $δ_{e}, δ_{e}^{'} \leq 0,$ then T_n (A) = T_e (A).

If $δ_{e}, δ_{e}^{'} > 0,$ $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{n} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{n}^{'} = \frac{7}{8} l_{e}^{'} + \frac{1}{8} u_{e}^{'} - \frac{1}{16} x_{e}^{'} - \frac{1}{16} y_{e}^{'}, \\ u_{n}^{'} = \frac{1}{8} l_{e}^{'} + \frac{7}{8} u_{e}^{'} + \frac{1}{16} x_{e}^{'} + \frac{1}{16} y_{e}^{'}, \\ x_{n}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} + \frac{5}{8} x_{e}^{'} - \frac{3}{8} y_{e}^{'}, \\ y_{n}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} - \frac{3}{8} x_{e}^{'} + \frac{5}{8} y_{e}^{'} . \end{matrix}$ (28)

If $δ_{e}, δ_{e}^{'} > 0$ , $x_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{n} = 0, \\ y_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{n}^{'} = \frac{7}{8} l_{e}^{'} + \frac{1}{8} u_{e}^{'} - \frac{1}{16} x_{e}^{'} - \frac{1}{16} y_{e}^{'}, \\ u_{n}^{'} = \frac{1}{8} l_{e}^{'} + \frac{7}{8} u_{e}^{'} + \frac{1}{16} x_{e}^{'} + \frac{1}{16} y_{e}^{'}, \\ x_{n}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} + \frac{5}{8} x_{e}^{'} - \frac{3}{8} y_{e}^{'}, \\ y_{n}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} - \frac{3}{8} x_{e}^{'} + \frac{5}{8} y_{e}^{'} . \end{matrix}$ (29)

If $δ_{e}, δ_{e}^{'} > 0$ , $y_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{n} = 0, \\ l_{n}^{'} = \frac{7}{8} l_{e}^{'} + \frac{1}{8} u_{e}^{'} - \frac{1}{16} x_{e}^{'} - \frac{1}{16} y_{e}^{'}, \\ u_{n}^{'} = \frac{1}{8} l_{e}^{'} + \frac{7}{8} u_{e}^{'} + \frac{1}{16} x_{e}^{'} + \frac{1}{16} y_{e}^{'}, \\ x_{n}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} + \frac{5}{8} x_{e}^{'} - \frac{3}{8} y_{e}^{'}, \\ y_{n}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} - \frac{3}{8} x_{e}^{'} + \frac{5}{8} y_{e}^{'} . \end{matrix}$ (30)

If $δ_{e}, δ_{e}^{'} > 0$ , $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ and $x_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{n} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{n}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} y_{e}^{'}, \\ u_{n}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} y_{e}^{'}, \\ x_{n}^{'} = 0, \\ y_{n}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} y_{e}^{'} . \end{matrix}$ (31)

If $δ_{e}, δ_{e}^{'} > 0$ , $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{n} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{n}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} x_{e}^{'}, \\ u_{n}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} x_{e}^{'}, \\ x_{n}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} x_{e}^{'}, \\ y_{n}^{'} = 0 . \end{matrix}$ (32)

If $δ_{e}, δ_{e}^{'} > 0$ , $x_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{n} = 0, \\ y_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{n}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} y_{e}^{'}, \\ u_{n}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} y_{e}^{'}, \\ x_{n}^{'} = 0, \\ y_{n}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} y_{e}^{'} . \end{matrix}$ (33)

If $δ_{e}, δ_{e}^{'} > 0$ , $x_{e} < \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{n} = 0, \\ y_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{n}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} x_{e}^{'}, \\ u_{n}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} x_{e}^{'}, \\ x_{n}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} x_{e}^{'}, \\ y_{n}^{'} = 0 . \end{matrix}$ (34)

If $δ_{e}, δ_{e}^{'} > 0$ , $y_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{n} = 0, \\ l_{n}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} y_{e}^{'}, \\ u_{n}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} y_{e}^{'}, \\ x_{n}^{'} = 0, \\ y_{n}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} y_{e}^{'} . \end{matrix}$ (35)

If $δ_{e}, δ_{e}^{'} > 0$ , $y_{e} < \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{n} = 0, \\ l_{n}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} x_{e}^{'}, \\ u_{n}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} x_{e}^{'}, \\ x_{n}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} x_{e}^{'}, \\ y_{n}^{'} = 0 . \end{matrix}$ (36)

If $δ_{e} > 0, δ_{e}^{'} \leq 0$ and $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ , then ${\begin{matrix} l_{n} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{n} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{n}^{'} = l_{e}^{'}, \\ u_{n}^{'} = u_{e}^{'}, \\ x_{n}^{'} = x_{e}^{'}, \\ y_{n}^{'} = y_{e}^{'} . \end{matrix}$ (37)

If δ_e > 0, $δ_{e}^{'} \leq 0$ and $x_{e} < \frac{3}{8} δ_{e}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{n} = 0, \\ y_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{n}^{'} = l_{e}^{'}, \\ u_{n}^{'} = u_{e}^{'}, \\ x_{n}^{'} = x_{e}^{'}, \\ y_{n}^{'} = y_{e}^{'} . \end{matrix}$ (38)

If δ_e > 0, $δ_{e}^{'} \leq 0$ and $y_{e} < \frac{3}{8} δ_{e}$ , then ${\begin{matrix} l_{n} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{n} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{n} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{n} = 0, \\ l_{n}^{'} = l_{e}^{'}, \\ u_{n}^{'} = u_{e}^{'}, \\ x_{n}^{'} = x_{e}^{'}, \\ y_{n}^{'} = y_{e}^{'} . \end{matrix}$ (39)

Proof. (1) Let δ_e ≤ 0 and $δ_{e}^{'} \leq 0$ , i.e. x_e + y_e ≤ 2 (u_e - l_e) and $x_{e}^{'} + y_{e}^{'} \leq 2 (u_{e}^{'} - l_{e}^{'})$ . Equation (21) shows x_e ≥ 0, y_e ≥ 0, $x_{e}^{'} \geq 0$ and $y_{e}^{'} \geq 0 .$ Hence we conclude $(l_{e}, u_{e}, x_{e}, y_{e}, l_{e}^{'}, u_{e}^{'}, x_{e}^{'}, y_{e}^{'}) \in Ω$ , where Ω is defined as Equation (23). This implies T_n (A) = T_e (A).

(2) Let δ_e > 0 and $δ_{e}^{'} > 0$ . By the case (1) of Lemma 4.2, we have x + y = 2 (u - l) and x′ + y′ = 2 (u′ - l′). Applying Schwarz’s inequality, we get $\begin{matrix} [\frac{1}{2} (l - l_{e})^{2} + \frac{1}{2} (u - u_{e})^{2} + \frac{1}{24} (x - x_{e})^{2} \\ + \frac{1}{24} (y - y_{e})^{2} + \frac{1}{2} (l^{'} - l_{e}^{'})^{2} + \frac{1}{2} (u^{'} - u_{e}^{'})^{2} \\ + \frac{1}{24} (x^{'} - x_{e}^{'})^{2} + \frac{1}{24} (y^{'} - y_{e}^{'})^{2}] \\ \cdot [{(2 \sqrt{2})}^{2} + {(- 2 \sqrt{2})}^{2} + {(\sqrt{24})}^{2} + {(\sqrt{24})}^{2} \\ + {(2 \sqrt{2})}^{2} + {(- 2 \sqrt{2})}^{2} + {(\sqrt{24})}^{2} + {(\sqrt{24})}^{2}] \\ \geq [x + y - 2 (u - l) - (x_{e} + y_{e} - 2 (u_{e} - l_{e})) \\ {+ (x^{'} + y^{'} - 2 (u^{'} - l^{'})) - (x_{e}^{'} + y_{e}^{'} - 2 (u_{e}^{'} - l_{e}^{'}))]}^{2} \\ = (δ_{e} + δ_{e}^{'})^{2} . \end{matrix}$

The equality holds iff $\begin{matrix} \frac{\frac{1}{\sqrt{2}} (l - l_{e})}{2 \sqrt{2}} = \frac{- \frac{1}{\sqrt{2}} (u - u_{e})}{2 \sqrt{2}} \\ = \frac{\frac{1}{\sqrt{24}} (x - x_{e})}{\sqrt{24}} = \frac{\frac{1}{\sqrt{24}} (2 u - 2 l - x - y_{e})}{\sqrt{24}} \\ = \frac{\frac{1}{\sqrt{2}} (l^{'} - l_{e}^{'})}{2 \sqrt{2}} = \frac{- \frac{1}{\sqrt{2}} (u^{'} - u_{e}^{'})}{2 \sqrt{2}} \\ = \frac{\frac{1}{\sqrt{24}} (x^{'} - x_{e}^{'})}{\sqrt{24}} = \frac{\frac{1}{\sqrt{24}} (2 u^{'} - 2 l^{'} - x^{'} - y_{e}^{'})}{\sqrt{24}} . \end{matrix}$ Hence we obtain Equation (28). Since $x_{e} \geq \frac{3}{8} δ_{e}$ , then $x_{n} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e} = - \frac{3}{8} δ_{e} + x_{e} \geq 0$ . If $y_{e} \geq \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then y_n, $x_{n}^{'}$ and $y_{n}^{'} \geq 0$ . Additionally, x_n + y_n = 2 (u_n - l_n) and $x_{n}^{'} + y_{n}^{'} = 2 (u_{n}^{'} - l_{n}^{'})$ . So $(l_{n}, u_{n}, x_{n}, y_{n}, l_{n}^{'}, u_{n}^{'}, x_{n}^{'}, y_{n}^{'}) \in Ω$ .

(3) Since the solution (l, u, x, y, l′, u′, x′, y′) of Problem 4.1 is in the boundary of Ω, by Equation (2) and $x_{e} < \frac{3}{8} δ_{e}$ , then x = 0 or y = 0. If y = 0, by (20) we have $\begin{matrix} y_{e} & \geq & y_{e} - 2 (u_{e} - l_{e}) = δ_{e} - x_{e} \\ > & δ_{e} - \frac{3}{16} (δ_{e} + δ_{e}^{'}) > \frac{6}{16} δ_{e} > x_{e} . \end{matrix}$ Applying above inequality, we have $\begin{matrix} f (l, u, x, 0, l^{'}, u^{'}, x^{'}, y^{'}) \\ > \frac{1}{2} (l - l_{e})^{2} + \frac{1}{2} (u - u_{e})^{2} \\ + \frac{1}{24} [x^{2} - 2 {xy}_{e} + x_{e}^{2} + y_{e}^{2}] \\ + \frac{1}{2} (l^{'} - l_{e}^{'})^{2} + \frac{1}{2} (u^{'} - u_{e}^{'})^{2} \\ + \frac{1}{24} [(x^{'} - x_{e}^{'})^{2} + (y^{'} - y_{e}^{'})^{2}] \\ = f (l, u, 0, x, l^{'}, u^{'}, x^{'}, y^{'}) . \end{matrix}$ Hence f (l, u, x, 0, l′, u′, x′, y′) is not minimal, which is a contradiction. Therefore x = 0. By the case (1) of Lemma 4.2, we obtain y = 2u - 2l. Substituting into Equation (22) by x = 0, y = 2u - 2l and y′ = 2u′ - 2l′ - x′, we get the function f (l, u, 0, 2u - 2l, l′, u′, x′, 2u′ - 2l′ - x′) . Thus we find partial derivatives and then solve system of equations. The solution of the system of equations is Equation (29).

(4) Similar to the proof of case (3), we may get Equation (30).

(5) Similar to the proof of case (3), we may get Equation (31).

(6) Similar to the proof of case (3), we may get Equation (32).

(7) Let $x_{e} < \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ . By the case (3) and (5) of Theorem 4.1, we have x = 0 and x′ = 0. By the case (1) of Lemma 4.2, we get the function y = 2u - 2l and y′ = 2u′ - 2l′. Substituting into Equation (22) by x = 0, y = 2u - 2l, x′ = 0 and y′ = 2u′ - 2l′, we obtain f (l, u, 0, 2u - 2l, l′, u′, 0, 2u′ - 2l′) . Thus we find partial derivatives and then solve system of equations. The solution of the system of equations is Equation (33).

(8) Similar to the proof of case (7), we may get Equation (34).

(9) Similar to the proof of case (7), we may get Equation (35).

(10) Similar to the proof of case (7), we may get Equation (36).

(11) Let δ_e > 0. By the case (1) of Lemma 4.2, we get x + y = 2 (u - l). Substituting into Equation (22) by y = 2u - 2l - x, we get the function f (l, u, x, 2u - 2l - x, l′, u′, x′, y′) . Thus we find partial derivatives and then solve system of equations. The solution of the system of equations is Equation (37).

(12) If δ_e > 0 and $x_{e} < \frac{3}{8} δ_{e}$ , we have x = 0 and y = 2u - 2l. Substituting into (22) by x = 0 and y = 2u - 2l, we obtain the function f (l, u, 0, 2u - 2l, l′, u′, x′, y′) . Thus we find partial derivatives and then solve system of equations. The solution of the system of equations is Equation (38).

(13) Similar to the proof of case (12), we may get Equation (39). □

Remark 4.1. T_e (A) ∈ IF (R), then $δ_{e}^{'} \leq δ_{e}$ . Hence Theorem 4.1 has only thirteen cases.

Because triangular intuitionistic approximation is particular trapezoidal intuitionistic approximation, we may obtain immediately the following conclusion by applying the proof of Theorem 4.1.

Theorem 4.2. Let A = 〈μ_A, ν_A〉 ∈ IF (R), T_e (A) = 〈μ_{T
_e}, ν_{T
_e}〉 be extended tarpezoidal intuitionistic approximation of A, and T_▵ (A) = 〈μ_▵, ν_▵〉 be triangular intuitionistic approximation of A, where μ_▵ = μ_▵ (l_m, u_m ; x_m, y_m) and $1 - ν_{▵} = (1 - ν_{▵}) (l_{m}^{'}, u_{m}^{'};$ $x_{m}^{'}, y_{m}^{'}) .$ δ_e and $δ_{e}^{'}$ are given by Equations (26) and (27), respectively. Then T_▵ (A) can be calculated in the following cases:

If $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{m} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{m} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{m} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{m}^{'} = \frac{7}{8} l_{e}^{'} + \frac{1}{8} u_{e}^{'} - \frac{1}{16} x_{e}^{'} - \frac{1}{16} y_{e}^{'}, \\ u_{m}^{'} = \frac{1}{8} l_{e}^{'} + \frac{7}{8} u_{e}^{'} + \frac{1}{16} x_{e}^{'} + \frac{1}{16} y_{e}^{'}, \\ x_{m}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} + \frac{5}{8} x_{e}^{'} - \frac{3}{8} y_{e}^{'}, \\ y_{m}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} - \frac{3}{8} x_{e}^{'} + \frac{5}{8} y_{e}^{'} . \end{matrix}$ (40)

If $x_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{m} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{m} = 0, \\ y_{m} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{m}^{'} = \frac{7}{8} l_{e}^{'} + \frac{1}{8} u_{e}^{'} - \frac{1}{16} x_{e}^{'} - \frac{1}{16} y_{e}^{'}, \\ u_{m}^{'} = \frac{1}{8} l_{e}^{'} + \frac{7}{8} u_{e}^{'} + \frac{1}{16} x_{e}^{'} + \frac{1}{16} y_{e}^{'}, \\ x_{m}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} + \frac{5}{8} x_{e}^{'} - \frac{3}{8} y_{e}^{'}, \\ y_{m}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} - \frac{3}{8} x_{e}^{'} + \frac{5}{8} y_{e}^{'} . \end{matrix}$ (41)

If $y_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'}, y_{e}^{'} \geq \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{m} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{m} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{m} = 0, \\ l_{m}^{'} = \frac{7}{8} l_{e}^{'} + \frac{1}{8} u_{e}^{'} - \frac{1}{16} x_{e}^{'} - \frac{1}{16} y_{e}^{'}, \\ u_{m}^{'} = \frac{1}{8} l_{e}^{'} + \frac{7}{8} u_{e}^{'} + \frac{1}{16} x_{e}^{'} + \frac{1}{16} y_{e}^{'}, \\ x_{m}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} + \frac{5}{8} x_{e}^{'} - \frac{3}{8} y_{e}^{'}, \\ y_{m}^{'} = - \frac{3}{4} l_{e}^{'} + \frac{3}{4} u_{e}^{'} - \frac{3}{8} x_{e}^{'} + \frac{5}{8} y_{e}^{'} . \end{matrix}$ (42)

If $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ and $x_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{m} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{m} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{m} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{m}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} y_{e}^{'}, \\ u_{m}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} y_{e}^{'}, \\ x_{m}^{'} = 0, \\ y_{m}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} y_{e}^{'} . \end{matrix}$ (43)

If $x_{e}, y_{e} \geq \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{7}{8} l_{e} + \frac{1}{8} u_{e} - \frac{1}{16} x_{e} - \frac{1}{16} y_{e}, \\ u_{m} = \frac{1}{8} l_{e} + \frac{7}{8} u_{e} + \frac{1}{16} x_{e} + \frac{1}{16} y_{e}, \\ x_{m} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} + \frac{5}{8} x_{e} - \frac{3}{8} y_{e}, \\ y_{m} = - \frac{3}{4} l_{e} + \frac{3}{4} u_{e} - \frac{3}{8} x_{e} + \frac{5}{8} y_{e}, \\ l_{m}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} x_{e}^{'}, \\ u_{m}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} x_{e}^{'}, \\ x_{m}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} x_{e}^{'}, \\ y_{m}^{'} = 0 . \end{matrix}$ (44)

If $x_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{m} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{m} = 0, \\ y_{m} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{m}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} y_{e}^{'}, \\ u_{m}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} y_{e}^{'}, \\ x_{m}^{'} = 0, \\ y_{m}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} y_{e}^{'} . \end{matrix}$ (45)

If $x_{e} < \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} y_{e}, \\ u_{m} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} y_{e}, \\ x_{m} = 0, \\ y_{m} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} y_{e}, \\ l_{m}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} x_{e}^{'}, \\ u_{m}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} x_{e}^{'}, \\ x_{m}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} x_{e}^{'}, \\ y_{m}^{'} = 0 . \end{matrix}$ (46)

If $y_{e} < \frac{3}{8} δ_{e}$ and $x_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{m} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{m} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{m} = 0, \\ l_{m}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} y_{e}^{'}, \\ u_{m}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} y_{e}^{'}, \\ x_{m}^{'} = 0, \\ y_{m}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} y_{e}^{'} . \end{matrix}$ (47)

If $y_{e} < \frac{3}{8} δ_{e}$ and $y_{e}^{'} < \frac{3}{8} δ_{e}^{'}$ , then ${\begin{matrix} l_{m} = \frac{4}{5} l_{e} + \frac{1}{5} u_{e} - \frac{1}{10} x_{e}, \\ u_{m} = \frac{1}{5} l_{e} + \frac{4}{5} u_{e} + \frac{1}{10} x_{e}, \\ x_{m} = - \frac{6}{5} l_{e} + \frac{6}{5} u_{e} + \frac{2}{5} x_{e}, \\ y_{m} = 0, \\ l_{m}^{'} = \frac{4}{5} l_{e}^{'} + \frac{1}{5} u_{e}^{'} - \frac{1}{10} x_{e}^{'}, \\ u_{m}^{'} = \frac{1}{5} l_{e}^{'} + \frac{4}{5} u_{e}^{'} + \frac{1}{10} x_{e}^{'}, \\ x_{m}^{'} = - \frac{6}{5} l_{e}^{'} + \frac{6}{5} u_{e}^{'} + \frac{2}{5} x_{e}^{'}, \\ y_{m}^{'} = 0 . \end{matrix}$ (48)

5 Algorithms and examples of trapezoidal intuitionistic approximations

In Section 4, we give a method of representations of trapezoidal and triangular intuitionistic approximations. Applying Equation (2), we may get their α-cut forms. In the following, we will give the algorithms for computing the α-cut representations of trapezoidal and triangular intuitionistic approximations.

Let A = 〈μ_A, ν_A〉 ∈ IF (R), where the α-cuts of μ_A and 1 - ν_A are (μ_A) _α = [(μ_A) _L (α) , (μ_A) _U (α)] , (1 - ν_A) _α = [(1 - ν_A) _L (α) , (1 - ν_A) _U (α)] . Let us adopt the following notation (see [17]): $\begin{matrix} φ & = & \int_{0}^{1} (3 α - 1) \\ [(μ_{A})_{U} (α) - (μ_{A})_{L} (α)] d α, \end{matrix}$ (49) $\begin{matrix} ψ & = & \int_{0}^{1} (3 α - 1) \\ [{(1 - ν_{A})}_{U} (α) - {(1 - ν_{A})}_{L} (α)] d α . \end{matrix}$ (50) Then by Equations (12)–(19), we get δ_e = x_e + y_e - 2 (u_e - l_e) = -4φ, $δ_{e}^{'} = x_{e}^{'} + y_{e}^{'} - 2 (u_{e}^{'} - l_{e}^{'}) = - 4 ψ .$ The proof of case (2) in Theorem 4.1 shows that trapezoidal intuitionistic approximation in case (2) is triangular. Let (μ_{T
_n}) _α = [s₁ - (1 - α) s₂, s₁ + (1 - α) s₃], $(1 - ν_{T_{n}})_{α} = [s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ (see [5]). By Equations (2), (41) and (12)–(19), we get $s_{1} = \int_{0}^{1} (3 α - 1) [(μ_{A})_{L} (α) + (μ_{A})_{U} (α)] d α,$ (51) $s_{2} = \frac{3}{2} s_{1} + 3 \int_{0}^{1} (α - 1) (μ_{A})_{L} (α) d α,$ (52) $s_{3} = - \frac{3}{2} s_{1} - 3 \int_{0}^{1} (α - 1) (μ_{A})_{U} (α) d α,$ (53) $\begin{matrix} s_{1}^{'} = \int_{0}^{1} (3 α - 1) [(1 - ν_{A})_{L} (α) \\ + (1 - ν_{A})_{U} (α)] d α, \end{matrix}$ (54) $s_{2}^{'} = \frac{3}{2} s_{1}^{'} + 3 \int_{0}^{1} (α - 1) (1 - ν_{A})_{L} (α) d α,$ (55) $s_{3}^{'} = - \frac{3}{2} s_{1}^{'} - 3 \int_{0}^{1} (α - 1) (1 - ν_{A})_{U} (α) d α .$ (56) Thus we obtain the representation of T_n (A) = 〈μ_{T
_n}, ν_{T
_n}〉. Similarly, we may obtain other α-cuts representations of T_n (A) = 〈μ_{T
_n}, ν_{T
_n}〉. Therefore we obtain the following algorithm according to Theorem 4.1.

Algorithm 5.1. A = 〈μ_A, ν_A〉 ∈ IF (R) and T_n (A) = 〈μ_{T
_n}, ν_{T
_n}〉 be tarpezoidal intuitionistic approximation of A:

Step 1. Compute Equations (49) and (50). If φ ≥ 0 and ψ ≥ 0, then (μ_{T
_n}) _α = [t₁ + (t₂ - t₁) α, t₄ - (t₄ - t₃) α] and $(1 - ν_{T_{n}})_{α} = [t_{1}^{'} + (t_{2}^{'} - t_{1}^{'}) α, t_{4}^{'} - (t_{4}^{'} - t_{3}^{'}) α]$ , where μ_{T
_n} and 1 - ν_{T
_n} can be computed by Equations (4)–(11). Thus we obtain T_n (A).

Step 2. If φ < 0 and ψ < 0, then compute s₁, s₂, s₃, $s_{1}^{'}, s_{2}^{'}$ and $s_{3}^{'}$ by Equations (51)–(56). If s₂ ≥ 0, s₃ ≥ 0, $s_{2}^{'} \geq 0$ and $s_{3}^{'} \geq 0$ , then (μ_{T
_n}) _α=[s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and $(1 - ν_{T_{n}})_{α} = [s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ .

Step 3. If s₂ < 0, $s_{2}^{'} \geq 0$ and $s_{3}^{'} \geq 0$ , then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{n}})_{α} = [s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ , where $\begin{matrix} w_{1} & = & \frac{4}{5} \int_{0}^{1} (μ_{A})_{L} (α) d α \\ + \frac{2}{5} \int_{0}^{1} (3 α - 1) (μ_{A})_{U} (α) d α, \end{matrix}$ (57) $w_{2} = 0,$ (58) $\begin{matrix} w_{3} & = & - 4 w_{1} \\ + 2 \int_{0}^{1} [(μ_{A})_{L} (α) + (μ_{A})_{U} (α)] d α . \end{matrix}$ (59)

Step 4. If s₃ < 0, $s_{2}^{'} \geq 0$ and $s_{3}^{'} \geq 0$ ,then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃]and (1 - ν_{T
_n}) _α = $[s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ , where $\begin{matrix} w_{1} & = & \frac{4}{5} \int_{0}^{1} (μ_{A})_{U} (α) d α \\ + \frac{2}{5} \int_{0}^{1} (3 α - 1) (μ_{A})_{L} (α) d α, \end{matrix}$ (60) $w_{2} = 4 w_{1} - 2 \int_{0}^{1} [(μ_{A})_{L} (α) + (μ_{A})_{U} (α)] d α,$ (61) $w_{3} = 0 .$ (62)

Step 5. If s₂ ≥ 0, s₃ ≥ 0 and $s_{2}^{'} < 0$ , then (μ_{T
_n}) _α = [s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and (1 - ν_{T
_n}) _α = $[w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where $\begin{matrix} w_{1}^{'} & = & \frac{4}{5} \int_{0}^{1} (1 - ν_{A})_{L} (α) d α \\ + \frac{2}{5} \int_{0}^{1} (3 α - 1) (1 - ν_{A})_{U} (α) d α, \end{matrix}$ (63) $w_{2}^{'} = 0,$ (64) $\begin{matrix} w_{3}^{'} & = & - 4 w_{1}^{'} \\ + 2 \int_{0}^{1} [(1 - ν_{A})_{L} (α) + (1 - ν_{A})_{U} (α)] d α . \end{matrix}$ (65)

Step 6. If s₂ ≥ 0, s₃ ≥ 0 and $s_{3}^{'} < 0$ , then (μ_{T
_n}) _α = [s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and (1 - ν_{T
_n}) _α = $[w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where $\begin{matrix} w_{1}^{'} & = & \frac{4}{5} \int_{0}^{1} (1 - ν_{A})_{U} (α) d α \\ + \frac{2}{5} \int_{0}^{1} (3 α - 1) (1 - ν_{A})_{L} (α) d α, \end{matrix}$ (66) $\begin{matrix} w_{2}^{'} & = & 4 w_{1}^{'} \\ - 2 \int_{0}^{1} [(1 - ν_{A})_{L} (α) + (1 - ν_{A})_{U} (α)] d α, \end{matrix}$ (67) $w_{3}^{'} = 0 .$ (68)

Step 7. If s₂ < 0 and $s_{2}^{'} < 0$ , then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{n}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (57), (58) and (59); $w_{1}^{'}$ , $w_{2}^{'}$ and $w_{3}^{'}$ can be computed by Equations (63), (64) and (65).

Step 8. If s₂ < 0 and $s_{3}^{'} < 0$ , then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{n}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (57), (58) and (59); $w_{1}^{'}$ , $w_{2}^{'}$ and $w_{3}^{'}$ can be computed by Equations (66), (67) and (68).

Step 9. If s₃ < 0 and $s_{2}^{'} < 0$ , then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{n}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (60), (61) and (62); $w_{1}^{'}$ , $w_{2}^{'}$ , $w_{3}^{'}$ can be computed by Equations (63), (64) and (65).

Step 10. If s₃ < 0 and $s_{3}^{'} < 0$ , then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and (1 - ν_{T
_n}) _α = $[w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (60), (61) and (62); $w_{1}^{'}$ , $w_{2}^{'}$ , $w_{3}^{'}$ can be computed by Equations (66), (67) and (68).

Step 11. If φ < 0 and ψ ≥ 0, then compute s₁, s₂, s₃ by Equations (51)–(53). If s₂ ≥ 0 and s₃ ≥ 0, then (μ_{T
_n}) _α = [s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and $(1 - ν_{T_{n}})_{α} = [t_{1}^{'} + (t_{2}^{'} - t_{1}^{'}) α, t_{4}^{'} - (t_{4}^{'} - t_{3}^{'}) α]$ , where $t_{1}^{'}$ – $t_{4}^{'}$ can be computed by Equations (8)–(11).

Step 12. If s₂ < 0, then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{n}})_{α} = [t_{1}^{'} + (t_{2}^{'} - t_{1}^{'}) α, t_{4}^{'} - (t_{4}^{'} - t_{3}^{'}) α]$ , where w₁, w₂, w₃ can be computed by Equations (57), (58) and (59); $t_{1}^{'}$ – $t_{4}^{'}$ can be computed by Equations (8)–(11).

Step 13. If s₃ < 0, then (μ_{T
_n}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{n}})_{α} = [t_{1}^{'} + (t_{2}^{'} - t_{1}^{'}) α, t_{4}^{'} - (t_{4}^{'} - t_{3}^{'}) α]$ , where w₁, w₂, w₃ can be computed by Equations (59), (60) and (61); $t_{1}^{'}$ – $t_{4}^{'}$ can be computed by Equations (8)–(11).

In the same vein, we may get the following algorithm for computing the triangular intuitionistic approximation of an intuitionistic fuzzynumber.

Algorithm 5.2. Let A = 〈μ_A, ν_A〉 ∈ IF (R), and T_▵ (A) = 〈μ_{T
_▵}, ν_{T
_▵}〉 be triangular intuitionistic approximation of A:

Step 1. Compute s₁, s₂, s₃, $s_{1}^{'}, s_{2}^{'}$ and $s_{3}^{'}$ by Equations (51)–(56). If s₂ ≥ 0, s₃ ≥ 0, $s_{2}^{'} \geq 0$ and $s_{3}^{'} \geq 0$ , then (μ_{T
_▵}) _α=[s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and $(1 - ν_{T_{▵}})_{α} = [s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ .

Step 2. If s₂ < 0, $s_{2}^{'} \geq 0$ and $s_{3}^{'} \geq 0$ , then (μ_{T
_▵}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and (1 - ν_{T
_▵}) _α = $[s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (57), (58) and (59).

Step 3. If s₃ < 0, $s_{2}^{'} \geq 0$ and $s_{3}^{'} \geq 0$ , then (μ_{T
_▵}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and (1 - ν_{T
_▵}) _α = $[s_{1}^{'} - (1 - α) s_{2}^{'}, s_{1}^{'} + (1 - α) s_{3}^{'}]$ . where w₁, w₂ and w₃ can be computed by Equations (60), (61) and (62).

Step 4. If s₂ ≥ 0, s₃ ≥ 0 and $s_{2}^{'} < 0$ , then (μ_{T
_▵}) _α = [s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and (1 - ν_{T
_▵}) _α = $[w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where $w_{1}^{'}$ , $w_{2}^{'}$ , $w_{3}^{'}$ can be computed by Equations (63), (64) and (65).

Step 5. If s₂ ≥ 0, s₃ ≥ 0 and $s_{3}^{'} < 0$ , then (μ_{T
_n}) _α = [s₁ - (1 - α) s₂, s₁ + (1 - α) s₃] and (1 - ν_{T
_n}) _α = $[w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where $w_{1}^{'}$ , $w_{2}^{'}$ and $w_{3}^{'}$ can be computed by Equations (66), (67) and (68).

Step 6. If s₂ < 0 and $s_{2}^{'} < 0$ , then (μ_{T
_▵}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{▵}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (57), (58) and (59); $w_{1}^{'}$ , $w_{2}^{'}$ and $w_{3}^{'}$ can be computed by Equations (63), (64) and (65).

Step 7. If s₂ < 0 and $s_{3}^{'} < 0$ , then (μ_{T
_▵}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{▵}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (57), (58) and (59); $w_{1}^{'}$ , $w_{2}^{'}$ and $w_{3}^{'}$ can be computed by Equations (66), (67) and (68).

Step 8. If s₃ < 0 and $s_{2}^{'} < 0$ , then (μ_{T
_▵}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{▵}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (60), (61) and (62); $w_{1}^{'}$ , $w_{2}^{'}$ , $w_{3}^{'}$ can be computed by Equations (63), (64) and (65).

Step 9. If s₃ < 0 and $s_{3}^{'} < 0$ , then (μ_{T
_▵}) _α = [w₁ - (1 - α) w₂, w₁ + (1 - α) w₃] and $(1 - ν_{T_{▵}})_{α} = [w_{1}^{'} - (1 - α) w_{2}^{'}, w_{1}^{'} + (1 - α) w_{3}^{'}]$ , where w₁, w₂ and w₃ can be computed by Equations (60), (61) and (62); $w_{1}^{'}$ , $w_{2}^{'}$ , $w_{3}^{'}$ can be computed by Equations (66), (67) and (68).

It is difficult to rank intuitionistic fuzzy numbers A and B using the ranking methods of intuitionistic fuzzy numbers in [18]. So, it is necessary to find the trapezoidal intuitionistic approximations of intuitionistic fuzzy numbers.

Example 5.1. Let’s discuss the trapezoidal intuitionistic approximations T_n of the following intuitionistic fuzzy numbers:

A = 〈μ_A, ν_A〉, where $(μ_{A})_{α} = [- 1 + α^{2}, 1 - \sqrt{α}]$ and $(1 - ν_{A})_{α} = [- 1, 1 - \sqrt{α}]$ .

B = 〈μ_B, ν_B〉, where $(μ_{B})_{α} = [\sqrt{α} - 1, 1 - \sqrt{α}]$ and $(1 - ν_{B})_{α} = [α - 2, 1 - \sqrt{α}]$ .

Let’s compute φ and ψ .

(1) By applying Equations (49) and (50), we get $φ (μ_{A}) = \frac{1}{20} > 0,$ $ψ (1 - ν_{A}) = \frac{7}{15} > 0 .$ Therefore, by Equations (4)–(11) (in step 1 of Algorithm 5), we have $t_{1} = - \frac{7}{6}$ , $t_{2} = - \frac{1}{6}$ , $t_{3} = - \frac{1}{15}$ , $t_{4} = \frac{11}{15}$ , $t_{1}^{'} = - 1$ , $t_{2}^{'} = - 1$ , $t_{3}^{'} = - \frac{1}{15}$ and $t_{4}^{'} = \frac{11}{15}$ . Hence $(μ_{T_{n} (A)})_{α} = [- \frac{7}{6} + α, \frac{11}{15} - \frac{4}{5} α],$ $(1 - ν_{T_{n} (A)})_{α} = [- 1, \frac{11}{15} - \frac{4}{5} α] .$ Thus T_n (A) = 〈μ_{T_n(A)}, ν_{T_n(A)}〉 equals the trapezoidal intuitionistic approximation of A.

(2) By applying Equations (49) and (50), we have $φ (μ_{B}) = - \frac{41}{15} < 0, ψ (1 - ν_{B}) = - \frac{7}{15} > 0 .$ Hence by Equations (51)–(53) and (4)–(7) (in step 11 of Algorithm 5), we get s₁ = 0, $s_{2} = \frac{7}{30} > 0$ , $s_{3} = \frac{7}{30} > 0$ , $t_{1}^{'} = - 2$ , $t_{2}^{'} = - 1$ , $t_{3}^{'} = - \frac{1}{15}$ and $t_{4}^{'} = \frac{11}{15}$ , So $(μ_{T_{n} (B)})_{α} = [- \frac{7}{30} + \frac{7}{30} α, \frac{7}{30} - \frac{7}{30} α],$ $(1 - ν_{T_{n} (B)})_{α} = [- 2 + α, \frac{11}{15} - \frac{4}{5} α] .$ Thus T_n (B) = 〈μ_{T_n(B)}, ν_{T_n(B)}〉 equals the trapezoidal intuitionistic approximation of B.

It is directly derived from Definitions 2.3 and 2.4 that $L (T_{n} (A)) = \frac{5}{9} < L (T_{n} (B)) = \frac{497}{720},$ i.e. T_n (A) < T_n (B). Hence we can conclude A < B according to the ranking of approximations.

Example 5.2. Let’s compute triangular intuitionistic approximation T_▵ of intuitionistic fuzzy numbers A and B in Example 5.1.

By computing Equations (51)–(56) (in step 1 of Algorithm 5), we obtain $s_{1} = - \frac{7}{60} < 0$ , $s_{2} = \frac{43}{40} > 0$ , $s_{3} = \frac{7}{8} > 0$ , $s_{1}^{'} = - \frac{8}{15} < 0$ , $s_{2}^{'} = \frac{7}{10} > 0$ and $s_{3}^{'} = \frac{3}{2} > 0$ . Hence ${(μ_{T_{▵} (A)})}_{α} = [- \frac{143}{120} + \frac{43}{40} α, \frac{91}{120} - \frac{7}{8} α],$ ${(1 - ν_{T_{▵} (A)})}_{α} = [- \frac{37}{30} + \frac{7}{10} α, \frac{29}{30} - \frac{3}{2} α] .$ Thus T_▵ (A) = 〈μ_{T_▵(A)}, ν_{T_▵(A)}〉 is the triangular intuitionistic approximation of A.

By computing Equations (51)–(56) (in step 1 of Algorithm 5), we obtain s₁ = 0, $s_{2} = \frac{7}{30} > 0$ , $s_{3} = \frac{7}{30} > 0$ , $s_{1}^{'} = - \frac{8}{15} < 0$ , $s_{2}^{'} = \frac{17}{10} > 0$ and $s_{3}^{'} = \frac{3}{2} > 0$ . Hence ${(μ_{T_{▵} (B)})}_{α} = [- \frac{7}{30} + \frac{7}{30} α, \frac{7}{30} - \frac{7}{30} α],$ ${(1 - ν_{T_{▵} (B)})}_{α} = [- \frac{67}{30} + \frac{17}{10} α, \frac{29}{30} - \frac{3}{2} α] .$ Thus T_▵ (B) = 〈μ_{T_▵(B)}, ν_{T_▵(B)}〉 is the triangular intuitionistic approximation of B.

It is directly derived from Definitions 2.3 and 2.4 that $L (T_{▵} (A)) = \frac{7061}{14400} < L (T_{▵} (B)) = \frac{9744}{14400},$ i.e. T_▵ (A) < T_▵ (B). Hence we can conclude A < B according to the ranking of approximations. The results of ranking order are identical in Examples 5.1 and 5.2.

6 Conclusion

In this paper, the representations and algorithms of approximating intuitionistic fuzzy numbers by trapezoidal intuitionistic fuzzy numbers are mainly discussed. The results of approximations are applied on ranking of intuitionistic fuzzy numbers in the examples. The approaches of approximating intuitionistic fuzzy numbers by trapezoidal fuzzy numbers with condition of preserving the expected interval will be investigated in the near future.

Footnotes

Acknowledgment

This work is supported by the National Natural Science Foundation of China (61374118).

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