On the orders of OWA operators

Abstract

In this paper, we propose a partial ordering ⪯ on the set of ordered weighted averaging (OWA) operators. Based on this relation ⪯, we introduce the negation, conjunction and disjunction operations, and establish a bounded De Morgan lattice equipped with an involutive negation for OWA operators. Finally, we develop a similarity measure between OWA operators based on the ordering ⪯.

Keywords

OWA operators orders lattices

1 Introduction

Ordered weighted averaging (OWA) operators were introduced by Yager [14]. Afterwards, various types of OWA operators have been proposed and a lot of successful applications have been carried out [1 , 15]. Now, OWA operator has become a theoretic basis and an important information fusion tool for decision-making designing. As we all know, for any OWA operator Agg, it has $min (x) \leq Agg (x) \leq max (x)$ for all x = (x₁, x₂, ⋯ , x_n) ∈ [0, 1] ⁿ. However, in general, how to compare two OWA operators? Some authors have introduced new measures for OWA operators. Yager [14] proposed the orness/andness measure for OWA operators, which can be viewed as a normed expected value of OWA operators. Yager and Filev [16] defined the dispersion (also known as entropy) measure for OWA operators. Jin and Mesiar [6] proposed some distance measures for OWA operators based on accumulated entries. In this paper, we first consider the ordering problem of OWA operators, i.e., for any two OWA operators Agg₁ and Agg₂, whether the inequality ${Agg}_{1} (x) \leq {Agg}_{2} (x)$ holds for all x ∈ [0, 1] ⁿ, in which we say Agg₂ is stronger than Agg₁. It is worth pointing that the ordering problem in t-norms and fuzzy implications have been well studied [2, 9]. After giving the order of OWA operators, we then apply it to define a similarity measure between OWA operators.

This paper is organized as follows. In Section 2, we briefly review the coordinatewise order, lexicographical order and OWA operators. In Section 3, we show that both the coordinatewise order and lexicographical order are not suitable for the ordering of weighted averaging (WA) and OWA operators. In Section 4, we introduce a new order for OWA operators and give the necessary and sufficient condition of Agg₁ (x) ≤ Agg₂ (x) , ∀ x ∈ [0, 1] ⁿ. In Section 5, we study the algebraic structures on the set of OWA operators based on this new order. In Section 6, we study the similarity of OWA operators. In Section 7, we summarize our main results.

2 Preliminaries

2.1 Orders

Let L be a set, ≤ is a partial order on L if it has the following properties [3, 5]: for any α, β, γ ∈ L,

Reflexivity: α ≤ α

Antisymmetry: α ≤ β and β ≤ α ⇒ α = β

Transitivity: α ≤ β and β ≤ γ ⇒ α ≤ γ

Then (L, ≤) is called a partially ordered set. Moreover, a partially ordered set L is a totally ordered set (other names are clain and linearly ordered set) if it also has the following property: for any α, β ∈ L

Comparability: α ≤ β or β ≤ α

This condition means any two elements of L are comparable.

Let L₁, ⋯ , L_n be partially (or totally) ordered sets. The coordinatewise order on the Cartesian product L₁ × ⋯ × L_n is defined by $(α_{1}, α_{2}, \dots, α_{n}) \leq_{c} (β_{1}, β_{2}, \dots, β_{n})$ (1) $\Leftrightarrow (\forall i) α_{i} ⪯ β_{i},$

The lexicographic order on the Cartesian product L₁ × ⋯ × L_n is defined by $\begin{matrix} (α_{1}, α_{2}, \dots, α_{n}) \leq_{l} (β_{1}, β_{2}, \dots, β_{n}) \\ \Leftrightarrow (\forall i) α_{i} = β_{i} \end{matrix}$ (2) $or (\exists j) α_{1} = β_{1}, \dots, α_{j - 1} = β_{j - 1}, α_{j} < β_{j}$

2.2 OWA operators

Let I = [0, 1], an n-dimension vector w = (w₁, w₂, ⋯ , w_n) is a weighting vector if has the following properties

(w₁, w₂, ⋯ , w_n) ∈ Iⁿ;

$\sum_{j = 1}^{n} w_{j} = 1$ .

For a fixed weighting vector w, the associated weighted averaging (WA) operator WA : Iⁿ → I is defined by

$WA (x_{1}, x_{2}, \dots, x_{n}) = \sum_{j = 1}^{n} w_{j} x_{j} .$ (3) The associated OWA operator f : Iⁿ → I is defined by [14]

$f (x_{1}, x_{2}, \dots, x_{n}) = \sum_{j = 1}^{n} w_{j} x_{(j)},$ (4) where (x₍₁₎, x₍₂₎, ⋯ , x_(n)) is a permutation of (x₁, x₂, ⋯ , x_n) satisfying x₍₁₎ ≥ x₍₂₎ ≥ ⋯ ≥ x_(n).

The following are some special cases of OWA aggregations and their associated weighting vectors [14, 15]:

Maximum operator: w^⊤ = (1, 0, ⋯ , 0);

Minimum operator: w_⊥ = (0, 0, ⋯ , 1);

Average operator: $w_{ave} = (\frac{1}{n}, \frac{1}{n}, \dots, \frac{1}{n}, \frac{1}{n})$ ;

Window-OWA operators w_win: $w_{i} = {\begin{matrix} 0, & i < k, \\ \frac{1}{m}, & k \leq i < k + m, \\ 0, & i \geq k + m . \end{matrix}$

Orlike S-OWA operator w_OS with α ∈ I:

$w_{i} = {\begin{matrix} \frac{1}{n - 1} (1 - α) + α, & i = 1, \\ \frac{1}{n - 1} (1 - α), & 2 \leq i \leq n . \end{matrix}$

Andlike S-OWA operator w_AS with β ∈ I:

$w_{i} = {\begin{matrix} \frac{1}{n - 1} (1 - β), & 1 \leq i \leq n - 1, \\ \frac{1}{n - 1} (1 - β) + β, & i = n . \end{matrix}$

3 Coordinatewise order and lexicographical order on WA and OWA operators

Coordinatewise order and lexicographical order are two orders on Iⁿ. In this section, we demonstrate that the both orders are not suitable for the ordering of WA and OWA operators.

3.1 Coordinatewise order on WA and OWA operators

Both WA and OWA operators are determined by its weighting vector. We can define the order of two Given two WA (or OWA) operators by the coordinatewise order of their associated weighting vectors. Because for any two weighting vectors w = (w₁, w₂, ⋯ , w_n) and $w^{'} = (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'})$ , we have $(\forall i) x_{i} \leq y_{i} \Rightarrow \sum_{j = 1}^{n} w_{j} x_{j} \leq \sum_{j = 1}^{n} w_{j}^{'} y_{j},$ (5) $\forall (x_{1}, \dots, x_{n}) \in I^{n} .$ $(\forall i) x_{i} \leq y_{i} \Rightarrow \sum_{j = 1}^{n} w_{j} x_{(j)} \leq \sum_{j = 1}^{n} w_{j}^{'} y_{(j)},$ (6) $\forall (x_{1}, \dots, x_{n}) \in I^{n} .$

However, the coordinatewise order of w and w′ has the following problem.

Theorem 1. Let $(w_{1}, w_{2}, \dots, w_{n}), (w_{1}^{'}, w_{2}^{'}, \dots,$ $w_{n}^{'}) \in I^{n}$ and $\sum_{j = 1}^{n} w_{j} = \sum_{j = 1}^{n} w_{j}^{'} = 1$ , then

$(\forall i) w_{j} \leq w_{j}^{'} \Leftrightarrow (\forall i) w_{j} = w_{j}^{'} .$ (7)

Proof. (⇐): It is clear.

(⇒): We prove it by using reduction to absurdity. $w_{j} \leq w_{j}^{'}$ means $w_{j} = w_{j}^{'}$ or $w_{j} < w_{j}^{'}$ . If there exists some i such that $w_{i} < w_{i}^{'}$ , then $\begin{matrix} \begin{matrix} \sum_{j = 1}^{n} w_{j}^{'} & = (\sum_{j = 1, j \neq i}^{n} w_{j}^{'}) + w_{i}^{'} \\ > (\sum_{j = 1, j \neq i}^{n} w_{j}) + w_{i} \\ = \sum_{j = 1}^{n} w_{j} \\ = 1 . \end{matrix} \end{matrix}$

This’s a contradiction of $\sum_{j = 1}^{n} w_{j}^{'} = 1$ . □

From above Theorem 1, for any two weighting vectors w and w′, we have $w \leq_{c} w^{'} \Leftrightarrow w = w^{'} .$ Thus the coordinatewise order of weighting vectors is not suitable for order of WA and OWA operators.

Similar to the orders of t-norm and fuzzy implications in [2, 9], we can define the order of WA operators by the following:

Definition 1. If, for two n-ary WA operators WA₁ and WA₂, the inequality

${WA}_{1} (x_{1}, x_{2}, \dots, x_{n}) \leq {WA}_{2} (x_{1}, x_{2}, \dots, x_{n})$ (8) holds for all (x₁, x₂, ⋯ , x_n) ∈ Iⁿ, then we say WA₁ is weaker than WA₂ and denote it as WA₁ ≤ WA₂.

Unfortunately, Definition 1 of order on WA operators is equivalent to the coordinatewise order of their associated weighting vectors.

Theorem 2. For any two n-ary WA operators WA₁ with the weighting vector w and WA₂ with the weighting vector w′. WA₁ ≤ WA₂ if and only if w = w′.

Proof. (⇐): It is clear.

(⇒): Consedering the following special cases $(0, \dots, 0, x_{i} = 1, 0, \dots, 0) \in I^{n}, i = 1, 2, \dots, n .$

By WA₁ ≤ WA₂, we get

WA₁ (1, 0, 0, . . . , 0) ≤ WA₂ (1, 0, 0, . . . , 0)

WA₁ (0, 1, 0, . . . , 0) ≤ WA₂ (0, 1, 0, . . . , 0)

⋯

WA₁ (0, 0, 0, . . . , 1) ≤ WA₂ (0, 0, 0, . . . , 1).

From these inequalities, we have

$w_{1} \leq w_{1}^{'}$

$w_{2} \leq w_{2}^{'}$

⋯

$w_{n} \leq w_{n}^{'}$ .

Then by Theorem 1, we have w = w′. □

From above two Theorems 1 and 2, for any two weighting vectors w and w′, Eq. (5) can be improved to $\begin{matrix} \sum_{j = 1}^{n} w_{j} x_{j} \leq \sum_{j = 1}^{n} w_{j}^{'} y_{j}, \forall (x_{1}, \dots, x_{n}) \in I^{n} \\ \Leftrightarrow & (\forall i) x_{i} \leq y_{i} (i . e ., w \leq_{c} w^{'}) \end{matrix}$ (9) $\Leftrightarrow (\forall i) x_{i} = y_{i} (i . e ., w = w^{'}) .$

3.2 Lexicographical order on WA and OWA operators

First, the lexicographical order on the set of weighting vectors is a little different from that in Eq. (2)

Lemma 1. If w = (w₁, w₂, ⋯ , w_n) and $w^{'} = (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'})$ be two weighting vectors. Then $\begin{matrix} (w_{1}, w_{2}, \dots, w_{n}) \leq_{l} (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'}) \\ \Leftrightarrow (\forall i) w_{i} = w_{i}^{'} \end{matrix}$ (10) $\begin{matrix} or (\exists 1 \leq j \leq n - 1) w_{1} = w_{1}^{'}, \dots, \\ w_{j - 1} = w_{j - 1}^{'}, w_{j} < w_{j}^{'} . \end{matrix}$

Proof. (⇐) is clear.

(⇒) Because $\sum_{j = 1}^{n} w_{j} = \sum_{j = 1}^{n} w_{j}^{'} = 1$ , then It is impossible to the case with $w_{1} = w_{1}^{'}, \dots, w_{n - 1} = w_{n - 1}^{'}, w_{n} < w_{n}^{'}$ . □

From above result, w ≤ _lw′ means w = w′ or w < _lw′, i.e., there exists 1 ≤ j ≤ n - 1 such that

$w_{1} = w_{1}^{'}, \dots, w_{j - 1} = w_{j - 1}^{'}, w_{j} < w_{j}^{'} .$

Then we show that the lexicographical order used on WA operators has the following problem.

Theorem 3. For any two n-ary WA operators WA₁ with the weighting vector w₁ = (w₁, w₂, ⋯ , w_n) and WA₂ with the weighting vector $w_{2} = (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'})$ . If w₁ < _lw₂, then there exists (x₁, x₂, ⋯ , x_n) ∈ Iⁿ such that WA₁ (x₁, x₂, ⋯ , x_n) > WA₂ (x₁, x₂, ⋯ , x_n).

Proof. w₁ < _lw₂ means there exists 1 ≤ i ≤ n - 1 such that $w_{1} = w_{1}^{'}, \dots, w_{i - 1} = w_{i - 1}^{'}, w_{i} < w_{i}^{'} .$

Since $\sum_{j = 1}^{n} w_{j} = \sum_{j = 1}^{n} w_{j} = 1$ , then there exist i < j ≤ n such that $w_{j} > w_{j}^{'}$ . By selecting the special (0, ⋯ , 0, x_j = 1, 0, ⋯ , 0) ∈ Iⁿ, we get

$\begin{matrix} \begin{matrix} {WA}_{1} (0, \dots, 0, x_{j} = 1, 0, \dots, 0) \\ = w_{j} \\ > w_{j}^{'} \\ = {WA}_{2} (0, \dots, 0, x_{j} = 1, 0, \dots, 0) . \end{matrix} \end{matrix}$ □

As we all known, for any two weighting vectors w and w′, we have

$\begin{matrix} w <_{c} w^{'} \\ \Rightarrow \sum_{j = 1}^{n} w_{j} x_{j} < \sum_{j = 1}^{n} w_{j}^{'} x_{j}, \exists (x_{1}, \dots, x_{n}) \in I^{n} . \end{matrix}$ (11)

But from above Theorem 3, we also have

$\begin{matrix} w <_{c} w^{'} \\ \Rightarrow \sum_{j = 1}^{n} w_{j} x_{j} > \sum_{j = 1}^{n} w_{j}^{'} x_{j}, \exists (x_{1}, \dots, x_{n}) \in I^{n} . \end{matrix}$ (12)

Thus the lexicographical order of weighting vectors is not suitable for the ordering of their associated WA operators.

Similarly, the lexicographical order of weighting vectors is not suitable for the ordering of their associated OWA operators, as seen in the following example.

Example 1. Let w = (0.7, 0.1, 0.2) and w′ = (0.6, 0.3, 0.1), then for x = (1, 0, 0), we have wx^T = 0.7 > 0.6 = w′x^T; but for y = (1, 1, 0), we have wy^T = 0.8 < 0.9 = w′y^T.

4 A new Order on OWA operators

We propose a new order on the set of OWA operators in this section.

First, we can define the order of OWA operators by the following:

Definition 2. If, for two n-ary OWA operators OWA₁ and OWA₂, the inequality

${OWA}_{1} (x_{1}, x_{2}, \dots, x_{n}) \leq {OWA}_{2} (x_{1}, x_{2}, \dots, x_{n})$ (13) holds for all (x₁, x₂, ⋯ , x_n) ∈ Iⁿ, then we say OWA₁ is weaker than OWA₂ and denote it as OWA₁ ≤ OWA₂. Moreover, we shall denote OWA₁ < OWA₂ whenever OWA₁ ≤ OWA₂ and OWA₁ ≠ OWA₂, i.e., OWA₁ ≤ OWA₂ and there exists (x₁, x₂, ⋯ , x_n) ∈ Iⁿ such that

${OWA}_{1} (x_{1}, x_{2}, \dots, x_{n}) < {OWA}_{2} (x_{1}, x_{2}, \dots, x_{n}) .$ (14)

Remark 1. Let n < m, for any n-ary OWA operator OWA₁ with the weighting vector w₁ = (w₁, w₂, ⋯ , w_n), we can define a m-ary OWA operator OWA₂ by the weighting vector $w_{2} = (w_{1}, w_{2}, \dots, w_{n}, \underset{m - n}{\underset{︸}{0, \dots, 0}})$ Then we have OWA₂ (x₁, x₂, ⋯ , x_n, x_n+1, ⋯ , x_m)= OWA₁ (x₁, x₂, ⋯ , x_n) for all (x₁, x₂, ⋯ , x_m) ∈ I^m. By this method, we can compare an n-ary OWA operator and a m-ary OWA operator whenever n ≠ m. For convenience, we only consider the set of all n-ary OWA operators.

In above section 3.2, we know that the similar method is not suitable for defining order on WA operators (see Definition 1 and Theorem 3). But we demonstrate here that it is feasible for defining order on OWA operators.

Theorem 4. For any two n-ary OWA operators OWA₁ with the weighting vector w and OWA₂ with the weighting vector w′. OWA₁ ≤ OWA₂ if and only if

$\sum_{j = 1}^{k} w_{j} \leq \sum_{j = 1}^{k} w_{j}^{'}$ (15) for all k = 1, 2, ⋯ , n - 1.

Proof. (⇒): If OWA₁ ≤ OWA₂. Then by selecting some special (x₁, x₂, ⋯ , x_n) ∈ Iⁿ, we have

OWA₁ (1, 0, 0, . . . , 0) ≤ OWA₂ (1, 0, 0, . . . , 0)

OWA₁ (1, 1, 0, . . . , 0) ≤ OWA₂ (1, 1, 0, . . . , 0)

⋯

OWA₁ (1, 1, 1, . . . , 1) ≤ OWA₂ (1, 1, 1, . . . , 1).

From these inequalities, we get

$w_{1} \leq w_{1}^{'}$

$w_{1} + w_{2} \leq w_{1}^{'} + w_{2}^{'}$

⋯

$w_{1} + w_{2} + \dots + w_{n} \leq w_{1}^{'} + w_{2}^{'} + \dots + w_{n}^{'}$ .

Thus Eq. (15) holds.

(⇐): This can be proved by the recurrence method. If Eq. (15) does not hold for some 1 ≤ k ≤ n - 1, i.e., $\sum_{j = 1}^{k} w_{j} > \sum_{j = 1}^{k} w_{j}^{'}$ . Then consider $x = (\underset{k}{\underset{︸}{1, \dots, 1}}, 0, \dots, 0)$ we have ${OWA}_{1} (x) = \sum_{j = 1}^{k} w_{j} > \sum_{j = 1}^{k} w_{j}^{'} = {OWA}_{2} (x)$ . □

From above two Theorem, for any two weighting vectors w and w′, Eq. (6) can be improved to

Definition 3. For any two weighting vector w and w′. Define w ⪯ w′ if Eq. (15) holds for all k = 1, 2, ⋯ , n - 1.

Thus OWA₁ ≤ OWA₂ is equivalent to w ⪯ w′ where w and w′ are weighting vectors of OWA₁ and OWA₂ respectively.

Lemma 2. For any two weighting vector w = (w₁, w₂, ⋯ , w_n) and $w^{'} = (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'})$ , we have $w ⪯ w^{'} \Rightarrow w \leq_{l} w^{'},$ (17) but not vice versa.

Proof. (w ⪯ w′ ⇒ w ≤ _lw′): From Eq. (15), we get

$w_{1} \leq w_{1}^{'}$

$w_{1} + w_{2} \leq w_{1}^{'} + w_{2}^{'}$

⋯

$w_{1} + w_{2} + \dots + w_{n} \leq w_{1}^{'} + w_{2}^{'} + \dots + w_{n}^{'}$ .

By $w_{1} \leq w_{1}^{'}$ , if $w_{1} < w_{1}^{'}$ thus w ≤ _lw′ holds; else if $w_{1} = w_{1}^{'}$ , by combing $w_{1} + w 2 \leq w_{1}^{'} + w_{2}^{'}$ , we have $w_{2} \leq w_{2}^{'}$ . Similarly, if $w_{2} < w_{2}^{'}$ thus w ≤ _lw′ holds; else if $w_{2} = w_{2}^{'}$ , by combing $w_{1} + w 2 + w_{3} \leq w_{1}^{'} + w_{2}^{'} + w_{3}^{'}$ , we have $w_{3} \leq w_{3}^{'}$ . So we can get w ≤ _lw′ by recurrence method.

(w ≤ _lw′ ≱ w ⪯ w′): See Example 1. □

Let W be the set of all weighting vectors of dimension n. ⪯ on W is a special case of ≤_l on W, moreover, it is a relationship between ≤_c and ≤_l, i.e., for any w, w′ ∈ W, we have $w = w^{'} \Leftrightarrow w \leq_{l} w^{'} \Rightarrow w ⪯ w^{'} \Rightarrow w \leq_{l} w^{'},$ (18) note that w ⪯ w′ ≱ w ≤ _lw′ and w ≤ _lw′ ≱ w ⪯ w′.

Example 2. In Example 1, w = (0.7, 0.1, 0.2) and w′ = (0.6, 0.3, 0.1), then w′ ≤ _lw since $w_{1}^{'} = 0.6 < 0.7 = w_{1}$ , but w′ ̸ w since $w_{1}^{'} + w_{2}^{'} = 0.6 + 0.3 > 0.7 + 0.1 = w_{1} + w_{2}$ . Then we cannot get $\sum_{j = 1}^{3} w_{j} x_{(j)} \leq \sum_{j = 1}^{3} w_{j}^{'} x_{(j)}, \forall (x_{1}, x_{2}, x_{3}) \in I^{3}$ .

If w = (0.7, 0.2, 0.1) and w′ = (0.6, 0.3, 0.1), then w′ ⪯ w since 0.6 < 0.7 and 0.6 + 0.3 ≤ 0.7 + 0.2. Then, for any x₍₁₎ ≥ x₍₂₎ ≥ x₍₃₎, we have 0.7x₍₁₎ + 0.2x₍₂₎ + 0.1x₍₃₎ - (0.6x₍₁₎ + 0.3x₍₂₎ + 0.1x₍₃₎) =0.1 (x₍₁₎ - x₍₂₎) ≥0. Thus we get $\sum_{j = 1}^{3} w_{j} x_{(j)} \leq \sum_{j = 1}^{3} w_{j}^{'} x_{(j)}, \forall (x_{1}, x_{2}, x_{3}) \in I^{3}$ .

Further, we demonstrate that ⪯ on W is suitable for the ordering of OWA operators, but it is not suitable for the ordering of WA operators, see the following example:

Example 3. See the above example, w = (0.7, 0.2, 0.1) and w′ = (0.6, 0.3, 0.1), we have w′ ⪯ w For x = (1, 0, 0), we have wx^T = 0.7 > 0.6 = w′x^T; for y = (0, 1, 0), we have wy^T = 0.2 < 0.3 = w′y^T.

Theorem 5. If $α \geq \frac{1}{n}$ in Orlike S-OWA operator and $β \geq \frac{1}{n}$ in Andlike S-OWA operator, then

$w_{*} ⪯ w_{AS} ⪯ w_{ave} ⪯ w_{OS} ⪯ w^{*}$ (19)

If $α \leq \frac{1}{n}$ in Orlike S-OWA operator and $β \leq \frac{1}{n}$ in Andlike S-OWA operator, then

$w_{*} ⪯ w_{OS} ⪯ w_{ave} ⪯ w_{AS} ⪯ w^{*}$ (20)

Moreover, if k ≥ 2, then Window-OWA operators are weaker than average operator, i.e., w_win ⪯ w_ave.

Proof. We only give the proof of w_OS ⪯ w_ave For any $β \geq \frac{1}{n}$ , $\frac{1}{n - 1} (1 - β) \leq \frac{1}{n}$ , then their sums of first k components have the relation $k (\frac{1}{n - 1} (1 - β)) \leq k (\frac{1}{n})$ for all k = 1, 2, ⋯ , n - 1. Thus w_OS ⪯ w_ave. □

This is very interesting, for example, when $β \geq \frac{1}{n}$ we have $Andlike S - OWA (x) \leq Ave (x)$ for all x ∈ Iⁿ.

Theorem 6. The relation ⪯ on W is a partial order but not a total order.

Proof. The relation ⪯ on W is the order of OWA operators in Definition 2. For any three n-ary OWA operators OWA₁, OWA₂ and OWA₃, and all x ∈ Iⁿ,

(1): OWA₁ (x) = OWA₁ (x)

(2): if OWA₁ (x) ≤ OWA₂ (x) and OWA₂ (x) ≤ OWA₁ (x), then OWA₁ (x) = OWA₂ (x).

(3): if OWA₁ (x) ≤ OWA₂ (x) and OWA₂ (x) ≤ OWA₃ (x), then OWA₁ (x) = OWA₃ (x).

(4): ⪯ on W is not a total order. See the example w = (0.7, 0.1, 0.2) and w′ = (0.6, 0.3, 0.1), in which w w′ and w′ w. □

We have explored the orders of WA and OWA operators based on the Definitions 1 and 2 respectively. The results can be summarized as follows (also see Table 1).

Table 1

Orders on WA and OWA operators

	WA	OWA
Coordinatewise order	×	×
Lexicographical order	×	×
The proposed order	×	√

(i) Coordinatewise order is not suitable for the ordering of WA and OWA operators (see Theorems 1 and 2);

(ii) Lexicographical order is not suitable for the ordering of WA and OWA operators (see Theorem 3 and Example 1);

(iii) The proposed order is suitable for the ordering of OWA operators (see Theorem 4).

5 Lattice of OWA operators

Since it is meaningful to study the algebraic structures on the set of OWA operators based on the ⪯ on W, we explore an alternative lattice over OWA operators and examine its properties in this section.

5.1 Negation on OWA operators

First, we define a negation operation on W. The standard fuzzy negation is N (x) =1 - x, ∀ x ∈ I. If it is used directly as h′ (w₁, w₂, ⋯ , w_n) = (1 - w₁, 1 - w₂, ⋯ , 1 - w_n), it would not be closed in Iⁿ for n ≥ 3. For example, while h′ (0.3, 0.3, 0.4) = (0.7, 0.7, 0.6) is not a weighting vector , this problem can be improved in the following way.

Lemma 3. For any weighting vector (w₁, w₂, ⋯ , w_n), let $h (w_{1}, w_{2}, \dots, w_{n}) = (\frac{1 - w_{1}}{n - 1}, \dots, \frac{1 - w_{n - 1}}{n - 1}, \frac{1 - w_{n}}{n - 1})$ . Then f has the following properties:

(i) h (w₁, ⋯ , w_n-1, w_n) is a weighting vector;

h is an inverted order mapping, i.e., $\begin{matrix} (w_{1}, w_{2}, \dots, w_{n}) ⪯ (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'}) \\ \Rightarrow & h (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'}) ⪯ h (w_{1}, w_{2}, \dots, w_{n}) . \end{matrix}$ (21)

Proof. We only give the proof of (ii). $(w_{1}, w_{2}, \dots, w_{n}) ⪯ (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'})$ means

$w_{1} \leq w_{1}^{'}$

$w_{1} + w_{2} \leq w_{1}^{'} + w_{2}^{'}$

⋯

$w_{1} + w_{2} + \dots + w_{n - 1} \leq w_{1}^{'} + w_{2}^{'} + \dots + w_{n - 1}^{'}$ .

Then we have

$1 - w_{1} \geq 1 - w_{1}^{'}$

$2 - (w_{1} + w_{2}) \geq 2 - (w_{1}^{'} + w_{2}^{'})$

⋯

$n - 1 - (w_{1} + w_{2} + \dots + w_{n - 1}) \geq n - 1 - (w_{1}^{'} + w_{2}^{'} + \dots + w_{n - 1}^{'})$ .

Thus

$\frac{1 - w_{1}}{n - 1} \geq \frac{1 - w_{1}^{'}}{n - 1}$

$\frac{1 - w_{1}}{n - 1} + \frac{1 - w_{2}}{n - 1} \geq \frac{1 - w_{1}^{'}}{n - 1} + \frac{1 - w_{2}^{'}}{n - 1}$

⋯

$\frac{1 - w_{1}}{n - 1} + \dots + \frac{1 - w_{n - 1}}{n - 1} \geq \frac{1 - w_{1}^{'}}{n - 1} + \dots + \frac{1 - w_{n - 1}^{'}}{n - 1}$ .

Then

$(\frac{1 - w_{1}^{'}}{n - 1}, \dots, \frac{1 - w_{n - 1}^{'}}{n - 1}, \frac{1 - w_{n}^{'}}{n - 1}) ⪯ (\frac{1 - w_{1}}{n - 1}, \dots, \frac{1 - w_{n - 1}}{n - 1}, \frac{1 - w_{n}}{n - 1})$ i.e.,

$h (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'}) ⪯ h (w_{1}, w_{2}, \dots, w_{n}) .$ □

Then $h (0.3, 0.3, 0.4) = (\frac{0.7}{2}, \frac{0.7}{2}, \frac{0.6}{2}) = (0.35, 0.35, 0.3)$ becomes a weighting vector. But for the function h, it still remains some problems as following.

Remark 2. For n ≥ 3, we have

$h (1, 0, \dots, 0) = (0, \frac{1}{n - 1}, \dots, \frac{1}{n - 1}) \neq (0, 0, \dots, 1)$ ;

$h (0, 0, \dots, 1) = (\frac{1}{n - 1}, \dots, \frac{1}{n - 1}, 0) \neq (1, 0, \dots, 0)$ ;

Because $0 \leq \frac{1 - x}{n - 1} \leq \frac{1}{n - 1}, \forall x \in I$ , then we have $h (w) \in [0, \frac{1}{n - 1}]^{n}$ for any w ∈ Iⁿ.

Lemma 4. For any weighting vector w = (w₁, w₂, ⋯ , w_n), let ¬ (w₁, w₂, ⋯ , w_n) = (w_n, ⋯ , w₂, w₁). Then ¬ has the following properties:

(i) ¬w^⊤ = w_⊥;

(ii) ¬w_⊥ = w^⊤;

(iii) ¬ (¬ w) = w;

(iv) ¬ is an inverted order mapping, i.e., $\begin{matrix} w ⪯ w^{'} \\ \Rightarrow & \neg w^{'} ⪯ \neg w . \end{matrix}$ (22)

Proof. (i)-(iii) are clear.

(iv) w ⪯ w′ means

$w_{1} \leq w_{1}^{'}$

$w_{1} + w_{2} \leq w_{1}^{'} + w_{2}^{'}$

⋯

$w_{1} + w_{2} + \dots + w_{n - 1} \leq w_{1}^{'} + w_{2}^{'} + \dots + w_{n - 1}^{'}$ .

Then we have

$1 - w_{1} \geq 1 - w_{1}^{'}$

$1 - (w_{1} + w_{2}) \geq 1 - (w_{1}^{'} + w_{2}^{'})$

⋯

$1 - (w_{1} + w_{2} + \dots + w_{n - 1}) \geq 1 - (w_{1}^{'} + w_{2}^{'} + \dots + w_{n - 1}^{'})$ .

Since $\sum_{j = 1}^{n} w_{j} = \sum_{j = 1}^{n} w_{j}^{'} = 1$ , we have

$w_{2} + w_{3} + \dots + w_{n} \geq w_{2}^{'} + w_{3}^{'} + \dots + w_{n}^{'}$

$w_{3} + \dots + w_{n} \geq w_{3}^{'} + \dots + w_{n}^{'}$

⋯

$w_{n} \geq w_{n}^{'}$ .

Thus $(w_{n}^{'}, \dots, w_{2}^{'}, w_{1}^{'}) ⪯ (w_{n}, \dots, w_{2}, w_{1})$ , i.e., ¬w′ ⪯ ¬ w . □

From above result, ¬ is a strong negation on the partially ordered set (W, ⪯).

For any two weighting vectors w = (w₁, w₂, ⋯ , w_n) and $w^{'} = (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'})$ , we have

$\sum_{j = 1}^{n} w_{j} x_{(j)} \leq \sum_{j = 1}^{n} w_{j}^{'} x_{(j)}, \forall (x_{1}, \dots, x_{n}) \in I^{n}$

if and only if

$\sum_{j = 1}^{n} w_{n - j + 1}^{'} x_{(j)} \leq \sum_{j = 1}^{n} w_{n - j + 1}^{'} x_{(j)}, \forall (x_{1}, \dots, x_{n}) \in I^{n}$ . More precisely, we have the following equation,

$\begin{matrix} \sum_{j = 1}^{n} w_{j} x_{(j)} \leq \sum_{j = 1}^{n} w_{j}^{'} x_{(j)}, \forall (x_{1}, \dots, x_{n}) \in I^{n} \\ \Leftrightarrow & (\forall k) \sum_{j = 1}^{k} w_{j} \leq \sum_{j = 1}^{k} w_{j}^{'} (i . e ., w ⪯ w^{'}) \end{matrix}$ (23) $\begin{matrix} \Leftrightarrow & (\forall k) \sum_{j = 1}^{k} w_{n - j + 1}^{'} \leq \sum_{j = 1}^{k} w_{n - j + 1} (i . e ., \neg w^{'} ⪯ \neg w) \\ \Leftrightarrow & \sum_{j = 1}^{n} w_{n - j + 1}^{'} x_{(j)} \leq \sum_{j = 1}^{n} w_{n - j + 1}^{'} x_{(j)}, \forall (x_{1}, \dots, x_{n}) \in I^{n} . \end{matrix}$

5.2 Conjunction and disjunction on OWA operators

Second, we define the conjunction and disjunction operations on W.

Definition 4. For any two weighting vectors w and w′, the conjunction and disjunction of w and w′ are, respectively, given by $w \land w^{'}$ (24) $\begin{matrix} = (w_{1} \land w_{1}^{'}, \dots, w_{n - 1} \land w_{n - 1}^{'}, 1 - (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'})), \\ w \lor w^{'} \end{matrix}$ (25) $= (1 - (\sum_{j = 2}^{n} w_{j} \land w_{j}^{'}), w_{2} \land w_{2}^{'}, \dots, w_{n} \land w_{n}^{'}) .$

Lemma 5. For any two weighting vectors w and w′,

w ∧ w′ and w ∨ w′ are weighting vectors,

w ∧ w′ ⪯ w and w ⪯ w ∨ w′.

Proof. (1) First, we have $w_{1} \land w_{1}^{'}, \dots, w_{n - 1} \land w_{n - 1}^{'} \in I$ and $0 \leq (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'}) \leq (\sum_{j = 1}^{n - 1} w_{j}) \leq 1$ , then $1 - (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'}) \in I$ . Second, $w_{1} \land w_{1}^{'} + \dots + w_{n - 1} \land w_{n - 1}^{'} + 1 - (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'}) = (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'}) + 1 - (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'}) = 1$ . Thus w ∧ w′ is a weighting vector. Similarly, we can show that w ∨ w′ is a weighting vector.

(2) First, we give the proof of w ∧ w′ ⪯ w. For any j = 1, 2, ⋯ , n - 1, $w_{j} \land w_{j}^{'} \leq w_{j}$ , then

$w_{1} \land w_{1}^{'} \leq w_{1}$

$w_{1} \land w_{1}^{'} + w_{2} \land w_{2}^{'} \leq w_{1} + w_{2}$

⋯

$w_{1} \land w_{1}^{'} + w_{2} \land w_{2}^{'} + \dots + w_{n - 1} \land w_{n - 1}^{'} \leq w_{1} + w_{2} + \dots + w_{n - 1}$ .

Thus w ∧ w′ ⪯ w.

Second, we give the proof of w ⪯ w ∨ w′. For any j = 1, 2, ⋯ , n - 1, $w_{j} \land w_{j}^{'} \leq w_{j}$ , then

$w_{n} \land w_{n}^{'} \leq w_{n}$

$w_{n} \land w_{n}^{'} + w_{n - 1} \land w_{n - 1}^{'} \leq w_{n} + w_{n - 1}$

⋯

$w_{n} \land w_{n}^{'} + w_{n - 1} \land w_{n - 1}^{'} + \dots + w_{2} \land w_{2}^{'} \leq w_{n} + w_{n - 1} + \dots + w_{2}$ .

Thus ¬ (w ∨ w′) ⪯ ¬ w. By Lemma 4 (iii) and (iv), we get w ⪯ w ∨ w′. □

Moreover, the negation ¬, conjunction ∧ and disjunction ∨ operations on W have the following properties.

Theorem 7. For any three weighting vectors

$\begin{matrix} w = (w_{1}, w_{2}, \dots, w_{n}), \\ w^{'} = (w_{1}^{'}, w_{2}^{'}, \dots, w_{n}^{'}), \\ w^{″} = (w_{1}^{″}, w_{2}^{″}, \dots, w_{n}^{″}), \end{matrix}$

the following properties hold:

w ∧ w = w, w ∨ w = w,

w ∧ w′ = w′ ∧ w, w ∨ w′ = w′ ∨ w,

w ∨ 1 = 1 andw ∧ 1 = (w₁, 0, . . . , 0, 1 - w₁),

where 1 = (1, 0, ⋯ , 0).

w ∧ 0 = 0 and w ∨ 0 = (1 - w_n, 0, . . . , 0, w_n),

where 0 = (0, 0, ⋯ , 1).

(w ∧ w′) ∧ w″ = w ∧ (w′ ∧ w″),

(w ∨ w′) ∨ w″ = w ∨ (w′ ∨ w″),

¬ (w ∧ w′) = ¬ w ∨ ¬ w′,

¬ (w ∨ w′) = ¬ w ∧ ¬ w′.

Proof. (1)-(5) are clear. We only give the proof of (6): $\begin{matrix} \neg w \lor \neg w^{'} \\ = & (w_{n}, w_{n - 1}, \dots, w_{1}) \lor (w_{n}^{'}, w_{n - 1}^{'}, \dots, w_{1}^{'}) \\ = & (1 - (\sum_{j = 1}^{n - 1} w_{j} \land w_{j}^{'}), w_{n - 1} \land w_{n - 1}^{'}, \dots, w_{1} \land w_{1}^{'}) \\ = & \neg (w \land w^{'}) . \end{matrix}$ $\begin{matrix} \neg w \land \neg w^{'} \\ = & (w_{n}, w_{n - 1}, \dots, w_{1}) \land (w_{n}^{'}, w_{n - 1}^{'}, \dots, w_{1}^{'}) \\ = & (w_{n} \land w_{n}^{'}, \dots, w_{2} \land w_{2}^{'}, 1 - (\sum_{j = 2}^{n} w_{j} \land w_{j}^{'})) \\ = & \neg (w \lor w^{'}) . □ \end{matrix}$

Example 4. If two OWA operator are given as follows:

$\begin{matrix} {OWA}_{1} (x_{1}, x_{2}, x_{3}) = 0.7 x_{(1)} + 0.1 x_{(2)} + 0.2 x_{(3)}, \\ {OWA}_{2} (x_{1}, x_{2}, x_{3}) = 0.6 x_{(1)} + 0.3 x_{(2)} + 0.1 x_{(3)} . \end{matrix}$ By the concepts of ∧ and ∨ in Definition 4, then

$\begin{matrix} ({OWA}_{1} \land {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \\ = 0.6 x_{(1)} + 0.1 x_{(2)} + 0.3 x_{(3)}, \\ ({OWA}_{1} \lor {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \\ = 0.8 x_{(1)} + 0.1 x_{(2)} + 0.1 x_{(3)} . \end{matrix}$ Since 0.6 < 0.7 and 0.6 + 0.1 < 0.7 + 0.1, then (0.6, 0.1, 0.3) ⪯ (0.7, 0.1, 0.2). Thus we have $({OWA}_{1} \land {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \leq {OWA}_{1} (x_{1}, x_{2}, x_{3}),$ for all (x₁, x₂, x₃) ∈ I³.

Since 0.7 < 0.8 and 0.7 + 0.1 < 0.8 + 0.1, then (0.7, 0.1, 0.2) ⪯ (0.8, 0.1, 0.1). Thus we have ${OWA}_{1} (x_{1}, x_{2}, x_{3}) \leq ({OWA}_{1} \lor {OWA}_{2}) (x_{1}, x_{2}, x_{3}),$ for all (x₁, x₂, x₃) ∈ I³.

By the concept of ¬ in Lemma 4, then

$\begin{matrix} \neg {OWA}_{1} (x_{1}, x_{2}, x_{3}) = 0.2 x_{(1)} + 0.1 x_{(2)} + 0.7 x_{(3)}, \\ \neg {OWA}_{2} (x_{1}, x_{2}, x_{3}) = 0.1 x_{(1)} + 0.3 x_{(2)} + 0.6 x_{(3)} . \end{matrix}$

Moreover, we have

$\begin{matrix} (\neg {OWA}_{1} \land \neg {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \\ = & 0.1 x_{(1)} + 0.1 x_{(2)} + 0.8 x_{(3)}, \\ (\neg {OWA}_{1} \lor \neg {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \\ = & 0.3 x_{(1)} + 0.1 x_{(2)} + 0.6 x_{(3)}, \end{matrix}$ and

$\begin{matrix} \neg ({OWA}_{1} \land {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \\ = 0.3 x_{(1)} + 0.1 x_{(2)} + 0.6 x_{(3)}, \\ \neg ({OWA}_{1} \lor {OWA}_{2}) (x_{1}, x_{2}, x_{3}) \\ = 0.1 x_{(1)} + 0.1 x_{(2)} + 0.8 x_{(3)} . \end{matrix}$

Clearly, (¬ OWA₁ ∧ ¬ OWA₂) = ¬ (OWA₁ ∨ OWA₂) and (¬ OWA₁ ∨ ¬ OWA₂) = ¬ (OWA₁ ∧ OWA₂)

Theorem 8. W, ∧ , ∨ , ¬ , 0, 1) is a bounded De Morgan lattice.

Theorem 9. (W, ∧ , ∨ , ¬ , 0, 1) is not a distributive lattice.

Proof. Let w = (0.1, 0.3, 0.6), w′ = (0.4, 0.1, 0.5), and w″ = (0.2, 0.7, 0.1), then $\begin{matrix} \begin{matrix} (w \land w^{'}) \lor (w \land w^{″}) \\ = ((0.1, 0.3, 0.6) \land (0.4, 0.1, 0.5)) \\ \lor ((0.1, 0.3, 0.6) \land (0.2, 0.7, 0.1)) \\ = (0.1, 0.1, 0.8) \lor (0.1, 0.3, 0.6) \\ = (0.3, 0.1, 0.6) . \end{matrix} \end{matrix}$ and $\begin{matrix} \begin{matrix} w \land (w^{'} \lor w^{″}) \\ = (0.1, 0.3, 0.6) \land ((0.4, 0.1, 0.5) \lor (0.2, 0.7, 0.1)) \\ = (0.1, 0.3, 0.6) \land (0.8, 0.1, 0.1) \\ = (0.1, 0.1, 0.8) . \end{matrix} \end{matrix}$

Thus (w ∧ w′) ∨ (w ∧ w″) ≠ w ∧ (w′ ∨ w″). □

From above Theorems, we can see that (W, ∧ , ∨ , ¬ , 0, 1) does not satisfy the condition of distributivity. So we consider a weaker condition, as the modularity condition is weaker than the distributivity condition.

Theorem 10. (W, ∧ , ∨ , ¬ , 0, 1) does not satisfy the condition of modularity.

Proof. Let w = (0.1, 0.3, 0.6), w′ = (0.4, 0.1, 0.5), and w″ = (0.2, 0.7, 0.1), then (w ∧ w′) ∨ (w ∧ w″) = (0.3, 0.1, 0.6) from above Theorem. Moreover,

$\begin{array}{l} w \land (w^{'} \lor (w \land w^{″})) \\ = (0.1, 0.3, 0.6) \\ \land ((0.4, 0.1, 0.5) \lor ((0.1, 0.3, 0.6) \land (0.2, 0.7, 0.1))) \\ = (0.1, 0.3, 0.6) \land ((0.4, 0.1, 0.5) \lor (0.1, 0.3, 0.6)) \\ = (0.1, 0.3, 0.6) \land (0.4, 0.1, 0.5) \\ = (0.1, 0.1, 0.8) . \end{array}$

Thus (w ∧ w′) ∨ (w ∧ w″) ≠ w ∧ (w′ ∨ (w ∧ w″)). □

6 Similarity of OWA operators

In this section we describe the degree of similarity of OWA operators based on the ordering ⪯.

Definition 5. Let W be the set of all weighting vectors of dimension n, a function S : W × W → [0, 1] is called a similarity measure of W, if for any w, w′, w″ ∈ W,

0 ≤ S (w, w′) ≤1,

S (w, w′) =1 if and only if w = w′,

S (w, w′) = S (w′, w),

If w ⪯ w′ ⪯ w″ then S (w, w″) ≤ S (w, w′) and S (w, w″) ≤ S (w′, w″).

Based on above definition of similarity measure of W, we give the following function: for any w, w′, ∈ W,

$S (w, w^{'}) = \sum_{k = 1}^{n - 1} \frac{1 - | \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} |}{n - 1} .$ (26)

Theorem 11. The function defined in Eq. 26 is a similarity measure of W.

Proof. (S1) Since $| \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} | \in [0, 1]$ then S (w, w′) ∈ [0, 1]. Moreover, if w^* = (1, 0, . . . , 0) and w_* = (0, 0, . . . , 1), then $S (w^{*}, w_{*}) = \sum_{k = 1}^{n - 1} \frac{1 - | 1 - 0 |}{n - 1} = 0$ . Thus the degree of similarity between Maximum operator and Minimum operator is zero.

(S2) If w = w′ then $\sum_{j = 1}^{k} w_{j} = \sum_{j = 1}^{k} w_{j}^{'}$ for all k = 1, 2, . . . , n - 1. Then $S (w, w^{'}) = \sum_{k = 1}^{n - 1} \frac{1 - 0}{n - 1} = 1$ . If w ≠ w′, then there exist k such that $\sum_{j = 1}^{k} w_{j} \neq \sum_{j = 1}^{k} w_{j}^{'}$ . Then $| \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} | > 0$ .

(S3) Since $| \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} | = | \sum_{j = 1}^{k} w_{j}^{'} - \sum_{j = 1}^{k} w_{j} |$ . Then $\begin{matrix} \begin{matrix} S (w, w^{'}) & = \sum_{k = 1}^{n - 1} \frac{1 - | \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} |}{n - 1} \\ = \sum_{k = 1}^{n - 1} \frac{1 - | \sum_{j = 1}^{k} w_{j}^{'} - \sum_{j = 1}^{k} w_{j} |}{n - 1} \\ = S (w^{'}, w) . \end{matrix} \end{matrix}$

(S4) If w ⪯ w′ ⪯ w″ then $0 \leq \sum_{j = 1}^{k} w_{j} \leq \sum_{j = 1}^{k} w_{j}^{'} \leq \sum_{j = 1}^{k} w_{j}^{″} \leq 1$ for all k = 1, 2, . . . , n - 1. Then $| \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} | \leq | \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} |$ for all k = 1, 2, . . . , n - 1. Then $\begin{matrix} \begin{matrix} S (w, w^{'}) & = \sum_{k = 1}^{n - 1} \frac{1 - | \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{'} |}{n - 1} \\ \geq \sum_{k = 1}^{n - 1} \frac{1 - | \sum_{j = 1}^{k} w_{j} - \sum_{j = 1}^{k} w_{j}^{″} |}{n - 1} \\ = S (w, w^{″}) . \end{matrix} \end{matrix}$ Similarly, we can obtain S (w′, w″) ≥ S (w, w″) □

Example 5. Let w = (0.4, 0.5, 0.1) w′ = (0.5, 0.4, 0.1) and w″ = (0.6, 0.2, 0.2). It is clear that w ⪯ w′ ⪯ w″. $\begin{matrix} \begin{matrix} S (w, w^{'}) \\ = \frac{1 - | 0.4 - 0.5 |}{3 - 1} + \frac{1 - | (0.4 + 0.5) - (0.5 + 0.4) |}{3 - 1} = 0.45, \\ S (w, w^{″}) \\ = \frac{1 - | 0.4 - 0.6 |}{3 - 1} + \frac{1 - | (0.4 + 0.5) - (0.6 + 0.2) |}{3 - 1} = 0.85, \\ S (w^{'}, w^{″}) \\ = \frac{1 - | 0.5 - 0.6 |}{3 - 1} + \frac{1 - | (0.5 + 0.4) - (0.6 + 0.2) |}{3 - 1} = 0.9 . \end{matrix} \end{matrix}$ Clearly, we have S (w, w″) ≤ S (w, w′) and S (w, w″) ≤ S (w′, w″).

7 Conclusion

In this paper, the ordering problem of OWA operators has been investigated, and our conclusion can be summarized as following:

Both the coordinatewise order ≤_c and lexicographic order ≤_l are not suitable for WA and OWA operators.

Order of two OWA operators in Definition 2 is equivalent to ⪯ of their associated weighting vectors, i.e.,

$\begin{matrix} \sum_{j = 1}^{n} w_{j} x_{(j)} \leq \sum_{j = 1}^{n} w_{j}^{'} x_{(j)}, \forall (x_{1}, \dots, x_{n}) \in I^{n} \\ \Leftrightarrow (\forall k) \sum_{j = 1}^{k} w_{j} \leq \sum_{j = 1}^{k} w_{j}^{'} . \end{matrix}$

⪯ is a relationship between ≤_c and ≤_l, i.e., $w \leq_{l} w^{'} \Rightarrow w ⪯ w^{'} \Rightarrow w \leq_{l} w^{'},$ for any w, w′ ∈ W.

(W, ∧ , ∨ , ¬ , 0, 1) is a bounded De Morgan lattice.

(W, ⪯), i.e., (OWA, ≤) is a partially ordered set, but not a totally ordered set.

(W, ∧ , ∨ , ¬) is not a distributive lattice. Moreover, it does not satisfy the condition of modularity.

A similarity measure of W is given based on the ordering ≤.

Our conclusion is contributing to a deeper understanding on the ordering problem of OWA operators. Meanwhile, we realise that the ordering of OWA operators is only considered on [0, 1] in this paper. A further discussion about the ordering of OWA operators on other aspects, such as convex poset [8] or complete lattices [11], are also meaningful in the future research. Other problems, such as how to define t-norms for OWA operators, are also worthwhile for investigating based on our proposed order of OWA operators.

Footnotes

Acknowledgments

This project was supported by the National Science Foundation of China under Grant No. 62006168 and Zhejiang Provincial Natural Science Foundation of China under Grant No. LQ21A010001.

References

Amarante

, Mm-OWA: A generalization of OWA operators, IEEE Trans Fuzzy Syst 26(4) (2018), 2099–2106.

Baczynski

and Jayaram

, Fuzzy implications, Heidelberg: Springer, 2008.

Davey

B.A.

and Priestley

H.A.

, Introduction to lattices and order, Cambridge university press, 2002.

Emrouznejad

and Marra

, Ordered weighted averaging operators 1988–2014: A citation-based literature survey, Int J Intell Syst 29 (2014), 994–1014.

Gratzer

, General Lattice Theory, New York: Academic, 1978.

Jin

and Mesiar

, The metric space of ordered weighted average operators with distance based on accumulated entries, Int J Intell Syst 32(7) (2017), 665–675.

Jin

, Kalina

, Mesiar

and Qian

, On some properties and comparative analysis for different OWA Monoids, Int J Intell Syst 32(11) (2017), 1115–1135.

Jin

, Mesiar

and Yager

, Ordered Weighted Averaging Aggregation on Convex Poset, IEEE Trans Fuzzy Syst 27(3) (2019), 612–617.

Klement

E.P.

, Mesiar

and Pap

, Triangular Norms, Kluwer, Dordrecht, 2000.

10.

Llamazares

, Generalizations of Weighted Means and OWA Operators by Using Unimodal Weighting Vectors, IEEE Trans Fuzzy Syst 28(9) (2020), 1961–1970.

11.

Lizasoain

and Moreno

, OWA operators defined on complete lattices, Fuzzy Sets Syst 224 (2013), 36–52.

12.

Mesiar

, Stupnanova

and Yager

R.R.

, Generalizations of OWAOperators, IEEE Trans Fuzzy Syst 23(6) (2015), 2154–2162.

13.

Mesiar

, Sipeky

, Gupta

and Jin

, Aggregation of OWA operators, IEEE Trans Fuzzy Syst 26(1) (2017), 284–291.

14.

Yager

R.R.

, On ordered weighted averaging aggregation operators in multi-criteria decision making, IEEE Trans Syst Man Cyber net B 18(1) (1998), 183–190.

15.

Yager

R.R.

, Families of OWA operators, Fuzzy sets and systems 59(2) (1993), 125–148.

16.

Yager

R.R.

and Filev

D.P.

, Induced ordered weighted averaging operators, IEEE Trans Syst Man Cyber net B 29(2) (1999), 141–150.