TODIM method for evaluating the service quality of boutique tourist scenic spot with 2-tuple linguistic information

Abstract

Recently, the TODIM (an acronym in Portuguese for Interactive Multi-criteria Decision Making) approach, which can characterize the decision makers’ psychological behaviors under risk, has been introduced to handle multiple attribute decision making (MADM) problems. Moreover, 2-tuple linguistic term set is an effective tool for depicting uncertainty of the MADM problems. In this paper, we will extend the TODIM method to the MADM with the 2-tuple linguistic information. Firstly, the definition and distance of 2-tuple linguistic information are briefly introduced, and the steps of the classical TODIM method for MADM problems are presented. Then, on the basis of the classical TODIM method, the extended TODIM method is proposed to deal with MADM problems in which the attribute values are in the 2-tuple linguistic information, and its significant characteristic is that it can fully consider the decision makers’ bounded rationality which is a real action in decision making. Finally, a numerical example for evaluating the service quality of boutique tourist scenic spot is proposed to verify the developed approach and its practicality and effectiveness.

Keywords

Multiple attribute decision making 2-tuple linguistic variables TODIM method prospect theory service quality

1 Introduction

Multi-criteria group decision-making (MCGDM) problems are wide-spread in real-life decision-making situations, especially with the increasing complexity of the socio-economic environment [1, 2]. In reality, decision-making information is usually uncertain or fuzzy, due to the complexity of things and in recognition of the limitations of decision makers. In order to thoroughly describe fuzzy information, Herrera and Martinez [3] proposed the 2-tuple linguistic model, composed of a linguistic term and a real number, to represent assessment information in a way that can effectively avoid information loss. Consequently, 2-tuple linguistic MCGDM problems have captured the attention of many researchers in recent years [4 –6]. 2-tuple linguistic information can describe the fuzziness of decision-making information, while it seems imperfect and inaccurate to deal with information in terms of randomness. In fact, randomness and fuzziness are the most important and fundamental of all kinds of uncertainty [7, 8]. For instance, for a linguistic decision-making problem, decision maker A may think that 75% fulfillment of a task is “good”, while decision maker B may hold that less than 80% fulfillment of the same task cannot be considered to be “good” with the same linguistic term scale. So in such a way, when considering the degree of certainty of an element belonging to a qualitative concept in a specific universe, it is more feasible to allow a stochastic disturbance of the membership degree encircling a determined central value than to allow a fixed number [9, 10]. Fortunately, the cloud model can easily overcome this weakness and make decision-making processes more realistic. The cloud model, which is a quantitative and qualitative uncertainty conversion model proposed by Professor Deyi Li based on traditional fuzzy set theory and probability statistics [11], has distinct advantages in terms of dealing with vague and random decision-making information. It not only easily characterizes the concept of uncertainty in the natural language, but also reflects the intrinsic connection between randomness and fuzziness. Therefore, the cloud model can be used to depict the randomness of 2-tuple linguistic information. To the best of the authors’ knowledge, however, research on converting 2-tuple linguistic variables into clouds has not been reported in the existing literature. The decision information in some practical MAGDM situations may be unquantifiable due to its nature, or cannot be precisely assessed in a quantitative form, but may be assessed in a qualitative one. Thus, it may take the form of linguistic variables [12], such as “poor”, “fair”, and “very good”. To utilize linguistic variables, a pre-defined linguistic assessment set is needed. Unfortunately, the traditional linguistic assessment set is discrete. So in many cases, the decision information provided by DMs may not match any of the original linguistic phrases in the linguistic assessment sets, resulting in loss of information. To overcome these limitations, Herrera and Martinez [3] introduced the 2-tuple linguistic representation model of which the significant advantage is to be continuous in its domain. Therefore, it can express any counting of information in the universe of the discourse. Recently, the 2-tuple linguistic model has been widely studied. Dong, et al. [13] developed two different models based on linguistic 2-tuples to address term sets that are not uniformly and symmetrically distributed. Truck [14] stressed a comparison between the 2-tuple semantic model and the 2-tuple symbolic model, and then proved that links can be made between them. Zhu et al. [15] utilized two 2-tuples in a 2-dimension linguistic lattice implication algebra to represent a 2-dimension linguistic label for more precise computing and aggregating 2-dimension linguistic information. Xu et al. [16] proposed a four-way procedure to estimate missing preference values when dealing with acceptable incomplete 2-tuple fuzzy linguistic preference relations in group decision-making. Gong et al. [17] established an optimization model of group consensus of 2-tuple linguistic preferential relations. In addition, the 2-tuple linguistic variable has been applied to many practical MCGDM problems such as supplier selection [18, 19], material selection [20], site selection [21], emergency response capacity evaluation [22] and in-flight service quality evaluation [23].

In this paper, we will extend the TODIM method to the MADM with the 2-tuple linguistic information. Firstly, the definition and distance of 2-tuple linguistic information are briefly introduced, and the steps of the classical TODIM method for MADM problems are presented. Then, on the basis of the classical TODIM method, the extended TODIM method is proposed to deal with MADM problems in which the attribute values are in the 2-tuple linguistic information, and its significant characteristic is that it can fully consider the decision makers’ bounded rationality which is a real action in decision making. Furthermore, we extend the above results to interval neutrosophic environment. Finally, a numerical example for evaluating the service quality of boutique tourist scenic spot is proposed to verify the developed approach and its practicality and effectiveness.

2 Preliminaries

Herrera [24, 25] first introduced the 2-tuple fuzzy linguistic approach for overcoming the drawback of the classical computational models, which include the semantic model and symbolic model. The 2-tuple linguistic model is a kind of new information processing method. It takes 2-tuple to represent linguistic assessment information and carry out operation. The basic concept of linguistic 2-tuple is symbolic translation. The 2-tuple linguistic representation and computational model has received more and more attention since its appearance.

In the following, we shall introduce the definition of the 2-tuple linguistic representation and computational model.

Let S ={ s_i |i = 1, 2, ⋯ , t } be a linguistic term set with odd cardinality. Any label, s_i represents a possible value for a linguistic variable, and it should satisfy the following characteristics [24, 25]:

(1) The set is ordered: s_i > s_j, if i > j; (2) Max operator: max(s_i, s_j) = s_i, if s_i ≥ s_j; (3) Min operator: min(s_i, s_j) = s_i, if s_i ≤ s_j. For example, S can be defined as $\begin{matrix} S = {s_{1} = extremely poor (EP), s_{2} = very poor (VP), \\ s_{3} = poor (P), s_{4} = medium (M), s_{5} = good (G), \\ s_{6} = very good (VG), s_{7} = extremely good (EG)} \end{matrix}$

Herrera and Martinez [24, 25] developed the 2-tuple fuzzy linguistic representation model based on the concept of symbolic translation. It is used for representing the linguistic assessment information by means of a 2-tuple (s_i, α_i), where s_i is a linguistic label from predefined linguistic term set S and α_i is the value of symbolic translation, and α_i ∈ [- 0.5, 0.5) .

Definition 1. [24, 25] Let β be the result ofan aggregation of the indices of a set of labels assessed in a linguistic term set S, i.e., the result of a symbolic aggregation operation, β ∈ [1, t], being t the cardinality of S. Let i = round (β) and α = β - i be two values, such that, i ∈ [1, t] and α ∈ [- 0.5, 0.5) then α is called a symbolic translation.

Definition 2. [24, 25] Let S ={ s₁, s₂, ⋯ , s_t } be a linguistic term set and β ∈ [1, t] is a number value representing the aggregation result of linguistic symbolic. Then the function Δ used to obtain the 2-tuple linguistic information equivalent to β is defined as: $Δ : [1, t] \to S \times [- 0.5, 0.5)$ (1) $Δ (β) = {\begin{matrix} s_{i}, i = round (β) \\ α = β - i, α \in [- 0.5, 0.5) \end{matrix}$ (2) where round(.) is the usual round operation, s_i has the closest index label to β and α is the value of the symbolic translation.

Definition 3. [24, 25] Let S ={ s₁, s₂, ⋯ , s_t } be a linguistic term set and (s_i, α_i) be a 2-tuple. There is always a function Δ^-1 can be defined, such that, from a 2-tuple (s_i, α_i) it return its equivalent numerical value β ∈ [1, t] ⊂ R, which is. $Δ^{- 1} : S \times [- 0.5, 0.5) \to [1, t]$ (3) $Δ^{- 1} (s_{i}, α) = i + α = β$ (4)

From Definitions 1 and 2, we can conclude that the conversion of a linguistic term into a linguistic 2-tuple consists of adding a value 0 as symbolic translation: $Δ (s_{i}) = (s_{i}, 0)$ (5)Definition 4. [24, 25] Let (s_k, a_k) and (s_l, a_l) be two 2-tuple, they should have the following properties:

If k < l then (s_k, a_k) is smaller than (s_l, a_l);

If k = l then

if a_k = a_l then (s_k, a_k), (s_l, a_l) represents the same information;

if a_k < a_l then (s_k, a_k) is smaller than (s_l, a_l); if a_k > a_l then (s_k, a_k) is bigger than (s_l, a_l).

Definition 5. Let δ₁ = (r₁, a₁) and δ₂ = (r₂, a₂) be two 2-tuple, then we can get the normalized Hamming distance: $d (δ_{1}, δ_{2}) = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) |$ (5)

Theorem 1. Let δ₁ = (r₁, a₁) , δ₂ = (r₂, a₂) and δ₃ = (r₃, a₃) be three 2-tuple, the Hamming distance d (δ₁, δ₂) has the following properties:

if d (δ₁, δ₂) = 0, then δ₁ = δ₂;

d (δ₁, δ₂) = d (δ₂, δ₁);

d (δ₁, δ₂) + d (δ₂, δ₃) ≥ d (δ₁, δ₃);

0 ≤ d (δ₁, δ₂) ≤ 1.

Proof. (P1) if d (l₁, l₂) = 0, then l₁ = l₂ $\begin{matrix} d (δ_{1}, δ_{2}) = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) | \\ \Rightarrow | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) | = 0 \\ \Rightarrow Δ^{- 1} (r_{1}, a_{1}) = Δ^{- 1} (r_{2}, a_{2}) \Rightarrow δ_{1} = δ_{2} \end{matrix}$

That means δ₁ = δ₂, so

if d (δ₁, δ₂) =0, then δ₁ = δ₂ is right.

d (δ₁, δ₂) = d (δ₂, δ₁) $d (δ_{1}, δ_{2}) = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) |$ $= \frac{1}{t} | Δ^{- 1} (r_{2}, a_{2}) - Δ^{- 1} (r_{1}, a_{1}) | = d (δ_{2}, δ_{1})$

So we complete the proof.□

d (δ₁, δ₂) = d (δ₂, δ₁) is hold.

d (δ₁, δ₂) + d (δ₂, δ₃) ≥ d (δ₁, δ₃)

\begin{matrix} d (δ_{1}, δ_{3}) = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{3}, a_{3}) | \\ = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) + Δ^{- 1} (r_{2}, a_{2}) \\ - Δ^{- 1} (r_{3}, a_{3}) | \\ \leq \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) | + | Δ^{- 1} (r_{2}, a_{2}) \\ - Δ^{- 1} (r_{3}, a_{3}) | \\ = d (δ_{1}, δ_{2}) + d (δ_{2}, δ_{3}) \end{matrix}

So we complete the proof. (P3) d (δ₁, δ₂) + d (δ₂, δ₃) ≥ d (δ₁, δ₃) is hold. $(P 4) 0 \leq d (δ_{1}, δ_{2}) \leq 1$ $\begin{matrix} d (δ_{1}, δ_{2}) = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) | \\ 0 \leq Δ^{- 1} (r_{1}, a_{1}) \leq t, 0 \leq Δ^{- 1} (r_{2}, a_{2}) \leq t \\ \Rightarrow 0 \leq | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) | \leq t \\ \Rightarrow 0 \leq \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) | \leq 1 \\ \Rightarrow 0 \leq d (δ_{1}, δ_{2}) \leq 1 \end{matrix}$

So we complete the proof. (P4) 0 ≤ d (δ₁, δ₂) ≤ 1 is hold.

3 The origin TODIM method

TODIM method [26] which based on prospect theory (PT) is to consider the subjectivity of DM’s behaviors, which provides the dominance of each alternative over others by some particular operation formulas; can be more reasonable and scientific in the application of MAGDM problems [27 –29].

Assume that {l₁, l₂, … l_m } be a group of alternatives, {c₁, c₂, … c_n } be a list of criteria with weighting vector be {w₁, w₂, … w_n }, thereby satisfying w_i ∈ [0, 1] and $\sum_{i = 1}^{n} w_{i} = 1$ . Construct a decision matrix l = [d_ij] _m×n where d_ij means the estimate results of the alternative l_i (i = 1, 2, …, m) based on the criterion c_j (j = 1, 2, …, n). Suppose that w_jk = w_j/ - w_k be relative weight of c_j to c_t where w_k = max(w_j) k, j = 1, 2, …, n. The traditional TODIM method decision making steps can be summarized as follows:

Step 1. Normalize l = [d_ij] _m×n into $l^{'} = {[d_{ij}^{'}]}_{m \times n}$ .

For benefit attributes: $l_{ij}^{'} = {(r_{ij}, a_{ij})}^{'} = (r_{ij}, a_{ij});$

For cost attributes: $l_{ij}^{'} = {(r_{ij}, a_{ij})}^{'} = t - (r_{ij}, a_{ij}) .$

Step 2. Calculate the dominance degree of l_i over each alternative l_t based on c_j.let β be the attenuation factor of the losses. Then $δ (l_{i}, l_{t}) = \sum_{j = 1}^{n} l_{j} (l_{i}, l_{t}) (i, t = 1, 2, \dots, m)$ (6)

$\begin{matrix} φ_{j} (l_{i}, l_{t}) = \\ {\begin{matrix} \sqrt{w_{jk} (d_{ij} - d_{tj}) / . - \sum_{j = 1}^{n} w_{jk}} if d_{ij} - d_{tj} > 0 \\ 0 if d_{ij} - d_{tj} = 0 \\ - \frac{1}{β} \sqrt{(\sum_{j = 1}^{n} w_{jk}) (d_{ij} - d_{tj}) / . - w_{jk}} if d_{ij} - d_{tj} < 0 \end{matrix} \end{matrix}$ (7) where φ_j (l_i, l_t) (d_ij - d_tj > 0) means gain and φ_j (l_i, l_t) (d_ij - d_tj < 0 ) indicates loss.

Step 3. Compute the overall value of l_i by the formula (8): $δ (l_{i}) = \frac{\sum_{t = 1}^{m} δ (l_{i}, l_{t}) - min_{i} {\sum_{t = 1}^{m} δ (l_{i}, l_{t})}}{max_{i} {\sum_{t = 1}^{m} δ (l_{i}, l_{t})} - min_{i} {\sum_{t = 1}^{m} δ (l_{i}, l_{t})}}$ (8)

Step 4. To choose the best alternative by rank the values of δ (l_i), the alternative with maximum value is the best choice.

4 The TODIM method with 2-tuple linguistic information

Assume that {l₁, l₂, … l_m } be a group of alternatives, {d₁, d₂, … d_λ } be a list of experts with weighting vector be {v₁, v₂, … v_t }, and {c₁, c₂, … c_n } be a list of criteria with weighting vector be {w₁, w₂, … w_n }, thereby satisfying w_i ∈ [0, 1] , v_i ∈ [0, 1] and $\sum_{i = 1}^{n} w_{i} = 1, \sum_{i = 1}^{t} v_{i} = 1$ . Construct a decision matrix $l^{λ} = {[l_{ij}^{λ}]}_{m \times n}$ where $l_{ij}^{λ} = (r_{ij}^{λ}, a_{ij}^{λ})$ means the estimate results of the alternative l_i (i = 1, 2, …, m) based on the criterion c_j (j = 1, 2, …, n) by expert d^λ. Let w_jk = w_j/ - w_k (0 ≤ w_jk ≤ 1) be relative weight of c_j to c_t where w_k = max(w_j) (k, j = 1, 2, …, n).

Consider both the 2-tuple linguistic information and traditional TODIM method which based on prospect theory (PT), we try to propose a 2-tuple linguistic TODIM method to solve MAGDM problems effectively. The model can be depicted as follows:

Step 1. Normalize $φ^{λ} = {[r_{ij}^{λ}]}_{m \times n}$ into ${(φ^{'})}^{λ} = {[{(r_{ij}^{'})}^{λ}]}_{m \times n}$ , and calculate the value of w_jk = w_j/ - w_k (0 ≤ w_jk ≤ 1), w_k = max(w_j) (k, j = 1, 2, …, n) .

For benefit attributes: $l_{ij}^{'} = {(r_{ij}, a_{ij})}^{'} = (r_{ij}, a_{ij})$ ;

For cost attributes: $l_{ij}^{'} = {(r_{ij}, a_{ij})}^{'} = t - (r_{ij}, a_{ij})$ .

Step 2. According to the computing results of relative weight w_jk, we can calculate the dominance degree of $l_{i}^{λ}$ over each alternative $l_{t}^{λ}$ based on c_j by expert d_λ. let β be the attenuation factor of the losses. Then

$\begin{matrix} φ_{j}^{λ} (l_{i}, l_{t}) = \\ {\begin{matrix} \sqrt{w_{jk} d (l_{ij}^{λ} - l_{tj}^{λ}) / . - \sum_{j = 1}^{n} w_{jk}} if l_{ij}^{λ} - l_{tj}^{λ} > 0 \\ 0 if l_{ij}^{λ} - l_{tj}^{λ} = 0 \\ - \frac{1}{β} \sqrt{(\sum_{j = 1}^{n} w_{jk}) d (l_{ij}^{λ} - l_{tj}^{λ}) / . - w_{jk}} if l_{ij}^{λ} - l_{tj}^{λ} < 0 \end{matrix} \end{matrix}$ (9) $d (l_{1}, l_{2}) = \frac{1}{t} | Δ^{- 1} (r_{1}, a_{1}) - Δ^{- 1} (r_{2}, a_{2}) |$ (10) where $φ_{j}^{λ} (l_{i}, l_{t}) (l_{ij}^{λ} - l_{tj}^{λ} > 0)$ means gain and $φ_{j}^{λ} (l_{i}, l_{t}) (l_{ij}^{λ} - l_{tj}^{λ} < 0)$ indicates loss, and based on Definition 5, $d (l_{ij}^{λ} - l_{tj}^{λ})$ means the normalized Hamming distance between $l_{ij}^{λ}$ and $l_{tj}^{λ}$ .

Next we construct a matrix model of dominance degree $φ_{j}^{λ} = {[φ_{j}^{λ} (l_{i}, l_{t})]}_{m \times m}$ under criteria c_j by expert d_λ to express the Equation (9) more clearly. $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \dots & l_{m} \end{matrix} \\ φ_{j}^{λ} (l_{i}, l_{t}) = \begin{matrix} l_{1} \\ l_{2} \\ ⋮ \\ l_{m} \end{matrix} [\begin{matrix} 0 & φ_{j}^{λ} (l_{1}, l_{2}) & \dots & φ_{j}^{λ} (l_{1}, l_{m}) \\ φ_{j}^{λ} (l_{2}, l_{1}) & 0 & \dots & φ_{j}^{λ} (l_{2}, l_{m}) \\ ⋮ & ⋮ & \dots & ⋮ \\ φ_{j}^{λ} (l_{m}, l_{1}) & φ_{j}^{λ} (l_{i}, l_{t}) & \dots & 0 \end{matrix}], \\ j = 1, 2, \dots, n \end{matrix}$ (11)

Step 3. Compute overall dominance degree $φ_{j}^{λ} = {[φ_{j}^{λ} (l_{i}, l_{t})]}_{m \times m}$ to get the matrix model φ^λ = [φ^λ (l_i, l_t)] _m×m. $φ^{λ} (l_{i}, l_{t}) = \sum_{j = 1}^{n} φ_{j}^{λ} (l_{i}, l_{t}) (i, t = 1, 2, \dots, m)$ (12) $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \dots & l_{m} \end{matrix} \\ φ^{λ} (l_{i}, l_{t}) = \begin{matrix} l_{1} \\ l_{2} \\ ⋮ \\ l_{m} \end{matrix} [\begin{matrix} 0 & φ^{λ} (l_{1}, l_{2}) & \dots & φ^{λ} (l_{1}, l_{m}) \\ φ^{λ} (l_{2}, l_{1}) & 0 & \dots & φ^{λ} (l_{2}, l_{m}) \\ ⋮ & ⋮ & \dots & ⋮ \\ φ^{λ} (l_{m}, l_{1}) & φ^{λ} (l_{i}, l_{t}) & \dots & 0 \end{matrix}] \end{matrix}$ (13)

Step 4. Calculate the overall dominance δ (l_i, l_t) based on the expert weighting vector {v₁, v₂, … v_t } and the results of Equation (13). $δ (l_{i}, l_{t}) = \sum_{j = 1}^{λ} v_{λ} φ^{λ} (l_{i}, l_{t}) (i, t = 1, 2, \dots, m)$ (14)

The overall dominance δ (l_i, l_t) matrix can be constructed by formula (25) as follows: $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \dots & l_{m} \end{matrix} \\ δ_{j} (l_{i}, l_{t}) = \begin{matrix} l_{1} \\ l_{2} \\ ⋮ \\ l_{m} \end{matrix} [\begin{matrix} 0 & δ_{j} (l_{1}, l_{2}) & \dots & δ_{j} (l_{1}, l_{m}) \\ δ_{j} (l_{2}, l_{1}) & 0 & \dots & δ_{j} (l_{2}, l_{m}) \\ ⋮ & ⋮ & \dots & ⋮ \\ δ_{j} (l_{m}, l_{1}) & δ_{j} (l_{i}, l_{t}) & \dots & 0 \end{matrix}], \\ j = 1, 2, \dots, n \end{matrix}$ (15)

Step 5. Compute the overall value of l_i by the formula (16): $δ (l_{i}) = \frac{\sum_{t = 1}^{m} δ (l_{i}, l_{t}) - min_{i} {\sum_{t = 1}^{m} δ (l_{i}, l_{t})}}{max_{i} {\sum_{t = 1}^{m} δ (l_{i}, l_{t})} - min_{i} {\sum_{t = 1}^{m} δ (l_{i}, l_{t})}}$

Step 6. To choose the best alternative by rank the values of δ (l_i), the alternative with maximum value is the best choice.

5 Numerical example

5.1 Calculating steps based on MAGDM problems

In this section, we provide a numerical example for evaluating the service quality of boutique tourist scenic spot by using 2-tuple linguistic TODIM method. Assume that five possible boutique tourist scenic spots l_i (i = 1, 2, 3, 4, 5) to be selected and four criteria to assess these boutique tourist scenic spots: ding172c₁ is the human factors in construction projects; ding173c₂ is the building materials and equipment factors; ding174c₃ is the management factors; ding175c₄ is the environmental factors. The five possible boutique tourist scenic spots φ_i (i = 1, 2, 3, 4, 5) are to be evaluated with 2-tuple linguistic information with the four criteria by three experts d^k (criteria weight ω₁ = (0.4, 0.1, 0.3, 0.2), experts weight ω₂ = (0.3, 0.4, 0.3) .), which are listed in Tables 1 –3.

Table 1
2-tuple linguistic information decision matrix by d¹

c₁ c₂ c₃ c₄

l ₁ (s₄,0) (s₂,0) (s₁,0) (s₃,0)

l ₂ (s₂,0) (s₃,0) (s₁,0) (s₂,0)

l ₃ (s₅,0) (s₂,0) (s₁,0) (s₄,0)

l ₄ (s₃,0) (s₄,0) (s₅,0) (s₃,0)

l ₅ (s₂,0) (s₄,0) (s₁,0) (s₂,0)

	c₁	c₂	c₃	c₄
l ₁	(s₄,0)	(s₂,0)	(s₁,0)	(s₃,0)
l ₂	(s₂,0)	(s₃,0)	(s₁,0)	(s₂,0)
l ₃	(s₅,0)	(s₂,0)	(s₁,0)	(s₄,0)
l ₄	(s₃,0)	(s₄,0)	(s₅,0)	(s₃,0)
l ₅	(s₂,0)	(s₄,0)	(s₁,0)	(s₂,0)

Table 2

2-tuple linguistic information decision matrix by d²

	c₁	c₂	c₃	c₄
l ₁	(s₂,0)	(s₄,0)	(s₂,0)	(s₃,0)
l ₂	(s₂,0)	(s₁,0)	(s₃,0)	(s₄,0)
l ₃	(s₃,0)	(s₁,0)	(s₄,0)	(s₂,0)
l ₄	(s₄,0)	(s₃,0)	(s₂,0)	(s₅,0)
l ₅	(s₂,0)	(s₁,0)	(s₂,0)	(s₄,0)

Table 3

2-tuple linguistic information decision matrix by d³

	c₁	c₂	c₃	c₄
l ₁	(s₁,0)	(s₂,0)	(s₃,0)	(s₂,0)
l ₂	(s₂,0)	(s₄,0)	(s₂,0)	(s₁,0)
l ₃	(s₃,0)	(s₄,0)	(s₁,0)	(s₁,0)
l ₄	(s₂,0)	(s₅,0)	(s₄,0)	(s₄,0)
l ₅	(s₃,0)	(s₃,0)	(s₂,0)	(s₁,0)

Step 1. Calculate the value of w_jk = w_j/w_k(0 ≤ w_jk ≤ 1), w_k = max(w_j) (k, j = 1, 2, …, n). $\begin{matrix} w_{k} = max (0.4, 0.1, 0.3, 0.2) = 0.4 \\ w_{jk} = w_{j} / w_{k} = {(1.00, 0.25, 0.75, 0.50)}^{T} \end{matrix}$

Step 2. According to the computing results of relative weight w_jk, we can calculate the dominance degree of $l_{i}^{λ}$ over each alternative l_t based on c_j by λth experts. The operation results are listed as follows. (β = 2.4)

For expert d₁, the dominance degree $l_{i}^{1}$ can be calculated: $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{1}^{1} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 3651 - 0 . 26900 . 2582 - 0 . 3804 \\ - 0 . 38040 . 0000 - 0 . 4658 - 0 . 26900 . 0000 \\ 0 . 25820 . 44720 . 00000 . 36510 . 4472 \\ - 0 . 26900 . 2582 - 0 . 38040 . 00000 . 2582 \\ - 0 . 38040 . 0000 - 0 . 4658 - 0 . 26900 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{2}^{1} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 53790 . 0000 - 0 . 7607 - 0 . 7607 \\ 0 . 12910 . 00000 . 1291 - 0 . 5379 - 0 . 5379 \\ 0 . 0000 - 0 . 53790 . 0000 - 0 . 7607 - 0 . 7607 \\ 0 . 18260 . 12910 . 18260 . 00000 . 0000 \\ 0 . 18260 . 12910 . 18260 . 00000 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{3}^{1} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 00000 . 0000 - 0 . 62110 . 0000 \\ 0 . 00000 . 00000 . 0000 - 0 . 62110 . 0000 \\ 0 . 00000 . 00000 . 0000 - 0 . 62110 . 0000 \\ 0 . 44720 . 44720 . 44720 . 00000 . 4472 \\ 0 . 00000 . 00000 . 0000 - 0 . 62110 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{4}^{1} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 1826 - 0 . 38040 . 00000 . 1826 \\ - 0 . 38040 . 0000 - 0 . 5379 - 0 . 38040 . 0000 \\ 0 . 18260 . 25820 . 00000 . 18260 . 2582 \\ 0 . 0000 - 0 . 38040 . 18260 . 0000 - 0 . 3804 \\ - 0 . 38040 . 0000 - 0 . 5379 - 0 . 38040 . 0000 \end{matrix}] \end{matrix}$

For expert d₂, the dominance degree $l_{i}^{2}$ can becalculated: $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{1}^{2} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 0000 - 0 . 2690 - 0 . 38040 . 0000 \\ 0 . 00000 . 0000 - 0 . 2690 - 0 . 38040 . 0000 \\ 0 . 25820 . 25820 . 0000 - 0 . 26900 . 2582 \\ 0 . 36510 . 36510 . 25820 . 00000 . 3651 \\ 0 . 00000 . 0000 - 0 . 2690 - 0 . 38040 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{2}^{2} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 22360 . 22360 . 12910 . 2236 \\ - 0 . 93170 . 00000 . 0000 - 0 . 76070 . 0000 \\ - 0 . 93170 . 00000 . 0000 - 0 . 76070 . 0000 \\ - 0 . 53790 . 18260 . 18260 . 00000 . 1826 \\ - 0 . 93170 . 00000 . 0000 - 0 . 76070 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{3}^{2} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 3106 - 0 . 43920 . 00000 . 0000 \\ 0 . 22360 . 0000 - 0 . 31060 . 22360 . 2236 \\ 0 . 31620 . 22360 . 00000 . 31620 . 3162 \\ 0 . 0000 - 0 . 3106 - 0 . 43920 . 00000 . 0000 \\ 0 . 0000 - 0 . 3106 - 0 . 43920 . 00000 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{4}^{2} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 38040 . 1826 - 0 . 5379 - 0 . 3804 \\ 0 . 18260 . 00000 . 2582 - 0 . 38040 . 0000 \\ - 0 . 3804 - 0 . 53790 . 0000 - 0 . 6588 - 0 . 5379 \\ 0 . 25820 . 18260 . 31620 . 00000 . 1826 \\ 0 . 18260 . 00000 . 2582 - 0 . 38040 . 0000 \end{matrix}] \end{matrix}$

For expert d₃, the dominance degree $l_{i}^{3}$ can be calculated: $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{1}^{3} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 2690 - 0 . 3804 - 0 . 2690 - 0 . 3804 \\ 0 . 25820 . 0000 - 0 . 26900 . 0000 - 0 . 2690 \\ 0 . 36510 . 25820 . 00000 . 25820 . 0000 \\ - 0 . 26900 . 00000 . 25820 . 00000 . 2582 \\ 0 . 36510 . 25820 . 00000 . 25820 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{2}^{3} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 7607 - 0 . 7607 - 0 . 9317 - 0 . 5379 \\ 0 . 18260 . 00000 . 0000 - 0 . 53790 . 1291 \\ 0 . 18260 . 00000 . 0000 - 0 . 53790 . 1291 \\ 0 . 22360 . 12910 . 12910 . 00000 . 1826 \\ 0 . 1291 - 0 . 5379 - 0 . 5379 - 0 . 76070 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{3}^{3} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 22360 . 3162 - 0 . 31060 . 2236 \\ - 0 . 31060 . 00000 . 2236 - 0 . 43920 . 0000 \\ - 0 . 4392 - 0 . 31060 . 0000 - 0 . 5379 - 0 . 3106 \\ 0 . 22360 . 31620 . 38730 . 00000 . 3162 \\ - 0 . 31060 . 00000 . 2236 - 0 . 43920 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ_{4}^{3} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 00000 . 18260 . 1826 - 0 . 53790 . 1826 \\ - 0 . 38040 . 00000 . 0000 - 0 . 65880 . 0000 \\ - 0 . 38040 . 00000 . 0000 - 0 . 65880 . 0000 \\ 0 . 25820 . 31620 . 31620 . 00000 . 3162 \\ - 0 . 38040 . 00000 . 0000 - 0 . 65880 . 0000 \end{matrix}] \end{matrix}$

Step 3. Compute overall dominance degree $φ_{j}^{λ} =$ $[φ_{j}^{λ} (l_{i}, l_{t})]_{m \times m}$ to get the matrix φ^λ = [φ^λ (l_i, l_t)] _m×m. $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ^{1} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 0 . 0098 - 0 . 6493 - 1 . 1237 - 0 . 9585 \\ - 0 . 6316 0 . 0000 - 0 . 8747 - 1 . 8084 - 0 . 5379 \\ 0 . 4400 0 . 1675 0 . 0000 - 0 . 8341 - 0 . 0553 \\ 0 . 3608 0 . 4541 0 . 4320 0 . 0000 0 . 3250 \\ - 0 . 5782 0 . 1291 - 0 . 8212 - 1 . 2705 0 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ^{2} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 4673 - 0 . 3020 - 0 . 7892 - 0 . 1568 \\ - 0 . 52550 . 0000 - 0 . 3213 - 1 . 29780 . 2236 \\ - 0 . 7376 - 0 . 05610 . 0000 - 1 . 37230 . 0365 \\ 0 . 08540 . 41970 . 31780 . 00000 . 7303 \\ - 0 . 7491 - 0 . 3106 - 0 . 4500 - 1 . 52150 . 0000 \end{matrix}] \end{matrix}$ $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ φ^{3} = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 6235 - 0 . 6423 - 2 . 0491 - 0 . 5121 \\ - 0 . 25020 . 0000 - 0 . 0454 - 1 . 6359 - 0 . 1399 \\ - 0 . 2718 - 0 . 05240 . 0000 - 1 . 4764 - 0 . 1815 \\ 0 . 43650 . 76161 . 09080 . 00001 . 0732 \\ - 0 . 1967 - 0 . 2797 - 0 . 3143 - 1 . 60050 . 0000 \end{matrix}] \end{matrix}$

Step 4. Calculate the overall dominance δ (l_i, l_t) based on the expert weighting vector (0.3, 0.4, 0.3) and the results of φ^λ = [φ^λ (l_i, l_t)] _m×m. $\begin{matrix} \begin{matrix} l_{1} & l_{2} & \begin{matrix} l_{3} & l_{4} \end{matrix} & l_{5} \end{matrix} \\ δ (l_{i}, l_{t}) = \begin{matrix} l_{1} \\ l_{2} \\ \begin{matrix} l_{3} \\ l_{4} \end{matrix} \\ l_{5} \end{matrix} [\begin{matrix} 0 . 0000 - 0 . 3710 - 0 . 5083 - 1 . 2675 - 0 . 5039 \\ - 0 . 47470 . 0000 - 0 . 4045 - 1 . 5524 - 0 . 1139 \\ - 0 . 24440 . 01210 . 0000 - 1 . 2421 - 0 . 0564 \\ 0 . 27340 . 53260 . 58400 . 00000 . 7116 \\ - 0 . 5321 - 0 . 1694 - 0 . 5206 - 1 . 46990 . 0000 \end{matrix}] \end{matrix}$

Step 5. Compute the overall value of l_i by the formula (16): $\begin{matrix} δ (l_{1}) = 0.0086, δ (l_{2}) = 0.0305, δ (l_{3}) = 0.2422 \\ δ (l_{4}) = 1.0000, δ (l_{5}) = 0 . 0000 . \end{matrix}$

Step 6. To choose the best alternative by rank the values of δ (l_i), the alternative with maximum value is the best choice. According to step 5, the ranking of l_i is l₄ > l₃ > l₂ > l₁ > l₅, and it’s clear that the best choice is l₄.

6 Conclusion

In our article, we proposed the 2-tuple linguistic TODIM method based on the fundamental theories of 2-tuple linguistic sets and origin TODIM model. Firstly, we briefly introduce the definition, operation laws and the distance calculating method of 2-tuple linguistic model. Then, the calculating steps of the origin TODIM model are simply presented. Thereafter, we extend the origin TODIM model to the2-tuple linguistic TODIM method, in our proposed method; it’s more reasonable and scientific for considering the subjectivity of DM’s behaviors and the dominance of each alternative over others. Finally, a numerical example for evaluating the service quality of boutique tourist scenic spot has been proposed to illustrate the new method and some comparisons are also conducted to further illustrate advantages of the new method.

In the future, the application of the proposed models and methods of 2-tuple linguistic can be investigated in the MAGDM problems, risk analysis and many other uncertain and fuzzy environments [30 –43].

Footnotes

Acknowledgment

This work is sponsored by the Social Science Foundation of Shaanxi Education Department, A Study of Optimization of Interpretive System in Tourist Destinations in Shaanxi (2013JK0301).

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TODIM method for evaluating the service quality of boutique tourist scenic spot with 2-tuple linguistic information

Abstract

Keywords

1 Introduction

2 Preliminaries

5.1 Calculating steps based on MAGDM problems

Table 1 2-tuple linguistic information decision matrix by d1 c1 c2 c3 c4 l 1 (s4,0) (s2,0) (s1,0) (s3,0) l 2 (s2,0) (s3,0) (s1,0) (s2,0) l 3 (s5,0) (s2,0) (s1,0) (s4,0) l 4 (s3,0) (s4,0) (s5,0) (s3,0) l 5 (s2,0) (s4,0) (s1,0) (s2,0)

Footnotes

Acknowledgment

References

Table 1
2-tuple linguistic information decision matrix by d¹

c₁ c₂ c₃ c₄

l ₁ (s₄,0) (s₂,0) (s₁,0) (s₃,0)

l ₂ (s₂,0) (s₃,0) (s₁,0) (s₂,0)

l ₃ (s₅,0) (s₂,0) (s₁,0) (s₄,0)

l ₄ (s₃,0) (s₄,0) (s₅,0) (s₃,0)

l ₅ (s₂,0) (s₄,0) (s₁,0) (s₂,0)