An extended VIKOR method for decision making problem with linguistic intuitionistic fuzzy numbers based on some new operational laws and entropy

2.1 Linguistic intuitionistic fuzzy numbers

Chen [5] put forward the conception of linguistic intuitionistic fuzzy number (LIFN), in which the membership degree and the non-membership degree are presented by linguistic variables [7, 46]. The definition is shown as follows:

Definition 1. [5] Let s _α ,  s _β  ∈ S_[0,t], (S_[0,t] is a linguistic term set with odd cardinality) and γ = (s _α ,  s _β ), where α,  β ∈ [0, t], and α + β ∈ [0, t], then γ is called the linguistic intuitionistic fuzzy number (LIFN). s _α is the membership degree and s _β is the non-membership degree, usually, s _π is called the indeterminacy degree or hesitation degree of γ, we also have π ∈ [0, t] ,  π = t - α - β.

For convenience, suppose Γ_[0,t] is the set of all LIFNs.

Chen [5] also introduced some operational laws of LIFNs as follows:

Definition 2. [5] If γ₁ = (s _{α 1} ,  s _{β 1} ) and γ₂ = (s _{α 2} ,  s _{β 2} ) ∈ Γ_[0,t] are two LIFNs, λ is a real number, λ > 0, then γ 1 ⊕ γ 2 = ( s α 1 + α 2 - α 1 α 2 t , s β 1 β 2 t )(1) γ 1 ⊗ γ 2 = ( s α 1 α 2 t , s β 1 + β 2 - β 1 β 2 t )(2) λ γ = λ ( s α , s β ) = ( s t - t ( 1 - α t ) λ , s t ( β t ) λ )(3) γ λ = ( s α , s β ) λ = ( s t ( α t ) λ , s t - t ( 1 - β t ) λ )(4)

In addition, the complement operation of LIFNs can be presented as ( s α , s β ) c = ( s β , s α )(5)

Then, Liu and Qin [25] introduced the Euclidean distance of LIFNs which is shown as follows:

Definition 3. [25] If γ₁ = (s _{α 1} ,  s _{β 1} ) and γ₂ = (s _{α 2} ,  s _{β 2} ) ∈ Γ_[0,t] are two LIFNs, then the Euclidean distance between γ₁ and γ₂ is defined as follows: d ( γ 1 , γ 2 ) = 1 2 t ( | α 1 - α 2 | + | β 1 - β 2 | )(6)

The VIKOR is one useful method to solve the MADM problems, it can consider both the “group utility” and “individual regret”. The decision making problem can be expressed as follows:

Suppose there are m alternatives which are expressed as (X₁,  X₂, …,  X _m ), and there are n attributes which are presented as (A₁,  A₂, …,  A _n ), the evaluation value of alternative X _i with respect to attribute A _j is expressed by x _ij ,  i = 1,  2, …,  m .  j = 1,  2, … n. Suppose the x j * expresses the virtual positive ideal value and the x j - expresses virtual negative ideal value under the attribute A _j . w = (w₁,  w₂, …,  w _n )  ^T is the attribute weight vector which satisfies w i ∈ [ 0 , 1 ] , ∑ i = 1 n w i = 1 . The compromise ranking by VIKOR method is based on the form of L _p -metric [28].

L pi = { ∑ j = 1 n [ ( f j * - f ij ) ( f j * - f j - ) ] p } 1 / - p 1 ≤ p ≤ ∞ ; i = 1 , 2 , … , m .(7)

In the VIKOR method, the maximum group utility can be presented by min S _i , and minimum individual regret can be presented by min R _i , where S _i  = L_1,i, and R _i  = L_∞,i.

The steps of the VIKOR method can be described as follows:

Step 1. Normalize the decision matrix

Step 2. Compute the virtual positive ideal x j * and the virtual negative ideal x j - values under the attribute A _j , we have x j * = max i x ij , x j - = min i x ij(8)

Step 3. Compute the group utility value S _i and individual regret value R _i (i = 1,  2, …,  m) shown as follows. S i = ∑ j = 1 n w j ( x j * - x ij ) / ( x j * - x j - ) ,(9) R i = max j w j ( x j * - x ij ) / ( x j * - x j - )(10)

Step 4. Compute the values Q _i (i = 1,  2, …,  m) according to the following formulas: Q i = v ( S i - S * ) ( S - - S * ) + ( 1 - v ) ( R i - R * ) ( R - - R * )(11)

Where S * = min i S i , S - = max i S i , R * = min i R i , R - = max i R i , v is the balance parameter of decision strategy which can balance the factors between group utility and individual regret. When v > 0.5, it explains that considering “the maximum group utility” is more that considering “the minimum individual regret” and vice versa. (In our research, we suppose that v > 0.5, the “minimum individual regret” and “the maximum group utility” are the same important.)

Step 5. Rank all the alternatives. According to the values S, R and Q, we will get three ranking results, and then we can obtain a set of the compromisesolutions.

Step 6. Obtain a compromise solution X⁽¹⁾, which is in the first position of all ranking alternatives produced by the value Q (i.e., the alternative is with minimum value Q) if it meets the following two conditions:

Condition 1. Acceptable advantage: Q ( X ( 2 ) ) - Q ( X ( 1 ) ) ≥ 1 m - 1 , where Q (X⁽²⁾) is the Q value in the second position of all ranking alternatives produced by the value Q, and m is the number ofalternatives.

Condition 2. Acceptable stability. Alternative X⁽¹⁾ must also be in the first position of all ranking alternatives produced by the value by S and R.

If one of above two conditions is not met, then we will get a collection of compromise alternatives and not one compromise solution.

If Condition 2 is not met, then we can get that alternatives X⁽¹⁾ and X⁽²⁾ should be compromise solutions.

Based on the above analysis, we know that the best solution is the one with the minimum Q value when the conditions 1 and 2 are met, and when one of two conditions is not met, the compromise solutions may be have more than one.

3.1 Some new operational laws of linguistic intuitionistic fuzzy numbers

Chen [5] proposed the some operational laws of LIFNs but not including the subtraction and division operations, these two operations of the LIFNs are necessary to solve the MADM problems, so we propose the subtraction and division operations for LIFNs as follows:

Definition 6. For any two given LIFNs γ₁ = (s _{α 1} ,  s _{β 1} ), γ₂ = (s _{α 2} ,  s _{β 2} ) ∈ Γ_[0,t], the subtraction γ₁ ! γ₂ and division γ₁ % γ₂ can be defined as follows:

(1) γ₁ ! γ₂ = (s _{α 1}  ! s _{α 2} ,  s _{β 2}  ! s _{β 2} ), where

s α 1 ! s α 2 = { s t ( α 1 t - α 2 t 1 - α 2 t ) , if α 1 ≥ α 2 and β 1 ≤ β 2 and β 2 ≥ 0 and β 1 π 2 ≤ β 2 π 1 0 , otherwise = { s t α 1 - t α 2 t - α 2 , if α 1 ≥ α 2 and β 1 ≤ β 2 and β 2 ≥ 0 and β 1 π 2 ≤ β 2 π 1 0 , otherwise(12) and

s β 2 ! s β 2 = { t ( β 1 t β 2 t ) , if α 1 ≥ α 2 and β 1 ≤ β 2 and β 2 ≥ 0 and β 1 π 2 ≤ β 2 π 1 t , otherwise = { s t β 1 β 2 , if α 1 ≥ α 2 and β 1 ≤ β 2 and β 2 ≥ 0 and β 1 π 2 ≤ β 2 π 1 t , otherwise(13)

(2) γ₁ % γ₂ = (s _{α 1}  % s _{α 2} ,  s _{β 1}  % s _{β 2} ), where

s α 1 % s α 2 = { s t ( α 1 t α 2 t ) , if α 1 ≥ α 2 and β 1 ≤ β 2 and β 2 ≥ 0 and β 1 π 2 ≤ β 2 π 1 0 , otherwise = { s t α 1 α 2 , if α 1 ≤ α 2 and β 1 ≥ β 2 and β 2 ≥ 0 and β 1 π 2 ≥ β 2 π 1 0 , otherwise(14) and

s β 1 % s β 2 = { t ( β 1 t - β 2 t 1 - β 2 t ) , if α 1 ≥ α 2 and β 1 ≤ β 2 and β 2 ≥ 0 and β 1 π 2 ≤ β 2 π 1 t , otherwise = { s t β 1 - t β 2 t - β 2 , if α 1 ≤ α 2 and β 1 ≥ β 2 and β 2 ≥ 0 and β 1 π 2 ≥ β 2 π 1 t , otherwise(15)

From definition 6, we can see the subtraction and division operations for LIFNs are still LIFNs.

Example 1. Suppose γ₁ = (s₇, s₁) ,  γ₂ = (s₆, s₂) ∈ Γ_[0,8] are two given LIFNs, and then we have

γ 1 ! γ 2 = ( s 8 * 7 - 8 * 6 8 - 6 , s 8 * 1 2 ) = ( s 4 , s 4 ) ;

In the following, we give two functions to compare two LIFNs.

Definition 7. Let γ = (s _α ,  s _β ) ∈ Γ_[0,t], the score function of LIFN can be express as following: Ls ( γ ) = α + t - β 2(16)

Definition 8. Let γ = (s _α ,  s _β ) ∈ Γ_[0,t], the accuracy function of LIFN can be express as following: Lh ( γ ) = α + β 2(17)

Definition 9. Let γ₁ = (s _{α 1} ,  s _{β 1} ) , γ₂ = (s _{α 2} , s _{β 2} ) ∈ Γ_[0,t], then

If Ls (γ₁) < Ls (γ₂), then γ₁ is smaller than γ₂, denoted by γ₁ ≺ γ₂;

Definition 10. If γ₁ = (s _{α 1} , s _{β 1} ) and γ₂ = (s _{α 2} , s _{β 2} ) ∈ Γ_[0,t] are two LIFNs,

γ₁ ⊆ γ₂ if and only if α 1 ≤ α 2 , β 1 ≥ β 2(18)

In this section, we propose linguistic intuitionistic fuzzy entropy.

Let M ={ γ _i  〈 s _α  (γ _i ) ,  s _β  (γ _i ) 〉 |γ _i  ∈ Γ_[0, t]  }, the mapping E (M) is called as entropy if E satisfies the following conditions:

Condition 1. E (M) = 0 if and only if M is γ _i  = (s _t , s₀) or γ _i  = (s₀,  s _t ) ∈ Γ_[0,t];

Condition 2. E (M) = 1 if and only if s _α  (γ _i ) = s _β  (γ _i ) for each γ _i  ∈ Γ_(0,t);

Condition 3. For any N ∈ Γ_[0,t] if M ⊇ N when s _α  (N) ≥ s _β  (N) for every γ _i  ∈ Γ_[0,t], or M ⊆ N when s _α  (N) ≤ s _β  (N) for every γ _i  ∈ Γ_[0,t], then E (M) ≤ E (N);

Condition 4. E (M) = E (M ^c ).

Theorem 2. Let γ _i  ∈ Γ_[0,t] be a universe, M ={ γ _i  〈 s _α  (γ _i ) , s _β  (γ _i ) 〉 |γ _i  ∈ Γ_[0,t]  }, the entropy of LIFNs is defined as follows:

E ( M ) = 1 n ∑ i = 1 n ( 1 - | α ( γ i ) t - β ( γ i ) t | + π ( γ i ) t 1 + π ( γ i ) t ) × t = 1 n ∑ i = 1 n ( t - | α ( γ i ) - β ( γ i ) | + π ( γ i ) t + π ( γ i ) )(20)

Proof.

Condition 1. M is γ _i  = (s _t ,  s₀) or γ _i  = (s₀,  s _t ) ∈ Γ_[0,t], then E (M) = 0;

If E (M) = 0, since 0 ≤ |α (γ _i ) - β (γ _i ) | ≤ t ≤ t + π (γ _i ), so t - | α ( γ i ) - β ( γ i ) | + π ( γ i ) t + π ( γ i ) ≥ 0 , we can get t - | α ( γ i ) - β ( γ i ) | + π ( γ i ) t + π ( γ i ) = 0 because of E (M) = 0, namely, |α (γ _i ) - β (γ _i ) | = t + π (γ _i ), so |α (γ _i ) - β (γ _i ) | = t, we can get (s _α ,  s _β ) = (s _t ,  s₀) or (s _α ,  s _β ) = (s₀,  s _t ).

Condition 2. If s _α  (γ _i ) = s _β  (γ _i ) for each γ _i  ∈ Γ_(0,t), it’s obvious that E (M) = 1;

If E (M) = 1, since 0 ≤ |α (γ _i ) - β (γ _i ) | ≤ t ≤ t + π (γ _i ), so t - |α (γ _i ) - β (γ _i ) | + π (γ _i ) = t + π (γ _i ) must be established, we can get |α (γ _i ) - β (γ _i ) | = 0, so, α (γ _i ) = β (γ _i ) namely, s _α  (γ _i ) = s _β  (γ _i ).

Condition 3. Since M ⊇ N when s _α  (N) ≥ s _β  (N) for every γ _i  ∈ Γ_[0,t], that is s _α  (N) ≤ s _α  (M) and s _β  (N) ≥ s _β  (M). Since t - α _M  ≥ 0, β _M  - t ≤ 0, we have β _N  (t - α _M ) ≥ β _M  (t - α _M ) and α _N  (β _M  - t) ≥ α _M  (β _M  - t). Hence β _N  (t - α _M ) + α _N  (β _M  - t) ≥ β _M  (t - α _M ) + α _M  (β _M  - t). It follows that β _N  (t - α _M ) + α _N  (β _M  - t) + α _M  - β _M  ≥ 0.

Thus (t - α _M ) (2t - α _N  - β _N ) ≤ (t - α _N )   (2t - α _M  - β _M ), which implies t - | α ( M ) - β ( M ) | + π ( M ) t + π ( M ) ≤ t - | α ( N ) - β ( N ) | + π ( N ) t + π ( N ) , thus E (M) ≤ E (N).

Similarity, when M ⊆ N and s _α  (N) ≤ s _β  (N) for every γ _i  ∈ Γ_[0,t], we can also prove that E (M) ≤ E (N).

Therefore, E (M) ≤ E (N).

Condition 4. For the two LIFNs M and M ^c , |α _M  - β _M | = |β _{M ^c}  - α _{M ^c} | = |α _{M ^c}  - β _{M ^c} |, so it’s obvious that the condition 4 is kept.

From above four conditions, we can know that the linguistic intuitionistic fuzzy entropy can consider the uncertainty both from fuzziness and from the lack of information based on LIFNs, when LIFNs are (s _t ,  s₀) or (s₀,  s _t ), the entropy value is smallest; When LIFNs meet s _α  (γ _i ) = s _β  (γ _i ) for each γ _i  ∈ Γ_(0,t), the information of LIFNs is most fuzzy and when γ _i  = (s₀,  s₀) γ _i  ∈ Γ_(0,t), the information of LIFNs lack all the information, the entropy value is largest; For given a LIFN, the more fuzzy of the information, and the more information missing, the greater of the entropy value. So the defined linguistic intuitionistic fuzzy entropy can measure not only the fuzziness but also the intuitionism of LIFNs.

5.1 Description of the example

In this section, we will give an example to show the proposed method. Suppose there is a decision making problem of selecting the best global supplier which is described as follows:

A manufacturing company desires to select a best global supplier for its most critical parts used in assembling process. Suppose that X ={ x₁,  x₂,  x₃,  x₄ } is a set of four potential global suppliers (i.e., alternatives) under consideration and C ={ c₁,  c₂,  c₃,  c₄,  c₅ } is a set of attributes, where (c₁,  c₂,  c₃,  c₄,  c₅) stand for “overall cost of the production”, “quality of the production”, “service performance of supplier”, “supplier’s profile”, “risk factor”, respectively, and the attribute weight unknown. The four alternatives x _i  (1, …, 4) are to be evaluated by using the LIFNs based on the linguistic term set: S = { s 0 = extremely poor , s 1 = very poor , s 2 = poor , s 3 = slightly poor , s 4 = fair , s 5 = slightly good , s 6 = good , s 7 = very good , s 8 = extremely good }

Then we can construct the decision matrix R = (γ _ij ) _4×5 shown in Table 2.

Step 1. Normalize the decision matrix.

Since “overall cost of the production”, “risk factor” are the cost attribute values, “quality of the production”, “service performance of supplier”, “supplier’s profile” are the benefit attribute values, we transform the cost attribute values to benefit type according to formula (21), and get R ˜ = [ ( s 3 , s 4 ) ( s 5 , s 2 ) ( s 3 , s 4 ) ( s 5 , s 2 ) ( s 4 , s 3 ) ( s 5 , s 2 ) ( s 4 , s 3 ) ( s 5 , s 2 ) ( s 5 , s 2 ) ( s 5 , s 2 ) ( s 6 , s 1 ) ( s 5 , s 2 ) ( s 6 , s 1 ) ( s 5 , s 2 ) ( s 5 , s 2 ) ( s 5 , s 2 ) ( s 6 , s 1 ) ( s 6 , s 1 ) ( s 6 , s 1 ) ( s 6 , s 1 ) ]

Step 2. Obtain the virtual positive ideal solution and the virtual negative ideal solution by Equations (22–25), and we can get X * = { γ ˜ 1 * , γ ˜ 2 * , γ ˜ 3 * , γ ˜ 4 * , γ ˜ 5 * } = { ( s 6 , s 1 ) , ( s 6 , s 1 ) , ( s 6 , s 1 ) , ( s 6 , s 1 ) , ( s 6 , s 1 ) } X - = { γ ˜ 1 - , γ ˜ 2 - , γ ˜ 3 - , γ ˜ 4 - , γ ˜ 5 - } = { ( s 3 , s 4 ) , ( s 4 , s 3 ) , ( s 3 , s 4 ) , ( s 5 , s 3 ) , ( s 4 , s 3 ) }

Step 3. Calculate the entropy of every attribute E j = 1 m ∑ i = 1 m e ij by formula (20), where j = 1,  2, …,  5. We can get E 1 = 0.6667 , E 2 = 0.6667 , E 3 = 0.6111 , E 4 = 0.6111 , E 5 = 0.6667

Step 4. Calculate the weight value of each attribute by Equation (26), we can get w 1 = 0.1930 , w 2 = 0.1930 , w 3 = 0.2105 , w 4 = 0.2105 , w 5 = 0.1930

Step 5. Compute S _i and R _i , by Equations (27, 28). and we have S 1 = w 1 | γ ˜ 1 * - γ ˜ 11 | | γ ˜ 1 * - γ ˜ 1 - | + w 2 | γ ˜ 2 * - γ ˜ 12 | | γ ˜ 2 * - γ ˜ 2 - | + w 3 | γ ˜ 3 * - γ ˜ 13 | | γ ˜ 3 * - γ ˜ 3 - | + w 4 | γ ˜ 4 * - γ ˜ 14 | | γ ˜ 4 * - γ ˜ 4 - | + w 5 | γ ˜ 5 * - γ ˜ 15 | | γ ˜ 5 * - γ ˜ 5 - | = ( s 6.015 , s 1.537 ) ; S 3 = w 1 | γ ˜ 1 * - γ ˜ 31 | | γ ˜ 1 * - γ ˜ 1 - | + w 2 | γ ˜ 2 * - γ ˜ 32 | | γ ˜ 2 * - γ ˜ 2 - | + w 3 | γ ˜ 3 * - γ ˜ 33 | | γ ˜ 3 * - γ ˜ 3 - | + w 4 | γ ˜ 4 * - γ ˜ 34 | | γ ˜ 4 * - γ ˜ 4 - | + w 5 | γ ˜ 5 * - γ ˜ 35 | | γ ˜ 5 * - γ ˜ 5 - | = ( s 3.644 , s 3.844 ) ; S 4 = w 1 | γ ˜ 1 * - γ ˜ 41 | | γ ˜ 1 * - γ ˜ 1 - | + w 2 | γ ˜ 2 * - γ ˜ 42 | | γ ˜ 2 * - γ ˜ 2 - | + w 3 | γ ˜ 3 * - γ ˜ 43 | | γ ˜ 3 * - γ ˜ 3 - | + w 4 | γ ˜ 4 * - γ ˜ 44 | | γ ˜ 4 * - γ ˜ 4 - | + w 5 | γ ˜ 5 * - γ ˜ 45 | | γ ˜ 5 * - γ ˜ 5 - | = ( s 1.004 , s 6.664 ) . R 1 = max 5 { w 1 | γ ˜ 1 * - γ ˜ 11 | | γ ˜ 1 * - γ ˜ 1 - | , w 2 | γ ˜ 2 * - γ ˜ 12 | | γ ˜ 2 * - γ ˜ 2 - | , w 3 | γ ˜ 3 * - γ ˜ 13 | | γ ˜ 3 * - γ ˜ 3 - | , w 4 | γ ˜ 4 * - γ ˜ 14 | | γ ˜ 4 * - γ ˜ 4 - | , w 5 | γ ˜ 5 * - γ ˜ 15 | | γ ˜ 5 * - γ ˜ 5 - | } = ( s 2.339 , s 5.300 ) ; R 2 = max 5 { w 1 | γ ˜ 1 * - γ ˜ 21 | | γ ˜ 1 * - γ ˜ 1 - | , w 2 | γ ˜ 2 * - γ ˜ 22 | | γ ˜ 2 * - γ ˜ 2 - | , w 3 | γ ˜ 3 * - γ ˜ 23 | | γ ˜ 3 * - γ ˜ 3 - | , w 4 | γ ˜ 4 * - γ ˜ 24 | | γ ˜ 4 * - γ ˜ 4 - | , w 5 | γ ˜ 5 * - γ ˜ 25 | | γ ˜ 5 * - γ ˜ 5 - | } = ( s 2.003 , s 5.707 ) ; R 3 = max 5 { w 1 | γ ˜ 1 * - γ ˜ 31 | | γ ˜ 1 * - γ ˜ 1 - | , w 2 | γ ˜ 2 * - γ ˜ 32 | | γ ˜ 2 * - γ ˜ 2 - | , w 3 | γ ˜ 3 * - γ ˜ 33 | | γ ˜ 3 * - γ ˜ 3 - | , w 4 | γ ˜ 4 * - γ ˜ 34 | | γ ˜ 4 * - γ ˜ 4 - | , w 5 | γ ˜ 5 * - γ ˜ 35 | | γ ˜ 5 * - γ ˜ 5 - | } = ( s 1.964 , s 5.828 ) ; R 4 = max 5 { w 1 | γ ˜ 1 * - γ ˜ 41 | | γ ˜ 1 * - γ ˜ 1 - | , w 2 | γ ˜ 2 * - γ ˜ 42 | | γ ˜ 2 * - γ ˜ 2 - | , w 3 | γ ˜ 3 * - γ ˜ 43 | | γ ˜ 3 * - γ ˜ 3 - | , w 4 | γ ˜ 4 * - γ ˜ 44 | | γ ˜ 4 * - γ ˜ 4 - | , w 5 | γ ˜ 5 * - γ ˜ 45 | | γ ˜ 5 * - γ ˜ 5 - | } = ( s 1.004 , s 6.664 ) .

Step 6. Compute the values Q _i  (i = 1, 2, 3, 4) by Equation (29) (suppose v = 0.5), we have Q 1 = ( s 6.683 , s 1.223 ) , Q 2 = ( s 5.539 , s 2.362 ) ,(30) Q 3 = ( s 4.460 , s 3.756 ) , Q 4 = ( s 0 , s 8 )(31)

Step 7. Calculate the score function of S _i , R _i and Q _i  (i = 1, 2, 3, 4 .), and we get Ls ( S 1 ) = 6.239 , Ls ( S 2 ) = 5.397 , Ls ( S 3 ) = 3.900 , Ls ( S 4 ) = 1.170 Ls ( R 1 ) = 2.520 , Ls ( R 2 ) = 2.148 , Ls ( R 3 ) = 2.068 , Ls ( R 4 ) = 1.170 ; Ls ( Q 1 ) = 6.730 , Ls ( Q 2 ) = 5.589 , Ls ( Q 3 ) = 4.352 , Ls ( Q 4 ) = 0 .

Step 8. Rank the alternatives.

We firstly gave the ranking results by the values Ls (S _i ), Ls (R _i ) and Ls (Q _i ) ,  i = 1, 2, 3, 4, the smaller the values Ls (S _i ), Ls (R _i ) and Ls (Q _i ) ,  i = 1, 2, 3, 4, the better the alternatives are. The results are listed in Table 3.

Step 9. Obtain the compromise ranking results. Firstly, we rank the alternatives by Ls (Q) in decreasing order, the alternative with first position is x₄ with Ls (Q₄) = 0, and alternative x₄ is also the best rankings by Ls (S) and Ls (R), which satisfies the condition 2. x₃ is the alternative with second position with Ls (Q₃) = 4.352.

As MQ = 1/(m-1) = 1/(4-1) = 0.333333, so Ls ( Q 4 ) - Ls ( Q 1 ) = 4.352 ≻ 0.333 .

Which is not satisfied the condition 1.

So, x₄ is the only compromise solution.