Pairwise opposite matrix and its cognitive prioritization operators: comparisons with pairwise reciprocal matrix and analytic prioritization operators

Abstract

The pairwise reciprocal matrix (PRM) of the analytic hierarchy/network process has been investigated by many scholars. However, there are significant queries about the appropriateness of using the PRM to represent the pairwise comparison. This research proposes a pairwise opposite matrix (POM) as the ideal alternative with respect to the human linguistic cognition of the rating scale of the paired comparison. Several cognitive prioritization operators (CPOs) are proposed to derive the individual utility vector (or priority vector) of the POM. Not only are the rigorous mathematical proofs of the new models demonstrated, but solutions of the CPOs are also illustrated by the presentation of graph theory. The comprehensive numerical analyses show how the POM performs better than the PRM. POM and CPOs, which correct the fallacy of the PRM associated with its prioritization operators, should be the ideal solutions for multi-criteria decision-making problems in various fields.

Appendix

A.1. Proofs

Proposition 1 Proof:

• ∑ _i ∑ _j b _ij = ∑ _i ∑ _j>i (b _ij + b _ji ) + 0. I = 0, where I is the identity matrix and b _ij +b _ji =0. □

Proposition 2 Proof:

• The proof is trivial. Let b _ij =v _i −v _j , b _ik =v _i −v _k , b _jk =v _j −v _k , then

  b _ik +b _jk =(v _i −v _k )−(v _j −v _k )=v _i −v _j =b _ij , ∀k∈(1, …, n). □

Proposition 3 Proof:

• For B _j ={b _1j , …, b _nj }^T, then B _j ^T={b _1j , …, b _nj }.

  Since B _i ={b _i1, …, b _in } and B _j ^T={b _1j , …, b _nj }, then B _i +B _j ^T={b _i1+b _1j , …, b _in +b _nj }.

  Case 1: B is perfectly accordant:

  As b _ik +b _kj =b _ij , k∈(1, …, n), then b _ij ∈{b _i1+b _1j , …, b _in +b _nj }.

  Thus B _i +B _j ^T−b _ij =0 × e ⁿ, where n-identity row e ⁿ is the row of n identity elements, for example

  Thus {d _ij }={0}. AI=0.

  Case 2: B is not accordant:

  Let B _i +B _j ^T={b _i1+b _1j , …, b _in +b _nj }={b ₁′, …, b _n ′}, and then

  B _i +B _j ^T−b _ij ={b ₁′, …, b _n ′}−b _ij .

  To normalize the above form, then 1/κ(B _i +B _j ^T−b _ij )=1/κ({b ₁′, …, b _n ′}−b _ij ).

  Next,

  
  

  Thus,

  Under this condition, if AI⩽0.1, B is defined to be unsatisfactory. If AI>0.1, B is defined to be unsatisfactory. □

Proposition 4 Proof:

• Let a scale item of the rating scale be the form α _i ∈X̄, and α _i =i′κ/τ, i′ =−τ, …, −1,0,1, …, τ. If b _ik =α _s and b _jk =α _t , then α _s −α _t =(s−t)κ/τ.

  Since s, t∈{−τ, …, −1,0,1, …, τ} and s≠t, then α _s −α _t =(s−t)κ/τ.

  This follows Min({(s−t)})=−2τ and Max({(s−t)})=2τ.

  In other words, (s−t)∈{−2τ, −2τ+1, …, −τ…0,…,τ…,2τ−1,2τ}.

  If (s−t)>τ or (s−t)<τ, then (b _ik −b _jk )=(α _s −α _t )>κ or (b _ik −b _jk )=(α _s −α _t )<κ.

  As Max (X̄)=α _τ =τκ/τ=κ and Min (X̄)=α _−τ =−τκ/τ=−κ, that is X̄=[−κ, …, κ] and (b _ik −b _jk )=(α _s −α _t )>κ or (b _ik −b _jk )=(α _s −α _t )<κ , then an outbound error exists.

  Otherwise, that is−τ⩽(s−t)⩽τ, the POM is within boundary.

  If this happens, then

  

Theorem 1 Proof:

• B≅B̃

  B+V≅B̃+V=n × V, which is written explicitly,

  
  
  

  Thus,

  
  
  

  Thus the solution is found. □

Theorem 2 Proof:

• This theorem can be proved by the Gaussian elimination method in which the augmented matrix [E _i ¦b _i ] is transformed to the reduced row echelon form [I¦ω _i ]. Explicitly,

  
  
  

  Thus, the general form is as shown in Equation (16). □

Proposition 5 Proof:

• For an accordant matrix, there should be a solution set in the linear systems set L=(L ₁, …, L _n }, which can be represented by the augmented matrices. The row reduction of L returns a row echelon form. If the row echelon form contains a vector (0 … 0¦c) where c is a constant, this means the matrix has no solution. Otherwise, L has a unique solution.

  It can be proved that when ω _i ≠ω _j , ∀i, j=1, …, n, the row echelon form must contain a vector (0 … 0¦c). This means that L has no unique solution and the matrix is discordant. Otherwise, L has unique solution set and A is the accordant matrix. □

Theorem 3 Proof:

• Let ω _i =(ω _i1, …, ω _in ), i=1, …, n. Then

  
  
  
  
  
  

  which is divided into two cases:

  Case 1:

  As b _ik =−b _ki ,

  

  Case 2:

  

  To combine both cases,

  

Proposition 6 Proof:

• The objective function minimizes the sum of {Δ _ij ²}, that is Min ∑ _i=1 ⁿ∑ _j=1 ⁿΔ _ij ². This gives ∑ _i=1 ⁿ∑ _j=i+1 ⁿΔ _ij ²=∑ _i=1 ⁿ∑ _j=1 ^jΔ _ij ², that is ∑ _i=1 ⁿ∑ _j=i+1 ⁿ(b _ij −v _i +v _j )²=∑ _i=1 ⁿ∑ _j=1 ^j(b _ij −v _i +v _j )²=Δ̄. Thus, ∑ _i=1 ⁿ∑ _j=i ⁿ(b _ij −v _i +v _j )²=∑ _i=1 ⁿ∑ _j=i+1 ⁿ(b _ij −v _i +v _j )²⁺∑ _i=1 ⁿ∑ _j=1 ^j(b _ij −v _i +v _j )²=2Δ̄. Hence, this proposition holds. □

Theorem 4 Proof:

• To have the closed form solution, the partial differentiation of Δ̄ with respective to all v _k ∈V is derived and is of the form:

  

  Then the linear system {L _i } is solved for V. Therefore,

  
  
  

  Let and Thus

  
  

  Case 1:

  

  Case 2:

  

  Case 3:

  
  

  To combine the three cases,

  

  Since ∂²Δ̄/∂v _k =2(n−1)>0, Δ̄ is convex.

  As L _k =δ²Δ̄/δv _k =0, k=1, 2 ,…, n, there exists a minimal v _k , k=1, 2 ,…, n.

  The values can be solved by a linear system. Thus the augmented matrix [E¦b] is of the form:

  

  V is solved by Gaussian elimination of [E¦b], in which the final row is added to other rows. Thus,

  

  To divide the above system by n, the reduced row echelon form is

  

  Interestingly, the result is the same as the row average plus the normal utility. □

Proposition 7 Proof:

• For ∀i∈{1, …, n}, if v _i ⩾0, then (1/n∑ _j=1 ⁿ b _ij ). If (1/n∑ _j=1 ⁿ b _ij ), then Max (ℵ)⩽κ since b _ij ∈ℵ.