# Solving a System of Linear Equations Based on Z-Numbers to Determinate the Market Balance Value

Authors: Zeinab Motamedi Pour; Tofigh Allahviranloo; Mozhdeh Afshar Kermani; Saeid Abbasbandy
Journal: Advances in Fuzzy Systems
 (2023)

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## Abstract

In this article, a general linear equations system with Z-number’s data is introduced. Since the nature of Z-numbers has two parameters, namely, reliability and fuzziness, it is difficult to find the exact solution to these systems. Therefore, a numerical procedure for calculating the solution is designed. The proposed method is illustrated with some applied examples. Determining the value of the market balance is one of the examined examples.


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## Body

## 1. Introduction

Many problems in various fields, such as engineering and economics, lead to the solution of linear equations systems. Since, in the real world, some parameters are presented inaccurately and colloquially, we must focus on solving linear equations systems with fuzzy data or Z-number data. In recent years, many authors have studied fuzzy linear equations systems (see [1–3]), but fuzzy numbers do not contain reliable information as Z-numbers. The concept of Z-numbers was introduced as a new topic by Professor Zadeh in 2011 [4]. The Z-number has a greater ability to describe human knowledge than the classical fuzzy numbers. This concept is able to deal with uncertain information effectively in a decision, and it is indicated by the symbol Z = (A, B).Here, a Z-number is defined as a pair fuzzy number in the formA,B, where A denotes a fuzzy subset of the constraints including x values and the second component, B, is a measure of reliability (certainty) of the first component. In addition, the contract x,A,B can also be introduced as a Z-value. Semantically, using the value proposition developed in [5], the variable x can be interpreted in such a way that the probability that x takes a value of A is equal to B. It means as follows:(1)ProbxisAisB.It should be mentioned that these numbers are directly related to quantity; it can be observed that they have more ability to describe the uncertainty environment than fuzzy numbers that do not have this feature. In this regard, several researchers have conducted many studies in various fields by utilizing Z-numbers [4–12].In this article, we first introduce the linear equations system of Z-numbers and then propose a numerical method to solve it numerically. Finally, we suppose the balance value of the market as a linear equations system with Z-number’s data and then determine the balance value of the market. The rest of the article is as follows.Section2 contains the basic concepts that are used in the continuation of the article. In Section 3, a linear equations system with Z-number’s data is introduced, and a method is presented for solving it. Section 4 provides several examples, one of them shows the use of a linear equation system to determine the balance value of the market. Finally, Section 6 contains the conclusions and suggestions.


## 2. Preliminaries

In this section, the required explanations about fuzzy sets and Z-numbers are provided from [13].Definition 1.
Fuzzy sets A fuzzy SetA˜ is defined on a universe X may be introduced as follows:(2)A=<x,μA˜x>xεX,where μA˜:X⟶0,1 is the membership function A. The membership value μA˜x describes the degree of belongingness of xεX in A˜.Definition 2.
A triangular fuzzy numberA˜ can be defined by a triplet a1,a2,a3, where the membership can be determined as follows: A triangular fuzzy number A˜  =  a1,a2,a3 can be considered as in the following equation (3).(3)μA˜x=0,x∈−∞,a1,x−a1a2−a1,x∈a1,a2,a3−xa3−a2,x∈a2,a3,0,x∈a3,+∞,.Definition 3.
The parametric form of a fuzzy numberA˜ is a pair of functions A_,A¯ in which the functions A_r and A¯r apply to values 0≤r≤1 under the following conditions:(1)
The functionA_ is a nondecreasing continuous function bounded on [0, 1] from left(2)
The functionA¯ is a nonincreasing continuous function bounded on [0, 1] from left(3)
A_r≤A¯r for 0≤r≤1
In the above definition, ifA_r and A¯r are linear functions, the parametric representation of a triangular fuzzy number is obtained.Definition 4 (Definition of Z-number).
Zadeh introduced the idea of Z-numbers with an unknown variablex. A Z-number is a pair of fuzzy numbers A˜,B˜ [14]. Here, A˜ is a fuzzy subset of constraints that x can have and B˜ is a reliability of the A˜ component. Zadeh also defined x,A˜,B˜ as a Z-valuation and indicated that this value means that x equals A˜,B˜. According to the suggestion of Zadeh, this valuation for Z was observed as a constraint in x and interpreted as follows:(4)ProbxisA˜isB˜.
Actually, this means that(5)Rx:xisA˜⟶Possx=u=μA˜u,(6)PxisA˜=∫RμA˜uPxuduisB˜.
Here,μA is the membership function of the fuzzy set A˜ and u is a value of x. Pxu is the probability density function of x, and Px=u is the probability function of x. When we do not know the basic probability distribution, it is clear from these data that the probability distribution is a fuzzy number.Definition 5.
Let a fuzzy setA˜ be defined on a universe X and may be given as follows: A˜=x;µA˜xx∈X, where µA˜:X⟶0;1 is the membership function A˜. The membership value µA˜x describes the degree of belongingness of x∈X in A˜. The expected value of a fuzzy set is denoted as follows:(7)EA˜x=∫XxμA˜xdx,which is not the same as the meaning of the expectation of probability space. It can be considered as the information strength supporting the fuzzy set A˜. In the following section, we will introduce the proposed approach according to the same expectation of fuzzy numbers [1].Definition 6.
The weighted Z-number is denoted as follows:(8)Z˜α=x,μA˜αxμA˜α=αμA˜x,x∈0,1.
Note that,α represents the weight of the reliability component of Z-number.Theorem 1 (Reference [4]).
(9)EA˜αx=αEA˜x;x∈X,(10)s.t.μA˜αx=αμA˜x;xεX.Proof.
(11)EA˜αx=∫XxμA˜αxdx=∫XαxμA˜xdx=α∫XxμA˜xdx=αEA˜x.


## 3. System of Linear Equations Based on Z-Numbers

Definition 7.
The followingn×nsystem.of.linear.equations is called a Z-system of linear equations and is abbreviated as ZSLE.(12)∑j=1naijxjZ=yiZ,i=1,…,n,where the matrix A=aij is a real matrix, and also, .Z is the symbol for valuing Z.
In this article, the aim is to introduce a method to solve ZSLE (12); so using Definition 4, we can rewrite equations system (12) as follows:(13)∑j=1naijx˜jr,Bj=y˜ir,Ri,i=1,…,n.
Here,Bj and Ri is reliability of the x˜jr and y˜ir, respectively.
Now, we limit this problem to the following two cases:Case 1.
The reliability (probability) functionsBj and Ri are known and are defined as follows:(14)Bj=1−e−λxj,and also(15)Ri=1−e−λyi.Case 2.
The criteria of reliability (probability) functionsBj and Ri are not known, and only their value is available. We propose a numerical method for both of the above cases, and in the numerical results section of an example, we compare the results obtained from both methods.


### 3.1. The Proposed Method for SolvingZ-System of Linear Equations: Case 1

Definition 8.
A fuzzy vectorxjZ=x1Z,…,xnZT=x˜1r,B1,…,x˜nr,BnT is called a solution for a ZSLE (12) whenever it holds in the equations system. If aij>0, we consider the ZSLE (12) as follows:(16)∑j=1naijxjZ=∑j=1naijx˜jr,Bj.
Consider the parametric formxjZ=x˜jr,Bj. So,(17)∑j=1naijx˜jr,Bj=y˜ir,Ri.
Therefore, to solve a ZSLE, a system of ordinary linear equations2n×2n must be solved, in which the column of values in the right side of system is yiZ=y1Z,…,ynZT=y˜1r,R1,…,y˜nr,RnT. By using Definition 6, we show the new 2n×2n system as follows:(18)s11x_1r,Bx_1+⋯+s1nx_nr,Bx_n+s1,n+1x¯1r,Bx¯1+⋯+s1,2n−x¯nr,Bx¯n=y_1α,⋮sn1x_1r,Bx_1+⋯+snnx_nr,Bx_n+sn,n+1−x¯1r,Bx¯1+⋯+sn,2n−x¯nr,Bx¯n=y_nα,sn+1,1x_1r,Bx_1+⋯+sn+1,nx_nr,Bx_n+sn+1,n+1−x¯1r,Bx¯1+⋯+sn+1,2n−x¯nr,Bx¯n=−y¯1α,⋮s2n1x_1r,Bx_1+⋯+s2n,nx_nr,Bx_n+s2n,n+1−x¯1r,Bx¯1+⋯+s2n,2n−x¯nr,Bx¯n=−y¯nα,where sij is defined as follows:(19)aij≥0⇒sij=si+n,j+n=aij,aij<0⇒si+n,j=si,j+n=aij,1≤i≤n,1≤j≤n.
The other components of the “S” that were not initialized above are set to zero. In this case, we will have(20)A=D+C.
Using the system matrix symbol, (14) is written as follows:(21)Su_jr,B1j=Y_ir,R1i,where sij≥0, for 1≤i,j≤2n, and(22)x˜r,B=u_1r,B11,…,u_nr,B1n,x¯1r,B21,…,x¯nr,B2nT,where B=B11,…,B1n,B21,…,B2n and(23)Y˜r,R=Y¯1r,R11,…,Y¯nr,R1n,Y¯1r,R21,…,Y¯nr,R2nT,where R=R11,…,R1n,R21,…,R2n.
Thus,(24)DCCDu_r,B1−x¯r,B2=Y_r,R1−Y¯r,R2,where the matrices B and C contain the positive and negative elements of the matrix A, respectively.
It is clear in the systemSx˜r,B=Y˜r,R that whenever the matrix S is invertible, the response vector of the system is calculated as x˜r,B=S−1Y˜r,R. The following theorem can be useful in this regard [1].Theorem 2 (see [7]).
MatrixS is invertible if and only if D+C and D−C matrices are invertible.Definition 9.
Suppose the answerx˜αr=u_jαr,−x¯jαr,1≤j≤n, r is unique.
ujαr=u_jαr,u¯jαr1≤j≤n where(25)u¯jαr=minx¯jαr,x¯jαr,x¯jα1,u¯jαr=maxx¯jαr,x¯jαr,x¯jα1,is called a Zα answer for the system SX˜α=Y˜α. If X˜jα=x_jαr,x¯jαr1≤j≤n and the Z˜α are triangular, then u_jαr=x¯jαr and u¯jαr=x¯jαr, for all values 1≤j≤n, and u¯jαr is called a strong Zα answer for the system.


### 3.2. The Proposed Method for SolvingZ-System of Linear Equations: Case 2

In this section, suppose thatB˜, the probability distribution function of the confidence functions R˜, is specified in (3).SupposeR˜i=r1i,r2i,r3i. Now, using Definition 6, the right side of (3) can be rewritten as follows:(26)y˜ir,R˜i=y˜ir,r1i,r2i,r3i,where α is calculated as follows Assumingy˜ir=p1i,p2i,p3i:Using Definition6, we have(27)y˜iα=p1i,p2i,p3i,r1i,r2i,r3i=r^1i,r^2i,r^3i,where r^1i,r^2i,r^3i=αp1i,αp2i,αp3i. The α calculation algorithm is given in [10].Using (13), (26), and (27), we can rewrite the system of (2) as follows:(28)∑j=1naijxjZ=r^1i,r^2i,r^3i,i=1,…,n,where r^1i,r^2i,r^3i∈RF, is given and determining the unknown Z-valuation vector xjZ including xjZ=x˜jr,B˜j=x˜jr,βμx˜j is the purpose. Assuming x˜jr=q1j,q2j,q3j. Using the Definition 6, we have(29)x˜jr,βμx˜j=q^1j,q^2j,q^3j=x˜jβ,where q^1j,q^2j,q^3j=βq1j,βq2j,βq3j, and β is calculated similarly to α.Definition 10.
A fuzzy vectorxjZ=x1Z,…,xnZt=x˜1r,βμx˜1,…,x˜nr,βμx˜nt=x˜1β,…,x˜nβt is called a solution for a ZSLE (12) whenever it holds in the equations system. If aij>0, we consider the ZSLE (12) as follows:(30)∑j=1naijx˜jβr=b^1i,b^2i,b^3i.
Consider the parametric formx˜jβr=x˜jβr,x¯jβr; so,(31)∑j=1naijx_jβr=y^ir,∑j=1naijx¯jβr=y^¯ir.
Therefore, to solve a ZSLE, a system of ordinary linear equations2n×2n must be solved, in which the column of values in the right side of the system is y_1β,…,y¯nβ,y¯1β,…,y¯nβT. We show the new 2n×2n system as follows:(32)s11x_1β+s12x_2β+⋯+s1nx_nβ+s1,n+1x¯1β+⋯+s1,nx¯nβ=y_1α,⋮sn1x_1β+sn2x_2β+⋯+snnx_nβ+sn,n+1−x¯1β+⋯+sn,2n−x¯nβ=y_nα,sn+1,1x_1β+sn+1,2x_2β+⋯+sn+1,nx_nβ+sn+1,n+1−x¯1β+⋯+sn+1,2n−x¯nβ=−y¯1α,⋮s2n1x_1β+s2n2x_2β+⋯+s2n,nx_nβ+s2n,n+1−x¯1β+⋯+s2n,2n−x¯nβ=−y¯nα,in which sij are defined as follows:(33)aij≥0⇒sij=si+n,j+n=aij,aij<0⇒si+n,j=si,j+n=aij,1≤i≤n,1≤j≤n,and whenever sij are not specified above, then they are equal to zero. In this case, we will have(34)A=D+C.
Using the system matrix symbol, (30) is written as follows:(35)SX˜β=Y˜α,where sij≥0, for 1≤i,j≤2n and X˜β=x_1β,…,x_nβ,x¯1β,…,x¯nβT and Y˜α=y_1α,…,y_nα,y¯1α,…,y¯nαT; thus,(36)DCCDX_β−X¯β=Y_α−Y¯α,where the matrices B and C contain the positive and negative elements of the matrix A, respectively.
It is clear in the systemSX˜β=Y˜α that whenever the matrix S is invertible, the response vector of the system is calculated as X˜β=S−1Y˜α. The following theorem can be useful in this regard.Definition 11 (Reference [15]).
IfX=u_1β,x_2β,…,x_nβ,x¯1β,x¯2β,…,x¯nβT satisfies in SX=b and for each 1≤j≤n, the inequalities x_jβ≤x¯jβ holds, then the solution X=x_1β,x_2β,…,x_nβ,x¯1β,x¯2β,…,x¯nβT is called a strong solution of the SX=b.


### 3.3. Jacobi Method

LetAXZ=bZ be a square system of n linear equations, where(37)A=a11a12…a1na21a22…a2n⋮⋮⋱⋮an1an2…ann,XZ=x1Zx2Z⋮xnZ,bZ=b1Zb2Z⋮bnZ.A can be decomposed into a diagonal component D, a lower triangular part L, and an upper triangular part U and given as follows: A=D+L+U, where D=a110⋮00a22⋮0……⋱…00⋮0, and L+U=0a21⋮an1a120⋮an2……⋱…a1na2n⋮0.Then, the solution is obtained iteratively via(38)Xk+1Z=D−1b−L+UXkZ,where XkZ is the kth approximation or iteration of XZ and Xk+1Z is the next or(k + 1)th iteration of X. The element-based formula is written as follows:(39)xik+1Z=1aiibi−∑j≠iaijxjkZ,i=1,…,n.The computation ofxik+1Z requires each element in xjkZ except where i=j.


## 3.1. The Proposed Method for SolvingZ-System of Linear Equations: Case 1

Definition 8.
A fuzzy vectorxjZ=x1Z,…,xnZT=x˜1r,B1,…,x˜nr,BnT is called a solution for a ZSLE (12) whenever it holds in the equations system. If aij>0, we consider the ZSLE (12) as follows:(16)∑j=1naijxjZ=∑j=1naijx˜jr,Bj.
Consider the parametric formxjZ=x˜jr,Bj. So,(17)∑j=1naijx˜jr,Bj=y˜ir,Ri.
Therefore, to solve a ZSLE, a system of ordinary linear equations2n×2n must be solved, in which the column of values in the right side of system is yiZ=y1Z,…,ynZT=y˜1r,R1,…,y˜nr,RnT. By using Definition 6, we show the new 2n×2n system as follows:(18)s11x_1r,Bx_1+⋯+s1nx_nr,Bx_n+s1,n+1x¯1r,Bx¯1+⋯+s1,2n−x¯nr,Bx¯n=y_1α,⋮sn1x_1r,Bx_1+⋯+snnx_nr,Bx_n+sn,n+1−x¯1r,Bx¯1+⋯+sn,2n−x¯nr,Bx¯n=y_nα,sn+1,1x_1r,Bx_1+⋯+sn+1,nx_nr,Bx_n+sn+1,n+1−x¯1r,Bx¯1+⋯+sn+1,2n−x¯nr,Bx¯n=−y¯1α,⋮s2n1x_1r,Bx_1+⋯+s2n,nx_nr,Bx_n+s2n,n+1−x¯1r,Bx¯1+⋯+s2n,2n−x¯nr,Bx¯n=−y¯nα,where sij is defined as follows:(19)aij≥0⇒sij=si+n,j+n=aij,aij<0⇒si+n,j=si,j+n=aij,1≤i≤n,1≤j≤n.
The other components of the “S” that were not initialized above are set to zero. In this case, we will have(20)A=D+C.
Using the system matrix symbol, (14) is written as follows:(21)Su_jr,B1j=Y_ir,R1i,where sij≥0, for 1≤i,j≤2n, and(22)x˜r,B=u_1r,B11,…,u_nr,B1n,x¯1r,B21,…,x¯nr,B2nT,where B=B11,…,B1n,B21,…,B2n and(23)Y˜r,R=Y¯1r,R11,…,Y¯nr,R1n,Y¯1r,R21,…,Y¯nr,R2nT,where R=R11,…,R1n,R21,…,R2n.
Thus,(24)DCCDu_r,B1−x¯r,B2=Y_r,R1−Y¯r,R2,where the matrices B and C contain the positive and negative elements of the matrix A, respectively.
It is clear in the systemSx˜r,B=Y˜r,R that whenever the matrix S is invertible, the response vector of the system is calculated as x˜r,B=S−1Y˜r,R. The following theorem can be useful in this regard [1].Theorem 2 (see [7]).
MatrixS is invertible if and only if D+C and D−C matrices are invertible.Definition 9.
Suppose the answerx˜αr=u_jαr,−x¯jαr,1≤j≤n, r is unique.
ujαr=u_jαr,u¯jαr1≤j≤n where(25)u¯jαr=minx¯jαr,x¯jαr,x¯jα1,u¯jαr=maxx¯jαr,x¯jαr,x¯jα1,is called a Zα answer for the system SX˜α=Y˜α. If X˜jα=x_jαr,x¯jαr1≤j≤n and the Z˜α are triangular, then u_jαr=x¯jαr and u¯jαr=x¯jαr, for all values 1≤j≤n, and u¯jαr is called a strong Zα answer for the system.


## 3.2. The Proposed Method for SolvingZ-System of Linear Equations: Case 2

In this section, suppose thatB˜, the probability distribution function of the confidence functions R˜, is specified in (3).SupposeR˜i=r1i,r2i,r3i. Now, using Definition 6, the right side of (3) can be rewritten as follows:(26)y˜ir,R˜i=y˜ir,r1i,r2i,r3i,where α is calculated as follows Assumingy˜ir=p1i,p2i,p3i:Using Definition6, we have(27)y˜iα=p1i,p2i,p3i,r1i,r2i,r3i=r^1i,r^2i,r^3i,where r^1i,r^2i,r^3i=αp1i,αp2i,αp3i. The α calculation algorithm is given in [10].Using (13), (26), and (27), we can rewrite the system of (2) as follows:(28)∑j=1naijxjZ=r^1i,r^2i,r^3i,i=1,…,n,where r^1i,r^2i,r^3i∈RF, is given and determining the unknown Z-valuation vector xjZ including xjZ=x˜jr,B˜j=x˜jr,βμx˜j is the purpose. Assuming x˜jr=q1j,q2j,q3j. Using the Definition 6, we have(29)x˜jr,βμx˜j=q^1j,q^2j,q^3j=x˜jβ,where q^1j,q^2j,q^3j=βq1j,βq2j,βq3j, and β is calculated similarly to α.Definition 10.
A fuzzy vectorxjZ=x1Z,…,xnZt=x˜1r,βμx˜1,…,x˜nr,βμx˜nt=x˜1β,…,x˜nβt is called a solution for a ZSLE (12) whenever it holds in the equations system. If aij>0, we consider the ZSLE (12) as follows:(30)∑j=1naijx˜jβr=b^1i,b^2i,b^3i.
Consider the parametric formx˜jβr=x˜jβr,x¯jβr; so,(31)∑j=1naijx_jβr=y^ir,∑j=1naijx¯jβr=y^¯ir.
Therefore, to solve a ZSLE, a system of ordinary linear equations2n×2n must be solved, in which the column of values in the right side of the system is y_1β,…,y¯nβ,y¯1β,…,y¯nβT. We show the new 2n×2n system as follows:(32)s11x_1β+s12x_2β+⋯+s1nx_nβ+s1,n+1x¯1β+⋯+s1,nx¯nβ=y_1α,⋮sn1x_1β+sn2x_2β+⋯+snnx_nβ+sn,n+1−x¯1β+⋯+sn,2n−x¯nβ=y_nα,sn+1,1x_1β+sn+1,2x_2β+⋯+sn+1,nx_nβ+sn+1,n+1−x¯1β+⋯+sn+1,2n−x¯nβ=−y¯1α,⋮s2n1x_1β+s2n2x_2β+⋯+s2n,nx_nβ+s2n,n+1−x¯1β+⋯+s2n,2n−x¯nβ=−y¯nα,in which sij are defined as follows:(33)aij≥0⇒sij=si+n,j+n=aij,aij<0⇒si+n,j=si,j+n=aij,1≤i≤n,1≤j≤n,and whenever sij are not specified above, then they are equal to zero. In this case, we will have(34)A=D+C.
Using the system matrix symbol, (30) is written as follows:(35)SX˜β=Y˜α,where sij≥0, for 1≤i,j≤2n and X˜β=x_1β,…,x_nβ,x¯1β,…,x¯nβT and Y˜α=y_1α,…,y_nα,y¯1α,…,y¯nαT; thus,(36)DCCDX_β−X¯β=Y_α−Y¯α,where the matrices B and C contain the positive and negative elements of the matrix A, respectively.
It is clear in the systemSX˜β=Y˜α that whenever the matrix S is invertible, the response vector of the system is calculated as X˜β=S−1Y˜α. The following theorem can be useful in this regard.Definition 11 (Reference [15]).
IfX=u_1β,x_2β,…,x_nβ,x¯1β,x¯2β,…,x¯nβT satisfies in SX=b and for each 1≤j≤n, the inequalities x_jβ≤x¯jβ holds, then the solution X=x_1β,x_2β,…,x_nβ,x¯1β,x¯2β,…,x¯nβT is called a strong solution of the SX=b.


## 3.3. Jacobi Method

LetAXZ=bZ be a square system of n linear equations, where(37)A=a11a12…a1na21a22…a2n⋮⋮⋱⋮an1an2…ann,XZ=x1Zx2Z⋮xnZ,bZ=b1Zb2Z⋮bnZ.A can be decomposed into a diagonal component D, a lower triangular part L, and an upper triangular part U and given as follows: A=D+L+U, where D=a110⋮00a22⋮0……⋱…00⋮0, and L+U=0a21⋮an1a120⋮an2……⋱…a1na2n⋮0.Then, the solution is obtained iteratively via(38)Xk+1Z=D−1b−L+UXkZ,where XkZ is the kth approximation or iteration of XZ and Xk+1Z is the next or(k + 1)th iteration of X. The element-based formula is written as follows:(39)xik+1Z=1aiibi−∑j≠iaijxjkZ,i=1,…,n.The computation ofxik+1Z requires each element in xjkZ except where i=j.


## 4. Determining the Balance Value of the Market by the Z-System of Linear Equations

First, some of the concepts that are required for this section are stated. Then, the amount of market equilibrium in environmentZ is determined by using a system of linear equations.Definition 12.
Demand Individual demand is the number of goods that the purchaser is willing to purchase because of its price and the stability of other factors in a certain period [16]. Of course, it should be noted that there is a demand for services such as passenger transport services (see [16] for more details). Need and demand are different, in which a lot of goods and services may be required but may not be demand for them. For example, a person may need a plane but does not demand it. Some of our needs become demand due to price and income (see [16] for more details).
The amount of demand for goods is affected by the following factors:(40)Qxd=FPx,I,Py,T,Ax,Ed,…,,where Px is the price of the good x, Py is the price of other goods, I is the consumer income or budget, T is the taste and preference of the consumer that can be created due to his needs, in which the source of these needs can be due to social customs and habits or, above all, the product of his values and beliefs, Ax are the advertisements for goods, and Ed represents the factor of price expectations is demand so that consumer demand is influenced by his expectations of either availability or nonavailability of goods in the future, as along with his prediction of future price trends of this product.
Note that if the price of goods is not in the above equation, we keep other factors affecting demand constantly. Hence,(41)Qxd=fPx.
This equation is called the demand function, which indicates the relationship between the price of a good and the amount of demand for the same good, assuming that other factors are constant. Table1 reports the individual’s demand for good x and the different quantities of commodity, which the consumer wants to purchase at various prices other than good x, assuming that other factors are constant. Moreover, Figure 1 illustrates the demand curve based on Table 1.
Dx is demand curve of x in Figure 1. According to Table 1, the equation of individual demand for good x is QX=100−2Px (by assuming that other factors are constant).Table 1
The demands.
PxQxd01001080206030404020500Figure 1
The demand curve.Definition 13.
Supply Individual supply is considered as the number of goods that the seller is willing to offer in the market for a certain period of time because of their prices and the stability of other factors [6]. The supply of a good is defined as the maximum quantity of that good, which the seller offers based on the price of that good in the market. Of course, it should be noted that there are service supplies in the economy, such as the supply of taxi services (see [16] for more details). Apart from the price factor Px that is a substantial variable affecting supply, there are some important factors such as production total costs (TCs). TC includes the price of production institutions and so on, Py is the price of related goods, T is the level of technology or technical knowledge, and Es is the factor of supply price expectations, which are effective to supply a product. Overall, the supply is related to profit; it means if profit increases, supply also increases, and vice versa. Meanwhile, profit also depends on the above factors. Therefore, the general form of the individual supply function can be described as follows:(42)Qxs=FPx,I,TC,Py,T,Es,….
If in the above equation (42), we assume other factors are constant except for the price of goods, we have the following equation (43):(43)Qxs=fPx.
This equation is called the supply function. As a result, it can be concluded that the supply function is a function that shows the relationship between the price of a good and the supply of the same good, assuming that other factors are constant. It is worth noting that this function can be interpreted in two ways:(1)
The minimum price that the supplier is willing to offer the goods(2)
It shows the maximum amount offered for each price
In the following, Table2 lists the supply table of a person for a good x and different quantities of that commodity that he is willing to sell at various prices other than the good x, assuming that other factors are constant. Moreover, Figure 2 depicts the supply cure according to Table 2.
According to Table2, the supply equation of a person for a good x is equal to the equation of QX=100−2Px (assuming that other factors are constant).Table 2
The supply table.
PxQxs501020154020602580Figure 2
The supply curve.Definition 14.
Balance A balance is defined as a state in which there is no motivation, stimulus, or force to change it. If we are not at balance, we tend to change [16]. The condition for market balance is that the supply equals demand; in other words, excess-demand or excess-supply would be equal to zero. Geometrically, as the balance occurs at the intersection of supply and demand curves, the balance price and value can be achieved [5]. In Figure 3, E is the balance point between Px∗ and Qx∗, i.e., the price and the balance value in Px∗isQxd=Qxs . Note that P1 is not a balance price because there is an excess-supply ES=Qxs−Qxd at this price. In other words, it is an incentive for suppliers to lower the price of their goods so that their goods can be sold. Besides, as there is excess-demand ED=Qxd−Qxs in P0, there is an incentive to raise prices. In Figure 3, the balance quantity of the goods x can be seen.
According to the supply and demand functions, the balance price and quantity of the goodX are obtained using the market balance condition as follows:
At the meantime, as the balance condition requiresSx=Dx, therefore, we have(44)−20+4Px=100−2Px⇒Px=20,(45)Qx=100−220=60.
Thus, the balance condition is generated between the price and quantity of the goodX, as illustrated in Figure 4.Figure 3
The balance quantity of the goodsx.Figure 4
The generated balance condition.


## 5. Determining the Balance Value of the Market inZ Environment Using the Linear Equations System

Assume that supply and demand are linear functions of price as follows:(46)qd=a∗p+b,qs=c∗p+d,where qs and qd, respectively, are the demand and supply values and they are equal. Price parameter p and real numbers a and c are definite, and coefficients b and d must be determined.Case 3.
Suppose there are some inaccurate parameters in the relationship between supply and demand that are given as follows:(47)qdZ=pZ+3+2r,7−2r,Rd,qsZ=−3pZ+9+4r,17−4r,Rs,where the coefficients b and d are given as Z-numbers. .Z is the symbol for valuing Z.
Also, according to the contents of the first case from Section3, Rd and Rs are considered as follows:(48)Rd=1−e−λ3+2r−e−λ7−2r,Rs=1−e−λ9+4r−e−λ17−4r.
Now, ifx1Z and x2Z are considered supply (demand) and price values, respectively, then by equating supply and demand values, a set of linear equations Z must be solved as follows:(49)x1Z=x2Z+3+2r,7−2r,1−e−λ3+2r−e−λ7−2r,x1Z=−3x2Z+9+4r,17−4r,1−e−λ9+4r−e−λ17−4r.
By rewriting the above equations, the following system is obtained as follows:(50)x1Z−x2Z=3+2r,7−2r,1−e−λ3+2r−e−λ7−2r,x1Z+3x2Z=9+4r,17−4r,1−e−λ9+4r−e−λ17−4r.
So, we have(51)1−113x1Zx2Z=3+2r,7−2r,1−e−λ3+2r−e−λ7−2r9+4r,17−4r,1−e−λ9+4r−e−λ17−4r.
Using relation (13), we can convert relation (51) as follows:(52)1−113x˜1r,B˜1x˜2r,B˜2=3+2r,7−2r,1−e−λ3+2r−e−λ7−2r9+4r,17−4r,1−e−λ9+4r−e−λ17−4r.
Using Definition8, we can convert relation (52) as follows:(53)1−113x¯1r,x¯1r,B˜1x¯2r,x¯2r,B˜2=3+2r,7−2r,1−e−λ3+2r−e−λ7−2r9+4r,17−4r,1−e−λ9+4r−e−λ17−4r,Theorem 3.
IfX and Y are two independent random variables with probability density quasi functions, fy, fx, then the probability density quasi functions for random variable Z=X+Y is(54)fzt=fXt∗fYt=∫−∞∞fXxfYt−xdx.To solve the system (53), the S matrix of extended coefficients is first obtained as follows:(55)1−e−λx_1r.e−λ−x¯2r=1−e−λ3+2r,1−e−λx_1r.e−λ3u_2r=1−e−λ7−2r,1−e−λx¯1r.e−λ−x_2r=1−e−λ9+4r,1−e−λx¯1r.e−λ3x¯2r=1−e−λ17−4r.So,(56)e−λx_1r+−x¯2r=e−λ3+2r,e−λx_1r+3u_2r=e−λ7−2r,e−λx¯1r+−x_2r=e−λ9+4r,e−λx¯1r+3x¯2r=e−λ17−4r.So,(57)u_1r+−x¯2r=3+2r,u_1r+3u_2r=7−2r,x¯1r+−u_2r=9+4r,x¯1r+3x¯2r=17−4r.So, using (57), we have(58)S=100−113000−1100013.Also,(59)S−1=1.1250−0.1250−0.37500.3750−0.37500.37500.1250−0.1250−0.37500.37501.1250−0.12500.1250−0.1250−0.37500.3750.Using the Jacobi method mentioned in Section3 to solve the system of linear equations, the answer to the problem can be calculated as follows:(60)x˜r,B˜=S−1b˜=6+r1+r8−r3−r.So,(61)u_1r=min6+r,8−r,7,7=6+r,Bu_1=e−λu_1r=e−λ6+r,u¯1r=max6+r,8−r,7,7=8−r,Bu¯1=e−λu¯1r=e−λ8+r,u_2r=min1+r,3−r,2,2=1+r,Bu_2=e−λu_2r=e−λ1+r,u¯2r=max1+r,3−r,2,2=3−r,Bu¯2=e−λu¯2r=e−λ3−r,U˜=6+r1+r8−r3−r.So,(62)UZ=6+r,8−r,R11+r,3−r,R2,where(63)R1=1−e−λ6+r−e−λ8+r,R2=1−e−λ1+r−e−λ3−r.So,(64)UZ=6+r,8−r,1−e−λ6+r−e−λ8+r1+r,3−r,1−e−λ1+r−e−λ3−r.In Table3, different values of UZ for λ=1 and r=0,0.1,…1 can be seen.In Figure5, the constraint of U˜1 and U˜2 can be seen. Also, in Figures 6 and 7, the confidence diagram can be seen for U˜1 and U˜2, respectively.Table 3
The different values ofUZ for λ=1 and r=0,0.1,…,1.
rU˜1,R1U˜2,R20((6, 8), 0.99)((1, 3), 0.68)0.1((6.1, 7.9), 0.998)((1.1, 2.9), 0.72)0.2((6.2, 7.8), 0.9983)((1.2, 2.8), 0.75)0.3((6.3, 7.7), 0.9986)((1.3, 2.7), 0.79)0.4((6.4, 7.6), 0.9988)((1.4, 2.6), 0.82)0.5((6.5, 7.5), 0.999)((1.5, 2.5), 0.85)0.6((6.6, 7.4), 0.9992)((1.6, 2.4), 0.88)0.7((6.7, 7.3), 0.9994)((1.7, 2.3), 0.91)0.8((6.8, 7.2), 0.9996)((1.8, 2.2), 0.94)0.9((6.9, 7.1), 0.9998)((1.9, 2.1), 0.97)1((7, 7), 1)((2, 2), 1)Figure 5
Chart of the restriction part ofU˜1 and U˜2.Figure 6
The confidence chart ofU˜1.Figure 7
The confidence chart ofU˜2.Case 4.
Suppose there are some inaccurate parameters in the relationship between supply and demand that are given as follows:(65)qdZ=pZ+3+2r,7−2r,0.677,0.777,0.877,qsZ=−3pZ+9+4r,17−4r,0.677,0.777,0.877,where the coefficients b and d are given as Z-numbers, and also, .Z is the symbol for valuing Z. Now, if x1Z and x2Z are considered supply (demand) and price values, respectively, then by equating supply and demand values, we have a Z-system of linear equations that must be solved as follows:(66)x1Z=x2Z+3+2r,7−2r,0.677,0.777,0.877,x1Z=−3x2Z+9+4r,17−4r,0.677,0.777,0.877.
By rewriting the above equations, the following system is obtained as follows:(67)x1Z−x2Z=3+2r,7−2r,0.677,0.777,0.877,x1Z+3x2Z=9+4r,17−4r,0.677,0.777,0.877.
So, we have(68)1−113x1Zx2Z=3+2r,7−2r,0.677,0.777,0.8779+4r,17−4r,0.677,0.777,0.877.
Using relation (16) and (17), write the right side of relation (47) as follows:(69)1−113x1Zx2Z=2.64+3.76r,6.16−1.76r7.92+3.52r,14.96−3.52r=y1αy2α,α=0.777.
Using relation (18), we can convert relation (52) as follows:(70)1−113x1βx2β=2.64+3.76r,6.16−1.76r7.92+3.52r,14.96−3.52r.
To solve the system (52), S, the matrix of extended coefficients is first obtained as follows:(71)x¯1β+−x¯2β=2.64+3.76r,x¯1β+3x¯2β=6.16−1.76r,x¯1β+−u_2β=7.92+3.52r,x¯1β+3x¯2β=14.96−3.52r.
So,(72)S=100−113000−1100013.
Also,(73)S−1=1.1250−0.1250−0.37500.3750−0.37500.37500.1250−0.1250−0.37500.37501.1250−0.12500.1250−0.1250−0.37500.3750.
With suppose(74)2.64+1.76r,6.16−1.76r7.92+3.52r,14.96−3.52r=b˜10.777b˜20.777.
Using the Jacobi method mentioned in Subsection3.3, to solve the system of linear equations, the answer to the problem can be calculated as follows:(75)X˜β=S−1b˜0.777=0.88r+5.280.88r+0.887.04−0.88r2.64−0.88r.
So,(76)u_1γr=min0.88r+5.28,7.04−0.88r,6.16,6.16=0.88r+5.28,u¯1γr=max0.88r+5.28,7.04−0.88r,6.16,6.16=7.04−0.88r,u_2γr=min0.88r+0.88,2.64−0.88r,1.76,1.76=0.88r+0.88,u¯2γr=max0.88r+0.88,2.64−0.88r,1.76,1.76=2.64−0.88r,where we suppose γ=β, then(77)U˜γ=0.88r+5.280.88r+0.887.04−0.88r2.64−0.88r.
In the special case, supposeβ=α.
Therefore, the amount of fuzzy demand based onZ-numbers (also fuzzy supply based on Z-numbers) is 5.28,7.04, that is, Z˜0.777-valuation (it is equivalent to 6+r,8−r,0.777 as a Z-valuation); also, the amount of fuzzy price based on Z-numbers is 0.88,2.64, that is, Z˜0.777-valuation, (it is equivalent to 1+r,3−r,0.777 as a Z-valuation).
We now compare the results obtained from the two proposed methods.
Comparison ofu˜1r and u˜2r result using the two proposed methods is given in Tables 4–7.
In many real issues, especially in economic issues, the decision maker has to make decisions in situations where the assumption of information in definite circumstances is out of the question. One of these issues in economics occurs when the objective is to determine the equilibrium value of supply and demand. In these cases, a linear equations system is created to determine the amount of supply (demand) and price. Here, due to uncertainty, a system of linear equations withZ-value was created and solved using two proposed methods. The results obtained from the two methods were compared with different landings. Since many real problems, especially economic problems, can be experienced in a system of linear equations involving Z-number with more than two variables. Next, two linear equation systems including Z-numbers 3×3 and 5×5 are simulated and solved.Table 4
Comparison ofu˜1r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((6, 8), 1)((6, 8), 0.777)0.2((6.2, 7.8), 1)((6.2, 7.8), 0.777)0.4((6.4, 7.6), 1)((6.4, 7.6), 0.777)0.6((6.6, 7.4), 1)((6.6, 7.4), 0.777)0.8((6.8, 7.2), 1)((6.8, 7.2), 0.777)1((7, 7), 1)((7, 7), 0.777)Table 5
Comparison ofu˜2r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((1, 3), 1)((1, 3), 0.777)0.2((1.2, 2.8), 1)((1.2, 2.8), 0.777)0.4((1.4, 2.6), 1)((1.4, 2.6), 0.777)0.6((1.6, 2.4), 1)((1.6, 2.4), 0.777)0.8((1.8, 2.2), 1)((1.8, 2.2), 0.777)1((2, 2), 1)((2, 2), 0.777)Table 6
Comparison ofu˜1r result using the two proposed methods λ=0.05.
rThe first proposed methodλ=0.05The second proposed method0((6, 8), 0.495)((6, 8), 0.777)0.2((6.2, 7.8), 0.495)((6.2, 7.8), 0.777)0.4((6.4, 7.6), 0.495)((6.4, 7.6), 0.777)0.6((6.6, 7.4), 0.495)((6.6, 7.4), 0.777)0.8((6.8, 7.2), 0.495)((6.8, 7.2), 0.777)1((7, 7), 0.495)((7, 7), 0.777)Table 7
Comparison ofu˜2r result using the two proposed methods λ=0.05.
rThe first proposed methodλ=0.05The second proposed method0((1, 3), 0.8187)((1, 3), 0.777)0.2((1.2, 2.8), 0.8187)((1.2, 2.8), 0.777)0.4((1.4, 2.6), 0.8187)((1.4, 2.6), 0.777)0.6((1.6, 2.4), 0.8187)((1.6, 2.4), 0.777)0.8((1.8, 2.2), 0.8187)((1.8, 2.2), 0.777)1((2, 2), 0.8187)((2, 2), 0.777)Example 1.
Consider the3×3 Z-system.(78)2x1Z=−x2Z−3x3Z+11+8r,27−8r,R1,4x1Z=−x2Z+x3Z+−23+10r,−5−8r,R2,−x1Z=−3x2Z−x3Z+10+5r,27−12r,R3.Case 5.
We solve the problem by the first proposed method.
According to the contents of Case5 from Section 3, R1, R2,andR3 are considered as follows:(79)R1=e−λ11+8r−e−λ27−8r,R2=e−λ−23+10r−e−λ−5−8r,R3=e−λ10+5r−e−λ27−12r.
So,(80)2x1Z=−x2Z−3x3Z+11+8r,27−8r,e−λ11+8r−e−λ27−8r,4x1Z=−x2Z+x3Z+−23+10r,−5−8r,e−λ−23+10r−e−λ−5−8r,−x1Z=−3x2Z−x3Z+10+5r,27−12r,e−λ10+5r−e−λ27−12r.
By rewriting the above equations, the following system is obtained as follows:(81)2x1Z+x2Z+3x3Z=11+8r,27−8r,e−λ11+8r−e−λ27−8r,4x1Z+x2Z−x3Z=−23+10r,−5−8r,e−λ−23+10r−e−λ−5−8r,−x1Z+3x2Z+x3Z=10+5r,27−12r,e−λ10+5r−e−λ27−12r.
So, we have(82)21341−1−131x1Zx2Zx3Z=11+8r,27−8r,e−λ11+8r−e−λ27−8r−23+10r,−5−8r,e−λ−23+10r−e−λ−5−8r10+5r,27−12r,e−λ10+5r−e−λ27−12r.
Using relation (13), we can convert relation (82) as follows:(83)21341−1−131x˜1r,B˜1x˜2r,B˜2x˜3r,B˜3=11+8r,27−8r,e−λ11+8r−e−λ27−8r−23+10r,−5−8r,e−λ−23+10r−e−λ−5−8r10+5r,27−12r,e−λ10+5r−e−λ27−12r.
Using Definition8, we can convert relation (52) as follows:(84)21341−1−131x_1r,x¯1r,B˜1x_2r,x¯2r,B˜2x_3r,x¯3r,B˜3=11+8r,27−8r,e−λ11+8r−e−λ27−8r−23+10r,−5−8r,e−λ−23+10r−e−λ−5−8r10+5r,27−12r,e−λ10+5r−e−λ27−12r.
ConsiderB˜1,B˜2, and B˜3 as follows:(85)B˜1=e−λx_1r−e−λx¯1r,B˜2=e−λx_2r−e−λx¯2r,B˜3=e−λx_3r−e−λx¯3r.
To solve the system (84), S the matrix of extended coefficients is first obtained as follows:(86)2x_1r+x_2r+3u_3r=11+8r,4x_1r+x_2r+−x¯3r=27−8r,−x¯1r+3x_2r+x_3r=−23+10r,2x¯1r+x¯2r+3x¯3r=−5−8r,4x¯1r+x¯2r+−x_3r=10+5r,−x_1r+3x¯2r+x¯3r=27−12r.
So,(87)e−λ2x_1r.e−λx_2r.e−λ3x_3r=e−λ11+8r,e−λ4u_1r.e−λu_2r.e−λ−x¯3r=e−λ27−8r,e−λ−x¯1r.e−λ3u_2r.e−λu_3r=e−λ−23+10r,e−λ2x¯1r.e−λx¯2r.e−λ3x¯3r=e−λ−5−8r,e−λ4x¯1r.e−λx¯2r.e−λ−u_3r=e−λ10+5r,e−λ−u_1r.e−λ3x¯2r.e−λx¯3r=e−λ27−12r.
So,(88)e−λ2u_1r+u_2r3u_3r=e−λ11+8re−λ4u_1r+u_2r+−x¯3r=e−λ27−8re−λ−x¯1r+3u_2r+u_3r=e−λ−23+10re−λ2x¯1r+x¯2r+3x¯3r=e−λ−5−8re−λ4x¯1r+x¯2r+−u_3r=e−λ10+5re−λ−u_1r+3x¯2r+x¯3r=e−λ27−12r.
So, using (86), we have(89)S=21300041000−1031−10000021300−1410−100031.
Also,(90)S−1=0.00700.2448−0.08390.0839−0.0629−0.00070−0.09180.03760.35140.02360.0760−0.03320.35930.1757−0.0612−0.06380.01660.01570.0839−0.0629−0.0070−0.00700.2448−0.08390.02360.0760−0.0332−0.09180.03760.3514−0.06380.01660.01570.3593−0.01757−0.0612.
Using the Jacobi method mentioned in Section 3 to solve the system of linear equations, the answer to the problem can be calculated as follows:(91)x˜r,B˜=S−1b˜=2r−4r+1r+6−1−r5−3r8−r.
So,(92)u_1r=min2r−4,−1−r,−2,−2=2r−4Bu_1=e−λu_1r=e−λ2r−4u¯1r=max2r−4,−1−r,−2,−2=2r−4=−1−rBu¯1=e−λu¯1r=e−λ−1−ru_2r=min1+r,5−3r,2,2=1+rBu_2=e−λu_2r=e−λ1+ru¯2r=max1+r,5−3r,2,2=5−3rBu¯2=e−λu¯2r=e−λ5−3ru_3r=minr+6,8−r,7,7=6+rBu_2=e−λu_3r=e−λ6+ru¯3r=maxr+6,8−r,7,7=8−rBu¯2=e−λu¯3r=e−λ8−rU˜=2r−41+r6+r−r−15−3r8−r.
So,(93)UZ=2r−4,−r−1,R11+r,5−3r,R26+r,8−r,R3,where(94)R1=1−e−λ2r−4−e−λ−r−1,R2=1−e−λ1+r−e−λ5−3r,R3=1−e−λ6+r−e−λ8−r.
So,(95)UZ=2r−4,−r−1,1−e−λ2r−4−e−λ−r−11+r,5−3r,1−e−λ1+r−e−λ5−3r6+r,8−r,1−e−λ6+r−e−λ8−r.
In Table8, different values of UZ for λ=1 and r=0,0.1,…1 can be seen.
In Figure8, the constraint of U˜1, U˜2, and U˜3 can be seen. Also, in Figures 9–11, the confidence diagram can be seen for U˜1, U˜2, and U˜3, respectively.Table 8
The different values ofUZ for λ=0.05 and r=0,0.1,…,1.
rU˜1,R1U˜2,R2U˜3,R30−4,−1,0.821,5,0.826,8,0.920.2−3.6,−1.2,0.861.2,4.4,0.866.2.8,0.930.4−3.2,−1.4,0.891.4,3.8,0.896.4,7.6,0.950.6−2.8,−1.6,0.931.6,3.2,0.936.6,7.4,0.970.8−2.4,−1.8,0.961.8,2.6,0.966.8,7.2,0.981−2,−2,12,2,17,7,1Figure 8
Chart of the restriction part ofU˜1, U˜2, and U˜3.Figure 9
The confidence chart ofU˜1,U˜2,U˜3.Figure 10
Chart of the restriction part ofU˜1, U˜2, U˜3, U˜4, and U˜5.Figure 11
The confidence chart ofU˜1, U˜2, U˜3, U˜4, and U˜5.Case 6.
We solve the problem by the second proposed method.(96)2qdZ=−pZ−3eZ+11+8r,27−8r,0.6,0.7,.08,4qsZ=−pZ+eZ+−23+10r,−5−8r,0.6,0.7,.08,−qkZ=−3pZ−eZ+10+5r,27−12r,0.6,0.7,.08.
So,(97)2x1Z=−x2Z−3x3Z+11+8r,27−8r,0.6,0.7,.08,4x1Z=−x2Z+x3Z+−23+10r,−5−8r,0.6,0.7,.08,−x1Z=−3x2Z−x3Z+10+5r,27−12r,0.6,0.7,.08.
By rewriting the above equations, the following system is obtained:(98)2x1Z+x2Z+3x3Z=11+8r,27−8r,0.6,0.7,.08,4x1Z+x2Z−x3Z=−23+10r,−5−8r,0.6,0.7,.08,−x1Z+3x2Z+x3Z=10+5r,27−12r,0.6,0.7,.08.
So, we have(99)21341−1−131x1Zx2Zx3Z=11+8r,27−8r,0.6,0.7,.08−23+10r,−5−8r,0.6,0.7,.0810+5r,27−12r,0.6,0.7,.08.
Using relation (16) and (17), write the right side of relation (52) as follows:(100)21341−1−131x1Zx2Zx3Z=9.13+6.64r,22−6.64r−19.09+8.3r,−4.15−6.64r8.3+4.15r,22.41−9.96r=y1αy2α,α=0.7.
Using relation (18), we can convert relation (52) as follows:(101)21341−1−131x1βx2βx3β=9.13+6.64r,22.41−6.64r−19.09+8.3r,−4.15−6.64r8.3+4.15r,22.41−9.96r.
To solve the system (96), S, the matrix of extended coefficients is first obtained as follows:(102)2x_1β+x_2β+3x_3β=9.13+6.64r4x_1β+x_2β+−x¯3β=22−6.64r−x¯1β+3x_2β+x_3β=−19.09+8.3r2x¯1β+x¯2β+3x¯3β=−4.15−6.64r4x¯1β+x¯2β+−u_3β=8.3+4.15r−x_1β+3x¯2β+x¯3β=22.41−9.96r.
So,(103)S=21300041000−1031−10000021300−1410−100031.
Also,(104)S−1=0.00700.2448−0.08390.0839−0.0629−0.00070−0.09180.03760.35140.02360.0760−0.03320.35930.1757−0.0612−0.06380.01660.01570.0839−0.0629−0.0070−0.00700.2448−0.08390.02360.0760−0.0332−0.09180.03760.3514−0.06380.01660.01570.3593−0.01757−0.0612.
With suppose(105)9.13+6.64r,22.41−6.64r−19.09+8.3r,−4.15−6.64r8.3+4.15r,22.41−9.96r=b˜10.7b˜20.7b˜30.7.
Using the Jacobi method mentioned in Section3 to solve the system of linear equations, the answer to the problem can be calculated as follows:(106)X˜β=S−1b˜0.7=1.66r−3.320.83r+0.830.83r+4.98−0.83r−0.834.15−2.49r6.64−0.83r.
So,(107)u_1γr=min1.66r−3.32,−0.83r−0.83,−1.66,−1.66=1.66r−3.32u¯1γr=max1.66r−3.32,−0.83r−0.83,−1.66,−1.66=−0.83r−0.83u_2γr=min0.83r+0.83,4.15−2.49r,1.66,1.66=0.83r+0.83u¯2γr=max0.83r+0.83,4.15−2.49r,1.66,1.66=4.15−2.49ru_3γr=min0.83r+4.98,6.64−0.83r,5.81,5.81=0.83r+4.98u¯3γr=max0.83r+4.98,6.64−0.83r,5.81,5.81=6.64−0.83r,where we suppose γ=β.(108)U˜γ=1.66r−3.320.83r+0.830.83r+4.98−0.83r−0.834.15−2.49r6.64−0.83r.
In the special case, supposeβ=α.
Therefore, the amount of fuzzy demand based onZ-numbers (also fuzzy supply based on Z-numbers) is −3.32,−0.83, that is, Z˜0.7-valuation (it is equivalent to −4+2r,−1−r,0.7 as a Z-valuation). Also, the amount of fuzzy price based on Z-numbers is 0.83,4.15, that is, Z˜0.7-valuation (it is equivalent to 1+r,5−3r,0.7 as a Z-valuation) and the amount of fuzzy price based on Z-numbers is −0.83,6.64, that is, Z˜0.7-valuation (it is equivalent to 6+r,8−r,0.7 as a Z-valuation).
We now compare the results that are obtained from the two proposed methods. Comparison ofu˜1r,u˜2r, and u˜3r result using the two proposed methods can be seen in Tables 9–11.Table 9
Comparison ofu˜1r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((−4, −1), 1)((−4, −1), 0.7)0.2((−3.6, −1.2), 1)((−3.6, −1.2), 0.7)0.4((−3.2, −1.4), 1)((−3.2, −1.4), 0.7)0.6((−2.8, −1.6), 1)((−2.8, −1.6), 0.7)0.8((−2.4, −1.8), 1)((−2.4, −1.8), 0.7)1((−2, −2), 1)((−2, −2), 0.7)Table 10
Comparison ofu˜2r result using the two proposed methods (λ=0).
rThe first proposed methodλ=0The second proposed method0((1, 2), 1)((1, 2), 0.7)0.2((1.2, 4.4), 1)((1.2, 4.4), 0.7)0.4((1.4, 3.8), 1)((1.4, 3.8), 0.7)0.6((1.6, 3.2), 1)((1.6, 3.2), 0.7)0.8((1.8, 2.6), 1)((1.8, 2.6), 0.7)1((2, 2), 1)((2, 2), 0.7)Table 11
Comparison ofu˜3r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((6, 8), 1)((6, 8), 0.7)0.2((6.2, 7.8), 1)((6.2, 7.8), 0.7)0.4((6.4, 7.6), 1)((6.4, 7.6), 0.7)0.6((6.6, 7.4), 1)((6.6, 7.4), 0.7)0.8((6.8, 7.2), 1)((6.8, 7.2), 0.7)1((7, 7), 1)((7, 7), 0.7)Example 2.
Consider the5×5Z-system:(109)6qdZ=−pZ−3eZ+fZ−6gZ+1+r,3−r,R1,5qsZ=−9pZ−eZ−2fZ−3gZ+6+r,8−r,R2,2qkZ=−3pZ−9eZ−2fZ−3gZ+5+r,7−r,R3,−qhZ=−pZ−3eZ−8fZ−3gZ+3+r,5−r,R4,qsZ=−2pZ−2eZ−2fZ−9gZ+2+r,4−r,R5.Case 7.
We solve the problem by the first proposed method.
According to the contents of the first case from Section3, R1, R2,R3,R4,R5 are considered as follows:(110)R1=e−λ1+r−e−λ3−r,R2=e−λ6+r−e−λ8−r,R3=e−λ5+r−e−λ7−r,R4=e−λ3+r−e−λ5−r,R5=e−λ2+r−e−λ4−r.
The set of linear equationsZ must be solved as follows:(111)6x1Z=−x2Z−3x3Z+x4Z−6x5Z+1+r,3−r,e−λ1+r−e−λ3−r,5x1Z=−9x2Z−x3Z−2x4Z−3x5Z+6+r,8−r,e−λ6+r−e−λ8−r,2x1Z=−3x2Z−9x3Z−2x4Z−3x5Z+5+r,7−r,e−λ5+r−e−λ7−r,−x1Z=−x2Z−3x3Z−8x4Z−3x5Z+3+r,5−r,e−λ3+r−e−λ5−r,x1Z=−2x2Z−2x3Z−2x4Z−9gx5Z+2+r,4−r,e−λ2+r−e−λ4−r.
By rewriting the above equations, the following system is obtained as follows:(112)6x1Z+x2Z+3x3Z−x4Z+6x5Z=1+r,3−r,e−λ1+r−e−λ3−r,5x1Z+9x2Z+x3Z+2x4Z+3x5Z=6+r,8−r,e−λ6+r−e−λ8−r,2x1Z+3x2Z+9x3Z+2x4Z+3x5Z=5+r,7−r,e−λ5+r−e−λ7−r,−x1Z+x2Z+3x3Z+8x4Z+3x5Z=3+r,5−r,e−λ3+r−e−λ5−r,x1Z+2px2Z+2x3Z+2x4Z+9x5Z=2+r,4−r,e−λ2+r−e−λ4−r.
So, we have(113)652193319−122633−1112328139x1Zx2Zx3Zx4Zx5Z=1+r,3−r,e−λ1+r−e−λ3−r6+r,8−r,e−λ6+r−e−λ8−r5+r,7−r,e−λ5+r−e−λ7−r3+r,5−r,e−λ3+r−e−λ5−r2+r,4−r,e−λ2+r−e−λ4−r.
Using relation (13), we can convert relation (113) as follows:(114)613−165912323923−1138312219x˜1r,B˜1x˜2r,B˜2x˜3r,B˜3x˜4r,B˜4x˜5r,B˜5=1+r,3−r,e−λ1+r−e−λ3−r6+r,8−r,e−λ6+r−e−λ8−r5+r,7−r,e−λ5+r−e−λ7−r3+r,5−r,e−λ3+r−e−λ5−r2+r,4−r,e−λ2+r−e−λ4−r.
Using Definition8, we can convert relation (114) as follows:(115)613−165912323923−1138312219x_1r,x¯1r,B˜1x_2r,x¯2r,B˜2x_3r,x¯3r,B˜3x_4r,x¯4r,B˜4x_5r,x¯5r,B˜5=1+r,3−r,e−λ1+r−e−λ3−r6+r,8−r,e−λ6+r−e−λ8−r5+r,7−r,e−λ5+r−e−λ7−r3+r,5−r,e−λ3+r−e−λ5−r2+r,4−r,e−λ2+r−e−λ4−r.
ConsiderB˜1, B˜2,B˜3,B˜4,andB˜5 as follows:(116)B˜1=e−λx_1r−e−λx¯1rB˜2=e−λx_2r−e−λx¯2rB˜3=e−λx_3r−e−λx¯3rB˜4=e−λx_4r−e−λx¯4rB˜5=e−λx_5r−e−λx¯5r.
To solve the system (115), S, the matrix of extended coefficients is first obtained as follows:(117)6x_1r+x_2r+3x_3r+−x¯4r+6x_5r=1+r,5x_1r+9x_2r+x_3r+2x_4r+3x_5r=3−r,2x_1r+3x_2r+9x_3r+2x_4r+3x_5r=6+r,−x¯1r+x_2r+3x_3r+8x_4r+3x_5r=8−r,x_1r+2x_2r+2x_3r+x_4r+9x_5r=5+r,6x¯1r+x¯2r+3x¯3r+−x_4r+6x¯5r=7−r,5x¯1r+9x¯2r+x¯3r+2x¯4r+3x¯5r=3+r,2x¯1r+3x¯2r+9x¯3r+2x¯4r+3x¯5r=5−r,−x_1r+x¯2r+3x¯3r+8x¯4r+3x¯5r=2+r,x¯1r+2x¯2r+2x¯3r+x¯4r+9x¯5r=4−r.
So,(118)e−λ6x_1r⋅e−λx_2r⋅e−λ3_x3r⋅e−λ−x¯4r⋅e−λ6x_5r=e−λ1+r,e−λ5x_1r⋅e−λ9x_2r⋅e−λx_3r⋅e−λ2x_4r⋅e−λ3x_5r=e−λ3−r,e−λ2x_1r⋅e−λ3x_2r⋅e−λ9x_3r⋅e−λ2x_4r⋅e−λ3x_5r=e−λ6+r,e−λ−x¯1r⋅e−λx_2r⋅e−λ3x_3r⋅e−λ8x_4r⋅e−λ3x_5r=e−λ8−r,e−λx_1r⋅e−λ2x_2r⋅e−λ2x_3r⋅e−λx_4r⋅e−λ9x_5r=e−λ5+r,e−λ6x_1r⋅e−λx¯2r⋅e−λ3x¯3r⋅e−λx_4r⋅e−λ6x¯5r=e−λ7−r,e−λ5x¯1r⋅e−λ9x¯2r⋅e−λx¯3r⋅e−λ2x¯4r⋅e−λ3x¯5r=e−λ3+r,e−λ2x¯1r⋅e−λ3x¯2r⋅e−λ9x¯3r⋅e−λ2x¯4r⋅e−λ3x¯5r=e−λ5−r,e−λ−x_1r⋅e−λx¯2r⋅e−λ3x¯3r⋅e−λ8x¯4r⋅e−λ3x¯5r=e−λ2+r,e−λx¯1r⋅e−λx¯2r⋅e−λ2x¯3r⋅e−λx¯4r⋅e−λ9x¯5r=e−λ4−r.
So,(119)e−λ6x_1r+x_2r+3x_3r+−x¯4r+6x_5r=e−λ1+r,e−λ5x_1r+9x_2r+x_3r+2x_4r+3x_5r=e−λ3−r,e−λ2x_1r+3x_2r+9x_3r+2x_4r+3x_5r=e−λ6+r,e−λ−x¯1r+x_2r+3x_3r+8x_4r+3x_5r=e−λ8−r,e−λx_1r+2x_2r+2x_3r+x_4r+9x_5r=e−λ5+r,e−λ6x¯1r+x¯2r+3x¯3r+−x_4r+6x¯5r=e−λ7−r,e−λ5x¯1r+9x¯2r+x¯3r+2x¯4r+3x¯5r=e−λ3+r,e−λ2x¯1r+3x¯2r+9x¯3r+2x¯4r+3x¯5r=e−λ5−r,e−λ−x_1r+x¯2r+3x¯3r+8x¯4r+3x¯5r=e−λ2+r,e−λx¯1r+2x¯2r+2x¯3r+x¯4r+9x¯5r=e−λ4−r.
So, using (117), we have(120)S=61306000−105912300000239230000001383−100001221900000000−106130600000591230000023923−10000013830000012319.
Also,(121)S−1=0.19120.0195−0.04550.0233−0.12660.00740.0018−0.00920.0272−0.0115−0.11010.10910.0291−0.04080.0410−0.0094−0.00150.0065−0.01630.0100−0.0109−0.03590.1297−0.0219−0.0167−0.0042−0.00050.0014−0.00200.00320.01740.0073−0.04390.1434−0.04720.02700.0028−0.00710.0055−0.01840.0037−0.0192−0.0254−0.00460.1250−0.0008−0.00010.00000.00040.00040.00740.0018−0.00920.0272−0.01150.19120.1912−0.04550.0233−0.1266−0.0094−0.00150.0065−0.01630.0100−0.11010.10910.0291−0.04080.0410−0.0042−0.00050.0014−0.00200.0032−0.0109−0.03590.1297−0.0219−0.01670.02700.0028−0.00710.0055−0.01840.01740.0073−0.04390.1434−0.0472−0.0008−0.00010.00000.00040.00040.0037−0.0192−0.0254−0.00460.1250.
Using the Jacobi method mentioned in Section3, to solve the system of linear equations, the answer to the problem can be calculated as follows:(122)x˜r,B˜=S−1b˜=0.046214r−0.0409030.613045+0.038839r0.319445+0.04646r0.185413+0.0671091r0.001054+0.079564r0.051525−0.046214r0.69072−0.038839r0.412365−0.04646r0.3196317−0.067109r0.158073−0.079564r.
So,(123)u_1r=min0.046214r−0.040903,0.051525−0.046214r,0.005311,0.005311=0.046214r−0.040903,Bu_1=e−λu_1r=e−λ0.046214r−0.040903,u¯1r=max0.046214r−0.040903,0.051525−0.046214r,0.005311,0.005311=0.051525−0.046214r,Bu¯1=e−λu¯1r=e−λ0.051525−0.046214r,u_2r=min0.613045+0.038839r,0.69072−0.038839r,0.651884,0.651884=0.613045+0.038839r,Bu_2=e−λu_2r=e−λ0.613045+0.038839r,u¯2r=max0.613045+0.038839r,0.69072−0.038839r,0.651884,0.651884=0.69072−0.038839r,Bu¯2=e−λu¯2r=e−λ0.69072−0.038839r,u_3r=min0.319445+0.04646r,0.412365−0.04646r,0.365905,0.365905=0.319445+0.04646r,Bu_3=e−λu_3r=e−λ0.319445+0.04646r,u¯3r=max0.319445+0.04646r,0.412365−0.04646r,0.365905,0.365905=0.412365−0.04646r,Bu¯3=e−λu¯3r=e−λ0.412365−0.04646r,u_4r=min0.185413+0.0671091r,0.3196317−0.067109r,0.2525,0.2525=0.185413+0.0671091r,Bu_4=e−λu_4r=e−λ0.185413+0.0671091r,u¯4r=max0.185413+0.0671091r,0.3196317−0.067109r,0.2525,0.2525=0.3196317−0.067109r,Bu¯4=e−λu¯4r=e−λ0.3196317−0.067109r,u_5r=min0.001054+0.079564r,0.158073−0.079564r,0.0785,0.0875=0.001054+0.079564r,Bu_5=e−λu_5r=e−λ0.001054+0.079564r,u¯5r=max0.001054+0.079564r,0.158073−0.079564r,0.0785,0.0875=0.158073−0.079564r,Bu¯5=e−λu¯5r=e−λ0.158073−0.079564r.
So,(124)U˜=0.046214r−0.0409030.613045+0.038839r0.319445+0.04646r0.185413+0.0671091r0.001054+0.079564r0.051525−0.046214r0.69072−0.038839r0.412365−0.04646r0.3196317−0.067109r0.158073−0.079564r.
So,(125)UZ=0.046214r−0.040903,0.051525−0.046214r,R10.613045+0.038839r,0.69072−0.038839r,R20.319445+0.04646r,0.412365−0.04646r,R30.185413+0.0671091r,0.3196317−0.067109r,R40.001054+0.079564r,0.158073−0.079564r,R5,where(126)R1=1−e−λ0.046214r−0.040903−e−λ0.051525−0.046214r,R2=1−e−λ0.613045+0.038839r−e−λ0.69072−0.038839r,R3=1−e−λ0.319445+0.04646r−e−λ0.412365−0.04646r,R4=1−e−λ0.185413+0.0671091r−e−λ0.3196317−0.067109r,R5=1−e−λ0.001054+0.079564r−e−λ0.158073−0.079564r.
In Table12, different values of UZ for λ=1 and r=0,0.1,…1 can be seen.
In Figure10, the constraint of U˜1, U˜2, U˜3, U˜3, and U˜5 can be seen. Also, in Figure 11, the confidence diagram can be seen for U˜1, U˜2, U˜3, U˜3, and U˜5, respectively.Table 12
The different values ofUZ for λ=0.05 and r=0,0.1,…,1.
rU˜1,R1U˜2,R2U˜3,R3U˜4,R4U˜5,R50−0.04,0.05,0.900.61,0.69,0.950.31,0.41,0.90.18,0.31,0.890.001,0.15,0.850.2−0.03,0.04,0.920.62,0.68,0.960.32,0.40,0.940.19,0.30,0.910.016,0.14,0.880.4−0.02,0.03,0.940.62,0.67,0.970.33,0.39,0.960.21,0.29,0.930.03,0.12,0.9130.6−0.01,0.02,0.960.63,0.66,0.980.34,0.38,0.970.22,0.27,0.950.04,0.11,0.940.8−0.003,0.01,0.980.64,0.65,0.990.35,0.37,0.980.23,0.26,0.970.06,0.09,0.9710.005,0.005,10.65,0.65,10.360.36,10.25,0.25,10.08,0.07,1Case 8.
We solve the problem by the second proposed method.(127)6qdZ+pZ+3eZ−fZ+6gZ=1+r,3−r,0.5,0.6,0.7,5qsZ+9pZ+eZ+2fZ+3gZ=6+r,8−r,0.5,0.6,0.7,2qkZ+3pZ+9eZ+2fZ+3gZ=5+r,7−r,0.5,0.6,0.7,−qhZ+pZ+3eZ+8fZ+3gZ=3+r,5−r,0.5,0.6,0.7,qsZ+2pZ+2eZ+2fZ+9gZ=2+r,4−r,0.5,0.6,0.7.
So,(128)6x1Z=−x2Z−3x3Z+x4Z−6x5Z+1+r,3−r,0.5,0.6,0.7,5x1Z=−9x2Z−x3Z−2x4Z−3x5Z+6+r,8−r,0.5,0.6,0.7,2x1Z=−3x2Z−9x3Z−2x4Z−3x5Z+5+r,7−r,0.5,0.6,0.7,−x1Z=−x2Z−3x3Z−8x4Z−3x5Z+3+r,5−r,0.5,0.6,0.7,x1Z=−2x2Z−2x3Z−2x4Z−9gx5Z+2+r,4−r,0.5,0.6,0.7.
By rewriting the above equations, the following system is obtained:(129)6x1Z+x2Z+3x3Z−x4Z+6x5Z=1+r,3−r,0.5,0.6,0.7,5x1Z+9x2Z+x3Z+2x4Z+3x5Z=6+r,8−r,0.5,0.6,0.7,2x1Z+3x2Z+9x3Z+2x4Z+3x5Z=5+r,7−r,0.5,0.6,0.7,−x1Z+x2Z+3x3Z+8x4Z+3x5Z=3+r,5−r,0.5,0.6,0.7,x1Z+2px2Z+2x3Z+2x4Z+9x5Z=2+r,4−r,0.5,0.6,0.7.
So, we have(130)613−165912323923−1138312219x1Zx2Zx3Zx4Zx5Z=1+r,3−r,0.5,0.6,0.76+r,8−r,0.5,0.6,0.75+r,7−r,0.5,0.6,0.73+r,5−r,0.5,0.6,0.72+r,4−r,0.5,0.6,0.7.
Using relation (27) and (28), write the right side of relation (130) as follows:(131)613−165912323923−1138312219x1Zx2Zx3Zx4Zx5Z=0.77+0.77r,2.31−0.77r4.62+0.77,6.16−0.77r3.85+0.77r+5.39−0.77r2.31+0.77r,3.85−0.77r1.54+0.77r,3.08−0.77r=y1αy2α,α=0.6.
Using relation (29), we can convert relation (131) as follows:(132)613−165912323923−1138312219x1βx2βx3βx4βx5β=0.77+0.77r,2.31−0.77r4.62+0.77,6.16−0.77r3.85+0.77r+5.39−0.77r2.31+0.77r,3.85−0.77r1.54+0.77r,3.08−0.77r.
To solve the system (127), S, the matrix of extended coefficients is first obtained as follows:(133)6x_1r+x_2r+3x_3r+−x¯4r+6x_5r=0.77+0.77r,5x_1r+9x_2r+x_3r+2x_4r+3x_5r=2.31−0.77r,2x_1r+3x_2r+9x_3r+2x_4r+3x_5r=4.62+0.77,−x¯1r+x_2r+3x_3r+8x_4r+3x_5r=6.16−0.77r,x_1r+2x_2r+2x_3r+x_4r+9x_5r=3.85+0.77r,6x¯1r+x¯2r+3x¯3r+−x_4r+6x¯5r=5.39−0.77r,5x¯1r+9x¯2r+x¯3r+2x¯4r+3x¯5r=2.31+0.77r,2x¯1r+3x¯2r+9x¯3r+2x¯4r+3x¯5r=3.85−0.77r,−x_1r+x¯2r+3x¯3r+8x¯4r+3x¯5r=1.54+0.77r,x¯1r+2x¯2r+2x¯3r+x¯4r+9x¯5r=3.08−0.77r.
So,(134)S=61306000−105912300000239230000001383−100001221900000000−106130600000591230000023923−10000013830000012319.
Also, suppose(135)S−1=0.19120.0195−0.04550.0233−0.12660.00740.0018−0.00920.0272−0.0115−0.11010.10910.0291−0.04080.0410−0.0094−0.00150.0065−0.01630.0100−0.0109−0.03590.1297−0.0219−0.0167−0.0042−0.00050.0014−0.00200.00320.01740.0073−0.04390.1434−0.04720.02700.0028−0.00710.0055−0.01840.0037−0.0192−0.0254−0.00460.1250−0.0008−0.00010.00000.00040.00040.00740.0018−0.00920.0272−0.01150.19120.1912−0.04550.0233−0.1266−0.0094−0.00150.0065−0.01630.0100−0.11010.10910.0291−0.04080.0410−0.0042−0.00050.0014−0.00200.0032−0.0109−0.03590.1297−0.0219−0.01670.02700.0028−0.00710.0055−0.01840.01740.0073−0.04390.1434−0.0472−0.0008−0.00010.00000.00040.00040.0037−0.0192−0.0254−0.00460.1250,0.77+0.77r,5.39−0.77r2.31−0.77r,2.31+0.77r4.62+0.77,3.85−0.77r6.16−0.77r,1.54+0.77r3.85+0.77r,3.08−0.77r=b˜10.6b˜20.6b˜30.6b˜40.6b˜50.6.
Using the Jacobi method mentioned in Section3, to solve the system of linear equations, the answer to the problem can be calculated as follows:(136)X˜β=S−1b˜0.6=0.035585r−0.031496,0.039674−0.035585r0.029907r+0.472045,0.531858−0.029907r0.035774r+0.245973,0.317522−0.035774r0.051674r+0.142768,0.246116−0.0516740.061264r−0.000812,0.121717−0.061264r.
So,(137)u_1γr=min0.035585r−0.031496,0.039674−0.035585r,0.004,0.004=0.035585r−0.031496,u¯1γr=max0.035585r−0.031496,0.039674−0.035585r,0.004,0.004=0.039674−0.035585r,u_2γr=min0.029907r+0.472045,0.531858−0.029907r,0.501,0.501=0.029907r+0.472045,u¯2γr=max0.029907r+0.472045,0.531858−0.029907r,0.501,0.501=0.531858−0.029907r,u_3γr=min0.035774r+0.245973,0.317522−0.035774r,0.281,0.281=0.035774r+0.245973,u¯3γr=max0.035774r+0.245973,0.317522−0.035774r,0.281,0.281=0.317522−0.035774r,u_4γr=min0.051674r+0.142768,0.246116−0.051674,0.194,0.194=0.051674r+0.142768,u¯4γr=max0.051674r+0.142768,0.246116−0.051674r,0.194,0.194=0.246116−0.051674r,u_5γr=min0.061264r−0.000812,0.121717−0.061264r,0.06,0.06=0.061264r−0.000812,u¯5γr=max0.061264r−0.000812,0.121717−0.061264r,0.06,0.06=0.121717−0.061264r.
Here, we supposeγ=β.(138)U˜γ=0.035585r−0.0314960.029907r+0.4720450.035774r+0.2459730.051674r+0.1427680.061264r−0.0008120.039674−0.035585r0.531858−0.029907r0.245973−0.035774r0.246116−0.051674r0.121717−0.061264r.
In the special case, supposeβ=α.
Therefore, the amount of fuzzy demand based onZ-numbers (also fuzzy supply based on Z-numbers) is −0.031496,0.039674, that is, Z˜10.6-valuation (it is equivalent to −0.0409+0.0462r,0.0515−0.0462r,0.6 as a Z-valuation).
Also, the amount of fuzzy price based on Z-numbers is0.472045,0.531858, that is, Z˜20.6-valuation (it is equivalent to 0.613+0.0388r,0.6907−0.0389r,0.6 as a Z-valuation).(139)0.245973,0.245973isZ˜30.6−valuation,it.is.equivalent.to0.1854+0.0464r,0.4123−0.1805r,0.6asaZ−valuation,0.142768,0.246116isZ˜40.6−valuation,it.is.equivalent.to0.1854+0.0671r,0.3196−0.0671r,0.6asaZ−valuation,−0.000812,0.121717isZ˜50.6−valuation,it.is.equivalent.to−0.001+0.795r,0.1580−0.0795r,0.6asaZ−valuation.
We now compare the results obtained from the two proposed methods. Comparison ofu˜1r,u˜2r,u˜3r,u˜4r, and u˜5r result using the two proposed methods is given in Tables 13–17.Table 13
Comparison ofu˜1r result using the two proposed methods (λ=0).
rThe first proposed method (λ=0)The second proposed method0((−0.0409, 0.908), 0.9)((−0.0409, 0.0515), 0.6)0.2((−0.031, 0.926), 0.92)((−0.03166, 0.04226), 0.6)0.4((−0.022, 0.944), 0.94)((−0.02242, 0.03302), 0.6)0.6((−0.013, 0.963), 0.96)((−0.01318, 0.02378), 0.6)0.8((−0.003, 0.981), 0.98)((−0.00394, 0.01454), 0.6)1((0.005, 0.005), 1)((0.0053, 0.0053), 0.6)Table 14
Comparison ofu˜2r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((0.613, 0.69), 0.95)((0.613, 0.6907), 0.6)0.2((0.620, 0.68), 0.96)((0.62076, 0.68292), 0.6)0.4((0.628, 0.67), 0.97)((0.62852, 0.67514), 0.6)0.6((0.636, 0.66), 0.98)((0.63628, 0.66736), 0.6)0.8((0.644, 0.65), 0.99)((0.64404, 0.65958), 0.6)1((0.6510.651), 1)((0.6518, 0.6518), 0.6)Table 15
Comparison ofu˜3r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((0.31, 0.41), 0.93)((0.1854, 0.4123), 0.6)0.2((0.32, 0.40), 0.94)((0.19468, 0.3762), 0.6)0.4((0.33, 0.39), 0.96)((0.20396, 0.3401), 0.6)0.6((0.34, 0.38), 0.97)((0.21324, 0.304), 0.6)0.8((0.35, 0.37), 0.98)((0.22252, 0.2679), 0.6)1((0.36, 0.36), 1)((0.2318, 0.2318), 0.6)Table 16
Comparison ofu˜4r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((0.18, 0.31), 0.89)((0.1854, 0.3196), 0.6)0.2((0.19, 0.30), 0.91)((0.19882, 0.30618), 0.6)0.4((0.21, 0.29), 0.93)((0.21224, 0.29276), 0.6)0.6((0.22, 0.27), 0.95)((0.22566, 0.27934), 0.6)0.8((0.23, 0.26), 0.97)((0.23908, 0.26592), 0.6)1((0.25, 0.25), 1)((0.2525, 0.2525), 0.6)Table 17
Comparison ofu˜5r result using the two proposed methods λ=0.
rThe first proposed methodλ=0The second proposed method0((0.001, 0.158), 0.85)((−0.001, 0.158), 0.6)0.2((0.016, 0.142), 0.88)((0.158, 0.1421), 0.6)0.4((0.032, 0.126), 0.91)((0.317, 0.1262), 0.6)0.6((0.048, 0.110), 0.94)((0.476, 0.1103), 0.6)0.8((0.064, 0.094), 0.97)((0.635, 0.0944), 0.6)1((0.080, 0.078), 1)((0.794, 0.0785), 0.6)


## 6. Conclusion and Recommendations

One of the most important uses of the Z-number arithmetic is to study linear equations systems based on Z-numbers. So far, no method has been provided to solve such systems, or if it has been provided, it is few.In this article, we present a numerical method for solving a linear system based on Z-numbers. To show the capability of the proposed method for different systems, we simulated and solved several examples of linear equations includingZ-numbers.Also, we solved an economic problem (determining market balance values) that was modeled to use a linear equation system including Z-numbers. Determining the price and quantity in the market mechanism is the purpose of solving these systems. In the future, we can introduce the full Z-system of linear equations (FZSLE) and provide a solution to it that is not covered in this article.In the future, the proposed method can be used in topics such as “Failure Mode and Effect Analysis” that Huang et al. using T-spherical fuzzy maximizing deviation and combined comparison solution methods [17]; also, fuzzy decision-making models in [14, 18] can be solved by the proposed method by assumption Z-numbers instead of fuzzy numbers and extending fuzzy linear equations systems to Z-system of linear equations.



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