# An Application of Interval Arithmetic for Solving Fully Fuzzy Linear Systems with Trapezoidal Fuzzy Numbers

Authors: Esmaeil Siahlooei; Seyed Abolfazl Shahzadeh Fazeli
Journal: Advances in Fuzzy Systems
 (2018)

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## Abstract

We present a method for solving fully fuzzy linear systems using interval aspects of fuzzy numbers. This new method uses a decomposition technique to convert a fully fuzzy linear system into two types of decomposition in the form of interval matrices. It finds the solution of a fully fuzzy linear system by using interval operations. This new method uses interval arithmetic and two new interval operations⊖ and ⊘. These new operations, which are inverses of basic interval operations + and ×, will be presented in the middle of this paper. Some numerical examples are given to illustrate the ability of proposed methods.


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## Body

## 1. Introduction

In various fields of sciences, in solving real-life problems, where the system of linear equations is noticed, there are situations where the values of the parameters cannot be stated exactly, but their estimation or some bounds on them can be measured. Modeling a lot of these problems leads to a fuzzy linear system because the inexact kind of real numbers can be modeled in fuzzy numbers.Fuzzy linear systems have been studied by several authors. Friedman et al. [1] proposed a general model for solving an n×n fuzzy linear system whose coefficient matrix is crisp and the right-hand side column is an arbitrary fuzzy vector. Allahviranloo [2] proposed solution of a fuzzy linear system by using an iterative method and later suggested various numerical methods to solve fuzzy linear systems [3]. Some methods for solving these systems can be found in [4–7]. Later, a system, called fully fuzzy linear system (FFLS), is introduced wherein elements in the coefficient matrix, the right-hand side vector, and the vector of unknowns are fuzzy numbers.This system is solved in [8, 9] using decomposition of the coefficient matrix. Some other methods used iterative methods for solving FFLS [3, 10, 11]. Many researchers studied FFLS and numerical techniques to solve them [12–16]. Kochen et al. proposed a method for solving square and nonsquare FFLS with trapezoidal fuzzy numbers in 2016 [17]. In 2017, Edalatpanah reviewed some iterative methods for solving FFLS [18]. Recently, Allahviranloo and Babakordi introduced a new method for solving an extension of FFLS that has two coefficient matrices [19]. A new method for solving FFLS with hexagonal fuzzy numbers is introduced by Malkawi et al. in 2018 [20].In Section2, we present some basics about fuzzy numbers and α-cut arithmetic. Some basics of interval arithmetic, new interval operations and some properties of interval numbers, and also our method for solving FFLS are investigated in Section 3. In Section 4, some examples are solved to show the effectiveness of the proposed method and concluding remarks are contained in Section 5.


## 2. Preliminaries

In this section we review fuzzy numbers andα-cut arithmetic on them. α-cut method is a method for performing interval operations like addition, multiplication, and subtraction on fuzzy numbers. α-cut fuzzy arithmetic is described in detail in [21].Let us consider an arbitrary trapezoidal fuzzy numberu=(u1,u2,u3,u4), where u1,u2,u3,u4∈R and u1≤u2≤u3≤u4. The membership function μu of u is defined as follows:(1)μux=0u1>xx-u1u2-u1u1≤x≤u21u2≤x≤u3u4-xu4-u3u3≤x≤u40u4<x.Two trapezoidal fuzzy numbers u and v are said to be equal, if and only if ui=vi for 1≤i≤4.Given a trapezoidal fuzzy numberu and membership function μu and a real number α∈{0,1}, then the α-cut of a fuzzy number u is an interval, which is defined as(2)uα=uα̲,uα¯,where uα̲, uα¯ are(3)uα̲=inf⁡x∈R∣μux≥α,(4)uα¯=sup⁡x∈R∣μux≥α.Note that (5)u=⋃α∈0,1uα.The basic arithmetic of two fuzzy numbers u and v is discussed in [21–23] based on interval arithmetic(6)u+vα=uα+vα=uα̲,uα¯+vα̲,vα¯,(7)u-vα=uα-vα=uα̲,uα¯-vα̲,vα¯,(8)uvα=uα×vα=uα̲,uα¯×vα̲,vα¯.Clearly, for a trapezoidal fuzzy number u=(u1,u2,u3,u4), we have u0=[u1,u4], u1=[u2,u3].A matrix (vector)A is called a fuzzy matrix (fuzzy vector), if at least one element of matrix (vector) A is a fuzzy number. A matrix A is called a fully fuzzy matrix, whose elements of A are fuzzy numbers. An α-cut of an n×m matrix A is an interval matrix with the same dimension whose elements are α-cut of elements of matrix A.


## 3. Materials and Methods

### 3.1. Interval Arithmetic

In this section, we briefly explain interval arithmetic and notion. Also, we present a new operation on interval arithmetic.Aninterval numberâ is the bounded, closed subset of real numbers, which is indicated by a hat, defined by â=[a̲,a¯], where a̲,a¯∈R and a̲≤a¯. Let IR denote the set of such intervals. An interval â is adegenerate interval when a̲=a¯.Infimum andsupremum of an interval number â are a̲ and a¯, respectively. Let a∈â, say a∈â when a̲≤a≤a¯.Let∘ denote a binary operation on real space R. For â,b̂∈IR(9)â∘b̂=a∘b∣a∈â,b∈b̂.Then, for basic arithmetic operations {+,-,×,/} in IR, we have(10)â+b̂=a̲+b̲,a¯+b¯(11)â-b̂=a̲-b¯,a¯-b̲(12)â×b̂=a̲b̲,a¯b¯a̲≥0,b̲≥0a¯b̲,a¯b¯a̲≥0,b̲<0<b¯a¯b̲,a̲b¯a̲≥0,b¯≤0a̲b¯,a¯b¯a̲<0<a¯,b̲≥0a¯b̲,a̲b̲a̲<0<a¯,b¯≤0a̲b¯,a¯b̲a¯≤0,b̲≥0a̲b¯,a̲b̲a¯≥0,b̲<0<b¯a¯b¯,a̲b̲a¯≤0,b¯≤0a¯b¯,a̲b̲a¯≤0,b¯≤0min⁡a¯b̲,a̲b¯,max⁡a̲b̲,a¯b¯a̲<0<a¯,b̲<0<b¯=min⁡a̲b̲,a̲b¯,a¯b̲,a¯b¯,max⁡a̲b̲,a̲b¯,a¯b̲,a¯b¯.(13)âb̂=â×1b̂=â×1b¯,1b̲where0∉b̂IR+ is the set of allimproper intervals which is defined in [24] as follows: (14)IR+=â=a̲,a¯∣a̲,a¯∈R,a̲≥a¯.The set(15)IR∗=â=a̲,a¯∣a̲,a¯∈Rcontains all interval numbers and improper interval numbers calledgeneral interval. The basic arithmetic of general intervals is defined similar to intervals, but the × operation has some differences that are explained in [25].By aninterval matrix, we mean a matrix whose elements are interval numbers. Let Â be an interval matrix with elements Âij and let B be a matrix with real elements Bij; we consider B∈Â if Bij∈Âij for all i and j. Basic arithmetic operations on interval matrices are defined similar to real matrices [26].


### 3.2. New Interval Operations

Letâ,b̂∈IR∗; we define two new operations ⊖ and ⊘ as follows:(16)â⊖b̂=a̲-b̲,a¯-b¯,(17)â⊘b̂=-∞,+∞a̲=a¯=0,b̲=b¯=00,0a̲=a¯=0,b̲≠0orb¯≠0undefined∃x≠0,x∈â,b̲=b¯=0undefined0∉a,0∈ba̲b̲,a¯b¯a̲≥0,b̲>00,a¯b¯a̲=0,b̲=0a̲b¯,a¯b¯a̲≤0≤a¯,b̲≥0a̲b¯,a¯b̲a¯≤0,b̲>0a̲b¯,0a¯=0,b̲=0â⊘-b̂b¯≤0a̲b¯,a¯b¯a̲≥0,b̲<0<b¯max⁡a̲b¯,a¯b̲,min⁡a¯b¯,a̲b̲a̲≤0≤a¯,b̲<0<b¯a¯b̲,a̲b̲a¯≤0,b̲<0<b¯The operation ⊖ is defined and used in some articles as an operation in extended interval arithmetic [27]. But, we use ⊖ in a different way. We illustrate that the operation ⊖ can act as an inverse of + in interval arithmetic.Theorem 1.
Consideringâ,b̂∈IR, the interval equation â+x̂=b̂ has a unique solution x̂=b̂⊖â and x̂∈IR∗.Proof.
(18) a ̂ + x ̂ = b ̂ ⇒ a ̲ , a ¯ + x ̲ , x ¯ = b ̲ , b ¯ ⇒ a ̲ + x ̲ = b ̲ a ¯ + x ¯ = b ¯ I a ̲ + x ̲ = b ̲ ⇒ x ̲ = b ̲ - a ̲ I I a ¯ + x ¯ = b ¯ ⇒ x ¯ = b ¯ - a ¯ (19) I , I I , 16 ⇒ x ̂ = b ̂ ⊖ a ̂Corollary 2.
Supposingx̂=b̂⊖â and x̂∈IR∗ and x̂∉IR, the equation â+x̂=b̂ has no solution in IR but has exactly one solution in IR∗.Corollary 3.
Ifâ,b̂∈IR and ĉ=â+b̂, then b̂ can be computed by ĉ⊖â uniquely. On the other hand, the operation ⊖ is the inverse operation of + in interval arithmetic.Theorem 4.
Consideringâ,b̂∈IR, the interval equation â×x̂=b̂ has a solution x̂=b̂⊘â, where x̂∈IR∗.Remark 5.
Solutionx̂ is not usually unique; e.g., the interval equation [-1,1]×x̂=[-2,2] has solutions [-1,2],[-2,0],[-2,2]. Actually this equation has unlimited solutions. All intervals [u,2] and [-2,v] where -2≤u≤0 and 0≤v≤2 can be solution of this equation.
In cases thatx̂ is not unique, we consider x̂ as the longest interval as possible. In the above example, x̂ will be [-2,2].Proof.
Letâ×x̂=b̂; Table 1 shows sign of b̂ related to signs of â and x̂ and also shows formulas of multiplication result.
Now, givenâ,b̂∈IR such that â×x̂=b̂, a¯≤0 and b̲≤0≤b¯. From Table 1, it is deduced that 0∈x̂ and b̂=[a¯x̲,a¯x¯]. Therefore (20)âx̂=b̂⇔b̂=a̲x¯,a̲x̲⇔b̲=a̲x¯⇔x¯=b̲x̲b¯=a̲x̲⇔x̲=b¯x̲⇔x̂=b̂⊘â.In this case, x̂ is calculated uniquely by operation ⊘. Other cases, except cases with condition 0∈â, are provable as above.
In cases0≤a̲ and 0≥a¯ among three possible choices, we could choose the appropriate interval according to the signs of b̲ and b¯, while in case 0∈â all three possible choices are similar in terms of signs of b̲ and b¯. To conquer this problem in case 0∈â, we choose the longest interval as the solution from three possibilities 0≤x̲, 0∈x̂, and 0≥x¯ which is 0∈x̂.
Now, letx̂ be an interval where 0∈x̂ and let x̂ be the solution of â×x̂=b̂; then b̂=[min⁡{a¯x̲,a̲x¯},max⁡{a̲x̲,a¯x¯}]. It follows that(21)b̲=min⁡a¯x̲,a̲x¯⇒b̲≤a¯x̲⇒x̲≥b̲a¯,(22)b¯=max⁡a̲x̲,a¯x¯⇒b¯≥a̲x̲⇒x̲≥b¯a̲.Then x̲≥max{b̲/a¯,b¯/a̲}. If x̲≠max{b̲/a¯,b¯/a̲}, then x̲>max{b̲/a¯,b¯/a̲}, and a¯x̲>max{b̲/,a¯b¯/a̲}≥b̲. This contradicts with b̲=min{a¯x̲,a̲x¯}. Hence(23)x̲=max⁡b̲a¯,b¯a̲.Similarly, x¯ is obtained as(24)x¯=min⁡b¯a¯,b̲a̲.It is observed simply that x̂ is equal to b̂⊘â.
In addition, two special cases should be considered separately:(i)
Ifâ=[0,0] and b̂≠[0,0], there is no x̂ such that â×x̂=b̂.(ii)
Ifâ=b̂=[0,0], then x̂ is free and expression â×x̂=b̂ is consistent for all x̂, and x̂ is considered as unbounded interval [-∞,+∞] that is the longest possible interval satisfying â×x̂=b̂.Table 1
Interval multiplication.
a ̂ × x ̂ = b ̂ 0 ≤ a ̲ 0 ∈ a ̂ 0 ≥ a ¯  0 ≤ x ̲ [ a _  x _ ,  a ¯  x ¯ ] [ a _ x ¯ ,  a ¯  x ¯ ] [ a _ x ¯ ,  a ¯ x _ ] [ + , + ] [ - , + ] [ - , - ]  0 ∈ x ̂ [ a ¯ x _ ,  a ¯  x ¯ ] [ min ⁡ a ¯ x _ , a _ x ¯ , max ⁡ a _  x _ , a ¯  x ¯ ] [ a _ x ¯ ,  a _  x _ ] - , + [ - , + ] [ - , + ]  0 ≥ x ¯ [ a ¯ x _ ,  a _ x ¯ ] [ a ¯ x _ ,  a _  x _ ] [ a ¯  x ¯ ,  a _  x _ ] [ - , - ] [ - , + ] [ + , + ]Corollary 6.
Ifx̂=b̂⊘â, x̂∈IR∗, and x̂∉IR, then the equation â×x̂=b̂ has no solution in IR but has solution in IR∗.Corollary 7.
Ifâ,b̂∈IR and ĉ=â×b̂, then b̂ can be computed by ĉ⊘â.The operation⊘ is the inverse operation of × in equation â×x̂=b̂ where 0∉x̂. In case 0∈x̂ the solution â×x̂=b̂ is not unique, but ẑ=b̂⊘â is the longest interval such that â×ẑ=b̂. In other words, any ŷ∈IR that satisfies â×ŷ=b̂ is a subset of ẑ.


### 3.3. Some Properties of Interval Numbers

Interval arithmetic and some of their properties over interval numbers and matrices are discussed in [26]. We note some of them here. For any â,b̂,ĉ∈IR,(25)â+b̂=b̂+â,â+b̂±ĉ=â+b̂±ĉ,âb̂=b̂â.Lemma 8.
For allâ,b̂∈IR, (â+b̂)⊖ĉ=â+(b̂⊖ĉ).Proof.
(26) a ^ + b ^ ⊖ c ^ = a _ + b _ - c _ , a ¯ + b ¯ - c ¯ = a _ + b _ - c _ , a ¯ + b ¯ - c ¯ = a ^ + b ^ ⊖ c ^ .All these properties are correct for interval matrices as well.


### 3.4. A Method for Solving a Fully Fuzzy Linear System

Consider ann×n FFLS(27)a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮an1x1+an2x2+⋯+annxn=bn,which is written as(28)Ax=bwhere the coefficient matrix A=(aij)n×n is the fuzzy matrix of trapezoidal fuzzy numbers, and x=(xi)n×1,b=(bi)n×1 are column vectors of trapezoidal fuzzy numbers.Allahviranloo et al. [14] defined the following solution sets for FFLS (27):(i) United trapezoidal fuzzy solution set:(29)UTFSS=x∣A1x1=b1,A0x0=b0,(ii) Tolerable trapezoidal fuzzy solution set:(30)TTFSS=x∣A1x1=b1,A0x0⊆b0,(iii) Controllable trapezoidal fuzzy solution set:(31)CTFSS=x∣A1x1=b1,A0x0⊇b0.Note that A1=[aij1]n×n and A0=[aij0]n×n are interval matrices, and b1=[bi1]n×1, b0=[bi0]n×1 are known right-hand side interval vectors, and x1=[xi1]n×1 and x0=[xi0]n×1 are unknown interval vectors.Now, we aim to propose a practical method to obtain the suitable solution of an FFLS. The suitable solution is defined in [14] as solution x∈UTFSS. To find such a solution x∈UTFSS, it is sufficient to solve the following two interval linear systems:(32)A0x0=b0(33)A1x1=b1.Now, consider the n×n linear system as(34)â11x̂1+â12x̂2+⋯+â1nx̂n=b̂1â21x̂1+â22x̂2+⋯+â2nx̂n=b̂2⋮ân1x̂1+ân2x̂2+⋯+ânnx̂n=b̂nwhere âij,b̂i∈IR for all (1≤i,j≤n) and x̂i are unknown intervals. The interval linear system is represented as(35)Âx̂=b̂Note that two systems (32) and (33) are in the form (34). Using operations ⊖ and ⊘, we can propose a new method to decompose an interval matrix Â to two interval matrices, similar to LU decomposition method [28]. Then, one can solve systems (32) and (33)) using this decomposition method.Assume thatÂ∈IRn×n, and there exist two interval matrices L̂,Û∈IRn×n such that Â=L̂×Û, where L̂ and Û are lower triangular and upper triangular matrices, respectively. Then using this decomposition method, we can solve the interval linear system (34).Consider the system(36)Â=L̂×Û,where Ân×n is a known interval matrix, L̂n×n is an unknown lower triangular interval matrix, and Ûn×n is an unknown upper triangular interval matrix. In this system, there exist n2 linear interval equations, and there are n2+n unknown variables. This system extends to n2+n linear equations with n2+n unknowns by adding n linear equations liî=1for1≤i≤n.To solve this linear system, first, we set all diagonal elements ofL̂ with [1,1]. Then, using reversing operations, all elements of L̂ and Û can be computed by solving linear interval equations âij=∑k=1nl̂ik×ûkj. To solve such linear equations, we need to consider two cases:Case 1 (i≤j).
(37) a ̂ i j = ∑ k = 1 i - 1 l ̂ i k × u ̂ k j + l ̂ i i × u ̂ i j = ∑ k = 1 i - 1 l ̂ i k × u ̂ k j + u ̂ i jwhich results in(38)âij=∑k=1i-1l̂ik×ûkj+ûij⇒ûij=âij⊖∑k=1i-1l̂ik×ûkj.Case 2 (i>j).
(39) a ̂ i j = ∑ k = 1 j - 1 l ̂ i k × u ̂ k j + l ̂ i j × u ̂ j jwhich results in(40)âij=∑k=1j-1l̂ik×ûkj+l̂ij×ûjj⇒l̂ij×ûjj=âij⊖∑k=1j-1l̂ik×ûkj⇒l̂ij=âij⊖∑k=1j-1l̂ik×ûkj⊘ûjj.Using these results, we can compute interval LU decomposition for an interval matrix. The Algorithm ILU_Decomposition computes such interval LU decomposition and is given in Algorithm 1.
If at least one ofu̲ii or u¯ii for 0≤i≤n-1 is [0,0], then a dividing-by-zero error occurs, and decomposition is not possible. Otherwise, there exist L̂,Û∈IR∗n×n such that Â=L̂Û.Algorithm 1:Computing the interval LU decomposition.
( 1 ) functionILU_Decomposition A ̂ , L ̂ , U ̂ , n (i) Input:Â∈IRn×n (ii) Output:L̂,Û∈IRn×n such that Â=L̂Û ( 2 )  for i = 1 to n do ( 3 )  l ̂ i i ← 1,1 ( 4 )  for r = 1 to n do ( 5 )  i ← r ( 6 )  for j = i to n do ( 7 )  s u m ← ∑ k = 1 j - 1 l ̂ i k × u ̂ k j ( 8 )  h ← a ̂ i j ⊖ s u m ( 9 )  u ̂ i j ← h ( 10 )  j ← r ( 11 )  for i = j + 1 to n do ( 12 )  s u m ← ∑ k = 1 i - 1 l ̂ i k × u ̂ k j ( 13 )  u ̂ i j ← a ̂ i j ⊖ s u m ( 14 )  l ̂ i j ← s u m ⊘ u ̂ j j return L ̂ and Û If L∉IRn×n or U∉IRn×n there is no decomposition in I R n × n but has decomposition in (IR∗)n×n.Example 9.
Consider the following3×3 interval matrix:(41)Â=1,40.5,3-3,20.5,30.5,4.5-3,3-3,2-3,3-2,4.25.Initially, assign 1,1 to l11̂,l22̂,l33̂, and initialize all other elements of L̂ and Û to [0,0]. Then, compute the rows of Û and columns of L̂ by annotating Algorithm 1. Compute the rows and columns index of Û and L̂, following the first loop (lines (4) to (14)).In caser=1,(42)L̂=1,10,00,00.5,0.751,10,0-0.75,0.50,01,1,Û=1,40.5,3-3,20,00,00,00,00,00,0.In case r=2,(43)L̂=1,10,00,00.5,0.751,10,0-0.75,0.5-0.333,0.6661,1,Û=1,40.5,3-3,20,00.25,2.25-0.75,1.50,00,00,0.In case r=3,(44)L̂=1,10,00,00.5,0.751,10,0-0.75,0.5-0.333,0.6661,1,Û=1,40.5,3-3,20,00.25,2.25-0.75,1.50,00,00,1.It is observed simply that Â=L̂×Û.Now, to solveÂx̂=b̂ for x̂ with Â=L̂Û, we have(45)Âx̂=b̂⇒L̂Ûx̂=b̂⇒L̂ŷ=b̂(46)Âx̂=b̂⇒L̂Ûx̂=b̂⇒Ûx̂=ŷTo solve Âx̂=b̂ we do the following steps:(1)
DecomposeÂ into two matrices L̂ and Û.(2)
SolveL̂ŷ=b̂ for ŷ.(3)
SolveÛx̂=ŷ for x̂.Step(1) can be performed by Algorithm 1 and step (2) can be performed as follows:(47)1,100…0l̂211,10…0⋮⋮⋱⋮l̂n-11l̂n-12…1,10l̂n1l̂n2……1.1×ŷ1ŷ2⋮ŷn-1ŷn=b̂1b̂2⋮b̂n-1b̂n⇒b̂i=ŷi+∑j=1i-1l̂ij×ŷj⇒ŷi=b̂i⊖∑j=1i-1l̂ij×ŷj,and finally, step (3) can be performed as(48)û11û12û13…ûn10û22û23…û2n⋮⋮⋱⋮00…ûn-1n-1ûn-1n00…0ûnn×x̂1x̂2⋮x̂n-1x̂n=ŷ1ŷ2⋮ŷn-1ŷn⇒x̂i=ûii×x̂i+∑j=i+1nûij×ŷj⇒x̂i=ŷi⊖∑j=i+1nûij×ŷj⊘ûii.Algorithm 2 shows our approach to solve a system of linear interval equation Âx̂=b̂.Algorithm 2:Solving linear system of intervals using interval LU decomposition.
( 1 ) functionILU_Solve A ̂ , b ̂ , n , x ̂ (i) Input:Â∈IRn×n,b̂∈IRn (ii) Output:x̂∈IRn:Âx̂=b̂ ( 2 )  call ILU_DecompositionÂ,L̂,Û,n ( 3 )  for i = 1 to n do ( 4 )  y ̂ i = b ̂ i ⊖ ∑ j = 1 i - 1 l ̂ i j × y ̂ j ( 5 )  for i = n to 1 do ( 6 )  x ̂ i = y ̂ i ⊖ ∑ j = i + 1 n u ̂ i j × y ̂ j ⊘ u ̂ i i return x ̂Example 10.
Considering the interval matrixÂ in Example 9 and the following b̂, we aim to solve Âx̂=b̂ for x̂.(49)b̂=-50.00,47.50-52.50,48.75-47.50,51.25.In Example 9, ILU decomposition of Â is obtained. So, we need to solve L̂ŷ=b̂ and Ûx̂=ŷ. By solving L̂ŷ=b̂ and Ûx̂=ŷ, ŷ and x̂ are obtained, respectively, as follows:(50)ŷ=-50,47.5-15,13.1251-1.8750,5andx̂=-5.00,7.50-5.00,2.50-1.875,5.00,


## 3.1. Interval Arithmetic

In this section, we briefly explain interval arithmetic and notion. Also, we present a new operation on interval arithmetic.Aninterval numberâ is the bounded, closed subset of real numbers, which is indicated by a hat, defined by â=[a̲,a¯], where a̲,a¯∈R and a̲≤a¯. Let IR denote the set of such intervals. An interval â is adegenerate interval when a̲=a¯.Infimum andsupremum of an interval number â are a̲ and a¯, respectively. Let a∈â, say a∈â when a̲≤a≤a¯.Let∘ denote a binary operation on real space R. For â,b̂∈IR(9)â∘b̂=a∘b∣a∈â,b∈b̂.Then, for basic arithmetic operations {+,-,×,/} in IR, we have(10)â+b̂=a̲+b̲,a¯+b¯(11)â-b̂=a̲-b¯,a¯-b̲(12)â×b̂=a̲b̲,a¯b¯a̲≥0,b̲≥0a¯b̲,a¯b¯a̲≥0,b̲<0<b¯a¯b̲,a̲b¯a̲≥0,b¯≤0a̲b¯,a¯b¯a̲<0<a¯,b̲≥0a¯b̲,a̲b̲a̲<0<a¯,b¯≤0a̲b¯,a¯b̲a¯≤0,b̲≥0a̲b¯,a̲b̲a¯≥0,b̲<0<b¯a¯b¯,a̲b̲a¯≤0,b¯≤0a¯b¯,a̲b̲a¯≤0,b¯≤0min⁡a¯b̲,a̲b¯,max⁡a̲b̲,a¯b¯a̲<0<a¯,b̲<0<b¯=min⁡a̲b̲,a̲b¯,a¯b̲,a¯b¯,max⁡a̲b̲,a̲b¯,a¯b̲,a¯b¯.(13)âb̂=â×1b̂=â×1b¯,1b̲where0∉b̂IR+ is the set of allimproper intervals which is defined in [24] as follows: (14)IR+=â=a̲,a¯∣a̲,a¯∈R,a̲≥a¯.The set(15)IR∗=â=a̲,a¯∣a̲,a¯∈Rcontains all interval numbers and improper interval numbers calledgeneral interval. The basic arithmetic of general intervals is defined similar to intervals, but the × operation has some differences that are explained in [25].By aninterval matrix, we mean a matrix whose elements are interval numbers. Let Â be an interval matrix with elements Âij and let B be a matrix with real elements Bij; we consider B∈Â if Bij∈Âij for all i and j. Basic arithmetic operations on interval matrices are defined similar to real matrices [26].


## 3.2. New Interval Operations

Letâ,b̂∈IR∗; we define two new operations ⊖ and ⊘ as follows:(16)â⊖b̂=a̲-b̲,a¯-b¯,(17)â⊘b̂=-∞,+∞a̲=a¯=0,b̲=b¯=00,0a̲=a¯=0,b̲≠0orb¯≠0undefined∃x≠0,x∈â,b̲=b¯=0undefined0∉a,0∈ba̲b̲,a¯b¯a̲≥0,b̲>00,a¯b¯a̲=0,b̲=0a̲b¯,a¯b¯a̲≤0≤a¯,b̲≥0a̲b¯,a¯b̲a¯≤0,b̲>0a̲b¯,0a¯=0,b̲=0â⊘-b̂b¯≤0a̲b¯,a¯b¯a̲≥0,b̲<0<b¯max⁡a̲b¯,a¯b̲,min⁡a¯b¯,a̲b̲a̲≤0≤a¯,b̲<0<b¯a¯b̲,a̲b̲a¯≤0,b̲<0<b¯The operation ⊖ is defined and used in some articles as an operation in extended interval arithmetic [27]. But, we use ⊖ in a different way. We illustrate that the operation ⊖ can act as an inverse of + in interval arithmetic.Theorem 1.
Consideringâ,b̂∈IR, the interval equation â+x̂=b̂ has a unique solution x̂=b̂⊖â and x̂∈IR∗.Proof.
(18) a ̂ + x ̂ = b ̂ ⇒ a ̲ , a ¯ + x ̲ , x ¯ = b ̲ , b ¯ ⇒ a ̲ + x ̲ = b ̲ a ¯ + x ¯ = b ¯ I a ̲ + x ̲ = b ̲ ⇒ x ̲ = b ̲ - a ̲ I I a ¯ + x ¯ = b ¯ ⇒ x ¯ = b ¯ - a ¯ (19) I , I I , 16 ⇒ x ̂ = b ̂ ⊖ a ̂Corollary 2.
Supposingx̂=b̂⊖â and x̂∈IR∗ and x̂∉IR, the equation â+x̂=b̂ has no solution in IR but has exactly one solution in IR∗.Corollary 3.
Ifâ,b̂∈IR and ĉ=â+b̂, then b̂ can be computed by ĉ⊖â uniquely. On the other hand, the operation ⊖ is the inverse operation of + in interval arithmetic.Theorem 4.
Consideringâ,b̂∈IR, the interval equation â×x̂=b̂ has a solution x̂=b̂⊘â, where x̂∈IR∗.Remark 5.
Solutionx̂ is not usually unique; e.g., the interval equation [-1,1]×x̂=[-2,2] has solutions [-1,2],[-2,0],[-2,2]. Actually this equation has unlimited solutions. All intervals [u,2] and [-2,v] where -2≤u≤0 and 0≤v≤2 can be solution of this equation.
In cases thatx̂ is not unique, we consider x̂ as the longest interval as possible. In the above example, x̂ will be [-2,2].Proof.
Letâ×x̂=b̂; Table 1 shows sign of b̂ related to signs of â and x̂ and also shows formulas of multiplication result.
Now, givenâ,b̂∈IR such that â×x̂=b̂, a¯≤0 and b̲≤0≤b¯. From Table 1, it is deduced that 0∈x̂ and b̂=[a¯x̲,a¯x¯]. Therefore (20)âx̂=b̂⇔b̂=a̲x¯,a̲x̲⇔b̲=a̲x¯⇔x¯=b̲x̲b¯=a̲x̲⇔x̲=b¯x̲⇔x̂=b̂⊘â.In this case, x̂ is calculated uniquely by operation ⊘. Other cases, except cases with condition 0∈â, are provable as above.
In cases0≤a̲ and 0≥a¯ among three possible choices, we could choose the appropriate interval according to the signs of b̲ and b¯, while in case 0∈â all three possible choices are similar in terms of signs of b̲ and b¯. To conquer this problem in case 0∈â, we choose the longest interval as the solution from three possibilities 0≤x̲, 0∈x̂, and 0≥x¯ which is 0∈x̂.
Now, letx̂ be an interval where 0∈x̂ and let x̂ be the solution of â×x̂=b̂; then b̂=[min⁡{a¯x̲,a̲x¯},max⁡{a̲x̲,a¯x¯}]. It follows that(21)b̲=min⁡a¯x̲,a̲x¯⇒b̲≤a¯x̲⇒x̲≥b̲a¯,(22)b¯=max⁡a̲x̲,a¯x¯⇒b¯≥a̲x̲⇒x̲≥b¯a̲.Then x̲≥max{b̲/a¯,b¯/a̲}. If x̲≠max{b̲/a¯,b¯/a̲}, then x̲>max{b̲/a¯,b¯/a̲}, and a¯x̲>max{b̲/,a¯b¯/a̲}≥b̲. This contradicts with b̲=min{a¯x̲,a̲x¯}. Hence(23)x̲=max⁡b̲a¯,b¯a̲.Similarly, x¯ is obtained as(24)x¯=min⁡b¯a¯,b̲a̲.It is observed simply that x̂ is equal to b̂⊘â.
In addition, two special cases should be considered separately:(i)
Ifâ=[0,0] and b̂≠[0,0], there is no x̂ such that â×x̂=b̂.(ii)
Ifâ=b̂=[0,0], then x̂ is free and expression â×x̂=b̂ is consistent for all x̂, and x̂ is considered as unbounded interval [-∞,+∞] that is the longest possible interval satisfying â×x̂=b̂.Table 1
Interval multiplication.
a ̂ × x ̂ = b ̂ 0 ≤ a ̲ 0 ∈ a ̂ 0 ≥ a ¯  0 ≤ x ̲ [ a _  x _ ,  a ¯  x ¯ ] [ a _ x ¯ ,  a ¯  x ¯ ] [ a _ x ¯ ,  a ¯ x _ ] [ + , + ] [ - , + ] [ - , - ]  0 ∈ x ̂ [ a ¯ x _ ,  a ¯  x ¯ ] [ min ⁡ a ¯ x _ , a _ x ¯ , max ⁡ a _  x _ , a ¯  x ¯ ] [ a _ x ¯ ,  a _  x _ ] - , + [ - , + ] [ - , + ]  0 ≥ x ¯ [ a ¯ x _ ,  a _ x ¯ ] [ a ¯ x _ ,  a _  x _ ] [ a ¯  x ¯ ,  a _  x _ ] [ - , - ] [ - , + ] [ + , + ]Corollary 6.
Ifx̂=b̂⊘â, x̂∈IR∗, and x̂∉IR, then the equation â×x̂=b̂ has no solution in IR but has solution in IR∗.Corollary 7.
Ifâ,b̂∈IR and ĉ=â×b̂, then b̂ can be computed by ĉ⊘â.The operation⊘ is the inverse operation of × in equation â×x̂=b̂ where 0∉x̂. In case 0∈x̂ the solution â×x̂=b̂ is not unique, but ẑ=b̂⊘â is the longest interval such that â×ẑ=b̂. In other words, any ŷ∈IR that satisfies â×ŷ=b̂ is a subset of ẑ.


## 3.3. Some Properties of Interval Numbers

Interval arithmetic and some of their properties over interval numbers and matrices are discussed in [26]. We note some of them here. For any â,b̂,ĉ∈IR,(25)â+b̂=b̂+â,â+b̂±ĉ=â+b̂±ĉ,âb̂=b̂â.Lemma 8.
For allâ,b̂∈IR, (â+b̂)⊖ĉ=â+(b̂⊖ĉ).Proof.
(26) a ^ + b ^ ⊖ c ^ = a _ + b _ - c _ , a ¯ + b ¯ - c ¯ = a _ + b _ - c _ , a ¯ + b ¯ - c ¯ = a ^ + b ^ ⊖ c ^ .All these properties are correct for interval matrices as well.


## 3.4. A Method for Solving a Fully Fuzzy Linear System

Consider ann×n FFLS(27)a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮an1x1+an2x2+⋯+annxn=bn,which is written as(28)Ax=bwhere the coefficient matrix A=(aij)n×n is the fuzzy matrix of trapezoidal fuzzy numbers, and x=(xi)n×1,b=(bi)n×1 are column vectors of trapezoidal fuzzy numbers.Allahviranloo et al. [14] defined the following solution sets for FFLS (27):(i) United trapezoidal fuzzy solution set:(29)UTFSS=x∣A1x1=b1,A0x0=b0,(ii) Tolerable trapezoidal fuzzy solution set:(30)TTFSS=x∣A1x1=b1,A0x0⊆b0,(iii) Controllable trapezoidal fuzzy solution set:(31)CTFSS=x∣A1x1=b1,A0x0⊇b0.Note that A1=[aij1]n×n and A0=[aij0]n×n are interval matrices, and b1=[bi1]n×1, b0=[bi0]n×1 are known right-hand side interval vectors, and x1=[xi1]n×1 and x0=[xi0]n×1 are unknown interval vectors.Now, we aim to propose a practical method to obtain the suitable solution of an FFLS. The suitable solution is defined in [14] as solution x∈UTFSS. To find such a solution x∈UTFSS, it is sufficient to solve the following two interval linear systems:(32)A0x0=b0(33)A1x1=b1.Now, consider the n×n linear system as(34)â11x̂1+â12x̂2+⋯+â1nx̂n=b̂1â21x̂1+â22x̂2+⋯+â2nx̂n=b̂2⋮ân1x̂1+ân2x̂2+⋯+ânnx̂n=b̂nwhere âij,b̂i∈IR for all (1≤i,j≤n) and x̂i are unknown intervals. The interval linear system is represented as(35)Âx̂=b̂Note that two systems (32) and (33) are in the form (34). Using operations ⊖ and ⊘, we can propose a new method to decompose an interval matrix Â to two interval matrices, similar to LU decomposition method [28]. Then, one can solve systems (32) and (33)) using this decomposition method.Assume thatÂ∈IRn×n, and there exist two interval matrices L̂,Û∈IRn×n such that Â=L̂×Û, where L̂ and Û are lower triangular and upper triangular matrices, respectively. Then using this decomposition method, we can solve the interval linear system (34).Consider the system(36)Â=L̂×Û,where Ân×n is a known interval matrix, L̂n×n is an unknown lower triangular interval matrix, and Ûn×n is an unknown upper triangular interval matrix. In this system, there exist n2 linear interval equations, and there are n2+n unknown variables. This system extends to n2+n linear equations with n2+n unknowns by adding n linear equations liî=1for1≤i≤n.To solve this linear system, first, we set all diagonal elements ofL̂ with [1,1]. Then, using reversing operations, all elements of L̂ and Û can be computed by solving linear interval equations âij=∑k=1nl̂ik×ûkj. To solve such linear equations, we need to consider two cases:Case 1 (i≤j).
(37) a ̂ i j = ∑ k = 1 i - 1 l ̂ i k × u ̂ k j + l ̂ i i × u ̂ i j = ∑ k = 1 i - 1 l ̂ i k × u ̂ k j + u ̂ i jwhich results in(38)âij=∑k=1i-1l̂ik×ûkj+ûij⇒ûij=âij⊖∑k=1i-1l̂ik×ûkj.Case 2 (i>j).
(39) a ̂ i j = ∑ k = 1 j - 1 l ̂ i k × u ̂ k j + l ̂ i j × u ̂ j jwhich results in(40)âij=∑k=1j-1l̂ik×ûkj+l̂ij×ûjj⇒l̂ij×ûjj=âij⊖∑k=1j-1l̂ik×ûkj⇒l̂ij=âij⊖∑k=1j-1l̂ik×ûkj⊘ûjj.Using these results, we can compute interval LU decomposition for an interval matrix. The Algorithm ILU_Decomposition computes such interval LU decomposition and is given in Algorithm 1.
If at least one ofu̲ii or u¯ii for 0≤i≤n-1 is [0,0], then a dividing-by-zero error occurs, and decomposition is not possible. Otherwise, there exist L̂,Û∈IR∗n×n such that Â=L̂Û.Algorithm 1:Computing the interval LU decomposition.
( 1 ) functionILU_Decomposition A ̂ , L ̂ , U ̂ , n (i) Input:Â∈IRn×n (ii) Output:L̂,Û∈IRn×n such that Â=L̂Û ( 2 )  for i = 1 to n do ( 3 )  l ̂ i i ← 1,1 ( 4 )  for r = 1 to n do ( 5 )  i ← r ( 6 )  for j = i to n do ( 7 )  s u m ← ∑ k = 1 j - 1 l ̂ i k × u ̂ k j ( 8 )  h ← a ̂ i j ⊖ s u m ( 9 )  u ̂ i j ← h ( 10 )  j ← r ( 11 )  for i = j + 1 to n do ( 12 )  s u m ← ∑ k = 1 i - 1 l ̂ i k × u ̂ k j ( 13 )  u ̂ i j ← a ̂ i j ⊖ s u m ( 14 )  l ̂ i j ← s u m ⊘ u ̂ j j return L ̂ and Û If L∉IRn×n or U∉IRn×n there is no decomposition in I R n × n but has decomposition in (IR∗)n×n.Example 9.
Consider the following3×3 interval matrix:(41)Â=1,40.5,3-3,20.5,30.5,4.5-3,3-3,2-3,3-2,4.25.Initially, assign 1,1 to l11̂,l22̂,l33̂, and initialize all other elements of L̂ and Û to [0,0]. Then, compute the rows of Û and columns of L̂ by annotating Algorithm 1. Compute the rows and columns index of Û and L̂, following the first loop (lines (4) to (14)).In caser=1,(42)L̂=1,10,00,00.5,0.751,10,0-0.75,0.50,01,1,Û=1,40.5,3-3,20,00,00,00,00,00,0.In case r=2,(43)L̂=1,10,00,00.5,0.751,10,0-0.75,0.5-0.333,0.6661,1,Û=1,40.5,3-3,20,00.25,2.25-0.75,1.50,00,00,0.In case r=3,(44)L̂=1,10,00,00.5,0.751,10,0-0.75,0.5-0.333,0.6661,1,Û=1,40.5,3-3,20,00.25,2.25-0.75,1.50,00,00,1.It is observed simply that Â=L̂×Û.Now, to solveÂx̂=b̂ for x̂ with Â=L̂Û, we have(45)Âx̂=b̂⇒L̂Ûx̂=b̂⇒L̂ŷ=b̂(46)Âx̂=b̂⇒L̂Ûx̂=b̂⇒Ûx̂=ŷTo solve Âx̂=b̂ we do the following steps:(1)
DecomposeÂ into two matrices L̂ and Û.(2)
SolveL̂ŷ=b̂ for ŷ.(3)
SolveÛx̂=ŷ for x̂.Step(1) can be performed by Algorithm 1 and step (2) can be performed as follows:(47)1,100…0l̂211,10…0⋮⋮⋱⋮l̂n-11l̂n-12…1,10l̂n1l̂n2……1.1×ŷ1ŷ2⋮ŷn-1ŷn=b̂1b̂2⋮b̂n-1b̂n⇒b̂i=ŷi+∑j=1i-1l̂ij×ŷj⇒ŷi=b̂i⊖∑j=1i-1l̂ij×ŷj,and finally, step (3) can be performed as(48)û11û12û13…ûn10û22û23…û2n⋮⋮⋱⋮00…ûn-1n-1ûn-1n00…0ûnn×x̂1x̂2⋮x̂n-1x̂n=ŷ1ŷ2⋮ŷn-1ŷn⇒x̂i=ûii×x̂i+∑j=i+1nûij×ŷj⇒x̂i=ŷi⊖∑j=i+1nûij×ŷj⊘ûii.Algorithm 2 shows our approach to solve a system of linear interval equation Âx̂=b̂.Algorithm 2:Solving linear system of intervals using interval LU decomposition.
( 1 ) functionILU_Solve A ̂ , b ̂ , n , x ̂ (i) Input:Â∈IRn×n,b̂∈IRn (ii) Output:x̂∈IRn:Âx̂=b̂ ( 2 )  call ILU_DecompositionÂ,L̂,Û,n ( 3 )  for i = 1 to n do ( 4 )  y ̂ i = b ̂ i ⊖ ∑ j = 1 i - 1 l ̂ i j × y ̂ j ( 5 )  for i = n to 1 do ( 6 )  x ̂ i = y ̂ i ⊖ ∑ j = i + 1 n u ̂ i j × y ̂ j ⊘ u ̂ i i return x ̂Example 10.
Considering the interval matrixÂ in Example 9 and the following b̂, we aim to solve Âx̂=b̂ for x̂.(49)b̂=-50.00,47.50-52.50,48.75-47.50,51.25.In Example 9, ILU decomposition of Â is obtained. So, we need to solve L̂ŷ=b̂ and Ûx̂=ŷ. By solving L̂ŷ=b̂ and Ûx̂=ŷ, ŷ and x̂ are obtained, respectively, as follows:(50)ŷ=-50,47.5-15,13.1251-1.8750,5andx̂=-5.00,7.50-5.00,2.50-1.875,5.00,


## 4. Numerical Results

Example 11.
Consider the following FFLS which is solved in [8, 9, 29]:(51)18,19,2010.5,12,13.55.5,6,6.21.9,2,2.13.9,4,4.41.3,1.5,1.71.9,2,2.21.9,2,2.34.4,4.5,4.6x11,x12,x13x21,x22,x23x31,x32,x33=1469.3,1897,2433.2358,434,543447.2,535.5,667.4First, we solve 1-cut system which is obtained as follows:(52)19126241.5224.5x12x22x32=1897434535.5.The 1-cut crisp solution will be (37,62,75)T.
Now, we solve0-cut system which is obtained as follows:(53)18,2010.5,13.55.5,6.21.9,2.13.9,4.41.3,1.71.9,2.21.9,2.34.4,4.6x11,x13x21,x23x31,x33=1469.3,2433.2358,543447.,667.4.The left-hand side matrix decomposed into the following two matrices:(54)L̂=1,10,00,00.1050,0.10561,10,00.1055,0.11010.2732,0.28361,1,Û=18,2010.5,13.55.5,6.20,02.7916,2.98260.7194,1.04900,00,03.6154,3.6314We solve equations L̂ŷ=b̂ and Ûx̂=ŷ and the solutions are(55)ŷ=1469.2,2433.3203.2,288.4234.4,321,x̂=29.0993,49.992856.0765,65.581464.8558,88.3868.Finally, solution of System (53) is obtained as(56)x=(29.0993,37,49.9928)(56.0765,62,65.5814)(64.8558,75,88.3868)Example 12.
Consider a3×3 FFLS:(57)Ax=bwhere(58)A=-7.2,-1.9,-1.9,5.4-4.4,-2.9,1,5-0.7,0,0.2,1.7-0.2,0,0.5,1-26,6.7,8.4,34.7-23,-3,15.9,24.3-9,-1.1,0.4,3.9-6.7,-3.5,0,1.1-17.4,3.5,4.1,23.2-23,-5.9,5,17.3-9,-1.7,0.7,5.5-7,-2,2,6.2-24,-4.6,0,18-15,-6.8,5.6,25-8,-0.5,1.4,12.7-11,-2,1,8.4,(59)b=-13.9,-2.6,1.3,12.3-63.7,-7.4,13.0,70.6-51,-4.9,7.4,54.8-67.8,-7.9,4.3,58.6.This system is converted to two interval systems A0x0=b0 and A1x1=b1 as follows:(60)-7.2,5.4-4.4,5-0.7,1.7-0.2,1-26,34.7-23,24.3-9,3.9-6.7,1.1-17.4,23.2-23,17.3-9,5.5-7,6.2-24,18-15,25-8,12.7-11,8.4x0=-13.9,12.3-63.7,70.6-51.0,54.8-67.8,58.6.(61)-1.9,-1.9-2.9,10,0.20,0.56.7,8.4-3,15.9-1.1,0.4-3.5,03.5,4.1-5.9,5-1.7,0.7-2,2-4.6,0-6.8,5.6-0.5,1.4-2,1x0=-2.6,1.3-7.4,13.0-4.9,7.4-7.9,4.3.Interval matrix A0 is decomposed to L̂0,Û0:(62)L̂0=1,10,00,00,0-4.8184,-0.64841,10,00,0-3.2230,0.7073-2.2224,1.01021,10,0-1.2122,3.3335-0.1101,2.7049-2.4038,0.89451,1,Û0=-7.2,5.4-4.4,5-0.7,1.7-0.2,10,01.0885,3.0808-0.8088,0.5272-1.8817,0.13640,00,0-2.3497,1.4468-1.8764,1.31120,00,00,0-1.5464,0.1876.Algorithm 2 solves L̂0Û0x0=b0 and obtains(63)x0=-0.4024,1.5463-0.0229,0.0851-0.4682,0.8773-1.5900,2.0357.Similarly, interval matrix A1 is decomposed to L̂1,Û1 as follows:(64)L̂1=1,10,00,00,0-4.4256,-3.52531,10,00,0-2.1571,-1.8455-1.2222,-0.88801,10,00,2.42610.1666,1.0327-0.6347,0.13031,1,Û1=-1.9001,-1.9-2.9,10,0.20010,0.50,01.4142,3.0642-0.1799,0.4037-1.2872,00,00,0-0.7857,0.4715-0.9224,0.43440,00,00,0-0.3958,-0.7927.
Then,x1 is obtained from L̂1Û1x1=b1 by Algorithm 2 as(65)x1=-0.2608,1.0021-0.0145,0.0538-0.3374,0.6323-0.9569,1.2251.Finally, x is found as(66)-0.4024,-0.2608,1.0021,1.5463-0.0229,-0.0145,0.0538,0.0851-0.4682,-0.3374,0.6323,0.8773-1.5900,-0.9569,1.2251,2.0357.


## 5. Conclusion

In the present paper, we proposed a new method for solving the fully fuzzy linear system with trapezoidal fuzzy numbers. This problem is converted into two interval linear systems; then by introducing a new interval method based on decomposition, we solved these systems. Our method is using two inverse operations⊖ and ⊘ which are presented in this paper for the first time. Numerical results demonstrated the efficiency of our new method.



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*Source: 2104343-2018-07-11.xml*