# An Analytical Approach to Evaluating Nonmonotonic Functions of Fuzzy Numbers

Authors: Arthur Seibel; Josef Schlattmann
Journal: Advances in Fuzzy Systems
 (2015)

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## Abstract

This paper presents a novel analytical approach to evaluating continuous, nonmonotonic functions of independent fuzzy numbers. The approach is
based on a parametricα-cut representation of fuzzy numbers and allows for the inclusion of parameter uncertainties into mathematical models.


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## Body

## 1. Introduction

This paper continues the research from our previous work [1]. In [1], we formulated a practical analytical approach to evaluating continuous, monotonic functions of independent fuzzy numbers, which is based on an alternative formulation of the extension principle [2]. In this paper, we extend this approach to general, nonmonotonic functions of independent fuzzy numbers. For the theoretical background, we assume the reader is familiar with Sections 2 and 4 from [1].


## 2. Preliminaries

In the remainder of this paper, we will use two typical fuzzy numbers in engineering, the triangular and the Gaussian fuzzy number. The definitions are provided below.Definition 1.
Thetriangular fuzzy number (TFN) [3] (1)x~=tfn(x¯,τL,τR)is defined by the membership function(2)μx~(x)=1+x-x¯τL,x¯-τL≤x≤x¯,1-x-x¯τR,x¯<x≤x¯+τR,where x¯ denotes themodal value, τL denotes theleft-hand, and τR denotes theright-hand spread of x~. If τL=τR, the TFN is calledsymmetric. Its α-cuts x(α)=[xL(α),xR(α)] result from the inverse functions of (2) with respect to x: (3)xL(α)=x¯-τL(1-α),0<α≤1,xR(α)=x¯+τR(1-α),0<α≤1.Definition 2.
TheGaussian fuzzy number (GFN) [3] (4)x~=gfn(x¯,σL,σR)is defined by the membership function (5)μx~(x)=exp⁡-12x-x¯σL2,x≤x¯,exp⁡-12x-x¯σR2,x>x¯,where x¯ denotes themodal value,σL denotes theleft-hand, and σR denotes theright-hand standard deviation of x~. If σL=σR, the GFN is calledsymmetric. Its α-cuts x(α)=[xL(α),xR(α)] result in (6)xL(α)=x¯-σL-2ln⁡(α),0<α≤1,xR(α)=x¯+σR-2ln⁡(α),0<α≤1.


## 3. Analytical Approach

For evaluating continuous, nonmonotonic functions of independent fuzzy numbers, the authors in [4, 5] suggest including the extreme points as constant profiles into the computation. However, this is not enough and can lead to erroneous results, as was pointed out in [6]. More specifically, all permutations of the interval boundaries of xm(α), m=1,…,n, with the components of the extreme points have to be considered as well.Basically, our analytical approach can be classified into two parts depending on the monotonicity off: a general and an extended part.


### 3.1. General Part

If the functionf is nonmonotonic in all xm, m=1,…,n, we can obtain the analytical solution as follows.(1)
Evaluate the functionf for all the 2n permutations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=1,…,n. For example, if n=2, then compute (7)yLL(α)=f(x1L(α),x2L(α)),yLR(α)=f(x1L(α),x2R(α)),yRL(α)=f(x1R(α),x2L(α)),yRR(α)=f(x1R(α),x2R(α)).(2)
Evaluate the functionf for all the 2ns combinations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=1,…,n, with the components of the extreme points (x1,r∗,…,xn,r∗), r=1,…,s. For example, if n=3 and s=1, then compute (8)yL11(α)=f(x1L(α),x2,1∗,x3,1∗),yR11(α)=f(x1R(α),x2,1∗,x3,1∗),y1L1(α)=f(x1,1∗,x2L(α),x3,1∗),y1R1(α)=f(x1,1∗,x2R(α),x3,1∗),y11L(α)=f(x1,1∗,x2,1∗,x3L(α)),y11R(α)=f(x1,1∗,x2,1∗,x3R(α)).(3)
If(x1,r∗,…,xn,r∗)∈supp(x~1)×⋯×supp(x~n) for certain r∈{1,…,s}, compute the corresponding y∗r=f(x1,r∗,…,xn,r∗).(4)
Plot all solution candidates in the same diagram.(5)
Finally, starting from the modal pointy¯=f(x¯1,…,x¯n) at α=1, construct the maximum envelope formed by the possible solution candidates for α→0 under the condition of convexity.Thisgeneral part of our approach requires a total of maximum 2n+(2n+1)s and minimum 2n+2ns function evaluations. It can be viewed as an analytical version of thelevel interval algorithm [6].Example 1.
The functionf1:R2→R with (9)y1=f1(x1,x2)=x12+x22-5x1-x2shall be evaluated for the two fuzzy numbers x~1=tfn(2,2,3) and x~2=tfn(2,2,2). Since (10)∂f1∂x1=2x1-5,∂f1∂x2=2x2-1,the function f1 is nonmonotonic in both x1 and x2 in the domain supp(x~1)×supp(x~2)=(0,5)×(0,4) with one (global) extremum at (x1,1∗,x2,1∗)=(2.5,0.5)∈(0,5)×(0,4). Hence, the general part of our approach should be applied. The solution candidates for y1(α) are (11)y1LL(α)=f1(x1L(α),x2L(α))=8α2-12α,y1LR(α)=f1(x1L(α),x2R(α))=8α2-24α+12,y1RL(α)=f1(x1R(α),x2L(α))=13α2-17α,y1RR(α)=f1(x1R(α),x2R(α))=13α2-29α+12,y1L1(α)=f1(x1L(α),x2,1∗)=4α2-10α-0.25,y1R1(α)=f1(x1R(α),x2,1∗)=9α2-15α-0.25,y11L(α)=f1(x1,1∗,x2L(α))=4α2-2α-6.25,y11R(α)=f1(x1,1∗,x2R(α))=4α2-14α+5.75,y1∗1=f1(x1,1∗,x2,1∗)=-6.5.
We can see from their plots in Figure1 that, starting from the modal point at α=1, the left branch of the maximum envelope, illustrated by the gray area, is formed by y1RL for 1≥α>0.83¯, by y11L for 0.83¯≥α>0.25, and by y1∗1 for 0.25≥α>0, where the value 0.83¯ corresponds to the intersection point between y1RL and y11L and the value 0.25 to the intersection point between y11L and y1∗1. Its right branch, on the other hand, is entirely formed by y1LR. Hence, the α-cuts y1(α)=[y1L(α),y1R(α)] of y~1 are (12)y1L(α)=-6.5,0<α≤0.25,4α2-2α-6.25,0.25<α≤0.83¯,13α2-17α,0.83¯<α≤1,y1R(α)=8α2-24α+12,0<α≤1.
Withy1L(0.25)=-6.5, y1L(0.83¯)=-5.138¯, y1L(1)=-4=y1R(1), and y1R(0)=12, the membership function of y~1 yields (13)μy~1(y)=14+144y+26,-6.5<y≤-5.138¯,1726+12652y+289,-5.138¯<y≤-4,32-142y+12,-4<y<12.Figure 1
Solution candidates from Example1.


### 3.2. Extended Part

Let the continuous functionf be (strictly) monotonic increasing in xi, i=1,…,k, (strictly) monotonic decreasing in xj, j=1,…,l, monotonic in xp, p=1,…,q, and nonmonotonic in xt, t=1,…,u, in the domain of interest, with k+l+q+u=n. Then, the analytical solution can be obtained as follows.(1)
Evaluate the functionf for xiL(α), i=1,…,k, and xjR(α), j=1,…,l, including all the 2q+u permutations of the interval boundaries of xp(α)=[xpL(α),xpR(α)], p=1,…,q, and xt(α)=[xtL(α),xtR(α)], t=1,…,u, to compute the monotonic solution candidates for yL(α).(2)
Evaluate the functionf for xiR(α), i=1,…,k, and xjL(α), j=1,…,l, including all the 2q+u permutations of the interval boundaries of xp(α)=[xpL(α),xpR(α)], p=1,…,q, and xt(α)=[xtL(α),xtR(α)], t=1,…,u, to compute the monotonic solution candidates for yR(α).(3)
Evaluate the functionf for xiL(α), i=1,…,k, and xjR(α), j=1,…,l, including all the us combinations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=k+l+q+1,…,n, with the components of the extreme points (x1,r∗,…,xu,r∗), r=1,…,s, to compute the nonmonotonic solution candidates for yL(α).(4)
Evaluate the functionf for xiR(α), i=1,…,k, and xjL(α), j=1,…,l, including all the us combinations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=k+l+q+1,…,n with the components of the extreme points (x1,r∗,…,xu,r∗), r=1,…,s, to compute the nonmonotonic solution candidates for yR(α).(5)
Plot all solution candidates in the same diagram.(6)
Finally, starting from the modal pointy¯=f(x¯1,…,x¯n) at α=1, construct the maximum envelope formed by the possible solution candidates for α→0 under the condition of convexity.Thisextended part of our approach requires a total of 2q+u+1+2us function evaluations.Example 2.
Now, the functionf2:R2→R with (14)y2=f2(x1,x2)=x12+x22-5x1shall be evaluated for the two fuzzy numbers from Example 1. Since (15)∂f2∂x1=2x1-5,∂f2∂x2=2x2>0,the function f2 is nonmonotonic in x1 with one (global) extremum at x1,1∗=2.5 and (strictly) monotonic increasing in x1 in the domain supp(x~1)×supp(x~2)=(0,5)×(0,4). Hence, the extended part of our approach should be applied. The monotonic solution candidates for y2L(α) are (16)y2LL(α)=f2(x1L(α),x2L(α))=8α2-10α,y2RL(α)=f2(x1R(α),x2L(α))=13α2-15α,and for y2R(α), (17)y2LR(α)=f2(x1L(α),x2R(α))=8α2-26α+16,y2RR(α)=f2(x1R(α),x2R(α))=13α2-31α+16.The nonmonotonic solution candidate for y2L(α) is (18)y21L(α)=f2(x1,1∗,x2L(α))=4α2-6.25,and for y2R(α), (19)y21R(α)=f2(x1,1∗,x2R(α))=4α2-16α+9.75.
We can see from their plots in Figure2 that, starting from the modal point at α=1, the left branch of the maximum envelope is formed by y2RL for 1≥α>0.83¯ and by y21L for 0.83¯≥α>0, where the value 0.83¯ corresponds to their intersection point. Its right branch, on the other hand, is entirely formed by y2LR. Hence, the α-cuts y2(α)=[y2L(α),y2R(α)] of y~2 are (20)y2L(α)=4α2-6.25,0<α≤0.83¯,13α2-15α,0.83¯<α≤1,y2R(α)=8α2-26α+16,0<α≤1.
Withy2L(0)=-6.25, y2L(0.83¯)=-3.472¯, y2L(1)=-2=y2R(1), and y2R(0)=16, the membership function of y~2 yields (21)μy~2(y)=144y+25,-6.25<y≤-3.472¯,1526+12652y+225,-3.472¯<y≤-2,138-188y+41,-2<y<16.Figure 2
Solution candidates from Example2.


## 3.1. General Part

If the functionf is nonmonotonic in all xm, m=1,…,n, we can obtain the analytical solution as follows.(1)
Evaluate the functionf for all the 2n permutations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=1,…,n. For example, if n=2, then compute (7)yLL(α)=f(x1L(α),x2L(α)),yLR(α)=f(x1L(α),x2R(α)),yRL(α)=f(x1R(α),x2L(α)),yRR(α)=f(x1R(α),x2R(α)).(2)
Evaluate the functionf for all the 2ns combinations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=1,…,n, with the components of the extreme points (x1,r∗,…,xn,r∗), r=1,…,s. For example, if n=3 and s=1, then compute (8)yL11(α)=f(x1L(α),x2,1∗,x3,1∗),yR11(α)=f(x1R(α),x2,1∗,x3,1∗),y1L1(α)=f(x1,1∗,x2L(α),x3,1∗),y1R1(α)=f(x1,1∗,x2R(α),x3,1∗),y11L(α)=f(x1,1∗,x2,1∗,x3L(α)),y11R(α)=f(x1,1∗,x2,1∗,x3R(α)).(3)
If(x1,r∗,…,xn,r∗)∈supp(x~1)×⋯×supp(x~n) for certain r∈{1,…,s}, compute the corresponding y∗r=f(x1,r∗,…,xn,r∗).(4)
Plot all solution candidates in the same diagram.(5)
Finally, starting from the modal pointy¯=f(x¯1,…,x¯n) at α=1, construct the maximum envelope formed by the possible solution candidates for α→0 under the condition of convexity.Thisgeneral part of our approach requires a total of maximum 2n+(2n+1)s and minimum 2n+2ns function evaluations. It can be viewed as an analytical version of thelevel interval algorithm [6].Example 1.
The functionf1:R2→R with (9)y1=f1(x1,x2)=x12+x22-5x1-x2shall be evaluated for the two fuzzy numbers x~1=tfn(2,2,3) and x~2=tfn(2,2,2). Since (10)∂f1∂x1=2x1-5,∂f1∂x2=2x2-1,the function f1 is nonmonotonic in both x1 and x2 in the domain supp(x~1)×supp(x~2)=(0,5)×(0,4) with one (global) extremum at (x1,1∗,x2,1∗)=(2.5,0.5)∈(0,5)×(0,4). Hence, the general part of our approach should be applied. The solution candidates for y1(α) are (11)y1LL(α)=f1(x1L(α),x2L(α))=8α2-12α,y1LR(α)=f1(x1L(α),x2R(α))=8α2-24α+12,y1RL(α)=f1(x1R(α),x2L(α))=13α2-17α,y1RR(α)=f1(x1R(α),x2R(α))=13α2-29α+12,y1L1(α)=f1(x1L(α),x2,1∗)=4α2-10α-0.25,y1R1(α)=f1(x1R(α),x2,1∗)=9α2-15α-0.25,y11L(α)=f1(x1,1∗,x2L(α))=4α2-2α-6.25,y11R(α)=f1(x1,1∗,x2R(α))=4α2-14α+5.75,y1∗1=f1(x1,1∗,x2,1∗)=-6.5.
We can see from their plots in Figure1 that, starting from the modal point at α=1, the left branch of the maximum envelope, illustrated by the gray area, is formed by y1RL for 1≥α>0.83¯, by y11L for 0.83¯≥α>0.25, and by y1∗1 for 0.25≥α>0, where the value 0.83¯ corresponds to the intersection point between y1RL and y11L and the value 0.25 to the intersection point between y11L and y1∗1. Its right branch, on the other hand, is entirely formed by y1LR. Hence, the α-cuts y1(α)=[y1L(α),y1R(α)] of y~1 are (12)y1L(α)=-6.5,0<α≤0.25,4α2-2α-6.25,0.25<α≤0.83¯,13α2-17α,0.83¯<α≤1,y1R(α)=8α2-24α+12,0<α≤1.
Withy1L(0.25)=-6.5, y1L(0.83¯)=-5.138¯, y1L(1)=-4=y1R(1), and y1R(0)=12, the membership function of y~1 yields (13)μy~1(y)=14+144y+26,-6.5<y≤-5.138¯,1726+12652y+289,-5.138¯<y≤-4,32-142y+12,-4<y<12.Figure 1
Solution candidates from Example1.


## 3.2. Extended Part

Let the continuous functionf be (strictly) monotonic increasing in xi, i=1,…,k, (strictly) monotonic decreasing in xj, j=1,…,l, monotonic in xp, p=1,…,q, and nonmonotonic in xt, t=1,…,u, in the domain of interest, with k+l+q+u=n. Then, the analytical solution can be obtained as follows.(1)
Evaluate the functionf for xiL(α), i=1,…,k, and xjR(α), j=1,…,l, including all the 2q+u permutations of the interval boundaries of xp(α)=[xpL(α),xpR(α)], p=1,…,q, and xt(α)=[xtL(α),xtR(α)], t=1,…,u, to compute the monotonic solution candidates for yL(α).(2)
Evaluate the functionf for xiR(α), i=1,…,k, and xjL(α), j=1,…,l, including all the 2q+u permutations of the interval boundaries of xp(α)=[xpL(α),xpR(α)], p=1,…,q, and xt(α)=[xtL(α),xtR(α)], t=1,…,u, to compute the monotonic solution candidates for yR(α).(3)
Evaluate the functionf for xiL(α), i=1,…,k, and xjR(α), j=1,…,l, including all the us combinations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=k+l+q+1,…,n, with the components of the extreme points (x1,r∗,…,xu,r∗), r=1,…,s, to compute the nonmonotonic solution candidates for yL(α).(4)
Evaluate the functionf for xiR(α), i=1,…,k, and xjL(α), j=1,…,l, including all the us combinations of the interval boundaries of xm(α)=[xmL(α),xmR(α)], m=k+l+q+1,…,n with the components of the extreme points (x1,r∗,…,xu,r∗), r=1,…,s, to compute the nonmonotonic solution candidates for yR(α).(5)
Plot all solution candidates in the same diagram.(6)
Finally, starting from the modal pointy¯=f(x¯1,…,x¯n) at α=1, construct the maximum envelope formed by the possible solution candidates for α→0 under the condition of convexity.Thisextended part of our approach requires a total of 2q+u+1+2us function evaluations.Example 2.
Now, the functionf2:R2→R with (14)y2=f2(x1,x2)=x12+x22-5x1shall be evaluated for the two fuzzy numbers from Example 1. Since (15)∂f2∂x1=2x1-5,∂f2∂x2=2x2>0,the function f2 is nonmonotonic in x1 with one (global) extremum at x1,1∗=2.5 and (strictly) monotonic increasing in x1 in the domain supp(x~1)×supp(x~2)=(0,5)×(0,4). Hence, the extended part of our approach should be applied. The monotonic solution candidates for y2L(α) are (16)y2LL(α)=f2(x1L(α),x2L(α))=8α2-10α,y2RL(α)=f2(x1R(α),x2L(α))=13α2-15α,and for y2R(α), (17)y2LR(α)=f2(x1L(α),x2R(α))=8α2-26α+16,y2RR(α)=f2(x1R(α),x2R(α))=13α2-31α+16.The nonmonotonic solution candidate for y2L(α) is (18)y21L(α)=f2(x1,1∗,x2L(α))=4α2-6.25,and for y2R(α), (19)y21R(α)=f2(x1,1∗,x2R(α))=4α2-16α+9.75.
We can see from their plots in Figure2 that, starting from the modal point at α=1, the left branch of the maximum envelope is formed by y2RL for 1≥α>0.83¯ and by y21L for 0.83¯≥α>0, where the value 0.83¯ corresponds to their intersection point. Its right branch, on the other hand, is entirely formed by y2LR. Hence, the α-cuts y2(α)=[y2L(α),y2R(α)] of y~2 are (20)y2L(α)=4α2-6.25,0<α≤0.83¯,13α2-15α,0.83¯<α≤1,y2R(α)=8α2-26α+16,0<α≤1.
Withy2L(0)=-6.25, y2L(0.83¯)=-3.472¯, y2L(1)=-2=y2R(1), and y2R(0)=16, the membership function of y~2 yields (21)μy~2(y)=144y+25,-6.25<y≤-3.472¯,1526+12652y+225,-3.472¯<y≤-2,138-188y+41,-2<y<16.Figure 2
Solution candidates from Example2.


## 4. Engineering Application

In order to illustrate the analytical approach in a more practical context, we consider a linear system with one degree of freedom consisting of a block with massm moving on a smooth surface as shown in Figure 3. The block is connected to a wall via a linear spring with spring constant k. This system is governed by the following linear, homogeneous ordinary differential equation of second order with constant coefficients [7]: (22)x¨+ω2x=0.Here, (23)ω=kmdenotes thenatural frequency of the system. The general solution of (22) is given by (24)xt=x0cos⁡ωt+x˙0ωsinωt,where x0=x(0) and x˙0=x˙(0) denote the initial conditions.Figure 3
Mass-spring system.We assumex0 and ω to be uncertain, both described by fuzzy numbers. More specifically, the uncertain initial position is modeled by the (symmetric) triangular fuzzy number (25)x~0=tfn(1,0.5,0.5)cmand the uncertain natural frequency by the (symmetric) Gaussian fuzzy number (26)ω~=gfn(1,0.05,0.05)Hz.Furthermore, we assume x˙0=0. We are interested in the uncertain position of the mass after one period (t=2π).Since(27)∂x∂x0=cos⁡(2πω)≥0,∂x∂ω=-2πx0sin(2πω),x is (strictly) monotonic increasing in x0 and nonmonotonic in ω in the domain supp(x~0)×supp(ω~)∩R+2 with an infinite number of local extrema ωi∗=i/2, i∈N0. Hence, the extended part of our approach should be applied. The monotonic solution candidates for xL(α) are (28)xLL(α)=xLR(α)=(1-0.5(1-α))cos⁡(0.1π-2ln⁡(α)),and for xR(α), (29)xRL(α)=xRR(α)=(1+0.5(1-α))cos⁡(0.1π-2ln⁡(α)).The nonmonotonic solution candidates for xL(α), on the other hand, are (30)xLj(α)=-1-0.51-α,j=2l+1,l∈N0,xLk(α)=+1-0.51-α,k=2l+2,l∈N0,and for xR(α), (31)xRj(α)=-1+0.51-α,j=2l+1,l∈N0,xRk(α)=+1+0.51-α,k=2l+2,l∈N0.We can see from their plots in Figure4 that the left branch of the maximum envelope is formed by xLL=xLR and the right branch by xRk. Hence, the α-cuts x(α)=[xL(α),xR(α)] of x~ are (32)xL(α)=(1-0.5(1-α))cos⁡(0.1π-2ln⁡(α)),xR(α)=1+0.5(1-α).Since xL(α) in (32) is not invertible with respect to α, it is not possible to give an analytical expression for the membership function of x~. However, the α-cuts and the membership function are both equivalent representations of a fuzzy number. The inverted plots of (32) are illustrated in Figure 5.Figure 4
Solution candidates forx(α).Figure 5
Membership function ofx~.


## 5. Conclusions

We extended our analytical approach from [1] to general, nonmonotonic functions of independent fuzzy numbers. It is based on an α-cut formulation of the extension principle and allows for the inclusion of parameter uncertainties into mathematical models.In further research activities, the influence of interdependency may be a subject of investigation.



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