Study of Educational Information Resource Download Quality with Optimal Symmetrical Interval Solution of Fuzzy Relation Inequality in the Format of a System of Differential Equations

Abstract

The min–max fuzzy relation inequalities are currently considered for representing the place-to-place (P2P) education knowledge, including resource sharing from one terminal to another. One terminal is the acceptor—receiving information—while the other terminal is the sink resource for educational information sharing, acting like an extractor. In the current manuscript, the idea of sharing educational information is established in the form of a dynamical system in which the unknown quantities represent the quality of downloading educational resources on different terminals. The download quality, measured in bits per second (bps), has been converted to a fuzzy format as it oscillates from low to high. Every solution of the min–max dynamical model is surely an optimal interval approach in the corresponding terminal-to-terminal network sharing system. Such a solution implies the stability of the interval solution with fluctuations from the minimum (low) to maximum (high) values of the interval. Furthermore, like the objective function in the linear programming and stability of the system, we study the system with the maximum fluctuation for a given solution in the form of download quality educational informative resources. Further, the solution will be treated in optimal relative local regions (MRO) and global regions (MAO). Bi-approaches are constructed to solve these maximal symmetrical interval fuzzy solutions for our analysis. The illustrations show that the bi-approaches are valid and effective for the studied model.

1. Introduction

Every relation among different sets or terminals can be easily converted to a fuzzy format or Boolean form by using the version of mathematical algebra of matrices $F^f={\left(b_{ij}^f\right)}_{i\times j}$, where ${\left(b_{ij}^f\right)}_{i\times j}\in [0,1]$ and $i=1,2\dots m,j=1,2\dots n$. Surely, any classical format relation is the proper version of fuzzy relations. Therefore, the generalized (fuzzy) relation may be written in an equality form, as well as in an inequality form, under the linear matrix format given by

$$\mathbb{B}\circ y=c$$

and

$$c\le \mathbb{B}\circ y\le C$$

through any fuzzy relation $\mathbb{B}$. ${\mathbb{B}}_{i\times j},y_{i\times 1},b_{j\times 1}$ are matrices in the format of fuzziness. $\mathbb{B}\circ y\in [c,C]$ is the interval of quality, and $y_{i\times 1}$ is the required quantity to be computed, representing the quality of downloading (receiving or extracting) educational informative resources among different sources or terminals. Download quality is therefore accounted for well in [1,2,3,4], acting like a membership function in fuzzy algebra, and it may be low or high. For the high and low values, the best representation is membership mapping in fuzzy algebra among any of qualities associated with real-world phenomena.

Fuzzy relation equalities and inequalities were widely discussed after the consideration of E. Sanchez [5], with uses in medical diagnosis [6]. Previously, the fuzzy relations were also applied in the processing of images for their coding and decoding [7], image reconstruction [8], and compression phenomena [9,10]. Further, in one study on the application of a system based on fuzzy relations, the resolution schemes and their associated optimality problems were considered alongside the classical linear equations, which may be treated by the simplex method.

A different question arises for the complete solution of a fuzzy relation system, including for the scheme of the solution, the maximal and minimal values, and the range of these values. For answering such question, scholars have introduced many resolution schemes for the fuzzy relation equations with max-T composition [11,12,13,14]. T is for a triangle norm with continuity, comprising addition and product operators. We consider the same approach for a fuzzy inequality equation as that for a fuzzy inequality relation system with max-T composition [15,16]. Further, finding a solution for the whole set is a complex task but not impossible. The system which is consistent for the max-T fuzzy relation equation system commonly has more numbers of minimal solution increasing exponentially. Such roots may be highly related to the specific covering system of the fuzzy relation system [17,18,19,20,21]. Furthermore, the covering-set scheme may be generalized and used for computing the fuzzy relation system programming with the max-T composed of addition and multiplication [22,23]. In the real sense, the whole solution set is not important, but a specific solution may be applied to certain real-world problems. Finding out a few specific roots for fuzzy relation equations and inequality-optimized models, which are beneficial for administrative fields, is of great importance. The different formats of optimized problems with the max-T fuzzy relation with some subjective conditions were investigated by resolution schemes of concerning polynomials [16,24,25,26,27,28].

Previously, max-T fuzzy relation equations or inequality-based systems with an addition–min composition were applied to foodstuff permutations [3,4,27,29]. The basic needs of humans were categorized by bi-groups of fuzzy relation inequalities. Using some feasible domain, the scholars further studied an optimal root of the given fuzzy relation, encompassing non-linear [3] or linear [4,29] programming, as an optimal technique for financier decision making. This approach, based on inequalities, was also used for a wireless communication system designed to transport information from one place to another [16,28,30,31]. In 2013, S. Freson et al. [32] introduced the bipolar fuzzy relation equation, with uses in markets for advertising worthy goods via solving linear and non-linear optimized problems for optimal control. Recently, Guo et al. [33,34] identified the uses of addition–min fuzzy relation inequalities in supply chains with unique minimal, finite maximal, and strong roots [34]. Recently, the max–min fuzzy relation inequality system was used to formulate the Peer-to-Peer (P2P) educational information resource sharing system [35,36,37]. Let us say we have a system of differential equations representing the P2P educational information sharing problem:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=(b_{11}\wedge y_1)\vee (b_{12}\wedge y_2)\vee (b_{13}\wedge y_3)\vee \dots \vee (b_{1n}\wedge y_n),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=(b_{21}\wedge y_1)\vee (b_{22}\wedge y_2)\vee (b_{23}\wedge y_3)\vee \dots \vee (b_{2n}\wedge y_n),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=(b_{31}\wedge y_1)\vee (b_{32}\wedge y_2)\vee (b_{33}\wedge y_3)\vee \dots \vee (b_{3n}\wedge y_n),\hfill \\ ⋮\hfill \\ {\displaystyle \frac{d{y}_n}{dt}}=(b_{m1}\wedge y_1)\vee (b_{m2}\wedge y_2)\vee (b_{m3}\wedge y_3)\vee \dots \vee (b_{mn}\wedge y_n),\hfill \\ y_1\left(0\right)=y_{1_0},y_2\left(0\right)=y_{2_0},y_3\left(0\right)=y_{3_0},\dots y_n\left(0\right)=y_{n_0}.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(1)

where $b_{ij}$ is the band among any two sources or a terminal sharing data from i to j for the direction source, with download quality, $y_n$, represented in the sense of receiving information in fuzzy format with initial speed or quality $y_n\left(0\right)=y_{n_0}$. System (1) can be written in an addition–multiplication format as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=\left(b_{11}{y}_1\right)+\left(b_{12}{y}_2\right)+\left(b_{13}{y}_3\right)+\dots +\left(b_{1n}{y}_n\right),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=\left(b_{21}{y}_1\right)+\left(b_{22}{y}_2\right)+\left(b_{23}{y}_3\right)+\dots +\left(b_{2n}{y}_n\right),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=\left(b_{31}{y}_1\right)+\left(b_{32}{y}_2\right)+\left(b_{33}{y}_3\right)+\dots +\left(b_{3n}{y}_n\right),\hfill \\ ⋮\hfill \\ {\displaystyle \frac{d{y}_n}{dt}}=\left(b_{m1}{y}_1\right)+\left(b_{m2}{y}_2\right)+\left(b_{m3}{y}_3\right)+\dots +\left(b_{mn}{y}_n\right),\hfill \\ y_1\left(0\right)=y_{1_0},y_2\left(0\right)=y_{2_0},y_3\left(0\right)=y_{3_0},\dots y_n\left(0\right)=y_{n_0}.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(2)

The quality of downloading or giving data of information among any two sources will lie in the interval form having maximal and minimal values. Therefore, the fuzzy relation system (1) can be written in interval form as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{c}_1\le {\displaystyle \frac{d{y}_1}{dt}}\le C_1,\hfill \\ c_2\le {\displaystyle \frac{d{y}_2}{dt}}\le C_2,\hfill \\ c_3\le {\displaystyle \frac{d{y}_3}{dt}}\le C_3,\hfill \\ ⋮\hfill \\ c_n\le {\displaystyle \frac{d{y}_n}{dt}}\le C_n.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(3)

or

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{c}_1\le (b_{11}\wedge y_1)\vee (b_{12}\wedge y_2)\vee (b_{13}\wedge y_3)\vee \dots \vee (b_{1n}\wedge y_n)\le C_1,\hfill \\ c_2\le (b_{21}\wedge y_1)\vee (b_{22}\wedge y_2)\vee (b_{23}\wedge y_3)\vee \dots \vee (b_{2n}\wedge y_n)\le C_2,\hfill \\ c_3\le (b_{31}\wedge y_1)\vee (b_{32}\wedge y_2)\vee (b_{33}\wedge y_3)\vee \dots \vee (b_{3n}\wedge y_n)\le C_3,\hfill \\ ⋮\hfill \\ c_n\le (b_{m1}\wedge y_1)\vee (b_{m2}\wedge y_2)\vee (b_{m3}\wedge y_3)\vee \dots \vee (b_{mn}\wedge y_n)\le C_n.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(4)

Proceeding further, we will examine the optimal interval solution for the quality of resource sharing information system.

2. Background Materials

The optimized symmetrical interval solution of system (1) with required work will be highlighted in this section. As a basic necessity, we give some fundamental concepts and results on system (1) in this preliminary section. Let us write the complete solution set along with the indexing sets as

$$\begin{array}{c}\hfill \begin{array}{cc}& P=\{y\in {[0,1]}^m|c_i^t\le ({\mathbb{B}}_i\circ y_i^t)\le C_i^t,\forall i\in I\},\hfill \end{array}\end{array}$$

(5)

where $\mathbb{B}={\left(b_{ij}\right)}_{m\times n}$, $y=(y_1,y_2,y_3,\dots y_n)$, $b_i=b_1,b_2,b_3,\dots b_m$, $c_i=c_1,c_2,c_3,\dots c_m$ and “∘” show the max–min fuzzy composition. Further, $c_i^t$ is the minimum value, while $C_i^t$ is the maximum value of quality $y_i^t$ of downloading from $b_{ij}$ terminals. The quality of downloading lies in the range of $[c_i^t,C_i^t].$

$$I=\{1,2,3,\dots m\}$$

and

$$J=\{1,2,3,\dots n\}$$

are the indexing sets.

Theorem 1

([35,36]). Let $y\in {[0,1]}^m$ be any vector; then, $y\in P$ if

(i) 

$b_{ij}\wedge y_j\le C_i,j\in J;$

(ii) 

$b_{ik}\wedge y_k\le C_i,k\in J.$

Proposition 1.

Consider $y^{\prime },y^{\prime \prime }\in P$ to be the solution of system (1), and if $y^{\prime }\le y^{\prime \prime }$, then $[y^{\prime },y^{\prime \prime }]\subseteq P.$ Further, consider

$$\begin{array}{c}\hfill b_{ij}\diamond C_i=\left\{\begin{array}{c}{C}_i,if{b}_{ij}>C_i,\hfill \\ 1,if{b}_{ij}\le C_i,i\in I,j\in J.\hfill \end{array}\right.\end{array}$$

(6)

Then, we choose vector $\widehat{y}=({\widehat{y}}_1,\widehat{y_2},\widehat{y_3},\dots ,\widehat{y_n})$ as follows:

$$\begin{array}{c}\hfill {\widehat{y}}_j=\underset{i\in I}{min}\{b_{ij}\diamond C_i\},\forall j\in J\end{array}$$

(7)

Theorem 2

([35,36]). System (1) will have a solution if $\widehat{y}\in P$, and $\widehat{y}\in P$ is the unique maximal solution of the given system.

For the complete solution of the considered system, we developed a path-based approach. Write

$$\begin{array}{c}\hfill {\widehat{J}}_i=\{j\in J:b_{ij}\wedge {\widehat{y}}_j\ge c_i\}\forall i\in I.\end{array}$$

(8)

Let

$$S={\widehat{J}}_1\times {\widehat{J}}_2\times {\widehat{J}}_3\times \dots \times {\widehat{J}}_n$$

and $j^S=(j_1,j_2,j_3,\dots j_n)\in S$ be the path for problem (1), where S is the set of all paths which will be consistent if $S\ne \varnothing$. By this path, we take a vector

$$y^{j^S}=y_1^{j^S},y_2^{j^S},y_3^{j^S},\dots ,y_n^{j^S}$$

as follows

$$\begin{array}{c}\hfill y_k^{j^S}=\left\{\begin{array}{c}0,if{I}_k^{j^S}=\varnothing \hfill \\ \underset{i\in I_k^{j^S}}{\bigvee }\left(c_i\right),if{I}_k^{j^S}\ne \varphi \ne \varnothing \forall k\in J.\hfill \end{array}\right.\end{array}$$

(9)

where $I_k^{j^S}=\{i\in I:j_i=k,k\in J\}$. By this, we can write Proposition 2 and Theorem 3.

Proposition 2

([35,36]). For any solution path, $j^S\in S$, and vector $y^{j^S}=y_1^{j^S},y_2^{j^S},y_3^{j^S},\dots y_n^{j^S}$ $I_k^{j^S}=\{i\in I:j_i=k,k\in J\}$$y^{j^S}\in P$ is the solution of system (1).

Definition 1.

Consider $j^S\in S$ to be any of the solution paths; then, $y^{j^S}$ is called the quasi-minimum solution of problem (1).

Furthermore, the whole solution set of problem (1), denoted by P, will be all quasi-minimum solutions and the maximum solutions as given in the following theorem:

Theorem 3

([35,36]). If problem (1) has consistency, then we can write

$$P=\bigcup _{j^S\in S}[y^{j^S},\widehat{y}]$$

with S being the path set of problem (1), while $\widehat{y}$ is the maximal solution.

Definition 2.

Consider $y_0=(y_0^1,y_0^2,y_0^3\dots y_0^n)\in P$ to be the initial solution in a P2P educational resource sharing system given by the operator for system (1). We chose random and bi-directional alteration ${\delta }_j^b$ in $y_0^j\in y_0.$ For the initial solution $y_0$, if

$$(y_0^1,y_0^2,\dots y_0^{j-1},y_0^j-{\delta }_b^j,y_0^{j+1},\dots ,y_0^n)$$

and

$$(y_0^1,y_0^2,\dots y_0^{j-1},y_0^j+{\delta }_b^j,y_0^{j+1},\dots ,y_0^n)$$

are solutions of (1), then ${\delta }_b^j$ is a symmetrical absolute oscillation for $y_0^j\in y_0$

Definition 3.

For the $y_0$ solution, if $\delta =({\delta }_1,{\delta }_1,{\delta }_2,\dots ,{\delta }_n)\in {[0,1]}^n|[y_0-\delta ,y_0+\delta ]\subseteq P,$ then $[y_0-\delta ,y_0+\delta ]$ is the symmetrical interval solution of problem (1) with repspect to $y_0,$ where

$$y_0\pm \delta =(y_0^1\pm {\delta }_1,y_0^2\pm {\delta }_2,\dots ,y_0^n\pm {\delta }_n)$$

Definition 4.

Consider $[y_0-\delta ,y_0+\delta ]$ as the symmetrical interval solution of problem (1) with respect to $y_0,$; then,

$${\delta }_{min}^A={\delta }_1\wedge {\delta }_2\wedge \dots \wedge {\delta }_n$$

is the absolute oscillation related to $y_0$ with respect to $[y_0-\delta ,y_0+\delta ].$

Definition 5.

If $y_0^j>0$ for $j=1,2,3\dots n$, then

$${\delta }_{min}^R={\displaystyle \frac{{\delta }_1}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n}{y_0^n}}$$

is called a relative oscillation for $y_0$ with respect to $[y_0-\delta ,y_0+\delta ].$ Further, if there exists some $j^r\in J$ for which $y_0^{j^r}=0$, then ${\delta }_{min}^A=0$ and ${\delta }_{min}^R=0$

Example 1.

Let our fuzzy relation system with an addition–multiplication composition be as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=\left(b_{11}{y}_1\right)+\left(b_{12}{y}_2\right)+\left(b_{13}{y}_3\right)+\left(b_{14}{y}_4\right),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=\left(b_{21}{y}_1\right)+\left(b_{22}{y}_2\right)+\left(b_{23}{y}_3\right)+\left(b_{24}{y}_4\right),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=\left(b_{31}{y}_1\right)+\left(b_{32}{y}_2\right)+\left(b_{33}{y}_3\right)+\left(b_{34}{y}_4\right),\hfill \\ {\displaystyle \frac{d{y}_4}{dt}}=\left(b_{41}{y}_1\right)+\left(b_{42}{y}_2\right)+\left(b_{43}{y}_3\right)+\left(b_{44}{y}_4\right),\hfill \\ y_0^1=0.7,y_0^2=0.4,y_0^3=0.6,y_0^4=0.7.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(10)

Convert (10) to an inequality system as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{c}_1\le {\displaystyle \frac{d{y}_1}{dt}}\le C_1,\hfill \\ c_2\le {\displaystyle \frac{d{y}_2}{dt}}\le C_2,\hfill \\ c_3\le {\displaystyle \frac{d{y}_3}{dt}}\le C_3,\hfill \\ c_4\le {\displaystyle \frac{d{y}_4}{dt}}\le C_4.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(11)

Alternatively,

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.45\le (0.5\wedge y_1)\vee (0.6\wedge y_2)\vee (0.7\wedge y_3)\vee 0.9\wedge y_4) \le 0.92,\hfill \\ 0.43\le (0.8\wedge y_1)\vee (0.9\wedge y_2)\vee (0.3\wedge y_3)\vee 0.5\wedge y_4) \le 0.85,\hfill \\ 0.39\le (0.3\wedge y_1)\vee (0.4\wedge y_2)\vee (0.7\wedge y_3)\vee 0.6\wedge y_4) \le 0.88,\hfill \\ 0.41\le (0.6\wedge y_1)\vee (0.5\wedge y_2)\vee (0.6\wedge y_3)\vee 0.8\wedge y_4) \le 0.87.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(12)

System (12) is consistent with initial solution $y_0=(0.7,0.4,0.6,0.7).$ Let us take any of the four interval solutions below:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{y}^1=[\underline{y^1},\overline{y^1}]=[(0.6,0.3,0.5,0.6),(0.8,0.5,0.7,0.8)],\hfill \\ y^2=[\underline{y^2},\overline{y^2}]=[(0.4,0.3,0.5,0.5),(1.0,0.5,0.7,0.9)],\hfill \\ y^3=[\underline{y^3},\overline{y^3}]=[(0.6,0.3,0.5,0.4),(0.8,0.5,0.7,1)],\hfill \\ y^4=[\underline{y^4},\overline{y^4}]=[(0.4,0.2,0.4,0.5),(1,0.6,0.8,0.9)].\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(13)

For a symmetric interval solution of (10), we write $y^1,y^2,y^3,y^4$ as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{y}^1=[y_0^1-{\delta }_1,y_0^1+{\delta }_1],where{\delta }_1=(0.1,0.1,0.1,0.1),\hfill \\ y^2=[y_0^2-{\delta }_2,y_0^2+{\delta }_2],where{\delta }_2=(0.3,0.1,0.1,0.2),\hfill \\ y^3=[y_0^3-{\delta }_3,y_0^3+{\delta }_3],where{\delta }_3=(0.1,0.1,0.1,0.3),\hfill \\ y^4=[y_0^4-{\delta }_4,y_0^4+{\delta }_4],where{\delta }_4=(0.3,0.2,0.2,0.2).\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(14)

It is verified by putting the values of $[\underline{y^1},\overline{y^1}],[\underline{y^2},\overline{y^2}],[\underline{y^4},\overline{y^4}]$ in system (10), which shows that they are solutions of the proposed system. But $[\underline{y^3},\overline{y^3}]$ is a not solution and, hence, not an interval symmetric solution. Further, we may easily compute the absolute and relative oscillation of the symmetrical interval solution $y^1,y^2,y^4.$ As $y^1=[y_0^1-{\delta }_1,y_0^1+{\delta }_1]$, the absolute oscillation for solution is

$${\delta }_{min}^A=0.1\wedge 0.1\wedge 0.1\wedge 0.1=0.1,$$

while the Relative oscillation for the solution is

$${\delta }_{min}^R={\displaystyle \frac{0.1}{0.7}}\wedge {\displaystyle \frac{0.1}{0.4}}\wedge {\displaystyle \frac{0.1}{0.6}}\wedge {\displaystyle \frac{0.1}{0.7}}=0.142$$

As $y^2=[y_0^2-{\delta }_2,y_0^2+{\delta }_2]$, the absolute oscillation for the solution is

$${\delta }_{min}^A=0.3\wedge 0.1\wedge 0.1\wedge 0.2=0.1,$$

while the relative oscillation for the solution is

$${\delta }_{min}^R={\displaystyle \frac{0.3}{0.7}}\wedge {\displaystyle \frac{0.1}{0.4}}\wedge {\displaystyle \frac{0.1}{0.6}}\wedge {\displaystyle \frac{0.2}{0.7}}=0.167.$$

As $y^4=[y_0^4-{\delta }_4,y_0^4+{\delta }_4]$, the absolute oscillation for the solution is

$${\delta }_{min}^A=0.3\wedge 0.2\wedge 0.2\wedge 0.2=0.2,$$

while the relative oscillation for the solution is

$${\delta }_{min}^R={\displaystyle \frac{0.3}{0.7}}\wedge {\displaystyle \frac{0.2}{0.4}}\wedge {\displaystyle \frac{0.2}{0.6}}\wedge {\displaystyle \frac{0.2}{0.7}}=0.28.$$

Further, from Example 1, for one solution $y^0$, there exists infinite symmetric interval solutions. As the interval symmetric solution can be absolute or relative and become wider and wider, the interval solution will be more and more stable. Therefore, we are interested in the biggest optimal absolute or relative oscillation for the solution. For the required method, we have developed a technique for the optimal symmetric interval solution.

3. Method for Optimal Absolute Oscillated (MAO) Symmetrical Interval Solution

In this section, we provide some definitions, theorems, and a proposition related to the aforementioned method, along with some proof and derivations.

Definition 6.

If $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]\subseteq P$ is the symmetrical interval solution of problem (1) , and let $[y_0-{\delta }^A,y_0+{\delta }^A]\subseteq P$ so it will always hold that

$${\delta }^{\ast A}\ge {\delta }^A.$$

Then, $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is said to be the MAO-type absolute optimal symmetrical interval solution with oscillation of magnitude ${\delta }_{min}^{\ast A}.$ Further, ${\delta }_{min}^{\ast A}={\delta }_1^{\ast A}\wedge {\delta }_2^{\ast A}\wedge \dots \wedge {\delta }_n^{\ast A}$ is the absolute oscillation with respect to $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ and ${\delta }_{min}^A={\delta }_1^A\wedge {\delta }_2^A\wedge \dots \wedge {\delta }_n^A$ with respect to $[y_0-{\delta }^A,y_0+{\delta }^A].$

Definition 7.

If $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]\subseteq P$ is the symmetrical interval solution of problem (1) , and let $[y_0-{\delta }^R,y_0+{\delta }^R]\subseteq P$ so it will always hold that

$${\delta }^{\ast R}\ge {\delta }^R.$$

Then, $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]$ is said to be the MRO-type optimal symmetrical interval solution with oscillation of magnitude ${\delta }_{min}^{\ast R}.$ Further, ${\delta }_{min}^{\ast R}={\displaystyle \frac{{\delta }_1^{\ast R}}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2^{\ast R}}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n^{\ast R}}{y_0^n}}$ is the absolute oscillation with respect to $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]$ and ${\delta }_{min}^R={\displaystyle \frac{{\delta }_1^R}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2^R}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n^R}{y_0^n}}$ with respect to $[y_0-{\delta }^R,y_0+{\delta }^R].$

By Definitions 6 and 7, we write that if

(i)

${\delta }^{\ast A}=({\delta }_1^{\ast A},{\delta }_2^{\ast A},\dots ,{\delta }_n^{\ast A})$ is an optimal solution of (1)

$$max{\delta }_{min}^A={\delta }_1\wedge {\delta }_2\wedge \dots \wedge {\delta }_n$$

$$s.t[y_0-\delta ,y_0+\delta ]\subseteq P,\delta \in {[0,1]}^n,$$

then $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is an MAO-form optimal symmetrical interval solution of (1) with respect to $y_0$, and the optimal oscillation is ${\delta }_{min}^{\ast A}={\delta }_1^{\ast A}\wedge {\delta }_2^{\ast A}\wedge \dots \wedge {\delta }_n^{\ast A}$.

(ii)

If ${\delta }^{\ast R}=({\delta }_1^{\ast R},{\delta }_2^{\ast R},\dots ,{\delta }_n^{\ast R})$ is an optimal solution of (1)

$$max{\delta }_{min}^R={\displaystyle \frac{{\delta }_1}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n}{y_0^n}}$$

$$s.t[y_0-\delta ,y_0+\delta ]\subseteq P,\delta \in {[0,1]}^n,$$

then $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]$ is nb MRO-form optimal symmetrical interval solution of (1) with respect to $y_0$, and the optimal oscillation is ${\delta }_{min}^{\ast R}={\delta }_1^{\ast R}\wedge {\delta }_2^{\ast R}\wedge \dots \wedge {\delta }_n^{\ast R}$

For the existence of the MAO- and MRO-based optimal symmetrical interval solutions, we must proceed with the following propositions.

Proposition 3.

If problem (1) has a solution with $y_0\in P$ having minimal solution ${\widehat{y}}^A$, then ${\widehat{y}}^A$ is an MAO-based maximal symmetrical interval solution.

Proof. 

Let the set of all minimal solutions be

$$\begin{array}{c}\hfill \widehat{P}=\left\{y^{j^s}\right|s\in S\}\end{array}$$

(15)

and

$$\begin{array}{c}\hfill {\widehat{P}}^{y_0}=\{\widehat{y}=({\widehat{y}}_1,{\widehat{y}}_2,\dots ,{\widehat{y}}_n)\in \widehat{P}|\widehat{y}\le y_0\}.\end{array}$$

(16)

By Theorem 3, as $y_0\in P$, this implies there will lie ${\widehat{y}}_0\in \widehat{P}|y_0\in [{\widehat{y}}_0,\widehat{y}].$ Therefore, ${\widehat{y}}_0\in {\widehat{P}}^{y_0}$, and this implies that ${\widehat{P}}^{y_0}\ne \varnothing .$ Further, ${\widehat{P}}^{y_0}\subseteq \widehat{P},$ which is bounded ${\widehat{P}}^{y_0}\le \infty .$ As ${\widehat{P}}^{y_0}\ne \varnothing$ and ${\widehat{P}}^{y_0}\le \infty$ there lies, ${\widehat{y}}_0\in {\widehat{P}}^{y_0}$ such that

$$\begin{array}{c}\hfill (y_0^1-{\widehat{y}}^{1A})\wedge (y_0^2-{\widehat{y}}^{2A})\wedge \dots \wedge (y_0^n-{\widehat{y}}^{nA})=\underset{\widehat{y}\in {\widehat{P}}^{y_0}}{max}\{(y_0^1-{\widehat{y}}^1)\wedge (y_0^2-{\widehat{y}}^2)\wedge \dots \wedge (y_0^n-{\widehat{y}}^n)\}.\end{array}$$

(17)

Write

$$\begin{array}{c}\hfill {\delta }_J^{\ast A}=({\widehat{y}}^j-y_0^j)\wedge (y_0^j-{\widehat{y}}^{jA})\forall j\in J\end{array}$$

(18)

It is obvious that $0\le {\delta }_j^{\ast A}\le 1$ for any $j\in J$. Therefore, we have ${\delta }^{\ast A}=({\delta }_1^{\ast A},{\delta }_2^{\ast A},\dots ,{\delta }_n^{\ast A})\in {[0,1]}^n$. Further, from Equation (18), we also have

$$\begin{array}{c}\hfill {\delta }_J^{\ast A}\le {\widehat{y}}^j-y_0^j,{\delta }_J^{\ast A}\le y_0^j-{\widehat{y}}^{jA}\forall j\in J.\end{array}$$

(19)

This implies that

$$\begin{array}{c}\hfill {\widehat{y}}^{jA}\le y_0^j-{\delta }_j^{\ast A}\le y_0^j+{\delta }_j^{\ast A}\le {\widehat{y}}^j\forall j\in J\end{array}$$

(20)

or

$$\begin{array}{c}\hfill {\widehat{y}}^A\le y_0-{\delta }^{\ast A}\le y_0+{\delta }^{\ast A}\le \widehat{y}.\end{array}$$

(21)

Hence, we have $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]\subseteq [{\widehat{y}}^A,\widehat{y}]\subseteq P.$ By Definition 3, $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is a symmetrical interval solution of problem (1). From Definition 4, the absolute oscillation for $y_0$ with respect to $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is

$$\begin{array}{c}\hfill {\delta }_{min}^{\ast A}={\delta }_1^{\ast A}\wedge {\delta }_2^{\ast A}\wedge \dots \wedge {\delta }_n^{\ast A}\end{array}$$

(22)

From Equations (18) and (22), we obtain

$$\begin{array}{c}\hfill {\delta }_{min}^{\ast A}=\underset{j\in J}{min}\{{\widehat{y}}^j-y_0^j\}\wedge \underset{j\in J}{min}\{y_0^j-{\widehat{y}}^{jA}\}.\end{array}$$

(23)

Further, we will show that $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is an MAO-based maximal symmetric interval solution. Let $[y_0-\delta ,y_0+\delta ]\subseteq P$ be any symmetrical interval solution of (1) with $\delta =({\delta }_1,{\delta }_2,\dots {\delta }_n)\in {[0,1]}^n.$$[y_0-\delta ,y_0+\delta ]\subseteq P$ shows that $y_0-\delta ,y_0+\delta \in P$. By Theorem 3, there will be a minimal solution ${\widehat{y}}_0\in \widehat{P}$ such that $y_0-\delta \in [{\widehat{y}}_0,\widehat{y}].$ As $\delta \in {[0,1]}^n$ and $\widehat{y}$ is the maximal solution, we have

$$\begin{array}{c}\hfill {\widehat{y}}_0\le y_0-\delta \le y_0\le y_0-\delta \le \widehat{y}.\end{array}$$

(24)

By the above equation, ${\widehat{y}}_0\in {\widehat{P}}^{y_0}.$ Choosing Equations (16) and (23), we obtain

$$\begin{array}{ccc}\hfill {\delta }_{min}& =& {\delta }_1\wedge {\delta }_2\wedge \dots \wedge {\delta }_n\hfill \\ & \le & (y_0^1-{\widehat{y}}_0^1)\wedge (y_0^2-{\widehat{y}}_0^2)\wedge \dots \wedge (y_0^n-{\widehat{y}}_0^n)\hfill \\ & \le & \underset{\widehat{y}\in {\widehat{S}}^{y_0}}{max}\{(y_0^1-{\widehat{y}}^1)\wedge (y_0^2-{\widehat{y}}^2)\wedge \dots \wedge (y_0^n-{\widehat{y}}^n)\}\hfill \\ & =& (y_0^1-{\widehat{y}}^{1A})\wedge (y_0^2-{\widehat{y}}^{2A})\wedge \dots \wedge (y_0^n-{\widehat{y}}^{nA})\hfill \\ & =& \underset{j\in J}{min}\{y_0^j-{\widehat{y}}^{jA}\}.\hfill \end{array}$$

(25)

By (24), we have

$$\begin{array}{ccc}\hfill {\delta }_{min}& =& {\delta }_1\wedge {\delta }_2\wedge \dots \wedge {\delta }_n\hfill \\ & \le & ({\widehat{y}}^1-y_0^1)\wedge ({\widehat{y}}^2-y_0^2)\wedge \dots \wedge ({\widehat{y}}^n-y_0^n)\hfill \\ & =& \underset{j\in J}{min}\{{\widehat{y}}^j-y_0^j\}.\hfill \end{array}$$

(26)

From (23), (25), and (26), we have

$$\begin{array}{c}\hfill {\delta }_{min}\le \underset{j\in J}{min}\{y_0^j-{\widehat{y}}^{jA}\}\wedge \{{\widehat{y}}^j-y_0^j\}={\delta }_{min}^{\ast A}.\end{array}$$

(27)

Now by Definition 6, the $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is an MAO-based optimal absolute oscillation with a maximal value of ${\delta }_{min}^{\ast A}$. □

Proposition 4.

If problem (1) has a solution with $y_0\in P$ with a minimal solution ${\widehat{y}}^R$, then ${\widehat{y}}^R$ is an MRO-based maximal symmetrical interval solution.

Proof. 

The proof is similar to that of the MAO-based symmetric interval solution. □

4. Solving the Maximum Absolute Oscillation and the Corresponding MAO-Based Optimal Symmetric Interval Solution

As we have stated earlier, the optimal symmetrical interval solution may not be unique. Therefore, for the stability of the given solution, we chose another MAO-based optimal interval solution $[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]$, satisfying $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]\subseteq [y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}].$ Order the parameters $c_1,c_2,\dots c_m$ as

$$\begin{array}{c}\hfill c_1\ge c_2\ge \dots \ge c_m.\end{array}$$

(28)

Define the indices with absolute maximal conservation $\{J_1^{AC},J_2^{AC},\dots ,J_m^{AC}\}$ where

$$\begin{array}{c}\hfill J_1^{AC}=arg\underset{J\in J_1^{y_0}}{max}\{y_0^J-c_1\}.\end{array}$$

(29)

For more generalization, we an write

$$\begin{array}{c}\hfill J_i^{AC}=\left\{\begin{array}{c}\begin{array}{c}arg\underset{J\in J_i^{y_0}}{max}\{y_0^J-c_i\},if\{J_1^{AC},J_2^{AC},\dots ,J_{i-1}^{AC}\}\cap J_{y_0}^i=\varnothing \hfill \\ min\{J_1^{AC},J_2^{AC},\dots ,J_{i-1}^{AC}\}\cap J_{y_0}^i,if\{J_1^{AC},J_2^{AC},\dots ,J_{i-1}^{AC}\}\cap J_{y_0}^i\ne \varnothing .\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(30)

By Theorem 3, given in [38], it is easily verified that $J_i^{AC}\in J_{y_0}^i\subseteq J_{\ge }^i$; hence, we can write

$$\begin{array}{c}\hfill J_1^{AC}=arg\underset{J\in J_1^{y_0}}{max}\{y_0^J-c_1\}.\end{array}$$

(31)

and

$$\begin{array}{c}\hfill (J_1^{AC},J_2^{AC},\dots ,J_m^{AC})\in J_{\ge }^1\times J_{\ge }^2\times \dots \times J_{\ge }^m=P\end{array}$$

(32)

is the path for system (1). So, we conclude that $(J_1^{AC},J_2^{AC},\dots ,J_m^{AC})$ is the conservative way for problem (1). The solution for the aforementioned path may be developed as follows: let

$$\begin{array}{c}\hfill I_{AC}^j=\{i\in I:j_{AC}^i=j\}\forall j\in J.\end{array}$$

(33)

and

$$\begin{array}{c}\hfill y_i^{AC}=\left\{\begin{array}{c}\begin{array}{c}0,if{I}_{AC}^j=\varnothing \hfill \\ \underset{i\in I_{AC}^j}{\bigvee }{c}_i,if{I}_{AC}^j\ne \varnothing .\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(34)

Then, we can find a vector $y^{AC}=(y_1^{AC},y_2^{AC}\dots y_n^{AC})$ which is the solution of the aforementioned conservative path as given in Proposition 2.

Proposition 5.

Regarding the conservative path given in equations (29) and (30), if for any $j^{\prime }\in J,$ and if $I_{AC}^{j^{\prime }}\ne \varnothing ,$ then $I_A^{j^{\prime }}\ne \varnothing .$ Further, if $I_{AC}^{j^{\prime }}\ne \varnothing$, it will then always hold that $min{I}_A^{j^{\prime }}\le min{I}_{AC}^{j^{\prime }}.$

Proof. 

The proof may be seen in [38]. □

Theorem 4.

Consider $y^{AC}$ given in (33) and (34), then $y^{AC}\le y^A.$

Proof. 

Choose any $j^{\prime }\in J.$ Now, if $I^{AC}=\varnothing$, then, by (30), $y_{j^{\prime }}^{AC}=0\le y_{j^{\prime }}^A$, and if $I^{AC}\ne \varnothing$, then by Proposition 1 $I^A\ne \varnothing$ and $min{I}_{j^{\prime }}^A\le min{I}_{j^{\prime }}^{AC}.$ Further, consider $min{I}_{j^{\prime }}^A=i^A$ and $min{I}_{j^{\prime }}^{AC}=i^{AC}$, assuming $c_1\ge c_2\ge \dots \ge c_m$ then, by Equation (40) in [38] and (30), it is implied that

$$\begin{array}{c}\hfill y_{j^{\prime }}^{AC}=\underset{i\in I_{j^{\prime }}^{AC}}{\bigvee }{c}_i=c_{i^{AC}}\le c_{i^a}=\underset{i\in I_{j^{\prime }}^A}{\bigvee }{c}_i=y_{j^{\prime }}^A\end{array}$$

(35)

By supposition of $j^{\prime }$, we say that $y^{AC}\le y^A.$ □

Now, by solution $y^{AC}$ of the conservative path $(J_1^{AC},J_2^{AC},\dots ,J_m^{AC})$ we define $({\delta }^{\ast AC}={\delta }_1^{\ast AC},{\delta }_2^{\ast AC},\dots ,{\delta }_n^{\ast AC})$ where

$$\begin{array}{c}\hfill {\delta }_j^{\ast AC}=({\widehat{y}}^j-y_0^j)\wedge (y_0^j-y_0^{AC}),\forall j\in J.\end{array}$$

(36)

Further, we write

$$\begin{array}{c}\hfill {\delta }_{min}^{\ast AC}={\delta }_1^{\ast AC}\wedge {\delta }_2^{\ast AC}\wedge \dots \wedge {\delta }_n^{\ast AC}\end{array}$$

(37)

and

$$\begin{array}{c}\hfill [y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}].\end{array}$$

(38)

On the basis of the above representation, we show in the next theorem that $[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]$ is also an MAO-based maximal interval solution for problem (1).

Theorem 5.

$[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]$ is an MAO maximal interval solution of problem (1), satisfying $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]\subseteq [y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]$

Proof. 

By Theorem 4 and Theorem 4 in [38], $y^{AC}$ is a solution of problem (1), satisfying $y^{AC}\le y^A\le y_0.$ So

$$\begin{array}{c}\hfill y_0^j-y_j^A\le y_0^j-y_j^{AC}\forall j\in J.\end{array}$$

(39)

Now, by (37)

$$\begin{array}{c}\hfill {\delta }_j^{\ast A}\le {\delta }_j^{\ast AC}\forall j\in J.\end{array}$$

(40)

This implies that ${\delta }^{\ast A}\le {\delta }^{\ast AC}$ Hence, $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]\subseteq [y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}].$ In addition, ${\delta }_j^{\ast AC}=({\widehat{y}}^j-y_0^j)\wedge (y_0^j-y_0^{AC}),\forall j\in J.$ So, by (37), we have

$$\begin{array}{c}\hfill y_0^j-{\delta }_j^{\ast AC}\ge y_0^j-(y_0^j-y_j^{AC})=y_j^{AC}\forall j\in J\end{array}$$

(41)

and

$$\begin{array}{c}\hfill y_0^j+{\delta }_j^{\ast AC}\le y_0^j+(y_0^j-y_j^{AC})={\widehat{y}}_j\forall j\in J.\end{array}$$

(42)

This implies that $[y_0-{\delta }_j^{\ast AC},y_0+{\delta }_j^{\ast AC}]\subseteq [y_j^{AC},{\widehat{y}}_j]\forall j\in J.$ Consequently, $[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]\subseteq [y^{AC},\widehat{y}]\subseteq P$ is the symmetrical solution in the interval format, and the given maximal oscillation is ${\delta }_{min}^{AC}$. It is derived that $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]$ is the MAO-based interval solution for (1) with maximal oscillation ${\delta }_{min}^{\ast A}.$ Additionally, $[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]$ is a symmetrical interval solution with maximum oscillation ${\delta }_{min}^{\ast AC},$ implying that ${\delta }_{min}^{\ast AC}\le {\delta }_{min}^{\ast A}.$ However, from (37) and (40), one can write

$$\begin{array}{c}\hfill {\delta }_{min}^{\ast A}={\delta }_1^{\ast A}\wedge {\delta }_2^{\ast A}\wedge \dots \wedge {\delta }_n^{\ast A}\le {\delta }_1^{\ast AC}\wedge {\delta }_2^{\ast AC}\wedge \dots \wedge {\delta }_n^{\ast AC}={\delta }_{min}^{AC}\end{array}$$

(43)

Both the above inequalities imply that ${\delta }_{min}^{\ast A}={\delta }_{min}^{\ast AC}$, which, again, implies that ${\delta }_{min}^{\ast AC}$ is the optimal oscillation. Hence, consequently, $[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]$ is an MAO-type maximal absolute interval solution of problem (1), satisfying $[y_0-{\delta }^{\ast A},y_0+{\delta }^{\ast A}]\subseteq [y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}].$ □

5. Solving the Maximum Relative Oscillation and the Corresponding MRO-Based Optimal Symmetric Interval Solution

The MRO-based optimal symmetrical interval solution may not be unique. Therefore, for the stability of the given solution, we chose another MRO-based optimal interval solution $[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]$ satisfying $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]\subseteq [y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}].$ Ordering the parameters $c_1,c_2,\dots c_m$ as

$$\begin{array}{c}\hfill c_1\ge c_2\ge \dots \ge c_m.\end{array}$$

(44)

Let us define the indices with relative maximal conservation $\{J_1^{RC},J_2^{RC},\dots ,J_m^{RC}\}$ where

$$\begin{array}{c}\hfill J_1^{AC}=arg\underset{J\in J_1^{y_0}}{max}\{{\displaystyle \frac{y_0^J-c_1}{y_0^J}}\}.\end{array}$$

(45)

For more generalization, we can write

$$\begin{array}{c}\hfill J_i^{AC}=\left\{\begin{array}{c}\begin{array}{c}arg\underset{J\in J_i^{y_0}}{max}\{{\displaystyle \frac{y_0^J-c_i}{y_0^J}}\},if\{J_1^{RC},J_2^{RC},\dots ,J_{i-1}^{RC}\}\cap J_{y_0}^i=\varnothing \hfill \\ min\{J_1^{RC},J_2^{RC},\dots ,J_{i-1}^{RC}\}\cap J_{y_0}^i,if\{J_1^{RC},J_2^{RC},\dots ,J_{i-1}^{RC}\}\cap J_{y_0}^i\ne \varnothing .\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(46)

By Theorem 4 given in [38], it is easily verified that $J_i^{RC}\in J_{y_0}^i\subseteq J_{\ge }^i,$ and hence we can write the following:

$$\begin{array}{c}\hfill J_1^{RC}=arg\underset{J\in J_1^{y_0}}{max}\{{\displaystyle \frac{y_0^J-c_1}{y_0^J}}\}.\end{array}$$

(47)

and

$$\begin{array}{c}\hfill (J_1^{RC},J_2^{RC},\dots ,J_m^{RC})\in J_{\ge }^1\times J_{\ge }^2\times \dots \times J_{\ge }^m=P\end{array}$$

(48)

is the path for system (1). So, we conclude that $(J_1^{RC},J_2^{RC},\dots ,J_m^{RC})$ is the conservative way for problem (1). The solution for the aforementioned path may be developed as follows: let

$$\begin{array}{c}\hfill I_{RC}^j=\{i\in I:j_{RC}^i=j\}\forall j\in J.\end{array}$$

(49)

and

$$\begin{array}{c}\hfill y_i^{RC}=\left\{\begin{array}{c}\begin{array}{c}0,if{I}_{RC}^j=\varnothing \hfill \\ \underset{i\in I_{RC}^j}{\bigvee }{c}_i,if{I}_{RC}^j\ne \varnothing .\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(50)

Then, we can find a vector $y^{RC}=(y_1^{RC},y_2^{RC}\dots y_n^{RC})$ which is the solution of the aforementioned conservative path as given in the Proposition 2.

Proposition 6.

Regarding the conservative path given in Equations (45) and (47), if for any $j^{\prime }\in J,$ if $I_{RC}^{j^{\prime }}\ne \varnothing ,$ then $I_R^{j^{\prime }}\ne \varnothing .$ Further, if $I_{RC}^{j^{\prime }}\ne \varnothing$, it will then always hold that $min{I}_R^{j^{\prime }}\le min{I}_{RC}^{j^{\prime }}.$

Proof. 

The proof may be seen in [38]. □

Theorem 6.

Consider $y^{RC}$ given in (49) and (50); then, $y^{RC}\le y^R.$

Proof. 

Choose any $j^{\prime }\in J.$ Now, if $I^{RC}=\varnothing$, then, by (46), $y_{j^{\prime }}^{RC}=0\le y_{j^{\prime }}^R$, and if $I^{RC}\ne \varnothing$, then by proposition 1, $I^R\ne \varnothing$ and $min{I}_{j^{\prime }}^R\le min{I}_{j^{\prime }}^{RC}.$ Further, consider $min{I}_{j^{\prime }}^R=i^R$ and $min{I}_{j^{\prime }}^{RC}=i^{RC}$, assuming $c_1\ge c_2\ge \dots \ge c_m$, by Equation (40) in [38] and (46), implies

$$\begin{array}{c}\hfill y_{j^{\prime }}^{RC}=\underset{i\in I_{j^{\prime }}^{RC}}{\bigvee }{c}_i=c_{i^{RC}}\le c_{i^a}=\underset{i\in I_{j^{\prime }}^R}{\bigvee }{c}_i=y_{j^{\prime }}^R\end{array}$$

(51)

By supposition of $j^{\prime }$, we say that $y^{RC}\le y^R.$ □

Now, by solution $y^{RC}$ of the conservative path $(J_1^{RC},J_2^{RC},\dots ,J_m^{RC})$ we define $({\delta }^{\ast RC}={\delta }_1^{\ast RC},{\delta }_2^{\ast RC},\dots {\delta }_n^{\ast RC})$ where

$$\begin{array}{c}\hfill {\delta }_j^{\ast RC}=({\widehat{y}}^j-y_0^j)\wedge (y_0^j-y_0^{RC}),\forall j\in J.\end{array}$$

(52)

Further, we write

$$\begin{array}{c}\hfill {\delta }_{min}^{\ast RC}={\displaystyle \frac{{\delta }_1^{\ast RC}}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2^{\ast RC}}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n^{\ast RC}}{y_0^n}}\end{array}$$

(53)

and

$$\begin{array}{c}\hfill [y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}].\end{array}$$

(54)

On the basis of the above representation, we show in the next theorem that $[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]$ is also an MRO-based maximal interval solution for problem (1).

Theorem 7.

$[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]$ is an MRO maximal interval solution of problem (1), satisfying $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]\subseteq [y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]$

Proof. 

By Theorem 4 and Theorem 4 in [38], $y^{RC}$ is a solution of problem (1), satisfying $y^{RC}\le y^R\le y_0.$ So,

$$\begin{array}{c}\hfill y_0^j-y_j^R\le y_0^j-y_j^{RC}\forall j\in J.\end{array}$$

(55)

Now, by (53)

$$\begin{array}{c}\hfill {\delta }_j^{\ast R}\le {\delta }_j^{\ast RC}\forall j\in J\end{array}$$

(56)

This implies that ${\delta }^{\ast R}\le {\delta }^{\ast RC}$ Hence, $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]\subseteq [y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}].$ Additionally, ${\delta }_j^{\ast RC}=({\widehat{y}}^j-y_0^j)\wedge (y_0^j-y_0^{RC}),\forall j\in J.$ So, by (53), we have

$$\begin{array}{c}\hfill y_0^j-{\delta }_j^{\ast RC}\ge y_0^j-(y_0^j-y_j^{RC})=y_j^{RC}\forall j\in J.\end{array}$$

(57)

and

$$\begin{array}{c}\hfill y_0^j+{\delta }_j^{\ast RC}\le y_0^j+(y_0^j-y_j^{RC})={\widehat{y}}_j\forall j\in J.\end{array}$$

(58)

This implies that $[y_0-{\delta }_j^{\ast RC},y_0+{\delta }_j^{\ast RC}]\subseteq [y_j^{RC},{\widehat{y}}_j]\forall j\in J.$ Consequently, $[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]\subseteq [y^{RC},\widehat{y}]\subseteq P$ is the symmetrical solution in the interval format with given relative maximal oscillation ${\delta }_{min}^{RC}$. It is derived that $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]$ is the MRO-based interval solution for (1) with maximal oscillation ${\delta }_{min}^{\ast R}.$ In addition, $[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]$ is a symmetrical interval solution with relative maximum oscillation ${\delta }_{min}^{\ast RC},$ implying that ${\delta }_{min}^{\ast RC}\le {\delta }_{min}^{\ast R}.$ However, from (53) and (56), one can write

$$\begin{array}{c}\hfill {\delta }_{min}^{\ast R}={\displaystyle \frac{{\delta }_1^{\ast R}}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2^{\ast R}}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n^{\ast R}}{y_0^n}}\le {\displaystyle \frac{{\delta }_1^{\ast AC}}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2^{\ast AC}}{y_0^2}}\wedge \dots \wedge {\displaystyle \frac{{\delta }_n^{\ast AC}}{y_0^n}}={\delta }_{min}^{RC}\end{array}$$

(59)

Both the above inequalities imply that ${\delta }_{min}^{\ast R}={\delta }_{min}^{\ast RC}$, which, again, implies that ${\delta }_{min}^{\ast RC}$ is the optimal relative oscillation. Hence, consequently, $[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]$ is the MRO-type maximal relative interval solution of problem (1), satisfying $[y_0-{\delta }^{\ast R},y_0+{\delta }^{\ast R}]\subseteq [y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}].$ □

6. Numerical Examples

In this section, we provide two examples on two different approaches of optimal fuzzy interval solution. The first example is an MAO (method of absolute oscillation)-based optimal interval solution, while the second one is based on the MRO (method of relative oscillation)-based interval solution.

6.1. MAO-Based Solution

Here, we chose a system of differential equations representing any P2P network with five terminals for sharing educational resources with each other. We are interested in maximizing the quality of downloading educational resources in the form of fuzzy interval solutions (fluctuations or oscillations), as it may vary from time to time depending upon the strength of the network signal. The strength of download quality is measured in bps (bits per second). The system is given as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=(0.4\wedge y_1)\vee (0.5\wedge y_2)\vee (0.9\wedge y_3)\vee (0.7\wedge y_4)\vee (0.8\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=(0.8\wedge y_1)\vee (0.4\wedge y_2)\vee (0.3\wedge y_3)\vee (0.6\wedge y_4)\vee (0.7\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=(0.4\wedge y_1)\vee (0.6\wedge y_2)\vee (0.5\wedge y_3)\vee (0.7\wedge y_4)\vee (0.9\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_4}{dt}}=(0.5\wedge y_1)\vee (0.9\wedge y_2)\vee (0.6\wedge y_3)\vee (0.3\wedge y_4)\vee (0.4\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_5}{dt}}=(0.2\wedge y_1)\vee (0.6\wedge y_2)\vee (0.9\wedge y_3)\vee (0.3\wedge y_4)\vee (0.8\wedge y_5),\hfill \\ y_1\left(0\right)=0.30,y_2\left(0\right)=0.35,y_3\left(0\right)=0.40,y_4\left(0\right)=0.30,y_5\left(0\right)=0.34\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(60)

where numerical data are the band among any two sources or terminals sharing data from i to j for the direction source, with the download quality being $y_n$ in the sense of receiving information in fuzzy format with initial speed or quality. System (60) can be written an addition–multiplication format as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=(0.4\times y_1)+(0.5\times y_2)+(0.9\times y_3)+(0.7\times y_4)+(0.8\times y_5),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=(0.8\times y_1)+(0.4\times y_2)+(0.3\times y_3)+(0.6\times y_4)+(0.7\times y_5),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=(0.4\times y_1)+(0.6\times y_2)+(0.5\times y_3)+(0.7\times y_4)+(0.9\times y_5),\hfill \\ {\displaystyle \frac{d{y}_4}{dt}}=(0.5\times y_1)+(0.9\times y_2)+(0.6\times y_3)+(0.3\times y_4)+(0.4\times y_5),\hfill \\ {\displaystyle \frac{d{y}_5}{dt}}=(0.2\times y_1)+(0.6\times y_2)+(0.9\times y_3)+(0.3\times y_4)+(0.8\times y_5),\hfill \\ y_1\left(0\right)=0.30,y_2\left(0\right)=0.35,y_3\left(0\right)=0.40,y_4\left(0\right)=0.30,y_5\left(0\right)=0.34\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(61)

The quality of downloading or giving data or information among any two sources will lie in the interval form having maximal and minimal values. Therefore, the fuzzy relation system (60) can be written in interval form as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.30\le {\displaystyle \frac{d{y}_1}{dt}}\le 0.83,\hfill \\ 0.35\le {\displaystyle \frac{d{y}_2}{dt}}\le 0.92,\hfill \\ 0.40\le {\displaystyle \frac{d{y}_3}{dt}}\le 0.88,\hfill \\ 0.30\le {\displaystyle \frac{d{y}_4}{dt}}\le 0.89,\hfill \\ 0.34\le {\displaystyle \frac{d{y}_5}{dt}}\le 0.91.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(62)

or

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.30\le (0.4\wedge y_1)\vee (0.5\wedge y_2)\vee (0.9\wedge y_3)\vee (0.7\wedge y_4)\vee (0.8\wedge y_5)\le 0.83,\hfill \\ 0.35\le (0.8\wedge y_1)\vee (0.4\wedge y_2)\vee (0.3\wedge y_3)\vee (0.6\wedge y_4)\vee (0.7\wedge y_5)\le 0.92,\hfill \\ 0.40\le (0.4\wedge y_1)\vee (0.6\wedge y_2)\vee (0.5\wedge y_3)\vee (0.7\wedge y_4)\vee (0.9\wedge y_5)\le 0.88,\hfill \\ 0.30\le (0.5\wedge y_1)\vee (0.9\wedge y_2)\vee (0.6\wedge y_3)\vee (0.3\wedge y_4)\vee (0.4\wedge y_5)\le 0.89,\hfill \\ 0.34\le (0.2\wedge y_1)\vee (0.6\wedge y_2)\vee (0.9\wedge y_3)\vee (0.3\wedge y_4)\vee (0.8\wedge y_5)\le 0.91.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(63)

For the given $y_0=(0.72,0.6,0.4,0.66,0.58)$, find the MAO-based maximal, symmetrical interval solution of problem (63). For the solution, we write the system (63) in descending form relative to $c_i$ as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.40\le (0.4\wedge y_1)\vee (0.6\wedge y_2)\vee (0.5\wedge y_3)\vee (0.7\wedge y_4)\vee (0.9\wedge y_5)\le 0.88,\hfill \\ 0.35\le (0.8\wedge y_1)\vee (0.4\wedge y_2)\vee (0.3\wedge y_3)\vee (0.6\wedge y_4)\vee (0.7\wedge y_5)\le 0.92,\hfill \\ 0.34\le (0.2\wedge y_1)\vee (0.6\wedge y_2)\vee (0.9\wedge y_3)\vee (0.3\wedge y_4)\vee (0.8\wedge y_5)\le 0.91,\hfill \\ 0.30\le (0.4\wedge y_1)\vee (0.5\wedge y_2)\vee (0.9\wedge y_3)\vee (0.7\wedge y_4)\vee (0.8\wedge y_5)\le 0.83,\hfill \\ 0.30\le (0.5\wedge y_1)\vee (0.9\wedge y_2)\vee (0.6\wedge y_3)\vee (0.3\wedge y_4)\vee (0.4\wedge y_5)\le 0.89.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(64)

Now, we follow the steps for the MAO-based solution.

 1. 

Using (6) and (9), we find the following potential optimal solution:

$$\widehat{y}=(1,0.89,0.83,1,0.88)$$

 2. 

Testing all the values in system (64) satisfies all the inequalities of (64), so this is a solution of (64). Using Theorem 2, we conclude that (64) is consistent, having solution $\widehat{Y}=(1,0.89,0.83,1,0.88)$.

 3. 

With respect to the given solution $y_0=(0.72,0.6,0.4,0.66,0.58),$ we calculate the index set $J_{y_0}^1,J_{y_0}^2,\dots J_{y_0}^5$ by Equation (34) in [38] as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{b}_{11}\wedge y_0^1=(0.4\wedge 0.72)=0.4\ge 0.4=c_1,\hfill \\ b_{12}\wedge y_0^2=(0.6\wedge 0.6)=0.6\ge 0.4=c_1,\hfill \\ b_{13}\wedge y_0^3=(0.5\wedge 0.4)=0.4\ge 0.4=c_1,\hfill \\ b_{14}\wedge y_0^4=(0.7\wedge 0.66)=0.66\ge 0.4=c_1,\hfill \\ b_{15}\wedge y_0^5=(0.9\wedge 0.58)=0.58\ge 0.4=c_1.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(65)

So we have $J_{y_0}^1=\{1,2,3,4,5\},$ Similarly we can calculate $J_{y_0}^2=\{1,3,4,5\},$$J_{y_0}^3=\{1,2,4,5\},$$J_{y_0}^4=\{1,2,3,4,5\},$$J_{y_0}^5=\{1,2,3,4,5\}.$

 4. 

By the above index set, we can calculate the conservative optimal indices $J_1^{AC},J_2^{AC},\dots J_5^{AC}$ by (29) and (30) as follows:

$$\underset{j\in J_{y_0}^1}{max}\{y_0^j-c_1\}$$

$$=max\{y_0^1-c_1,y_0^2-c_1,y_0^3-c_1,y_0^4-c_4,y_0^5-c_5\}$$

$$=max\{0.72-0.40,0.6-0.40,0.4_0-0.4,0.66-0.40,0.58-0.40\}$$

$$=0.72-0.40=y_0^1-c_1$$

So, we have $J_1^{AC}=arg{max}_{j\in j_{y_0}^1}=1$. Next, by 80 $\left\{J_1^{AC}\right\}\cap J_{y_0}^2=\left\{1\right\}\cap \{1,3,4,5\}=\varnothing \ne \varnothing$; therefore, we have $J_2^{AC}=1.$ Further, $\{J_1^{AC},J_2^{AC}\}\cap J_{y_0}^3=\{1,1\}\cap \{1,2,4,5\}=\left\{1\right\}\ne \varnothing$ So $J_3^{AC}=1.$ Next, $\{J_1^{AC},J_2^{AC},J_3^{AC}\}\cap J_{y_0}^4=\{1,1,1\}\cap \{1,2,3,4,5\}=\left\{1\right\}\ne \varnothing$ So, $J_4^{AC}=1.$ Similarly, $J_5^{AC}=1.$ Therefore, the conservative path is $(J_1^{AC},J_2^{AC},J_3^{AC},J_4^{AC},J_5^{AC})=(1,1,1,1,1).$

 5. 

In this step, we find the indices $I_1^{AC},I_2^{AC},\dots ,I_5^{AC}$ from the obtained conservative path. Therefore, $I_1^{AC}=\{1,2,3,4,5\}$ because i = 1,2,3,4,5 $J_i^{AC}=1$; further, $I_2^{AC}=\varnothing ,I_3^{AC}=\varnothing ,I_4^{AC}=\varnothing ,I_5^{AC}=\varnothing ,$ because i=2,3,4,5 $J_i^{AC}\ne 1$.

 6. 

Additionally, $I_2^{AC}=\varphi ,I_3^{AC}=\varphi ,I_4^{AC}=\varphi ,I_5^{AC}=\varphi ,$. Therefore, $y_2^{AC},y_3^{AC},y_4^{AC},y_5^{AC}=\varphi$ by (32). Further, by (32) we can write

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{y}_1^{AC}=\underset{i\in I_1^{AC}}{\bigvee }{c}_i=c_1\vee c_2\vee c_3\vee c_4\vee c_5=0.40\vee 0.35\vee 0.34\vee 0.30\vee 0.30=0.40.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(66)

Therefore, the vector $y^{AC}=(0.40,0,0,0,0).$

 7. 

Let

$$\widehat{y}=(1,0.89,0.83,1,0.88),$$

$$y_0=(0.72,0.6,0.4,0.66,0.58),$$

$$y^{AC}=(0.40,0,0,0,0).$$

Now, we can find ${\delta }_1^{\ast AC},{\delta }_2^{\ast AC},\dots {\delta }_5^{\ast AC}$ by (36)

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\delta }_1^{\ast AC}=({\widehat{y}}^1-y_0^1)\wedge (y_0^1-{\widehat{y}}^1)=(1-0.72)\wedge (0.72-0.40)=0.28,\hfill \\ {\delta }_1^{\ast AC}=({\widehat{y}}^1-y_0^1)\wedge (y_0^1-{\widehat{y}}^1)=(0.89-0.6)\wedge (0.6-0)=0.29,\hfill \\ {\delta }_1^{\ast AC}=({\widehat{y}}^1-y_0^1)\wedge (y_0^1-{\widehat{y}}^1)=(0.83-0.4)\wedge (0.4-0)=0.40,\hfill \\ {\delta }_1^{\ast AC}=({\widehat{y}}^1-y_0^1)\wedge (y_0^1-{\widehat{y}}^1)=(1-0.66)\wedge (0.66-0)=0.34,\hfill \\ {\delta }_1^{\ast AC}=({\widehat{y}}^1-y_0^1)\wedge (y_0^1-{\widehat{y}}^1)=(0.88-0.58)\wedge (0.58-0)=0.30.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(67)

So we find ${\delta }^{\ast AC}=({\delta }_1^{\ast AC},{\delta }_2^{\ast AC},\dots ,{\delta }_5^{\ast AC})=(0.28,0.29,0.40,0.34,0.30)$

 8. 

Next, we can find the absolute maximal oscillation of $y_0$ based on the values obtained in step 7.

$${\delta }_{min}^{\ast AC}={\delta }_1^{\ast AC}\wedge {\delta }_2^{\ast AC}\wedge {\delta }_3^{\ast AC}\wedge {\delta }_4^{\ast AC}\wedge {\delta }_5^{\ast AC}$$

$${\delta }_{min}^{\ast AC}=0.28\wedge 0.29\wedge 0.40\wedge 0.34\wedge 0.30$$

$${\delta }_{min}^{\ast AC}=0.28$$

So, the MAO-based maximal symmetric interval solution with respect to $y_0=(0.72,0.6,0.4,0.66,0.58)$ is

$$[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]=[(0.44,0.32,0.12,0.38,0.30),(1,0.88,0.68,0.94,0.86)]$$

Furthermore, if we take the case with $n\ne m$, then one solution $y_6=0.8$ will remain constant, which means it will be not fluctuating in the interval fuzzy solution. As y lies both in the derivative column left side of Equation (60) and also in the right side of said equation in the max–min fuzzy relation, Equation (60) will be as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=(0.4\times y_1)+(0.5\times y_2)+(0.9\times y_3)+(0.7\times y_4)+(0.8\times y_5)+(0.7\times 0.8=y_6),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=(0.8\times y_1)+(0.4\times y_2)+(0.3\times y_3)+(0.6\times y_4)+(0.7\times y_5)+(0.8\times 0.8=y_6),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=(0.4\times y_1)+(0.6\times y_2)+(0.5\times y_3)+(0.7\times y_4)+(0.9\times y_5)+(0.4\times 0.8=y_6),\hfill \\ {\displaystyle \frac{d{y}_4}{dt}}=(0.5\times y_1)+(0.9\times y_2)+(0.6\times y_3)+(0.3\times y_4)+(0.4\times y_5)+(0.7\times 0.8=y_6),\hfill \\ {\displaystyle \frac{d{y}_5}{dt}}=(0.2\times y_1)+(0.6\times y_2)+(0.9\times y_3)+(0.3\times y_4)+(0.8\times y_5)+(0.6\times 0.8=y_6),\hfill \\ y_1\left(0\right)=0.30,y_2\left(0\right)=0.35,y_3\left(0\right)=0.40,y_4\left(0\right)=0.30,y_5\left(0\right)=0.34\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(68)

Here, $y_6$ only acts as an extractor (sink) for all the five terminal acceptors. In this case, applying the same procedure as for the above example, the MAO-based maximal symmetric interval solution with respect to $y_0=(0.72,0.6,0.4,0.66,0.58)$ is the same as before. The interval solution is

$$[y_0-{\delta }^{\ast AC},y_0+{\delta }^{\ast AC}]=[(0.44,0.32,0.12,0.38,0.30),(1,0.88,0.68,0.94,0.86)].$$

6.2. MRO-Based Solution

Here, again, we chose a system of differential equations representing any P2P network with five terminals for sharing educational resources with each other. We are aiming to maximize the quality of downloading educational resources in the form of fuzzy interval solutions (fluctuations or oscillations) as it may vary from time to time depending upon the strength of the network signal. The system is given as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=(0.4\wedge y_1)\vee (0.5\wedge y_2)\vee (0.9\wedge y_3)\vee (0.7\wedge y_4)\vee (0.8\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=(0.8\wedge y_1)\vee (0.4\wedge y_2)\vee (0.3\wedge y_3)\vee (0.6\wedge y_4)\vee (0.7\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=(0.4\wedge y_1)\vee (0.6\wedge y_2)\vee (0.5\wedge y_3)\vee (0.7\wedge y_4)\vee (0.9\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_4}{dt}}=(0.5\wedge y_1)\vee (0.9\wedge y_2)\vee (0.6\wedge y_3)\vee (0.3\wedge y_4)\vee (0.4\wedge y_5),\hfill \\ {\displaystyle \frac{d{y}_5}{dt}}=(0.2\wedge y_1)\vee (0.6\wedge y_2)\vee (0.9\wedge y_3)\vee (0.3\wedge y_4)\vee (0.8\wedge y_5),\hfill \\ y_1\left(0\right)=0.30,y_2\left(0\right)=0.35,y_3\left(0\right)=0.40,y_4\left(0\right)=0.30,y_5\left(0\right)=0.34\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(69)

where numerical data are the band among any two sources or terminals sharing data from i to j for the direction source, with quality of downloading as $y_n$ in the sense of receiving information in fuzzy format with initial speed or quality. System (69) can be written in an addition–multiplication format as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\displaystyle \frac{d{y}_1}{dt}}=(0.4\times y_1)+(0.5\times y_2)+(0.9\times y_3)+(0.7\times y_4)+(0.8\times y_5),\hfill \\ {\displaystyle \frac{d{y}_2}{dt}}=(0.8\times y_1)+(0.4\times y_2)+(0.3\times y_3)+(0.6\times y_4)+(0.7\times y_5),\hfill \\ {\displaystyle \frac{d{y}_3}{dt}}=(0.4\times y_1)+(0.6\times y_2)+(0.5\times y_3)+(0.7\times y_4)+(0.9\times y_5),\hfill \\ {\displaystyle \frac{d{y}_4}{dt}}=(0.5\times y_1)+(0.9\times y_2)+(0.6\times y_3)+(0.3\times y_4)+(0.4\times y_5),\hfill \\ {\displaystyle \frac{d{y}_5}{dt}}=(0.2\times y_1)+(0.6\times y_2)+(0.9\times y_3)+(0.3\times y_4)+(0.8\times y_5),\hfill \\ y_1\left(0\right)=0.30,y_2\left(0\right)=0.35,y_3\left(0\right)=0.40,y_4\left(0\right)=0.30,y_5\left(0\right)=0.34\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(70)

The quality of downloading or giving data or information among any two sources will lie in the interval form having maximal and minimal values. Therefore, the fuzzy relation system (69) can be written interval form as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.30\le {\displaystyle \frac{d{y}_1}{dt}}\le 0.83,\hfill \\ 0.35\le {\displaystyle \frac{d{y}_2}{dt}}\le 0.92,\hfill \\ 0.40\le {\displaystyle \frac{d{y}_3}{dt}}\le 0.88,\hfill \\ 0.30\le {\displaystyle \frac{d{y}_4}{dt}}\le 0.89,\hfill \\ 0.34\le {\displaystyle \frac{d{y}_5}{dt}}\le 0.91.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(71)

or

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.30\le (0.4\wedge y_1)\vee (0.5\wedge y_2)\vee (0.9\wedge y_3)\vee (0.7\wedge y_4)\vee (0.8\wedge y_5)\le 0.83,\hfill \\ 0.35\le (0.8\wedge y_1)\vee (0.4\wedge y_2)\vee (0.3\wedge y_3)\vee (0.6\wedge y_4)\vee (0.7\wedge y_5)\le 0.92,\hfill \\ 0.40\le (0.4\wedge y_1)\vee (0.6\wedge y_2)\vee (0.5\wedge y_3)\vee (0.7\wedge y_4)\vee (0.9\wedge y_5)\le 0.88,\hfill \\ 0.30\le (0.5\wedge y_1)\vee (0.9\wedge y_2)\vee (0.6\wedge y_3)\vee (0.3\wedge y_4)\vee (0.4\wedge y_5)\le 0.89,\hfill \\ 0.34\le (0.2\wedge y_1)\vee (0.6\wedge y_2)\vee (0.9\wedge y_3)\vee (0.3\wedge y_4)\vee (0.8\wedge y_5)\le 0.91.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(72)

For the given $y_0=(0.5,0.4,0.3,0.6,0.2)$, find the MRO-based maximal, symmetrical interval solution of problem (72). For the solution, we write system (72) in descending form relative to $c_i$ as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}0.40\le (0.4\wedge y_1)\vee (0.6\wedge y_2)\vee (0.5\wedge y_3)\vee (0.7\wedge y_4)\vee (0.9\wedge y_5)\le 0.88,\hfill \\ 0.35\le (0.8\wedge y_1)\vee (0.4\wedge y_2)\vee (0.3\wedge y_3)\vee (0.6\wedge y_4)\vee (0.7\wedge y_5)\le 0.92,\hfill \\ 0.34\le (0.2\wedge y_1)\vee (0.6\wedge y_2)\vee (0.9\wedge y_3)\vee (0.3\wedge y_4)\vee (0.8\wedge y_5)\le 0.91,\hfill \\ 0.30\le (0.4\wedge y_1)\vee (0.5\wedge y_2)\vee (0.9\wedge y_3)\vee (0.7\wedge y_4)\vee (0.8\wedge y_5)\le 0.83,\hfill \\ 0.30\le (0.5\wedge y_1)\vee (0.9\wedge y_2)\vee (0.6\wedge y_3)\vee (0.3\wedge y_4)\vee (0.4\wedge y_5)\le 0.89.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(73)

Now, follow the steps for MRO-based solution.

 1. 

Using (6) and (9), find the potential optimal solution as follows:

$$\widehat{y}=(1,0.89,0.83,1,0.88)$$

 2. 

Through testing all the values in system (73), we have satisfied all the inequalities of (73), so this is a solution of (73). Using Theorem 2, we conclude that (73) is consistent, having solution $\widehat{y}=(1,0.89,0.83,1,0.88)$.

 3. 

With respect to the given solution $y_0=(0.5,0.4,0.3,0.6,0.2),$ we calculate the index set $J_{y_0}^1,J_{y_0}^2,\dots J_{y_0}^5$ by Equation (34) in [38] as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{b}_{11}\wedge y_0^1=(0.4\wedge 0.5)=0.4\ge 0.4=c_1,\hfill \\ b_{12}\wedge y_0^2=(0.6\wedge 0.4)=0.4\ge 0.4=c_1,\hfill \\ b_{13}\wedge y_0^3=(0.5\wedge 0.3)=0.3\le 0.4=c_1,\hfill \\ b_{14}\wedge y_0^4=(0.7\wedge 0.6)=0.6\ge 0.4=c_1,\hfill \\ b_{15}\wedge y_0^5=(0.9\wedge 0.2)=0.2\le 0.4=c_1.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(74)

So, we have $J_{y_0}^1=\{1,2,4\},$ and similarly, we can calculate $J_{y_0}^2=\{1,2,4\},$$J_{y_0}^3=\{2,4\},$$J_{y_0}^4=\{1,2,3,4\},$$J_{y_0}^5=\{1,2,3,4\}.$

 4. 

By the above index set, we can calculate the conservative optimal indices $J_1^{RC},J_2^{RC},\dots J_5^{RC}$ by (45) and (46) as follows:

$$\underset{j\in J_{y_0}^1}{max}\{{\displaystyle \frac{y_0^j-c_1}{y_0^j}}\}$$

$$=max\{{\displaystyle \frac{y_0^1-c_1}{y_0^1}},{\displaystyle \frac{y_0^2-c_1}{y_0^2}},{\displaystyle \frac{y_0^4-c_1}{y_0^4}}\}$$

$$=max\{{\displaystyle \frac{0.5-0.40}{0.5}},{\displaystyle \frac{0.40-0.40}{0.40}},{\displaystyle \frac{0.6-0.4}{0.6}}\}$$

$$={\displaystyle \frac{0.6-0.4}{0.6}}={\displaystyle \frac{y_0^4-c_1}{y_0^4}}$$

So, we have $J_1^{RC}=arg{max}_{j\in j_{y_0}^1}=4$. Next, by (46) $\left\{J_1^{RC}\right\}\cap J_{y_0}^2=\left\{4\right\}\cap \{1,2,4\}=\left\{4\right\}\ne \varnothing$, we have $J_2^{RC}=4.$ Further, $\{J_1^{RC},J_2^{RC}\}\cap J_{y_0}^3=\{4,4\}\cap \{2,4\}=\left\{4\right\}\ne \varnothing$. So, $J_3^{RC}=4.$ Next, $\{J_1^{RC},J_2^{RC},J_3^{RC}\}\cap J_{y_0}^4=\{4,4,4\}\cap \{1,2,3,4\}=\left\{4\right\}\ne \varnothing$. So, $J_4^{RC}=4.$ Similarly, $J_5^{RC}=4.$ Therefore, the conservative path is $(J_1^{RC},J_2^{RC},J_3^{RC},J_4^{RC},J_5^{RC})=(4,4,4,4,4).$

 5. 

In this step, we find the indices $I_1^{RC},I_2^{RC},\dots I_5^{RC}$ from the obtained conservative path. Therefore, $I_4^{RC}=\{1,2,3,4,5\}$ because i = 1,2,3,4,5 $J_i^{RC}=4$; further, $I_1^{RC}=\varnothing ,I_2^{RC}=\varnothing ,I_3^{RC}=\varnothing ,I_5^{RC}=\left\{\right\},$ because i = 1,2,3,5 $J_i^{RC}\ne 4$.

 6. 

Also, $I_1^{RC}=\varphi ,I_2^{RC}=\varphi ,I_3^{RC}=\varphi ,I_5^{RC}=\varphi$. Therefore, $y_1^{RC},y_2^{RC},y_3^{RC},y_5^{RC}=\varphi$ by (48). Further, by (48), we can write

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{y}_4^{RC}=\underset{i\in I_4^{RC}}{\bigvee }{c}_i=c_1\vee c_2\vee c_3\vee c_4\vee c_5=0.40\vee 0.35\vee 0.34\vee 0.30\vee 0.30=0.40.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(75)

Therefore, we have the vector $y^{RC}=(0,0,0,0.4,0).$

 7. 

Consider

$$\widehat{y}=(1,0.89,0.83,1,0.88),$$

$$y_0=(0.5,0.4,0.3,0.6,0.2),$$

$$y^{RC}=(0,0,0,0.4,0).$$

Now, we can find ${\delta }_1^{\ast RC},{\delta }_2^{\ast RC},\dots {\delta }_5^{\ast RC}$ by (52)

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\begin{array}{c}{\delta }_1^{\ast RC}=({\widehat{y}}^1-y_0^1)\wedge (y_0^1-{\widehat{y}}_1^{RC})=(1-0.5)\wedge (0.5-0)=0.5,\hfill \\ {\delta }_2^{\ast AC}=({\widehat{y}}^2-y_0^2)\wedge (y_0^2-{\widehat{y}}_2^{RC})=(0.89-0.4)\wedge (0.4-0)=0.4,\hfill \\ {\delta }_3^{\ast AC}=({\widehat{y}}^3-y_0^3)\wedge (y_0^3-{\widehat{y}}_3^{RC})=(0.83-0.3)\wedge (0.3-0)=0.30,\hfill \\ {\delta }_4^{\ast AC}=({\widehat{y}}^4-y_0^4)\wedge (y_0^4-{\widehat{y}}_4^{RC})=(1-0.6)\wedge (0.6-0.4)=0.20,\hfill \\ {\delta }_5^{\ast AC}=({\widehat{y}}^5-y_0^5)\wedge (y_0^5-{\widehat{y}}_5^{RC})=(0.88-0.2)\wedge (0.2-0)=0.20.\hfill \end{array}\hfill \end{array}\right.\end{array}$$

(76)

So, we find ${\delta }^{\ast RC}=({\delta }_1^{\ast RC},{\delta }_2^{\ast RC},\dots {\delta }_5^{\ast RC})=(0.5,0.4,0.3,0.2,0.2)$

 8. 

Next, we can find the relative maximal oscillation of $y_0$ based on the values obtained in step 7.

$${\delta }_{min}^{\ast RC}={\displaystyle \frac{{\delta }_1^{\ast AC}}{y_0^1}}\wedge {\displaystyle \frac{{\delta }_2^{\ast AC}}{y_0^2}}\wedge {\displaystyle \frac{{\delta }_3^{\ast AC}}{y_0^3}}\wedge {\displaystyle \frac{{\delta }_4^{\ast AC}}{y_0^4}}\wedge {\displaystyle \frac{{\delta }_5^{\ast AC}}{y_0^5}}$$

$${\delta }_{min}^{\ast AC}={\displaystyle \frac{0.5}{0.5}}\wedge {\displaystyle \frac{0.4}{0.4}}\wedge {\displaystyle \frac{0.30}{0.3}}\wedge {\displaystyle \frac{0.2}{0.6}}\wedge {\displaystyle \frac{0.2}{0.2}}$$

$${\delta }_{min}^{\ast AC}=0.3$$

So, the MRO-based maximal symmetric interval solution with respect to $y_0=(0.5,0.4,0.3,0.6,0.2)$ is

$$[y_0-{\delta }^{\ast RC},y_0+{\delta }^{\ast RC}]=[(0.17,0.07,0,0.3,0),(0.8,0.7,0.6,0.9,0.5)]$$

7. Concluding Remarks

The system of differential equations shown herein represents a resource sharing educational system among different terminals in any P2P network. The download quality, measured in bits per second (bps), associated with downloading educational resources may vary and fluctuate between minimum and maximum values; therefore, it is represented in a min–max fuzzy relation inequality format. We have successfully studied the proposed system for an optimal interval symmetrical solution with some given solutions. The obtained solution was tested for both absolute and relative maximal oscillation for the given solution. It was also verified that the chosen system is consistent; therefore, it both an absolute and relative maximal solution exist. Both the algorithms of an MAO- and MRO-based optimal solution were developed with the help of definitions, theorems, propositions, etc. The said algorithms were verified with numerical examples, showing the effectiveness of the methods. Such analysis may be extended to a variety of P2P educational resource sharing systems. This research may also be applied to two-terminal fuzzy relation systems with different resolution techniques like two-terminal educational institutions in which both act as the reservoir and sink.