On solving fuzzy rough multiobjective integer linear fractional programming problem

Abstract

In solving real life fractional programming problem, we often face the state of uncertainty as well as hesitation due to various uncontrollable factors. To overcome these limitations, the fuzzy rough approach is applied to this problem. In this paper, an efficient method is proposed for solving fuzzy rough multiobjective integer linear fractional programming problem where all the variables and parameters are fuzzy rough numbers. Here, the fuzzy rough multiobjective problem transformed into an equivalent multiobjective integer linear fractional programming problem. Furthermore, from the obtained problem, five crisp multiobjective integer linear fractional programming problems are constructed and the resultant problems are solved as a crisp integer linear programming problem by using Dinkelbach concept. Finally, the effectiveness of the proposed procedure is illustrated through numerical examples.

Keywords

Fractional programming integer programming problem triangular fuzzy rough numbers fuzzy rough linear programming

1 Introduction

The linear fractional programming problemhas significant applications in various areas of life such as production planning, financial sector, health care, and all engineering fields. The main aim of this problem is maximize/minimize a ratio of physical and/or economical linear functions subject to linear constraints. In the literature, several methods (see [2, 18]) have been recommended to solve integer linear fractional programming (ILFP) problems.

Authors [6, 11] proposed methods for finding efficient solutions to multiobjective linear fractional programming problems. In the actual applications, a model involves many parameters whose values are given by experts. However, both experts and decision makers frequently do not know the value of those parameters due to various uncontrollable factors.

From this point of view, some researchers [1, 12] have used rough interval to deal with the imprecision of parameters and proposed methods to solve the linear programming problems with rough interval. Hamazehee et al. [12] introduced a new class of Linear Programming (LP) problems in which some or all of the coefficients are rough interval data and proposed a method to compute the range of optimal values. E. Ammar and M. Muamer. [1] proposed a method to solve linear fractional programming problem with rough interval coefficients in the objective function.

Many researchers [7 , 19] dealt with the imprecision of parameters by using the concept of fuzzy set theory [3, 21] and represented some or all of the parameters as fuzzy numbers. So, the linear fractional programming is generalized to fuzzy linear fractional programming (FLFP). Authors [8, 14] proposed methods for solving linear fractional programming problems in which the coefficients in the objective function are assumed to be fuzzy numbers.While others [7, 19] proposed methods for solving linear fractional programming problem in which all the parameters are represented by fuzzy numbers and the variables are represented by real numbers. In [17], an algorithm to solve fuzzy multi objective integer linear fractional programming (FMOILFP) problem has been described. The basic idea of the computational phase of the algorithm is based mainly upon a modified Isbell-Marlow method together with the brunch and bound technique.

A few researchers [13 , 20] have proposed methods to deal with the linear fractional programming problems in which all the parameters as well as variables are represented as fuzzy numbers, called fully fuzzy linear fractional programming (FFLFP) problems.

This study will demonstrate that it is more effective to propose and use fuzzy rough linear fractional programming (FRLFP) to solve many real world problems where uncertain parameters exist, this is because the FRLFP solution method can generate fuzzy rough solutions, respectively corresponding to the two levels of uncertain information (lower and upper approximation fuzzy numbers) and thus avoid losing uncertain information.

The concept of “fuzzy rough number” will be introduced to represent uncertain information of all parameters and decision variables, and the associated solution method will be presented to solve fuzzy rough multiobjective integer linear fractional programming (FRMOILFP) problem providing uncertain solutions.

After the modeling formulation, a case study will be provided for demonstrating its applicability and advantages upon the approach.

This paper is organized as follows: In Section 2 some basic definitions and some arithmetic results are presented. In Section 3, formulation of FRMOILFP problem and application for solving FRMOILFP problem are established. Analgorithm solution for FRMOILFP problem is proposed in Section 4. In Section 5, advantages of the proposed method are discussed. In Section 6, two numerical examples are given to illustrate the theory developed in this paper. Finally, the conclusion part is present in Section 7.

2 Preliminaries

2.1 Triangular fuzzy number

The following are some definitions of the basic arithmetic operators and partial ordering relations on triangular fuzzy numbers based on the function principle which can be established in [3 , 21] and are used in Section 3.

Definition 2.1. A fuzzy number $\tilde{A}$ is a triangular fuzzy number denoted by (a₁, a₂, a₃) where a₁, a₂ and a₃ are real numbers and its membership function $μ_{\tilde{A}} (x)$ is given below: $μ_{\tilde{A}} (x) = {\begin{matrix} \frac{(x - a_{1})}{(a_{2} - a_{1})} for a_{1} \leq x \leq a_{2} \\ \frac{(a_{3} - x)}{(a_{3} - a_{2})} for a_{2} \leq x \leq a_{3} \\ 0 otherwise \end{matrix}$

Definition 2.2. Let (a₁, a₂, a₃) and (b₁, b₂, b₃) be two positive triangular fuzzy numbers, then

(a₁, a₂, a₃) ⊕ (b₁, b₂, b₃) = (a₁ + b₁, a₂ + b₂, a₃ + b₃).

(a₁, a₂, a₃) ⊖ (b₁, b₂, b₃) = (a₁ - b₃, a₂ - b₂, a₃ - b₁).

k (a₁, a₂, a₃) = (ka₁, ka₂, ka₃) for k > 0.

k (a₁, a₂, a₃) = (ka₃, ka₂, ka₁) for k < 0.

(a₁, a₂, a₃) ⊗ (b₁, b₂, b₃) = (a₁ × b₁, a₂ × b₂, a₃ × b₃).

$(a_{1}, a_{2}, a_{3}) ø (b_{1}, b_{2}, b_{3}) = (\frac{a_{1}}{b_{3}}, \frac{a_{2}}{b_{2}}, \frac{a_{3}}{b_{1}})$ .

Let F (R) be the set of all real triangular fuzzy numbers.

Definition 2.3. Let $\tilde{A} = (a_{1}, a_{2}, a_{3})$ and $\tilde{B} = (b_{1}, b_{2}, b_{3})$ be in F (R), then

$\tilde{A} \approx \tilde{B} if a_{i} = b_{i}, i = 1, 2, 3 .$

$\tilde{A} ≺ \tilde{B} if a_{i} < b_{i}, i = 1, 2, 3 .$

$\tilde{A} ≻ \tilde{B} if a_{i} > b_{i}, i = 1, 2, 3 and$ $\tilde{A} ≻ 0 if a_{i} > 0, i = 1, 2, 3$ .

2.2 Fuzzy rough interval

In this section, the definitions of fuzzy rough interval, fuzzy rough numbers, triangular fuzzy rough numbers and basic operations for triangular fuzzy rough number are presented [9, 15].

Definition 2.4. Let X be denote a compact set of real numbers. A fuzzy rough interval ${\tilde{X}}^{R}$ is defined as ${\tilde{X}}^{R} = [{\tilde{X}}^{LAI} : {\tilde{X}}^{UAI}]$ where ${\tilde{X}}^{LAI}$ and ${\tilde{X}}^{UAI}$ are fuzzy set called lower and upper approximation fuzzy numbers of ${\tilde{X}}^{R}$ with ${\tilde{X}}^{LAI} \subseteq {\tilde{X}}^{UAI}$ .

Definition 2.5. A fuzzy rough number ${\tilde{A}}^{R}$ is a convex normalized fuzzy rough interval of the real line $R$ whose membership function is piecewise continuous.

Definition 2.6. A fuzzy rough number ${\tilde{A}}^{R}$ is a triangular fuzzy rough number denoted by ${\tilde{A}}^{R} = [(a^{LL}$ , a^M, a^UL) : (a^LU, a^M, a^UU)] where a^LU, $a^{LL}, a^{M}, a^{UL} and a^{UU} \in R$ such that a^LU ≤ a^LL ≤ a^M ≤ a^UL ≤ a^UU and the membership function can be defined as: $μ_{{\tilde{A}}^{R}} (x) = {\begin{matrix} μ_{{\tilde{A}}^{L}} (x) = {\begin{matrix} \frac{x - a^{LL}}{a^{M} - a^{LL}} a^{LL} \leq x \leq a^{M} \\ \frac{a^{UL} - x}{a^{UL} - a^{M}} a^{M} \leq x \leq a^{UL} \\ 0 otherwise \end{matrix} \\ μ_{{\tilde{A}}^{U}} (x) = {\begin{matrix} \frac{x - a^{LU}}{a^{M} - a^{LU}} a^{LU} \leq x \leq a^{M} \\ \frac{a^{UU} - x}{a^{UU} - a^{M}} a^{M} \leq x \leq a^{UU} \\ 0 otherwise \end{matrix} \end{matrix}$

Note that ${\tilde{A}}^{L} = (a^{LL}, a^{M}, a^{UL}), {\tilde{A}}^{U} = (a^{LU}$ , a^M, a^UU) and ${\tilde{A}}^{L} \tilde{\subseteq} {\tilde{A}}^{U}$ . Where $μ_{{\tilde{A}}^{L}} (x)$ and $μ_{{\tilde{A}}^{U}} (x)$ are membership functions of lower and upper approximation triangular fuzzy number respectively.

The membership function of triangular fuzzy rough number is shown in Fig. 1.

Fig.1

Membership function of the triangular fuzzy rough number.

Definition 2.7. Let ${\tilde{A}}^{R} \geq 0 and {\tilde{B}}^{R} \geq 0$ be two fuzzy rough intervals, then

${\tilde{A}}^{R} \oplus {\tilde{B}}^{R} = [({\tilde{A}}^{L} \oplus {\tilde{B}}^{L}) : ({\tilde{A}}^{U} \oplus {\tilde{B}}^{U})]$ .

${\tilde{A}}^{R} ⊖ {\tilde{B}}^{R} = [({\tilde{A}}^{L} ⊖ {\tilde{B}}^{L}) : ({\tilde{A}}^{U} ⊖ {\tilde{B}}^{U})]$ .

${\tilde{A}}^{R} \otimes {\tilde{B}}^{R} = [({\tilde{A}}^{L} \otimes {\tilde{B}}^{L}) : ({\tilde{A}}^{U} \otimes {\tilde{B}}^{U})]$ .

${\tilde{A}}^{R} ø {\tilde{B}}^{R} = [({\tilde{A}}^{L} ø {\tilde{B}}^{L}) : ({\tilde{A}}^{U} ø {\tilde{B}}^{U})]$ .

2.3 Integer linear fractional programming problem

The general form of integer linear fractional programming (ILFP) problem is discussed as follows: $\begin{matrix} \begin{matrix} Max Z (x) = \frac{N (x)}{D (x)} = \frac{c^{T} x + α}{d^{T} x + β} \\ Subject to : \\ x \in S = {x \in R^{n} : Ax \leq B, x \geq 0 and integer} \end{matrix}} \end{matrix}$ (2.1)

where c, d ∈ Rⁿ, α, β ∈ R, B ∈ R^m, A ∈ R^m×n and D (x) >0

Theorem 2.1. The solution x^* is an optimal solution of the (ILFP) problem (2.1) if and only if Max {N (x) - Z^*D (x) ∀ x ∈ S} =0 where $Z^{*} = \frac{N (x^{*})}{D (x^{*})} = Max \frac{N (x)}{D (x)}$ .

Proof. Let x^* be an optimal solution of problem (2.1), then $Z^{*} = \frac{N (x^{*})}{D (x^{*})} \geq \frac{N (x)}{D (x)} for allx \in S$

Hence

N (x) - Z^*D (x) ≤0

N (x) - Z^*D (x^*) =0

From (1) Max {N (x) - Z^*D (x) ∀ x ∈ S} =0. From (2) the maximum is taken on at x^*.

Thus the first part of the proof is finished. Now let x^* be a solution of the problem: $Max {N (x) - Z^{*} D (x) \forall x \in S} = 0 where$ $Z^{*} = \frac{N (x^{*})}{D (x^{*})}, then$ $\begin{matrix} N (x) - Z^{*} D (x) \leq 0 for all x \in S \\ \Rightarrow Z^{*} = \frac{N (x^{*})}{D (x^{*})} \geq \frac{N (x)}{D (x)} . \end{matrix}$

This leads to that Z^* is maximum value of problem (2.1) and x^* is optimal solution of problem (2.1). Thus the theorem is proved.

2.4 Multiobjective integer linear fractional programming problem

The general multiobjective integer linear fractional programming (MOILFP) problem may be written as: $\begin{matrix} Max (Z_{1} (x), Z_{2} (x), . . . . . ., Z_{K} (x)) \\ Subject to : \\ x \in S = {x \in R^{n} : Ax \leq B, x \geq 0 and integer} \end{matrix}}$ (2.2) $\begin{matrix} where Z_{i} (x) = \frac{N_{i} (x)}{D_{i} (x)} = \frac{c_{i}^{T} x + α_{i}}{d_{i}^{T} x + β_{i}}, c_{i}, d_{i} \in R^{n}, \\ α_{i}, β_{i} \in R, B \in R^{m}, A \in R^{m \times n} and D_{i} (x) > 0 \\ for all i = 1, 2, . . . . . . . ., k . \end{matrix}$

Definition 2.8. (Efficient solution of MOILFP);

A point $\bar{x} \in R^{n}$ is an efficient solution of MOILFP problem if and only if there does not exist another $x \in R^{n}$ such that $\frac{N_{i} (x)}{D_{i} (x)} \geq \frac{N_{i} (\bar{x})}{D_{i} (\bar{x})}$ , for all i and

$\frac{N_{i} (x)}{D_{i} (x)} > \frac{N_{i} (\bar{x})}{D_{i} (\bar{x})}$ , for at least on i.

Theorem 2.2. If $\bar{x}$ is an optimal solution of $Max {\sum_{i = 1}^{k} w_{i} (N_{i} (x) - Z_{i}^{*} D_{i} (x)), x \in S} where$ $Z_{i}^{*} = \frac{N_{i} (x^{*})}{D_{i} (x^{*})} = Max \frac{N_{i} (x)}{D_{i} (x)}, for all i = 1, 2, . . . . . ., k$ $and w_{i} \in W = {w_{i} \in R^{n} : w_{i} \geq 0, \sum_{i = 1}^{k} w_{i} = 1}, then$ $\bar{x}$ is an efficient solution of MOILFP problem (2.2).

Proof. Let $x_{i}^{*}$ and $Z_{i}^{*}$ be the global maximum points and values of each objective function of (MOILFP) respectively, that is, $Z_{i}^{*} = \frac{N_{i} (x^{*})}{D_{i} (x^{*})} = Max \frac{N_{i} (x)}{D_{i} (x)} for all i, x \in S .$

Thus, $\frac{N_{i} (x^{*})}{D_{i} (x^{*})} \geq \frac{N_{i} (x)}{D_{i} (x)} or D_{i} (x) \frac{N_{i} (x^{*})}{D_{i} (x^{*})} \geq N_{i} (x)$ or $N_{i} (x) - Z_{i}^{*} D_{i} (x) \leq 0$ for all i = 1, 2, . . . , k and for all x ∈ S.

Again, let $\bar{x}$ be an optimal solution of the problem: $Max {\sum_{i = 1}^{k} w_{i} (N_{i} (x) - Z_{i}^{*} D_{i} (x)), x \in S}$

Therefore, $\sum_{i = 1}^{k} w_{i} (N_{i} (x) - Z_{i}^{*} D_{i} (x)) \leq \sum_{i = 1}^{k} w_{i} (N_{i} (\bar{x}) - Z_{i}^{*} D_{i} (\bar{x}))$ $\leq \sum_{i = 1}^{k} w_{i} Max (N_{i} (x) - Z_{i}^{*} D_{i} (x)) \leq 0; for x \in S$

From these inequalities, $N_{i} (x) - Z_{i}^{*} D_{i} (x)) \leq N_{i} (\bar{x}) - Z_{i}^{*} D_{i} (\bar{x}) \leq 0$ $or D_{i} (x) [\frac{N_{i} (x)}{D_{i} (x)} - Z_{i}^{*}] \leq D_{i} (\bar{x}) [\frac{N_{i} (\bar{x})}{D_{i} (\bar{x})} - Z_{i}^{*}]$ $or [\frac{N_{i} (x)}{D_{i} (x)} - Z_{i}^{*}] \leq \frac{D_{i} (\bar{x})}{D_{i} (x)} [\frac{N_{i} (\bar{x})}{D_{i} (\bar{x})} - Z_{i}^{*}] .$

Both via Theorem (2.1) and the inequality $D_{i} (\bar{x}) / D_{i} (x) \geq 1$ , one can write that $[\frac{N_{i} (x)}{D_{i} (x)} - Z_{i}^{*}] \leq [\frac{N_{i} (\bar{x})}{D_{i} (\bar{x})} - Z_{i}^{*}] and$ $[\frac{N_{i} (x)}{D_{i} (x)}] \leq [\frac{N_{i} (\bar{x})}{D_{i} (\bar{x})}] for all i .$

Hence, $\bar{x}$ is an efficient solution of the MOILFP problem (2.2).

Now, if $\bar{x}$ is not an efficient solution of the MOILFP problem, then there exist x ∈ S such that, $\frac{N_{i} (\bar{x})}{D_{i} (\bar{x})} \leq \frac{N_{i} (x)}{D_{i} (x)} for all i and \frac{N_{j} (\bar{x})}{D_{j} (\bar{x})} < \frac{N_{j} (x)}{D_{j} (x)}$ for at least one j.

Then, $\frac{N_{i} (\bar{x})}{D_{i} (\bar{x})} \leq \frac{N_{i} (x)}{D_{i} (x)} \leq Z_{i}^{*}$ $or N_{i} (\bar{x}) - Z_{i}^{*} D_{i} (\bar{x}) \leq N_{i} (x) - Z_{i}^{*} D_{i} (x) for all i$ $and \frac{N_{j} (\bar{x})}{D_{j} (\bar{x})} < \frac{N_{j} (x)}{D_{j} (x)} < Z_{j}^{*}$ $or N_{j} (\bar{x}) - Z_{j}^{*} D_{j} (\bar{x}) \leq N_{j} (x) - Z_{j}^{*} D_{j} (x)$ for at least one j.

Summing the k inequalities, $\sum_{i = 1}^{k} (N_{i} (\bar{x}) - Z_{i}^{*} D_{i} (\bar{x})) \leq \sum_{i = 1}^{k} (N_{i} (x) - Z_{i}^{*} D_{i} (x))$

$\begin{matrix} or \sum_{i = 1}^{k} w_{i} (N_{i} (\bar{x}) - Z_{i}^{*} D_{i} (\bar{x})) \leq \sum_{i = 1}^{k} w_{i} (N_{i} (x) \\ - Z_{i}^{*} D_{i} (x)); \end{matrix}$ which contradicts that is an optimal solution of the problem: $Max {\sum_{i = 1}^{k} w_{i} (N_{i} (x) - Z_{i}^{*} D_{i} (x)), x \in S} .$

Hence the theorem is proved.

3 Problem formulation

The multiobjective integer linear fractional programming problems with fully fuzzy rough coefficients and variables (FRMOILFP) are defined as follows:

$\begin{matrix} Max {\tilde{Z}}_{i}^{R} (x) = \frac{{\tilde{N}}_{i}^{R} (x)}{{\tilde{D}}_{i}^{R} (x)} = \frac{{\tilde{c}}_{i}^{R} {\tilde{x}}^{R} \oplus {\tilde{α}}_{i}^{R}}{{\tilde{d}}_{i}^{R} {\tilde{x}}^{R} \oplus {\tilde{β}}_{i}^{R}} \\ Subject to : \\ {\tilde{A}}^{R} {\tilde{x}}^{R} \leq {\tilde{B}}^{R} \\ {\tilde{x}}^{R} \geq {\tilde{0}}^{R} and interger, i = 1, 2, . . ., k \geq 2 \end{matrix}}$ (3.1) where ${\tilde{c}}_{i}^{R}, {\tilde{d}}_{i}^{R}$ are (1 × n) positive vectors of fuzzy rough intervals, ${\tilde{α}}_{i}^{R}, {\tilde{β}}_{i}^{R}$ are positive fuzzy rough intervals, ${\tilde{A}}^{R}$ is (m × n) positive matrix of fuzzy rough intervals coefficients of constraints, ${\tilde{B}}^{R}$ is (m × 1) positive column of fuzzy rough intervals and ${\tilde{x}}^{R} = ({\tilde{x}}_{1}^{R}, {\tilde{x}}_{2}^{R}, . . ., {\tilde{x}}_{n}^{R})$ is (n × 1) column of all decision variables.

The problem (3.1) can be written as: $\begin{matrix} Max [{\tilde{Z}}_{i}^{L} (x) : {\tilde{Z}}_{i}^{U} (x)] = \frac{[{\tilde{c}}_{i}^{L} {\tilde{x}}^{L} \oplus {\tilde{α}}_{i}^{L} : {\tilde{c}}_{i}^{U} {\tilde{x}}^{U} \oplus {\tilde{α}}_{i}^{U}]}{[{\tilde{d}}_{i}^{L} {\tilde{x}}^{L} \oplus {\tilde{β}}_{i}^{L} : {\tilde{c}}_{i}^{U} {\tilde{x}}^{U} \oplus {\tilde{β}}_{i}^{U}]} \\ Subject to : \\ [{\tilde{A}}^{L} {\tilde{x}}^{L} : {\tilde{A}}^{U} {\tilde{x}}^{U}] \leq [{\tilde{B}}^{L} : {\tilde{B}}^{U}] \\ [{\tilde{x}}^{L} : {\tilde{x}}^{U}] \geq [{\tilde{0}}^{L} : {\tilde{0}}^{U}] and interger, \\ i = 1, 2, . . ., k \geq 2 \end{matrix}}$ (3.2)

Now using the operations of fuzzy rough interval: $\begin{matrix} Max [{\tilde{Z}}_{i}^{L} (x) : {\tilde{Z}}_{i}^{U} (x)] = [\frac{{\tilde{c}}_{i}^{L} {\tilde{x}}^{L} \oplus {\tilde{α}}_{i}^{L}}{{\tilde{d}}_{i}^{L} {\tilde{x}}^{L} \oplus {\tilde{β}}_{i}^{L}} : \frac{{\tilde{c}}_{i}^{U} {\tilde{x}}^{U} \oplus {\tilde{α}}_{i}^{U}}{{\tilde{d}}_{i}^{U} {\tilde{x}}^{U} \oplus {\tilde{β}}_{i}^{U}}] \\ Subject to : \\ [{\tilde{A}}^{L} {\tilde{x}}^{L} : {\tilde{A}}^{U} {\tilde{x}}^{U}] \leq [{\tilde{B}}^{L} : {\tilde{B}}^{U}] \\ [{\tilde{x}}^{L} : {\tilde{x}}^{U}] \geq [{\tilde{0}}^{L} : {\tilde{0}}^{U}] and interger, \\ i = 1, 2, . . ., k \geq 2 \end{matrix}}$ (3.3)

Let us assume that ${\tilde{x}}^{L}, {\tilde{x}}^{U}, {\tilde{c}}_{i}^{L}, {\tilde{c}}_{i}^{U}, {\tilde{α}}_{i}^{L}, {\tilde{α}}_{i}^{U}$ , ${\tilde{β}}_{i}^{U}, {\tilde{β}}_{i}^{L}, {\tilde{β}}_{i}^{U}, {\tilde{d}}_{i}^{L},$ ${\tilde{d}}_{i}^{U}, {\tilde{A}}^{L}, {\tilde{A}}^{U}, {\tilde{B}}^{L}, {\tilde{B}}^{U}, {\tilde{Z}}_{i}^{L}, {\tilde{Z}}_{i}^{U}$ are triangular fuzzy numbers. Therefore, the problem (3.3) can be written as: $\begin{matrix} Max [(Z_{i}^{LL}, Z_{i}^{M}, Z_{i}^{UL}) : (Z_{i}^{LU}, Z_{i}^{M}, Z_{i}^{UU}) \\ Subject to : \\ (A^{LU}, A^{M}, A^{UU}) \otimes (x^{LU}, x^{M}, x^{UU}) \leq (B^{LU}, B^{M}, B^{UU}) \\ (A^{LL}, A^{M}, A^{UL}) \otimes (x^{LL}, x^{M}, x^{UL}) \leq (B^{LL}, B^{M}, B^{UL}) \\ (x^{LL}, x^{M}, x^{UL}), (x^{LU}, x^{M}, x^{UU}) \geq 0 and interger \\ i = 1, 2, . . . . . ., k \geq 2 \end{matrix}}$ (3.4) $\begin{matrix} - 24 ptwhere (Z_{i}^{LL}, Z_{i}^{M}, Z_{i}^{UL}) = \\ - 24 pt \frac{(p_{i}^{LL}, q_{i}^{M}, r_{i}^{UL}) \otimes (x_{i}^{LL}, x_{i}^{M}, x_{i}^{UL}) \oplus (α_{i}^{LL}, α_{i}^{M}, α_{i}^{UL})}{(m_{i}^{LL}, u_{i}^{M}, v_{i}^{UL}) \otimes (x_{i}^{LL}, x_{i}^{M}, x_{i}^{UL}) \oplus (β_{i}^{LL}, β_{i}^{M}, β_{i}^{UL})} \end{matrix}$ $\begin{matrix} - 24 ptand (Z_{i}^{LU}, Z_{i}^{M}, Z_{i}^{UU}) = \\ - 24 pt \frac{(p_{i}^{LU}, q_{i}^{M}, r_{i}^{UU}) \otimes (x_{i}^{LU}, x_{i}^{M}, x_{i}^{UU}) \oplus (α_{i}^{LU}, α_{i}^{M}, α_{i}^{UU})}{(m_{i}^{LU}, u_{i}^{M}, v_{i}^{UU}) \otimes (x_{i}^{LU}, x_{i}^{M}, x_{i}^{UU}) \oplus (β_{i}^{LU}, β_{i}^{M}, β_{i}^{UU})} \end{matrix}$

The problem (3.4) is equivalent to the following crisp multiobjective problem $Max (Z_{i}^{LL}, Z_{i}^{UL}, Z_{i}^{M}, Z_{i}^{UU}, Z_{i}^{LU})$ Subject to: $\begin{matrix} A^{LL} x^{LL} \leq B^{LL}, A^{UL} x^{UL} \leq B^{UL}, A^{M} x^{M} \leq B^{M} \\ A^{LU} x^{LU} \leq B^{LU}, A^{UU} x^{UU} \leq B^{UU} \\ x^{LL}, x^{UL}, x^{M}, x^{LU}, x^{UU} \geq 0 and interger, \\ i = 1, 2, . . . . . ., k \geq 2 . \end{matrix}$ where $z_{i}^{LL} = \frac{p_{i}^{LL} x^{LL} + α_{i}^{LL}}{v_{i}^{UL} x^{UL} + β_{i}^{UL}}$ , $Z_{i}^{UL} = \frac{r_{i}^{UL} x^{UL} + α_{i}^{UL}}{m_{i}^{LL} x^{LL} + β_{i}^{LL}}, Z_{i}^{M} = \frac{q_{i}^{M} x^{M} + α_{i}^{M}}{u_{i}^{M} x^{M} + β_{i}^{M}},$ $Z_{i}^{UU} = \frac{r_{i}^{UU} x^{UU} + α_{i}^{UU}}{m_{i}^{LU} x^{LU} + β_{i}^{LU}}, Z_{i}^{LU} = \frac{p_{i}^{LU} x^{LU} + α_{i}^{LU}}{v_{i}^{UU} x^{UU} + β_{i}^{UU}} .$

From the above problem five multiobjective integer linear fractional programming (MOILFP) problems will be constructed as follows: $P_{1} : {\begin{matrix} Max Z_{i}^{M} = \frac{q_{i}^{M} x^{M} + α_{i}^{M}}{u_{i}^{M} x^{M} + β_{i}^{M}}, i = 1, 2, . . ., k \geq 2 \\ Subject to : \\ A^{M} x^{M} \leq B^{M} \\ x^{M} \geq 0 and integer . \end{matrix}$ $P_{2} : {\begin{matrix} Max Z_{i}^{UU} = \frac{r_{i}^{UU} x^{UU} + α_{i}^{UU}}{m_{i}^{LU} x^{LU} + β_{i}^{LU}}, i = 1, 2, . . ., k \geq 2 \\ Subject to : \\ A^{LU} x^{LU} \leq B^{LU}, A^{UU} x^{UU} \leq B^{UU} \\ x^{LU}, x^{UU} \geq 0 and integer . \end{matrix}}$ $P_{3} : {\begin{matrix} Max Z_{i}^{LU} = \frac{p_{i}^{LU} x^{LU} + α_{i}^{LU}}{v_{i}^{UU} x^{UU} + β_{i}^{UU}}, i = 1, 2, . . ., k \geq 2 \\ Subject to : \\ A^{LU} x^{LU} \leq B^{LU}, A^{UU} x^{UU} \leq B^{UU} \\ x^{LU}, x^{UU} \geq 0 and integer \end{matrix}}$ $P_{4} : {\begin{matrix} Max Z_{i}^{UL} = \frac{r_{i}^{UL} x^{UL} + α_{i}^{UL}}{m_{i}^{LL} x^{LL} + β_{i}^{LL}}, i = 1, 2, . . ., k \geq 2 \\ Subject to : \\ A^{LL} x^{LL} \leq B^{LL}, A^{UL} x^{UL} \leq B^{UL} \\ x^{LL}, x^{UL} \geq 0 and integer \end{matrix}}$ $P_{5} : {\begin{matrix} Max Z_{i}^{LL} = \frac{p_{i}^{LL} x^{LL} + α_{i}^{LL}}{v_{i}^{UL} x^{UL} + β_{i}^{UL}}, i = 1, 2, . . ., k \geq 2 \\ Subject to : \\ A^{LL} x^{LL} \leq B^{LL}, A^{UL} x^{UL} \leq B^{UL} \\ x^{LL}, x^{UL} \geq 0 and integer \end{matrix}}$

The problems (P_i, i = 1, 2, 3, 4, 5) will be converted into the equivalent ILP problems $(P_{i}^{'}, i = 1, 2, 3, 4, 5)$ by Theorem (2.2) as follows: $P_{1}^{'} : {\begin{matrix} Max (\sum_{i = 1}^{k} w_{i} [q_{i}^{M} x^{M} + α_{i}^{M} - Z_{i}^{M *} (u_{i}^{M} x^{M} + β_{i}^{M})]) \\ Subject to : \\ A^{M} x^{M} \leq B^{M} \\ x^{M} \geq 0 and integer \end{matrix}$ $P_{2}^{'} : {\begin{matrix} Max (\sum_{i = 1}^{k} \begin{matrix} w_{i} [r_{i}^{UU} X^{UU} + α_{i}^{UU} \\ - Z_{i}^{UU *} (m_{i}^{LU} x^{LU} + β_{i}^{LU})]) \end{matrix} \\ Subject to : \\ A^{LU} x^{LU} \leq B^{LU}, A^{UU} x^{LU} \leq B^{UU} \\ x^{LU}, x^{UU} \geq 0 and integer \end{matrix}}$ $P_{3}^{'} : {\begin{matrix} Max (\sum_{i = 1}^{k} \begin{matrix} w_{i} [p_{i}^{LU} x^{LU} + α_{i}^{LU} \\ - Z_{i}^{LU *} (v_{i}^{UU} x^{UU} + β_{i}^{UU})]) \end{matrix} \\ Subject to : \\ A^{LU} x^{LU} \leq B^{LU}, A^{UU} x^{LU} \leq B^{UU} \\ x^{LU}, x^{UU} \geq 0 and integer \end{matrix}}$ $P_{4}^{'} : {\begin{matrix} Max (\sum_{i = 1}^{k} \begin{matrix} w_{i} [r_{i}^{UL} x^{UL} + α_{i}^{UL} \\ - Z_{i}^{UL *} (m_{i}^{LL} x^{LL} + β_{i}^{LL}) \end{matrix} \\ Subject to : \\ A^{LL} x^{LL} \leq B^{LL}, A^{UL} x^{UL} \leq B^{UL} \\ x^{LL}, x^{UL} \geq 0 and integer \end{matrix}}$ $P_{5}^{'} : {\begin{matrix} Max (\sum_{i = 1}^{k} \begin{matrix} w_{i} [p_{i}^{LL} x^{LL} + α_{i}^{LL} \\ - Z_{i}^{LL *} (v_{i}^{UL} x^{UL} + β_{i}^{UL})]) \end{matrix} \\ Subject to : \\ A^{LL} x^{LL} \leq B^{LL}, A^{UL} x^{UL} \leq B^{UL} \\ x^{LL}, x^{UL} \geq 0 and integer \end{matrix}}$ $w \in W = {w_{i} : \sum_{i = 1}^{k} w_{i} = 1, w_{i} \geq 0 i = 1, 2, . ., k} .$

Finally, by using the following theorems the efficient solution of the given FRMOILFP problem will be obtained.

Theorem 3.1. If $\bar{x} \in R^{n}$ is an optimal solution of ILP problems $(P_{i}^{'}, i = 1, 2, 3, 4, 5)$ then $\bar{x} \in R^{n}$ is an efficient solution of the corresponding MOILFP problems (P_i, i = 1, 2, 3, 4, 5)

Proof. By Theorem (2.2), the proof is trivial.

Theorem 3.2. Let ${\bar{x}}^{M}, {\bar{x}}^{UU}, {\bar{x}}^{LU}, {\bar{x}}^{UL} and {\bar{x}}^{LL} \in R^{n}$ be an efficient solutions of P₁, P₂, P₃, P₄ and P₅ respectively, then

${\tilde{x}}^{R} = [({\bar{x}}^{LL}, {\bar{x}}^{M}, {\bar{x}}^{UL}) : ({\bar{x}}^{LU}, {\bar{x}}^{M}, {\bar{x}}^{UU})]$ is an efficient fuzzy rough solution of the problem (3.1).

Proof. Let ${\tilde{y}}^{R} = [(y^{LL}, y^{M}, y^{UL}) : (y^{LU}, y^{M}, y^{UU})]$ be a feasible solution of the problem (3.1). Clearly, y^M, y^UU, y^LU, y^UL and y^LL are feasible solutions of P₁, P₂, P₃, P₄ and P₅ respectively,

Now, since ${\bar{x}}^{M}, {\bar{x}}^{UU}, {\bar{x}}^{LU}, {\bar{x}}^{UL} and {\bar{x}}^{LL}$ be an efficient solutions of P₁, P₂, P₃, P₄ and P₅ respectively, then

Z_i $({\bar{x}}^{M})$ ≥ Z_i (y^M), Z_i $({\bar{x}}^{UU})$ ≥ Z_i (y^UU), Z_i $({\bar{x}}^{LU})$ ≥ Z_i (y^LU), Z_i $({\bar{x}}^{UL})$ ≥ Z_i (y^UL), and Z_i $({\bar{x}}^{LL})$ ≥ Z_i (y^LL) for all i = 1, 2, . . ., k and Z_i $({\bar{x}}^{M})$ > Z_i (y^M),

Z_j $({\bar{x}}^{UU})$ > Z_j (y^UU), Z_j $({\bar{x}}^{LU})$ > Z_j (y^LU), Z_j $({\bar{x}}^{UL})$ > Z_j (y^UL), and Z_j $({\bar{x}}^{LL})$ > Z_j (y^LL) for at least one j.

This implies that $Z_{i}^{*} ({\tilde{x}}^{R}) \geq Z_{i}^{*} ({\tilde{y}}^{R})$ for all

i = 1, 2, . . . , k and $Z_{i}^{*} ({\tilde{x}}^{R}) > Z_{i}^{*} ({\tilde{y}}^{R})$ for at least one j.

Therefore, ${\tilde{x}}^{R} = [({\bar{x}}^{LL}, {\bar{x}}^{M}, {\bar{x}}^{UL}) : ({\bar{x}}^{LU}, {\bar{x}}^{M}$ , ${\bar{x}}^{UU})$ is an efficient fuzzy rough solution to the given problem (3.1).

4 Algorithm solution for FRMOILFP problem

The algorithm for solving FRMOILFP problem can be constructed as follows:

Step 1. Convert the problem to the form of FRMOILFP problem (3.4).

Step 2. Transfer the problem (3.4) to five problems with forms, P₁, P₂, P₃, P₄ and P₅, which are MOILFP problems.

Step 3. Find the maximum value of each objective function of the problems P₁, P₂, P₃, P₄ and $P_{5} as : (Z_{i})^{*} = \frac{N_{i} (x^{*})}{D_{i} (x^{*})} = {Max}_{x \in X} \frac{N_{i} (x^{)}}{D_{i} (x^{)}} .$

Step 4. Use the weighting method to convert each problem P₁, P₂, P₃, P₄ and P₅ with a single objective in the form $P_{1}^{'}, P_{2}^{'}, P_{3}^{'}, P_{4}^{'}$ and $P_{5}^{'}$ respectively.

Step 5. Find the optimal solution of each integer linear programming ILP problems $P_{1}^{'}, P_{2}^{'}, P_{3}^{'}$ , $P_{4}^{'} and P_{5}^{'}$ .

Step 6. Using the results of step 5, obtain an efficient solution to the given FRMOILFP problem by the Theorems (3.1) and (3.2) with objective value: ${\tilde{Z}}_{i}^{R} (\bar{x}) = [(Z_{i}^{LL} (\bar{x}), Z_{i}^{M} (\bar{x}), Z_{i}^{UL} (\bar{x}))$ : $(Z_{i}^{LU} (\bar{x}), Z_{i}^{M} (\bar{x}), Z_{i}^{UU} (\bar{x}))],$ $i = 1, 2, . . . . . ., k .$

5 Advantages of the proposed method

A fuzzy rough linear fractional programming (FRLFP) is slightly similar to the existing fuzzy linear fractional programming (FLFP) [13 , 20] in terms of the capability in handling complex uncertainties, it is necessary to compare their effectiveness in dealing with uncertainty problems. By comparing the objective values of FRLFP (lower and upper fuzzy number) with that of FLFP, it is illustrated that all FLFP objective values fall into the upper objective values of the FRLFP, indicating the reliability of FRLFP in searching for all efficient solutions. In other words, the upper objective values always cover all objective values for the same problem whatever the solution method is used. Comparisons between efficient solutions of decision variables from FRLFP and FLFP also indicate that all efficient solutions obtained from FLFP are involved in the upper efficient solution of FRLFP. It has to be mentioned that, the efficient solutions obtained from FLFP show only the lower efficient solutions of FRLFP, indicating that numerous information of uncertain parameters will be lost in the optimization process. However, the solution method of FRLFP can generate efficient solutions in the form of two fuzzy numbers (lower efficient solution and upper efficient solution), which embeds more information derived from the uncertain modeling inputs. Therefore, less information would be lost within the computation process of FRLFP than FLFP.

6 Numerical examples

Example 6.1. Let us consider the following FRMOILFP problem: $Max (\begin{matrix} Z_{1} (x) = \frac{{\tilde{8}}^{R} {\tilde{x}}_{1}^{R} \oplus {\tilde{7}}^{R} {\tilde{x}}_{2}^{R}}{{\tilde{4}}^{R} {\tilde{x}}_{1}^{R} \oplus {\tilde{9}}^{R}}, \\ Z_{2} (x) = \frac{{\tilde{4}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{5}}^{R} {\tilde{x}}_{2}^{R}}{{\tilde{3}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{3}}^{R} {\tilde{x}}_{2}^{R} + {\tilde{9}}^{R}} \end{matrix})$ Subject to: $\begin{matrix} \begin{matrix} [(2, 3, 4) : (1, 3, 5)] {\tilde{x}}_{1}^{R} \oplus [(3, 4, 4) : (2, 4, 4)] \\ {\tilde{x}}_{2}^{R} \leq [(50, 60, 70) : 40, 60, 80)] \\ [(4, 5, 6) : (3, 5, 7)] {\tilde{x}}_{1}^{R} \oplus [(3, 4, 5) : (2, 4, 5)] \\ {\tilde{x}}_{2}^{R} \leq [(60, 75, 80) : (55, 75, 90)] \\ {\tilde{x}}_{1}^{R}, {\tilde{x}}_{2}^{R} \geq {\tilde{0}}^{R} and integer . \end{matrix} \end{matrix}$

Where ${\tilde{8}}^{R} = [(7, 8, 9) : (6, 8, 10)], {\tilde{7}}^{R} = [(6, 7, 8) : (5, 7, 9)],$ ${\tilde{4}}^{R} = [(3, 4, 5) : [(2, 4, 6)], {\tilde{9}}^{R} = [(8, 9, 10) : (7, 9, 11)],$ ${\tilde{5}}^{R} = [(4, 5, 6) : (3, 5, 7)], {\tilde{3}}^{R} = [2, 3, 4) : (1, 3, 5)] .$

Solution: Let us assume that: ${\tilde{X}}_{1}^{R} = [{\tilde{X}}_{1}^{L} : {\tilde{X}}_{1}^{U}] = [(x_{1}^{LL}, x_{1}^{M}, x_{1}^{UL}) : (x_{1}^{LU}, x_{1}^{M}, x_{1}^{UU})]$ ${\tilde{X}}_{2}^{R} = [{\tilde{X}}_{2}^{L} : {\tilde{X}}_{2}^{U}] = [(x_{2}^{LL}, x_{2}^{M}, x_{2}^{UL}) : (x_{2}^{LU}, x_{2}^{M}, x_{2}^{UU})]$

The above FRMOILFP problem can be converted into the following MOILFP problem: $Max (\begin{matrix} Z_{1}^{M} = \frac{8 x_{1}^{M} + 7 x_{2}^{M}}{4 x_{1}^{M} + 9}, Z_{2}^{M} = \frac{4 x_{1}^{M} + 5 x_{2}^{M}}{3 x_{1}^{M} + 3 x_{2}^{M} + 9} \\ Z_{1}^{UU} = \frac{10 x_{1}^{UU} + 9 x_{2}^{UU}}{2 x_{1}^{LU} + 7}, Z_{2}^{UU} = \frac{6 x_{1}^{UU} + 7 x_{2}^{UU}}{x_{1}^{LU} + x_{2}^{LU} + 7} \\ Z_{1}^{LU} = \frac{6 x_{1}^{LU} + 5 x_{2}^{LU}}{6 x_{1}^{UU} + 11}, Z_{2}^{LU} = \frac{2 x_{1}^{LU} + 3 x_{2}^{LU}}{5 x_{1}^{UU} + 5 x_{2}^{UU} + 11} \\ Z_{1}^{UL} = \frac{9 x_{1}^{UL} + 8 x_{2}^{UL}}{3 x_{1}^{LL} + 8}, Z_{2}^{UL} = \frac{5 x_{1}^{UL} + 6 x_{2}^{UL}}{2 x_{1}^{LL} + 2 x_{2}^{LL} + 8} \\ Z_{1}^{LL} = \frac{7 x_{1}^{LL} + 6 x_{2}^{LL}}{5 x_{1}^{UL} + 10}, Z_{2}^{LL} = \frac{3 x_{1}^{LL} + 4 x_{2}^{LL}}{4 x_{1}^{UL} + 4 x_{2}^{UL} + 10} \end{matrix})$

Subject to: $3 x_{1}^{M} + 4 x_{2}^{M} \leq 60, 5 x_{1}^{M} + 4 x_{2}^{M} \leq 75$

$5 x_{1}^{UU} + 4 x_{2}^{UU} \leq 80, 7 x_{1}^{UU} + 5 x_{2}^{UU} \leq 90$

$x_{1}^{LU} + 2 x_{2}^{LU} \leq 40, 3 x_{1}^{LU} + 2 x_{2}^{LU} \leq 55$

$\begin{matrix} 4 x_{1}^{UL} + 4 x_{2}^{UL} \leq 70, 6 x_{1}^{UL} + 5 x_{2}^{UL} \leq 80 \\ 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 50, 4 x_{1}^{LL} + 3 x_{2}^{LL} \leq 60 \end{matrix}$

$x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL}, x_{1}^{M}, x_{2}^{M}, x_{1}^{LU}, x_{2}^{LU}, x_{1}^{UU}, x_{2}^{UU} \geq 0$ and integer.

From the above problem, five multiobjective integer linear fractional programming problem can be constructed and solved as follows: $\begin{matrix} P_{1} : Max (\begin{matrix} Z_{1}^{M} = \frac{8 x_{1}^{M} + 7 x_{2}^{M}}{4 x_{1}^{M} + 9}, \\ Z_{2}^{M} = \frac{4 x_{1}^{M} + 5 x_{2}^{M}}{3 x_{1}^{M} + 3 x_{2}^{M} + 9} \end{matrix}) \\ Subject to : \\ 3 x_{1}^{M} + 4 x_{2}^{M} \leq 60, 5 x_{1}^{M} + 4 x_{2}^{M} \leq 75, \\ x_{1}^{M}, x_{2}^{M} \geq 0 and integer . \end{matrix}$

It is observed that $Z_{1}^{M} (x) \leq 11.66, Z_{2}^{M} (x) \leq 1.38$ .

This MOILFP problem is equivalent to the following ILP problem by Theorem (2.2): $\begin{matrix} P_{1}^{'} : Max (\begin{matrix} w_{1} [8 x_{1}^{M} + 7 x_{2}^{M} - 11.66 (4 x_{1}^{M} + 9)] + \\ w_{2} [4 x_{1}^{M} + 5 x_{2}^{M} - 1.38 (3 x_{1}^{M} + 3 x_{2}^{M} + 9)] \end{matrix}) \end{matrix}$ $\begin{matrix} Subject to : \\ 3 x_{1}^{M} + 4 x_{2}^{M} \leq 60, 5 x_{1}^{M} + 4 x_{2}^{M} \leq 75 \\ x_{1}^{M}, x_{2}^{M} \geq 0 and integer . \end{matrix}$

For w₁, w₂ = 0.5, this integer linear programming problem $(P_{1}^{'})$ has been solved using the branch and bound method and the optimal integer solution has been found $(x_{1}^{M} = 0, x_{2}^{M} = 15)$ and then this optimal integer solution is an efficient solution of MOILFP problem (P₁) by Theorem (3.1) with objective values $Z_{1}^{M} (\bar{x}) = 11.66$ and $Z_{3}^{M} (\bar{x}) = 1.38$ . $P_{2} : Max (\begin{matrix} Z_{1}^{UU} = \frac{10 x_{1}^{UU} + 9 x_{2}^{UU}}{2 x_{1}^{UU} + 7}, \\ Z_{2}^{UU} = \frac{6 x_{1}^{UU} + 7 x_{2}^{UU}}{x_{1}^{LU} + x_{2}^{LU} + 7} \end{matrix})$ Subject to: $\begin{matrix} \frac{10 x_{1}^{UU} + 9 x_{2}^{UU}}{2 x_{1}^{LU} + 7} \geq 11.66, \frac{6 x_{1}^{UU} + 7 x_{2}^{UU}}{x_{1}^{LU} + x_{2}^{LU} + 7} \geq 1.38 \\ 5 x_{1}^{UU} + 4 x_{2}^{UU} \leq 80, 7 x_{1}^{UU} + 5 x_{2}^{UU} \leq 90 \\ x_{1}^{LU} + 2 x_{2}^{LU} \leq 40, 3 x_{1}^{LU} + 2 x_{2}^{LU} \leq 55 \\ x_{1}^{UU} \geq 0, x_{2}^{UU} \geq 15 \\ x_{1}^{UU}, x_{2}^{UU}, x_{1}^{LU}, x_{2}^{LU} \geq 0 and integer . \end{matrix}$

It is observed that $Z_{1}^{UU} (x) \leq 23.14$ and $Z_{2}^{UU} (x) \leq 18$

This MOILFP problem is equivalent to the following ILP problem by Theorem (2.2): $P_{2}^{'} : Max (\begin{matrix} w_{1} [10 x_{1}^{UU} + 9 x_{2}^{UU} - 23.14 (2 x_{1}^{LU} + 7)] + \\ w_{2} [6 x_{1}^{UU} + 7 x_{2}^{UU} - 18 (x_{1}^{LU} + x_{2}^{LU} + 7)] \end{matrix})$ Subject to: $\begin{matrix} \frac{10 x_{1}^{UU} + 9 x_{2}^{UU}}{2 x_{1}^{LU} + 7} \geq 11.66, \frac{6 x_{1}^{UU} + 7 x_{2}^{UU}}{x_{1}^{LU} + x_{2}^{LU} + 7} \geq 1.38 \\ 5 x_{1}^{UU} + 4 x_{2}^{UU} \leq 80, 7 x_{1}^{UU} + 5 x_{2}^{UU} \leq 90 \\ x_{1}^{LU} + 2 x_{2}^{LU} \leq 40, 3 x_{1}^{LU} + 2 x_{2}^{LU} \leq 55 \\ x_{1}^{UU} \geq 0, x_{2}^{UU} \geq 15 \\ x_{1}^{LU} \leq 0, x_{2}^{LU} \leq 15 \\ x_{1}^{UU}, x_{2}^{UU}, x_{1}^{LU}, x_{2}^{LU} \geq 0 and integer . \end{matrix}$

For w₁, w₂ = 0.5, this integer linear programming problem $(P_{2}^{'})$ has been solved using the branch and bound method and the optimal integer solution has been found $(x_{1}^{LU}, x_{2}^{LU}, x_{1}^{UU} = 0 and x_{2}^{UU} = 18)$ and then this optimal integer solution is an efficient solution of MOILFP problem (P₂) by Theorem (3.1) with objective values $Z_{1}^{UU} (\bar{x}) = 23.14$ and $Z_{2}^{UU} (\bar{x}) = 18$ . $P_{3} : Max (\begin{matrix} Z_{1}^{LU} = \frac{6 x_{1}^{LU} + 5 x_{2}^{LU}}{6 x_{1}^{UU} + 11}, \\ Z_{2}^{LU} = \frac{2 x_{1}^{LU} + 3 x_{2}^{LU}}{5 x_{1}^{UU} + 5 x_{2}^{UU} + 11} \end{matrix})$ Subject to: $\frac{6 x_{1}^{LU} + 5 x_{2}^{LU}}{6 x_{1}^{UU} + 11} \leq 11.6, \frac{2 x_{1}^{LU} + 3 x_{2}^{LU}}{5 x_{1}^{UU} + 5 x_{2}^{UU} + 11} \leq 1.38$ $5 x_{1}^{UU} + 4 x_{2}^{UU} \leq 80, 7 x_{1}^{UU} + 5 x_{2}^{UU} \leq 90$ $x_{1}^{LU} + 2 x_{2}^{LU} \leq 40, 3 x_{1}^{LU} + 2 x_{2}^{LU} \leq 55$ $\begin{matrix} x_{1}^{UU} \geq 0, x_{2}^{UU} \geq 15, x_{1}^{LU} \leq 0, x_{2}^{LU} \leq 15 \\ x_{1}^{UU}, x_{2}^{UU}, x_{1}^{LU}, x_{2}^{LU} \geq 0 and integer . \end{matrix}$

Now, substituting $(x_{1}^{LU}, x_{2}^{LU}, x_{1}^{UU} = 0 and$ $x_{2}^{UU} = 18)$ in (P₃), then the efficient solution of MOILFP problem (P₃) is $(x_{1}^{LU}, x_{2}^{LU}, x_{1}^{UU} = 0 and x_{2}^{UU} = 18)$ with objective values $Z_{1}^{LU} (\tilde{x}) = 0 and Z_{2}^{LU} (\tilde{x}) = 0$ . $P_{4} : Max (\begin{matrix} Z_{1}^{UL} = \frac{9 x_{1}^{UL} + 8 x_{2}^{UL}}{3 x_{1}^{LL} + 8}, \\ Z_{2}^{UL} = \frac{5 x_{1}^{UL} + 6 x_{2}^{UL}}{2 x_{1}^{LL} + 2 x_{2}^{LL} + 8} \end{matrix})$ Subject to: $11.66 \leq \frac{9 x_{1}^{UL} + 8 x_{2}^{UL}}{3 x_{1}^{LL} + 8} \leq 23.14,$ $1.38 \leq \frac{5 x_{1}^{UL} + 6 x_{2}^{UL}}{2 x_{1}^{LL} + 2 x_{2}^{LL} + 8} \leq 18$ $\begin{matrix} 4 x_{1}^{UL} + 4 x_{2}^{UL} \leq 70, 6 x_{1}^{UL} + 5 x_{2}^{UL} \leq 80 \\ 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 50, 4 x_{1}^{LL} + 3 x_{2}^{LL} \leq 60 \end{matrix}$ $\begin{matrix} x_{1}^{UL} \geq 0, x_{2}^{UL} \geq 15 \\ x_{1}^{LL} \leq 0, x_{2}^{LL} \leq 15 \\ x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL} \geq 0 and integer . \end{matrix}$

It is observed that $Z_{1}^{UL} (x) \leq 16 and Z_{2}^{UL} (x) \leq 12 .$

This MOILFP problem is equivalent to the following ILP problem by Theorem (2.2): $P_{4}^{'} : Max (\begin{matrix} w_{1} [9 x_{1}^{UL} + 8 x_{2}^{UL} - 16 (3 x_{1}^{LL} + 8)] + \\ w_{2} [5 x_{1}^{UL} + 6 x_{2}^{UL} - 12 (2 x_{1}^{LL} + 2 x_{2}^{LL} + 8)] \end{matrix})$

Subject to: $11.66 \leq \frac{9 x_{1}^{UL} + 8 x_{2}^{UL}}{3 x_{1}^{LL} + 8} \leq 23.14$ $1.38 \leq \frac{5 x_{1}^{UL} + 6 x_{2}^{UL}}{2 x_{1}^{LL} + 2 x_{2}^{LL} + 8} \leq 18$ $\begin{matrix} 4 x_{1}^{UL} + 4 x_{2}^{UL} \leq 70, 6 x_{1}^{UL} + 5 x_{2}^{UL} \leq 80 \\ 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 50, 4 x_{1}^{LL} + 3 x_{2}^{LL} \leq 60 \end{matrix}$ $\begin{matrix} x_{1}^{UL} \geq 0, x_{2}^{UL} \geq 15 \\ x_{1}^{LL} \leq 0, x_{2}^{LL} \leq 15 \\ x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL} \geq 0 and integer . \end{matrix}$

For w₁, w₂ = 0.5, this integer linear programming problem $(P_{2}^{'})$ has been solved using the branch and bound method and the optimal integer solution has been found $(x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, = 0 and x_{2}^{UL} = 16)$ and then this optimal integer solution is an efficient solution of MOILFP problem (P₄) by Theorem (3.1) with objective value: $Z_{1}^{UL} (\tilde{x}) = 16 and Z_{2}^{UL} (\tilde{x}) = 12 .$ $P_{5} : Max (\begin{matrix} Z_{1}^{LL} = \frac{7 x_{1}^{LL} + 6 x_{2}^{LL}}{5 x_{1}^{UL} + 10}, \\ Z_{2}^{LL} = \frac{3 x_{1}^{LL} + 4 x_{2}^{LL}}{4 x_{1}^{UL} + 4 x_{2}^{UL} + 10} \end{matrix})$ Subject to: $0 \leq \frac{7 x_{1}^{LL} + 6 x_{2}^{LL}}{5 x_{1}^{UL} + 10} \leq 11.66,$ $0 \leq \frac{3 x_{1}^{LL} + 4 x_{2}^{LL}}{4 x_{1}^{UL} + 4 x_{2}^{UL} + 10} \leq 1.38$ $\begin{matrix} 4 x_{1}^{UL} + 4 x_{2}^{UL} \leq 70, 6 x_{1}^{UL} + 5 x_{2}^{UL} \leq 80 \\ 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 50, 4 x_{1}^{LL} + 3 x_{2}^{LL} \leq 60 \end{matrix}$ $\begin{matrix} x_{1}^{UL} \geq 0, x_{2}^{UL} \geq 15 \\ x_{1}^{LL} \leq 0, x_{2}^{LL} \leq 15 \\ x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL} \geq 0 and integer . \end{matrix}$

Now, substituting $(x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL} = 0 and x_{2}^{UL} = 16)$ in (P₅), then the efficient solution of MOILFP problem (P₅) is $(x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL} = 0 and x_{2}^{UL} = 16)$ with objective values $Z_{1}^{LL} (\bar{x}) = 0 and Z_{2}^{LL} (\bar{x}) = 0$

Then by Theorem (3.2) the efficient solution of original FRMOILFP problem is: ${\tilde{X}}_{1}^{R} = [(0, 0, 0) : (0, 0, 0)],$ ${\tilde{X}}_{2}^{R} = [(0, 15, 16) : (0, 15, 18)]$ with objective values: ${\tilde{Z}}_{1}^{R} = [(0, 11.66, 16) : (0, 11.66, 23.14)]$ , ${\tilde{Z}}_{2}^{R} = [(0, 1.38, 12) : (0, 1.38, 18)]$ .

The membership functions of ${\tilde{Z}}_{1}^{R}$ and ${\tilde{Z}}_{2}^{R}$ are shown in Figs. 2 and 3 respectively.

Fig.2

Membership function of ${\tilde{Z}}_{1}^{R}$ .

Fig.3

Membership function of ${\tilde{Z}}_{2}^{R}$ .

Example 6.2. Solve the following FRMOILFP problem: $Max (\begin{matrix} {\tilde{Z}}_{1}^{R} (x) = \frac{{\tilde{5}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{7}}^{R} {\tilde{x}}_{2}^{R}}{{\tilde{4}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{3}}^{R} {\tilde{x}}_{2}^{R} + {\tilde{7}}^{R}}, \\ {\tilde{Z}}_{2}^{R} (x) = \frac{{\tilde{6}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{5}}^{R} {\tilde{x}}_{2}}{{\tilde{2}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{7}}^{R} {\tilde{x}}_{2}^{R} + {\tilde{2}}^{R}}, \\ {\tilde{Z}}_{3}^{R} (x) = \frac{{\tilde{8}}^{R} {\tilde{x}}_{1}^{R} \oplus {\tilde{7}}^{R} {\tilde{x}}_{2}^{R}}{{\tilde{4}}^{R} {\tilde{x}}_{1}^{R} \oplus {\tilde{9}}^{R}} \end{matrix})$ Subject to: ${\tilde{3}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{4}}^{R} {\tilde{x}}_{2}^{R} {\tilde{\leq}}^{R} {\tilde{28}}^{R}$

${\tilde{6}}^{R} {\tilde{x}}_{1}^{R} + {\tilde{3}}^{R} {\tilde{x}}_{2}^{R} {\tilde{\leq}}^{R} {\tilde{24}}^{R}$

${\tilde{x}}_{1}^{R} \geq {\tilde{1}}^{R}$

${\tilde{x}}_{1}^{R} and {\tilde{x}}_{2}^{R} {\tilde{\geq}}^{R} {\tilde{0}}^{R} and integer .$

Where $\begin{matrix} {\tilde{5}}^{R} = [(2, 5, 6) : (1, 5, 7)], {\tilde{1}}^{R} = [(0, 1, 2) : (0, 1, 13)], \\ {\tilde{7}}^{R} = [(5, 7, 9) : (3, 7, 10)], {\tilde{4}}^{R} = [(3, 4, 6) : (2, 4, 7)], \\ {\tilde{3}}^{R} = [(2, 3, 4) : (1, 3, 5)], {\tilde{6}}^{R} = [(5, 6, 7) : (3, 6, 8)], \\ {\tilde{2}}^{R} = [(2, 2, 3) : (2, 2, 5)], {\tilde{8}}^{R} = [(6, 8, 10) : (5, 8, 12)], \\ {\tilde{9}}^{R} = [(8, 9, 10) : (6, 9, 12)], \\ {\tilde{28}}^{R} = [(10, 28, 50) : (9, 28, 65)] and \\ {\tilde{24}}^{R} = [(15, 24, 45) : (10, 24, 62)] . \end{matrix}$

Solution: Let us assume that: $\begin{matrix} - 24 pt {\tilde{X}}_{1}^{R} = [{\tilde{X}}_{1}^{L} : {\tilde{X}}_{1}^{U}] = [(x_{1}^{LL} x_{1}^{M}, x_{1}^{UL}) : (x_{1}^{LU} x_{1}^{M}, x_{1}^{UU})] \\ - 24 pt {\tilde{X}}_{2}^{R} = [{\tilde{X}}_{2}^{L} : {\tilde{X}}_{2}^{U}] = [(x_{2}^{LL} x_{2}^{M}, x_{2}^{UL}) : (x_{2}^{LU} x_{2}^{M}, x_{2}^{UU})] \end{matrix}$

The above FRMOILFP problem can be converted into the following MOILFP problem: $Max (\begin{matrix} Z_{1}^{M} = \frac{5 x_{1}^{M} + 7 x_{2}^{M}}{4 x_{1}^{M} + 3 x_{2}^{M} + 7}, Z_{2}^{M} = \frac{6 x_{1}^{M} + 5 x_{2}^{M}}{2 x_{1}^{M} + 7 x_{2}^{M} + 2}, \\ Z_{3}^{M} = \frac{8 x_{1}^{M} + 7 x_{2}^{M}}{4 x_{1}^{M} + 9}, Z_{1}^{UU} = \frac{7 x_{1}^{UU} + 10 x_{2}^{UU}}{2 x_{1}^{LU} + x_{2}^{LU} + 3}, \\ Z_{2}^{UU} = \frac{8 x_{1}^{UU} + 7 x_{2}^{UU}}{x_{1}^{LU} + 3 x_{2}^{LU} + 2}, Z_{3}^{UU} = \frac{12 x_{1}^{UU} + 10 x_{2}^{UU}}{2 x_{1}^{LU} + 6}, \\ Z_{1}^{LU} = \frac{x_{1}^{LU} + 3 x_{2}^{LU}}{7 x_{1}^{UU} + 5 x_{2}^{UU} + 10}, Z_{2}^{UL} = \frac{3 x_{1}^{LU} + x_{2}^{LU}}{5 x_{1}^{UU} + 10 x_{2}^{UU} + 5}, \\ Z_{3}^{LU} = \frac{5 x_{1}^{LU} + 3 x_{2}^{LU}}{7 x_{1}^{UU} + 12}, Z_{1}^{UL} = \frac{7 x_{1}^{UL} + 9 x_{2}^{UL}}{3 x_{1}^{LL} + 2 x_{2}^{LL} + 5}, \\ Z_{2}^{UL} = \frac{7 x_{1}^{UL} + 6 x_{2}^{UL}}{x_{1}^{LL} + 5 x_{2}^{LL} + 2}, Z_{3}^{UL} = \frac{10 x_{1}^{UL} + 9 x_{2}^{UL}}{3 x_{1}^{LL} + 8}, \\ Z_{1}^{LL} = \frac{2 x_{1}^{LL} + 5 x_{2}^{LL}}{6 x_{1}^{UL} + 4 x_{2}^{UL} + 9}, Z_{2}^{LL} = \frac{5 x_{1}^{LL} + 2 x_{2}^{LL}}{3 x_{1}^{UL} + 9 x_{2}^{UL} + 3}, \\ Z_{3}^{LL} = \frac{6 x_{1}^{LL} + 5 x_{2}^{LL}}{6 x_{1}^{UL} + 10} \end{matrix})$ Subject to: $3 x_{1}^{M} + 4 x_{2}^{M} \leq 28, 6 x_{1}^{M} + 3 x_{2}^{M} \leq 24,$

$x_{1}^{M} \geq 1$

$5 x_{1}^{UU} + 7 x_{2}^{UU} \leq 65, 8 x_{1}^{UU} + 5 x_{2}^{UU} \leq 62, x_{1}^{UU} \geq 3$

$x_{1}^{LU} + 2 x_{2}^{LU} \leq 9, 3 x_{1}^{LU} + x_{2}^{LU} \leq 10, x_{1}^{LU} \geq 0$

$4 x_{1}^{UL} + 6 x_{2}^{UL} \leq 50, 7 x_{1}^{UL} + 4 x_{2}^{UL} \leq 45, x_{1}^{UL} \geq 2$

$2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 10, 5 x_{1}^{LL} + 2 x_{2}^{LL} \leq 15, x_{1}^{LL} \geq 0$

$x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL}, x_{1}^{M}, x_{2}^{M}, x_{1}^{LU}, x_{2}^{LU}, x_{1}^{UU}, x_{2}^{UU} \geq 0$ and integer.

From the above problem, five multiobjective integer linear fractional programming problem can be constructed and solved as follows: $P_{1} : Max (\begin{matrix} Z_{1}^{M} = \frac{5 x_{1}^{M} + 7 x_{2}^{M}}{4 x_{1}^{M} + 3 x_{2}^{M} + 7}, \\ Z_{2}^{M} = \frac{6 x_{1}^{M} + 5 x_{2}^{M}}{2 x_{1}^{M} + 7 x_{2}^{M} + 2}, \\ Z_{3}^{M} = \frac{8 x_{1}^{M} + 7 x_{2}^{M}}{4 x_{1}^{M} + 9} . \end{matrix})$ Subject to: $3 x_{1}^{M} + 4 x_{2}^{M} \leq 28, 6 x_{1}^{M} + 3 x_{2}^{M} \leq 24, x_{1}^{M} \geq 1$ $- 80 {ptx}_{1}^{M}, x_{2}^{M} \geq 0 and integer .$

An ILP problem, which is equivalent to the $P_{1}^{'}$ , is constructed according to the proposed algorithm as follows: $P_{1}^{'} : Max {w_{1} [5 x_{1}^{M} + 7 x_{2}^{M} - Z_{1}^{M *} (4 x_{1}^{M} + 3 x_{2}^{M} + 7)]$ $+ w_{2} [6 x_{1}^{M} + 5 x_{2}^{M} - Z_{2}^{M *} (2 x_{1}^{M} + 7 x_{2}^{M} + 2)]$ $+ w_{3} [8 x_{1}^{M} + 7 x_{2}^{M} - Z_{3}^{M *} (4 x_{1}^{M} + 9)]}$ Subject to: $3 x_{1}^{M} + 4 x_{2}^{M} \leq 28, 6 x_{1}^{M} + 3 x_{2}^{M} \leq 24, x_{1}^{M} \geq 1$ $x_{1}^{M}, x_{2}^{M} \geq 0 and integer .$

For w₁, w₂ = 0.25 and w₃ = 0.5, the optimal solution of $P_{1}^{'}$ is $x_{1}^{M} = 1, x_{2}^{M} = 6 .$

Hence, an efficient solution of the P₁ is $x_{1}^{M} = 1,$ $x_{2}^{M} = 6$ with $Z_{1}^{M} = 1.6, Z_{2}^{M} = 0.818, Z_{3}^{M} = 3.846$ $P_{2} : Max (\begin{matrix} Z_{1}^{UU} = \frac{7 x_{1}^{UU} + 10 x_{2}^{UU}}{2 x_{1}^{LU} + x_{2}^{LU} + 3}, \\ Z_{2}^{UU} = \frac{8 x_{1}^{UU} + 7 x_{2}^{UU}}{x_{1}^{LU} + 3 x_{2}^{LU} + 2}, \\ Z_{3}^{UU} = \frac{12 x_{1}^{UU} + 10 x_{2}^{UU}}{2 x_{1}^{LU} + 6} . \end{matrix})$

Subject to: $\begin{matrix} 5 x_{1}^{UU} + 7 x_{2}^{UU} \leq 65, 8 x_{1}^{UU} + 5 x_{2}^{UU} \leq 62, \\ x_{1}^{UU} \geq 3, x_{1}^{LU} + 2 x_{2}^{LU} \leq 9, 3 x_{1}^{LU} + x_{2}^{LU} \leq 10, \\ x_{1}^{LU} \geq 0, x_{1}^{UU}, x_{2}^{UU}, x_{1}^{LU}, x_{2}^{LU} \geq 0 and integer . \end{matrix}$

An ILP problem, which is equivalent to the P₂, is constructed according to the proposed algorithm as follows: $\begin{matrix} P_{2}^{'} : Max {w_{1} [7 x_{1}^{UU} + 10 x_{2}^{UU} \\ - Z_{1}^{UU *} (2 x_{1}^{LU} + x_{2}^{LU} + 3)] + w_{2} [8 x_{1}^{UU} + 7 x_{2}^{UU} \\ - Z_{2}^{UU *} (x_{1}^{LU} + 3 x_{2}^{LU} + 2)] + w_{3} [12 x_{1}^{UU} + 10 x_{2}^{UU} \\ - Z_{3}^{UU *} (2 x_{1}^{LU} + 6)]} \end{matrix}$

For w₁, w₂ = 0.25 and w₃ = 0.5, the optimal solution of $P_{2}^{'}$ is $x_{1}^{LU} = 0, x_{2}^{LU} = 0, x_{1}^{UU} = 3$ , $x_{2}^{UU} = 7,$

Hence, an efficient solution of P₂ is $x_{1}^{LU} = 0,$

$x_{2}^{LU} = 0, x_{1}^{UU} = 3, x_{2}^{UU} = 7$ with $Z_{1}^{UU *} (x) = 32.33,$ $Z_{2}^{UU} (x) = 36.5$ , $Z_{3}^{UU *} (x) = 17.66$ $P_{3} : Max (\begin{matrix} Z_{1}^{LU} = \frac{x_{1}^{LU} + 3 x_{2}^{LU}}{7 x_{1}^{UU} + 5 x_{2}^{UU} + 10}, \\ Z_{2}^{LU} = \frac{3 x_{1}^{LU} + x_{2}^{LU}}{5 x_{1}^{UU} + 10 x_{2}^{UU} + 5}, \\ Z_{3}^{LU} = \frac{5 x_{1}^{LU} + 3 x_{2}^{LU}}{7 x_{1}^{UU} + 12} . \end{matrix})$ Subject to: $\begin{matrix} 5 x_{1}^{UU} + 7 x_{2}^{UU} \leq 65, 8 x_{1}^{UU} + 5 x_{2}^{UU} \leq 62, \\ x_{1}^{UU} \geq 3, x_{1}^{LU} + 2 x_{2}^{LU} \leq 9, 3 x_{1}^{LU} + x_{2}^{LU} \leq 10, \\ x_{1}^{LU} \geq 0, x_{1}^{UU}, x_{2}^{UU}, x_{1}^{LU}, x_{2}^{LU} \geq 0 and integer . \end{matrix}$

Now, substituting $x_{1}^{LU} = 0, x_{2}^{LU} = 0, x_{1}^{UU} = 3$ $x_{2}^{UU} = 7$ in P₃, the efficient solution of P₃ is $x_{1}^{LU} = 0, x_{2}^{LU} = 0, x_{1}^{UU} = 3, x_{2}^{UU} = 7$ with $Z_{1}^{LU *} (x) = 0$ , $Z_{2}^{LU *} (x) = 0$ , $Z_{3}^{LU *} (x) = 0$ $P_{4} : Max (\begin{matrix} Z_{1}^{UL} = \frac{7 x_{1}^{UL} + 9 x_{2}^{UL}}{3 x_{1}^{LL} + 2 x_{2}^{LL} + 5}, \\ Z_{2}^{UL} = \frac{7 x_{1}^{UL} + 6 x_{2}^{UL}}{x_{1}^{LL} + 5 x_{2}^{LL} + 2}, \\ Z_{3}^{UL} = \frac{10 x_{1}^{UL} + 9 x_{2}^{UL}}{3 x_{1}^{LL} + 8} . \end{matrix})$

Subject to: $4 x_{1}^{UL} + 6 x_{2}^{UL} \leq 50, 7 x_{1}^{UL} + 4 x_{2}^{UL} \leq 45, x_{1}^{UL} \geq 2$ $\begin{matrix} 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 10, 5 x_{1}^{LL} + 2 x_{2}^{LL} \leq 15, \\ x_{1}^{LL} \geq 0, x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL} \geq 0 and integer . \end{matrix}$

An ILP problem, which is equivalent to the P₄, is constructed according to the proposed algorithm as follows: $\begin{matrix} P_{4}^{'} : Max {w_{1} [7 x_{1}^{UL} + 9 x_{2}^{UL} - Z_{1}^{UL *} (3 x_{1}^{LL} + 2 x_{2}^{LL} + 5)] \\ + w_{2} [7 x_{1}^{UL} + 6 x_{2}^{UL} - Z_{2}^{UL *} (x_{1}^{LL} + 5 x_{2}^{LL} + 2)] \\ + w_{3} [10 x_{1}^{UL} + 9 x_{2}^{UL} - Z_{3}^{UL *} (3 x_{1}^{LL} + 8)]} \end{matrix}$

Subject to: $4 x_{1}^{UL} + 6 x_{2}^{UL} \leq 50, 7 x_{1}^{UL} + 4 x_{2}^{UL} \leq 45, x_{1}^{UL} \geq 2,$ $\begin{matrix} 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 10, 5 x_{1}^{LL} + 2 x_{2}^{LL} \leq 15, x_{1}^{LL} \geq 0, \\ x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{1}^{UL} \geq 0 and integer . \end{matrix}$

For w₁, w₂ = 0.25 and w₃ = 0.5, the optimal solution of $P_{4}^{'}$ is $x_{1}^{LL} = 0, x_{2}^{LL} = 0, x_{1}^{UL} = 2 .$ $- 100 {ptx}_{2}^{UL} = 7$

Hence, an efficient solution of P₄ is $x_{1}^{LL} = 0$ , $x_{2}^{LL} = 0, x_{1}^{UL} = 2, x_{2}^{UL} = 7,$ with $Z_{1}^{UL *} = 15.4$ , $Z_{2}^{UL *} = 28, Z_{3}^{UL *} = 10.375$ . $P_{5} : Max (\begin{matrix} Z_{1}^{LL} = \frac{2 x_{1}^{LL} + 5 x_{2}^{LL}}{6 x_{1}^{UL} + 4 x_{2}^{UL} + 9}, \\ Z_{2}^{LL} = \frac{5 x_{1}^{LL} + 2 x_{2}^{LL}}{3 x_{1}^{UL} + 9 x_{2}^{UL} + 3}, \\ Z_{3}^{LL} = \frac{6 x_{1}^{LL} + 5 x_{2}^{LL}}{6 x_{1}^{UL} + 10} \end{matrix})$

Subject to: $\begin{matrix} 4 x_{1}^{UL} + 6 x_{2}^{UL} \leq 50, 7 x_{1}^{UL} + 4 x_{2}^{UL} \leq 45, \\ x_{1}^{UL} \geq 2, 2 x_{1}^{LL} + 3 x_{2}^{LL} \leq 10, 5 x_{1}^{LL} + x_{2}^{LL} \leq 15, \\ x_{1}^{LL} \geq 0, \\ x_{1}^{LL}, x_{2}^{LL}, x_{1}^{UL}, x_{2}^{UL} \geq 0 and integer . \end{matrix}$

Now, substituting $x_{1}^{LL}$ = 0, $x_{2}^{LL}$ = 0, $x_{1}^{UL}$ = 3, $x_{2}^{UL}$ = 7 in P₅, the efficient solution of P₅ is $x_{1}^{LL}$ = 0, $x_{2}^{LL}$ = 0, $x_{1}^{UL}$ = 3, $x_{2}^{UL}$ = 7 with $Z_{1}^{LL *}$ (x) = 0, $Z_{2}^{LL *}$ (x) = 0, $Z_{3}^{LL *}$ (x) = 0.

Finally, by Theorem (3.2) the efficient solution of original FRMOILFP problem is:

${\tilde{x}}_{1}^{R} = [(0, 1, 2) : (0, 1, 3)], {\tilde{x}}_{2}^{R} = [(0, 6, 7) : (0, 6,$ 7)] with the objective values $\begin{matrix} {\tilde{Z}}_{1}^{R} = [(0, 1.6, 15.4) : (0, 1.6, 32.33)] \\ {\tilde{Z}}_{2}^{R} = [(0, 0.818, 32.5) : (0, 0.818, 36.5)] \\ {\tilde{Z}}_{3}^{R} = [(0, 3.846, 10.375) : (0, 3.846, 17.66)] \end{matrix}$

By comparing the results of the proposed method with existing methods, we can conclude that all objective values of the existing methods fall into the upper objective values of our proposed method, indicating the reliability of FRLFP in searching for all efficient solutions. In other words, the upper objective values always cover all objective values for the same problem whatever the solution method is used. Therefore, less information would be lost within the computation process of the proposed method than existing methods.

7 Conclusion

In this paper, a method is proposed to obtained the efficient solution of the fuzzy rough multiobjective integer linear fractional programming (FRMOILFP) problem, In the discussed method, FRMOILFP problem is transformed to a crisp multiobjective integer linear fractional programming problem and the resultant problem is converted to five multiobjective integer linear fractional programming (MOILFP) problems and the resultant problems are solved as a crisp Integer linear programming ILP problem by using Dinkelbach method. The numerical results for the real life problems obtained to validity of the proposed method. The approach appears to be promising and computationally easy to implement.

Footnotes

Acknowledgments

The authors would like to thank the Editor-in-Chief and anonymous referees for the various suggestions which have led to an improvement in both the quality and the clarity of this paper.

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