70 

 This work is licensed under a Creative Commons Attribution 4.0 International License. 

 

  

 

 

A New Approach to Solving Linear Fractional Programming Problem 

with Rough Interval Coefficients in the Objective Function 

 

 

 

 

 

 

                                     
 

Abstract 

This paper presents a linear fractional programming problem (LFPP) with rough 

interval coefficients (RICs) in the objective function. It shows that the LFPP with RICs in the 

objective function can be converted into a linear programming problem (LPP) with RICs by 

using the variable transformations. To solve this problem, we will make two LPP with interval 

coefficients (ICs). Next, those four LPPs can be constructed under these assumptions; the LPPs 

can be solved by the classical simplex method and used with MS Excel Solver. There is also 

argumentation about solving this type of linear fractional optimization programming problem. 

The derived theory can be applied to several numerical examples with its details, but we show 

only two examples for promising. 

Keywords: Linear Fractional Programming, Linear Programming, Rough Interval Function, 

Rough Interval Coefficients, Interval Coefficients. 

1. Introduction 

The essential optimization is nonlinear programming; the linear fractional programming 

problem is one of them. The coefficient on the system can commonly not be established 

exactly in various linear programming problems. The interval technique, in which unknown 

coefficients are converted into intervals, is one way to address this programming problem. 

Because many real-world problems are expressed as fractional functions, fractional 

programming has gained much control. These issues frequently arise in situations involving 

actual capital return on investment versus required capital. The objective function of a linear 

fractional programming problem is highly valuable in construction, such as economic and 

company organization. Charnes and Cooper have presented several solutions to this problem. 

Ibn Al Haitham Journal for Pure and Applied Science 

Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index 

 

Doi: 10.30526/35.2.2809 

Article history: Received 1, Februry, 2022, Accepted,7 , March, 2022, Published in April 2022. 

 

Rebaz B. Mustafa 

Department of Mathematics, College 

of Education, Salahaddin University, 

Erbil, Iraq. 

rebaz.mustafa@epu.edu.iq 

Nejmaddin A. Sulaiman 

Department of Mathematics, College 

of Education, Salahaddin University, 

Erbil, Iraq. 

nejmadin.sulaiman@su.edu.krd 

https://creativecommons.org/licenses/by/4.0/
mailto:rebaz.mustafa@epu.edu.iq


Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

71 

Charnes and Cooper’s method [1]can convert a fractional programming problem into a linear 

programming problem, which is known as linear fractional programming. [2] the notion of a 

rough interval will be applied to model dual uncertain information of different parameters. 

The related solution method for rough interval fuzzy linear programming problems with dual 

uncertain solutions will be described. 

The rough set (RS) theory's main principle is that each ambiguous notion is substituted by a 

couple of exact conceptions known as the lower and upper approximations of the vague idea. 

For an ambiguous idea RS, a lower approximation includes complete items that surely belong 

to the model RS, and an upper approximation comprises all items that possibly belong to the 

model RS. In other words, the concept's lower approximation is the union of all fundamental 

concepts contained in it. The concept's upper approximation is the union of all fundamental 

concepts with a non-empty set intersection. Pawlak[3] introduced the rough set theory in 1982 

as a strategy for managing ambiguity and uncertainty together. 

Many techniques for solving fractional programming difficulties have emerged in recent 

decades. Some works on rough programming have been developed in the last decade. [4] and 

[5] developed a new sort of rough programming recently, where they established two solution 

concepts: surely optimum solution and possibly optimal solution. [6] proposed rough intervals 

(RI) to transact with partly uncertain or ill-defined relevant variables. The rough set ideas are 

adapted to represent constant variables. It is worth noting that, at first, rough sets could only 

handle discrete objects and couldn't express continuous values. Rough intervals are a subset 

of RSs. It satisfies all of the rough set's attributes and key ideas, such as the notions of upper 

and lower approximation[7]. [8] introduced a method to solve LFPP with ICs in the objective 

function by variable transformation. The initial problem is converted into a nonlinear 

programming problem, which is finally transformed into an LPP. [7] presented a solution of 

the fully rough interval coefficients of LPP. They constructed two-interval coefficients of LPP 

and found some new solutions. [9] suggested a new method to solve the rough interval linear 

fractional programming problem and introduced two possible kinds of formula transformation 

with its proof. They obtained optimal values and solutions for the initial LFPP with RICs. 

Khalifa[10] defined a new approach to solving LFPP with RICs in the objective function when 

the problem can be transformed into a series of LPPs with RI under some assumptions. We 

can be used our approach of multi-objective linear fractional programming problem or multi-

objective linear programming problem with RICs, when maybe changed the coefficients of 

examples from[11],[12], and [13] into RICs. 

In this paper, the concentration of our argumentation and investigation promotes an approach 

to finding surely and possibly optimal solutions to a linear fractional programming problem 

with rough interval coefficients in the objective function.  

The remainder of the paper is laid out as follows. Section 2 presents some fundamental 

understanding of rough intervals. The formulation and derivation of LFPP with RICs are 

described in section 3. Two numerical examples are put to the test in Section 4. Finally, in 

section 5, there are some concluding observations. 

  



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

72 

2. Definitions  

Definition: 2.1. Let 𝐸 denotes a compact set of real numbers. A rough interval 𝐴 ⊆ 𝐸 is 

defined  

𝐴 = (𝐴∗, 𝐴
∗ ), where 𝐴∗ ⊆  𝐴

∗, 𝐴∗ and 𝐴
∗are conventional lower and upper approximation 

intervals of𝐴, respectively. 

𝐴∗ = [𝑎∗
𝑙 , 𝑎∗

𝑢 ]and𝐴∗ = [𝑎∗𝑙 , 𝑎∗𝑢]. Where 𝑎∗
𝑙 , 𝑎∗

𝑢, 𝑎∗𝑙  and 𝑎∗𝑢 are real numbers. 

Note that the intervals 𝐴∗ and 𝐴
∗ are not the complement each other. 

The arithmetic operations on rough intervals are based on interval arithmetic. Some of these 

mathematical procedures will be explained as follows: 

Let 𝐴 = ([𝑎∗
𝑙 , 𝑎∗

𝑢], [ 𝑎∗𝑙 , 𝑎∗𝑢] ) and 𝐵 = ([𝑏∗
𝑙 , 𝑏∗

𝑢 ], [ 𝑏∗𝑙 , 𝑏∗𝑢 ] ) be two rough intervals. Then we 

have: 

𝐴 + 𝐵 =  ([𝑎∗
𝑙 + 𝑏∗

𝑙 , 𝑎∗
𝑢 + 𝑏∗

𝑢], [ 𝑎∗𝑙 + 𝑏∗𝑙 , 𝑎∗𝑢 + 𝑏∗𝑢] ) 

𝐴 − 𝐵 =  ([𝑎∗
𝑙 − 𝑏∗

𝑢, 𝑎∗
𝑢 − 𝑏∗

𝑙 ], [ 𝑎∗𝑙 − 𝑏∗𝑢, 𝑎∗𝑢 − 𝑏∗𝑙 ] ) 

𝐴 × 𝐵 =  ([𝑎∗
𝑙 × 𝑏∗

𝑙 , 𝑎∗
𝑢 × 𝑏∗

𝑢], [ 𝑎∗𝑙 × 𝑏∗𝑙 , 𝑎∗𝑢 × 𝑏∗𝑢 ] ) 

𝐴 ÷ 𝐵 =  ([𝑎∗
𝑙 ÷ 𝑏∗

𝑢, 𝑎∗
𝑢 ÷ 𝑏∗

𝑙 ], [ 𝑎∗𝑙 ÷ 𝑏∗𝑢, 𝑎∗𝑢 ÷ 𝑏∗𝑙 ]) 

Definition: 2.2. Function 𝑓: 𝑅𝑛 → 𝐴 is called a rough interval function with 𝑓(𝑥) =

(𝑓∗(𝑥), 𝑓
∗(𝑥)), where for every𝑥 ∈ 𝑅𝑛, 𝑓∗(𝑥), 𝑓

∗(𝑥) are lower and upper approximation 

interval valued functions. 

Definition: 2.3.  A feasible point 𝑥∗ ∈ 𝑆 is said to be an optimal solution of optimization 

problem LFPP with RICs, if there does not exist 𝑥 ∈ 𝑆, such that𝑓(𝑥∗) ≤ 𝑓(𝑥). Where 𝑆 =

{𝑥 ∈ 𝑅𝑛: 𝐴𝑥 ≤ 𝑏, 𝑥 ≥ 0} 

3. Formulation of the problem 

The generally extended form of a LFPP with RICs in the objective function is as follows: 

𝑀𝑎𝑥. 𝑍 =  
∑ {(𝐴∗𝑖 , 𝐴𝑖

∗)𝑥𝑖 +
𝑘
𝑖=1 (𝐴∗𝑘+1, 𝐴𝑘+1

∗ )}

∑ {(𝐵∗𝑖 , 𝐵𝑖
∗)𝑥𝑖 +

𝑘
𝑖=1 (𝐵∗𝑘+1, 𝐵𝑘+1

∗ )}
 

Subject to:                                                                                                                                           (1) 

∑ 𝐴𝑖 𝑥𝑖 ≤ 𝑏

𝑘

𝑖=1

 

For each  𝑥𝑖 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

Where 𝐴𝑖 , 𝑖 = 1,2, … , 𝑘, and 𝑏 both are m-dimensional constant column vectors.  



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

73 

Assume that∑ (𝐵∗𝑖 , 𝐵𝑖
∗)𝑥𝑖 +

𝑘
𝑖=1 (𝐵∗𝑘+1, 𝐵𝑘+1

∗ )  > 0, for all𝑥𝑖 ∈ 𝑆, 𝑖 = 1,2, … , 𝑘, where 𝑆 is the 

compact feasible region of the problem (1). 

In the deriving theory for solving the problem (1), we introduce the variable, t=
1

∑ (B∗i,Bi
∗)xi+

k
i=1 (B∗k+1,Bk+1

∗ )
 , to transform LFP problem with RICs in the objective function into 

a rough interval linear programming problem by the method of Charnes and Cooper[1], and 

then we have: 

𝑀𝑎𝑥. 𝑍 = ∑ (𝐴∗𝑖 , 𝐴𝑖
∗)𝑥𝑖 𝑡 +

𝑘
𝑖=1 (𝐴∗𝑘+1, 𝐴𝑘+1

∗ )𝑡  

Subject to: 

∑ (𝐵∗𝑖 , 𝐵𝑖
∗)𝑥𝑖 𝑡 +

𝑘
𝑖=1 (𝐵∗𝑘+1, 𝐵𝑘+1

∗ )𝑡 = 1                                                                                (2) 

∑ 𝐴𝑖 𝑥𝑖 𝑡 − 𝑏𝑡 ≤ 0
𝑘
𝑖=1   

𝑥𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

 By introducing variable 𝑢𝑖 = 𝑥𝑖 𝑡, 𝑖 = 1,2, … , 𝑘, problem (2) is transformed into the following 

equivalent problem:. 

𝑀𝑎𝑥. 𝑍 = ∑ (𝐴∗𝑖 , 𝐴𝑖
∗)𝑢𝑖 +

𝑘
𝑖=1 (𝐴∗𝑘+1, 𝐴𝑘+1

∗ )𝑡  

Subject to: 

∑ (𝐵∗𝑖 , 𝐵𝑖
∗)𝑢𝑖 +

𝑘
𝑖=1 (𝐵∗𝑘+1, 𝐵𝑘+1

∗ )𝑡 = 1                                      (3) 

∑ 𝐴𝑖 𝑢𝑖 − 𝑏𝑡 ≤ 0
𝑘
𝑖=1   

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

By Hamzehee et al. [7], we will build two linear programming with interval coefficients. One 

of these problems is linear programming, where all of it is coefficients are lower approximate 

to rough intervals. The other is linear programming where all of its coefficients are upper 

approximate of rough intervals. To solve problem (3), the following two linear programming 

with interval coefficients are discussed in the sequel. 

𝑀𝑎𝑥. 𝑍 = ∑ 𝐴∗𝑖 𝑢𝑖 +
𝑘
𝑖=1 𝐴∗𝑖+𝑘 𝑡  

Subject to: 

∑ 𝐵∗𝑖 𝑢𝑖 +
𝑘
𝑖=1 𝐵∗𝑖+𝑘 𝑡 = 1                                                          (4) 

∑ 𝐴𝑖 𝑢𝑖 − 𝑏𝑡 ≤ 0
𝑘
𝑖=1   

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

 

 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

74 

And  

𝑀𝑎𝑥. 𝑍 = ∑ 𝐴𝑖
∗𝑢𝑖 +

𝑘
𝑖=1 𝐴𝑖+𝑘

∗ 𝑡  

Subject to: 

∑ 𝐵𝑖
∗𝑢𝑖 +

𝑘
𝑖=1 𝐵𝑖+𝑘

∗ 𝑡 = 1                                                           (5) 

∑ 𝐴𝑖 𝑢𝑖 − 𝑏𝑡 ≤ 0
𝑘
𝑖=1   

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

In the beginning, we can derive problem (4). 

To solve the problem (4) an equivalent problem can be written: 

𝑀𝑎𝑥. 𝑍 = [𝑎∗1
𝑙 , 𝑎∗1

𝑢 ]𝑢1 + [𝑎∗2
𝑙 , 𝑎∗2

𝑢 ]𝑢2 + ⋯ + [𝑎∗𝑘
𝑙 , 𝑎∗𝑘

𝑢 ]𝑢𝑘 + [𝑎∗𝑘+1
𝑙 , 𝑎∗𝑘+1

𝑢 ]𝑡 

Subject to: 

[𝑏∗1
𝑙 , 𝑏∗1

𝑢 ]𝑢1 + [𝑏∗2
𝑙 , 𝑏∗2

𝑢 ]𝑢2 + ⋯ + [𝑏∗𝑘
𝑙 , 𝑏∗𝑘

𝑢 ]𝑢𝑘 + [𝑏∗𝑘+1
𝑙 , 𝑏∗𝑘+1

𝑢 ]𝑡 = 1                            

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0       

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

The linear combination of each region interval of the problem (4) yields the following 

problem: 

𝑀𝑎𝑥. 𝑍 = [𝛾1𝑎∗1
𝑙 + (1 − 𝛾1)𝑎∗1

𝑢 ]𝑢1 + ⋯ + [𝛾𝑘 𝑎∗𝑘
𝑙 + (1 − 𝛾𝑘 )𝑎∗𝑘

𝑢 ]𝑢𝑘 + [𝛾𝑘+1𝑎∗𝑘+1
𝑙 + (1 −

𝛾𝑘+1)𝑎∗𝑘+1
𝑢 ]𝑡  

Subject to: 

[𝛿1𝑏∗1
𝑙 + (1 − 𝛿1)𝑏∗1

𝑢 ]𝑢1 + ⋯ + [𝛿𝑘 𝑏∗𝑘
𝑙 + (1 − 𝛿𝑘 )𝑏∗𝑘

𝑢 ]𝑢𝑘 + [𝛿𝑘+1𝑏∗𝑘+1
𝑙 + (1 −

𝛿𝑘+1)𝑏∗𝑘+1
𝑢 ]𝑡 = 1                            

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0                                                                                                                                               

(6) 

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘, 𝛾𝑖 , 𝛿𝑖 ∈ [0,1], 𝑖 = 1,2, … , 𝑘 + 1. 

The equality constraint in problem (6) can be further reduced to: 

[𝛿1𝑢1(𝑏∗1
𝑙 − 𝑏∗1

𝑢 ) + ⋯ + 𝛿𝑘 𝑢𝑘 (𝑏∗𝑘
𝑙 − 𝑏∗𝑘

𝑢 ) + 𝛿𝑘+1𝑡(𝑏∗𝑘+1
𝑙 − 𝑏∗𝑘+1

𝑢 )] + 𝑏∗1
𝑢 𝑢1 + ⋯ + 𝑏∗𝑘

𝑢 𝑢𝑘 +

𝑏∗𝑘+1
𝑢 𝑡 = 1           (7) 

Since 𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘, 𝛿𝑖 ∈ [0,1], (𝑏∗𝑖
𝑢 − 𝑏∗𝑖

𝑙 ) ≥ 0, 𝑖 = 1,2, … , 𝑘 + 1. 

Therefore, (7) can be written as: 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

75 

1 ≤ 1 + [𝛿1𝑢1(𝑏∗1
𝑢 − 𝑏∗1

𝑙 ) + ⋯ + 𝛿𝑘 𝑢𝑘 (𝑏∗𝑘
𝑢 − 𝑏∗𝑘

𝑙 ) + 𝛿𝑘+1𝑡(𝑏∗𝑘+1
𝑢 − 𝑏∗𝑘+1

𝑙 )] ≤ 1 +

[𝑢1(𝑏∗1
𝑢 − 𝑏∗1

𝑙 ) + ⋯ + 𝑢𝑘 (𝑏∗𝑘
𝑢 − 𝑏∗𝑘

𝑙 ) + 𝑡(𝑏∗𝑘+1
𝑢 − 𝑏∗𝑘+1

𝑙 )]                                                                                                            

(8) 

By combining (7) and (8), the result is 

1 ≤ 𝑏∗1
𝑢 𝑢1 + ⋯ + 𝑏∗𝑘

𝑢 𝑢𝑘 + 𝑏∗𝑘+1
𝑢 𝑡 ≤ 1 + 𝑢1(𝑏∗1

𝑢 − 𝑏∗1
𝑙 ) + ⋯ + 𝑢𝑘 (𝑏∗𝑘

𝑢 − 𝑏∗𝑘
𝑙 ) + 𝑡(𝑏∗𝑘+1

𝑢 − 𝑏∗𝑘+1
𝑙 )                    

(9) 

This is then reduced to  

𝑏∗1
𝑢 𝑢1 + ⋯ + 𝑏∗𝑘

𝑢 𝑢𝑘 + 𝑏∗𝑘+1
𝑢 𝑡 ≥ 1                                                                                                                       

(10) 

And 

𝑏∗1
𝑙 𝑢1 + ⋯ + 𝑏∗𝑘

𝑙 𝑢𝑘 + 𝑏∗𝑘+1
𝑙 𝑡 ≤ 1                                                                                                                       

(11) 

Therefore, using (10) and (11), the equation (6) is converted into the following equation: 

𝑀𝑎𝑥. 𝑍 = [𝛾1𝑎∗1
𝑙 + (1 − 𝛾1)𝑎∗1

𝑢 ]𝑢1 + ⋯ + [𝛾𝑘 𝑎∗𝑘
𝑙 + (1 − 𝛾𝑘 )𝑎∗𝑘

𝑢 ]𝑢𝑘 + [𝛾𝑘+1𝑎∗𝑘+1
𝑙 + (1 −

𝛾𝑘+1)𝑎∗𝑘+1
𝑢 ]𝑡  

Subject to: 

𝑏∗1
𝑢 𝑢1 + ⋯ + 𝑏∗𝑘

𝑢 𝑢𝑘 + 𝑏∗𝑘+1
𝑢 𝑡 ≥ 1                                                (12) 

𝑏∗1
𝑙 𝑢1 + ⋯ + 𝑏∗𝑘

𝑙 𝑢𝑘 + 𝑏∗𝑘+1
𝑙 𝑡 ≤ 1                                                                                                                        

(12)                                         

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0                                                                                                                          

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘, 𝛾𝑖 ∈ [0,1], 𝑖 = 1,2, … , 𝑘 + 1. 

In addition, if we let the point (�̅�1, �̅�2, … , �̅�𝑘 , 𝑡̅) of the feasible region from the problem (12), 

with 𝛾𝑖 ∈ [0,1], (𝑎∗𝑖
𝑙 − 𝑎∗𝑖

𝑢 ) ≤ 0 for𝑖 = 1,2, … , 𝑘 + 1. Before the objective of the problem (12) 

can be written as: 

[𝛾1(𝑎∗1
𝑙 − 𝑎∗1

𝑢 )]�̅�1 + ⋯ + [𝛾𝑘 (𝑎∗𝑘
𝑙 − 𝑎∗𝑘

𝑢 )]�̅�𝑘 + [𝛾𝑘+1(𝑎∗𝑘+1
𝑙 − 𝑎∗𝑘+1

𝑢 )]𝑡̅ + 𝑎∗1
𝑢 �̅�1 + ⋯ +

𝑎∗𝑘
𝑢 �̅�𝑘 + 𝑎∗𝑘+1

𝑢 𝑡̅ ≥ [(𝑎∗1
𝑙 − 𝑎∗1

𝑢 )]�̅�1 + ⋯ + [(𝑎∗𝑘
𝑙 − 𝑎∗𝑘

𝑢 )]�̅�𝑘 + [(𝑎∗𝑘+1
𝑙 − 𝑎∗𝑘+1

𝑢 )]𝑡̅ +

𝑎∗1
𝑢 �̅�1 + ⋯ + 𝑎∗𝑘

𝑢 �̅�𝑘 + 𝑎∗𝑘+1
𝑢 𝑡̅  

= 𝑎∗1
𝑙 �̅�1 + ⋯ + 𝑎∗𝑘

𝑙 �̅�𝑘 + 𝑎∗𝑘+1
𝑙 𝑡 ̅  

The right-hand side of the above equality proves that 𝑎∗1
𝑙 , … , 𝑎∗𝑘

𝑙 , 𝑎∗𝑘+1
𝑙  is the surely lower 

limit of the interval coefficients in the objective function, and the same situation of the problem 

(12), with 𝛾𝑖 ∈ [0,1], (𝑎∗𝑖
𝑢 − 𝑎∗𝑖

𝑙 ) ≥ 0 for𝑖 = 1,2, … , 𝑘 + 1. Before the objective of the 

problem (12) can be written as: 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

76 

[𝛾1(𝑎∗1
𝑢 − 𝑎∗1

𝑙 )]�̅�1 + ⋯ + [𝛾𝑘 (𝑎∗𝑘
𝑢 − 𝑎∗𝑘

𝑙 )]�̅�𝑘 + [𝛾𝑘+1(𝑎∗𝑘+1
𝑢 − 𝑎∗𝑘+1

𝑙 )]𝑡̅ + 𝑎∗1
𝑙 �̅�1 + ⋯ +

𝑎∗𝑘
𝑙 �̅�𝑘 + 𝑎∗𝑘+1

𝑙 𝑡̅ ≥ [(𝑎∗1
𝑢 − 𝑎∗1

𝑙 )]�̅�1 + ⋯ + [(𝑎∗𝑘
𝑢 − 𝑎∗𝑘

𝑙 )]�̅�𝑘 + [(𝑎∗𝑘+1
𝑢 − 𝑎∗𝑘+1

𝑙 )]𝑡̅ +

𝑎∗1
𝑙 �̅�1 + ⋯ + 𝑎∗𝑘

𝑙 �̅�𝑘 + 𝑎∗𝑘+1
𝑙 𝑡̅  

= 𝑎∗1
𝑢 �̅�1 + ⋯ + 𝑎∗𝑘

𝑢 �̅�𝑘 + 𝑎∗𝑘+1
𝑢 𝑡 ̅  

The right hand side of the above equality proves that 𝑎∗1
𝑢 , … , 𝑎∗𝑘

𝑢 , 𝑎∗𝑘+1
𝑢  is the surely upper 

limit of the interval coefficient in the objective function. 

Now, the surely best optimum objective function is 𝑍 = 𝑎∗1
𝑢 𝑢1 + ⋯ + 𝑎∗𝑘

𝑢 𝑢𝑘 + 𝑎∗𝑘+1
𝑢 𝑡 and the 

surely worst optimum objective function is𝑍 = 𝑎∗1
𝑙 𝑢1 + ⋯ + 𝑎∗𝑘

𝑙 𝑢𝑘 + 𝑎∗𝑘+1
𝑙 𝑡. 

While, to get the surely worst optimum taken of the surely lower limit of the interval 

coefficients in the objective function, thus the conversions formula of LFP problems with 

rough interval coefficients in the objective function of the surely lower limit, can be written as 

follows: 

But in this case, assume that every constraint is smaller than or equal. 

 

 

 

𝑀𝑎𝑥. 𝑍 = 𝑎∗1
𝑙 𝑢1 + ⋯ + 𝑎∗𝑘

𝑙 𝑢𝑘 + 𝑎∗𝑘+1
𝑙 𝑡  

Subject to: 

𝑏∗1
𝑢 𝑢1 + ⋯ + 𝑏∗𝑘

𝑢 𝑢𝑘 + 𝑏∗𝑘+1
𝑢 𝑡 ≤ 1  

𝑏∗1
𝑙 𝑢1 + ⋯ + 𝑏∗𝑘

𝑙 𝑢𝑘 + 𝑏∗𝑘+1
𝑙 𝑡 ≤ 1                                                             (13) 

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0   

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

And while, to get the surely best optimum taken of the surely upper limit of the interval 

coefficients in the objective function, thus the conversions formula of LFP problems with 

rough interval coefficients in the objective function is 

𝑀𝑎𝑥. 𝑍 = 𝑎∗1
𝑢 𝑢1 + ⋯ + 𝑎∗𝑘

𝑢 𝑢𝑘 + 𝑎∗𝑘+1
𝑢 𝑡  

Subject to: 

𝑏∗1
𝑢 𝑢1 + ⋯ + 𝑏∗𝑘

𝑢 𝑢𝑘 + 𝑏∗𝑘+1
𝑢 𝑡 ≥ 1                                                

𝑏∗1
𝑙 𝑢1 + ⋯ + 𝑏∗𝑘

𝑙 𝑢𝑘 + 𝑏∗𝑘+1
𝑙 𝑡 ≤ 1                                                          (14)                                       

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0                                      

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 



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77 

The optimal solution (�̅�1, �̅�2, … , �̅�𝑘 , 𝑡)̅ of problems (13) and (14) is the same as the optimal 

solution of the original problem (4) which can be easily obtained by (𝑥1
∗, 𝑥2

∗, … , 𝑥𝑘
∗ ) =

(
𝑢1

�̅�
,

𝑢2

�̅�
, … ,

𝑢𝑘

�̅�
). 

For solving problem (5), using the same derivation and assumptions of problem (4), we can 

obtain problem (15) &(16), with possibly worst optimum, and best optimum. 

𝑀𝑎𝑥. 𝑍 = 𝑎1
∗𝑙 𝑢1 + ⋯ + 𝑎𝑘

∗𝑙 𝑢𝑘 + 𝑎𝑘+1
∗𝑙 𝑡  

Subject to: 

𝑏1
∗𝑢 𝑢1 + ⋯ + 𝑏𝑘

∗𝑢𝑢𝑘 + 𝑏𝑘+1
∗𝑢 𝑡 ≤ 1                                                

𝑏1
∗𝑙 𝑢1 + ⋯ + 𝑏𝑘

∗𝑙 𝑢𝑘 + 𝑏𝑘+1
∗𝑙 𝑡 ≤ 1                                                         (15) 

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0                                      

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

And 

𝑀𝑎𝑥. 𝑍 = 𝑎1
∗𝑢𝑢1 + ⋯ + 𝑎𝑘

∗𝑢𝑢𝑘 + 𝑎𝑘+1
∗𝑢 𝑡  

Subject to: 

𝑏1
∗𝑢 𝑢1 + ⋯ + 𝑏𝑘

∗𝑢𝑢𝑘 + 𝑏𝑘+1
∗𝑢 𝑡 ≥ 1                                                

𝑏1
∗𝑙 𝑢1 + ⋯ + 𝑏𝑘

∗𝑙 𝑢𝑘 + 𝑏𝑘+1
∗𝑙 𝑡 ≤ 1                                                     (16)                                      

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0                                      

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

Then to solve problem (13, 14, 15, 16) using the simplex routine, we obtain the optimal value 

and solution of the original problem (1).   

Next, will be confirmed that LFP problems with RICs in the objective function according to 

the fact that each fixed number 𝑛 can be equivalently written as the interval [𝑛, 𝑛], and also 

equivalently as the rough interval ([𝑛, 𝑛], [𝑛, 𝑛]), we claim that LFP is a special case of LFP 

with ICs in the objective function and the LFP with ICs in the objective function is a special 

case of LFP with RICs in the objective function.  

𝑀𝑎𝑥. 𝑍

=  
([𝑛1, 𝑛1], [𝑛1, 𝑛1])𝑥1 + ([𝑛2, 𝑛2], [𝑛2, 𝑛2])𝑥2 + ⋯ + ([𝑛𝑘 , 𝑛𝑘 ], [𝑛𝑘 , 𝑛𝑘 ])𝑥𝑘 + ([𝑛𝑘+1, 𝑛𝑘+1], [𝑛𝑘+1, 𝑛𝑘+1])

([ℎ1, ℎ1], [ℎ1, ℎ1])𝑥1 + ([ℎ2, ℎ2], [ℎ2, ℎ2])𝑥2 + ⋯ + ([ℎ𝑘 , ℎ𝑘 ], [ℎ𝑘 , ℎ𝑘 ])𝑥𝑘 + ([ℎ𝑘+1, ℎ𝑘+1], [ℎ𝑘+1, ℎ𝑘+1])
 

Subject to:                                                                                                                                                                                                   

(17) 

∑ 𝐴𝑖 𝑥𝑖 ≤ 𝑏

𝑘

𝑖=1

 



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78 

Proof: 

Because the RICs in the numerator of the objective function has a similar cost. The coefficients 

of the objective function at its optimum surely the best, surely the worst, possibly the best, and 

possibly the worst are the same optimum. By Charnes-Cooper’s method, problem (17) is 

transformed into the following problem. 

𝑀𝑎𝑥. 𝑍 =  𝑛1𝑢1 + 𝑛2𝑢2 + ⋯ + 𝑛𝑘 𝑢𝑘 + 𝑛𝑘+1𝑡  

Subject to: 

ℎ1𝑢1 + ⋯ + ℎ𝑘 𝑢𝑘 + ℎ𝑘+1𝑡 ≥ 1                                                

ℎ1𝑢1 + ⋯ + ℎ𝑘 𝑢𝑘 + ℎ𝑘+1𝑡 ≤ 1                                                

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0                                                              (18) 

𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘. 

The combination of the first two constraints causes the following problem: 

𝑀𝑎𝑥. 𝑍 =  𝑛1𝑢1 + 𝑛2𝑢2 + ⋯ + 𝑛𝑘 𝑢𝑘 + 𝑛𝑘+1𝑡  

Subject to: 

ℎ1𝑢1 + ⋯ + ℎ𝑘 𝑢𝑘 + ℎ𝑘+1𝑡 = 1                                                                                           

𝐴1𝑢1 + 𝐴2𝑢2 + ⋯ + 𝐴𝑘 𝑢𝑘 − 𝑏𝑡 ≤ 0, where 𝑢𝑖 , 𝑡 ≥ 0, 𝑖 = 1,2, … , 𝑘.                                             (19) 

4. Numerical Examples 

Example1: Consider the following LFPP with RICs in the objective function: 

𝑀𝑎𝑥. 𝑍 =
([1,3], [0.5,4])𝑥1 + ([2,4], [1,4.5])𝑥2

([0.5,1.5], [0.25,2])𝑥1 + ([0.5,1.5], [0.25,2])𝑥2 + ([1,3], [
1
3

, 3.5])
 

Subject to:                                                                                                                                         

𝑥1 − 𝑥2 ≥ 1, 2𝑥1 + 3𝑥2 ≤ 15, 𝑥1 ≥ 3, 𝑥1, 𝑥1 ≥ 0. 

Solution 

Problem (20) is transformed into rough interval linear programming problem by Charnes 

&Cooper method. 

We introduce 𝑡 =
1

([0.5,1.5],[0.25,2])𝑥1+([0.5,1.5],[0.25,2])𝑥2+([1,3],[
1

3
,3.5])

 

𝑀𝑎𝑥. 𝑍 = ([1,3], [0.5,4])𝑥1𝑡 + ([2,4], [1,4.5])𝑥2𝑡 + 0𝑡 

Subject to: 

(20) 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

79 

([0.5,1.5], [0.25,2])𝑥1𝑡 + ([0.5,1.5], [0.25,2])𝑥2𝑡 + ([1,3], [
1

3
, 3.5]) 𝑡 = 1 

−𝑥1𝑡 + 𝑥2𝑡 + 𝑡 ≤ 0, 2𝑥1𝑡 + 3𝑥2𝑡 − 15𝑡 ≤ 0, −𝑥1𝑡 + 3𝑡 ≤ 0, 𝑥1, 𝑥1, 𝑡 ≥ 0. 

Let𝑢𝑖 = 𝑥𝑖 𝑡, 𝑖 = 1,2. 

𝑀𝑎𝑥. 𝑍 = ([1,3], [0.5,4])𝑢1 + ([2,4], [1,4.5])𝑢2 + 0𝑡 

Subject to:                                                                                                                   (21)                     

([0.5,1.5], [0.25,2])𝑢1 + ([0.5,1.5], [0.25,2])𝑢2 + ([1,3], [
1

3
, 3.5]) 𝑡 = 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

Now, by Hamzehee et al. [7] problem (21) can be changed into two linear programming with 

interval coefficients (22) & (23). 

𝑀𝑎𝑥. 𝑍 = ([1,3])𝑢1 + ([2,4])𝑢2 + 0𝑡 

Subject to:                                                                                                                 (22)                        

([0.5,1.5])𝑢1 + ([0.5,1.5])𝑢2 + ([1,3])𝑡 = 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

 

𝑀𝑎𝑥. 𝑍 = ([0.5,4])𝑢1 + ([1,4.5])𝑢2 + 0𝑡 

Subject to:                                                                                                                (23) 

                          

([0.25,2])𝑢1 + ([0.25,2])𝑢2 + ([
1

3
, 3.5]) 𝑡 = 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

Problem (22) is converted to linear programming problems (24) & (25). 

𝑀𝑎𝑥. 𝑍 = 𝑢1 + 2𝑢2 + 0𝑡 

Subject to:                                                                                                       (24)                                     

0.5𝑢1 + 0.5𝑢2 + 𝑡 ≤ 1 

1.5𝑢1 + 1.5𝑢2 + 3𝑡 ≤ 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

 

 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

80 

 

𝑀𝑎𝑥. 𝑍 = 3𝑢1 + 4𝑢2 + 0𝑡 

Subject to:                                                                                                                       (25) 

                     

0.5𝑢1 + 0.5𝑢2 + 𝑡 ≤ 1 

1.5𝑢1 + 1.5𝑢2 + 3𝑡 ≥ 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

And problem (23) is converted to linear programming problems (26) & (27). 

 

𝑀𝑎𝑥. 𝑍 = 0.5𝑢1 + 𝑢2 + 0𝑡 

Subject to:                                                                                                                     (26)                      

0.25𝑢1 + 0.25𝑢2 +
1

3
𝑡 ≤ 1 

2𝑢1 + 2𝑢2 + 3.5𝑡 ≤ 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

 

𝑀𝑎𝑥. 𝑍 = 4𝑢1 + 4.5𝑢2 + 0𝑡 

Subject to:                                                                                                                 (27)                         

0.25𝑢1 + 0.25𝑢2 +
1

3
𝑡 ≤ 1 

2𝑢1 + 2𝑢2 + 3.5𝑡 ≥ 1 

−𝑢1 + 𝑢2 + 𝑡 ≤ 0, 2𝑢1 + 3𝑢2 − 15𝑡 ≤ 0, −𝑢1 + 3𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

Now we can solve problems (24), (25), (26) and (27). Using the simplex routine, we obtain tht 

0.7154 is surely the worst optimum, 5.17 is surely the best optimum, 0.2767 is possibly the 

worst optimum, and 13.85 is possibly the best optimum. The optimal solution of the original 

problem is (3.6, 2.6) with the objective value ([0.7154, 5.17], [0.2767, 13.85]) 

Example2: Consider the following LFPP with RICs in the objective function: 

𝑀𝑎𝑥. 𝑍 =
([3.5,4.5], [3,5])𝑥1 + ([2,3], [1,4])𝑥2 + ([8,10], [7,11])

([1,1.5], [0.5,2])𝑥1 + ([1.5,1.75], [1,2])𝑥2 + ([4.5,5.5], [4,6])
 

Subject to:                                                                                                                            (28) 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

81 

𝑥1 + 3𝑥2 ≤ 30, −𝑥1 + 2𝑥2 ≤ 5, 𝑥1, 𝑥2 ≥ 0. 

Solution 

Problem (28) is transformed into a rough interval linear programming problem by Charnes 

&Cooper method. 

We introduce 𝑡 =
1

([1,1.5],[0.5,2])𝑥1+([1.5,1.75],[1,2])𝑥2+([4.5,5.5],[4,6])
 

𝑀𝑎𝑥. 𝑍 = ([3.5,4.5], [3,5])𝑥1𝑡 + ([2,3], [1,4])𝑥2𝑡 + ([8,10], [7,11])𝑡 

Subject to: 

([1,1.5], [0.5,2])𝑥1𝑡 + ([1.5,1.75], [1,2])𝑥2𝑡 + ([4.5,5.5], [4,6])𝑡 = 1 

𝑥1𝑡 + 3𝑥2𝑡 − 30𝑡 ≤ 0, −𝑥1𝑡 + 2𝑥2𝑡 − 5𝑡 ≤ 0, 𝑥1, 𝑥2, 𝑡 ≥ 0. 

Let  𝑢𝑖 = 𝑥𝑖 𝑡, 𝑖 = 1,2. 

𝑀𝑎𝑥. 𝑍 = ([3.5,4.5], [3,5])𝑢1 + ([2,3], [1,4])𝑢2 + ([8,10], [7,11])𝑡 

Subject to:                                                                                                                            (29) 

([1,1.5], [0.5,2])𝑢1 + ([1.5,1.75], [1,2])𝑢2 + ([4.5,5.5], [4,6])𝑡 = 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

Now by Hamzehee et al. [7] problem (29) can be changed into two linear programming with 

interval coefficients (30) & (31). 

𝑀𝑎𝑥. 𝑍 = ([3.5,4.5])𝑢1 + ([2,3])𝑢2 + ([8,10])𝑡 

Subject to:                                                                                                                         (30) 

([1,1.5])𝑢1 + ([1.5,1.75])𝑢2 + ([4.5,5.5])𝑡 = 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

 

𝑀𝑎𝑥. 𝑍 = ([3,5])𝑢1 + ([1,4])𝑢2 + ([7,11])𝑡 

Subject to:                                                                                                                        (31) 

([0.5,2])𝑢1 + ([1,2])𝑢2 + ([4,6])𝑡 = 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

Problem (30) is converted to linear programming problems (32) & (33). 

𝑀𝑎𝑥. 𝑍 = 3.5𝑢1 + 2𝑢2 + 8𝑡 

Subject to:                                                                                                                        (32) 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

82 

 𝑢1 + 1.5𝑢2 + 4.5𝑡 ≤ 1 

1.5𝑢1 + 1.75𝑢2 + 5.5𝑡 ≤ 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

𝑀𝑎𝑥. 𝑍 = 4.5𝑢1 + 3𝑢2 + 10𝑡 

Subject to:                                                                                                                       (33) 

 𝑢1 + 1.5𝑢2 + 4.5𝑡 ≤ 1 

1.5𝑢1 + 1.75𝑢2 + 5.5𝑡 ≥ 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

And the problem (31) is converted to linear programming problems (34) & (35). 

𝑀𝑎𝑥. 𝑍 = 3𝑢1 + 𝑢2 + 7𝑡 

Subject to:                                                                                                                    (34) 

0.5𝑢1 + 𝑢2 + 4𝑡 ≤ 1 

2𝑢1 + 2𝑢2 + 6𝑡 ≤ 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

 

𝑀𝑎𝑥. 𝑍 = 5𝑢1 + 4𝑢2 + 11𝑡 

Subject to:                                                                                                                  (35) 

0.5𝑢1 + 𝑢2 + 4𝑡 ≤ 1 

2𝑢1 + 2𝑢2 + 6𝑡 ≥ 1 

𝑢1 + 3𝑢2 − 30𝑡 ≤ 0, −𝑢1 + 2𝑢2 − 5𝑡 ≤ 0, 𝑢1, 𝑢1, 𝑡 ≥ 0. 

Now, we can solve problems (32), (33), (34), and (35).  By using the simplex routine, we 

obtain that 2.2376 is surely the worst optimum, 4.2028 is surely the best optimum, 1.469 is 

possibly the worst optimum and 8.47 is possibly the best optimum. The optimal solution of 

the original problem is (30, 0) with the objective value ([2.2376, 4.2028], [1.469, 8.47]) 

5. Conclusion 

     In this study, we proposed a new approach to solving a linear fractional programming 

problem with rough interval coefficients in the objective function. In this technique, we 

changed rough intervals into ordinary intervals and utilized convex combinations of points 

instead of intervals. The basic problem is converted into linear programming using some 

techniques. We configure series linear programming problems and obtain optimal solutions 

and values by using the simplex method. For future studies, we can try to perform this 

technique and notion in any fractional programming. 



Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (2)2022 
 

83 

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