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In this paper, a modified method to find the efficient solutions of multi-objective linear fractional programming (MOLFP) problems is presented. While some of the previously proposed methods provide only one efficient solution to the MOLFP problem, this modified method provides multiple efficient solutions to the problem. As a result, it provides the decision makers flexibility to choose a better option from alternatives according to their financial position and their level of satisfaction of objectives. A numerical example is provided to illustrate the modified method and also a real life oriented production problem is modeled and solved.

Making decisions is part of our daily lives. A major concern is that almost all decision problems have multiple, usually conflicting criteria. Multi-objective programming (also known as multi-objective optimization, vector optimization, multi-criteria optimization, multi-attribute optimization or Pareto optimization) is an area of multiple criteria decision making, that is concerned with mathematical optimization problems involving more than one objective function to be optimized simultaneously.

In multi-objective analysis, linear fractional objectives are sometimes encountered. Fractional programming concerns with the optimization problem of one or several ratios of functions subject to some linear constraints. These ratios are quantities that measure the efficiency of system. When one or more objectives of a multi-objective programming problem are linear fractional i.e. ratio of two linear functions under some technological linear restrictions, then the problem is called “multi-objective linear fractional programming (MOLFP) problem”.

When there is more than one fractional objective function, it is difficult to talk about the optimal solutions of these problems. In the case when several fractional objective functions exist, the optimal solution for an objective function may not be an optimal solution for some other objective functions. Therefore, one needs to find the notion of the “best compromise solution”, also known as “non-dominated solution”, “efficient solution”, “Pareto optimal solution”, “Pareto efficient solution” etc. Thus the concept of optimality in the MOLFP problem is replaced with that of efficiency. A solution is called efficient if none of the objective functions can be improved in value without demeaning the value of any other objective. There may exist a good number of efficient solutions; as vectors cannot be ordered completely, all efficient solutions are equally good.

There exist several methodologies to solve MOLFP problem in the literature. Among them, few approaches have been reported in the early age. Kornbluth and Steuer [

In the recent years, some other approaches have been reported for solving MOLFP problems. Guzel and Sivri [

Though some approaches to find an efficient solution of MOLFP problem can be observed in the literature but hardly any method for finding multiple efficient solutions.

Here we concentrate on finding more than one (depending on the number of objectives) efficient solution of MOLFP problem by using the methods proposed by Dheyab [

Linear Fractional Programming (LFP) Problem: An LFP problem is defined as follows:

(LFP) Maximize z = N ( x ) D ( x ) = c T x + α d T x + β

subject to A x ≦ b

x ≧ 0

Here,

S = { x ∈ ℝ n | A x ≦ b , x ≧ 0 , b ∈ ℝ m } is the feasible set in decision space,

A is an m × n matrix,

x ∈ ℝ n and b ∈ ℝ m ; ( b ≧ 0 ),

c T , d T ∈ ℝ n and

α , β ∈ ℝ .

The denominator D ( x ) = d T x + β ≠ 0 for all x = ( x 1 , x 2 , ⋯ , x n ) ∈ S .

Multi-objective Linear Fractional Programming (MOLFP) Problem: An MOLFP problem is defined as follows:

(MOLFP) Maximize { Z ( x ) = ( z 1 ( x ) , z 2 ( x ) , ⋯ , z k ( x ) ) }

subject to A x ≦ b

x ≧ 0

Here,

S = { x ∈ ℝ n | A x ≦ b , x ≧ 0 , b ∈ ℝ m } is the feasible set in decision space,

A is an m × n matrix,

x ∈ ℝ n and b ∈ ℝ m ; ( b ≧ 0 ),

k ≥ 2 ,

z i ( x ) = c i T x + α i d i T x + β i = N i ( x ) D i ( x ) ; c i T , d i T ∈ ℝ n ; α i , β i ∈ ℝ ; for all i = 1 , 2 , ⋯ , k

and D i ( x ) = d i T x + β i > 0 , for all i = 1 , 2 , ⋯ , k , for all x ∈ S .

Efficient solution of MOLFP problem: A solution x ¯ ∈ S is an efficient solution of the problem (MOLFP) if and only if there is no x ∈ S such that z i ( x ) ≥ z i ( x ¯ ) for all i = 1 , 2 , ⋯ , k and z i ( x ) > z i ( x ¯ ) for at least one i.

Note that, for vectors x and y ;

x ≧ y implies x i ≧ y i for each i,

x ≥ y implies x i ≥ y i for i and x r > y r for at least one i = r , and x > y implies x i > y i for each i.

The complementary development method studied by Dheyab [

Later, Porchelvi et al. [

Step I: At first consider the first objective function z 1 ( x ) = N 1 ( x ) D 1 ( x ) , in which

N 1 ( x ) is the numerator function and D 1 ( x ) is the denominator function. The value of the objective function is taken as maximum ( max N 1 ( x ) ) for the numerator and minimum ( min D 1 ( x ) ) for the denominator function.

Step II: An LP problem is formulated as: max z ¯ 1 ( x ) subject to the constraints of the original problem, where the objective function z ¯ 1 ( x ) is formed by subtracting the denominator function ( D 1 ( x ) ) from the numerator function ( N 1 ( x ) ) i.e. z ¯ 1 ( x ) = N 1 ( x ) − D 1 ( x ) This LP problem is solved by regular simplex method.

Step III: The same LP formulation procedure is repeated for the second

objective function z 2 ( x ) = N 2 ( x ) D 2 ( x ) . This time, the LP problem is solved

including the maximization of prior objective function z ¯ 1 ( x ) as one of the constraints.

Step IV: Again the same LP formulation procedure is repeated for the third

objective function z 3 ( x ) = N 3 ( x ) D 3 ( x ) . And then, the LP problem is solved

including the maximization of prior objective function z ¯ 2 ( x ) as one of the constraints.

Step V: The same procedure is repeated until all the objective functions are optimized. Calculations up to this step provide one efficient solution of the problem (MOLFP). Reclamation of the values of max z i ( x ) is done by substituting the efficient solution into the original objective functions z i ( x ) .

Consider an MOLFP problem

max { z 1 ( x ) = − 3 x 1 + 2 x 2 x 1 + x 2 + 3 , z 2 ( x ) = 7 x 1 + x 2 5 x 1 + 2 x 2 + 1 , z 3 ( x ) = x 1 + 4 x 2 2 x 1 + 3 x 2 + 2 }

subject to x 1 − x 2 ≥ 1

2 x 1 + 3 x 2 ≤ 15

x 1 + 9 x 2 ≥ 9

x 1 ≥ 3

x 1 , x 2 ≥ 0

First consider the first objective function z 1 ( x ) = − 3 x 1 + 2 x 2 x 1 + x 2 + 3 and separate it

into sub-functions (numerator and denominator). Now as per the algorithm we have to construct an LP problem subject to the original constraints as follows:

max z ¯ 1 ( x ) = ( − 3 x 1 + 2 x 2 ) − ( x 1 + x 2 + 3 ) = − 4 x 1 + x 2 − 3

subject to x 1 − x 2 ≥ 1

2 x 1 + 3 x 2 ≤ 15

x 1 + 9 x 2 ≥ 9

x 1 ≥ 3

x 1 , x 2 ≥ 0.

Solving this LP problem by regular simplex method (for convenience we use Mathematica), we get the following optimal solution:

x 1 = 3 , x 2 = 2 with max z ¯ 1 = − 13.

Next, consider the second objective function z 2 ( x ) = 7 x 1 + x 2 5 x 1 + 2 x 2 + 1 .

According to the algorithm, we construct a new LP problem as follows:

max z ¯ 2 ( x ) = ( 7 x 1 + x 2 ) − ( 5 x 1 + 2 x 2 + 1 ) = 2 x 1 − x 2 − 1

subject to x 1 − x 2 ≥ 1

2 x 1 + 3 x 2 ≤ 15

x 1 + 9 x 2 ≥ 9

x 1 ≥ 3

4 x 1 − x 2 ≤ 10

x 1 , x 2 ≥ 0.

Here the set of constraints include the maximization of prior objective z ¯ 1 as a constraint, that is, the constraint 4 x 1 − x 2 ≤ 10 is obtained by simplifying − 4 x 1 + x 2 − 3 ≥ − 13 .

Solving this LP problem we can get the following optimal solution:

x 1 = 3 , x 2 = 2 with max z ¯ 2 = 3.

At last, we consider the third objective function z 3 ( x ) = x 1 + 4 x 2 2 x 1 + 3 x 2 + 2 and solve the following LP problem:

max z ¯ 3 ( x ) = ( x 1 + 4 x 2 ) − ( 2 x 1 + 3 x 2 + 2 ) = − x 1 + x 2 − 2

subject to x 1 − x 2 ≥ 1

2 x 1 + 3 x 2 ≤ 15

x 1 + 9 x 2 ≥ 9

x 1 ≥ 3

2 x 1 − x 2 ≥ 4

x 1 , x 2 ≥ 0.

Here the set of constraints include the maximization of prior objective z ¯ 2 as a constraint, that is, the constraint 2 x 1 − x 2 ≥ 4 is obtained by simplifying 2 x 1 − x 2 − 1 ≥ 3 .

Solving this LP problem we can get the following optimal solution:

x 1 = 3 , x 2 = 2 with max z ¯ 3 = − 3.

This gives an efficient solution for the given MOLFP problem which is,

x 1 = 3 , x 2 = 2 with max z 1 = − 5 8 , max z 2 = 23 20 , max z 3 = 11 14 .

For any MOLFP problem, there may exist a good number of efficient solutions. As vectors cannot be ordered completely, all efficient solutions are equally acceptable. It depends on the situation that which of the efficient solutions is preferable to the decision makers. Decision makers’ preference may depend on their financial position, time limit etc. So it is better to find more than one efficient solution for any MOLFP problem and provide decision maker facility to choose a better option from alternatives according to their level of satisfaction of objectives.

Keeping all these in mind we tried to find more than one efficient solution for an MOLFP problem. We modified the complementary method (developed by Porchelvi et al. [

The same problem (NE 1) using this modified method is solved here which gives three efficient solutions.

In this modified method, the objective functions are considered in different orders (in six possible ways).

So the following cases arise:

We proceed as follows:

Case 1: This case is same as the complementary development algorithm, which gives the following efficient solution for the given MOLFP problem:

x 1 = 3 , x 2 = 2 with max z 1 = − 5 8 , max z 2 = 23 20 , max z 3 = 11 14 .

To get more efficient solution we have to consider the following cases:

Case 2: Here in this case, first considering the second objective z 2 ( x ) we formulate and solve the LP problem: max z ¯ 2 ( x ) subject to the constraints of the original problem. Next consider the first objective function z 1 ( x ) and formulate the LP problem: max z ¯ 1 ( x ) subject to the original constraints in addition to the maximization of prior objective z ¯ 2 ( x ) as a constraint. After solving this LP problem, finally consider the third objective z 3 ( x ) . Then formulate and solve the LP problem: max z ¯ 3 ( x ) subject to the original constraints in addition to the maximization of prior objective z ¯ 1 ( x ) as a constraint. This

will give the following efficient solution:.

x 1 = 36 5 , x 2 = 1 5 with max z 1 = − 53 26 , max z 2 = 23 17 , max z 3 = 8 17 .

Proceeding in this manner, the remaining cases will give the following efficient solutions:

Case 3: x 1 = 18 5 , x 2 = 13 5 with max z 1 = − 14 23 , max z 2 = 139 121 , max z 3 = 14 17 ;

Case 4: x 1 = 3 , x 2 = 2 with max z 1 = − 5 8 , max z 2 = 23 20 , max z 3 = 11 14 ;

Case 5: x 1 = 3 , x 2 = 2 with max z 1 = − 5 8 , max z 2 = 23 20 , max z 3 = 11 14 ;

Case 6: x 1 = 36 5 , x 2 = 1 5 with max z 1 = − 53 26 , max z 2 = 23 17 , max z 3 = 8 17 .

Concluding the results we get the following three efficient solutions:

1. x 1 = 3 , x 2 = 2 with max z 1 = − 5 8 , max z 2 = 23 20 , max z 3 = 11 14 ; 2. x 1 = 36 5 , x 2 = 1 5 with max z 1 = − 53 26 , max z 2 = 23 17 , max z 3 = 8 17 ; 3. x 1 = 18 5 , x 2 = 13 5 with max z 1 = − 14 23 , max z 2 = 139 121 , max z 3 = 14 17 .

So our modified method provides multiple efficient solutions to the problem (NE 1) including the efficient solution (no. 1) obtained by the previous method.

Remark: Using this modified method, for an MOLFP problem with n objectives we can get at best n efficient solutions.

We have partially taken data for the following problem from S.K. Saha et al. [

Suppose an industry has Tk. 30,000,000/= by which it can produce six different products Dalda, Coconut oil, Mustard oil, Sunflower oil, Soybean oil and Palm oil. The net refined oil from per metric ton of dalda, coconut, mustard seeds, sunflower seeds, soybean crude oil and palm crude oil are respectively 300 kg, 400 kg, 400 kg, 980 kg, 970 kg and 980 kg; moreover the time needed are 4 days, 5 days, 6 days, 6 days, 3.5 days and 5 days respectively. The industry has a fixed establishment cost and time of Tk. 500,000/= and 20 days respectively. The management of industry wishes to produce maximum 600 metric tons of different types of oil. The cost for different raw materials to produce per metric ton crude oil/ seed in taka is given in

In addition, the industry has the following limitations on expenditures:

Maximum investment for crude oil/seeds is Tk. 20,000,000/=;

Maximum investment for transportation is Tk. 500,000/=;

Maximum investment for storage is Tk. 15,000/=;

Maximum investment for customs duties and vat is Tk. 6,000,000/=;

Maximum investment for chemicals is Tk. 50,000/=;

Maximum investment for electricity and gas is Tk. 120,000/=;

Maximum investment for maintenance is Tk. 30,000/=;

Maximum investment for labor is Tk. 200,000/=;

Maximum investment for management is Tk. 25,000/=;

Maximum investment for delivery is Tk. 10,000/=.

The management of industry wants to maximize the ratio of return on investment and maximize the ratio of return on time.

This leads to a multi-objective linear fractional programming problem.

• Selection of unknown variables

Let x 1 , x 2 , x 3 , x 4 , x 5 and x 6 be the metric tons of crude oil/seeds of dalda, coconut oil, mustard oil, sunflower oil, soybean oil and palm oil has to be refined respectively.

Name of products | Cost of crude oil/seeds | Transportation cost | Storage cost | Customs duties and vat | Chemical cost | Cost of electricity and gas | Maintenance cost | Labor cost | Management cost | Delivery cost | Return (per metric ton) |
---|---|---|---|---|---|---|---|---|---|---|---|

1) Dalda | 22,800 | 650 | 20 | 11,400 | 148 | 180 | 60 | 30 | 42 | 15 | 59,890 |

2) Coconut oil | 9200 | 630 | 22 | 3220 | - | 220 | 40 | 32 | 38 | 18 | 23,390 |

3) Mustard oil (seed) | 16,000 | 320 | 20 | 1800 | - | 200 | 35 | 28 | 36 | 16 | 30,750 |

4) Sunflower oil | 25,500 | 660 | 18 | 12,750 | 238 | 150 | 50 | 35 | 40 | 14 | 59,750 |

5) Soybean oil | 20,000 | 360 | 20 | 3250 | - | 100 | 30 | 26 | 37 | 17 | 40,700 |

6) Palm oil | 20,000 | 640 | 17 | 3000 | 135 | 160 | 45 | 20 | 35 | 18 | 59,435 |

• Identification of constraints

1) The management of industry wishes to produce maximum 600 metric tons different types of oil, which yields

0.3 x 1 + 0.4 x 2 + 0.4 x 3 + 0.98 x 4 + 0.97 x 5 + 0.98 x 6 ≤ 600 ;

2) The industry has maximum investment for crude oil/seeds Tk. 20,000,000/=, which results

22800 x 1 + 9200 x 2 + 16000 x 3 + 25500 x 4 + 20000 x 5 + 20000 x 6 ≤ 20000000 ;

Similarly,

3) For transportation,

650 x 1 + 630 x 2 + 320 x 3 + 660 x 4 + 360 x 5 + 640 x 6 ≤ 500000 ;

4) For storage,

20 x 1 + 22 x 2 + 20 x 3 + 18 x 4 + 20 x 5 + 17 x 6 ≤ 15000 ;

5) For customs duties and vat,

11400 x 1 + 3220 x 2 + 1800 x 3 + 12750 x 4 + 3250 x 5 + 3000 x 6 ≤ 6000000 ;

6) For chemicals,

148 x 1 + 238 x 4 + 135 x 6 ≤ 50000 ;

7) For electricity and gas,

180 x 1 + 220 x 2 + 200 x 3 + 150 x 4 + 100 x 5 + 160 x 6 ≤ 120000 ;

8) For maintenance,

60 x 1 + 40 x 2 + 35 x 3 + 50 x 4 + 30 x 5 + 45 x 6 ≤ 30000 ;

9) For labor,

30 x 1 + 32 x 2 + 28 x 3 + 35 x 4 + 26 x 5 + 20 x 6 ≤ 200000 ;

10) For delivery,

15 x 1 + 18 x 2 + 16 x 3 + 14 x 4 + 17 x 5 + 18 x 6 ≤ 10000 ;

11) For management,

42 x 1 + 38 x 2 + 36 x 3 + 40 x 4 + 37 x 5 + 35 x 6 ≤ 25000 .

We must assume that the variables x i ; i = 1 , 2 , ⋯ , 6 are not allowed to be negative. That means negative quantities of product cannot be produced.

• Identification of objectives

Using the given information we have,

TotalReturn ( inTk . ) = 59890 x 1 + 23390 x 2 + 30750 x 3 + 59750 x 4 + 40700 x 5 + 59435 x 6

TotalCost ( inTk . ) = 35345 x 1 + 13420 x 2 + 18455 x 3 + 39455 x 4 + 23840 x 5 + 24070 x 6 + 500000

TotalTime ( inhour ) = 96 x 1 + 120 x 2 + 144 x 3 + 144 x 4 + 84 x 5 + 120 x 6 + 480 .

So the objectives become

Maximize 59890 x 1 + 23390 x 2 + 30750 x 3 + 59750 x 4 + 40700 x 5 + 59435 x 6 35345 x 1 + 13420 x 2 + 18455 x 3 + 39455 x 4 + 23840 x 5 + 24070 x 6 + 500000

and

Maximize 59890 x 1 + 23390 x 2 + 30750 x 3 + 59750 x 4 + 40700 x 5 + 59435 x 6 96 x 1 + 120 x 2 + 144 x 3 + 144 x 4 + 84 x 5 + 120 x 6 + 480 .

Here each objective function is a ratio of two linear functions and all of the constraints are linear, so the problem can be modeled as the following MOLFP problem:

max { z 1 ( x ) = 59890 x 1 + 23390 x 2 + 30750 x 3 + 59750 x 4 + 40700 x 5 + 59435 x 6 35345 x 1 + 13420 x 2 + 18455 x 3 + 39455 x 4 + 23840 x 5 + 24070 x 6 + 500000 , z 2 ( x ) = 59890 x 1 + 23390 x 2 + 30750 x 3 + 59750 x 4 + 40700 x 5 + 59435 x 6 96 x 1 + 120 x 2 + 144 x 3 + 144 x 4 + 84 x 5 + 120 x 6 + 480 }

where, z 1 ( x ) represents return on investment and z 2 ( x ) represents return on time.

Subject to

0.3 x 1 + 0.4 x 2 + 0.4 x 3 + 0.98 x 4 + 0.97 x 5 + 0.98 x 6 ≤ 600

22800 x 1 + 9200 x 2 + 16000 x 3 + 25500 x 4 + 20000 x 5 + 20000 x 6 ≤ 20000000

650 x 1 + 630 x 2 + 320 x 3 + 660 x 4 + 360 x 5 + 640 x 6 ≤ 500000

20 x 1 + 22 x 2 + 20 x 3 + 18 x 4 + 20 x 5 + 17 x 6 ≤ 15000

11400 x 1 + 3220 x 2 + 1800 x 3 + 12750 x 4 + 3250 x 5 + 3000 x 6 ≤ 6000000

148 x 1 + 238 x 4 + 135 x 6 ≤ 50000

180 x 1 + 220 x 2 + 200 x 3 + 150 x 4 + 100 x 5 + 160 x 6 ≤ 120000

60 x 1 + 40 x 2 + 35 x 3 + 50 x 4 + 30 x 5 + 45 x 6 ≤ 30000

30 x 1 + 32 x 2 + 28 x 3 + 35 x 4 + 26 x 5 + 20 x 6 ≤ 200000

15 x 1 + 18 x 2 + 16 x 3 + 14 x 4 + 17 x 5 + 18 x 6 ≤ 10000

42 x 1 + 38 x 2 + 36 x 3 + 40 x 4 + 37 x 5 + 35 x 6 ≤ 25000

x i ≥ 0 ; i = 1 , 2 , ⋯ , 6.

To find efficient solutions of the problem we are going to consider the following two cases:

z 1 : represents return on investment, and z 2 : represents return on time.

Proceeding in according to the proposed modified method, we get the following two efficient solutions:

Solution 1: x 1 = 0 , x 2 = 0 , x 3 = 0 , x 4 = 0 , x 5 = 196.078 and x 6 = 370.37

with max z 1 = 2.1288 and max z 2 = 488.531

and

Solution 2: x 1 = 337.838 , x 2 = 0 , x 3 = 0 , x 4 = 0 , x 5 = 290.143 and x 6 = 0

with max z 1 = 1.65524 and max z 2 = 559.348 .

√ max z 1 for Solution 1 > max z 1 for Solution 2.

√ max z 2 for Solution 1 < max z 2 for Solution 2.

√ The increment in the maximum value of z 2 between two solutions is comparatively poor than that of z 1 .

√ If the management of industry is more concerned about cost than that of time, then they will choose Solution 1.

√ If there is shortage of time and the management has to fulfill their target within limited time then they will choose Solution 2.

√ If there is no time shortage and the management concentrate on the overall situation (keeping last observation in mind), then they can choose Solution 1.

In this work, we tried to find multiple solutions of MOLFP problem. We have elaborately explained the procedure with numerical example. Further, one application is also shown by discussing a real life-oriented problem.

Pramy, F.A. and Islam, M.A. (2017) Determining Efficient Solutions of Multi-Objective Linear Fractional Programming Problems and Application. Open Journal of Optimization, 6, 164-175. https://doi.org/10.4236/ojop.2017.64011