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Modeling of Gauss Elimination Technique and AHA Simplex Algorithm for Multi-objective Linear Programming Problems

This work was carried out in collaboration between both authors. Author SJ designed the study, performed the modeling of Gauss elimination technique for LPP, wrote the elimination procedure and wrote the first draft of the manuscript. Author AM managed the analyses of the study to reduce MOLPP into LPP by AHA algorithm ant then solved by Gauss elimination technique. Both the authors read and approved the final manuscript.


Introduction
Mathematical modeling is an important tool to describe the different characteristics of an observed phenomenon (a problem) and their interactivity as well as their dynamics through Mathematics in optimization. It provides us a way to simplify observed phenomenon and to reach at its computationally description. Mathematical modeling is a simplification of real world problems. This is an art of constructing problems from an application area into manageable mathematical formulations whose numerical and theoretical analysis provides guidance, perception and answers which are useful for generating other applications.
This is an interesting, innovative, interdisciplinary, growing and important subject for researchers since many interesting algorithm have been introduced by researchers and still have lots of possibilities to find more new algorithms which must be easy, appropriate and less time taking. Some algorithms have been computationally tested. Each algorithm or method has some merits and demerits. In the framework of each application, some algorithm or method seems more suitable than others.
Sen, C. [1] developed a new approach for MOLPP. According to his suggestion, a multi-objective function should be constructed under the limitation in such a way that the optimum value of individual problem is greater than zero. The solution of linear multi-objective programming problems had been carried out by Sulaiman, N. A. and Sadiq, G. W. [2] by using mean and median value. An optimal transformation technique to find the solution of MOLPP had been developed by Sulaiman, N. A. and O. Hamadameen, A. Q. [3]. Hamad-Amin A. O. [4] discussed an adaptive arithmetic average technique to solve MOLPP. Sulaiman, N. A. and Mustafa, R. B. [5] proposed harmonic mean technique to solve multi-objective linear programming problems. Nahar, S. and Alim, M. A. [6] proposed a new statistical averaging method to solve MOLPP. Different elimination techniques for linear programming problems have been studied earlier by Kohler [7], Williams [8], Kanniappan et al. [9], Sharma et al. [10]. Jain et al. [11,12] proposed Gauss as well as Modified Fourier elimination techniques for linear fractional programming problems. Further, Jain et al. [13,14] presented extended Gauss as well as extended Modified Fourier elimination technique for integer solution of linear fractional programming problem. Jain, S. [15] proposed the technique to find the solution of multi-objective linear programming problem by modeling of Gauss elimination technique. Jain, S. [16] proposed modeling of Fourier elimination technique for multi-objective fractional programming problem. Jain, S. [17] presented the modeling of Gauss elimination technique for multi-objective fractional programming problems; in which, he considered all the linear objective functions as constraints to solve MOLPP. Here, we have taken a multi-objective linear programming problem and applied the AHA simplex algorithm to solve linear programming problem for multi-objective functions. After that, the single LPP has been constructed by using various techniques (Chandra Sen, averaging techniques and new averaging techniques) from the MOLPP under consideration and then solved it by Gauss elimination technique. We solved the LPP thus formed by all these techniques one by one by Gauss elimination technique and the result thus obtained are compared with other existing techniques available in the literature and found the same answer. The optimal answer found by this technique is the same as the result obtained from either simplex method or by graphical method (in case of only two decision variables) or by AHA simplex algorithm. The AHA simplex algorithm had been introduced to solve Linear Programming Problem (LPP) by Ansari, A. H. [18]. Jain et al. [19] commented on the result obtained by Nahar, S. and Alim, M. A. [6] in the case of new harmonic average technique.
In the later section, description about methodology is presented followed by AHA simplex algorithm and Gauss elimination technique for inequalities. Further, problem formulation is described and after that various techniques to convert MOLPP into a single LPP has been discussed.

Methodology
The mathematical programming problem can be solved by various techniques available in the literature so far but two new techniques namely Gauss elimination method and AHA simplex algorithm are discussed here in details:

AHA simplex algorithm
Step 1: We have to construct as many separate linear programming problems as the number of objectives involved in the multi-objective linear programming problem under consideration.
Step 2: To apply AHA simplex algorithm, firstly we have to write each linear programming problem in the form given below: Step 3: The objective function of every linear program must be in maximization form. If it is not so, then we have to convert it into maximize form by using the standard result i.e., Maximize Z = -Minimize (Z ) Step 4: The requirement vector components i.e.,b must be positive always. If any one of b is not positive, then we have to make it positive by multiplying the corresponding inequality by -1.
Step 5: Now, we have to check the sign convention of all the decision variables involved in the objective inequality.
(i) The solution will be an optimal one if all -c ≥ 0 ; where Z = 0 andx = 0, ∀ j (ii) There is a requirement to improve the solution if at least one -c < 0, then go to the next step.
Step 6: Select the most negative value involved among all-c . Let it be for j = k. { ; ∀a > 0}, then the row in which a lies is said to be pivotal row; which is left blank in initial iteration. The intersection of pivotal row and pivotal column is known as pivotal element i.e. a .
Step 7: To construct the next table for iteration, proceed similarly as the process adopted for ordinary simplex algorithm.

Gauss elimination technique for inequalities
In Gauss elimination technique, we solve the system of simultaneous equation with the help of elimination of variables one by one and finally the above system reduces to upper triangular system of equations, which can be solved by back substitution. Here, we are applying Gauss elimination technique for a system of inequalities of the same nature i.e., either less than or equal to (≤) or greater than or equal to (≥). In this technique, the variables are eliminated by combining the inequalities in such a way that the inequalities and variables reduce in one iteration so at last there remains only one variable with one inequality remains. This last inequality gives value of last variable in bounded form and finally taking the value of last variable either maximum or minimum according to the objective function of LPP reduced from MOLPP. Finally, we get value of other variables by back substitution of value of the remaining variables [12].
For the sake of clarity and simplicity, we consider the system of n variables and m inequalities: To eliminate the first variable say x , multiply the first row by , , ………….., respectively and then subtract them from second, third and so on upto the last row. Then we get first iteration, which is ( ) ………………………………………………….
( ) Now after first iteration, we get (n-1) variables and (m-1) inequalities. Repeating this process or after (n-1) iterations we have only one variable remains. Finally, one can get the value of other variables by back substitution of the value of last variable. It is possible that some redundant constraints may present in the system.

Problem Formulation
Here we consider the Multi-objective linear programming problem as: where, A is a matrix of order m × n and S = {xϵ R |Ax ≤ b ,x ≥ 0 and bϵ R (x ≥ 0)} is a feasible set and s ≥ 2. Also,

Chandra Sen's technique
Now, as per Chandra Sen's technique, we can convert MOLPP into a single LPP as follows: where is non-zero for i = 1, 2, …… , s subject to the constraints given in the problem under consideration. Here may be positive or negative and denotes the value of objective function of i th objective.

New averaging techniques
Let m = Minimum {AA } where AA = |τ | and τ is the maximum value of Z for i = 1 , 2 , …., r. Also let m = Minimum {AL } where AL = |τ | and τ is the minimum value of Z for i = r + 1, . …., s.

New harmonic averaging technique
Harmonic average will be m = Hence, objective function becomes Max. Z = After that, by taking objective function as constraint and all constraints of the same sign of inequality, reduced form of MOLPP for Gauss elimination technique is as follows: Now, we have to combine the inequalities in such a way that the inequalities and the variables are reduced one by one in each iteration. If an absurd inequality 0 ≤ d is found at any stage, where d is a negative number then the given MOLPP has infeasible solution. Otherwise one can get the feasible solution.

Numerical Example
Solve the given multi-objective linear programming problem: (Samsun, N. et al (2017)) x , x ≥ 0 First of all, we have to solve each and every linear programming problem along with the given constraints to find the optimal value of each objective of MOLPP by any of the methods available in the literature. Here, we have applied AHA Simplex Algorithm to find the optimal value of each objective function. For the first linear program, the procedure is as under: Now, we apply AHA simplex algorithm to find the optimal solution.
The most negative coefficient in the first row is -2 which corresponds to the variable x . Hence the entering variable is x . Here minimum positive ratio is 3 which is coming at two positions. Select arbitrarily the fourth row. Therefore variable x enters into the fourth row. Pivot element is 1.
The most negative coefficient in the first row is -1 which corresponds to the variable x 1 . Hence the entering variable is x 1 . Here minimum positive ratio is 0, therefore variable x 1 enters into the second row. Pivot element is 1.
The most negative coefficient in the first row is -1 which corresponds to the variable x 3 . Hence the entering variable is x 3 . Here minimum positive ratio is 4, which is occurring from the two positions. Hence we have to select it arbitrarily. Let variable x 3 enters into the first row. Pivot element is 6.
Now, it can be observed that all the coefficients of x j in the objective inequality is either zero or positive. Therefore, this is an optimal solution. The optimal solution occurs at   We get different bounded values for Z. Out of these Z = 3.9999 is the only value that satisfies all the inequalities altogether. Hence Z = 3.9999. Now, putting Z = 3.9999 into the inequalities involved in second stage, we get different bounded values for variable . Out of these, = 3 is the only value that satisfies all the inequalities altogether. Hence = 3.Now putting Z = 3.9999 and = 3 into the inequalities involved in first stage, we get different bounded values for the variable . Out of these, = 4 is the only value that satisfies all the inequalities altogether. Hence = 4. We get different bounded values for Z. Out of these Z = 5.0141 is the only value that satisfies all the inequalities altogether. Hence Z = 5.0141. Now, putting Z = 5.0141 into the inequalities involved in second stage, we get different bounded values for variable . Out of these, = 3 is the only value that satisfies all the inequalities altogether. Hence = 3.Now putting Z = 5.0141 and = 3 into the inequalities involved in first stage, we get different bounded values for the variable . Out of these, = 4 is the only value that satisfies all the inequalities altogether. Hence = 4.
We get different bounded values for Z. Out of these Z = 6.37103 is the only value that satisfies all the inequalities altogether. Hence Z = 6.37103. Now, putting Z = 6.37103 into the inequalities involved in second stage, we get different bounded values for variable . Out of these, = 3 is the only value that satisfies all the inequalities altogether. Hence = 3.Now putting Z = 6.37103 and = 3 into the inequalities involved in first stage, we get different bounded values for the variable . Out of these, = 4 is the only value that satisfies all the inequalities altogether. Hence = 4. We get different bounded values for Z. Out of these Z = 9.7141 is the only value that satisfies all the inequalities altogether. Hence Z = 9.7141. Now, putting Z = 9.7141 into the inequalities involved in second stage, we get different bounded values for variable . Out of these, = 3 is the only value that satisfies all the inequalities altogether. Hence = 3.Now putting Z =9.7141 and = 3 into the inequalities involved in first stage, we get different bounded values for the variable . Out of these, = 4 is the only value that satisfies all the inequalities altogether. Hence = 4.
We get different bounded values for Z. Out of these Z = 9.81495 is the only value that satisfies all the inequalities altogether. Hence Z = 9.81495. Now, putting Z = 9.81495 into the inequalities involved in second stage, we get different bounded values for variable . Out of these, = 3 is the only value that satisfies all the inequalities altogether. Hence = 3.Now putting Z =9.81495 and = 3 into the inequalities involved in first stage, we get different bounded values for the variable . Out of these, = 4 is the only value that satisfies all the inequalities altogether. Hence = 4. We get different bounded values for Z. Out of these Z = 9.9164 is the only value that satisfies all the inequalities altogether. Hence Z = 9.9164. Now, putting Z = 9.9164 into the inequalities involved in second stage, we get different bounded values for variable . Out of these, = 3 is the only value that satisfies all the inequalities altogether. Hence = 3.Now putting Z =9.9164 and = 3 into the inequalities involved in first stage, we get different bounded values for the variable . Out of these, = 4 is the only value that satisfies all the inequalities altogether. Hence = 4.
The optimal solution of each reduced LPP thus formed by MOLPP discussed in numerical example by using AHA simplex algorithm and Gauss elimination technique is tabulated below:  It is obvious to verify that results are similar to the tabular values obtained from traditional simplex method or by Graphical method (in case of having only two variables) or by any other method available in the literature. Also it is very much clear from Table 12 that if we use new harmonic average technique to solve reduced LPP, then one can get the best optimal solution out of these techniques discussed above.

Conclusion
In this research paper, we used the AHA simplex algorithm and technique of Gauss elimination technique to solve MOLPP for the very first time. Different researchers used to solve traditional simplex method to solve MOLPP while we tried to apply different algorithms and techniques. It is observed that harmonic average technique possess dominance property than the techniques of Chandra Sen, Arithmetic averaging and Geometric averaging techniques. When we apply new statistical averaging techniques, the value of the objective function gets optimize as compared to statistical averaging techniques. New harmonic averaging techniques provide best optimize value of the objective function as compared to new arithmetic and new geometric averaging techniques.