Denominator Objective Restriction Method and ABS Algorithm to Solve Linear Fractional Programming
Problems
Adarsh Mangal1 and Saroj Sharma2
1Department of Mathematics,
Government Engineering College, Ajmer, INDIA.
2Research Scholar,
Bhagwant University, Ajmer, INDIA.
(Received on:July8,2016) ABSTRACT
In last few years ABS algorithm has been studied. ABS method has been used broadly to solve linear and non-linear system of equations subject to large no. of constraints and variables. In this paper, we are using Denominator Restriction Method and ABS Method to solve Linear Fractional Programming Problems. Firstly the fractional programming problem has been solved by Denominator Objective Restriction method and then fractional programming problem is reduced to linear programming problem by suitable substitutions and then only its solution can be find out provided that degeneracy has been treated correctly. The same has been verified by graphical method, traditional simplex method for fractional programming problem, Charnes-Copper method and C-approach.
AMS Subject Classification: 65F30.
Keywords:ABS algorithm, Denominator Objective Restriction, Fractional Programming Problem and Linear Programming Problem and C-approach.
INTRODUCTION
Linear programming is a technique for determining an optimum schedule for interdependent activities in view of available resources. It indicates the correct combination of various variable which can be best used to achieve the objective taking in account fully the practical limitations within which the problem must be solved. Linear programming deals with the optimization (maximum or minimum) of a function of variables known as objective
1947 that George Dantzig and associates found out a technique for solving military planning problems and then he also developed a powerful mathematical tool known as “simplex method” to solve linear programming problems.
Linear fractional programming is a technique used to solve the optimization of the fraction of the two linear function subject to set of linear inequalities and non-negativity constraints. The problem can be directly solved by starting with basic feasible solution and showing conditions for improving the current basic feasible solution. We shall establish the optimality criterion to test optimality of solution. Finally the problem can be solved by the method similar to simplex method of the linear programming. The method was discussed by W.H. Marlow and J.R. Isbel in 1956. It was used in military, programming games, in economics and financial analysis.
ABS method was introduced by Abaffy, Broyden and Spedicato for solving linear equations first in the form of basic ABS class, later generalized as scaled ABS class. The method have been used to solve determined or undetermined linear systems and later have been extended to linear least squares, non linear equations, optimization problems and integer (Diophantine) equations and linear programming problems. The class of ABS method unifies most existing method for solving linear system and provide variety of other ways of implementing a specific algorithm.
P. Pandian and M. Jayalakshmi proposed a method based on simplex method for solving linear fractional programming problem, known as denominator objective restriction method. Two linear programming problems are constructed from the given Linear Fractional Programming problem. One is of maximization type and the other of minimization type. These two problems are solved to attain optimal solution to the given LFP problem.
PRELIMINARIES
Let the fractional programming under consideration is of the form:-
(P) Maximize
Subject to 0
Where , , are 1 vectors, is an 1 vector and , are scalars.
Assumption:
Set of feasible solutions to problem (P), ; , 0 is non-empty and bounded.
Subject to , 0 and
Subject to , 0
The following two theorems are used in this method for connecting the solutions of the problem (P), problem (N) and the problem (D).
Theorem 1 : Let be optimal solution to the problem(N). If is a sequence of basic feasible solution to the problem (D), such that for all 0,1,2 … 1 and by the simplex method considering the solution as an initial feasible solution, then is an optimal solution to the problem (P).
Theorem 2 : Let be optimal solution to the problem(N). If is a sequence of basic feasible solution to the problem (D), such that for all 0,1,2 … 1 and is an optimal solution to the problem (D) by simplex method considering the solution
as an initial feasible solution, then is an optimal solution to the problem (P).
Now a new method namely, denominator restriction method is used for finding an optimal solution to the LFP problem (P).
The method proceeds as follows:-
Step 1: Construct tow single objective linear programming problem i.e.; the problem (N) and problem (D) from the given problem (P).
Step 2: Compute the optimal solution of the problem (N) by means to the simplex method. Let the optimal solution to the problem (N) be and . .
Step 3: Using the optimal table of the problem (N) as an initial simplex table for problem (D), continue to find a sequence of improved basic feasible solutions to the problem (D) and value of Z at each of the improved basic feasible solution by the simplex method.
Step 4(a): If for all 0,1,2 … 1 and for some ,
stop the computation process and go to step 5.
Step 4(b): If for all 0,1,2 … and is an optimal solution to the problem (D) for some , stop the computation process and then go to step 6.
Step 5: If is an optimal solution to the problem (P) and Max. by the theorem 1.
Step 6: is an optimal solution to the problem (P) and Max. by the theorem 2.
Here we are solving the fractional programming problem by the following methods:-
(i) Denominator Objective Restriction Method.
(ii) Charnes – Cooper Method (iii) ABS Algorithm
(iv) C - approach
(v) Simplex method for LFPP (vi) Graphical Method.
Consider the following linear fractional programming problem:
6 5
2 7
2 3
3 2 6
, 0 1. Denominator Objective Restriction Method
Minimum
Subject to 2 3
3 2 6 , 0
Following two problems can be obtained from the above given problem:
(N) Minimum 6 5
Subject to 2 3 3 2 6 , 0
⇒ 6 5
Subject to 2 3
3 2 6 , , , 0
(D) Maximum 2 7
Subject to 2 3 3 2 6 , 0
⇒ 2 0 7
Subject to 2 3
3 2 6
, , , 0
Now solving the (N) by the simplex method we get the following iteration tables:
as ∆ C Z 0
the optimal solution is 3
23
Minimum 4 6 5
6 5
and
51 40
Cj –6 –5 0 0
CB XB 1 2 1 2
0 1 1 2 1 0 3 3 ←
0 2 3 2 0 1 6 2
Zj 0 0 0 0
Cj – Zj –6 –5 0 0
↑
Cj –6 –5 0 0
F.R CB XB 1 2 1 2
–6 1 1 2 1 0 3 3
2
3 0 2 0 –4 –3 1 –3 3
4
Zj –6 –12 –6 0
Cj – Zj 0 7 6 0
↑
Cj –6 –5 0 0
CB XB 1 2 1 2
–6 1 1 0 1
2
1 2
3
2
–5 2 0 1 3
4
1 4
3
4
Zj –6 –5 3
4
7
4
Cj– Zj 0 0 3
4
7
4
Now by step 3 of the method we get
3 2
3 4
51 40
Now 0 3
2 15
14
⇒ The previous solution is optimal i.e.
3 2
3 4
51 40 2. Solution by Charnes-Cooper Method:
Charnes and Cooper gave simple technique in 1962 to solve a linear fractional programming problem. They reduced linear fractional programming problem into linear programming problem by some substitution:-
Let 1
2 7
Then the problem will be of the form:-
Minimum 6 5
Subject to 3 2 0
6 3 2 0 7 2 0 1
and , 0 0
Cj 2 0 0 0
CB XB 1 2 1 2
2 1 1 0 1
2
1 2
3
2 3 →
0 2 0 1 3
4
1 4
3
4 3
Zj 2 0 1 1
Cj – Zj 0 0 1 1
↑
Cj 2 0 0 0
CB XB 1 2 1 2
0 2 2 0 1 1 3
0 2 1
2 1 1
2
3 2
3
2
Zj 2 0 0 0
Cj– Zj 0 0 0 0
Now by introducing slack variables , 0 and an artificial variable 0 in above linear programming problem and for minimize the infeasibility form , from two phase simplex method we have
Simplex Table No. 1
Simplex Table No. 2
Simplex Table No. 3
Basic Variable constant
0 0 40/33 1 13/33 1/11
0 1 14/33 0 7/33 2/11
1 0 4/33 0 2/33 1/11
0 27/11 ↑ 0↓ 14/11 12/11 Simplex Table No. 4
From last table all value of – are positive and 0, so optimal solution of the fractional programming problem is
∴ 0.1; 0.15; 0.075
i.e.
Basic Variable 0 1 2 3 4 1 constant
3 3 1 2 1 0 0 0
4 6 3 2 0 1 0 0
1 7 2 0 0 0 1 1
0 6 5 0 0 0 0
7 ↑ 2 0 0 0 1 ↓ 1
Basic Variable 0 1 2 3 4 1 constant
3 0 13/7 2 1 0 3/7
4 0 33/7 2 0 1 6/7
1 1 2/7 0 0 0 1/7
0 6 5 0 0 0
0 0↑ 0 0 0↓ drop 1
Basic Variable 0 1 2 3 4 1 constant
2 0 0 1 3/40
1 0 1 0 3/20
0 1 0 0 1/10
0 0 0 81/40 . 475 51/40
3
20 10 3
2 3
40 10 3
4
6 5
2 7
51 40 3. Solution by ABS Method:
Minimum 0 6 5
Subject to 3 2 0
6 3 2 0 7 2 0 1 Now
3 1 2
6 3 2
7 2 0
0 0 1
Let the identity Matrix be
1 0 0
0 1 0
0 0 1
Calculate the first decision parameter 3 1 2
1 0 0
0 1 0
0 0 1
3 1 2 3 1 2
3 1 2 9 1 4 14 14 0
Compute the search vector
1 0 0
0 1 0
0 0 1
3 1 2
3 1 2
Then compute the step size
0 0 0
3 1 2
0 0 0
0
3 1 2
3 1 2
3 1 2
0 0 0
0 3 1 2
0 0
Now for i=2 solve the system again, update H i.e., 0
1 0 0
0 1 0
0 0 1
1 0 0
0 1 0
0 0 1
9 3 6
3 1 2
6 2 4
1 0 0
0 1 0
0 0 1
9/14 3/14 6/14
3/14 1/14 2/14
6/14 2/14 4/14
5/14 3/14 6/14
3/14 13/14 2/14
6/14 2/14 10/14
Compute the next search vector
5/14 3/14 6/14
3/14 13/14 2/14
6/14 2/14 10/14
6 3 2
9/14 17/14 22/14 Compute the next step size
0 0 0
6 3 2
0 0 0
0
6 3 2
9/14 17/14 22/14
9/14 17/14 22/14
0 0 0
0 0 0
0 0
Now for i=3 solve the system i.e., 0
5/14 3/14 6/14
3/14 13/14 2/14
6/14 2/14 10/14
9/14 17/14 22/14
9/14 17/14 22/14
5/14 3/14 6/14
3/14 13/14 2/14
6/14 2/14 10/14
81/854 153/854 198/854
153/854 289/854 374/854
198/854 374/854 126/854
224/854 336/854 168/854
336/854 504/854 252/854
168/854 252/854 126/854
224/854 336/854 168/854
336/854 504/854 252/854
168/854 252/854 126/854
7 2 0
2240/854 3360/854 1680/854
0 0 0
0 1
7 2 0
2240/854 3360/854 1680/854
2240/854 3360/854 1680/854
1 22400/854
2240/854 3360/854 1680/854
2240 22400
3360 22400
1680 22400
0.1 0.15 0.075
0.1; 0.15; 0.075
i.e. Minimum 0 6 5 6 0.15 5 0.075
0.90 .375 1.275 i.e.
0.15
0.1 1.5 3/2 0.075
0.1 0.075 3/4
⇒ 3/2, 3/4
6 3/2 5 3/4
2 3/2 7
51/40 4. Solved by ABS Method through C:
Basic ABS Algorithm for LPP Input Section
Enter size of Matrix A : Row : 3 Cols : 3
Enter Elements of Matrix A : Element [1] [1] : –3 Element [1] [2] : 1 Element [1] [3] : 2 Element [2] [1] : –6 Element [2] [2] : 3 Element [2] [3] : 2 Element [3] [1] : 7 Element [3] [2] : 2 Element [3] [3] : 0 Enter Elements of Matrix B :
Element [1] [1] : 0 Element [2] [1] : 0 Element [3] [1] : 1 Output Section :
1 : 0.000000 0.000000 0.000000 2 : 0.000000 0.000000 0.000000 3 : 0.100000 0.150000 0.075000 ∶ 0.100000 0.150000 0.075000
5. Ordinary Simplex Method for LFPP:
Minimum
Subject to 2 3
3 2 6 , 0
6 5 0 0
2 7 7 0
7 0
6 5 0 3, 0, 0
6 3 18
2 7 3, 0, 7 2 3 7
Cj 6 5 0 0
dj 2 0 0 0
1 2 1 2
0 0 1 1 2 1 0 3 3 out
0 0 2 3 2 0 1 6 2
Cj Zj1 6 5 0 0
dj Zj2 2 0 0 0
∆j Z2 Cj Zj1 Z1
dj Zj2
42
↑ in
35 0 0
Cj 6 5 0 0
dj 2 0 0 0
1 2 1 2
2 6 1 1 2 1 0 3 3/2
0 0 2 0 4 3 1 3 3/4 out
Cj Zj1 0 7 6 0
dj Zj2 0 4 2 0
∆j Z2 Cj Zj1 Z1 dj Zj2
0 19
↑ in
42 0
13
Cj 6 5 0 0
dj 2 0 0 0
2 6 1 0 1/2 1/2 3/2
0 5 0 1 3/4 1/4 3/4
C Z 0 0 3
4
7 4
d Z 0 0 1 1
∆ Z C Z Z
d Z
0 0 21 4
121/4
6 5 3/2, 3/4, 0 6 3
2 5 3
4 51/4
2 3/2, 3/4, 7 2 3
2 7 10
3/2, 3/4 6 5
2 7
51 40 6. Solution by Graphical Method:
6 5
2 7
Subject to 2 3 3 2 6 , 0
0, 1.5, 75
70
1.5, 0.75 51
40
2, 0 12 11
Out of these three feasible solutions, the optimal solution of the above LFPP is
1.5, 0.75 for which .
CONCLUSION
A fractional programming problem can be solved by the discussed all the methods but denominator objective restriction method and ABS algorithm are two new techniques to solve a fractional programming problem. We found that all the six methods provide same optimal solutions.
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