Ijcot v4p301

International Journal of Computer & Organization Trends – Volume 4 – January 2014

A New Approach for Easy Computation by using -Matrix for Solving Mixed Integer Linear Fractional Programming Problems V.Seerengasamy, Professor of Mathematics, PSNA College of Engineering and Technology, Dindigul-624 622, Tamil Nadu, India.

Dr.K.Jeyaraman, Dean of Science and Humanities, PSNA College of Engineering and Technology, Dindigul-624622, Tamil Nadu, India.

Abstract To minimize the computational effort needed in solving a Mixed Integer Linear Fractional Programming Problem, a new method has been proposed. Here we use  matrix for finding the solution of the mixed integer linear fractional programming problems.

Keywords: Mixed Integer Linear Fractional Programming Problems,  matrix and Promising variables.

1. Introduction: In real time situation some problems have non-negativity constraints of mixed type ( some variables real and some variables integers). Such type of problems known to be called Mixed Linear Fractional Programming Problems. To solve Mixed Integer Linear Fractional Programming Problems with reduced computational effort, a new method of approach has been proposed. In this method, among the decision variables, the variables which can enter into the basis are identified and ordered based on the maximum contribution to the objective function. The ordered decision variables one by one are allowed to enter into the basis by checking whether it is still giving an improved solution.

2. General Mixed Integer Linear Fractional Programming Problems in Matrix form The general form of Mixed Integer Linear Fractional Programming Problems is given by

ISSN: 2249-2593

Extremize Z =

C T X  c0 DT X  d 0

Subject to AX (  =  ) P 0 X  0 and some variables are integers Where  a11 a  21  a31  A mxn = . .  .   . a  m1

 x1  x   2  x3    X nx1 =  .  .   . x   n

 b1  b   2  b3    P0 =  .   .     .  b   m

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a12

a13

a 22

a 23

a32

a33

.

.

.

.

.

.

am2

a m3

. . . a1n  . . . a 2 n  . . . a3n   . . . .  . . . .   . . . .  . . . a mn 

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International Journal of Computer & Organization Trends – Volume 4 – January 2014 Let the columns corresponding to the matrix A be denoted by P 1 , P 2 , P 3 ……P n where

 a11   a12  a  a   21   22   a 31   a 32      P1 =  .  P 2 =  .   .   .       .   .  a  a   m1   m2   a13  a   23   a 33    P 3 =  .  ……………P n =  .     .  a   m3 

Step 2 : Perform phase I Step 3 : Perform phase II Step 4 : If the set J is empty then , Perform phase III Step 5 : stop.

Phase I - Ordering of Promising variables Step 1 . Using the intercepts of the decision variables along the respective axes with respect to the chosen basis a matrix is called  matrix is to be constructed.

A typical intercept for the j th variable, x j due to the

 b  i th the resource, b i is  i  a ij > 0  a ij   a1n  a   2n   a3n     .   .     .  a   mn 

C T = ( c 1 , c 2 , c 3 ……c n ),

The expanded form of  matrix is S1 S2 . .

Si .

x1 x2

DT = ( d1,

xj

d 2 ,d 3 ……d n ) and c 0 , d 0 are scalars.

3. Approach.

In the new approach three different phases are suggested to solve Mixed Integer Linear Fractional Programming Problems. In the first phase using the intercepts of the promising variables based on the different constraints along the respective axes are found out. Using the intercept matrix obtained for all the promising variables, a maximum of m variables are selected and arranged, where m is the number of constraints. Also the maximum value that each of the arranged variable can assume is found out. In the second phase the arranged variables are allowed to enter into the basis with an integer value which is less than the maximum value it assumes. These two phases are repeated till no variable could enter with an integer value. In phase III the value of most promising variable in the ordered set is retained and value of remaining variables are set to zero. Then steps of phase I and II are performed until improved solution is obtained.

The step by step procedure is given below.

Step 1 : Let iteration = 0

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xn

 b1 a  11  b1  a12  .   .  b1 a  1j  .  . b  1  a1n

b2 a 21 b2 a 22 .

. . . . . .

. b2 a2 j .

. .

. b2 a2n

. .

. . . . . .

bi ai1 bi ai 2 .

.

. . . . . .

. bi aij .

. .

. bi ain

. .

. . . . . .

Sm bm  a m1   bm  am 2  .   .  bm  a mj   .  .  bm   a mn 

Each row of the  matrix consists of m number of intercepts of the decision variable along their respective axes and each column consists of intercepts formed by the number of promising decision variables in each of the m constraints.

Step 2 . The minimum intercept and its position in each row of  matrix is find out. If there are more one minimum intercept then one of them is selected arbitrarily.

Step 3 . If the minimum intercept value is less than 1 then check the corresponding non-negative variables.

If the variable satisfy the non-negative constraints then allow the variable to enter into the basis.

Otherwise the variables are not considered for the process.

Step 4 . Multiply the minimum intercept of the variable corresponding to a row with the corresponding contribution coefficient in the

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International Journal of Computer & Organization Trends – Volume 4 – January 2014 objective function both in the numerator and denominator and the objective function value  c j x j  c0    is calculated. d x d  0   j j Step 5 . Repeat step 2 to 4 for all the row in the  matrix.

Step 6 . Let  = 0. J is a set consisting of the subscript of the promising variables.  c j x j  c0   value Step 7 . Select the variable whose  d x d  0   j j  c j x j  c0   occurs, is the largest. If the same largest  d x d  0   j j for more than one variable then the variable that has maximum contribution including the fractional value is taken as the promising variable. If that is also same then select any one arbitrarily.

Step 8 . Let it be x R .Then x R is selected as the promising variable.

Step 9. Increment  by 1.The subscript of the variable x R is stored as the l th element in set J. Step 10. The row corresponding to the variable x R as well as the other rows whose minimum occurs in the column at which the minimum for x R occurs are deleted.

Step 11. Step 7 to 10 are repeated till either all the rows or all the columns are deleted.

Step 12. The set of variables collected in Steps 7 to 10 are the ordered promising variables.

Let J= { Subscripts of the promising variables arranged in the descending  c j x j  c0   value }. order  d x d  0   j j Let  be the total number of elements in the set J. Phase II – Arranged variables are allowed to enter into the basis The arranged promising variables are allowed to enter into the basis one by one based on the entering criteria.The step by step procedure is given below.

Step 1. Let k  1 , X B is the solution vector and flag(=0) is the flag vector.

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0  0    0    flag =  .  .   . 0    n1

;

P 0old

= P0 Step 2. Iteration is incremented by 1.

Step 3. The k th element in the set J is selected and let it be j. Then the entering variable is x j .

Step 4. Computation of x j value.

The value with which x j can enter into the basis is computed by using the following formula (P )  1 = min { int { 0old i } ; ( P j ) i > 0 } ( Pj ) i i = 1,2,3….m Step 5. If (  1 < 1 and x j is non-negative variable ) then  =1 Compute v =   1 else if (  1  1 and  1 < 2 ) then  =1 Compute v = int (   1 ) else  is chosen between 0 to 1 ( Let  = 0.5 ) Compute v = int (   1 ) pos = k Step 6. Add v to the j th element of the vector X B Step 7. P 0 vector is modified using the relation ( P 0new ) i = ( P 0old ) i - (P j ) i v i = 1,2,….m Step 8. ( P 0old ) is replaced by ( P 0new ) Step 9. Let k = k + 1. If k > 1 then goto step 1.

Step 10. Let the k th element in the set J be q and the variable be x q For the variable x q ( P0old ) i } ; ( Pq )i > 0 } ( Pq ) i i = 1,2,….m u = 2 if u < 1 then go to Step 14.

Compute  2 = min{ int {

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International Journal of Computer & Organization Trends – Volume 4 – January 2014 Step 11. If ( u < 1 and x j is non-negative variable ) then  =1 flag = 1 else  = 0.5 c j v  c0 c q u  c0 Step 12 . If ( )>( ) d qu  d0 d jv  d0 if flag  1 v = int (  v ) else v= v else Replace x j by x q if flag  1 v = int (  u ) else v= u flse = 0 if v  0 goto Step 7.

Phase III – Determination of new ( improved ) solution vector to the Mixed Integer Linear Fractional Programming Problems Except for the most promising variable in the solution set obtained in phase II the values of remaining variables are set to zero. Taking this as starting solution, phase I and II are performed until improved solution is obtained. If there is no improvement the next promising variable value along with the most promising variable also is retained and the remaining basic variables made to zero. Phase III is repeated until the basic variables list exhausted.

4. Algorithm Stage I. The basic variables are arranged according to the descending order of their contribution to the objective function Step 1.  = 0, k =0, n 1 is is the number of basic variables having nonzero values in the solution Step 2. X is the solution vector obtained in phase II. Step 3. If  th element in X, ie X  > 0, then

Step 5. If  < n 1 then goto step 3.

Step 6. The array W is sorted in the descending order based on the 0 th column values of W Stage II. Finding the solution by assigning all the variable values except one in the basis to zero level.

Step 7. k = 0 Step 8.  = 0 Step 9. i = 0 Step 10. If i > k then J = W i1

 c x  c0   , let it be stored as k th Multiply     d  x  d 0  th row 0 column element of array W and  is stored as k th row 1 th column element of array W. k is incremented by one.

Step 4.  is incremented by one

Subject to the constraints 2x1+6.3x2 + x3≤ 11 9x1+ 6 x2+ 10 x3 ≤ 28 Where x 2 ≥ 0 and x1and x3 are integers

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Xj=0 Step 11. i is incremented by one Step 12. If i < n 1 then goto step 10 Step 13. Now P c is the current resource vector or ( RHS ) and corresponding objective function value Z 1 is calculated.

Stage III. Find the new solution Step 14. Use phase I and phase II and find the new solution X which is stored as Y(  ) and the corresponding objective function value Z 2 is stored as V (  ).

Step 15.  is incremented by one Step 16. If  < n 1 then goto step 9 Step 17. Find the largest of V (  ) and its position pos, where ( 0   < n 1 ), Let it be stored in Z 3 .

Step 18. If Z 3 > Z, then Replace X by Y ( pos) goto step 1. else if k < n 1 then increment k by 1. goto step 8.

5. Numerical Example Solve the following Integer linear Programming Problem.

Maximize Z =

Solution

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International Journal of Computer & Organization Trends – Volume 4 – January 2014

=

Z=

A= P0 =

, P1=

, P2 =

= 0.429

P0- new= P0-old – P j V =

, P3 =

_

=

Phase – I T

X=

T

,C = (5,8,6 ) , D = ( 3,2,4) ,C0 = 10 , To find Matrix dj cj xj

d0 = 50.

Phase - I To find Matrix dj cj xj

= = Arrangement of promising variables

J = {3 ,2}

Phase – II Arrangement of promising variables First element in J is x3 J = {2, 3} = minimum

Phase – II

= min {3.7, 1.2} = 1.2

V = int ( XB =

P0 =

XB =

First element in J is x2 = minimum = min {1.746, 4.667} = 1.746

Z=

V = int (

Second element in J is x2 = minimum = Since < 1, x2 is non-negative variable) V=( =

XB =

=

Z=

= 0.346

P0- new = P0-old – PjV =

_

XB =

_

=

=

=

=

Z=

= 0.538

P0- new= P0-old – P j V =

_

=

Phase – I To find Matrix dj

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= 0.517

P0- new= P0-old – P j V =

XB =

Second element in J is x3 = minimum = min {4.7, 2.2} = 2.2 V = int (

=

cj

xj

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International Journal of Computer & Organization Trends – Volume 4 – January 2014 REFERENCES: 1.

C.Audet, P.Hansen, B.Jaumard and G.Savard, Journal of Optimization theory and Application. Vol.93, No.2, (1997) 273-300.

x1 and x3 are integers so, the ordered set J={ x2 }

2.

A.I.Barros, J.B.G. Frenk, S.Schaible and S. Zhang, A new algorithm for generalized fractional programs Mathematical Programming 72 (1996), 2, 147-175.

Phase – II = minimum

3.

A.Charles, W.W. Cooper An explicit general solution in linear fractional programming, Vol.20 449-467 September 1973.

4.

Erik B.Bajalinov Linear Fractional programming theory, methods, Applications and Siftware.

5.

Fengquiyou & Ignacio Grossmam. Solving Mixed-Integer Linear Fractional Programming Problems with Dinkelbach’s Algorithm and MINLP methods

6.

Hamdy A.Taha, “Operations Research- An Introduction”, Seventh Edition, Prentice-Hall of India Private Limited, 2004

7.

Proceeding similarly we get the final solution.

The Optimal solution is x1 = 0, x2 = , x3 = 2

H.Ishii, T. Ibaraki and H.Mine, Fractional knapsack problems, Mathematical Programming 13 (1976), 3, 255271.

8.

Kanti Swarup, Gupta P.K.Manmohan, Research”, Sultan Chand and Sons,2010.

=

9.

G.KarthiKeyan,”Design of a new computer oriented algorithm to solve linear programming problems”, Ph.D., thesis, Alagappa University, India, May 2011.

=

= min {

, 0} = 0.

XB =

=

Z=

= 0.538

Phase III: 2 is retained for x3 and the remaining variables are set to zero, (ie) x1 = 0 and x2 = 0 Now P0 =

Max Z =

=

=

= 0.538.

6.Conclusion: In this a new approach to solve Mixed Integer Linear Fractional Programming problem has been discussed. The above algorithm rendered best optimal solution.

“Operations

10. Pant. J.C. Operation Research, Introduction to optimization, 7th edition 2008.

11. Er. Prem Kumar Gupta and Dr. D.S. Hira, Problems in Operation Research (Principles and solutions) S. Chand & company, Ram Nagar, New Delhi.

12. Stancu–Minasian, I.M. Fractional programming theory, methods and applications series, Mathematics and its Application Vol. 409 (1997)432p.

13. Suresh Chandra, M. Chandra Mohan, A note on integer linear fractional programming, Volume 27 (1980)171-174.

14. L.Vicente, G.Savard and S.Judics, Journal of Optimization Theory and Applications, 89, No.3 (1996) 597-614.

15. Wukfred Candler and Robert Townsley, Computers and Operations Research, 9(1982) 59-76

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