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 Is a iterative method used when number of variables are more than number of constraint s equation  Basic Solution  Given a linear program in standard.

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Presentation on theme: " Is a iterative method used when number of variables are more than number of constraint s equation  Basic Solution  Given a linear program in standard."— Presentation transcript:

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2  Is a iterative method used when number of variables are more than number of constraint s equation  Basic Solution  Given a linear program in standard form with n variables and m equations, a basis solution is obtained by setting n-m of the variables equal to zero and solving the constrain equation for other values of m variables

3  Steps Leading to the Simplex Method FormulateProblem as LP FormulateProblem Put In StandardForm StandardForm TableauForm TableauFormExecuteSimplexMethodExecuteSimplexMethod

4 MIN 2x 1 - 3x 2 - 4x 3 s. t. x 1 + x 2 + x 3 < 30 2x 1 + x 2 + 3x 3 > 60 x 1 - x 2 + 2x 3 = 20 x 1, x 2, x 3 > 0

5  An LP is in standard form when:  All variables are non-negative  All constraints are equalities  Putting an LP formulation into standard form involves:  Adding slack variables to “<“ constraints  Subtracting surplus variables from “>” constraints.

6  Problem in Standard Form MIN 2x 1 - 3x 2 - 4x 3 s. t. x 1 + x 2 + x 3 + s 1 = 30 2x 1 + x 2 + 3x 3 - s 2 = 60 x 1 - x 2 + 2x 3 = 20 x 1, x 2, x 3, s 1, s 2 > 0

7 SIMPLEX METHOD Nonbasic Variables One of the n-m variables set equal to zero to obtained a basic solution Basic Variables One of the m variables not required to equal zero in a basic solution Basic Feasible Solution A basis feasible solution corresponds to an extreme point, it should satisfy the non- negativity constraints

8  A set of equations is in tableau form if for each equation:  its right hand side (RHS) is non-negative, and  there is a basic variable. (A basic variable for an equation is a variable whose coefficient in the equation is +1 and whose coefficient in all other equations of the problem is 0.)  To generate an initial tableau form:  An artificial variable must be added to each constraint that does not have a basic variable.

9  Tableau Form  A required form in which the given linear problem must be written before starting the initial simplex tableau.  c1 c2 …………….. Cn  a11 a22 a33…..a1n b1  a21 a22 a33 ….a2n b2 .. .am1 am2 am3…amnbm cj=coefficient of objective function aij =coefficient associated with variable j in ith constraint. The column of aij matrix corresponds to basis variables bi=RHS of ith constraint. It shows the values of basis variables

10  Three different simplex method exist for three different type of constraints.  Constraints having less than equal to at RHS  Ax1+bx2 <=c  Constraints having greater or equal to at RHS  AX1+bx2 >=c  Constraints having equal to at RHS  AX1 +bx2=c

11 o Formulate a linear programming model of the problem o Add slack variable to each constraint to obtained standard form o Obtained a basic feasible solution by putting the nonbasis variable equal to zero ensuring that all of the nonbasis variables should be positive thus satisfying non-negativity constraints. o …….

12  Max 50x1+ 40x2  S.t  3x1+5x2+ 1s1<=150  1x2+ 1s2<=20  8x1+5x2 + 1s3<=300

13  Max 50x1+ 40x2 +0s1 +0s2 +0s3  S.t  3x 1 +5x 2 + 1s 1 =150  1x 2 + 1s 2 =20  8x 1 +5x 2 + 1s 3 =300  X 1,x 2,s 1,s 2,s 3 >=0

14  Chose non-basic variables : n-m= 2 variables  x 2 =0 s 1 =0  3X 1 =150  1s 2 =20  8X1+1s3=300  X1=50;x2=0;s1=0;s2=20;s3=-100  Not a feasible basic solution because s3=-100 does not satisfy the non-negativity condition

15  Lets put x1=0 x2= 0 so now the basic variables are s1,s2,s3  S1=150  S2=20  S3=300  X1=0  X2=0  It is feasible basic solution as all variables satisfy the non-negativity condition

16  For each constraint equation, the coefficient of one of m basic variables in that equation must be 1 and coefficient for all remaining basic variables in that equation must be 0  The coefficient for each basis variable must be 1 in only one constraint equation.  In Present case s1,s2,s3 are basic variables and x1 and x2 are nonbasis variable.

17  X 1 x 2 s 1 s 2 s 3  50 40 0 0 0  3 5 1 0 0 150  0 1 0 1 0 20  8 5 0 0 1 300 Objective function coefficient B column A matrix; m rows and n columns of coefficient of variable in constraint equation Each column of the basis variable is a unit vector, a row is associated with basis variable. Value of each basis variable is then given by bi value bi value in the row associated with basic variable S1=b1=150;s2=b2=20;s3=b3=300

18  Requires changing of set of basic variables  One of nonbasic variable is made basic  One of current basic variable is made non-basic  Add two New Columns to simplex tableau  One Column is labelled Basic ( in present case S1,S2,S3) , other column is labelled ‘Cb’ (coefficient of objective function for each of basic variables)  Add two Row Zj,Cj-Zj  Zj:is decrease in coefficient of objective function when one unit of variable corresponding to jth column of A matrix is brought in to basis.

19  Multiply the element in the cb column by the corresponding element in jth column of A matrix  Z1=0(3)+0(0)+0(8)=0  Z2=0(5)+0(1)+0(5)=0  Z3=0(1)+0(0)+0(0)=0  Z4=0(0)+0(1)+0(0)=0  Z5=0(0)+0(0)+0(1)=0 Value of objective function associated with current basic variable: It is obtained by multiplying the objective function coefficient in ‘Cb’ column by corresponding value of basic variable in ‘B’ Column. 0(150)+0(20)+0(300)=0

20  Basic cb X1 x2 s1 s2 s3  50 40 0 0 0  s1 0 3 5 1 0 0 150  s2 0 0 1 0 1 0 20  s3 0 8 5 0 0 1 300  Zj 0 0 0 0 0 0  Cj-Zj 50 40 0 0 0 The initial simplex tableau: X1=0;x2=0;s1=150 ;s2=20;s3=300, objective function value=0 Coefficient in X1 column indicate amount of decrease in current basic variables when nonbasic variable x1 is increased from 0 to 1. it is true for all other columns i.e. X2,X3. If X1 is increased by 1 then s1 is decreased by 3,s2=0,S3 by 8 keeping X2=0 If X2 is increased by 1 then s1 decreased by 5,s2 by 1,s3 by 5 keeping X1=0

21  Criteria for Entering a new variable into basis  Each unit of x1 increases the value of objective function by 50  Each unit x2 increased the value of objective function by 40  Looking in cj-zj row select the variable to enter the basis that will cause largest per unit improvement value in objective function.

22  Criteria for Removing a variable from current basis  If incoming basic variable is from column j  For each row i, compute the ratio bi/aij for each aij greater than zero. the basic variable that will be removed from the basis corresponds to minimum of these ratios.  150/3= 50; 300/8= 37.5  X1 is new basic variable, and variable associated with row 3 i.e s 3 will be leaving. Thus a 31 =8 corresponding to first column to is to enter the basic variable and basic variable corresponding to third row is leaving the basic variable. A 31 is called pivot element

23  Make the column of new basic variable a unit column, the column should look like exactly the column s3 the incoming nonbasic variable.  A11=0;a21=0;a31=1  Elementary operation  Convert the pivot column to 1 i.e set a31=1  Row operation multiple pivot row by 1/8->  1X1+5/8x2+ 0s1+0s2+1/8s3=75/2--- > new pivot roe  To set a11=0 multiply new pivot row by 3 and subtract it from row 1  (3x1-5x2+1s1)-(3x1+15/8x2+3/8s3)=150-225/2  0X1+25/8x2+1s1-3/8s3=75/2  a21 is already zero so do nt need to do any row operation

24  Basic cb X 1 x 2 s 1 s 2 s 3  50 40 0 0 0  s1 0 0 25 / 8 1 0 - 3 / 8 75 / 2  s2 0 0 0 0 1 0 20  X1 50 1 5/8 0 0 1/8 75/2  Zj 1875  Cj-Zj New solution after 1 iteration: S1=75/2 S2=20;s3=0 X1=75/2;X2=0 Obj function value: 0(75/2)+0(20)+50(75/2)=1875

25  Calcualte Zj and cj-zj again after first iteration  Z1=0(0)+0(0) +50(1)=50  Z2=0(25/8)+0(1)+50(5/8)=250/8  Z3=0(1)+0(0)+50(0)=0  z4= 0(0)+0(1)+50(0)=0  Z5=0(-3/8)+0(0)+50(1/8)=50/8

26  Basic cb X1 x2 s1 s2 s3  50 40 0 0 0  s1 0 0 25 / 8 1 0 -3 / 8 75 / 2  s2 0 0 0 0 1 0 20  X1 50 1 5 / 8 0 0 1 / 8 75 / 2  Zj 50 250 / 8 0 0 50 / 8 1875  Cj-Zj 0 70 / 8 0 0 - 50 / 8 Now again choose which variable will be used as new basis: checking the rule we select x2 because it has highest cofficient in cj-zj row

27  To determine which variable will removed from the basis when x2 enters we need to compute bi/aij. In this case j is 2 because X2 lies in second column  B1/a12=75/2*8/25=12  B2/a22=20/1=20  B3/a23=75/2*8/5=60  The smallest one is b1/a12 hence s1 will leave the basis. the pivot element is a21=25/8  Nonbasic variable X2 must be made basic which means we a21=1,a22=0 ;a23=0

28  Step1: multiply every element in row 1by 8/25 for a12=1  Step2: subtract new row from row 2 to make a22=0  Step3 multiply the new pivot row by 5/8 and subtract from row 3 to make a32=0

29  Basic cb X1 x2 s1 s2 s3  50 40 0 0 0  X2 40 0 1 8 / 25 0 - 3 / 25 12  s2 0 0 0 -8 / 25 1 3 / 25 8  X1 50 1 0 -5 / 25 0 5 / 25 30  Zj 50 40 1 4 / 5 0 26 / 5 1980  Cj-Zj 0 0 - 14 / 5 0 -26 / 5 Objfunction: 40(12)+0(8)+50(30)=1980 Third itertation is not possible becaz cj-zj is either zero or negative. Final solution: X1=30;X2=12;S1=0;S2=8;S3=0

30  Greater than or equal to constraint ( Artificial Variable technique)  Add an artificial variable in the >= constraint  Add an artificial variable to the objective function  Adding an artificial variable would violates the constraint and hence it should not be appear in final solution.  This is achieved by a very high penalty –M for maximization and +M for Minimization problem  Assume highest negative contribution to the profit due to this artificial variable, hence this variable should be removed ASAP. why

31  Create the initial tableau using the artificial table  Start the iteration of changing the basis variable to non basic variable  Check on every iteration if the artificial variable is equal to zero, if yes remove it from the tableau  Repeat the iteration without the artificial variable until cj-zj row has non positive number left.

32  Max 50x1+ 40x2  S.t  3x1+5x2+ <=150  1x2+ <=20  8x1+5x2 <=300  1x1+1x2>=25  X1,x2>=0

33  Max 50x1+ 40x2 +0s1 +0s2 +0s3  S.t  3x1+5x2+ 1s1=150  1x2+ 1s2=20  8x1+5x2 + 1s3=300  1x1+1x2+ - 1s4>=25  X1,x2,s1,s2,s3 >=0

34  X1=x2=0  X1=0;x2=0;s1=150;s2=20;s3=300;s4=-25  Is it basis solution ????  Addition of an artificial variable a4 in >=0 constraint  3x1+5x2+ 1s1=150  1x2+ 1s2=20  8x1+5x2 + 1s3=300  1x1+1x2+ -1s4 +a4 >=25

35  S1,s2,s3 and a4 appear in different constraint with coefficient of 1 and RHS is non negative both requirement of tableau has satisfied  Initial basic feasible solution  X1=x2=0;s1=150;s2=20;s3=300;s4=0;a4=25

36  Basic cb X1 x2 s1 s2 s3 s4 a4  50 40 0 0 0 0 -M  s1 0 3 5 1 0 0 0 0 150  s2 0 0 1 0 1 0 0 0 20  s3 0 8 5 0 0 1 0 0 300  a4 0 1 1 0 0 0 -1 0 25

37  Basic cb X1 x2 s1 s2 s3 s4 a4  50 40 0 0 0 0 -M  s1 0 3 5 1 0 0 0 0 150  s2 0 0 1 0 1 0 0 0 20  s3 0 8 5 0 0 1 0 0 300  a4 0 1 1 0 0 0 -1 0 25  Zj -M -M 0 0 0 M -M -25M  Cj-zj 50+M 40+M 0 0 0 -M 0 Is this is a basic feasible solution ????

38  X1=x2=s4=0 (Non basis)  S1=150;s2=20;s3=300;a4=25 (basic)  Which is basic and non basic variable to leave and enter the tableau.  Largest value in the cj-zj ???  X1 has to leave non basic to become the basis variable

39  Divide the ‘b’ column matrix with x1 column matrix and find the lowest value  150/3=50;300/8=37.5;25/1  a4 has to go to non basis variable.

40  Basic cb X1 x2 s1 s2 s3 s4 a4  50 40 0 0 0 0 -M  s1 0 0 2 1 0 0 3 -3 75  s2 0 0 1 0 1 0 0 0 20  s3 0 0 -3 0 0 1 8 -8 100  x1 50 1 1 0 0 0 -1 1 25  Zj 50 50 0 0 0 -50 50 1250  Cj-zj 0 -10 0 0 0 50 -M-50 A4=0, now drop its associated column from tableau as it has been eliminated. Now the simplex tableau is

41  Basic cb X1 x2 s1 s2 s3 s4  50 40 0 0 0 0  s1 0 0 2 1 0 0 3 75  s2 0 0 1 0 1 0 0 20  s3 0 0 -3 0 0 1 8 100  x1 50 1 1 0 0 0 -1 25  Zj 50 50 0 0 0 -50 1250  Cj-zj 0 -10 0 0 0 50 Start the new iteration find the new basic and non basic variable ???

42  S4 is new non basic variable as cj-zj is highest  Divide b column matrix with s4 column matrix  75/3=25;100/8=12.525/-1  This means s3 will be going to non basic  We need to find  A16=0;a26=0;a36=1;a46=0  Applying row operation

43  Basic cb X1 x2 s1 s2 s3 s4  50 40 0 0 0 0  s1 0 0 25/8 1 0 -3/8 0 75/2  s2 0 0 1 0 1 0 0 20  s4 0 0 -3/8 0 0 1/8 1 25/2  x1 50 1 5/8 0 0 1/8 0 75/2  Zj 50 250/8 0 0 50/8 0 1875  Cj-zj 0 70/8 0 0 -50/8 0 Require one more iteration Basis and non basic variable ????

44  Basic cb X1 x2 s1 s2 s3 s4  50 40 0 0 0 0  x2 40 0 1 8/25 0 -3/25 0 12  s2 0 0 0 -8/25 1 3/25 0 8  s4 0 0 -3/25 0 2/25 1 17  x1 50 1 0 -5/25 0 5/25 0 30  Zj 50 40 14/5 0 26/5 0 1980  Cj-zj 0 0 -14/5 0 -26/5 0 X2 and S1

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