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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 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

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

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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

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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.

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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

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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

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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.

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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

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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

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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 …….

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

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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

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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

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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

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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.

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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

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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.

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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

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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

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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

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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

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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

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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

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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

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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 ????

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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

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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.

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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

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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 ???

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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

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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 ????

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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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