The Simplex Procedure

Daniel B. Taylor

AAEC 5024

Department of Agricultural and Applied Economics

Virginia Tech

The Basic Model

Max Z= 3X1 +5x2

st X1 <=4

2X2 <=12

3x1 +2x2 <=18

Completing the Initialization Step

• Add slack (Si) variables so that the constraints may be specified as equality constraints

• Reformulate the objective function by moving all the terms to the left hand side of the equality sign – in part to make the interpretation of the solution more straight forward

The Model to Enter in the Simplex Tableau

Max Z -3X1 -5x2 -0S1 -0S2 -0S3 =0

st X1 +S1 =4

2X2 +S2 =12

3X1 +2X2 +S3 =18

The Simplex Tableau

• Construct the Simplex Tableau

Iter-ation

RN BV RHS bi/aij

Coefficient of

Begin to Fill out the Tableau

• The purpose of the first two columns is to give reference numbers to refer to when discussing the tableau

Iter-ation

RN BV RHS bi/aij

Coefficient of

Begin to Fill out the Tableau

• The purpose of the first two columns is to give reference numbers to refer to when discussing the tableau– The iteration column records the number of the

iteration you are performing– Conventionally the first tableau which really is

the last phase of the initialization step is labeled zero.

Iter-ation

RN BV RHS bi/aij

0

Coefficient of

Continue to Fill out the Tableau

• RN just stands for the row number.

• We label the objective function row “0”

Iter-ation

RN BV RHS bi/aij

0

0

Coefficient of

Continue to Fill out the Tableau

• RN just stands for the row number.

• We label the objective function row 0

• The remaining rows contain the constraints, and in this example are labeled 1-3

Iter-ation

RN BV RHS bi/aij

0

0 1

Coefficient of

Iter-ation

RN BV RHS bi/aij

0

0 1

2

Coefficient of

Iter-ation

RN BV RHS bi/aij

0

0 1

2

3

Coefficient of

“Coefficients of”

• Area of the Table

Iter-ation

RN BV RHS bi/aij

0

0 1

2

3

Coefficient of

“Coefficients of”

• Area of the Table

• Is where the decision making variables (Xj) and the slack variables (Si) are listed

Iter-ation

RN BV

X1

RHS bi/aij

0

0 1

2

3

Coefficient of

Iter-ation

RN BV

X1 X2

RHS bi/aij

0

0 1

2

3

Coefficient of

Iter-ation

RN BV

X1 X2 S1

RHS bi/aij

0

0 1

2

3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2

RHS bi/aij

0

0 1

2

3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0

0 1

2

3

Coefficient of

Basic Variables

• The column labeled BV just keeps track of the basic variables following each iteration

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0

0 1

2

3

Coefficient of

Basic Variables

• The column labeled BV just keeps track of the basic variables following each iteration

• Since there is not a basic variable in the objective function, we simply label the BV row “OBJ”

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ

0 1

2

3

Coefficient of

Basic Variables

• The column labeled BV just keeps track of the basic variables following each iteration

• Since there is not a basic variable in the objective function, we simply label the BV row “OBJ”

• In the initial tableau (0) the slack variables associated with each constraint are our basic variables

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ

0 1 S1

2

3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ

0 1 S1

2 S2

3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ

0 1 S1

2 S2

3 S3

Coefficient of

Right Hand Side

• The column labeled RHS contains the numbers on the right hand side of the equations in the linear programming problem

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ

0 1 S1

2 S2

3 S3

Coefficient of

Completing the Initialization Step

• Coefficients are taken from each equation and entered into the appropriate row of the tableau

• So for the first row, the objective function

The Model to Enter in the Simplex Tableau

Max Z -3X1 -5x2 -0S1 -0S2 -0S3 =0

st X1 +S1 =4

2X2 +S2 =12

3X1 +2X2 +S3 =18

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3

0 1 S1

2 S2

3 S3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5

0 1 S1

2 S2

3 S3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0

0 1 S1

2 S2

3 S3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0

0 1 S1

2 S2

3 S3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0

0 1 S1

2 S2

3 S3

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0

0 1 S1

2 S2

3 S3

Coefficient of

Completing the Initialization Step

• For the second row which is the first constraint

The Model to Enter in the Simplex Tableau

Max Z -3X1 -5x2 -0S1 -0S2 -0S3 =0

st X1 +S1 =4

2X2 +S2 =12

3X1 +2X2 +S3 =18

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0

0 1 S1 1 0 1 0 0 4

2 S2

3 S3

Coefficient of

Completing the Initialization Step

• For the second constraint

The Model to Enter in the Simplex Tableau

Max Z -3X1 -5x2 -0S1 -0S2 -0S3 =0

st X1 +S1 =4

2X2 +S2 =12

3X1 +2X2 +S3 =18

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0

0 1 S1 1 0 1 0 0 4

2 S2 0 2 0 1 0 12

3 S3

Coefficient of

Completing the Initialization Step

• For the third constraint

The Model to Enter in the Simplex Tableau

Max Z -3X1 -5x2 -0S1 -0S2 -0S3 =0

st X1 +S1 =4

2X2 +S2 =12

3X1 +2X2 +S3 =18

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0

0 1 S1 1 0 1 0 0 4

2 S2 0 2 0 1 0 12

3 S3 3 2 0 0 1 18

Coefficient of

Select the Entering Basic Variable

• Choose the most negative objective function coefficient

• Why?• Because with the reformulated objective

function that coefficient will increase the objective function value most rapidly

• The column of the entering basic variable is referred to as the pivot column

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0

0 1 S1 1 0 1 0 0 4

2 S2 0 2 0 1 0 12

3 S3 3 2 0 0 1 18

Coefficient of

Determine the Leaving Basic Variable

• Choose the minimum of the of the result of dividing the RHS coefficients by the coefficients in the pivot column:

(bi/aij) for aij>0

• Why the minimum? Otherwise the solution will either be infeasible or unbounded.

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 2 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

Coefficient of

Pivot Row

• The row selected for the leaving basic variable is referred to as the pivot row

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 2 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

Coefficient of

Pivot Number

• The number at the intersection of the pivot row and pivot column is referred to as the pivot number

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

Coefficient of

Number the Next Tableau

• Tableau Number 1

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

1

Coefficient of

Renumber the Rows

• 1

• 2

• 3

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0

1 1

2

3

Coefficient of

Write Down the Remaining Basic Variables

• S2 has left the basis as it was the basic variable in the pivot row

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ

1 1 S1

2

3 S3

Coefficient of

Write Down the Remaining Basic Variables

• S2 has left the basis as it was the basic variable in the pivot row

• X2 enters the basis replacing S2

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ

1 1 S1

2 X2

3 S3

Coefficient of

Prepare the Pivot Row to Perform Row Operations

• Divide the coefficients in the pivot row by the pivot number and write them down in the same row in the next tableau – tableau number 1

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ

1 1 S1

2 X2 0 1 0 1/2 0 6

3 S3

Coefficient of

Row Operations

• Now the idea is to use row operations to drive all of the other entries in the pivot column to zero, using the row that you just divided by 2 and moved down into tableau 1

• Remind any one of Gauss-Jordan reduction?

Row Operations

• Now the idea is to use row operations to drive all of the other entries in the pivot column to zero, using the row that you just divided by 2 and moved down into tableau 1

• Remind any one of Gauss-Jordan reduction?

In case you were wondering, you use this and only this row for the row operations on the other rows. The fact that you know what row to use for the operations coupled with the entering andleaving basic variable rules is what makes the simplex solution process “easy” – well I guess we can at least say straight forward in that you always know exactly what row operations to perform.

Row Operations

• Now the idea is to use row operations to drive all of the other entries in the pivot column to zero, using the row that you just divided by 2 and moved down into tableau 1

• Remind any one of Gauss-Jordan reduction?• Since the coefficient in row 1 is already zero all

you have to do is copy that row into tableau 1

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ

1 1 S1 1 0 1 0 0 4

2 X2 0 1 0 1/2 0 6

3 S3

Coefficient of

Work On Row Three

• Subtract 2 times the new row two from the old row 3 in tableau 0 and write down the results in the new row 3 in tableau 1

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ

1 1 S1 1 0 1 0 0 4

2 X2 0 1 0 1/2 0 6

3 S3 3 0 0 -1 1 6

Coefficient of

Complete the Iteration

• Add 5 times the new row 2 to the old row 0 in tableau 0 and write down the results in row 0 in tableau 1

• The iteration is complete because all entries in the old pivot column are now zero except for the old pivot number, which is 1

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30

1 1 S1 1 0 1 0 0 4

2 X2 0 1 0 1/2 0 6

3 S3 3 0 0 -1 1 6

Coefficient of

Start the Next Iteration

• Select the entering basic variable

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30

1 1 S1 1 0 1 0 0 4

2 X2 0 1 0 1/2 0 6

3 S3 3 0 0 -1 1 6

Coefficient of

Start the Next Iteration

• Select the entering basic variable

• X1

Start the Next Iteration

• Select the entering basic variable

• X1

• Calculate (bi/aij) for aij>0

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 3 0 0 -1 1 6 6/3=2

Coefficient of

Start the Next Iteration

• Select the entering basic variable

• X1

• Calculate (bi/aij) for aij>0

• Select the leaving basic variable

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 3 0 0 -1 1 6 6/3=2

Coefficient of

Start the Next Iteration

• Select the entering basic variable

• X1

• Calculate (bi/aij) for aij>0

• Select the leaving basic variable

• S3

Start the Next Iteration

• Select the entering basic variable

• X1

• Calculate (bi/aij) for aij>0

• Select the leaving basic variable

• S3

• The pivot number is 3

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

Coefficient of

Begin to Fill Out the Next Tableau

• Specify the iteration number (2)

• Write down the row numbers

• Specify the basic variables

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ

2 1 S1

2 X2

3 X1

Coefficient of

Prepare the Pivot Row to Perform Row Operations

• Divide the coefficients in the pivot row by the pivot number and write them down in the same row in the next tableau – tableau number 2

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ

2 1 S1

2 X2

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Prepare the Pivot Row to Perform Row Operations

• Divide the coefficients in the pivot row by the pivot number and write them down in the same row in the next tableau – tableau number 2

• Since the coefficient in row 2 is already zero all you have to do is copy that row into tableau 2

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ

2 1 S1

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Work On Row One

• Subtract 1 times the new row three from the old row 1 in tableau 1 and write down the results in the new row 1 in tableau 2

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Complete the Iteration

• Add 3 times the new row 3 to the old row 0 in tableau 1 and write down the results in row 0 in tableau 2

• The iteration is complete because all entries in the old pivot column are now zero except for the old pivot number, which is 1

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

You are Done!

• You have arrived at the optimal solution to the problem (assuming no math errors).

• Why?

• Because there are no negative objective function coefficients – thus no candidates for a leaving basic variable

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

You are Done!

• You have arrived at the optimal solution to the problem (assuming no math errors).

• Why?• Because there are no negative objective

function coefficients – thus no candidates for a leaving basic variable

• And your solution is feasible – because all RHS values are positive

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Interpretation of the Final Tableau

• The Objective Function Value is 36

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Interpretation of the Final Tableau

• The Objective Function Value is 36

• The Values of the basic variables are:– S1=2– X2=6– X1=2

• The non-basic variables are– S2=0– S3=0

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Interpretation of the Final Tableau• The Values of the basic variables are:

– S1=2

– X2=6

– X1=2

• The non-basic variables are– S2=0

– S3=0

• The shadow prices are:– 0 for constraint 1– 3/2 for constraint 2– 1 for constraint 3

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

The Slack Variable Matrix

• Remember you have essentially been using Gauss-Jordan reduction to solve the problem

• Among other things this matrix keeps track of the net effects (in mathematical terms) of the row operations that you have preformed

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of

Iter-ation

RN BV

X1 X2 S1 S2 S3

RHS bi/aij

0 OBJ -3 -5 0 0 0 0 NA

0 1 S1 1 0 1 0 0 4 NA

2 S2 0 22 0 1 0 12 12/2=6

3 S3 3 2 0 0 1 18 18/2=9

0 OBJ -3 0 0 5/2 0 30 NA

1 1 S1 1 0 1 0 0 4 4/1=4

2 X2 0 1 0 1/2 0 6 NA

3 S3 33 0 0 -1 1 6 6/3=2

0 OBJ 0 0 0 3/2 1 36

2 1 S1 0 0 1 1/3 -1/3 2

2 X2 0 1 0 1/2 0 6

3 X1 1 0 0 -1/3 1/3 2

Coefficient of