The Simplex Procedure Daniel B Taylor AAEC 5024

The Simplex Procedure Daniel B. Taylor AAEC 5024 Department of Agricultural and Applied Economics Virginia Tech

The Basic Model Max Z= 3 X 1 st X 1 3 x 1 +5 x 2 <=4 2 X 2 <=12 +2 x 2 <=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 -3 X 1 st -5 x 2 X 1 -0 S 2 -0 S 3 +S 1 2 X 2 3 X 1 -0 S 1 +2 X 2 =0 =4 +S 2 =12 +S 3 =18

The Simplex Tableau • Construct the Simplex Tableau

Coefficient of Iteration RN BV RHS bi/aij

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

Coefficient of Iteration RN BV RHS bi/aij

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.

Coefficient of Iteration 0 RN BV RHS bi/aij

Continue to Fill out the Tableau • RN just stands for the row number. • We label the objective function row “ 0”

Coefficient of Iteration RN 0 0 BV RHS bi/aij

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

Coefficient of Iteration RN 0 0 1 BV RHS bi/aij

Coefficient of Iteration RN 0 0 1 2 BV RHS bi/aij

Coefficient of Iteration RN 0 0 1 2 3 BV RHS bi/aij

“Coefficients of” • Area of the Table

Coefficient of Iteration RN 0 0 1 2 3 BV RHS bi/aij

“Coefficients of” • Area of the Table • Is where the decision making variables (Xj) and the slack variables (Si) are listed

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 bi/aij

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 X 2 bi/aij

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 X 2 S 1 bi/aij

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 X 2 S 1 S 2 bi/aij

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 X 2 S 1 S 2 S 3 bi/aij

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

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 X 2 S 1 S 2 S 3 bi/aij

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”

Coefficient of Iteration RN RHS X 1 0 0 BV 1 2 3 OBJ X 2 S 1 S 2 S 3 bi/aij

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

Coefficient of Iteration 0 RN BV RHS X 1 0 OBJ 1 S 1 2 3 X 2 S 1 S 2 S 3 bi/aij

Coefficient of Iteration 0 RN BV RHS X 1 0 OBJ 1 S 1 2 S 2 3 X 2 S 1 S 2 S 3 bi/aij

Coefficient of Iteration 0 RN BV RHS X 1 0 OBJ 1 S 1 2 S 2 3 S 3 X 2 S 1 S 2 S 3 bi/aij

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

Coefficient of Iteration 0 RN BV RHS X 1 0 OBJ 1 S 1 2 S 2 3 S 3 X 2 S 1 S 2 S 3 bi/aij

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 -3 X 1 st -5 x 2 X 1 -0 S 2 -0 S 3 +S 1 2 X 2 3 X 1 -0 S 1 +2 X 2 =0 =4 +S 2 =12 +S 3 =18

Coefficient of Iteration 0 RN BV RHS X 1 0 OBJ 1 S 1 2 S 2 3 S 3 -3 X 2 S 1 S 2 S 3 bi/aij

Coefficient of Iteration 0 RN BV 0 OBJ 1 S 1 2 S 2 3 S 3 RHS X 1 X 2 -3 -5 S 1 S 2 S 3 bi/aij

Coefficient of Iteration 0 RN BV 0 OBJ 1 S 1 2 S 2 3 S 3 RHS X 1 X 2 S 1 -3 -5 0 S 2 S 3 bi/aij

Coefficient of Iteration 0 RN BV 0 OBJ 1 S 1 2 S 2 3 S 3 RHS X 1 X 2 S 1 S 2 -3 -5 0 0 S 3 bi/aij

Coefficient of Iteration 0 RN BV 0 OBJ 1 S 1 2 S 2 3 S 3 RHS X 1 X 2 S 1 S 2 S 3 -3 -5 0 0 0 bi/aij

Coefficient of Iteration 0 RN BV 0 OBJ 1 S 1 2 S 2 3 S 3 RHS X 1 X 2 S 1 S 2 S 3 -3 -5 0 0 bi/aij

Completing the Initialization Step • For the second row which is the first constraint

The Model to Enter in the Simplex Tableau Max Z -3 X 1 st -5 x 2 X 1 -0 S 2 -0 S 3 +S 1 2 X 2 3 X 1 -0 S 1 +2 X 2 =0 =4 +S 2 =12 +S 3 =18

Coefficient of Iteration 0 RN BV RHS X 1 X 2 S 1 S 2 S 3 0 OBJ -3 -5 0 0 1 S 1 1 0 0 4 2 S 2 3 S 3 bi/aij

Completing the Initialization Step • For the second constraint

The Model to Enter in the Simplex Tableau Max Z -3 X 1 st -5 x 2 X 1 -0 S 2 -0 S 3 +S 1 2 X 2 3 X 1 -0 S 1 +2 X 2 =0 =4 +S 2 =12 +S 3 =18

Coefficient of Iteration 0 RN BV RHS X 1 X 2 S 1 S 2 S 3 0 OBJ -3 -5 0 0 1 S 1 1 0 0 4 2 S 2 0 1 0 12 3 S 3 bi/aij

Completing the Initialization Step • For the third constraint

The Model to Enter in the Simplex Tableau Max Z -3 X 1 st -5 x 2 X 1 -0 S 2 -0 S 3 +S 1 2 X 2 3 X 1 -0 S 1 +2 X 2 =0 =4 +S 2 =12 +S 3 =18

Coefficient of Iteration 0 RN BV RHS X 1 X 2 S 1 S 2 S 3 0 OBJ -3 -5 0 0 1 S 1 1 0 0 4 2 S 2 0 1 0 12 3 S 3 3 2 0 0 1 18 bi/aij

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

Coefficient of Iteration 0 RN BV RHS X 1 X 2 S 1 S 2 S 3 0 OBJ -3 -5 0 0 1 S 1 1 0 0 4 2 S 2 0 1 0 12 3 S 3 3 2 0 0 1 18 bi/aij

Determine the Leaving Basic Variable • Choose the minimum 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.

Coefficient of Iteration 0 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9

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

Coefficient of Iteration 0 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9

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

Coefficient of Iteration 0 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9

Number the Next Tableau • Tableau Number 1

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9

Renumber the Rows • 1 • 2 • 3

Coefficient of Iteration 0 RN X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 1 BV 1 2 3

Write Down the Remaining Basic Variables • S 2 has left the basis as it was the basic variable in the pivot row

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ 1 S 1 2 3 S 3

Write Down the Remaining Basic Variables • S 2 has left the basis as it was the basic variable in the pivot row • X 2 enters the basis replacing S 2

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ 1 S 1 2 X 2 3 S 3

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

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ 1 S 1 2 X 2 0 1/2 0 6 3 S 3

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 In case you were wondering, you use this and only this row for column toonzero, using the. The row youknow just the row operations the other rows. factthat you what row to use for operations coupled withinto the entering and 1 divided bythe 2 and moved down tableau leaving basic variable rules is what makes the simplex solution • Remind any. Ione reduction? process “easy” – well guessof we. Gauss-Jordan 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

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ 1 S 1 1 0 0 4 2 X 2 0 1/2 0 6 3 S 3

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

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ 1 S 1 1 0 0 4 2 X 2 0 1/2 0 6 3 S 3 3 0 0 -1 1 6

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

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 1 S 1 1 0 0 4 2 X 2 0 1/2 0 6 3 S 3 3 0 0 -1 1 6

Start the Next Iteration • Select the entering basic variable

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 1 S 1 1 0 0 4 2 X 2 0 1/2 0 6 3 S 3 3 0 0 -1 1 6

Start the Next Iteration • Select the entering basic variable • X 1

Start the Next Iteration • Select the entering basic variable • X 1 • Calculate (bi/aij) for aij>0

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2

Start the Next Iteration • • Select the entering basic variable X 1 Calculate (bi/aij) for aij>0 Select the leaving basic variable

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2

Start the Next Iteration • • • Select the entering basic variable X 1 Calculate (bi/aij) for aij>0 Select the leaving basic variable S 3

Start the Next Iteration • • • Select the entering basic variable X 1 Calculate (bi/aij) for aij>0 Select the leaving basic variable S 3 The pivot number is 3

Coefficient of Iteration 0 1 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2

Begin to Fill Out the Next Tableau • Specify the iteration number (2) • Write down the row numbers • Specify the basic variables

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 1 S 1 2 X 2 3 X 1

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 1 S 1 2 X 2 3 X 1 1 0 0 -1/3 2

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 1 S 1 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

Interpretation of the Final Tableau • The Objective Function Value is 36

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

Interpretation of the Final Tableau • The Objective Function Value is 36 • The Values of the basic variables are: – S 1=2 – X 2=6 – X 1=2 • The non-basic variables are – S 2=0 – S 3=0

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

Interpretation of the Final Tableau • The Values of the basic variables are: – S 1=2 – X 2=6 – X 1=2 • The non-basic variables are – S 2=0 – S 3=0 • The shadow prices are: – 0 for constraint 1 – 3/2 for constraint 2 – 1 for constraint 3

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

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

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2

Coefficient of Iteration 0 1 2 RN BV X 1 X 2 S 1 S 2 S 3 RHS bi/aij 0 OBJ -3 -5 0 0 NA 1 S 1 1 0 0 4 NA 2 S 2 0 1 0 12 12/2=6 3 S 3 3 2 0 0 1 18 18/2=9 0 OBJ -3 0 0 5/2 0 30 NA 1 S 1 1 0 0 4 4/1=4 2 X 2 0 1/2 0 6 NA 3 S 3 3 0 0 -1 1 6 6/3=2 0 OBJ 0 0 0 3/2 1 36 1 S 1 0 0 1 1/3 -1/3 2 2 X 2 0 1/2 0 6 3 X 1 1 0 0 -1/3 2