The Simplex Procedure

1
The Simplex Procedure

• Daniel B. Taylor
• AAEC 5024
• Department of Agricultural and Applied Economics
• Virginia Tech

2
The Basic Model
Max Z 3X1 5x2
st X1 lt4
2X2 lt12
3x1 2x2 lt18
3
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

4
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
5
The Simplex Tableau

• Construct the Simplex Tableau

6
Coefficient of

Iter-ation RN BV RHS bi/aij












7
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

8
Coefficient of

Iter-ation RN BV RHS bi/aij












9
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.

10
Coefficient of

Iter-ation RN BV RHS bi/aij

0










11
Continue to Fill out the Tableau

• RN just stands for the row number.
• We label the objective function row 0

12
Coefficient of

Iter-ation RN BV RHS bi/aij
0
0










13
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

14
Coefficient of

Iter-ation RN BV RHS bi/aij
0
0 1










15
Coefficient of

Iter-ation RN BV RHS bi/aij
0
0 1
2









16
Coefficient of

Iter-ation RN BV RHS bi/aij
0
0 1
2
3








17
Coefficients of

• Area of the Table

18
Coefficient of

Iter-ation RN BV RHS bi/aij
0
0 1
2
3








19
Coefficients of

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

20
Coefficient of

Iter-ation RN BV X1 RHS bi/aij
0
0 1
2
3








21
Coefficient of

Iter-ation RN BV X1 X2 RHS bi/aij
0
0 1
2
3








22
Coefficient of

Iter-ation RN BV X1 X2 S1 RHS bi/aij
0
0 1
2
3








23
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 RHS bi/aij
0
0 1
2
3








24
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0
0 1
2
3








25
Basic Variables

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

26
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0
0 1
2
3








27
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

28
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0 OBJ
0 1
2
3








29
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

30
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0 OBJ
0 1 S1
2
3








31
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0 OBJ
0 1 S1
2 S2
3








32
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0 OBJ
0 1 S1
2 S2
3 S3








33
Right Hand Side

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

34
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0 OBJ
0 1 S1
2 S2
3 S3








35
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

36
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
37
Coefficient of

Iter-ation RN BV X1 X2 S1 S2 S3 RHS bi/aij
0 OBJ -3
0 1 S1
2 S2
3 S3








38
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








39
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








40
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








41
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








42
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








43
Completing the Initialization Step

• For the second row which is the first constraint

44
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
45
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 1 0 1 0 0 4
2 S2
3 S3








46
Completing the Initialization Step

• For the second constraint

47
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
48
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 1 0 1 0 0 4
2 S2 0 2 0 1 0 12
3 S3








49
Completing the Initialization Step

• For the third constraint

50
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
51
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 1 0 1 0 0 4
2 S2 0 2 0 1 0 12
3 S3 3 2 0 0 1 18








52
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

53
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 1 0 1 0 0 4
2 S2 0 2 0 1 0 12
3 S3 3 2 0 0 1 18








54
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 aijgt0
• Why the minimum? Otherwise the solution will
  either be infeasible or unbounded.

55
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29








56
Pivot Row

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

57
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29








58
Pivot Number

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

59
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29








60
Number the Next Tableau

• Tableau Number 1

61
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29

1






62
Renumber the Rows

• 1
• 2
• 3

63
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0
1 1
2
3




64
Write Down the Remaining Basic Variables

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

65
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ
1 1 S1
2
3 S3




66
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

67
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ
1 1 S1
2 X2
3 S3




68
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

69
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ
1 1 S1
2 X2 0 1 0 1/2 0 6
3 S3




70
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?

71
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
and leaving 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.
72
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

73
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ
1 1 S1 1 0 1 0 0 4
2 X2 0 1 0 1/2 0 6
3 S3




74
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

75
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
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




76
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

77
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
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




78
Start the Next Iteration

• Select the entering basic variable

79
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
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




80
Start the Next Iteration

• Select the entering basic variable
• X1

81
Start the Next Iteration

• Select the entering basic variable
• X1
• Calculate (bi/aij) for aijgt0

82
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32




83
Start the Next Iteration

• Select the entering basic variable
• X1
• Calculate (bi/aij) for aijgt0
• Select the leaving basic variable

84
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32




85
Start the Next Iteration

• Select the entering basic variable
• X1
• Calculate (bi/aij) for aijgt0
• Select the leaving basic variable
• S3

86
Start the Next Iteration

• Select the entering basic variable
• X1
• Calculate (bi/aij) for aijgt0
• Select the leaving basic variable
• S3
• The pivot number is 3

87
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32




88
Begin to Fill Out the Next Tableau

• Specify the iteration number (2)
• Write down the row numbers
• Specify the basic variables

89
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
0 OBJ
2 1 S1
2 X2
3 X1
90
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

91
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
0 OBJ
2 1 S1
2 X2
3 X1 1 0 0 -1/3 1/3 2
92
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

93
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
0 OBJ
2 1 S1
2 X2 0 1 0 1/2 0 6
3 X1 1 0 0 -1/3 1/3 2
94
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

95
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
96
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

97
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
98
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

99
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
100
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

101
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
102
Interpretation of the Final Tableau

• The Objective Function Value is 36

103
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
104
Interpretation of the Final Tableau

• The Objective Function Value is 36
• The Values of the basic variables are
• S12
• X26
• X12
• The non-basic variables are
• S20
• S30

105
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
106
Interpretation of the Final Tableau

• The Values of the basic variables are
• S12
• X26
• X12
• The non-basic variables are
• S20
• S30
• The shadow prices are
• 0 for constraint 1
• 3/2 for constraint 2
• 1 for constraint 3

107
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
108
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

109
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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
110
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 2 0 1 0 12 12/26
3 S3 3 2 0 0 1 18 18/29
0 OBJ -3 0 0 5/2 0 30 NA
1 1 S1 1 0 1 0 0 4 4/14
2 X2 0 1 0 1/2 0 6 NA
3 S3 3 0 0 -1 1 6 6/32
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