1. Simplex Tableau Method

2. General Form of a Linear Programming (LP) Problem
MAX (or MIN): c1X1 + c2X2 + … + cnXn
Subject to: a11X1 + a12X2 + … + a1nXn <= b1 :
ak1X1 + ak2X2 + … + aknXn <= bk
am1X1 + am2X2 + … + amnXn = bm

3. The Simplex Method
To use the simplex method, we first convert all inequalities to equalities by adding slack variables to <= constraints and subtracting slack variables from >= constraints.
For example: ak1X1 + ak2X2 + … + aknXn <= bk
is converted to: ak1X1 + ak2X2 + … + aknXn + Sk = bk
And: ak1X1 + ak2X2 + … + aknXn >= bk
is converted to: ak1X1 + ak2X2 + … + aknXn - Sk = bk

4. Example
A company manufactures different electronic components for computers. Component A requires 2 hours of fabrication and 1 hour of assembly, component B require 3 hours of fabrication with 1 hour of assembly, and component C requires 2 hours of fabrication and 2 hours of assembly. The company has up to 1,000 labor hours for fabrication and 800 labor hours of assembly time, each week. If the profit on each component A, B, and C is $7, $8 and $10 respectively, how many of each component will produce the maximum the profit?

5. Setup the simplex Tableau
Maximize the Profit
Let:
X1 be the number of components of A
X2 be the number of components of B
X3 be the number of components of C
Maximize: P = 7X1+ 8 X2+ 10 X3
Subject to:
Fabrication time 2X1+ 3X2+ 2X3 <= 1000
Assembly time 1X1+ 1X2+2 X3 <= 800
X1, X2, X3 >= 0

6. Next build the Simplex Tableau by introducing Slack Variables S
Rewrite the objective function and make the constraints equality equations by adding positive slack variables S1 and S2
-7X1 - 8 X X3 + P = 0
2X1+ 3X2+ 2X3 + S1 = 1000
1X1+ 1X2+2X3+ S2= 800
From this, Produce the Simplex Tableau
See: Simplex Tableau Adobe pdf or Text doc.

7. Simplex Tableau Example
Setup the simplex Tableau
Maximize the Profit
Let:
X1 be the number of components of A
X2 be the number of components of B
X3 be the number of components of C
Maximize: P = 7X1+ 8 X2+ 10 X3
Subject to:
Fabrication time 2X1+ 3X2+ 2X3 <= 1000
Assembly time 1X1+ 1X2+2 X3 <= 800
X1, X2, X3 >= 0

8. Next build the Simplex Tableau by introducing Slack Variables S
Rewrite the objective function and make the constraints equality equations by adding positive slack variables S1 and S2
-7X1 - 8 X X3 + P = 0
2X1+ 3X2+ 2X3 + S1 = 1000
1X1+ 1X2+2X3+ S2= 8
Form the Produce P in the Simplex Tableau at the bottom. (optional)
Always (?) put slack variable equations first in Tableau and the objective function in the bottom row.
Tableau 1 X1 X2 X3 S1 S2 P
RHS 2 3 1
1000
800 -7 -8
-10

9. Next start reducing tableau to find the values for X1, X2, X3 and P the profit. Look at objective function row (P), find the largest negative number (-10) and this is the first pivot column.
Next, find the row S1 or S2 in which to pick the pivot cell. To do this divide the RHS number by the pivot column (X3) element values and the resulting smallest dividend value is the row that will be the pivot row. In this example cell value (S1 X3) is divided into cell value (S1 RHS), 1000/ 2 = 500, and cell value (S2 X3 ) is divided into cell value (S2 RHS), 800/ 2 = Our pivot (S2 X3) contains value 2 must be converted to value 1. First perform row operations on row 2 by taking ½ R2 –> R2 see Tableau 2
Tableau 2 X1 X2 X3 S1 S2 P
RHS 2 3 1
1000
400 -7 -8
-10

10. Next two row operations, use pivot row R2 to change R1 and R3 -2
Next two row operations, use pivot row R2 to change R1 and R3 -2*R2 + R1 –> R1 -2R R Yields R *R2 + R3 –> R3 10R R R These rows create next Tableau3 and all math operations were done on first pivot row 2. S1 and X3 are now the basic decision variables
Tableau 3 X1 X2 X3 S1 S2 P
RHS 1 2 -1
200
400 -2 -3 5
4000

11. Row reduction done; now look at objective function row (P), if there is any negative number pick the largest as the new pivot column (-3), which is X2
Next, find the row S1 or X3 in which to pick the pivot cell in column X2. To do this divide the RHS number by the pivot column (X2) element values and the resulting smallest dividend value is the row that will be the pivot row. In this example cell value (S1 X3) is divided into cell value (S1 RHS), 200/ 2 = 100, and cell value (X3) is divided into cell value (X3 RHS), 400/ ½ = 800.
The smallest value is in R1 and our second pivot (S1 X2) contains value 2 must be converted to value 1, by again doing row operations:
Take ½ R1 –> R1
Tableau 4 X1 X2 X3 S1 S2 P
RHS 1 -½
100
400 -2 -3 5
4000
-½*R1 + R2 –> R2
3*R1 + R3 –> R3
Tableau 5 X1 X2 X3 S1 S2 P
RHS 1 -½
100
1/4
-1/4
3/4
350
3/2
7/2
4300

12. Take 2*R1 –> R1 to get 1 in (X2 X1)
Row reduction done, now look at objective function row (P), if there is any negative number pick the as the new pivot column (-1/2), at least one more row operation.
Next, find the row X2 or X3 in which to pick the pivot cell. To do this divide the RHS number by the pivot column (X1) element values and the resulting smallest dividend value is the row that will be the pivot row. In this example cell value (X2 X1) is divided into cell value (X2 RHS), 100/½= 200, and cell value (X3) is divided into cell value (X3 RHS), 350/ 1/4 = 1400.
The new pivot element in (X2 X1) is ½ so turn that into 1 by doing the row operation
Take 2*R1 –> R1 to get 1 in (X2 X1)
Tableau 6 X1 X2 X3 S1 S2 P
RHS 1 2 -1
200
1/4
-1/4
3/4
350 -½
3/2
7/2
4300
Take
-1/4* R1 + R2 –> R2 to get 0 in (X3 X1)
½*R1 + R3 –> R3 to get 0 in (P X1)
Tableau 7 X1 X2 X3 S1 S2 P
RHS 1 2 -1
200 -½
300 3
4400

13. The objective function P does not have any negative numbers so the simplex method is complete. How to interpret the results.
Maximize P = 7X1 + 8 X X3
Subject to
Fabrication time 2X1 + 3X2 + 2X3 <= 1000
Assembly time 1X1 + 1X2 +2 X3 <= 800
X1, X2, X3 >= 0
The left tableau decision variables X1, X3 in column 1 equal the LHS column value and the
maximum profit P = 4400 when X1 = 200, X2 = 0, and X3 = 300.
Means make 200 components type A, 300 components type C, and do not make components type B. In addition, you will have 0 hours of slack time S1 for Fabrication
and 0 hours of slack time S2 for Assembly.

16. Simplex Method Summary
The simplex method operates by first identifying any basic feasible solution (or extreme point) for an LP problem, then moving to an adjacent extreme point, if such a move improves the value of the objective function.
When no adjacent extreme point has a better objective function value, the current extreme point is optimal and the simplex method terminates.
The process of moving from one extreme point to an adjacent one is accomplished by switching one of the basic variables with one of the nonbasic variables to create a new basic feasible solution that corresponds to the adjacent extreme point.