max: z = x_{1} + x_{2} s.t.: x_{1} + 2x_{2} ≤ 8 3x_{1} + 2x_{2} ≤ 12 x_{1} + 3x_{2} ≥ 3 < **** New "feature **** 

Constraint: x_{1} + x_{2} ≥ 5 (x_{1} ≥ 0, x_{2} ≥ 0 are implied) 
max: z = x_{1} + x_{2} s.t.: x_{1} + 2x_{2} ≤ 8 3x_{1} + 2x_{2} ≤ 12 x_{1} + 3x_{2} ≥ 3 
Rewrite the Linear Programming problem as a system of linear equations (=):
z = x_{1} + x_{2} ⇒ z  x_{1}  x_{2} = 0 x_{1} + 2x_{2} ≤ 8 ⇒ x_{1} + 2x_{2} + s_{1} = 8 3x_{1} + 2x_{2} ≤ 12 ⇒ 3x_{1} + 2x_{2} + s_{2} = 12 x_{1} + 3x_{2} ≥ 3 ⇒ x_{1} + 3x_{2}  s_{3} = 3 
Linear system: z  x_{1}  x_{2} = 0 x_{1} + 2x_{2} + s_{1} = 8 3x_{1} + 2x_{2} + s_{2} = 12 x_{1} + 3x_{2}  s_{3} = 3 
Note:



Big trouble:


I like the twophase Simplex method because it's more intuitive (so I'll teach you that :))
x_{1} x_{2} s_{1} s_{2} s_{3} z  RHS ===============================+==== 1 1 0 0 0 1  0 1 2 1 0 0 0  8 3 2 0 1 0 0  12 1 3 0 0 1 0  3 



x_{1} x_{2} s_{1} s_{2} s_{3} a_{1} z  RHS ====================================+==== 1 1 0 0 0 0 1  0 1 2 1 0 0 0 0  8 3 2 0 1 0 0 0  12 1 3 0 0 1 1 0  3 
The corresponding solution is:
z = 0 s_{1} = 8 s_{2} = 12 a_{1} = 3 < "Remove" this variable 


x_{1} x_{2} s_{1} s_{2} s_{3} a_{1} z  RHS
====================================+====
1 1 0 0 0 0 1  0 < Objective function
1 2 1 0 0 0 0  8
3 2 0 1 0 0 0  12
1 3 0 0 1 1 0  3

Fact:



is:
max: a_{1}  a_{2}  ...  a_{m} (Remember that: a_{1} ≥ 0 a_{2} ≥ 0 ... a_{m} ≥ 0 ) 
Because:

Now:

More importantly:


1. Set up an initial Simplex Tableau: 1a. Add a surplus variable s_{i} to each: "≤"constraint and "≥"constraint Note: surplus variables of "≥"constraint has negative sign ! 2. If there is one or more "≥"constraints then: 2a: Add an artificial variable a_{i} to each ≥ constraint 2b. Use the Simplex method to solve: max: a_{1}  a_{2}  ...  a_{n} Using same set of constraints Note: you need to fix the Simplex Tableau first (see example) 2c. When Simplex method terminates, replace the objective row of the Final Simplex Tableau by the original objective function 3. Use the Simplex method to solve the LP Note: you need to fix the Simplex Tableau first (see example) 

max: z = x_{1} + x_{2} s.t.: x_{1} + 2x_{2} ≤ 8 3x_{1} + 2x_{2} ≤ 12 x_{1} + 3x_{2} ≥ 3 
max: z  x_{1}  x_{2} = 0 s.t.: x_{1} + 2x_{2} + s_{1} = 8 3x_{1} + 2x_{2} + s_{2} = 12 x_{1} + 3x_{2} + s_{3} = 3 
Rewritten in table form:
x_{1} x_{2} s_{1} s_{2} s_{3} z  RHS ===============================+==== 1 1 0 0 0 1  0 1 2 1 0 0 0  8 3 2 0 1 0 0  12 1 3 0 0 1 0  3 
(which did not have a feasible initial basic solution)

