
2 x_{1} + 4 x_{2} = 16 3 x_{1} + 2 x_{2} = 12 
2 x_{1} + 4 x_{2} = 16 3 x_{1} + 2 x_{2} = 12 

Example: divide the first equation by 2 (= multiply first equation by 0.5)
2 x_{1} + 4 x_{2} = 16 (equation 1) 3 x_{1} + 2 x_{2} = 12 (equation 2) Manipulation: 0.5 × (2 x_{1} + 4 x_{2}) = 0.5 × (16) ==> x_{1} + 2 x_{2} = 8 Result: x_{1} + x_{2} = 8 (equation 1) 3 x_{1} + 2 x_{2} = 12 (equation 2) 

Example: subtract 3 × the first equation from the second equation
x_{1} + 2 x_{2} = 8 (row 1) 3 x_{1} + 2 x_{2} = 12 (row 2) (row 2) ===> 3 x_{1} + 2 x_{2} = 12 3 x (row 1) ===> 3 x_{1} + 6 x_{2} = 24  (subtract)  0 x_{1}  4 x_{2} = 12 

Note:





The basic solution corresponding to the new Simplex Tableau is:

Note:

T = an initial Simplex Tableau; // How: // Add surplus variables // to obtain a basic solution Find a pivot element p in T that // Discussed next makes the obj. function increase in value; while ( p can be found ) { T = Perform pivot operation on p in T // Discussed above Find a pivot element p in T that makes the obj. function increase in value; } 




Reason:

Example:

Note:


Example:
Explanation:


max: z = x_{1} + x_{2} s.t.: x_{1} + 2x_{2} ≤ 8 3x_{1} + 2x_{2} ≤ 12 x_{1} ≥ 0, x_{2} ≥ 0 
Add surplus variables: z  x_{1}  x_{2} = 0 x_{1} + 2x_{2} + s_{1} = 8 3x_{1} + 2x_{2} + s_{2} = 12 
This is the initial Simplex Tableau


Graphically:


Graphically:

DONE !!!
(Optimal solution found)