Metodo Simplex nD (Tablero)
El método de reemplazo y pivoteo tiene el problema de que llega a un punto en es que dificil reemplazar las variables, para esto vamos a tener una tablero que nos va especificar las ecuaciones en términos de una matriz, y vamos aplicar Gauss-Jordan para pivotear las variables.
z = 3xe + 2xi
x1 = 6 - xe - 2xi
x2 = 8 - 2xe - xi
x3 = 1 + xe - xi
x4 = 2 - xi
xe,xi >= 0
x1,x2,x3,x4 >= 0
Vamos despejar las ecuacions términos de sus valores constantes
0 = -3xe - 2xi + z
6 = xe + 2xi + x1
8 = 2xe + xi + x2
1 = -xe + xi + x3
2 = xi + x4
xe,xi >= 0
x1,x2,x3,x4 >= 0
| Base |
\(x_e\) |
\(x_i\) |
\(x_1\) |
\(x_2\) |
\(x_3\) |
\(x_4\) |
\(z\) |
LD |
| \(z\) |
-3 |
-2 |
0 |
0 |
0 |
0 |
1 |
0 |
| \(x_1\) |
1 |
2 |
1 |
0 |
0 |
0 |
0 |
6 |
| \(x_2\) |
2 |
1 |
0 |
1 |
0 |
0 |
0 |
8 |
| \(x_3\) |
-1 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
| \(x_4\) |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
2 |
Selección de la variable que entra es la menor negativa, iteramos hasta que todas las variables en la función objetivo tengan coeficiente positivo.
Entra xe
| Base |
\(x_e\) |
\(x_i\) |
\(x_1\) |
\(x_2\) |
\(x_3\) |
\(x_4\) |
\(z\) |
LD |
|
| \(z\) |
-3 |
-2 |
0 |
0 |
0 |
0 |
1 |
0 |
|
| \(x_1\) |
1 |
2 |
1 |
0 |
0 |
0 |
0 |
6 |
x1 = 0, xe = 6 |
| \(x_2\) |
2 |
1 |
0 |
1 |
0 |
0 |
0 |
8 |
x2 = 0, xe = 4 |
| \(x_3\) |
-1 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
x3 = 0, xe = -1 (No factible) |
| \(x_4\) |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
2 |
x4 = 0, 0 = 2 (inconsistente) |
| Sale de la base x2 |
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| Base |
\(x_e\) |
\(x_i\) |
\(x_1\) |
\(x_2\) |
\(x_3\) |
\(x_4\) |
\(z\) |
LD |
|
| \(z\) |
-3 |
-2 |
0 |
0 |
0 |
0 |
1 |
0 |
|
| \(x_1\) |
1 |
2 |
1 |
0 |
0 |
0 |
0 |
6 |
|
| \(x_e\) |
1 |
1/2 |
0 |
1/2 |
0 |
0 |
0 |
4 |
dividimos la fila entre 2 |
| \(x_3\) |
-1 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
|
| \(x_4\) |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
2 |
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| Operando queda |
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| Base |
\(x_e\) |
\(x_i\) |
\(x_1\) |
\(x_2\) |
\(x_3\) |
\(x_4\) |
\(z\) |
LD |
| \(z\) |
0 |
-1/2 |
0 |
3/2 |
0 |
0 |
1 |
12 |
| \(x_1\) |
0 |
3/2 |
1 |
-1/2 |
0 |
0 |
0 |
2 |
| \(x_e\) |
1 |
1/2 |
0 |
1/2 |
0 |
0 |
0 |
4 |
| \(x_3\) |
0 |
3/2 |
0 |
1/2 |
1 |
0 |
0 |
5 |
| \(x_4\) |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
2 |
- Variable que entra a la base: xi
- Variable que sale de la base:
| Base |
\(x_e\) |
\(x_i\) |
\(x_1\) |
\(x_2\) |
\(x_3\) |
\(x_4\) |
\(z\) |
LD |
|
| \(z\) |
0 |
-1/2 |
0 |
3/2 |
0 |
0 |
1 |
12 |
|
| \(x_1\) |
0 |
3/2 |
1 |
-1/2 |
0 |
0 |
0 |
2 |
x1 = 0, xi = 4/3 |
| \(x_e\) |
1 |
1/2 |
0 |
1/2 |
0 |
0 |
0 |
4 |
xe = 0, xi = 24/3 |
| \(x_3\) |
0 |
3/2 |
0 |
1/2 |
1 |
0 |
0 |
5 |
x3 = 0, xi = 10/3 |
| \(x_4\) |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
2 |
x4 = 0, xi = 6/3 |
| Sale de la base x1. |
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| Base |
\(x_e\) |
\(x_i\) |
\(x_1\) |
\(x_2\) |
\(x_3\) |
\(x_4\) |
\(z\) |
LD |
Operación |
| \(z\) |
0 |
0 |
1/3 |
4/3 |
0 |
0 |
1 |
38/3 |
Fila z + (1/2)×Fila xi |
| \(x_i\) |
0 |
1 |
2/3 |
-1/3 |
0 |
0 |
0 |
4/3 |
Fila x1 × (2/3) |
| \(x_e\) |
1 |
0 |
-1/3 |
2/3 |
0 |
0 |
0 |
10/3 |
Fila xe - (1/2)×Fila xi |
| \(x_3\) |
0 |
0 |
-1 |
1 |
1 |
0 |
0 |
3 |
Fila x3 - (3/2)×Fila xi |
| \(x_4\) |
0 |
0 |
-2/3 |
1/3 |
0 |
1 |
0 |
2/3 |
Fila x4 - Fila xi |
| Dado que todos los coeficiente de la función objetivo son positivos, el método termina. |
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| Observe que nos dio el mismo resultado que el anterior ejemplo. |
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