Ejercicios De Dualidad

Ejercicio 1.

Zmax=5x1+6x2

S.A.

x1+2x2=5
-x1+5x2≥3
4x1+7x2≤8x1 ;sin restriccion x2≥0


Solución:* Dual
Zmin=5y1+3y2+8y3+MA1+MA2

S.A.
y1-y2+4y3=5
2y1+5y2+7y3≥6

y1;sin restriccion y2≤0 y3≥0

* Primal
Zmin-5y1-3y2-8y3-MA1-MA2=0

S.A.
y1-y2+4y3+A1=5
2y1+5y2+7y3-H1+A2=6

* Tabla inicial
V.B. | Y1 | Y2 | Y3 | H1 | A1 | A2 | L.D. |
Z | –5 | –3 | –8 | 0 | –M |–M | 0 |
A1 | 1 | –1 | 4 | 0 | 1 | 0 | 5 |
A2 | 2 | 5 | 7 | –1 | 0 | 1 | 6 |

* Primera iteración
V.B. | Y1 | Y2 | Y3 | H1 | A1 | A2 | L.D. |
Z | –5+3M | –3+4M | –8+11M | –M | 0 | 0 | 11M |
A1 | 1 | –1 | 4 | 0 | 1 | 0 | 5 |
A2 | 2 | 5 | 7 | –1 | 0 | 1 | 6 |

* Segunda iteración
V.B. | Y1 | Y2 | Y3 | H1 | A1 | A2 | L.D. |
Z | -197-17M | 197-277M | 0 | -87+47M | 0 | 87-117M| 487+117M |
A1 | -17 | -277 | 0 | 47 | 1 | -47 | 117 |
Y3 | 27 | 57 | 1 | -17 | 0 | 17 | 67 |

* Ultima Tabla
V.B. | Y1 | Y2 | Y3 | H1 | A1 | A2 | L.D. |
Z | –3 | –5 | 0 | 0 | 2–M | –M | 10 |
H1 | -14 | -274 | 0 | 1 | 74 | –1 | 114 |
Y3 | 14 | -14 | 1 | 0 | 14 | 0 | 54 |

Conclusión:
Y1 = 0
Y2 = 0 Zmin = $10
Y3 = 54

Ejercicio 2.
Zmin=6x1+3x2
S.A.6x1-3x2+x3≥2
3x1+4x2+x3≥5

x1, x2, x3, ≥0

Solución:


* Dual
Zmax=2y1+5y2
S.A.
6y1+3y2≤6
-3y1+4y2≤3
y1+y2≤0

y1≥0 y2≥0

* Primal
Zmax-2y1-5y2=0
S.A.
6y1+3y2+H1=6
-3y1+4y2+H2=3
y1+y2+H3=0

* Tabla inicial (Primera iteración)
V.B. |Y1 | Y2 | H1 | H2 | H3 | L.D. |
Z | –2 | –5 | 0 | 0 | 0 | 0 |
H1 | 6 | 3 | 1 | 0 | 0 | 6 |
H2 | –3 | 4 | 0 | 1 | 0 | 3 |
H3 | 1 | 1 | 0 | 0 | 1 | 0 |

* Segunda iteración
V.B. | Y1 | Y2 | H1 | H2 | H3 | L.D. |
Z | -234 | 0 | 0 | 54 | 0 | 154 |
H1 | 334 | 0 | 1 | -34 | 0 | 154 |
H3 | 74 | 0 | 0 | -14 | 1 | -34 |
Y2 | -34 | 1 | 0 | 14 | 0 | 34 |

* Ultima Tabla
V.B. |Y1 | Y2 | H1 | H2 | H3 | L.D. |
Z | 0 | 0 | 2333 | 811 | 0 | 7011 |
H3 | 0 | 0 | -733 | -111 | 1 | -1711 |
Y1 | 1 | 0 | 433 | -111 | 0 | 511 |
Y2 | 0 | 1 | 111 | 211 | 0 | 1211 |
Conclusión:
Y1 = 511=0.45 Zmax = 7011= $6.36
Y2 = 1211=1.09

Ejercicio 3
Zmax=x1+x2
S.A.
2x1+x2=5
3x1-x2=6

x1, x2, →sin restricción
Solución:
* DualZmin=5y1+6y2+MA1+ MA2
S.A.
2y1+3y2=1
y1-y2=1

y1, y2, →sin restricción

* Primal
Zmin-5y1-6y2 -MA1-MA2=0
S.A.
2y1+3y2 + A1=1
y1-y2+A2=1

* Tabla inicial
V.B | Y1 | Y2 | A1 | A2 | L.D. |
Z | -5 | -6 | -M | -M | 0 |
A1 | 2 | 3 | 1 | 0 | 1 |
A2 | 1 | -1 | 0 | 1 | 1 |

* Primera iteración
V.B | Y1 | Y2 | A1 | A2 | L.D. |
Z | -5+3M | -6+2M | 0 | 0 | 2M |
A1 | 2 |3 | 1 | 0 | 1 |
A2 | 1 | -1 | 0 | 1 | 1 |

* Ultima tabla
V.B | Y1 | Y2 | A1 | A2 | L.D. |
Z | 0 | 32-52M | 52-32M | 0 | 52+12M |
Y1 | 1 | 32 | 12 | 0 | 12 |
A2 | 0 | -52 | -12 | 1 | 12 |

Conclusión:
Debido a que dentro del resultado se encuentra una artificial y la variable M en el L.D, NO TIENE SOLUCION FACTIBLE.


Ejercicio 4.
Zmin=5x1+2x2
S.A.
x1-x2≥3
2x1+x2≥5...