EXAMEN PARCIAL
http://uapjosemedinaj.blogspot.com/2014/10/examen-parcial.html
Ejercicio1
Maximizar Z=X1+3X2
Sujeto a :
X1+X2<= 2 …. (1)
-X1+X2<=4 …. (2)
X1;X2>=0
Cm,n= m!/n!(m-n)! = C4,2= 6
MAXIMIZAR Z = 16X1 + 15X2
Cálculos para hallar S1
Coeficiente: 40
0 - (40.0 * 0 ) = 0.0
40 - (40.0 * 1 ) = 0.0
31 - (40.0 * 0 ) = 31.0
1 - (40.0 * 0 ) = 1.0
0 - (40.0 * 0 ) = 0.0
0 - (40.0 * 1 ) = -40.0
124 - (40.0 * 3 ) = 4.0
Hallando fila que reemplazara
Hallando fila que sera reemplazada.
Maximizar Z=X1+3X2
Sujeto a :
X1+X2<= 2 …. (1)
-X1+X2<=4 …. (2)
X1;X2>=0
Cm,n= m!/n!(m-n)! = C4,2= 6
Variables no basicas
|
Variables basicas
|
Solucion basica
|
Punto de esquina solucion
|
Factible
|
Valor objetivo
|
|
(x1,x2)
|
(s1,s2)
|
(2,4)
|
A
|
Si
|
0
|
|
(x1,s1)
|
(s2,s2)
|
(2,2)
|
B
|
Si
|
6
|
|
(x1,s2)
|
(x2,s1)
|
(4,-2)
|
C
|
No
|
---
|
|
(x2,s1)
|
(x1,s2)
|
(2,0)
|
D
|
Si
|
2
|
|
(x2,s2)
|
(x1,s2)
|
(-4,6)
|
---
|
No
|
---
|
|
(s1,s2)
|
(x1,x2)
|
(-1,3)
|
---
|
No
|
---
|
|
Ejercicio3
Considere El Problema: (SIMPLEX)
MAXIMIZAR Z = 16X1 + 15X2
Sujeto a:
40X1 + 31X2 <= 124
-X1 + X2 <= 1
X1 <= 3
X1 ; X2 >= 0
Nuevos Datos:
Z = 16x1 +
15x2 + 0s1 + 0s2 + 0s3
40x1 +
31x2 + s1 = 124
-x1 + x2 + s2 = 1
x1 + s3 = 3
-x1 + x2 + s2 = 1
x1 + s3 = 3
Iteraciones:
1° Iteracion :
Básica
|
Z
|
x1
|
x2
|
s1
|
s2
|
s3
|
Solución
|
Z
|
1
|
-16
|
-15
|
0
|
0
|
0
|
0
|
S1
|
0
|
40
|
31
|
1
|
0
|
0
|
124
|
S2
|
0
|
-1
|
1
|
0
|
1
|
0
|
1
|
S3
|
0
|
1
|
0
|
0
|
0
|
1
|
3
|
Ahora
obtendremos la fila de pivote:
Básica
|
X1
|
Solución
|
Relación
|
S1
|
40
|
124
|
124/40 = 3.1
|
S2
|
-1
|
1
|
-----
|
S3
|
1
|
3
|
3/1 = 3
|
Tomaremos el menor valor
obtenido en la columna de Relación
Tomaremos el menor
valor obtenido en la columna de Relación.
En este caso el 3;
por lo tanto S3 sera la fila que salga y obtenemos el elemento pivote el cual
seria 1(Sombreado con color rojo en la anterior tabla tabla).
Ahora hallaremos la
nueva fila :
Nueva fila pivote =
Fila pivote actual / elemento pivote
Fila pivote actual
= [0 , 1 , 0 , 0 , 0 , 1 , 3 ]
Elemento pivote = 1
Cálculos nueva
Fila(S3)
0/1 = 0
1/1 = 1
0/1 = 0
0/1 = 0
0/1 = 0
1/1 = 1
3/1 = 3
0/1 = 0
1/1 = 1
0/1 = 0
0/1 = 0
0/1 = 0
1/1 = 1
3/1 = 3
Fila pivote obtenida
= [0 , 1 , 0 , 0 , 0 , 1 , 3 ]
Ahora hallar las
filas Z,s1,s2,x1
Hallando Z:
1 -
(-16 * 0 ) = 1
-16 - (-16 * 1
) = 0
-15 - (-16 * 0
) = -15
0 -
(-16 * 0 ) = 0
0 -
(-16 * 0 ) = 0
0 -
(-16 * 1 ) = 16
0 -
(-16 * 3 ) = 48
Fila obtenida (Z)
= [1 , 0 , -15 , 0 , 0 , 16 , 48 ]
Cálculos para hallar S1
Coeficiente: 40
0 - (40.0 * 0 ) = 0.0
40 - (40.0 * 1 ) = 0.0
31 - (40.0 * 0 ) = 31.0
1 - (40.0 * 0 ) = 1.0
0 - (40.0 * 0 ) = 0.0
0 - (40.0 * 1 ) = -40.0
124 - (40.0 * 3 ) = 4.0
Cálculos para S2
Coeficiente: -1
0 - (-1.0 * 0 ) = 0.0
-1 - (-1.0 * 1 ) = 0.0
1 - (-1.0 * 0 ) = 1.0
0 - (-1.0 * 0 ) = 0.0
1 - (-1.0 * 0 ) = 1.0
0 - (-1.0 * 1 ) = 1.0
1 - (-1.0 * 3 ) = 4.0
Coeficiente: -1
0 - (-1.0 * 0 ) = 0.0
-1 - (-1.0 * 1 ) = 0.0
1 - (-1.0 * 0 ) = 1.0
0 - (-1.0 * 0 ) = 0.0
1 - (-1.0 * 0 ) = 1.0
0 - (-1.0 * 1 ) = 1.0
1 - (-1.0 * 3 ) = 4.0
Básica
|
Z
|
x1
|
x2
|
s1
|
s2
|
s3
|
Solución
|
Z
|
1
|
0
|
-15
|
0
|
0
|
16
|
48
|
S1
|
0
|
0
|
31
|
1
|
0
|
-40
|
4
|
S2
|
0
|
0
|
1
|
0
|
1
|
1
|
4
|
x1
|
0
|
1
|
0
|
0
|
0
|
1
|
3
|
Hallando fila que reemplazara
Básica
|
x2
|
Solución
|
Relación
|
S1
|
31
|
4
|
4/31=0.12(Sale)
|
S2
|
1
|
4
|
4/1= 4
|
X1
|
0
|
3
|
No
|
Llenando una nueva
tabla con los valores obtenidos.
Básica
|
Z
|
x1
|
x2
|
s1
|
s2
|
s3
|
Solución
|
Z
|
1
|
0
|
0
|
0.48
|
0
|
-3.35
|
49.94
|
x2
|
0
|
0
|
1
|
0.03
|
0
|
-1.29
|
0.13
|
S2
|
0
|
0
|
0
|
-0.03
|
1
|
2.29
|
3.87
|
x1
|
0
|
1
|
0
|
0
|
0
|
1
|
3
|
Hallando fila que sera reemplazada.
Básica
|
s3
|
Solución
|
Relación
|
x2
|
-1.29
|
0.13
|
No
|
S2
|
2.29
|
3.87
|
1.69(Sale)
|
X1
|
1
|
3
|
3
|
Básica
|
Z
|
x1
|
x2
|
s1
|
s2
|
s3
|
Solución
|
Z
|
1
|
0
|
0
|
0.44
|
1.46
|
0
|
55.61
|
x2
|
0
|
0
|
1
|
0.01
|
0.57
|
0
|
2.31
|
S2
|
0
|
0
|
0
|
-0.01
|
0.44
|
1
|
1.69
|
x1
|
0
|
1
|
0
|
0.01
|
-0.44
|
0
|
1.31
|
Respuesta:
Z = 55.61
X1 = 1.31
X2 = 2.31
S3 = 1.69
X1 = 1.31
X2 = 2.31
S3 = 1.69





