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Libro 1

The document outlines a linear programming problem aimed at maximizing the objective function Z = 7X1 + 10X2 under certain constraints. It details the steps of the simplex method, including the transformation of inequalities into equations and the iterative process to find the optimal solution. The final results indicate Z = 58, X1 = 4, and X2 = 3.

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Gonzalo
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35 views5 pages

Libro 1

The document outlines a linear programming problem aimed at maximizing the objective function Z = 7X1 + 10X2 under certain constraints. It details the steps of the simplex method, including the transformation of inequalities into equations and the iterative process to find the optimal solution. The final results indicate Z = 58, X1 = 4, and X2 = 3.

Uploaded by

Gonzalo
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLSX, PDF, TXT or read online on Scribd
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Max Z= 7X1 + 10X2

Sujeto a (-X1)+ 3X2 <=6


7X1 + X2 <= 35
X1 >=0 , X2 >= 0 y enteras

(-X1)+3X2 <= 6 --------------> -X1 + 3X2 +S1 = 6


7X1 + X2 <= 35 ---------------> 7X1 + X2 +S2 = 35

Z -7X1 - 10X2 = 0

Z X1 X2 S1 S2 LD
Z 1 -7 -10 0 0 0
S1 0 -1 3 1 0 6
S2 0 7 1 0 1 35

Z X1 X2 S1 S2 LD
10 Z 1 -10 1/3 0 3 1/3 0 20
X2 0 - 1/3 1 1/3 0 2
-1 S2 0 7 1/3 0 - 1/3 1 33

Z X1 X2 S1 S2 LD
10 1/3 Z 1 0 0 2 19/22 1 9/22 66 1/2
1/3 X2 0 0 1 7/22 1/22 3 1/2
X1 0 1 0 - 1/22 3/22 4 1/2

Z= 66 1/2
X1= 3 1/2
X2 4 1/2
S1= 0
S2= 0

Z+ (63/22)S1 + (31/22 )S2 = 66 1/2


X2 +( 7/22)S1 + (1/22)S2 = 3 1/2
X1-(1/22)S1+(3/22)S2 = 4 1/2

X1 + (-1/22 )S1 + (3/22)S2 = 4 1/2


X1 + (-1 + 21/22)S1 + (0 + 3/22)S2 = (4 + 1/2)
X1 -S1 + 0S2 -4 = (-21/22)S1 + (-3/22)S2 + 1/2
restr. (-21/22)S1 + (-3/22)S2 <= -1/2 normalizar la inecuación
(-21/22)S1 + (-3/22)S2 +S3 <= -1/2

X2 +( 7/22)S1 + (1/22)S2 = 3 1/2


X2 +(0 + 7/22)S1 + (0 + 1/22)S2 = (3 + 1/2)
X2 + 0S1 + 0S2 -3 = (-7/22)S1 + (-1/22)S2 + 1/2

restr. (-7/22)S1 + (-1/22)S2 <= - 1/2 normalizar la inecuación


(-7/22)S1 + (-1/22)S2 + S3 <= - 1/2

Z X1 X2 S1 S2 S3
Z 1 0 0 2 19/22 1 9/22 0
X2 0 0 1 7/22 1/22 0
X1 0 1 0 - 1/22 3/22 0
S3 0 0 0 - 7/22 - 1/22 1
-9 -31

Z X1 X2 S1 S2 S3
-2 19/22 Z 1 0 0 0 1 9
- 7/22 X2 0 0 1 0 0 1
1/22 X1 0 1 0 0 1/7 - 1/7
S1 0 0 0 1 1/7 -3 1/7

X1 + (1/7)S2 + (-1/7)S3 = 4 4/7


X1 + (0 + 1/7)S2 + (-1 + 6/7)S3 = (4 + 4/7)
X1 + 0S2 -S3 -4 = (-1/7)S2 + (-6/7)S3 + 4/7

(-1/7)S2 + (-6/7)S3 + 4/7 <= 0 Normalizar


(-1/7)S2 + (-6/7)S3+ S4 <= -4/7

Z X1 X2 S1 S2 S3
Z 1 0 0 0 1 9
X2 0 0 1 0 0 1
X1 0 1 0 0 1/7 - 1/7
S1 0 0 0 1 1/7 -3 1/7
S4 0 0 0 0 - 1/7 - 6/7
-7 -10 1/2

Z X1 X2 S1 S2 S3
-1 Z 1 0 0 0 0 3
0 X2 0 0 1 0 0 1
- 1/7 X1 0 1 0 0 0 -1
- 1/7 S1 0 0 0 1 0 -4
S2 0 0 0 0 1 6
Z=58
X1=4
X2= 3
Cuociente

2
35

Cuociente

--
4 1/2

Cuociente
LD Cuociente
66 1/2
3 1/2
4 1/2
- 1/2

LD Cuociente
62
3
4 4/7
1 4/7

S4 LD Cuociente
0 62
0 3
0 4 4/7
0 1 4/7
1 - 4/7

S4 LD Cuociente
7 58
0 3
1 4
1 1
-7 4

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