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Ejercicio El 25

The document describes an optimization problem to maximize the objective function Z=40x1+30x2 subject to three constraints. It introduces slack variables to transform inequality constraints into equalities. The optimal solution is found to be x1=49/2, x2=21, with the maximum value of Z being 1610.

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23 views4 pages

Ejercicio El 25

The document describes an optimization problem to maximize the objective function Z=40x1+30x2 subject to three constraints. It introduces slack variables to transform inequality constraints into equalities. The optimal solution is found to be x1=49/2, x2=21, with the maximum value of Z being 1610.

Uploaded by

valentina10
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 = 40x1 +30x2

sujeta a: 0,4x1 +0,5x2 <=20 (2/5)x1+ (1/2)x2<=20


0,2x2 <=5 (1/5)x2<=5
0,6x1 +0,3x2 <=21 (3/5)x1+ (3/10)<=21
x1, x2>=0

CUANDO LAS RESTRICCIONES SEAN DE LA FORMA


<= SIEMPRE SUMAREMOS VARIABLE DE HOLGURA
TANTAS COMO RESTRICCIONES TENGAMOS.

(2/5)X1+(1/2)X2 +X3 =20


(1/5)X2 + X4 = 5
(3/5)X1 + (3/10)X2 +X5 =21

Max Z = 40x1 +30x2 +0X3 +0X4 +0X5 max Z -40x1 -30x2-0x3-0x4-0x5=0

Sol Bas.Inic Z x1 x2 x3
x3 0 2/5 1/2 1
x4 0 0 1/5 0
x5 0 3/5 2/7 0
Z 1 -40 -30 0

Sol Bas.Inic Z x1 x2 x3
x3 0 2/5 1/2 1
x4 0 0 1/5 0
x1 0 1 1/2 0
Z 1 -40 -30 0

Sol Bas.Inic Z x1 x2 x3
x3 0 0 2/7 1
x4 0 0 1/5 0
x1 0 1 1/2 0
Z 1 0 -10 0

Sol Bas.Inic Z x1 x2 x3
x2 0 0 1 3 1/2
x4 0 0 0 - 5/7
x1 0 1 0 -1 3/4
Z 1 0 0 35

x1 49/2 Z = 40*49/2 +30*21 +0+ 0*4/5+ 0 = 1610


x2 21
x3 0 (2/5)X1+(1/2)X2 +X3 =20
x4 4/5 0
x5 0 (1/5)X2 + X4 = 5
1

(3/5)X1 + (3/10)X2 +X5 =21

1
Restriccion x y Disp
1 2/5 1/2 20
2 1/5 5
3 3/5 2/7 21
/5)x1+ (1/2)x2<=20

/5)x1+ (3/10)<=21

ROJAS SON VARIABLES DE HOLGURA

x1=0
x2 = 0

ax Z -40x1 -30x2-0x3-0x4-0x5=0

x4 x5 Disp
0 0 20
1 0 5
0 1 21
0 0 0

x4 x5 Disp
0 0 20
1 0 5
0 1 2/3 35 1 2/3 -2/5+ 1a *40 +3a
0 0 0

x4 x5 Disp
0 - 2/3 6 *7/2
1 0 5
0 1 2/3 35
0 66 2/3 1400

x4 x5 Disp
0 -2 1/3 21 *-1/5 +2a *-1/2 + 3a *10 +4a
1 4/9 4/5
0 2 5/6 24 1/2
0 130/3 1610

+0+ 0*4/5+ 0 = 1610

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