P.l.metodo Dual- Gauss Y Jordan, Simplex, Metodo M, Metodo Algebraico.pdf 4v4j1y
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PROGRAMACIÓN LINEAL MÉTODO DUAL
(20/05/2015)
MAX Z=5X1+6X2
s.a X1+9X2≤60 2X1+3X2≤45 5X1-2X2≤20 X2≤30 X1,X2≥0 1.- X1+9X2=60 X1
P(0,0)
X2 0 60
6,67 0
0≤60 Verdadero
2.- 2X1+3X2=45 P(0,0) X1
X2 0 22,5
15 0
0≤45 Verdadero
3.- 5X1-2X2=20 P(0,0) X1
X2 0 4
-10 0
0≤20 Verdadero
-5X1-45X2=-300 s.o.
5X1-2X2=20 -47X2=280
Z=67,56 X1=6,36
X2=5.96 X2=5,96 H1=0 X1+9(5.96)=60
H2=14
60-53.64=6.36
H3=0
X1=6.36
H4=24
X2=5,96 Z=67,56
.- 6,36+9(5,96)+H1≤60 H1≤0 .- 2(6,36)+6(5,96)+H2≤45
H2≤14 .- 5(6,36)-2(5,96)+H3≤20 H3≤0 .-
H4≤24 PROBLEMA DUAL
Min Z=60Y1+45Y2+20Y3+30Y4 Limitado 1.- Y1+2Y2+5Y3≥5 2.- 9Y1+3Y2-2Y2+Y4≥6 Y1, Y2, Y3, Y4≥0 Y1 Y2=0 Y3 Y4=0 Y1+5Y3=5 9Y1-2Y3=6
-9y1-45y3=-45 9y1-2y3=6 +47Y3=+39 39
Y3= 47 39
Y1+5(47)=5 Y1=
235−195 47
40
= 47
40
39
Z=60 (47)+20(47) Z=
3180 47
Z=67,66 MÉTODO DE GAUSS-JORDÁN EJERCICIO 1 2x1+3X2+4X3=5 2x1+X2-5X3= 6
-4
-6 -8 = -10
3x1-2X2+4X3=1
4
+1 -5 = 6
-3 -9/2 -6 3 -2 1 - 3/2 + 2= 5/2 0 – 5 – 13 = -4
= -15/2
4
7
0 - 3/2 -39/2 =-6/5 1 +3/2
+2
= 5 /2
0- 13/2 -2 = -1/2
1 + 0 - 19/10 = 13/10 0 + 1 + 13/5 = 4/5
00 - 13/2 -169/10 =26/5 0 19/10 = +893/1490 11 +13/2 +2 = 13/10 =-1 /2 0 -19/10
0 + 0 – 149/10 =47/10
1 + 0 + 0 = 283/149 0 + 1 + 0 = -3/149 0 + 0 + 1 = 47/149
0
0
-13/5
=-611/745
0
1
13/5
=
4/5
X= 283/149 Y= -3/149 Z= 47/149
COMPROBACIÓN 2x1+3X2+4X3=5 2(283/149)+3(3/149)+4(47/149)
EJERCICIO 2
5=5
2x+5y-2z= 10 5x-2y+3z= 7
-5
4x+6y+8z=1
5
-4 4 1 + 5/2 - 1= 5 0 – 29/2 – 8 = -18
-25/2 5 = -25 -2
3= 7
-10 4
= -20
6
8
= 1
0 - 5/2 40/29 =-20/29 1 5/2
-1
= 5
0 - 4 12 = -19
1 + 0 + 11/29 = 55/29 0 + 1 + 16/29 = 36/29 0 + 0 – 284/29 =-407/29
1 + 0 + 0 = 693/149 0 + 1 + 0 = 32/71 0 + 0 + 1 = -407/284
0
4
-64/29 =144/29
1
-4
0
0 16/29 = -1628/2059
1
0 -16/29 = 36/29
12
=
-19
0
0
-11/29
= 4477/8236
1
0
11/29
= 55/29
X= 693/284
COMPROBACIÓN
Y= 32/71
2x+5y-2z= 10
Z=-407/284
2(693/284)+3(32/71)+4(407/284) 10 =10 METODO SIMPLEX (12/06/2015)
EJERCICIO N.- 1 ECUACIÓN
MAX. Z= 2X1 + X2 S.A 3X1 + X2 ≤ 6 X1 – X2 ≤ 2 X2 ≤ 3 Xi ≥ 0
Z – 2X1 – X2 - 0H1 – 0H2 – OH3 = 0 S.A 3X1 + X2 + H1 = 6 X1 + X2 + H2 = 12 X2 +H3 =3 Xi ≥ 0 ; Hj ≥ 0
FORMA ESTÁNDAR Z= 2X + X2 + 0H1 + 0H2 + OH3 S.A 3X1 + X2 + H1 = 6 X1 + X2 +H2 = 2 X2 +H3 = 3 Xi ≥ 0 ; Hj ≥ 0
VARIABLES BÁSICAS
VARIABLES
VALOR
Z
X1
X2
H1
H2
H3
Z
1
-2
-1
0
0
0
0
H1
0
3
1
1
0
0
6
H2
0
1
-1
0
1
0
2
H3
0
0
1
0
0
1
3
Z
1
0
- 1/3
2/3
0
0
4
X1
0
1
1/3
1/3
0
0
2
H2
0
0
-4/3
- 1/3
1
0
0
H3
0
0
1
0
0
1
3
Z
1
0
0
2/3
0
1/3
5
X2
0
0
1
0
0
1
3
X1
0
0
0
1/3
0
- 1/3
1
H2
0
0
0
- 1/3
1
4/3
4
VALOR ÓPTIMO
SOLUCIÓN ÓPTIMA
Z= 5
X1= 1 X2= 3 H1= 0 H2= 4 H3= 0
EJERCICIO N.-2 MAX. Z= 400A + 300B S.A 2A + B ≤ 60 A + 3B ≤ 40 A + B ≤ 30 A, B ≥ 0
FORMA ESTÁNDAR
ECUACIÓN
Z=400A + 300B + OH1 + 0H2 + 0H3 S.A 2A + B + 0H1 ≤ 60 A + 3B A+ B
Z- 400A - 300B - OH1 - 0H2 - 0H3 = 0 S.A 2A + B + 0H1 = 60 A + 3B + 0H2 = 40 A+ B + 0H3 = 30 A, B ≥ 0
+ 0H2 ≤ 40 + 0H3 ≤ 30 A, B ≥ 0
Z
A
B
H1
H2
A1
VALOR
Z
1
- 400
-300
0
0
0
0
H1
0
2
1
1
0
0
60
H2
0
1
3
0
1
0
40
H3
0
1
1
0
0
1
30
Z
1
0
100
0
0
400
1200
H1
0
0
-1
1
0
-2
0
H2
0
0
2
0
1
-1
10
A
0
1
1
0
0
1
30
VALOR ÓPTIMO
SOLUCIÓN ÓPTIMA
Z= 12000
A = 30 B = 0 H1= 0 H2= 10 H3= 0
PROBLEMA DUAL. Z= 60Y1+ 40Y2+ 30Y3 S.A 2Y1 + Y2 + Y3 ≥ 400 Y1 + 3Y2 + Y3 ≥ 300 Y1 ≥ 0
2Y1 + Y3 = 400 -Y1 - Y3 = -300 Y1 = 100 Y1 = 100
Z = 12000
Y3 = 200
(16/06/2015) EJERCICIO N.-1 MINIMIZAR ECUACIÓN Z = 3/2X + 2Y S.A 2X + 2Y ≤ 8 2X + 6Y ≥ 12 X, Y ≥ 0
Z – 3/2X –2Y - 0H1 – 0H2 – MA1 = 0 S.A 2X + 2Y + H1 =8 2X + 6Y + A1 – H2 = 12 X, Y, H1, H2, A1 ≥ 0
FORMA ESTÁNDAR Z= 3/2X + 2Y + 0H1 + 0H2 + MA1 S.A 2X + 2Y + H1 = 8 2X + 6Y +A1-H2 = 12 X, Y, H1, H2, A1 ≥ 0 Z – 3/2X – 2Y – OH1 – OH2 – MA1 = 0 2XM + 6MY -MH2 + MA1 = 12M Z –3/2X+2XM–2Y+6MY–OH1 –MH2 = 12M (-1) -Z + 3/2X-2XM +2Y-6MY +OH1 +MH2 = -12M -Z +X (3/2-2M) +Y (2-6M) +OH1 +MH2 = -12M 2X + 2Y + H1 =8 2X + 6Y + A1 – H2 = 12 X, Y, H1, H2, A1 ≥ 0
Z
X
Y
H1
H2
A1
VALOR
Z
-1
3/2-4M
2-6M
0
M
0
-12M
H1
0
2
2
1
0
0
8
A1
0
2
6
0
-1
1
12
Z
-1
5/6
0
0
1/3
-1/3 +M
-6
X
0
4/3
0
1
1/3
- 1/3
4
H1
0
1/3
1
0
-1/6
1/6
2
VALOR ÓPTIMO
SOLUCIÓN ÓPTIMA
Z= - 4 Z =
X=0
4
Y=2 H1= 4
H2= 0 EJERCICIO N.-2 MINIMIZAR Z = 4X + 5Y S.A 2X + 2Y ≤ 10 2X + 6Y ≥ 18 X+ Y =7 X, Y ≥ 0
FORMA ESTÁNDAR Z=4X + 5Y + 0H1 + 0H2 + MA1 + MA2 S.A 2X + 2Y + H1 = 10 2X + 6Y +A1-H2 = 18 X+Y + A2 = 7 X, Y, H1, H2, A1, A2 ≥ 0
ECUACIÓN Z – 4X – 5Y – OH1 – OH2 – MA1 – MA2 = 0 2XM + 6MY -MH2 + MA1 = 18M XM + YM + MA2 = 7M Z – 4X + 3XM – 5Y + 7MY – OH1 – MH2 = 25M (-1) -Z + 4X - 3XM + 5Y - 7MY + OH1 + MH2= -25M -Z +X (4 - 3M) +Y (5 - 7M) +OH1 +MH2 = -25M 2X + 2Y + H1 = 10 2X + 6Y +A1-H2 = 18 X+Y + A2 = 7 X, Y, H1, H2, A1, A2 ≥ 0
Z
X
Y
H1
H2
A1
A2
VALOR
Z
-1
4-3M
5-7M
0
M
0
0
-25M
H1
0
2
2
1
0
0
0
10
A1
0
2
6
0
-1
1
0
18
A2
0
-1
1
0
0
0
1
7
Z
-1
7/3-2/3M
0
0
5/6-1/6M
-5/6+6/7M
0
-15-4M
H1
0
4/3
0
1
1/3
-1/3
0
4
Y
0
1/3
1
0
-1/6
1/6
0
3
A2
0
2/3
0
0
1/6
-1/6
1
4
Z
-1
0
0
-7/4+1/2M
1/4
-1/4+M
0
-22-2M
X
0
1
0
3/4
1/4
-1/4
0
3
Y
0
0
1
-1/4
-1/4
1/4
0
2
A2
0
0
0
-1/2
0
0
1
2
VALOR ÓPTIMO -Z= -22-2M Z= 22 + 2M
SOLUCIÓN ÓPTIMA X= 3 Y= 2 H1= 0 H2= 4 A1= 0 A2= 2
NO TIENE SOLUCIÓN
MÉTODO ALGEBRAICO Maximizar F(x,y)=3x+2y RESTRICCIONES (1) 2x+y≤ 10 (2) x+y≥ 7 RESTRICCIONES DE NO NEGATIVIDAD (3) x, y0 Despeje de x 𝑥+𝑦 = 7 𝑥 =7−𝑦 𝑋 = 7−𝟒 𝑿=𝟑
Despeje de y 2(7 − 𝑦) + 𝑦 = 10 14 − 2𝑦 + 𝑦 = 10 −𝑦 = −4 𝒚=𝟒
MÉTODO GRÁFICO SISTEMAS ECUACIONES
(1)
(2)
2x+y= 10
x+y=7
x y
x y
0 10
0 7
5 0
7 0
COMPROBACIÓN
GRÁFICO
P(0,0)
P(0,0)
(1)
(2)
2(0)+1(0)≤10
1(0)+1(0)≤7
0≤10
0≥7
VERDAD
VERDAD
ARCO CONVEXO Punto
X
Y
Z
A
0
7
14
B
3
4
17
C
5
0
15
MÉTODO SIMPLEX F.O Z= 3X+2Y S.A 2X+Y≤10 X+Y≤7
NO NEGATIVIDAD X,Y≥0
FORMA ESTANDAR Z=3X+2Y+0H1+0H2 2X+Y+ H1
=10
X+Y
=7
- H2
X,Y, H1, H2, A1≥0
FORMA ECUACIÓN Z-3X-2Y-0H1-0H2 =0 2X+Y+ H1+0=10 X + Y+0- H2=7 X,Y, H1, H2, A1≥0
Z
X
Y
H1
H2
VALOR
Z
1
-3
-2
0
0
0
H1
0
2
1
1
0
10
A1
0
4
1
0
1
7
Z
X
Y
H1
H2
VALOR
Z
1
1
1 2
2 3
0
15
H1
0
1
1 2
1 2
0
5
X
0
0
1 2
−
1
2
1 2
1 2
0
-1
−
0
1
1
0
3
1
-3
−
1 2
0
-5
0
1
7
3 2
3 2
0
15
-2
0
0
0
Z
X
Y
H1
H2
VALOR
Z
1
0
0
1
1
17
H1
0
1
0
1
-1
3
X
0
0
1
-1
2
4
0
0
0
1
0
0
1 2 1 − 2 −
1 2
1 2 1 2
−
1 2
-1
-2
0
5
1
2
0
0
S.O F(X,Y)=17 S.A X=3 Y=4 H1=0 METODO DUAL H2=0 MIN f(X,Y)=10Y1+7Y2 S.A 2Y1+Y 2 ≥3 Y1+Y 2 ≥2
2Y1+Y2=3 - Y1-Y 2 =-3 Y1=1 Y2=1
COMPROBACIÓN Z=10(1)+7(1) Z=17
−
1 2
3 2
0
15
(20/05/2015)
MAX Z=5X1+6X2
s.a X1+9X2≤60 2X1+3X2≤45 5X1-2X2≤20 X2≤30 X1,X2≥0 1.- X1+9X2=60 X1
P(0,0)
X2 0 60
6,67 0
0≤60 Verdadero
2.- 2X1+3X2=45 P(0,0) X1
X2 0 22,5
15 0
0≤45 Verdadero
3.- 5X1-2X2=20 P(0,0) X1
X2 0 4
-10 0
0≤20 Verdadero
-5X1-45X2=-300 s.o.
5X1-2X2=20 -47X2=280
Z=67,56 X1=6,36
X2=5.96 X2=5,96 H1=0 X1+9(5.96)=60
H2=14
60-53.64=6.36
H3=0
X1=6.36
H4=24
X2=5,96 Z=67,56
.- 6,36+9(5,96)+H1≤60 H1≤0 .- 2(6,36)+6(5,96)+H2≤45
H2≤14 .- 5(6,36)-2(5,96)+H3≤20 H3≤0 .-
H4≤24 PROBLEMA DUAL
Min Z=60Y1+45Y2+20Y3+30Y4 Limitado 1.- Y1+2Y2+5Y3≥5 2.- 9Y1+3Y2-2Y2+Y4≥6 Y1, Y2, Y3, Y4≥0 Y1 Y2=0 Y3 Y4=0 Y1+5Y3=5 9Y1-2Y3=6
-9y1-45y3=-45 9y1-2y3=6 +47Y3=+39 39
Y3= 47 39
Y1+5(47)=5 Y1=
235−195 47
40
= 47
40
39
Z=60 (47)+20(47) Z=
3180 47
Z=67,66 MÉTODO DE GAUSS-JORDÁN EJERCICIO 1 2x1+3X2+4X3=5 2x1+X2-5X3= 6
-4
-6 -8 = -10
3x1-2X2+4X3=1
4
+1 -5 = 6
-3 -9/2 -6 3 -2 1 - 3/2 + 2= 5/2 0 – 5 – 13 = -4
= -15/2
4
7
0 - 3/2 -39/2 =-6/5 1 +3/2
+2
= 5 /2
0- 13/2 -2 = -1/2
1 + 0 - 19/10 = 13/10 0 + 1 + 13/5 = 4/5
00 - 13/2 -169/10 =26/5 0 19/10 = +893/1490 11 +13/2 +2 = 13/10 =-1 /2 0 -19/10
0 + 0 – 149/10 =47/10
1 + 0 + 0 = 283/149 0 + 1 + 0 = -3/149 0 + 0 + 1 = 47/149
0
0
-13/5
=-611/745
0
1
13/5
=
4/5
X= 283/149 Y= -3/149 Z= 47/149
COMPROBACIÓN 2x1+3X2+4X3=5 2(283/149)+3(3/149)+4(47/149)
EJERCICIO 2
5=5
2x+5y-2z= 10 5x-2y+3z= 7
-5
4x+6y+8z=1
5
-4 4 1 + 5/2 - 1= 5 0 – 29/2 – 8 = -18
-25/2 5 = -25 -2
3= 7
-10 4
= -20
6
8
= 1
0 - 5/2 40/29 =-20/29 1 5/2
-1
= 5
0 - 4 12 = -19
1 + 0 + 11/29 = 55/29 0 + 1 + 16/29 = 36/29 0 + 0 – 284/29 =-407/29
1 + 0 + 0 = 693/149 0 + 1 + 0 = 32/71 0 + 0 + 1 = -407/284
0
4
-64/29 =144/29
1
-4
0
0 16/29 = -1628/2059
1
0 -16/29 = 36/29
12
=
-19
0
0
-11/29
= 4477/8236
1
0
11/29
= 55/29
X= 693/284
COMPROBACIÓN
Y= 32/71
2x+5y-2z= 10
Z=-407/284
2(693/284)+3(32/71)+4(407/284) 10 =10 METODO SIMPLEX (12/06/2015)
EJERCICIO N.- 1 ECUACIÓN
MAX. Z= 2X1 + X2 S.A 3X1 + X2 ≤ 6 X1 – X2 ≤ 2 X2 ≤ 3 Xi ≥ 0
Z – 2X1 – X2 - 0H1 – 0H2 – OH3 = 0 S.A 3X1 + X2 + H1 = 6 X1 + X2 + H2 = 12 X2 +H3 =3 Xi ≥ 0 ; Hj ≥ 0
FORMA ESTÁNDAR Z= 2X + X2 + 0H1 + 0H2 + OH3 S.A 3X1 + X2 + H1 = 6 X1 + X2 +H2 = 2 X2 +H3 = 3 Xi ≥ 0 ; Hj ≥ 0
VARIABLES BÁSICAS
VARIABLES
VALOR
Z
X1
X2
H1
H2
H3
Z
1
-2
-1
0
0
0
0
H1
0
3
1
1
0
0
6
H2
0
1
-1
0
1
0
2
H3
0
0
1
0
0
1
3
Z
1
0
- 1/3
2/3
0
0
4
X1
0
1
1/3
1/3
0
0
2
H2
0
0
-4/3
- 1/3
1
0
0
H3
0
0
1
0
0
1
3
Z
1
0
0
2/3
0
1/3
5
X2
0
0
1
0
0
1
3
X1
0
0
0
1/3
0
- 1/3
1
H2
0
0
0
- 1/3
1
4/3
4
VALOR ÓPTIMO
SOLUCIÓN ÓPTIMA
Z= 5
X1= 1 X2= 3 H1= 0 H2= 4 H3= 0
EJERCICIO N.-2 MAX. Z= 400A + 300B S.A 2A + B ≤ 60 A + 3B ≤ 40 A + B ≤ 30 A, B ≥ 0
FORMA ESTÁNDAR
ECUACIÓN
Z=400A + 300B + OH1 + 0H2 + 0H3 S.A 2A + B + 0H1 ≤ 60 A + 3B A+ B
Z- 400A - 300B - OH1 - 0H2 - 0H3 = 0 S.A 2A + B + 0H1 = 60 A + 3B + 0H2 = 40 A+ B + 0H3 = 30 A, B ≥ 0
+ 0H2 ≤ 40 + 0H3 ≤ 30 A, B ≥ 0
Z
A
B
H1
H2
A1
VALOR
Z
1
- 400
-300
0
0
0
0
H1
0
2
1
1
0
0
60
H2
0
1
3
0
1
0
40
H3
0
1
1
0
0
1
30
Z
1
0
100
0
0
400
1200
H1
0
0
-1
1
0
-2
0
H2
0
0
2
0
1
-1
10
A
0
1
1
0
0
1
30
VALOR ÓPTIMO
SOLUCIÓN ÓPTIMA
Z= 12000
A = 30 B = 0 H1= 0 H2= 10 H3= 0
PROBLEMA DUAL. Z= 60Y1+ 40Y2+ 30Y3 S.A 2Y1 + Y2 + Y3 ≥ 400 Y1 + 3Y2 + Y3 ≥ 300 Y1 ≥ 0
2Y1 + Y3 = 400 -Y1 - Y3 = -300 Y1 = 100 Y1 = 100
Z = 12000
Y3 = 200
(16/06/2015) EJERCICIO N.-1 MINIMIZAR ECUACIÓN Z = 3/2X + 2Y S.A 2X + 2Y ≤ 8 2X + 6Y ≥ 12 X, Y ≥ 0
Z – 3/2X –2Y - 0H1 – 0H2 – MA1 = 0 S.A 2X + 2Y + H1 =8 2X + 6Y + A1 – H2 = 12 X, Y, H1, H2, A1 ≥ 0
FORMA ESTÁNDAR Z= 3/2X + 2Y + 0H1 + 0H2 + MA1 S.A 2X + 2Y + H1 = 8 2X + 6Y +A1-H2 = 12 X, Y, H1, H2, A1 ≥ 0 Z – 3/2X – 2Y – OH1 – OH2 – MA1 = 0 2XM + 6MY -MH2 + MA1 = 12M Z –3/2X+2XM–2Y+6MY–OH1 –MH2 = 12M (-1) -Z + 3/2X-2XM +2Y-6MY +OH1 +MH2 = -12M -Z +X (3/2-2M) +Y (2-6M) +OH1 +MH2 = -12M 2X + 2Y + H1 =8 2X + 6Y + A1 – H2 = 12 X, Y, H1, H2, A1 ≥ 0
Z
X
Y
H1
H2
A1
VALOR
Z
-1
3/2-4M
2-6M
0
M
0
-12M
H1
0
2
2
1
0
0
8
A1
0
2
6
0
-1
1
12
Z
-1
5/6
0
0
1/3
-1/3 +M
-6
X
0
4/3
0
1
1/3
- 1/3
4
H1
0
1/3
1
0
-1/6
1/6
2
VALOR ÓPTIMO
SOLUCIÓN ÓPTIMA
Z= - 4 Z =
X=0
4
Y=2 H1= 4
H2= 0 EJERCICIO N.-2 MINIMIZAR Z = 4X + 5Y S.A 2X + 2Y ≤ 10 2X + 6Y ≥ 18 X+ Y =7 X, Y ≥ 0
FORMA ESTÁNDAR Z=4X + 5Y + 0H1 + 0H2 + MA1 + MA2 S.A 2X + 2Y + H1 = 10 2X + 6Y +A1-H2 = 18 X+Y + A2 = 7 X, Y, H1, H2, A1, A2 ≥ 0
ECUACIÓN Z – 4X – 5Y – OH1 – OH2 – MA1 – MA2 = 0 2XM + 6MY -MH2 + MA1 = 18M XM + YM + MA2 = 7M Z – 4X + 3XM – 5Y + 7MY – OH1 – MH2 = 25M (-1) -Z + 4X - 3XM + 5Y - 7MY + OH1 + MH2= -25M -Z +X (4 - 3M) +Y (5 - 7M) +OH1 +MH2 = -25M 2X + 2Y + H1 = 10 2X + 6Y +A1-H2 = 18 X+Y + A2 = 7 X, Y, H1, H2, A1, A2 ≥ 0
Z
X
Y
H1
H2
A1
A2
VALOR
Z
-1
4-3M
5-7M
0
M
0
0
-25M
H1
0
2
2
1
0
0
0
10
A1
0
2
6
0
-1
1
0
18
A2
0
-1
1
0
0
0
1
7
Z
-1
7/3-2/3M
0
0
5/6-1/6M
-5/6+6/7M
0
-15-4M
H1
0
4/3
0
1
1/3
-1/3
0
4
Y
0
1/3
1
0
-1/6
1/6
0
3
A2
0
2/3
0
0
1/6
-1/6
1
4
Z
-1
0
0
-7/4+1/2M
1/4
-1/4+M
0
-22-2M
X
0
1
0
3/4
1/4
-1/4
0
3
Y
0
0
1
-1/4
-1/4
1/4
0
2
A2
0
0
0
-1/2
0
0
1
2
VALOR ÓPTIMO -Z= -22-2M Z= 22 + 2M
SOLUCIÓN ÓPTIMA X= 3 Y= 2 H1= 0 H2= 4 A1= 0 A2= 2
NO TIENE SOLUCIÓN
MÉTODO ALGEBRAICO Maximizar F(x,y)=3x+2y RESTRICCIONES (1) 2x+y≤ 10 (2) x+y≥ 7 RESTRICCIONES DE NO NEGATIVIDAD (3) x, y0 Despeje de x 𝑥+𝑦 = 7 𝑥 =7−𝑦 𝑋 = 7−𝟒 𝑿=𝟑
Despeje de y 2(7 − 𝑦) + 𝑦 = 10 14 − 2𝑦 + 𝑦 = 10 −𝑦 = −4 𝒚=𝟒
MÉTODO GRÁFICO SISTEMAS ECUACIONES
(1)
(2)
2x+y= 10
x+y=7
x y
x y
0 10
0 7
5 0
7 0
COMPROBACIÓN
GRÁFICO
P(0,0)
P(0,0)
(1)
(2)
2(0)+1(0)≤10
1(0)+1(0)≤7
0≤10
0≥7
VERDAD
VERDAD
ARCO CONVEXO Punto
X
Y
Z
A
0
7
14
B
3
4
17
C
5
0
15
MÉTODO SIMPLEX F.O Z= 3X+2Y S.A 2X+Y≤10 X+Y≤7
NO NEGATIVIDAD X,Y≥0
FORMA ESTANDAR Z=3X+2Y+0H1+0H2 2X+Y+ H1
=10
X+Y
=7
- H2
X,Y, H1, H2, A1≥0
FORMA ECUACIÓN Z-3X-2Y-0H1-0H2 =0 2X+Y+ H1+0=10 X + Y+0- H2=7 X,Y, H1, H2, A1≥0
Z
X
Y
H1
H2
VALOR
Z
1
-3
-2
0
0
0
H1
0
2
1
1
0
10
A1
0
4
1
0
1
7
Z
X
Y
H1
H2
VALOR
Z
1
1
1 2
2 3
0
15
H1
0
1
1 2
1 2
0
5
X
0
0
1 2
−
1
2
1 2
1 2
0
-1
−
0
1
1
0
3
1
-3
−
1 2
0
-5
0
1
7
3 2
3 2
0
15
-2
0
0
0
Z
X
Y
H1
H2
VALOR
Z
1
0
0
1
1
17
H1
0
1
0
1
-1
3
X
0
0
1
-1
2
4
0
0
0
1
0
0
1 2 1 − 2 −
1 2
1 2 1 2
−
1 2
-1
-2
0
5
1
2
0
0
S.O F(X,Y)=17 S.A X=3 Y=4 H1=0 METODO DUAL H2=0 MIN f(X,Y)=10Y1+7Y2 S.A 2Y1+Y 2 ≥3 Y1+Y 2 ≥2
2Y1+Y2=3 - Y1-Y 2 =-3 Y1=1 Y2=1
COMPROBACIÓN Z=10(1)+7(1) Z=17
−
1 2
3 2
0
15
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