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EL PROBLEMA DUAL
MAXIMIZAR
Z= 400A + 300B
2A + B ≤ 60
A + 3B ≤ 40
A + B ≤ 30
A, B ≥ 0
FORMA ESTÁNDAR
Z= 400A + 300B + 0H1 + 0H2 + 0H3
2A + B + H1 ≤ 60
A + 3B +H2 ≤ 40
A + B + H3 ≤ 30
A, B, H1, H2, H3 ≥ 0
FORMA CANÓNICA O DE ECUACIONES
Z - 400A - 300B - 0H1 - 0H2 - 0H3 = 0
2A + B + H1 = 60
A + 3B +H2 = 40
A + B + H3 = 30
A, B, H1, H2, H3 ≥ 0
TABLA SIMPLEX
VB Z A B H1 H2 H3 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 0 0 100 0 0 400 12000
H1 0 0 -1 1 0 -2 0
H2 0 0 2 0 1 -1 10
A 0 1 1 0 0 1 30
VE= A
VS= H3
PIVOTE=1
RESPUESTAS DEL PROBLEMA PRIMAL:
Solución Óptima Z= 12000
Valores Óptimos A=30 H1=0
B= 0 H2=10 H3=0
FORMULACIÓN DEL PROBLEMA DUAL
MINIMIZAR
Z= 60Y1 + 40Y2 + 30Y3 Y1=100
2Y1 + Y2 + Y3 ≥ 400 Y2= 0
Y1 + 3Y2 + Y3 ≥ 300 Y3= 200
Yi ≥ 0
COMPROBACIÓN
Z= 60(100)+40(0)+30(200)
Z= 6000 + 6000
Z= 12000
2Y1 + Y3 = 400
-Y1 - Y3 = 300
Y1 = 100
2(100) + Y3 = 400
Y3 = 400-200
Y3 = 200
MINIMIZAR
Z= 4X1 + 7X2
X1 ≤ 6
2X2 = 14
3X1 + 2X2 ≥ 20
Xi ≥ 0
FORMA ESTÁNDAR
Z= 4X1 + 7X2 +MA1 +MA2 + 0H1 + 0H2 (-1)
-Z= -4X1 - 7X2 -MA1 -MA2 - 0H1 - 0H2
X1 + H1 ≤ 6
2X2 + A1 =14 (-M)
3X1 + 2X2 + A2 – H2 ≥ 20 (-M)
X1, X2, A1, A2, H1, H2 ≥ 0
FORMA CANÓNICA O DE ECUACIONES
-Z + 4X1 + 7X2 +MA1 + MA2 +0H1+0H2 = 0
-2MX2 -MA1 = -14M
- 3MX1 -2MX2 -MA2 +M = -20M
-Z+ (-3M+4) X1+ (-4M+7) X2 +0H1+MH2= -34M
X1 + H1 = 6
2X2 + A1 = 14
3X1 + 2X2 + A2 -H2 = 20
X1, X2, A1, A2, H1, H2 ≥ 0
TABLA SIMPLEX
VB Z X1 X2 A1 A2 H1 H2 VALOR
Z -1 (-3M+4) (-4M+7) 0 0 0 M (-34M)
H1 0 1 0 0 0 1 0 6
A1 0 0 2 1 0 0 0 14
A2 0 3 2 0 1 0 -1 20
Z -1 (-3M+4) 0 (2M-7/2) 0 0 M (-6M-49)
H1 0 1 0 0 0 1 0 6
X2 0 0 1 1/2 0 0 0 7
A2 0 3 0 -1 1 0 -1 6
Z -1 0 0 (M-13/6) (M-4/3) 0 1 1/3 -57
H1 0 0 0 1/3 - 1/3 1 1/3 4
X2 0 0 1 1/2 0 0 0 7
X1 0 1 0 - 1/3 1/3 0 - 1/3 2
VE= X2
VS= A1
PIVOTE= 2
VE= X1
VS= A2
PIVOTE= 3
RESPUESTAS:
Solución Óptima -Z= -57 Z=57
Valores Óptimos X1=2 H1=4
X2=7 H2=0
FORMULACIÓN DEL PROBLEMA DUAL
MINIMIZAR
Z= 6Y1 + 14Y2 + 20Y3 Y1=0
Y1 + 3Y3 ≥ 4 Y2= 13/6
2Y2 + 2Y3 < > 7 Y3= 4/3
Yi ≥ 0
COMPROBACIÓN
Z= 6(0)+14(13/6)+20(4/3)
Z= 57

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PROBLEMA DUAL

  • 1. EL PROBLEMA DUAL MAXIMIZAR Z= 400A + 300B 2A + B ≤ 60 A + 3B ≤ 40 A + B ≤ 30 A, B ≥ 0 FORMA ESTÁNDAR Z= 400A + 300B + 0H1 + 0H2 + 0H3 2A + B + H1 ≤ 60 A + 3B +H2 ≤ 40 A + B + H3 ≤ 30 A, B, H1, H2, H3 ≥ 0 FORMA CANÓNICA O DE ECUACIONES Z - 400A - 300B - 0H1 - 0H2 - 0H3 = 0 2A + B + H1 = 60 A + 3B +H2 = 40 A + B + H3 = 30 A, B, H1, H2, H3 ≥ 0 TABLA SIMPLEX VB Z A B H1 H2 H3 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 0 0 100 0 0 400 12000 H1 0 0 -1 1 0 -2 0 H2 0 0 2 0 1 -1 10 A 0 1 1 0 0 1 30 VE= A VS= H3 PIVOTE=1
  • 2. RESPUESTAS DEL PROBLEMA PRIMAL: Solución Óptima Z= 12000 Valores Óptimos A=30 H1=0 B= 0 H2=10 H3=0 FORMULACIÓN DEL PROBLEMA DUAL MINIMIZAR Z= 60Y1 + 40Y2 + 30Y3 Y1=100 2Y1 + Y2 + Y3 ≥ 400 Y2= 0 Y1 + 3Y2 + Y3 ≥ 300 Y3= 200 Yi ≥ 0 COMPROBACIÓN Z= 60(100)+40(0)+30(200) Z= 6000 + 6000 Z= 12000 2Y1 + Y3 = 400 -Y1 - Y3 = 300 Y1 = 100 2(100) + Y3 = 400 Y3 = 400-200 Y3 = 200 MINIMIZAR Z= 4X1 + 7X2 X1 ≤ 6 2X2 = 14 3X1 + 2X2 ≥ 20 Xi ≥ 0 FORMA ESTÁNDAR Z= 4X1 + 7X2 +MA1 +MA2 + 0H1 + 0H2 (-1) -Z= -4X1 - 7X2 -MA1 -MA2 - 0H1 - 0H2 X1 + H1 ≤ 6 2X2 + A1 =14 (-M) 3X1 + 2X2 + A2 – H2 ≥ 20 (-M) X1, X2, A1, A2, H1, H2 ≥ 0 FORMA CANÓNICA O DE ECUACIONES
  • 3. -Z + 4X1 + 7X2 +MA1 + MA2 +0H1+0H2 = 0 -2MX2 -MA1 = -14M - 3MX1 -2MX2 -MA2 +M = -20M -Z+ (-3M+4) X1+ (-4M+7) X2 +0H1+MH2= -34M X1 + H1 = 6 2X2 + A1 = 14 3X1 + 2X2 + A2 -H2 = 20 X1, X2, A1, A2, H1, H2 ≥ 0 TABLA SIMPLEX VB Z X1 X2 A1 A2 H1 H2 VALOR Z -1 (-3M+4) (-4M+7) 0 0 0 M (-34M) H1 0 1 0 0 0 1 0 6 A1 0 0 2 1 0 0 0 14 A2 0 3 2 0 1 0 -1 20 Z -1 (-3M+4) 0 (2M-7/2) 0 0 M (-6M-49) H1 0 1 0 0 0 1 0 6 X2 0 0 1 1/2 0 0 0 7 A2 0 3 0 -1 1 0 -1 6 Z -1 0 0 (M-13/6) (M-4/3) 0 1 1/3 -57 H1 0 0 0 1/3 - 1/3 1 1/3 4 X2 0 0 1 1/2 0 0 0 7 X1 0 1 0 - 1/3 1/3 0 - 1/3 2 VE= X2 VS= A1 PIVOTE= 2 VE= X1 VS= A2 PIVOTE= 3 RESPUESTAS: Solución Óptima -Z= -57 Z=57 Valores Óptimos X1=2 H1=4 X2=7 H2=0
  • 4. FORMULACIÓN DEL PROBLEMA DUAL MINIMIZAR Z= 6Y1 + 14Y2 + 20Y3 Y1=0 Y1 + 3Y3 ≥ 4 Y2= 13/6 2Y2 + 2Y3 < > 7 Y3= 4/3 Yi ≥ 0 COMPROBACIÓN Z= 6(0)+14(13/6)+20(4/3) Z= 57