Your SlideShare is downloading. ×
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
5 ejercicios resueltos modelos de programacion lineal
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

5 ejercicios resueltos modelos de programacion lineal

1,453

Published on

Published in: Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
1,453
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
25
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Ejercicio 1.1 REQUERIMIENTO HORAS REQUERIMINETO HORAS HORAS DISPONIBLES MES M1 M2 ESTRUCTURA PRINCIPAL 4 2 1600 ALAMBRADO ELECTRICO 2.5 1 1200 ENSAMBLADO 4.5 1.5 1600 BENEFICIO UNITARIO 40.00 10.00X1: Número de equipos de prueba M1.X2: Número de equipos de prueba M2.Max Z = 40 X1 + 10 X2s.a 4 X1 + 2 X2 <= 1600 2.5 X1 + 1 X2 <= 1200 4.5 X1 + 1.5 X2 <= 1600Estándar:Min Z = - 40 X1 - 10 X2 + 0 X3 + 0X4 + 0X5 4 X1 + 2 X2 + X3 = 1600 2.5 X1 + 1 X2 + X4 = 1200 4.5 X1 + 1.5 X2 + X5 = 1600Canónico: X1 X2 X3 X4 X5 Z -40 -10 0 0 0 0 4 2 1 0 0 1600 2.5 1 0 1 0 1200 4.5 1.5 0 0 1 1600X1 = X2 = 0 , X3 = 1600 , X4 = 1200 , X5 = 1600 , Z = 0 X1 X2 X3 X4 X5 Z -40 -10 0 0 0 0 4 2 1 0 0 1600 5/2 1 0 1 0 1200 1 1/3 0 0 2/9 3200/9 X1 X2 X3 X4 X5 Z 0 10/3 0 0 80/9 128000/9 0 -2/3 -1 0 8/9 -1600/9 0 -1/6 0 -1 5/9 -2800/9 1 1/3 0 0 2/9 3200/9 X1 X2 X3 X4 X5 Z 0 10/3 0 0 80/9 128000/9 0 2/3 1 0 -8/9 1600/9 0 1/6 0 1 -5/9 2800/9 1 1/3 0 0 2/9 3200/9Solución:X1 = 3200/9 , X2 = 0 , X3 = 1600/9 , X4 = 2800/9 , X5 = 0 , Z = 128000/9
  • 2. Ejercicio 1.3 TIPO A TIPO B TOTAL BOMBONES DE LICOR 0.30 Kgr. 0.40 Kgr. 100 Kgr. BOMBONES DE NUEZ 0.50 Kgr. 0.20 Kgr. 120 Kgr. BOMBONES DE FRUTA 0.20 Kgr. 0.40 Kgr. 100 Kgr. UTILIDAD 120.00 90.00X1: Número de cajas de tipo A a fabricar.X2: Número de cajas de tipo B a fabricar.Max Z = 120 X1 + 90 X2s.a 0.3 X1 + 0.4 X2 <= 100 0.5 X1 + 0.2 X2 <= 120 0.2 X1 + 0.4 X2 <= 100Estándar:Min Z = - 120 X1 - 90 X2 + 0 X3 + 0X4 + 0X5 0.3 X1 + 0.4 X2 + X3 = 100 0.5 X1 + 0.2 X2 + X4 = 120 0.2 X1 + 0.4 X2 + X5 = 100Canónico: X1 X2 X3 X4 X5 Z -120 -90 0 0 0 0 3/10 4/10 1 0 0 100 5/10 2/10 0 1 0 120 2/10 4/10 0 0 1 100X1 = X2 = 0 , X3 = 100 , X4 = 120 , X5 = 100 , Z = 0 X1 X2 X3 X4 X5 Z 0 -42 0 240 0 28800 0 -7/25 -1 3/5 0 -28 1 2/5 0 10/5 0 240 0 -16/50 0 2/5 -1 -52 X1 X2 X3 X4 X5 Z 0 -42 0 240 0 28800 0 7/25 1 -3/5 0 28 1 2/5 0 10/5 0 240 0 16/50 0 -2/5 1 52 X1 X2 X3 X4 X5 Z 0 0 150 150 0 33000 0 1 25/7 -15/7 0 100 -1 0 10/7 -20/7 0 -200 0 0 8/7 -10/35 -1 -20 X1 X2 X3 X4 X5 Z
  • 3. 0 0 150 150 0 33000 0 1 25/7 -15/7 0 100 1 0 -10/7 20/7 0 200 0 0 -8/7 10/35 1 20Solución:X1 = 200 , X2 = 100 , X3 = 0 , X4 = 0 , X5 = 20 , Z = 33000
  • 4. Ejercicio 1.4 MAIZ LUPULO PRECIO CERVEZA RUBIA 5 lb 2 lb 5.00 CERVEZA NEGRA 2 lb 1 lb 2.00 TOTAL 60 lb 25 lbX1: Número de barriles de cerveza rubia a producir.X2: Número de barriles de cerveza negra a producir.Max Z = 5 X1 + 2 X2s.a 5 X1 + 2 X2 <= 60 2 X1 + 1 X2 <= 25Estándar:Min Z = - 5 X1 - 2 X2 + 0 X3 + 0X4 5 X1 + 2 X2 + X3 = 60 2 X1 + 1 X2 + X4 = 25Canónico: X1 X2 X3 X4 Z -5 -2 0 0 0 5 2 1 0 60 2 1 0 1 25X1 = X2 = 0 , X3 = 60 , X4 = 25 , Z = 0 X1 X2 X3 X4 Z -5 -2 0 0 0 1 2/5 1/5 0 12 2 1 0 1 25 X1 X2 X3 X4 Z 0 0 1 0 60 1 2/5 1/5 0 12 0 -1/5 2/5 -1 -1 X1 X2 X3 X4 Z 0 0 1 0 60 1 2/5 1/5 0 12 0 1/5 -2/5 1 1Solución:X1 = 12 , X2 = 0 , X3 = 0 , X4 = 1 , Z = 60
  • 5. Ejercicio 1.5 TIEMPO COCCION HUEVOS PRECIO TORTA DE CHOCOLATE 1/3 hora 4 1.00 TORTA DE VAINILLA 2/3 hora 1 0.50 TOTAL 8 horas 30X1: Número de tortas de chocolate a preparar.X2: Número de tortas de vainilla a preparar.Max Z = 1 X1 + 0.5 X2s.a 1/3 X1 + 2/3 X2 <= 8 4 X1 + 1 X2 <= 30Estándar:Min Z = - 1 X1 – 0.5 X2 + 0 X3 + 0X4 1/3 X1 + 2/3 X2 + X3 = 8 4 X1 + 1 X2 + X4 = 30Canónico: X1 X2 X3 X4 Z -1 -1/2 0 0 0 1/3 2/3 1 0 8 4 1 0 1 30X1 = X2 = 0 , X3 = 8 , X4 = 30 , Z = 0 X1 X2 X3 X4 Z -1 -1/2 0 0 0 1/3 2/3 1 0 8 1 1/4 0 1/4 30/4 X1 X2 X3 X4 Z 0 -1/4 0 1/4 30/4 0 -7/12 -1 1/12 -11/2 1 1/4 0 1/4 30/4 X1 X2 X3 X4 Z 0 -1/4 0 1/4 30/4 0 7/12 1 -1/12 11/2 1 1/4 0 1/4 30/4 X1 X2 X3 X4 Z 0 0 3/7 3/14 69/7 0 1 12/7 -1/7 66/7 -1 0 3/7 -2/7 -36/7 X1 X2 X3 X4 Z 0 0 3/7 3/14 69/7 0 1 12/7 -1/7 66/7 1 0 -3/7 2/7 36/7
  • 6. Solución:X1 = 36/7 , X2 = 66/7 , X3 = 0 , X4 = 0 , Z = 69/7
  • 7. Ejercicio 1.10 TULIPANES ROSAS JAZMINES PRECIO DISPOSICION 1 30 20 4 50.00 DISPOSICION 2 10 40 3 30.00 DISPOSICION 3 20 50 2 60.00 TOTAL 1000 800 100X1: Número de disposiciones tipo 1 a usar.X2: Número de disposiciones tipo 2 a usar.X3: Número de disposiciones tipo 3 a usar.Max Z = 50 X1 + 30 X2 + 60 X3s.a 30 X1 + 10 X2 + 20 X3 <= 1000 20 X1 + 40 X2 + 50 X3 <= 800 4 X1 + 3 X2 + 2 X3 <= 100Estándar:Min Z = - 50 X1 - 30 X2 - 60 X3 + 0X4 + 0X5 + 0 X6 30 X1 + 10 X2 + 20 X3 + X4 = 1000 20 X1 + 40 X2 + 50 X3 + X5 = 800 4 X1 + 3 X2 + 2 X3 + X6 = 100Canónico: X1 X2 X3 X4 X5 X6 Z -50 -30 -60 0 0 0 0 30 10 20 1 0 0 1000 20 40 50 0 1 0 800 4 3 2 0 0 1 100X1 = X2 = 0 , X3 = 0 , X4 = 1000 , X5 = 800 , X6 = 100 , Z = 0 X1 X2 X3 X4 X5 X6 Z -26 18 0 0 6/5 0 960 -22 6 0 -1 2/5 0 -680 2/5 4/5 1 0 1/50 0 16 -16/5 -7/5 0 0 2/50 -1 -68 X1 X2 X3 X4 X5 X6 Z -26 18 0 0 6/5 0 960 22 -6 0 1 -2/5 0 680 2/5 4/5 1 0 1/50 0 16 16/5 7/5 0 0 -2/50 1 68 X1 X2 X3 X4 X5 X6 Z -26 18 0 0 6/5 0 960 22 -6 0 1 -2/5 0 680 2/5 4/5 1 0 1/50 0 16 1 7/16 0 0 -1/80 5/16 85/4 X1 X2 X3 X4 X5 X6 Z 0 235/8 0 0 35/40 65/8 3025/2 0 125/8 0 -1 5/40 55/8 -425/2 0 -25/40 -1 0 -5/200 1/8 -75/10 1 7/16 0 0 -1/80 5/16 85/4
  • 8. X1 X2 X3 X4 X5 X6 Z 0 235/8 0 0 35/40 65/8 3025/2 0 -125/8 0 1 -5/40 -55/8 425/2 0 25/40 1 0 5/200 -1/8 75/10 1 7/16 0 0 -1/80 5/16 85/4Solución:X1 = 85/4 , X2 = 0 , X3 = 75/10 , X4 = 425/2 , X5 = 0 , X6 = 0 , Z = 3025/2

×