PROBLEMAS DE METODO-SIMPLEX
64
Maximizar Z = 13 X1 + 16 X2 SUJETA A: 5 X1 + 5 X2 <= 35 6 X1 + 9 X2 <= 54 X2 >= 2 Xj >= 0 5 X1 + 5 X2 + X3 = 35 6 X1 + 9 X2 + X4 = 54 X2 - X5 = 2 (11M + 13) X1 + (15M + 16) X2 + MX3 + MX4 - MX5 - 91M SUSTITUYENDO LA M POR 1000: MAXIMIZAR Z = 11013 X1 + 15016 X2 + 1000 X3 + 1000 X4 - 1000 X5 - 91000 SUJETA A: 5 X1 + 5 X2 + X3 <= 35 6 X1 + 9 X2 + X4 <= 54 X2 - X5 <= 2 Xj >= 0 5 X1 + 5 X2 + X3 + X6 = 35 6 X1 + 9 X2 + X4 + X7 = 54 X2 - X5 + X8 = 2 Cj θ V BAS X1 X2 X3 X4 0 7 X6 5 5 1 0 0 6 X7 6 9 0 1 0 2 X8 0 1 0 0 Ø -11013 -15016 -1000 -1000 Cj θ V BAS X1 X2 X3 X4 0 5 X6 5 0 1 0 0 4 X7 6 0 0 1 15016 - X2 0 1 0 0 Ø -11013 -1000 -1000 Cj θ V BAS X1 X2 X3 X4 0 3 X6 1.66667 0 1 -0.5556 -1000 6 X5 0.66667 0 0 0.11111 15016 9 X2 0.66667 1 0 0.11111 Ø -1669 -1000 557.333 Cj θ V BAS X1 X2 X3 X4 11013 X1 1 0 0.6 -0.3333 -1000 X5 0 0 -0.4 0.33333 15016 X2 0 1 -0.4 0.33333 Ø 1.4 1
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PROBLEMAS RESUELTOS DEL METODO SIMPLEX DLE LIBRO MÉTODOS CUANTITATIVOS PARA LA TOMA DE DECISIONES EN ADMINISTRAICON
Transcript of PROBLEMAS DE METODO-SIMPLEX
Maximizar Z = 13 X1 + 16 X2SUJETA A: 5 X1 + 5 X2 <= 35
6 X1 + 9 X2 <= 54X2 >= 2Xj >= 0
5 X1 + 5 X2 + X3 = 356 X1 + 9 X2 + X4 = 54X2 - X5 = 2
(11M + 13) X1 + (15M + 16) X2 + MX3 + MX4 - MX5 - 91MSUSTITUYENDO LA M POR 1000:
MAXIMIZAR Z = 11013 X1 + 15016 X2 + 1000 X3 + 1000 X4 - 1000 X5 - 91000SUJETA A: 5 X1 + 5 X2 + X3 <= 35
6 X1 + 9 X2 + X4 <= 54X2 - X5 <= 2Xj >= 0
5 X1 + 5 X2 + X3 + X6 = 356 X1 + 9 X2 + X4 + X7 = 54X2 - X5 + X8 = 2
Cj θ V BAS X1 X2 X3 X40 7 X6 5 5 1 00 6 X7 6 9 0 10 2 X8 0 1 0 0
Ø -11013 -15016 -1000 -1000
Cj θ V BAS X1 X2 X3 X40 5 X6 5 0 1 00 4 X7 6 0 0 1
15016 - X2 0 1 0 0Ø -11013 -1000 -1000
Cj θ V BAS X1 X2 X3 X40 3 X6 1.66667 0 1 -0.5556
-1000 6 X5 0.66667 0 0 0.1111115016 9 X2 0.66667 1 0 0.11111
Ø -1669 -1000 557.333
Cj θ V BAS X1 X2 X3 X411013 X1 1 0 0.6 -0.3333-1000 X5 0 0 -0.4 0.3333315016 X2 0 1 -0.4 0.33333
Ø 1.4 1
CTLFSNNOCSTEPPLT:X1 3X5 2X2 4
Z = 11013 *3 + 15016 * 4 + 1000 * 2 - 91000
MAXIMA: Z = 103
5 X1 + 5 X2 <= 356 X1 + 9 X2 <= 54X2 >= 2
5 X1 + 5 X2 + X3 + X6 = 356 X1 + 9 X2 + X4 + X7 = 54X2 - X5 + X8 = 2
Z = 13 X1 + 16 X2
103
11013 X1 + 15016 X2 + 1000 X3 + 1000 X4 - 1000 X5 - 91000
X5 X6 X7 X8 b0 1 0 0 350 0 1 0 54-1 0 0 1 2
1000
X5 X6 X7 X8 b5 1 0 -5 259 0 1 -9 36-1 0 0 1 2
-14016 15016
X5 X6 X7 X8 b0 1 -0.5556 0 51 0 0.11111 -1 40 0 0.11111 0 6
1557.33 1000
X5 X6 X7 X8 b0 0.6 -0.3333 0 31 -0.4 0.33333 -1 20 -0.4 0.33333 0 4
1001.4 1001 1000
5 X1 + 5 X2 <= 356 X1 + 9 X2 <= 54
5 X1 + 5 X2 + X3 + X6 = 356 X1 + 9 X2 + X4 + X7 = 54X2 - X5 + X8 = 2
Z = 13 X1 + 16 X2
Ejercicios Método Simplex Análitico complejo8.1
De manera separada, uméntese cada una de las siguientes restricciones y la funciónobjetivo para cada caso. Supóngase la siguiente función objetivo:
Maximizar: Z= 8X1 + 7X2 + 9X3a) 8X1 - 7X2 + 4X3 ≤ 28b) X1 + 9X2 ≤ 81c) X1 + 3X3 ≥ 51d) 7X2 + 4X3 ≥ 83e) 19X1 - 8X2 + X3 = 22f) 7X1 = 12X2
En cada una de las restricciones se le agrega una variable artificial
8X1 -7X2 + 4X3 + X4 = 28X1 + 9X2 + X5 = 81X1 + 3X3 -X6 = 517X2 + 4X3 - X7 = 8319X1 -8X2 + X3 = 227X1 -12X2 = 0
(36M + 8)X1 + (-18M + 7)X2 + (12M + 9)X3 + MX4 + MX5 + MX6 - 265M
M=1000
sustituyendo M
Maximizar Z=36008X1 -17093X2 + 12009X3 + 1000X4 + 1000X5 + 1000X6 - 265000
Sujeta a: 8X1 -7X2 + 4X3 + X4 ≤ 28X1 + 9X2 + X5 ≤ 81
X1 + 3X3 -X6 ≤ 517X2 + 4X3 - X7 ≤ 8319X1 -8X2 + X3 ≤ 227X1 -12X2 ≤ 0Xj ≥ 0
Se le agrega una variable de holgura
8X1 -7X2 + 4X3 + X4 + X8 = 28X1 + 9X2 + X5 + X9 = 81X1 + 3X3 -X6 + X10 = 517X2 + 4X3 - X7 + X11 = 8319X1 -8X2 + X3 + X12 = 227X1 -12X2 + X13 = 0
Cj θ V bas X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 b0 X8 8 -7 4 1 0 0 0 1 0 0 0 0 0 28 ≤0 X9 1 9 0 0 1 0 0 0 1 0 0 0 0 81 ≥0 X10 1 0 3 0 0 -1 0 0 0 1 0 0 0 51 =0 X11 0 7 4 0 0 0 -1 0 0 0 1 0 0 830 X12 19 -8 1 0 0 0 0 0 0 0 0 1 0 220 X13 7 -12 0 0 0 0 0 0 0 0 0 0 1 0
ᴓ ###
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 3.5 X8 8 -7 4 1 0 0 00 81 X9 1 9 0 0 1 0 00 51 X10 1 0 3 0 0 -1 00 - X11 0 7 4 0 0 0 -10 1.15789 X12 19 -8 1 0 0 0 00 0 X13 7 -12 0 0 0 0 0
ᴓ -36008 17093 -12009 -1000 -1000 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 5.2353 X8 0 -3.63158 3.5789473684 1 0 0 00 -1517 X9 0 9.42105 -0.052631579 0 1 0 00 16.911 X10 0 0.42105 2.9473684211 0 0 -1 00 20.75 X11 0 7 4 0 0 0 -1
36008 22 X1 1 -0.42105 0.0526315789 0 0 0 00 22 X13 0 -9.05263 -0.368421053 0 0 0 0
ᴓ 1931.74 -10113.84211 -1000 -1000 -1000
0 -1.01471 1.0000000001 0.27941 0 0 0-3.15789E-005
X8 X9 X10 X11 X12 X13 b1 0 0 0 0 0 280 1 0 0 0 0 810 0 1 0 0 0 510 0 0 1 0 0 830 0 0 0 1 0 220 0 0 0 0 1 0
X8 X9 X10 X11 X12 X13 b1 0 0 0 -0.42105 0 18.70 1 0 0 -0.05263 0 79.80 0 1 0 -0.05263 0 49.80 0 0 1 0 0 830 0 0 0 0.05263 0 1.160 0 0 0 -0.36842 1 -8.11
1895.16
0.279412 0 0 0 -0.11765 0 5.24
8 -- 6Resuélvase este problema de minimizaciónMinimizar: Z = 8X1 + 5X2 + 7X3Restricciones:
10X1 + 8X2 + 12X3 ≤ 1202X1+ 4X2 -3X3 ≥ 16Xi ≥ 0, i=1,2,3
Agregando variable artificial
10X1 + 8X2 + 12X3 + X4 = 1202X1+ 4X2 -3X3 - X5 = 16
Se agregan las M
(12M - 8)X1 + (12M - 5)X2 + (9M - 7)X3 + MX4 - MX5 - 136M
M=1000
Maximizar Z= 11992X1 + 11995X2 + 8993X3 + 1000X4 - 1000X5 - 136000Sujeta a: 10X1 + 8X2 + 12X3 + X4 ≤ 120
2X1+ 4X2 -3X3 - X5 ≤ 16
Variable de holgura
10X1 + 8X2 + 12X3 + X4 + X6 = 1202X1+ 4X2 -3X3 - X5 + X7 = 16
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 15 X6 10 8 12 1 0 1 00 4 X7 2 4 -3 0 -1 0 1
ᴓ -11992 -11995 -8993 -1000 1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 4.888889 X6 6 0 18 1 2 1 -2
11995 -5.33333 X2 0.5 1 -0.75 0 -0.25 0 0.25ᴓ -5994.5 -17989.25 -1000 -1998.75 2998.75
Cj θ V bas X1 X2 X3 X4 X5 X6 X78993 88 X3 0.333333333 0 1 0.0555556 0.1111111 0.055556 -0.111111
11995 184 X2 0.75 1 0 0.0416667 -0.166667 0.041667 0.166667ᴓ 1.916666667 -0.5972222 0.0555556 999.4028 999.9444
Cj θ V bas X1 X2 X3 X4 X5 X6 X71000 X4 6 0 18 1 2 1 -2
11995 X2 0.5 1 -0.75 0 -0.25 0 0.25ᴓ 5.5 10.75 1.25 1000 998.75
X1= 0 Zmín= 8(0)+5(4)+7(0)= 20X2= 4 10(0)+8(4)+12(0) ≤ 120 32X3= 0 2(0)+4(4)-3(0) ≥ 16 20X4= 88X5= 0 10(0)+8(4)+12(0)+88 = 120
2(0)+4(4)-3(0)-0 = 16
11992(0)+11995(4)+8993(0)+1000(88)-1000(0)-136000 = -2010(0)+8(4)+12(0)+88+0= 120
2(0)+4(4)-3(0)-0+0= 16 Z=-20
8 --7Resuélvase el siguiente problema de minimización y explíquese la soluciónMinimizar
Z=8X1 + 12X2 + 4X3Sujeta a:
X1 + X2 + X3 ≤ 302X1 + 3X2 + X3 ≥ 12Xi≥0, i=1,2,3
Agregando variable artificial
X1 + X2 + X3 + X4 = 302X1 + 3X2 + X3 - X5 = 12
Se agregan las M
(3M - 8)X1 + (4M - 12)X2 + (2M - 4)X3 + MX4 + MX5 - 32M
M=1000
MINIMIZAR Z= 2992X1 + 3988X2 + 1996X3 + 1000X4 - 1000X5 - 32000X1 + X2 + X3 + X4 ≤ 302X1 + 3X2 + X3 - X5 ≤ 12
Variable de holgura
X1 + X2 + X3 + X4 + X6 ≤ 302X1 + 3X2 + X3 - X5 + X7 ≤ 12
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 30 X6 1 1 1 1 0 1 00 4 X7 2 3 1 0 -1 0 1
ᴓ -2992 -3988 -1996 -1000 1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 39 X6 0.333333333 0 0.6666667 1 0.3333333 1 -0.333333
3988 - X2 0.666666667 1 0.3333333 0 -0.333333 0 0.333333ᴓ -333.333333 -666.66667 -1000 -329.3333 1329.333
Cj θ V bas X1 X2 X3 X4 X5 X6 X71000 X4 0.333333333 0 0.6666667 1 0.3333333 1 -0.3333333988 X2 0.666666667 1 0.3333333 0 -0.333333 0 0.333333
ᴓ 0 0 4 1000 996
X1= 0X2 0 Zmín= 8(0)+12(4)+4(0)= 48X3 0 (0)+(4)+(0)≤ 30X4 0 2(0)+3(4)+(0)≥ 12X5 0
2992(0)+3988(4)+1996(0)+1000(26)-1000(0)-42000 = -48
Zmín= -48
8 -- 8Resuélvase el siguiente problema y explíquese la solución
Maximizar: Z=10X1 + 8X2 + 14X3Sujeta a:
5X1 + 4X2 + 7X3 ≤ 702X1 + X2 + 3X3 ≥ 15Xi=0 i=1,2,3
Se le agrega la variable artificial
5X1 + 4X2 + 7X3 + X4 = 702X1 + X2 + 3X3 -X5 = 15
Se agregan las M
(7M + 10)X1 + (5M + 8)X2 + (10M + 14)X3 + X4M - X5M - 85M
M=1000
Se sustituye la M
7010X1 + 5008X2 + 10014X3 + 1000X4 -1000X5 - 85000
Maximizar: Z= 7010X1 + 5008X2 + 10014X3 + 1000X4 -1000X5 - 850005X1 + 4X2 + 7X3 + X4 ≤ 702X1 + X2 + 3X3 -X5 ≤ 15
Variable de holgura5X1 + 4X2 + 7X3 + X4 + X6 = 702X1 + X2 + 3X3 -X5 + X7 = 15
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 10 X6 5 4 7 1 0 1 00 5 X7 2 1 3 0 -1 0 1
ᴓ -7010 -5008 -10014 -1000 1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 15 X6 0.333333333 1.6666666667 0 1 2.3333333 1 -2.333333
10014 -15 X3 0.666666667 0.3333333333 1 0 -0.333333 0 0.333333ᴓ -334 -1670 -1000 -2338 3338
Cj θ V bas X1 X2 X3 X4 X5 X6 X7-1000 X5 0.142857143 0.7142857143 0 0.4285714 1 0.428571 -110014 X3 0.714285714 0.5714285714 1 0.1428571 0 0.142857 0
ᴓ 0 0 2 1002 1000
X1 0 Zmax= 10(0)+8(0)+14(10) = 140X2 0 5(0)+4(0)+7(10)≤ 70 = 70X3 10 2(0)+(0)+3(10)≥ 15 = 30X4 0X5 15 5(0)+4(0)+7(10)+(0)= 70 = 70
2(0)+(0)+3(10)-(15)= 15 = 15
7010(0)+5008(0)+10014(10)+1000(0)-1000(15)-85000 1405(0)+4(0)+7(10)+(0)≤ 70 702(0)+(0)+3(10)-(15)≤ 15 15
8 -- 9a) Resuélvase el siguiente problema con el método simplex y explíquese la respuesta
Maximizar Z= 3X1 + 4X2
Restricciones 5X1 + 10X2 ≤ 507X1 + 7X2 ≥ 498X1 - 2X2 ≤ 8X1,X2 ≥ 0
Se agrega la variable artificial
5X1 + 10X2 + X3 = 507X1 + 7X2 - X4 = 498X1 - 2X2 + X5 = 8
Se agrega las M
(20M + 3)X1 + (15M + 4)X2 + MX3 - MX4 + MX5 - 107M
M=1000
Se sustituye las M
20003X1 + 15004X2 + 1000X3 - 1000X4 + 1000X5 -107000
Se aumenta la variable de holgura
5X1 + 10X2 + X3 + x6 ≤ 507X1 + 7X2 - X4 + x7 ≤ 498X1 - 2X2 + X5 + x8 ≤ 8
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 10 X6 5 10 1 0 0 1 00 7 X7 7 7 0 -1 0 0 10 1 X8 8 -2 0 0 1 0 0
ᴓ -20003 -15004 -1000 1000 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 4 X6 0 11.25 1 0 -0.625 1 00 4.8 X7 0 8.75 0 -1 -0.875 0 1
20003 -4 X1 1 -0.25 0 0 0.125 0 0ᴓ -20004.75 -1000 1000 1500.375
Cj θ V bas X1 X2 X3 X4 X5 X6 X715004 X2 0 1 0.0888889 0 -0.055556 0.088889 0
0 X7 0 0 -0.7777778 -1 -0.388889 -0.777778 120003 X1 1 0 0.0222222 0 0.1111111 0.022222 0
ᴓ 778.2 1000 389 1778.2
X1= 2 Zmax= 3(2)+4(4) = 22X2= 4 5(2)+10(4)≤ 50 = 50X3= 0 7(2)+7(4)≥ 49 = 42X4= 0 8X1 - 2X2 ≤ 8X5= 0X6= 0
X7= 7X8= 0
b) Hágase una gráfica de este problema y sombreése la región factible
No tiene solución el problema porque no satisface las desigualdades.
8 -- 10Resuélvase el siguiente problema de PL con el método simplex y explíquese la soluciónMaximizar: Z= 10X1 + 8X2Restricciones:
7X1 + 4X2 ≥ 353X1 ≤ 2X2 + 3X1,X2 ≥ 0
Se le agrega la variable artificial
7X1 + 4X2 - X3 = 353X1 -2X2 + X4 = 3
Se le agregan las M
(10M + 10)X1 + (2M + 8)X2 - MX3 + MX4 - 38M
M=1000
Se sustituye M
10010X1 + 2008X2 - 1000X3 + 1000X4 -38000
Se agrega la variable de holgura
7X1 + 4X2 - X3 + X5 ≤ 353X1 -2X2 + X4 + X6 ≤ 3
Cj θ V bas X1 X2 X3 X4 X5 X6 b0 5 X5 7 4 -1 0 1 0 350 1 X6 3 -2 0 1 0 1 3
ᴓ -10010 -2008 1000 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 b0 3.230769 X5 0 8.6666666667 -1 -2.3333333 1 -2.333333 28
10010 -1.5 X1 1 -0.6666666667 0 0.3333333 0 0.333333 1ᴓ -8681.333333333 1000 2336.6667 3336.667
Cj θ V bas X1 X2 X3 X4 X5 X6 b2008 -28 X2 0 1 -0.1153846 -0.2692308 0.1153846 -0.269231 3.230769
10010 -41 X1 1 0 -0.0769231 0.1538462 0.0769231 0.153846 3.153846ᴓ -1.6923077 -0.6153846 1001.6923 999.3846
Las soluciones son ilimitadas ya que las θ se vuelen todas negativas
8 -- 11Resuélvase el siguiente problema y coméntese la soluciónMaximizar: Z= 5X1 + 8X2 + 3X3Restricciones:
X1 + X2 - 2X3 ≤ 30X1 + X2 + X3 ≤ 20X1 - 3X2 ≤ 60Xi 0, i=1,2,3
Se le agrega la variable artifical
X1 + X2 - 2X3 + X4 = 30X1 + X2 + X3 +X5 = 20X1 - 3X2 + X6 = 60
Se le agregan las M(3M + 5)X1 + (-1M + 8)X2 + (-1M + 3)X3 + X4M + X5M + X6M - 110M
M=1000
Se sustituyen las M
3005X1 - 992X2 - 997X3 + 1000X4 +1000X5 + 1000X6 - 110000
Se agrega la variable de holgura
X1 + X2 - 2X3 + X4 + X7 ≤ 30X1 + X2 + X3 +X5 + X8 ≤ 20X1 - 3X2 + X6 +X9 ≤ 60
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 30 X7 1 1 -2 1 0 0 10 20 X8 1 1 1 0 1 0 00 60 X9 1 -3 0 0 0 1 0
ᴓ -3005 992 997 -1000 -1000 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 10 X7 0 0 -3 1 -1 0 1
3005 - X1 1 1 1 0 1 0 00 - X9 0 -4 -1 0 -1 1 0
ᴓ 3997 4002 -1000 2005 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X71000 - X4 0 0 -3 1 -1 0 13005 - X1 1 1 1 0 1 0 0
0 40 X9 0 -4 -1 0 -1 1 0ᴓ 3997 1002 1005 -1000 1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X71000 - X4 0 0 -3 1 -1 0 1
3005 0.05 X1 1 1 1 0 1 0 01000 -0.1 X6 0 -4 -1 0 -1 1 0
ᴓ -3 2 5 1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X71000 X4 0 0 -3 1 -1 0 1-992 X2 1 1 1 0 1 0 01000 X6 4 0 3 0 3 1 0
ᴓ 3 5 8 1000
X1 0X2 20 Zmax= 5(0)+8(20)+3(0) = 160X3 0 (0)+(20)-2(0)≤30 = 20X4 10 (0)+(20)+(0)≤ 20 = 20X5 0 (0)-3(20)≤ 60 = -60X6 120X7 0 (0)+(20)-2(0)+(10)=30 = 30X8 0 (0)+(20)+(0)+(0)= 20 = 20X9 0 (0)-3(20)+(120)= 60 = 60
3005(0)-992(20)-997(0)+1000(10)+1000(0)+1000(120)-110000 =
(0)+(20)-2(0)+(10)+(0)≤ 30 = 30(0)+(20)+(0)+(0)+(0)≤ 20 = 20(0)-3(20)+(120)+(0)≤ 60 = 60
8 -- 12Resuélvase este problema de minimizaciónMinimizar Z= 4X1 + 6X2 + 4X3
Restricciones 3X1 + X2 + 4X3 ≥ 362X1 + 4X2 + X3 ≥ 40X1,X2 ≥ 0
Se agrega la variable artificial
3X1 + X2 + 4X3 - X4 = 362X1 + 4X2 + X3 - X5 = 40
Se agregan las M
(5M - 4)X1 + (5M - 6)X2 + (5M - 4)X3 - MX4 - MX5 -76M
M=1000
Se sustituye las M
4996X1 + 4994X2 + 4996X3 - 1000X4 - 1000X5 - 76000
Se agregan las variables de holgura
3X1 + X2 + 4X3 - X4 + X6 ≤ 362X1 + 4X2 + X3 - X5 + X7 ≤ 40
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 12 X6 3 1 4 -1 0 1 00 20 X7 2 4 1 0 1 0 1
ᴓ -4996 -4994 -4996 1000 1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X74996 36 X1 1 0.3333333333 1.3333333 -0.3333333 0 0.333333 0
0 4.8 X7 0 3.3333333333 -1.6666667 0.6666667 1 -0.666667 1ᴓ -3328.666666667 1665.3333 -665.33333 1000 1665.333
Cj θ V bas X1 X2 X3 X4 X5 X6 X74996 X1 1 0 1.5 -0.4 -0.1 0.4 -0.14994 X2 0 1 -0.5 0.2 0.3 -0.2 0.3
ᴓ 1 0.4 1998.6 999.6 998.6
X1= 10.4 Zmín= 4(10.4)+6(4.8)+4(0) = 70.4X2= 4.8 3(10.4)+(4.8)+4(0)≥36 = 36
X3= 0 2(10.4)+4(4.8)+(0)≥ 40 = 40X4= 0X5= 0 3(10.4)+(4.8)+4(0)-(0)= 36 = 36X6= 0 2(10.4)+4(4.8)+(0)-(0)= 40 = 40X7= 0
4996(10.4)+4994(4.8)+4996(0)-1000(0)-1000(0)-76000 = -70.4
3(10.4)+(4.8)+4(0)-(0)+(0)≤ 36 362(10.4)+4(4.8)+(0)-(0)+(0)≤ 40 40
8 -- 13Resuélvase el siguiente problema y expliquese la soluciónMaximizar: Z= 6X1 + 5X2
Restricciones: 3X1 + 9X2 ≤ 2710X1 + 3.75X2 ≤ 37.55X1 + 5X2 ≤ 25X1,X2 ≥ 0
Se agrega la variable artificial
3X1 + 9X2 + X3 = 2710X1 + 3.75X2 +X4 = 37.55X1 + 5X2 +X5 = 25
Se agregan las M
(18M + 6)X1 + (17.75M + 5)X2 + MX3 + MX4 + MX5 - 89.5
M=1000
Se sustituyen las M
18006X1 + 17755X2 + 1000X3 + 1000X4 + 1000X5 - 89500
Se agregan las variables de holgura
3X1 + 9X2 + X3 + X6 ≤ 2710X1 + 3.75X2 +X4 + X7 ≤ 37.55X1 + 5X2 +X5 + X8 ≤ 25
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 9 X6 3 9 1 0 0 1 00 3.75 X7 10 3.75 0 1 0 0 10 5 X8 5 5 0 0 1 0 0
ᴓ -18006 -17755 -1000 -1000 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 2 X6 0 7.875 1 -0.3 0 1 -0.3
18006 10 X1 1 0.375 0 0.1 0 0 0.10 2 X8 0 3.125 0 -0.5 1 0 -0.5
ᴓ -11002.75 -1000 800.6 -1000 1800.6
Cj θ V bas X1 X2 X3 X4 X5 X6 X717755 - X2 0 1 0.1269841 -0.0380952 0 0.126984 -0.03809518006 - X1 1 0 -0.047619 0.1142857 0 -0.047619 0.114286
0 0 X8 0 0 -0.3968254 -0.3809524 1 -0.396825 -0.380952ᴓ 397.1746 381.44762 -1000 1397.175 1381.448
Cj θ V bas X1 X2 X3 X4 X5 X6 X717755 X2 0 1 0.1269841 -0.0380952 0 0.126984 -0.03809518006 X1 1 0 -0.047619 0.1142857 0 -0.047619 0.1142861000 X5 0 0 -0.3968254 -0.3809524 1 -0.396825 -0.380952
ᴓ 0.3492063 0.4952381 1000.349 1000.495
X1= 3 Zmax= 6(3)+5(2) = 28X2= 2 3(3)+9(2)≤ 27 = 27X3= 0 10(3)+3.75(2)≤ 37.5 = 37.5X4= 0 5(3)+5(2)≤ 25 = 25X5= 0X6= 0 3(3)+9(2)+(0)= 27 = 27X7= 0 10(3)+3.75(2)+(0)= 37.5 = 37.5X8= 0 5(3)+5(2)+(0)= 25 = 25
18006(3)+17755(2)+1000(0)+1000(0)+1000(0)-89500 =
38(3)+9(2)+(0)+(0)≤ 27 =1083)+3.75(2)+(0)+(0)≤ 37.5 =5(3)+5(2)+(0)+(0)≤ 25 =
8 -- 14Resuélvase y coméntese la solución a este problemaMaximizar: Z= 8X1 + 4X2 + 6X3 ≤
≥Restricciones 4X1 + 3X2 + 5X3 ≤ 60
2X1 + X2 ≤ 30
Se agrega la variable artificial
4X1 + 3X2 + 5X3 + X4 = 602X1 + X2 + X5 = 30
Se agregan las M
(6M + 8)X1 + (4M + 4)X2 + (5M + 6)X3 + MX4 + MX5 - 90M
M=1000
Se sustituyen las M
6008X1 + 4004X2 + 5006X3 + 1000X4 + 1000X5 - 90000
Se agrega la variable de holgura
4X1 + 3X2 + 5X3 + X4 + X6 ≤ 602X1 + X2 + X5 + X7 ≤ 30
Cj θ V bas X1 X2 X3 X4 X5 X6 X70 15 X6 4 3 5 1 0 1 00 15 X7 2 1 0 0 1 0 1
ᴓ -6008 -4004 -5006 -1000 -1000
Cj θ V bas X1 X2 X3 X4 X5 X6 X76008 - X1 1 0.75 1.25 0.25 0 0.25 0
0 0 X7 0 -0.5 -2.5 -0.5 1 -0.5 1ᴓ 502 2504 502 -1000 1502
Cj θ V bas X1 X2 X3 X4 X5 X6 X76008 X1 1 0.75 1.25 0.25 0 0.25 0
1000 X5 0 -0.5 -2.5 -0.5 1 -0.5 1ᴓ 2 4 2 1002 1000
X1= 15 Zmax= 8(15)+4(0)+6(0) = 120X2= 0 4(15)+3(0)+5(0)≤ 60 = 60X3= 0 2(15)+(0)≤ 30 = 30X4= 0X5= 0 4(15)+3(0)+5(0)+(0)= 60 = 60X6= 0 2(15)+(0)+(0)= 30 = 30X7= 0
6008(15)+4004(0)+5006(0)+1000(0)+1000(0)-90000 =4(15)+3(0)+5(0)+(0)+(0)≤ 60 = 602(15) (0)+(0)+(0)≤ 30 = 30
8 -- 15En la AAA Welding, Joe está tratando de decidir cuántas amarras para trailer debe de hacer para hacer un metal de desperdicio. Tiene dos tipos demetal y puede hacer cualquiera de dos tipos de ganchos. En la tabla siguiente se proporcionan los datos necesarios.
Requerido para:
Metal Soldadura I Soldadura II DisponibleHierro Acanalado 5 5 35 unidades
Hierro Plano 6 9 54 unidades
Joe gana $13 por cada gancho de tipo 1 y $16 por cada gancho de tipo 2. Ya prometió hacer 2 ganchos del tipo 2. a) Resuélvase este problema con el método simplex.b) Le han ofrecido a Jeo hierro acanalado adicional a $2 por unidad. ¿Deberá de comprarlo?c) Construyase el problema dual y resuélvase.
b12016
b884
b4.888888897.66666667
b884
b3012
b264
b264
b7015
b355
b1510
≤
X8 b0 500 491 8
X8 b-0.625 45-0.875 420.125 1
2500.375
X8 b-0.05555556 4-0.38888889 70.11111111 2
1389
No tiene solución el problema porque no satisface las desigualdades.
X8 X9 b0 0 301 0 200 1 60
X8 X9 b-1 0 101 0 20-1 1 40
3005
X8 X9 b-1 0 101 0 20-1 1 40
2005
X8 X9 b-1 0 10
1 0 20-1 1 40
1005 1000
X8 X9 b-1 0 101 0 203 1 120
1008 1000
160
b3640
b1216
b10.44.8
X8 b0 270 37.51 25
X8 b0 15.750 3.751 6.25
X8 b0 20 3
1 0
X8 b0 20 31 0
1000
28
2737.525
b6030
b150
b15
0
120
En la AAA Welding, Joe está tratando de decidir cuántas amarras para trailer debe de hacer para hacer un metal de desperdicio. Tiene dos tipos de
8 -- 8Resuélvase el siguiente problema y explíquese la solución
Maximizar: Z=10X1 + 8X2 + 14X3Sujeta a:
5X1 + 4X2 + 7X3 ≤ 702X1 + X2 + 3X3 ≥ 15Xi=0 i=1,2,3
se le agrega la variable artificial5X1+4X2+7X3+X4=702X1+X2+3X3-X5=15
se agraegna las M
(7M+10)X1+(5M+8)X2+(10M+14)X3+MX4-X5M-M85
si M=1000
Z=7010X1+5008X2+1014X3+1000X4-100X5-85000
Maximizar Z=7010X1+5008X2+10014X3+1000X4-100X5-850005X1+4X2+7X3+X4≤70
Variable De Holgura 5X1+4X2+7X3+X4+x6=702X1+X2+3X3-X5+x7=15
Cj θ V bas X1 X2 X30 10 x6 5 4 70 5 x7 2 1 3
ᴓ -7010 -5008 -10014
Cj θ V bas X1 X2 X30 15 X6 0.33333333 1.66666667 0
10014 -15 X3 0.66666667 0.33333333 1ᴓ -334 -1670 0
Cj θ V bas X1 X2 X3-1000 X5 0.14285714 0.71428571 010014 X3 0.71428571 0.57142857 1
ᴓ 0 0 0
X1 0 Zmax= 10(0)+8(0)+14(10)X2 0 5(0)+4(0)+7(10)≤ 70
2X1+X2+3X3-X5≤15
X3 10 2(0)+(0)+3(10)≥ 15X4 0X5 15 5(0)+4(0)+7(10)+(0)= 70X6 2(0)+(0)+3(10)-(15)= 15X7
7010(0)+5008(0)+10014(10)+1000(0)-1000(15)-850005(0)+4(0)+7(10)+(0)≤ 702(0)+(0)+3(10)-(15)≤ 15
X4 X5 X6 X7 b1 0 1 0 700 -1 0 1 15
-1000 1000
X4 X5 X6 X7 b1 2.33333333 1 -2.33333333 350 -0.33333333 0 0.33333333 5
-1000 -2338 3338
X4 X5 X6 X7 b0.42857143 1 0.42857143 -1 150.14285714 0 0.14285714 0 10
2 0 1002 1000
= 140= 70
= 30
= 70= 15
7010(0)+5008(0)+10014(10)+1000(0)-1000(15)-85000 1407015
8 -- 9a) Resuélvase el siguiente problema con el método simplex y explíquese la respuesta
Maximizar Z= 3X1 + 4X2
Restricciones 5X1 + 10X2 ≤ 507X1 + 7X2 ≥ 498X1 - 2X2 ≤ 8X1,X2 ≥ 0
Se agraga la variable artificial 5X1 + 10X2+X3 = 507X1 + 7X2-X4 = 498X1 - 2X2+X5 =8
Se agregan las M
(20M+3)X1+(15M+4)X2+MX3-MX4+MX5-107M
Si M=1000
Se aumenta la variable de holgura 5X1 + 10X2+X3+X6 ≤ 507X1 + 7X2-X4+X7 ≤ 498X1 - 2X2+X5+X8 ≤ 8
Cj θ V bas X1 X2 X30 10 X6 5 10 10 7 X7 7 7 00 1 X8 8 -2 0
ᴓ -20003 -15004 -1000
Cj θ V bas X1 X2 X30 4 X6 0 11.25 10 4.8 X7 0 8.75 0
20003 -4 X1 1 -0.25 0ᴓ 0 -20004.75 -1000
Cj θ V bas X1 X2 X315004 X2 0 1 0.0888888889
0 X7 0 0 -0.777777777820003 X1 1 0 0.0222222222
ᴓ 0 0 778.2
x1 2x2 4 Zmax= 3(2)+4(4)
Z=20003X1 + 15004X2 + 1000X3 - 1000X4 + 1000X5 -107000
x3 0 5(2)+10(4)≤ 50x4 0 7(2)+7(4)≥ 49x5 0 8X1 - 2X2 ≤ 8x6 0x7 7x8 0
b) Hágase una gráfica de este problema y sombreése la región factible
No tiene solución el problema porque no satisface las desigualdades.
a) Resuélvase el siguiente problema con el método simplex y explíquese la respuesta
X4 X5 X6 X7 X8 b0 0 1 0 0 50-1 0 0 1 0 490 1 0 0 1 8
1000 -1000
X4 X5 X6 X7 X8 b0 -0.625 1 0 -0.625 45-1 -0.875 0 1 -0.875 420 0.125 0 0 0.125 1
1000 1500.375 0 0 0
X4 X5 X6 X7 X8 b0 -0.05555556 0.08888889 0 -0.05555556 4
-1 -0.38888889 -0.77777778 1 -0.38888889 70 0.11111111 0.02222222 0 0.11111111 2
1000 389 1778.2 0 1389
= 22
= 50= 42
No tiene solución el problema porque no satisface las desigualdades.
Resuélvase el siguiente problema de PL con el método simplex y explíquese la soluciónMaximizar: Z= 10X1 + 8X2Restricciones:
7X1 + 4X2 ≥ 353X1 ≤ 2X2 + 3X1,X2 ≥ 0
Se agrega la varable atrificial 7X1+4X2-X3=353X1-2X2+X4=3
Se le agregan las M
Z=(10M+10)X1+(2M+8)X2-MX3+MX4-38
M=1000
Z= 10010X1 + 2008X2 - 1000X3 + 1000X4 -38000
Agregamos la variable de holgura
Cj θ V bas X1 X2 X30 5 X5 7 4 -10 1 X6 3 -2 0
ᴓ -10010 -2008 1000
Cj θ V bas X1 X2 X30 3.23076923 X5 0 8.66666667 -1
10010 -1.5 X1 1 -0.66666667 0ᴓ 0 -8681.33333 1000
Cj θ V bas X1 X2 X32008 -28 X2 0 1 -0.11538462
10010 -41 X1 1 0 -0.07692308ᴓ 0 0 -1.69230769
7X1+4X2-X3+x5≤353X1-2X2+X4+x6≤3
Resuélvase el siguiente problema de PL con el método simplex y explíquese la solución
X4 X5 X6 b0 1 0 351 0 1 3
-1000
X4 X5 X6 b-2.33333333 1 -2.33333333 280.33333333 0 0.33333333 12336.66667 0 3336.66667
X4 X5 X6 b-0.26923077 0.11538462 -0.26923077 3.230769230.15384615 0.07692308 0.15384615 3.15384615
-0.61538462 1001.69231 999.384615
