INSTITUTO TECNOLOGICO SUPERIOR DE TAMAZUNCHALE INGENIERIA EN SISTEMA S ISTEMAS S COMPUTACIONALES INVESTIGACIÓN DE OPERACIONES ING.EDUARDO FRANCO AUSTRIA
INTEGRANTES: PERFECTO HERNÁNDEZ HERNÁNDEZ CESAREO EHRNÁNDEZ REYES MARTIN CÁRDENAS NUÑEZ ELIZABETH HERNANDEZ HERNANDEZ GRIZELDA MATEO MAXIMINO
TERCER SEMESTRE PERFECTO HERNÁNDEZ HERNÁNDEZ MARTIN CARDENAS NUÑEZ. CESAREO HERNÁNDEZ REYES GRIZELDA MATEO MAXIMINO ELIZABETH HERNÁNDEZ HERNÁNDEZ
M1
FECHA DE ENTREGA: 11/09/2014 11/09/2014 Página 1
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1.- Método Mé todo Grafico Grafico ............................................................................................................................. 3 SITUACIÓN 1 (Inzunza, López, de la Vega & Inzunza, 2012) ................................................ ........... 3 Situación 4 (González, 2003) ................................................................................. .......................... 7 SITUACIÓN 7 (Taha, 2012) .......................................................................................................... 11 SITUACIÓN 10................................................. .............................................. ............................. 14 2.- METODO SIMPLEX................................................................................................................... 17 SITUACIÓN 4 (Taha, 2012) .......................................................................................................... 17 SITUACIÓN 5 (Taha, 2012) .......................................................................................................... 20 SITUACIÓN 6 (Chapra & Canale, 2011) ........................................................................................ 21
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1.- Método Grafico siguientes ntes problemas pro blemas según se gún se solicita. solicita. Instrucciones: Resuelva los siguie
SITUACIÓN 1 (Inzunza, López, de la Vega & Inzunza, 2012) Wilson es una empresa fabricante de equipo deportivo y entre sus productos tiene dos guantes de béisbol béisbol:: para jugador jugador de cuadro cuadro y para receptor. receptor. El proceso consi consiste de tres operaciones operaciones básicas: básicas: corte y costura, detallado y empaque. Se dispone de 850, 325 y 150 horas, respectivamente. Ambos productos pasan por todo el proceso. Los requerimientos y la ganancia esperada por producto producto se mu muestran estran a conti continnuación:
Modelo Jugador de cuadro Receptor
Tiempo de producción Corte y Detallado Empaque costura 1.2 0.6 0.15 1.45 0.4 0.25
La empresa desea de sea maxi maximizar mizar la utili utilidad dad total. a) Identifique las variables
= . = . b) Plantee la función objetivo
= 15 +18 c) Plantee las restricciones Sujeto a:
1.2 +1.45 ≤850 0.6 +0.4 ≤325 0.15 +0.25 ≤150 + ≥ 0
Ganancia 15 18
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d) Aplique Aplique el método gráfi grá fico co
= 15 +18 1.2 +1.45 ≤850→1.2 +1.45 =850 0.6 +0.4 ≤326→0.6 +0.4 =325 0.15 +0.25 ≤150→0.15 +0.25 =150 1.2 +1.45 =850 Si = 0, = Sustituir = 0 en 1 1.20+1.45 =850 850 =586.20 = 1.45 Par ordenado (0,586.20) 2. 0.6 +0.9 =325 Si = 0, = Sustituir = 0 en 2 0.60+0.9 =325 = 325 = 812. 812.5 0.4 Par ordenado (0,812.5) 3. 0.15 +0.25 =150 Si = 0, = Sustituir = 0 en 3 0.150+0.25 =150 150 =600 = 0.25 Par ordenado ordenado (0,600) 1.
= 0 = = 0 en 1 1.2 +1.450=850 = 850 =708.33 1.2 Par ordenado (708.33,0) Si , Sustituir
= 0 = = 0 en 2 0.6 +0.90=325 = 325 =541.66 0.6 Par ordenado (541.66 ,0)
Si , Sustituir
= 0 = = 0 en 3 0.15 +0.250=150 150 =1000 = 0.15 Par ordenado (1000,0)
Si , Sustituir
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Grafica 1:
Evaluar vértice:
= 0,
= 586. 586.20 20 Sustituir y en la F.O Max =150 + 18586.20 = 1055 10551.1.6 1.2 +1.45 =8500.6 0.6 +0.4 =325 1.2 0.72 +0.87 =510 0.72 0.48 =390 .39 =120 = 120 307.69 69 . 39 = 307. Sustituir y en la F.O a)
= 307. 307.69 69 en 2 0.6 +0.4307.69 =325 0.6 +123.076=325 0.6 =325123.076 = 201.924 .6 =336.54 b) Sustituir
Max z = 15(336.54)+18(307.69)=5048.1+5538.42=10586.52
=541.66, = 0 Sustituir y en la F.O c)
Max z = 15(541.66)+18(0)=8124.9
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Resumen a) b) c)
= 0 =336.54 =541.66
=586.20 =307.69 = 0
= 10551.6 Max = 1058 10585.5.52 Max = 8124.9 Max
Nota: los los guant guantes es no se pueden pueden produci producirr en fracci fracción. ón.
Evaluar nuevos puntos a) , 500 Sustituir y en la F.O Max z = 15(0)+18(500)=9,000
=0 =
b) , 400 Sustituir y en la F.O Max z = 15(100)+18(400)=8,700
=200 =
d) , 300 Sustituir y en la F.O Max z = 15(300)+18(300)=9,900
=400 =
f) , 100 Sustituir y en la F.O Max z = 15(400)+18(100)=7800
c) , 400 Sustituir y en la F.O Max z = 15(200)+18(400)=10200 e) , 200 Sustituir y en la F.O Max z = 15(400)+18(200)=9600
=100 = =300 = =400 =
=500 =
g) , 0 Sustituir y en la F.O Max z = 15(500)+18(0)=7500 Resumen: a) b) c) d) e) f) g)
= 0 , = 100, =, = 300, = 400, = 400, = 5 0 00 ,
=500, = 400, =, =300, =200, =100, =0,
Max=9,000 Max Max = 8,700 =, Max=9900 Max=9600 Max=7800 Max=7500
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Comprobación: Si
=200, =400 1. 1.2200+1.45400≤850→820≤850 2. 0.6200+0.4400≤325→280≤325 3. 1.15200+0.25400≤150→130≤150 e) Conclusión:
Para obtener una ganancia máxima de 10,200 dólares se debe de fabricar una cantidad de 200 guantes para jugador de cuadro y 400 guantes para receptor. Quedando disponibles 30 horas, para corte – costura, costura, 45 horas para detallado y 20 horas para empaque.
Situación 4 (González, 2003) 2003 ) La asociación de estudiantes de una institución dispone de $100, 000 y ha pensado invertirlos en dos negocios. El primero le reporta una utilidad de $25 mensuales y el segundo $40 mensuales por cada $100 inverti nvertidos. dos. Debido Debido a ciertas ciertas condici condicion ones es impuestas puestas por la asambl asamblea ea de socios, socios, se debe invertir al menos el 25% del capital en el primer negocio y no más del 50% en el segundo. Además, la cantidad invertida en el último no debe ser mayor a 1.5 veces la cantidad invertida en el primero. Se pide plantear este problema como un modelo de programación lineal y resolver por el método gráfico. a) Identifique las variables
= 1
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c) Plantee las restricciones Sujeto a:
+ ≤ 1000 100000 00 ≥25000 ≤50000 ≤1.5 + ≥ 0 a) Aplique Aplique el método gráfi grá fico co
=.25 =.25 +0.40 + ≤100000→ + =100000 ≥ 2 5 → = 25 ≤ 5 0 → = 50 ≤1.5 →1.5 + = 0 100000 00 Si = 0, = 1. + = 1000 Si = 0, = Sustituir = 0 en 1 Sustituir = 0 en 1 +0=100000 0 + =100000 =100000 =100000 Par ordenado (100000,0) Par ordenado (0,100000)
=25000∴ = 0
=50000∴ =50000∴ = 0
2. Par ordenado (25000,0)
3. Si Par ordenado (0,50000)
1.5 + = 0 Si = 0, =
1.5 + =
4.
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= 20000 =0 1 .5 =20000
Par ordenado (20000,30000)
Grafica 2:
Evaluar vértices
=25000 =
A. , 0 Sustituir y en la F.O Max z = .25(25000)+.4(0)=6250
=2500 =25000 en 4 1.525000+ = 0 -37500+ =0 =37500. B. Sustituir
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Max z = 0.25(10000)+0.4(0)=25000
Resumen:
=25000, B. =25000, =33333.33, ,, D. = E. =100000 A.
C.
= 0, = 3750 37500,0, =50000, =50000, = = 0,
Max=6250 Max = 21250 Max=28333.33 = Max = 25000
Comprobación: Si
=50000, =50000 1. + ≤100000⟹50000+50000≤100000⟹100000≤100000 2. ≥25000⟹50000≥25000 3. ≤50000⟹50000≤50000 4. ≤1.5 ⟹50000≤1.550000⟹50000≤75000
Conclusión: Para obtener una ganancia máxima de $32500 se debe de invertir la cantidad de $50000 en
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SITUACIÓN SITUACIÓN 7 (Taha, 2012) 201 2) Una compañía fabrica dos productos, A y B. El volumen de ventas de A es por lo menos 80% de las ventas totales de A y B. Sin embargo, la compañía no puede vender más de 100 unidades de A por día. Ambos productos utilizan una materia prima, cuya disponibilidad diaria máxima es de 240 lb. Las tasas de consumo de la materia prima son de 2 lb por unidad de A y de 4 lb por unidad de B. Las utilidades de A y B son de $20 y $50, respectivamente. Determine la combinac combinaciió n óptima óptima de productos prod uctos para la compañía. compañía. a) Identifique las variables
= . = . b) Plantee la función objetivo
= 20 +50 c) Plantee las restricciones Sujeto a:
2 + 4 ≤240 ≤100 0.2 +0.8 ≤ , ≥ 0
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2 + 4 =2400 Si = 0, = Sustituir = 0 en 1 20+4 =240 = 240 4 = 60 1.
= 0 = = 0 en 1 2 +40=240 = 240 =120 2 Par ordenado ordenado (120,0)
Si , Sustituir
Par ordenado ordenado (0,60)
=100∴ = 0
2. Par ordenado ordenado (100,0)
0.2 +0.8 = 0 Si = 0, = Sustituir = 0 en 3 0.20+0.8 = 0 = 0.80 = 0 Par ordenado ordenado (0,0) 3.
Grafica 3:
= 0 = = 0 en 3 0.2 +0.80=0 0 =0 = 0.2 Par ordenado ordenado (0,0)
Si , Sustituir
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= 80 = 2
A. , 0 Sustituir y en la F.O Max z = 20(80)+50(20)=2600
2 + 4 =240 =100 Sustituir =100 en 1 2100+4 =240 4 =240200 = =10 Sustituir , en la F.O B.
Max =20(100)+50(10)=2500
=100, = 0
C. Sustituir y en la F.O Max z = 20(100)+50(0)=2000
Resumen:
=, B. =100, =100 A.
C.
=, =10, =0
= Max=2500 Max=2000
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SITUACIÓN 10 Supongamos que se cuenta con dos alimentos; pan y queso, cada uno de ellos contiene calorías y proteínas proteínas en dive diversas rsas proporcione proporciones. s. Un kil kilogram ogramoo de pan conti contiene 2000 calorí calorías as y 50 gram gramos os de proteínas proteínas,, un kil kilogram ogramoo de queso queso con conti tien enee 4000 calorí calorías as y 200 gram gramos os de proteínas proteínas.. Supongam Supongamos os que una dieta normal requiere cuando menos 6000 calorías y 200 gramos de proteínas diariamente. diariamente. Por lo lo tanto si el kilogra kilogram mo de pan cuesta $6 .00 y $21 $21.00 .00 el queso. ¿Qué ¿Q ué cantidad cantidad de pan y queso debemos d ebemos comprar para satisfacer satisfacer la dieta? dieta? a) Identifique las variables
= . = . b) Plantee la función función objetivo objetivo
= 6 +21
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2000 +4000 =6000 Si = 0, = Sustituir = 0 en 1 20000 +4000 =6000 = 6000 =1.5 4000 Par ordenado ordenado (0,1.5) 2. 50 +200 =200 Sustituir = 0 en 2 500 +200 =200 = 200 =1 200 Par ordenado ordenado (0,1) 1.
Grafica 4:
= 0 = = 0 en 1 2000 +40000=6000 = 6000 =3 2000 Par ordenado ordenado (3,0)
Si , Sustituir
= 0, = = 0 en 2 50 +2000=200 = 200 =4 50 Par ordenado ordenado (4,0)
Si Sustituir
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Evaluar vértices
= 0, =1.5 Sustituir y en la F.O
Max z = 6(0)+21(1.5)=31.5
= 2, =0.5 Sustituir y en la F.O
Max z = 6(2)+21(0.5)=22.5
= 4, = 0 Sustituir y en la F.O
Max z = 6(4)+21(0)=24
Resumen: a.
B. C.
= 4
= 0
=1.5 =. 0
Max=31.5 =. Max=24
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2.- METODO ME TODO SIMPLEX SITUACIÓN SITUACIÓN 4 (Taha, 2012) 201 2) Gutchi Company fabrica bolsos de mano, bolsos para rasuradora y mochilas. La elaboración incluye piel y materiales sintéticos, y la piel es la materia prima escasa. El proceso de producción requiere dos tipos de mano de obra calificada: costura y acabado. La siguiente tabla da la disponib disponib ilid ilid ad de los recursos, su consumo consumo por los tres productos p roductos y las utili utilidad dades es por uni unidad dad.. Recurso Piel (ft (ft2 ) Costur Cos turaa (h) Acabado (h)
Requ Re quee rimi rimiee ntos de recursos por unid unidad ad Bolsos de Bols os para para Mochilas mano rasuradora 2 1 3 2 1 2 1 .5 1
Disponibilidad diaria 42 40 45
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-22
-45
1
3
1
1
1
.5
1
-24
.
1
45
.
1
42 40
42 = 14 403 = 20 452 = 45 1
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= = + = + = + Variables básicas
Variables no básicas
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SITUACIÓN SITUACIÓN 5 (Taha, 2012) 201 2) La división de educación continua del Colegio Comunitario de Ozark ofrece un total de 30 cursos cada semestre. Por lo común, los cursos ofrecidos son de dos tipos: prácticos, como carpintería, procesami procesamiento ento de palabras palabras y manten anteniimiento ento autom automotri otrizz; y hum humani anistas como como histori historia, a, música sica y bell bellas artes. artes. Para satisf satisfacer acer las deman demandas das de la comun comuniidad, cada semestr semestree debe ofrecerse ofrecerse como como mínimo 10 cursos de cada tipo. La división estima que los ingresos producidos por el ofrecimiento de cursos prácticos y humanistas son aproximadamente de $1500 y $1000 por curso, respectivamente. Idee un ofrecimi ofrecimiee nto de cursos cursos óptimo óptimo para el colegio. colegio. Modelo matemático:
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SITUACIÓN SITUA CIÓN 6 (Chapra & Canale, 2011) 2011 ) Una compañía de electrónica produce transistores, resistores y chips de computadora. Cada transistor requiere cuatro unidades de cobre, una de zinc y dos de vidrio. Cada resistor requiere tres, tres y una unidades de dichos materiales, respectivamente, y cada chip de computadora requiere dos, d os, una una y tres uni unidad dades es de los materiales, materiales, respecti respe ctivamente. vamente. Los suministros de estos materiales varían de una semana a la otra, de modo que la compañía necesita determinar una corrida de producción diferente cada semana. Por ejemplo, cierta semana las cantidades disponibles de los materiales son 960 unidades de cobre, 510 unidades de zinc y 610 un uniidad dades es de vi vidrio. Cada unidad producida de transistores, resistores y chips le genera a la empresa una utilidad de $5, $2 y $7 dólares. ¿Qué ¿ Qué cantidad cantidad de transistores transistores,, resistores y chips chips debe deb e fabricar fabricar esta semana?
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Tabla simplex simplex
Solución
-5
-2
-7
0
0
0
0
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Restricciones:
4 + 3 + 2 ≤960 ⇒ 4 (415 2 ) +30+265≤960