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 "Optimizaci\[OAcute]n lineal y no lineal con ",
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Cell["\<\
M\[EAcute]todos de Optimizaci\[OAcute]n ADE. Pr\[AAcute]cticas. Dpto de \
Econom\[IAcute]a e Historia Econ\[OAcute]mica. Universidad de Salamanca.  \
\>", "Department",
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Cell["\<\
Guillermo S\[AAcute]nchez (guillermo@usal.es). \
\>", "Department",
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\[CapitalUAcute]ltima actualizaci\[OAcute]n: 2012-11-25\
\>", "Date",
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Funciones de optimizaci\[OAcute]n disponibles en Mathematica\
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Cell[TextData[{
 "Los problemas de optimizaci\[OAcute]n matematicamente consisten en lo \
siguiente: Para una funci\[OAcute]n de ",
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 " variables, el \[AAcute]rea que cumple con las restricciones del problema \
en una regi\[OAcute]n de un espacio ",
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 "\[Hyphen]dimensional donde cada desigualdad , o restricci\[OAcute]n, es un \
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 "\[Hyphen]dimensional,  que l\[IAcute]mita uno de los lados de la regi\
\[OAcute]n. "
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 "Para resolver este tipo de problemas con ",
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 " disponemos, entre otras,  de las funciones  ",
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 "."
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 ", {",
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 " encuentra, por m\[EAcute]todos n\[UAcute]mericos, el m\[AAcute]ximo de f \
que cumplen las condiciones especificadas en las restricciones  para las \
variables definidas. Tambien se puede utilizar para ",
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 " , definido no negativo {0\[LessSlantEqual]x, 0\[LessSlantEqual]y}, podemos \
calcular el m\[AAcute]ximo como sigue "
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NMaximize[{x + y, 0\[LessSlantEqual]x\[LessSlantEqual]1, 0\[LessSlantEqual]y\
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Cell["\<\
Maximize[{x + y, 0\[LessSlantEqual]x\[LessSlantEqual]1, 0\[LessSlantEqual]y\
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Cell["\<\
Vemos que el maximo est\[AAcute] en {x \[Rule] 1, y \[Rule] 2} y es 3. \
Podemos comprobarlo (recordemos que   /. se utiliza para reemplazamientos}\
\>", "Text",
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Cell["\<\
Observe que el problema anterior tambien podemos resolverlo utilizando \
formato ling\[UDoubleDot]isto escribiendo \[OpenCurlyDoubleQuote]Maximize x + \
y  with x<1 and y<2\[CloseCurlyDoubleQuote] u otra equivalente.\
\>", "Text",
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Cell["Podemos copiar la entrada y modificarla si es necesario.", "Text",
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Cell["\<\
El m\[IAcute]nimo para las mismas condiciones podemos calcularlo como sigue :\
\>", "Text",
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NMinimize[{x + y, 0\[LessSlantEqual]x<1, 0\[LessSlantEqual]y<2}, {x, y}]\
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Cell["\<\
Minimize[{x + y, 0\[LessSlantEqual]x<1, 0\[LessSlantEqual]y<2}, {x, y}]\
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Cell[TextData[{
 "Veamos el problema graficamente: representamos el plano",
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 ", ",
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 "}. Vemos que en el grafico el maximo esta en (1,2) y (0,0). Pruebe a pulsar \
con el rat\[OAcute]n sobre el centro del gr\[AAcute]fico y desplazelo para \
verlo sobre distinta perspectivas. El simbolo && correspode a la proposici\
\[OAcute]n l\[OAcute]gica Y (o AND). En este caso indica que deben cumplirse \
simultaneamente las dos restricciones indicadas. Pueden incluirse tantas \
restricciones como se desee."
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Vamos a mostrar con algunos ejemplos modelos t\[IAcute]picos de optimizaci\
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En algunos problemas de optimizaci\[OAcute]n se requiere que las soluciones \
sean enteras,  puede hacerse como sigue (en este caso solo se requiere que \
sea entera la variable y):\
\>", "Text",
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\[UAcute]mero de varibles pues simplifica enormente la entrada de datos."
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El problema anterior tiene la siguiente equivalencia en LinearProgramming\
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\tx + 2 y\t\t-> \t\tc: {1, 2}
\tRestricciones
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\tx + y \[GreaterEqual] 26    ->\t\tm2:  {1, 1}\tb2:  {26, 1}
\
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Puede parecer m\[AAcute]s complicada la notaci\[OAcute]n sin embargo, como se \
ha dicho, es mucho m\[AAcute]s r\[AAcute]pida en problemas de optimizaci\
\[OAcute]n lineal donde se utilicen muchas variables y restricciones\
\>", "Text",
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Planteamos la funci\[OAcute]n a optimizar o funci\[OAcute]n objetivo (en \
nuestro caso se trata de maximizar el beneficio)\
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Finalmente maximizamos la funci\[OAcute]n objetivo con sus correspondientes \
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\[DownQuestion]Como escribir de forma f\[AAcute]cil que todas las variables \
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En la mayor\[IAcute]a de aplicaciones pr\[AAcute]cticas de \
optimizaci\[OAcute]n en problemas ec\[OAcute]nomicos se requiere que todas \
las variables sean no negativas, esto es mayores o iguales a cero. Eso puede \
hacerse escribiendolas de una en una, como en el ejemplo anterior, o de forma \
mucho m\[AAcute]s sencilla como se muestra a continuaci\[OAcute]n (utilizamos \
el \[UAcute]ltimo ejemplo)\
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Cell["\<\
Queremos minimizar la funci\[OAcute]n x^2 + (y - 1)^2  con la restricci\
\[OAcute]n: x^2 + y^2 \[LessEqual] 4. \
\>", "Text",
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cuando se emplean algoritmos de c\[AAcute]lculo n\[UAcute]merico)"
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Cell["El problema anterior podemos representarlo graficamente. ", "Text",
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Cell["\<\
Pulse el interior de gr\[AAcute]fico y g\[IAcute]relo para identificar \
visualmente donde est\[AAcute] el m\[IAcute]nimo.\
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 " es un tipo de gr\[AAcute]fico muy apropiado para intentar ver por donde se \
encuentra el m\[IAcute]nimo. (La opci\[OAcute]n ",
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 " especifica el n\[UAcute]mero de contornos que queremos utilizar, si no se \
incluye la opci\[OAcute]n el programa utiliza unos valores por defecto)."
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Podemos ver que el minimo est\[AAcute] en (0, 1). Si se deja la \
opci\[OAcute]n por defecto los colores m\[AAcute]s oscuros representan los \
valores m\[AAcute]s peque\[NTilde]os y los m\[AAcute]s claros lo m\[AAcute]s \
elevados. Para entenderlo mejor comparelo con el gr\[AAcute]fico anterior. Si \
se desplaza por el gr\[AAcute]fico el cursor del rat\[OAcute]n por los \
limites entre contornos se ver\[AAcute] los valores que va tomando la funci\
\[OAcute]n. Los puntos unidos por la misma l\[IAcute]nea corresponden a \
aquellos en los que la funci\[OAcute]n toma el mismo valor.\
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Cell["Ejercicios sobre aplicaciones lineales", "Section",
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Cell["\<\
Ejercicio 1.- Fabricaci\[OAcute]n de juguetes \
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Una factor\[IAcute]a fabrica dos tipos de juguetes de madera:soldados y \
trenes. El beneficio que obtiene es de 3 euros por cada soldado y de 2 euros \
por cada tren.Un soldado necesita 2 horas de acabado y 1 hora de carpinter\
\[IAcute]a,mientras que cada tren necesita 1 hora de acabado y 1 hora de \
carpinter\[IAcute]a.Cada semana se dispone de 100 horas de acabado y 80 de \
carpinter\[IAcute]a. Por demanda del mercado no se venden m\[AAcute]s de 40 \
soldados cada semana.
a) \[DownQuestion]Cuantas unidades debemos de fabricar de cada tipo de \
juguete para maximizar el beneficio?\
\>", "Text",
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Cell["\<\
b) \[DownQuestion] Cu\[AAcute]nto estar\[IAcute]amos dispuestos a pagar por \
10 horas m\[AAcute]s de carpinter\[IAcute]a?\
\>", "Text"],

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Cell["\<\
Variables: X1, X2 : N\[UAcute]mero de soldados y trenes a fabricar\
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Cell["Funcion objetivo (a maximizar): 3 X1 + 2 X2", "Text",
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Restricciones:2 X1 + X2 \[LessEqual]100;  X1 + X2\[LessEqual] 80;  X1 \
\[LessEqual] 40; X1 ,X2\[GreaterEqual]0\
\>", "Text",
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Cell["\<\
Sol apdo. b). 
Es necesario modificar una de las restricciones -\[DownQuestion]cual?-y ver \
que beneficio se obtiene. Se compara con el obtenido en el apdo a) y de ah\
\[IAcute] puede deducirse directamente al soluci\[OAcute]n.\
\>", "Text"]
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Cell["\<\
Ejercicio 2 .- Lineas de fabricaci\[OAcute]n \
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Una empresa produce dos productos en dos l\[IAcute]neas de \
fabricaci\[OAcute]n diferentes.La primera l\[IAcute]nea tiene una capacidad \
de producci\[OAcute]n de 60 unidades y la segunda de 50. Producir el primer \
producto requiere 1 hora mientras que producir el segundo requiere 2 h.El \
total de horas disponibles es de 120. El beneficio que se obtiene de los dos \
productos es de 20 \[Euro] y 30 \[Euro] respectivamente.\
\>", "Text",
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Cell["\<\
a) \[DownQuestion]Cuantas unidades hay que fabricar de cada producto para que \
el beneficio sea m\[AAcute]ximo?. \[DownQuestion]Cual ser\[AAcute] el \
beneficio?\
\>", "Text",
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Cell["\<\
b) En la soluci\[OAcute]n encontrada: \[DownQuestion]Se consumen todas las \
horas de trabajo disponibles? \[DownQuestion]Se agota la capacidad de \
producci\[OAcute]n de las dos l\[IAcute]neas de producci\[OAcute]n?\
\>", "Text",
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Cell["\<\
a) MAX     20 X1 + 30 X2
Restricciones: X1 <=  60;    X2 <=  50; X1 + 2 X2 <= 120; X1, X2>=0\
\>", "Text",
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Cell["\<\
b)  \[DownQuestion]Se consumen todas las horas de trabajo disponibles? Si;  \
\[DownQuestion]Se agota la capacidad de producci\[OAcute]n de las dos l\
\[IAcute]neas de producci\[OAcute]n? No, en la l\[IAcute]nea 2 hay una \
holgura de 20 pues tiene una capacidad de 50 y consumimos 30.
\
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Cell[CellGroupData[{

Cell["Ejercicio 3.- Evacuaci\[OAcute]n", "Subsection",
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Cell["\<\
Una cat\[AAcute]strofe obliga a evacuar tres nucleos de poblaci\[OAcute]n \
(Salamanca, Avila y Zamora)  hacia dos campamentos de refugiados (A y B). La \
poblaci\[OAcute]n a evacuar de cada ciudad y las capacidades m\[AAcute]ximas \
de los campamentos se muestran en la tabla. La duraci\[OAcute]n, en horas, de \
cada trayecto tambien se muestra en la tabla.\
\>", "Text",
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Cell["\<\
Se pide: minimizar el tiempo total de evacuaci\[OAcute]n de la poblaci\
\[OAcute]n desde las ciudades a los campamentos de refugiados, sin exceder \
las capacidades estos. \
\>", "Text",
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Cell["\<\
a) Plantee matem\[AAcute]ticamente el problema (funci\[OAcute]n objetivo y \
restricciones) \
\>", "Text",
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 ":{1, 2, 3} indica la poblaci\[OAcute]n de origen {1 = Salamanca, 2 =Avila, \
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b) Indique cual es el tiempo total en horas que minimiza el trasporte\
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Cell["\<\
c) Complete la siguiente tabla con el n\[UAcute]mero de personas que enviaria \
desde ciudades a campamentos\
\>", "Text",
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La restriccion de que todas las variables deben ser no negativas puedo a\
\[NTilde]adirlas escribiendolas directamente en la funcion a optimizar  o mas \
facil (aunque no se ha explicado la instruccion Map, puede usar F1 para \
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ejercicios con la misma restriccion):\
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Cell["\<\
utilizando [[1]] extraemos el primer t\[EAcute]rmino de la lista anterior, es \
decir el total de horas sumando las empleadas por cada habitante. \
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Cell["\<\
Si queremos calcular el tiempo medio dividimos el total de horas entre el n\
\[UAcute]mero de habitantes desplazados. Para ello extraemos el segundo \
termino de la solucion (que es el resultado de cada una de las variables) y \
lo sustituimos empleando /. que significa \"reemplaza por\"\
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Cell["Ejercicio 4.- Problema de transporte ", "Subsection",
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Cell["\<\
La cadena de grandes almacenes CARREFA desea contratar el suministro de cajas \
de leche para los centros que tiene en Barcelona, Madrid, Sevilla y \
Valladolid. Para ello recibe oferta de tres envasadores: Leche Pascualo, \
Leche Simone y  Leche Pulova. Los precios son los siguientes:  Pascualo  3.00 \
\[Euro] por caja, Simone 2.80 \[Euro] por caja  y Pulova 2.70 \[Euro] por \
caja. El m\[AAcute]ximo n\[UAcute]mero de cajas que diariamente puede \
suministrar cada envasador se indica en la tabla adjunta, en la columna \
\"Suministro m\[AAcute]ximo\".  La demanda diaria de cada centro se muestra \
en la fila \"Demanda\".  Los costes de transporte, en \[Euro] por caja, se \
indican en la misma tabla. \
\>", "Text"],

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Cell["\<\
CARREFA desea calcular las cantidades que debe encargar a cada envasador para \
minimizar el coste total.\
\>", "Text"],

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Cell["a) Formular matematicamente el problema ", "Subsubsection"],

Cell[TextData[{
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Cell["\<\
Nota: Tenga en cuenta que el coste total de cada caja est\[AAcute] dado por \
el precio de compra m\[AAcute]s el de transporte.\
\>", "Text"],

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Funci\[OAcute]n objetivo: El minimo vendra dada sumando los costes de compra \
mas los costes de transporte \
\>", "Text",
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Los precios de compra  son: Pascualo 3.00 \[Euro] , Simone 2.80 \[Euro]y \
Pulova 2.70 \[Euro]  y los de trasporte son los que se muestran en la tabla\
\>", "Text",
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Sumo los costes de compra m\[AAcute]s los de trasporte para los diferentes \
tipos de leche\
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b) En la soluci\[OAcute]n \[OAcute]ptima: \[DownQuestion]Cual es el coste \
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Cell["\<\
d) \[DownQuestion]A que precio m\[AAcute]ximo deber\[IAcute]a vender Leche \
Pascualo para saturar su capacidad de envasado?\
\>", "Subsubsection",
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Cell["\<\
Las restricciones y las variables son las mismas independientemente del \
precio de la leche, luego no hay que escribirlas de nuevo\
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Los precios de compra iniciales son Pascualo 3.00 \[Euro] , Simone 2.80 \
\[Euro] y Pulova 2.70 \[Euro], en vez de escribirlos directamente en la funci\
\[OAcute]n de optimizaci\[OAcute]n lo definimos previamente {pa, sim, pul} = \
{3, 2.8, 2.7}; vamos ejecutando sucesivamente la celda bajando el precio (pa) \
hasta que se obtenga que Pascualo vende 2500 uds. \
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Cell["El mismo ejemplo utilizando LinearProgramming (LP)", "Subsubsection",
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Cell["\<\
Vamos a resolver el ejemplo anterior comparando  notaci\[OAcute]n utilizada \
on NMinimize con la de LinearProgramming (LP). En la pr\[AAcute]ctica esto no \
es necesario, de hecho es m\[AAcute]s sencillo realizarlo directamente \
utilzando la  natacion de LP, aqui lo realizamos a efectos did\[AAcute]cticos \
\>", "Text",
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Cell[TextData[{
 "Con LinearProgramming (LP) no es necesario definir explicitamente las \
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Cell["La funci\[OAcute]n a optimizar es fo", "Text",
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Cell[TextData[{
 "Para construir la funci\[OAcute]n de optimizaci\[OAcute]n hemos de utilizar \
los terminos agrupados que multiplican a cada variable. Por ejemplo: el coste \
total de ",
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pa) y as\[IAcute] para todas las variables"
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LinearProgramming asume que son todas las variables son no negativas, por \
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es  multiplicar la funci\[OAcute]n a optimizar por la soluci\[OAcute]n \
obtenida (utilizamos \".\" para multiplicar matrices)."
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Las restricciones y las variables son las mismas independientemente del \
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Los precios de venta son: Pascualo ? \[Euro] , Simone 2.80 \[Euro]y Pulova \
2.70 \[Euro] que puedo escribirlo como sigue\
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Pruebe a ir bajando el precio (pa)  hasta que obtenga que Pascualo vende 2500 \
uds. Podemos empezar por 3 \[Euro] y ir reduciendo 0.1 \[Euro]  hasta que \
obtenemos 2500 \[Euro]\
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Cell["\<\
 Ejemplos de aplicaciones de optimizaci\[OAcute]n no lineal\
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 "El consumo de leche en cierta regi\[OAcute]n sigue la funci\[OAcute]n ",
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 " renta diaria por familia en \[Euro]/d\[IAcute]a. El rango de validez de la \
expresi\[OAcute]n anterior es para las rentas familiares comprendidas entre \
20 y 200 \[Euro]/dia. El precio del L leche oscila entre 0.4 \[Euro] y 1.2 \
\[Euro] \[DownQuestion]Cual ser\[AAcute] el consumo m\[AAcute]ximo y m\
\[IAcute]nimo? "
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Cell["\<\
Procedemos de la forma habitual, definir variables, funci\[OAcute]n a \
optimizar y restricciones:\
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Cell["\<\
Podemos comprobar graficamente la validez de la soluci\[OAcute]n (pulsar \
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\[IAcute]mites de la curva correponden a las restricciones.\
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Cell[TextData[{
 "Otra forma de verlo es  utilizando ",
 StyleBox["ContourPlot.", "Input"],
 " Recuerdo que los colores mas suaves (en el gr\[AAcute]fico, los de la \
izquierda) indican valores mayores que los colores m\[AAcute]s oscuros (los \
de la dececha) que indican valores menores. Si pasamos con el rat\[OAcute]n \
la punta del cursor nos mostra los distintos valores de la funci\[OAcute]n \
c(r,t). As\[IAcute] podemos comprobar que la esquina superior izda es la de \
mayor valor (m\[AAcute]ximo) y la inferior derecha la del minimo valor. Los \
limites de la curva correponden a las restricciones."
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Cell["\<\
Ejemplo 2.- Optimizaci\[OAcute]n red electrica\
\>", "Subsection",
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Cell["\<\
El siguiente ejemplo (datos inventados) se pretende optimiar el uso de una \
red electrica.\
\>", "Text",
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Cell["\<\
Una empresa el\[EAcute]ctrica dispone de centrales el\[EAcute]ctricas (una \
nuclear, dos de gas y una de carb\[OAcute]n) cuyo beneficio por hora de \
funcionamiento depende del n\[UAcute]mero de horas que funcione al \
a\[NTilde]o. Estos valores se muestran en la tabla donde  t1, t2, t3, t4  \
indica las horas de funcionamiento de la central correspondiente. Cada \
central debe funcionar al a\[NTilde]o al menos 5000 h si es nuclear , 2000 h \
si es de carb\[OAcute]n, y 2500 h si es de gas. Ademas el conjunto de las \
centrales pueden llegar a funcionar conjuntamente un m\[AAcute]ximo de 20000 \
horas. Considerese un a\[NTilde]o normal, es decir de 8760 h.
\[DownQuestion]Cual es el beneficio anual esperado? \[DownQuestion]Cuantas \
horas debe funcionar cada central para que el beneficio anual se \
m\[AAcute]ximo \[DownQuestion]Cual sera este beneficio?\
\>", "Text",
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Cell[TextData[Cell[BoxData[GridBox[{
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Cell["Definimos las variables: ", "Text",
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Cell["Funci\[OAcute]n a optimizar", "Text",
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Cell["\<\
Finalmente podemos calcular la funci\[OAcute]n que m\[AAcute]ximiza el \
beneficio\
\>", "Text",
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El banco Aupacrisis  ofrece un producto de inversi\[OAcute]n din\[AAcute]mico \
a un plazo fijo de 7 a\[NTilde]os. Durante ese periodo  hay que ir \
traspasando el dinero entre tres fondos A, B y C. Cada uno de los fondos  da \
una rentabilidad que se c\[AAcute]lcula seg\[UAcute]n el tiempo que se esta \
en ese fondo (Ver tabla; ta, tb y tc son los tiempos de permanencia, en a\
\[NTilde]os, en el fondo A, B y C respectivamente y Co es el capital inicial \
que se invierte) con un tiempo m\[IAcute]nimo de permancia en cada fondo de \
un a\[NTilde]o. En el tercer fondo hay  que mantener la inversi\[OAcute]n un \
tiempo superior a la suma de los dos primeros periodos. La rentabilidad total \
al final del periodo de 7 a\[NTilde]os ser\[AAcute] la suma de las \
rentabilidades de cada intervalo . Dispongo de 100 000 \[Euro]. \
\[DownQuestion]Que tiempos de permancia en cada fondo debo elegir de forma \
que la rentabilidad acumulada en los 7 a\[NTilde]os sea m\[AAcute]xima?\
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la temperatura exterior, ",
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son ",
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\[DownQuestion]cual ser\[AAcute] el consumo por hora?"
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Los siguientes casos propuestos, y resueltos parcialmente, pueden ser \
requeridos por el profesor para utilizarlos como un dato m\[AAcute]s al \
evaluar la nota del alumno.\
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Un grupo asesor financiero tiene un monto de 10 000 000 \[Euro] para ser \
colocado en varios productos financieros. El cuadro siguiente describe las 5 \
categor\[IAcute]as, con su respectivo beneficio  y riesgo asociado.\
\>", "Text",
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El capital no invertido en alguna de estas categor\[IAcute]as, es colocado en \
deuda publica  (sin riesgo) y con inter\[EAcute]s del 3%. El objetivo para el \
grupo asesor es asignar el dinero a cada una de las categor\[IAcute]as para \
cumplir con las metas siguientes :
(a) Maximizar el beneficio por \[Euro] invertido
(b) Que el riesgo promedio no supere el 5% ( sobre el dinero invertido) 
(c) Invertir al menos 20% en pr\[EAcute]stamos comerciales
(d) El total invertido en 2nda Hipotecas y Prestamos Personales no podr\
\[AAcute] ser mayor que el invertido en 1era Hipotecas.\
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Cell["\<\
La solucion es el 11.2%, el beneficion esperado es 1 120 000 \[Euro], con una \
inversi\[OAcute]n del 40% en 1ra Hipoteca., 40% en prestamos personales y  \
20% en prestamos comerciales.\
\>", "Text",
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Cell["\<\
Como invertimos 10 000 000 \[Euro] el beneficio ser\[AAcute] (observe que se \
divide entre 100, pues el resultado anterior para expresarlo como fracci\
\[OAcute]n de 1  es: 11.2/100):\
\>", "Text",
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  3.533012793055067*^9}}],

Cell[CellGroupData[{

Cell[BoxData[
 RowBox[{"10000000", " ", 
  RowBox[{
   RowBox[{"sol", "[", 
    RowBox[{"[", "1", "]"}], "]"}], "/", "100"}], " ", 
  "\"\<\[Euro]\>\""}]], "Input",
 CellChangeTimes->{{3.499144407971778*^9, 3.49914445489666*^9}}],

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}, Open  ]]
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Cell[CellGroupData[{

Cell["Gestion de cartera ", "Subsection",
 CellChangeTimes->{{3.4949980413746815`*^9, 3.49499805078922*^9}, {
  3.494999399933387*^9, 3.494999434926388*^9}, {3.4949994683853016`*^9, 
  3.494999473034568*^9}, {3.494999936502076*^9, 3.4949999445895395`*^9}, {
  3.4949999890070796`*^9, 3.495000210990776*^9}, {3.495000255252308*^9, 
  3.4950002686700754`*^9}, {3.495000309544413*^9, 3.4950003304346085`*^9}, {
  3.495001065980679*^9, 3.49500109064809*^9}, {3.4950011437631283`*^9, 
  3.495001147323332*^9}, {3.5330543489921966`*^9, 3.5330543570730104`*^9}}],

Cell["\<\
Se trata de gestionar una cartera de inversion de 1 000 000 \[Euro] para un \
horizonte de 6 a\[NTilde]os.  En cada periodo se puede invertir en una o m\
\[AAcute]s de las opciones siguientes :
- Opci\[OAcute]n A en una Caja de Ahorro  con un interes anual de del 5%.
- Opci\[OAcute]n Y inversi\[OAcute]n mantenida (maturity) por 2 a\[NTilde]os, \
con un retorno total de 12% al final del perido si compramos ahora o 11% m\
\[AAcute]s adelante.
- Opci\[OAcute]n Z, inversi\[OAcute]n mantenida 3 a\[NTilde]os, y un retorno \
total de 18% al final del perido.
- Opci\[OAcute]n W inversi\[OAcute]n mantenida 4 a\[NTilde]os, y un retorno \
total de 24% al final del perido.
Se pueden hacer movimientos (dep\[OAcute]sitos /retiros) en la Caja de Ahorro \
en cualquier momento.
Se pueden comprar Opciones Y todos los a\[NTilde]os, salvo en el a\[NTilde]o \
3 .
Se pueden comprar Opciones Z todos los a\[NTilde]os, despu\[EAcute]s del 1er \
a\[NTilde]o .
La opci\[OAcute]n W, disponible en este momento, es una oportunidad \
\[UAcute]nica.
(Con el objetivo de simplificar, se asumir\[AAcute] que cada opci\[OAcute]n \
puede se comprada en cualquier \[OpenCurlyDoubleQuote]denominaci\[OAcute]n\
\[CloseCurlyDoubleQuote]).\
\>", "Text",
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  3.49500156864743*^9, 3.4950015692244625`*^9}}],

Cell["Graficamente el problema puede representarse como sigue :", "Text",
 CellChangeTimes->{{3.4950003814965286`*^9, 3.495000395728343*^9}}],

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Cell["\<\
Sea At la cantidad invertida en la opci\[OAcute]n A al inicio del \
a\[NTilde]o, y de manera similar Yt, Zt y Wt.  En un a\[NTilde]o dado, el \
monto de dinero que se traslada desde al a\[NTilde]o anterior, m\[AAcute]s el \
retorno de las colocaciones que llegan a madurez ese a\[NTilde]o, debe ser \
igual a lo invertido en nuevas colocaciones m\[AAcute]s el monto de dinero \
que se dej\[OAcute] para el a\[NTilde]o siguiente. \
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En t = 1 => A1 + Y1 + W1 = 1000000
En t = 2 => A2 + Y2 + Z2 = 1.05 A1
En t = 3 => A3 + Z3 = 1.05 A2 + 1.12 Y1
En t = 4 => A4 + Y4 + Z4 = 1.05 A3 + 1.11 Y2
En t = 5 => A5 + Y5 = 1.05 A4 + 1.18 Z2 + 1.24 W1
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No Negatividad : At, Yt, Zt y Wt \[GreaterEqual]0\
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Cell["\<\
Una empresa de suministros de bacalao realiza la distribuci\[OAcute]n desde \
un puerto pesquero a tres ciudades C1, C2 y C3. La demanda diaria de cada uno \
de estos ciudades se muestra en la fila \"Demanda\" (cantidad m\[AAcute]xima \
que puede llegar a vender) es funci\[OAcute]n del tiempo t que va desde que \
dispone el pescado en la lonja hasta que lo entrega en la ciudad \
correspondiente (obs\[EAcute]rvese que la demanda va disminuyendo con el \
tiempo). Debe entregarlo entre 2 y 4 horas. Por otro lado tiene unos costes \
de trasporte que depende del tiempo de entrega dados por la funci\[OAcute]n \
indicada en la tabla (el coste sera mayor mientras m\[AAcute]s r\[AAcute]pido \
se haga la entrega)   \[DownQuestion]C\[AAcute]lcula para que valor de t:{t1, \
t2, t3}, que representan el tiempo desde la lonja a C1, C2 y C3, que maximiza \
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Es decir, el beneficio m\[AAcute]ximo (1.52666 millones de \[Euro]) se \
optiene para los valores de de t indicados\
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Optimizaci\[OAcute]n de una planta de tratamiento de minerales\
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La siguiente ecuaci\[OAcute]n indica los kilogramos (r) de cobre que se \
obtienen mensualmente  en una planta de tratamiento de minerales:\
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donde u, v, y w representan respectivamente el consumo mensual de cal (en  \
toneladas),  mineral  (en toneladas, t) y mano de obra (en horas).   Los \
costes son: 1 t de cal 23\[LineSeparator] euros, 1 t de mineral  20 euros y 1 \
hora de mano de obra 23 euros. A esos precios sl mes se dispone como mucho \
3000 t de cal y 4000 t de mineral al mes y 2000 horas. \
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a) El precio de venta del  cobre es 36 euros/kg.   \[DownQuestion]Qu\[EAcute] \
valores deben tener u, v y w para maximizar el beneficio? \[DownQuestion]C\
\[UAcute]al ser\[AAcute] es beneficio?\
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b) El precio del cobre baja a 34 euros kg \[DownQuestion]Obtengo beneficio? \
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