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Ortega Froysa} + +\begin{document} +\maketitle +\pagebreak +\tableofcontents +\pagebreak + +\section{Ejercicio 5} +\noindent +{\bf ¿Los siguientes hosts son de la misma subred? ¿Por qué?} \\ +\\ +{\em A. Dirección 138.65.46.12 con Máscara 255.255.0.0 y Dirección 138.65.57.96 +con Máscara 255.240.0.0} + +Estas dos direcciones no pueden pertenecer a la misma red, ya que tienen +máscaras distintas. Esto se ve de manera más evidente cuando comparamos la +dirección de red de cada una en bits, que nos sale a: + +\begin{verbatim} +10001010.01000001.00000000.00000000 (138.65.0.0) +10001010.01000000.00000000.00000000 (138.64.0.0) +\end{verbatim} + +Por lo tanto, no son de la misma red.\\ +\\ +{\em B. Dirección 156.132.33.110 con Máscara 255.255.0.0 y Dirección +156.132.45.110 con Máscara 255.255.0.0} + +Sabemos ahora que ambas direcciones tienen la misma máscara: 255.255.0.0, que +tan sólo se fija en los primeros 16 bits (2 bytes) que son de red, mientras que +los demás identifican el terminal (i.e.\ {\em host}). Entonces, para que sean de +la misma red tienen que coincidir los primeros 16 bits de cada dirección. Dado +que 16 bits son 2 bytes, y las direcciones IP nos facilitan la lectura por byte, +podemos compararlo en decimal: 156.132 y 156.132 respectivamente. Estas dos son +iguales, y por lo tanto forman parte de la misma red. \\ +\\ +{\em C. Dirección 205.48.36.139/30 y Dirección 205.48.36.140/30} + +Como en el último apartado, estamos tratando de dos direcciones con la misma +máscara de red, pero en este caso la máscara es de 30 (i.e.\ 255.255.255.252). +De este modo, 30 bits identifican la red, y tan sólo 2 el terminal. Para +comprobar, conseguimos (por medio de una multiplicación binaria) la dirección de +red de cada dirección y comparamos si son iguales: + +\begin{verbatim} +a) +11001101.00110000.00100100.10001011 (dirección) +11111111.11111111.11111111.11111100 (máscara) +11001101.00110000.00100100.10001000 (red) +b) +11001101.00110000.00100100.10001100 (dirección) +11111111.11111111.11111111.11111100 (máscara) +11001101.00110000.00100100.10001100 (red) +\end{verbatim} + +Vemos que la dirección de red difiere en el trigésimo bit, que pertenece a la +red, lo cual indica que no son de la misma red. \\ +\\ +{\em D. La Dirección 178.25.211.0 con máscara 255.255.192.0 y la Dirección IP +178.25.214.0/18} + +Primero, vamos a convertir la máscara de la segunda dirección en una máscara que +sea más fácil de convertir a binario, de forma que /18 es igual a 255.255.192.0. +Vemos entonces que ambas máscaras son iguales. Sin meternos en binario, sabemos +que los dos primeros bytes tienen que coincidir, que es verdad que es 179.25 en +ambas direcciones. Del último byte no nos importa ya que este pertenece al +terminal y no a la red. Por lo tanto, comparamos si, en los bits del tercer byte +que pertenecen a la red, coinciden ambas direcciones: + +\begin{verbatim} +a) +11010011 = 211 (dirección) +11000000 = 192 (máscara) +11000000 = 192 (red) +b) +11010110 = 214 (dirección) +11000000 = 192 (máscara) +11000000 = 192 (red) +\end{verbatim} + +Ya que en ambos casos la dirección de red es igual, podemos concluir que +pertenecen a la misma red. + +\section{Ejercicio 6} +\noindent +{\bf +Nuestra empresa tiene la dirección IP pública 173.54.15.0, y se va a hacer +una reestructuración de departamentos. Nos piden que asignemos los siguientes +equipos a cada uno de ellos: + +\begin{itemize} + \item Dirección: 6 equipos + \item RR.HH.: 4 equipos + \item Producción: 53 equipos + \item Financiero: 18 equipos + \item Comercial y Marketing: 23 equipos + \item Informática: 8 equipos +\end{itemize} + +Diseña cuantas subredes consideres necesarias.} \\ +\\ +{\em Máscara de red.} + +Precisamos como mínimo 6 redes, pero como sólo podemos crear un número de redes +(cuando sean de igual tamaño) igual a una potencia de 2, tenemos que rondar para +arriba a 8 redes ($2^{3}$). Al ser $2^{3}$ sabemos que precisaremos 3 bits para +la red, y considerando que estamos trabajando con un entorno de tipo C, que ya +usa 24 bits para la red, el total de bits que usaremos para la red y subred +sería $24 + 3 = 27$, que nos da una máscara de 255.255.255.224. \\ +\\ +{\em Número de equipos máximo que se podrían incluir en cada subred.} + +Conociendo que se usan 27 bits para designar la red, es lógico que el número de +bits que se usan para los hosts son $32 - 27 = 5$. Sabiendo esto, y que estamos +trabajando con bits (y por lo tanto en binario), y restando las dos direcciones +especiales, nos sale que cada subred podrá contener un máximo de $2^{5} - 2 = +30$ equipos. \\ +\\ +{\em Dirección de cada subred. Direcciones IP válidas de cada subred y dirección +de difusión.} + +\begin{table}[h!] + \centering + \begin{tabular}{|p{2.5cm}|c|c|c|} + \hline + {\bf Red} & {\bf Dir. Red} & {\bf Dir. Difusión} & {\bf Dir. Válidas} \\ + \hline + Dirección & 173.54.15.0 & 173.54.15.31 & 173.54.15.1 -- 173.54.15.30 \\ + \hline + RR.HH. & 173.54.15.32 & 173.54.15.63 & 173.54.15.33 -- 173.54.15.62 \\ + \hline + Producción & 173.54.15.64 & 173.54.15.95 & 173.54.15.65 -- 173.54.15.94 + \\ \hline + Financiero & 173.54.15.96 & 173.54.15.127 & 173.54.15.97 -- + 173.54.15.126 \\ \hline + Comercial y Marketing & 173.54.15.128 & 173.54.15.159 & 173.54.15.129 -- + 173.54.15.158 \\ \hline + Informática & 173.54.15.160 & 173.54.15.191 & 173.54.15.161 -- + 173.54.15.190 \\ \hline + \end{tabular} +\end{table} + +\section{Ejercicio 7} +\noindent +{\bf Dadas las siguientes direcciones IP, crear el número de subredes que se +indica, detallando en cada caso.} \\ +\\ +{\bf a) Dirección 218.73.15.0 y 6 subredes} + +\begin{itemize} + {\em \item Máscara de red:} + + Conocemos que $\lceil log_{2}(6) \rceil = 3$, por lo tanto son 3 bits de + subred. Se trata de una clase C, donde hay 24 bits para la red; $24 + 3 + = 27$, y por tanto son 27 bits de red y 5 de host: 255.255.255.224. + + {\em \item Número de hosts máximo por subred} + + Conociendo que son 27 bits los que identifican la red, son $32 - 27 = 5$ + los que identifican el host. También tomamos en cuenta que existen dos + direcciones especiales (red y difusión). Por lo tanto puede admitir como + máximo $2^{5} - 2 = 30$ hosts. +\end{itemize} + +\begin{table}[h!] + \centering + \begin{tabular}{|c|c|c|c|} + \hline + {\bf Red} & {\bf Dir. Red} & {\bf Dir. Difusión} & {\bf Dir. Válidas} \\ + \hline + A & 218.73.15.0 & 218.73.15.31 & 218.73.15.1 -- 218.73.15.30 \\ \hline + B & 218.73.15.32 & 218.73.15.63 & 218.73.15.33 -- 218.73.15.62 \\ \hline + C & 218.73.15.64 & 218.73.15.95 & 218.73.15.65 -- 218.73.15.94 \\ \hline + D & 218.73.15.96 & 218.73.15.127 & 218.73.15.97 -- 218.73.15.126 \\ + \hline + E & 218.73.15.128 & 218.73.15.159 & 218.73.15.129 -- 218.73.15.158 \\ + \hline + F & 218.73.15.160 & 218.73.15.191 & 218.73.15.161 -- 218.73.15.190 \\ + \hline + \end{tabular} +\end{table} + +\noindent +{\bf b) Dirección 175.124.0.0 en 8 subredes} + +\begin{itemize} + {\em \item Máscara de red:} + + Conocemos que $\lceil log_{2}(8) \rceil = 3$, por lo tanto son 3 bits de + subred. Se trata de una clase B, donde hay 16 bits para la red; $16 + 3 + = 19$, y por tanto son 19 bits de red: 255.255.224.0. + + {\em \item Número de hosts máximo por subred} + + Conociendo que son 19 bits los que identifican la red, son $32 - 19 = + 13$ los que identifican el host. También tomamos en cuenta que existen + dos direcciones especiales (red y difusión). Por lo tanto puede admitir + como máximo $2^{13} - 2 = 8190$ hosts. +\end{itemize} + +\begin{table}[h!] + \centering + \begin{tabular}{|c|c|c|c|} + \hline + {\bf Red} & {\bf Dir. Red} & {\bf Dir. Difusión} & {\bf Dir. Válidas} \\ + \hline + A & 175.124.0.0 & 175.124.31.255 & 175.124.0.1 -- 175.124.30.255 \\ \hline + B & 175.124.32.0 & 175.124.63.255 & 175.124.32.1 -- 175.124.62.255 \\ \hline + C & 175.124.64.0 & 175.124.95.255 & 175.124.64.1 -- 175.124.94.255 \\ \hline + D & 175.124.96.0 & 175.124.127.255 & 175.124.96.1 -- 175.124.126.255 \\ + \hline + E & 175.124.128.0 & 175.124.159.255 & 175.124.128.1 -- 175.124.158.255 \\ + \hline + F & 175.124.160.0 & 175.124.191.255 & 175.124.160.1 -- 175.124.190.255 + \\ \hline + G & 175.124.192.0 & 175.124.123.255 & 175.124.192.1 -- 175.124.123.254 + \\ \hline + H & 175.124.224.0 & 175.124.255.255 & 175.124.224.1 -- 175.124.255.254 + \\ \hline + \end{tabular} +\end{table} + +\noindent +{\bf c) Dirección 121.67.0.0 en 15 subredes} + +\begin{itemize} + {\em \item Máscara de red:} + + Conocemos que $\lceil log_{2}(15) \rceil = 4$, por lo tanto son 3 bits de + subred. Se trata de una clase B, donde hay 16 bits para la red; $16 + 4 + = 20$, y por tanto son 20 bits de red: 255.255.240.0. + + {\em \item Número de hosts máximo por subred} + + Conociendo que son 19 bits los que identifican la red, son $32 - 20 = + 12$ los que identifican el host. También tomamos en cuenta que existen + dos direcciones especiales (red y difusión). Por lo tanto puede admitir + como máximo $2^{12} - 2 = 4094$ hosts. +\end{itemize} + +\begin{table}[h!] + \centering + \begin{tabular}{|c|c|c|c|} + \hline + {\bf Red} & {\bf Dir. Red} & {\bf Dir. Difusión} & {\bf Dir. Válidas} \\ + \hline + A & 121.67.0.0 & 121.67.15.255 & 121.67.0.1 -- 121.67.15.254 \\ \hline + B & 121.67.16.0 & 121.67.31.255 & 121.67.16.1 -- 121.67.31.254 \\ \hline + C & 121.67.32.0 & 121.67.47.255 & 121.67.32.1 -- 121.67.47.254 \\ \hline + D & 121.67.48.0 & 121.67.63.255 & 121.67.48.1 -- 121.67.63.254 \\ \hline + E & 121.67.64.0 & 121.67.79.255 & 121.67.64.1 -- 121.67.79.254 \\ \hline + F & 121.67.80.0 & 121.67.95.255 & 121.67.80.1 -- 121.67.95.254 \\ \hline + G & 121.67.96.0 & 121.67.111.255 & 121.67.96.1 -- 121.67.111.254 \\ + \hline + H & 121.67.112.0 & 121.67.127.255 & 121.67.112.1 -- 121.67.127.254 \\ + \hline + I & 121.67.128.0 & 121.67.143.255 & 121.67.128.1 -- 121.67.143.254 \\ + \hline + J & 121.67.144.0 & 121.67.159.255 & 121.67.144.1 -- 121.67.159.254 \\ + \hline + K & 121.67.160.0 & 121.67.175.255 & 121.67.160.1 -- 121.67.175.254 \\ + \hline + L & 121.67.176.0 & 121.67.191.255 & 121.67.176.1 -- 121.67.191.254 \\ + \hline + M & 121.67.192.0 & 121.67.207.255 & 121.67.192.1 -- 121.67.207.254 \\ + \hline + N & 121.67.208.0 & 121.67.223.255 & 121.67.208.1 -- 121.67.223.254 \\ + \hline + O & 121.67.224.0 & 121.67.239.255 & 121.67.224.1 -- 121.67.239.254 \\ + \hline + \end{tabular} +\end{table} + +\pagebreak + +\section{Derechos de Autor y Licencia} + +\noindent +Copyright \copyright\ \the\year\ Nicolás A. Ortega Froysa +\\ +Este documento se distribuye bajo los términos y condiciones de la licencia +Creative Commons Attribution No Derivatives 4.0 International. + +\end{document} diff --git a/1/PAR/assignments/t5-examen/t5-examen-nortega5879.toc b/1/PAR/assignments/t5-examen/t5-examen-nortega5879.toc new file mode 100644 index 0000000..2d77d5e --- /dev/null +++ b/1/PAR/assignments/t5-examen/t5-examen-nortega5879.toc @@ -0,0 +1,5 @@ +\babel@toc {spanish}{}\relax +\contentsline {section}{\numberline {1}Ejercicio 5}{3}{section.1}% +\contentsline {section}{\numberline {2}Ejercicio 6}{4}{section.2}% +\contentsline {section}{\numberline {3}Ejercicio 7}{5}{section.3}% +\contentsline {section}{\numberline {4}Derechos de Autor y Licencia}{8}{section.4}%