CCNA BASICS CHAPTER 4 ¾: IP ADDRESSING. CCNA BASICS SOME BASIC BASIS BASE 2 BASE 16 BASE 10.

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CCNA BASICS CHAPTER 4 ¾: IP ADDRESSING

Transcript of CCNA BASICS CHAPTER 4 ¾: IP ADDRESSING. CCNA BASICS SOME BASIC BASIS BASE 2 BASE 16 BASE 10.

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CCNA BASICS

CHAPTER 4 ¾:

IP ADDRESSING

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SOME BASIC BASIS

BASE 2

BASE 16

BASE 10

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IP ADDRESSING

- USED IN TCP/IP ENVIRONMETS

- IP ADDRESS IS UNIQUE

- A WAY TO LOCATE COMPUTERS WORLD WIDE

- SEGMENT ORIENTED

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RouterRouter

Portion of Routing Table192.168.1.0 255.255.255.0 192.168.1.1192.168.2.0 255.255.255.0 192.168.2.1192.168.3.0 255.255.255.0 192.168.3.1192.168.4.0 255.255.255.0 192.168.4.1192.168.5.0 255.255.255.0 192.168.5.1192.168.6.0 255.255.255.0 192.168.6.1192.168.7.0 255.255.255.0 192.168.7.1192.168.8.0 255.255.255.0 192.168.8.1

ROUTING

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Is destination local?Yes, add the destination MAC addressNo, add the Router’s MAC address

Always add the destination’s IP address

Is destination local?Yes, add the destination MAC addressNo, add the Router’s MAC address

Always add the destination’s IP address

Verify packetDecrease TTLIs destination local?

Yes, add the destination MAC addressNo, add another Router’s MAC address

Verify packetDecrease TTLIs destination local?

Yes, add the destination MAC addressNo, add another Router’s MAC address

Verify packetVerify IP addressSend the packet up to the next layer

Verify packetVerify IP addressSend the packet up to the next layer

Is destination local?Yes, add the destination MAC addressNo, add the router’s MAC address

Always add the destination’s IP address

Is destination local?Yes, add the destination MAC addressNo, add the router’s MAC address

Always add the destination’s IP address

Verify packet

Verify IP address

Send the packet up to the next layer

Verify packet

Verify IP address

Send the packet up to the next layer

Verify packetDecrease TTLIs destination local?

Yes, add the destination MAC addressNo, add another router’s MAC address

Router 2Router 2

Router 1Router 1

AA

BB

CC

DD

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192.168.1.0192.168.1.0 192.168.3.0192.168.3.0

192.168.1.100192.168.1.100 192.168.2.101192.168.2.101

192.168.2.100192.168.2.100 192.168.3.100192.168.3.100

192.168.2.0192.168.2.0

192.168.1.100192.168.1.100192.168.1.100192.168.1.100IPAddress

Host IDHost ID

Network IDNetwork ID

IP ADDRESSING

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NET ID

HOST ID

32 BITS IN DOTTED DECIMAL NOTATION

MASK

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ALL COMES OUT FROM THIS:

CLASS A: IP BEGINS WITH 0

CLASS B: IP BEGINS WITH 10

CLASS C: IP BEGINS WITH 110

CLASS D: IP BEGINS WITH 1110

CLASS E: IP BEGINS WITH 1111

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ww xx yy zz

Class A Network IDNetwork IDNetwork IDNetwork ID Host IDHost ID

Class B Network IDNetwork IDNetwork IDNetwork ID Host IDHost ID

Class C Network IDNetwork IDNetwork IDNetwork ID Host IDHost ID

IP CLASSES

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ww xx yy zz

Class A 255255255255 0 0 0 0 0 0

Class B 255 255255 255 255 255255 255 0 0 0 0

Class C 255 255 255255 255 255 255 255 255255 255 255 00

DEFAULT SUBNET MASKS

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GUIDELINES:

NETWORK ID CANNOT BE 127

NODES IN NET ID OR HOST ID CANNOT BE 255 (ALL 1´S)

NODES IN NET ID OR HOST ID CANNOT BE 0

HOST ID MUST BE UNIQUE TO THE NETWORK

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PRACTICE EXAM

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BINARY DECIMAL OCTAL HEXADECIMAL11101

17875

F23A345

110111001010101

10111101111101

574222

623647

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IP ADDRESS CLASS SUBNET MASK NETWORK ID HOST ID129.23.123.9245.5.9.123190.57.23.871.1.76.180.12.5.7224.234.210.78130.65.17.7490.34.90.3412.6.34.9120.128.129.130128.120.123.130254.8.8.8193.123.43.89

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CLASS N. OF NETS N. OF HOSTS RANGE OF NET ID´S FIRST 4 BITSABCDE

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CAN THIS BE IP´S ON TWO HOSTS ON THE SAME NET?

172.16.20.89172.16.89.20

10.23.30.12910.98.134.43

198.45.60.34198.45.61.35

200.200.98.3200.200.56.3

125.34.12.8125.215.34.6

127.0.0.1127.0.0.2

236.45.45.45236.45.45.45

120.23.56.12120.24.56.13

241.19.23.67241.19.23.100

176.99.123.0176.99.0.234

199.12.165.0199.12.165.1

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CHAPTER 4 5/6

SUBNETTING

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SUBNET MASK CODE

1S = Positions Rrepresenting Network or Subnet addresses

0S = Positions Representing host address

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Binary numbering convention

Position / value 128 64 32 16 8 4 2 1

Binary example:

Decimal Equivalent

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#1 1 X 2 2 02 2 X 2 4 23 4 X 2 8 64 8 X 2 16 145 16 X 2 32 306 32 X 2 64 627 64 X 2 128 1268 128 X 2 256 2549 256 X 2 512 510

10 512 X 2 1024 102211 1024 X 2 2048 204612 2048 X 2 4096 409413 4096 X 2 8192 819014 8192 X 2 16384 1638215 16384 X 2 32768 32766

The “bits to play with table”

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PRIMER OCTETO

SEGUNDO OCTETO

TERCER OCTETO

CUARTO OCTETO

IP HOST BINARIO 10101100 00010000 00000010 10100000MASK BINARIO 11111111 11111111 11111111 11000000SUBNET 10101100 00010000 00000010 10000000BROADCAST 10101100 00010000 00000010 10111111PRIMERO 10101100 00010000 00000010 10000001ULTIMO 10101100 00010000 00000010 10111110

1. CONVIERTA EL IP EN CUESTION A BINARIO ASI COMO LA MASCARA DE RED.

2. DIBUJE UNA LINEA IMAGINARIA EN EL ULTIMO 1 DE LA MASCARA DE RED DONDE TERMINAN 1 CONSECUTIVOS

3.COPIE EL IP HOST A SUBNET Y LLENE DE CEROS A PARTIR DE LA LINEA IMAGINARIA Y RELLENO DE 0 EL RESULTADO SERA SU SUBNET

4. EL SUBNET SE COPIA A BROADCAST Y RELLENA A PARTIR DE LA LINEA IMAGINARIA POR PUROS UNOS PARA GENERAR EL ULTIMO RANGO. BRODCAST

5. COPIA EL SUBNET A PRIMERO Y ADICIONA 1 AL ULTIMO BIT Y OBTIENE EL PRIMER ASIGNABLE.

6. COPIA EL BROADCAST Y LE RESTAS 1 O SEA LE PONES 0 AL ULTIMO BIT.

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0163248648096

112128144160176192208224240256

32 by 32 or 16 by 16 to find out ranges

0326496

128160192224256

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NO EXAM THIS TIME...

(THIS TIME!!!)

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NO EXAM ...BUT SOME EXERCISES!!

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WHICH OF THE FOLLOWING ARE HOST ASIGNABLE IP´S?

192.168.25.18/25

143.200.200.240/29

100.23.15.183/29

15.18.236.0/22

80.236.15.119/11

250.18.43.20/23

1.252.13.26/13

15.19.48.30/30

25.25.19.4/12

129.127.0.115/25

127.213.46.52/18

200.136.43.80/25

119.20.49.81/25

200.39.93.21/27

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170.16.224.18/20 170.16.224.112/22 170.16.231.18/20

170.16.231.39/21 170.16.247.99/19 170.16.234.19/21

170.16.232.39/21 170.16.235.39/21 170.16.230.39/21

170.16.230.29/21QUIEN SE VE?CAMBIA MASCARAS.

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170.16.177.50/23 170.16.177.235/27 170.16.177.180/23

170.16.177.231/29 170.16.177.75/22 170.16.183.49/21

170.16.176.43/21 170.16.183.15/22 170.16.177.199/21

170.16.177.3/22 170.16.180.90/22 170.16.180.44/20

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Asignar los 15 primeros hosts asignables a la red 195.28.99.0/27

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Asignar la mascara a la red 150.120.0.0 para tener 70 subredes o más con el mayor numero de hosts por subred.

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Asignar la mascara a la red 128.120.0.0 para tener al menos 400 hosts por subred y el maximo de subredes posibles

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Determinar el net id, el numero de hosts, el numero de subredes, broadcast y rango de ip’s validas alrededor de las siguientes ip’s

129.15.18.39/26

200.29.15.42/29

36.15.18.143/22

195.48.52.18/27

199.15.15.15/25

203.43.18.69/26

240.43.45.9/28

139.200.195.139/28

99.99.192.168/28

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Asignar ip’s a los hosts de la red 192.168.252.0 en el menor rango posible considerando el numero de maquinas con que se cuenta y que se prevee un crecimiento en este numero de hasta un 11% anual en los siguientes dos años. Se desea que las computadoras de cada area se vean entre si pero no se vean entre areas diferentes.

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Cuantos routers hay entre los siguientes servidores, considerando que estos pueden compartir informacion y que sus ip’s son los siguientes:

119.12.200.4/18

119.12.208.14/18

119.15.200.8/16

119.15.200.7/16

119.12.200.6/19

119.12.200.16/19

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Una maquina cuyo net id según su clase es 192.192.3.0 se encuentra en la sexta subred

asignable según su ip y submascara actuales. Al cambiar su mascara a 255.255.255.224, esta

maquina sera el decimo host asignable en su nuevo segmento, ( al mantener su actual ip). Cual es el ip

y la mascara antes del cambio de esta maquina?

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Inventa una dirección, la cual con cierta mascara se encuentre en la

duodécima subred válida y con otra mascara sea el trigesimosexto host

válido de su segmento