CCNA BASICS CHAPTER 4 ¾: IP ADDRESSING
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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