CCNA 4.14 TP Correction

All people seem to need data processing

( Application presentation session transport network data link physic )

host : end device ( PC Server )

switch : to connect many device

router : to connect two or more different networks

internet cable : straight cable and crossover cable

The straight through cable is used to connect:

 

• switch to router
• switch to pc/server

 

 

The crossover cable is used to connect:

 

• switch to switch
• router to router
• router to pc/server
• pc/server to pc/server

 

192.168.0.0 /16 ( 16个1 )

11111111.11111111.00000000.00000000

255.255.0.0

It means that the first address of this network ( the network address ) is : 192.128.00000000.00000000 => 192.168.0.0

The last address of this network ( the broadcast address ) is 192.16811111111.11111111 => 192.168.255.255

And the range of IP address for the host is from 192.168.00000000.00000001 => 192.168.0.1 to 192.168.11111111.11111110 => 192.168.255.254

Let’s devide this network into smaller subnetworks.

Always start by the biggest subnetwork which contains the biggest number of hosts..

Office subnetwork : 511 hosts

So we need 511 IP adderss for the hosts ( end device ) + 1 IP address for the network address +1IP address as a broadcast address = 511 + 1 + 1 = 513 IP addresses

2^9=512 <range for 513 IP addresses < 2^10=1024

1024 in binary is : 0000 0100 0000 0000

So for this subnetwork, we will use the following subnet mask : 11111111.11111111.11111100.00000000 => 255.255.252.0 or /22

So to create the Office subnetwork as a part os the big 192.168.0.0/16 network, we will start at the beginning os this big network.

So the network address of our Office subnetwork will be the first address of the big network : 192.168.0.0/22 ( the host part of the subnet mask is 00.00000000 )

The broadcast address will be : 192.168.3.255/22 ( The host part of the subnet mask is 11.11111111 )

So the range of IP addresses for the hosts will be : from 192.168.0.1/22 to 192.168.3.254/22

we will also book the last host IP address for the gateway ( =router ),so the gateway address of this  subnetwork is ; 192.168.3.254

Then, for the factory subnetwork : 

128(2^7) < 250 IPs =1=1 = 252 < 256(2^8)

So the subnet mask for this subnetwork is : 11111111.11111111.11111111.00000000 => 255.255.255.0 or /24

network address : 192.168.4.0/24 ( this is the next available address of the big network )

broadcast address : 192.168.4.255/24

IP range for hosts : 192.168.4.1/24 to 192.268.4.254/24

gateway is : 192.168.4.154/24

For the small subnetwork OfficeRTR < - > EdgeRTR :  

This is a PtP ( Point to Point ) connection,that means we connect only two devices together.So we need 2+1+1 = 4 IP addresses in this network

subnet mask : 11111111.11111111.11111111.11111100 => 255.255.252 or 30

network address : 192.168.5.4/30 ( this is the next available address of the big network )

 

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enable

configure terminal

interface gigabitEthernet

ip adress …(last IP address)

no shutdown

exit 

in

自动填充命令语句

重现之前用过的语句

Configure R1

 

en
conf t
hostname R1
no ip domain-lookup
enable secret class
banner motd #Do not touch my router#
line con 0
password cisco
exit
line vty 0 4
password cisco
exit
interface fastEthernet 0/0
ip address 192.168.1.1 255.255.255.0
no shutdown
exit
interface serial 0/0/0
ip address 192.168.2.1 255.255.255.0
clock rate 64000
exit
end
copy runnig-config startup-config

 

 

 

Configure R2 :

 

en
conf t
hostname R2
no ip domain-lookup
enable secret class
banner motd #Do not touch my router#
line con 0
password cisco
exit
line vty 0 4
password cisco
exit
interface fastEthernet 0/0
ip address 192.168.3.1 255.255.255.0
no shutdown
exit
interface serial 0/0/0
ip address 192.168.2.2 255.255.255.0
clock rate 64000
exit
end
copy runnig-config startup-config