COURSE



CCNA Syllabus 



ABOUT THE COURSE

It is difficult to configure a Router or a Switch or any device for a newbie in the networking industry, even experts face problems while configuring complex networks. Some people have a lack of knowledge some have limited devices and for some explanation are not clear. I will try to explain from scratch is this e-book. I solved vlsm on my client requirements so you will know how to deal with your clients. You can design a network for your company as well as for your clients. This e-book shows you a step by step configuration.





Table of Contents




26<= 2n – 2

26<= 25 – 2

26<= 32 – 2

26<= 30

Total number of network bits= 32

32 – n = 32 – 5 = 27 (CIDR value of Router 1 i.e. Given network)

i.e. 195.100.0.0/27

N.S.M = 11111111. 11111111. 11111111.11100000

                        = 255.255.255.224

Block Size        = 256 – N.S.M

                        = 256 – 224 = 32

Note: Add this block size in given network so that we can get our new network for Router 2.

so the new network is 192.100.0.32/?

Now find FVH, LVH and Broadcast address

Given Network19510000
First Valid Host-FVH19510001
Last Valid Host-LVH195100030
Broadcast195100031
New Network195100032



R2<= 14 hosts

14<= 2n – 2

14<= 24 – 2

14<= 16 – 2

14<= 14

Total number of network bits = 32

32 – n = 32 -4 = 28 (CIDR value of new network i.e. Router 2)

i.e. 195.100.0.32/28

N.S.M = 11111111. 11111111. 11111111.11110000

                        = 255.255.255.240

Block Size        = 256 – N.S.M

                        = 256 – 240 = 16

Note: Add this block size in given network so that we can get our new network for Router 1 and 2 (Virtual Network).

so the new network is 192.100.0.48/?

Now find FVH, LVH and Broadcast address

Given Network195100032
First Valid Host-FVH19510001
Last Valid Host-LVH195100046
Broadcast195100047
New Network195100048


Router 1 & 2 <= 2 hosts (we required only two IP to connect 2 routers)

2 <= 2n – 2

2 <= 22 – 2

2 <= 4 – 2

2 <= 2

Total number of network bits = 32

32 – n = 32 – 2 = 30 (CIDR of Virtual network (R1&R2)

i.e. 195.100.0.48/30

N.S.M = 11111111.11111111.11111111.11111100

= 255.255.255.252

Block Size        = 256 – N.S.M

= 256 – 252 = 4

Note: Add this block size in given network so that we can get our new network for Router 3. This network can be used in future.

so the new network is 192.100.0.52 /?

Now find FVH, LVH and Broadcast address

Given Network195100048
First Valid Host-FVH195100049
Last Valid Host-LVH195100050
Broadcast195100051
New Network195100052


Basic Configuration on Router 1


Router>enable
Router#configure terminal
Router(config)#hostname R1
R1(config)#interface fastEthernet 0/0
R1(config-if)#ip address 195.100.0.1  255.255.255.224
R1(config-if)#no shutdown
R1(config-if)#exit  

R1(config)#interface serial 0/0/0
R1(config-if)#ip address 195.100.0.49 255.255.255.252
R1(config-if)#clock rate 64000
R1(config-if)#no shutdown
R1(config-if)#exit
R1(config)#exit
R1#write


Basic Configuration on Router 2

 enable

Router#configure terminal
Router(config)#hostname R2
R2(config)#interface fastEthernet 0/0
R2(config-if)#ip address 195.100.0.33 255.255.255.240
R2(config-if)#no shutdown
R2(config-if)#exit  

R2(config)#interface serial 0/1/0
R2(config-if)#ip address 195.100.0.50 255.255.255.252
R2(config-if)#no shutdown
R2(config-if)#exit
R2(config)#exit
R2#write



Verification on Router 1 & 2

R1#show ip route     
195.100.0.0/24 is variably subnetted, 2 subnets, 2 masks
C       195.100.0.0/27 is directly connected, FastEthernet0/0
C       195.100.0.48/30 is directly connected, Serial0/0/0

R2#show ip route     
195.100.0.0/24 is variably subnetted, 2 subnets, 2 masks
C       195.100.0.32/28 is directly connected, FastEthernet0/0
C       195.100.0.48/30 is directly connected, Serial0/1/0