Basic Router Configuration On Cisco Packet Tracer

In this lab, I am gonna teach you Basic Router Configuration On Cisco Packet Tracer

Basic Router Configuration VLSM

Given network: 200.150.100.0

Required hosts on Router 1: 14

Required hosts on Router 2: 6

R1<= 14

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 given network i.e. Router 1)

i.e. 200.150.100.0/28

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 2.

So the new network is 200.150.100.16/?

Now find FVH, LVH, and broadcast address

Given Network	200	150	100	0
First Valid Host-FVH	200	150	100	1
Last Valid Host-LVH	200	150	100	14
Broadcast	200	150	100	15
New Network	200	150	100	16

R2<= 6

6<= 2ⁿ – 2

6<= 2³ – 2

6<= 8 – 2

6<= 6

Total number of network bits = 32

32 – n = 32 – 3 = 29 (CIDR value of new network i.e. Router 2)

i.e. 200.150.100.16/29

N.S.M = 11111111. 11111111. 11111111.11111000

                        = 255.255.255.248

Block size        = 256 – N.S.M

                        = 256 – 248 = 8

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 200.150.100.24/?

Now find FVH, LVH, and broadcast address

Given Network	200	150	100	16
First Valid Host-FVH	200	150	100	17
Last Valid Host-LVH	200	150	100	22
Broadcast	200	150	100	23
New Network	200	150	100	24

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. 200.150.100.24/30

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 200.150.100.28/?

Now find FVH, LVH and Broadcast address

Given Network	200	150	100	24
First Valid Host-FVH	200	150	100	25
Last Valid Host-LVH	200	150	100	26
Broadcast	200	150	100	27
New Network	200	150	100	28

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Basic Router Configuration on Router 1

Router>enable Router#configure terminal Router(config)#hostname R1 R1(config)#interface fastEthernet 0/0 R1(config-if)#ip address 200.150.100.1         255.255.255.240 R1(config-if)#no shutdown R1(config-if)#exit   R1(config)#interface serial 0/0/0 R1(config-if)#ip address 200.150.100.25       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 Router Configuration on Router 2

Router>enable Router#configure terminal Router(config)#hostname R2 R2(config)#interface fastEthernet 0/0 R2(config-if)#ip address 200.150.100.17       255.255.255.248 R2(config-if)#no shutdown R2(config-if)#exit   R2(config)#interface serial 0/1/0 R2(config-if)#ip address 200.150.100.26       255.255.255.252 R2(config-if)#no shutdown R2(config-if)#exit R2(config)#exit R2#write

Verification on Router 1

R1#show ip route      200.150.100.0/24 is variably subnetted, 2 subnets, 2 masks C       200.150.100.0/28 is directly connected, FastEthernet0/0 C       200.150.100.24/30 is directly connected, Serial0/0/0

Verification on Router 2

R2#show ip route      200.150.100.0/24 is variably subnetted, 2 subnets, 2 masks C       200.150.100.16/29 is directly connected, FastEthernet0/0 C       200.150.100.24/30 is directly connected, Serial0/1/0

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