Class C:-
192.0.0.0 : - Here last 0 is called Group no/Subnet No/Network No
11111111 11111111 11111111 00000000
here network bit is 24
no of host is 2 ka power 8 -2 =256-2=254
1 shows network bit
0 shows host bit
physical address(mac address) = 6 byte(48 bits)
0 is group no & 255 is broadcast no which never assign
Example:- 192.0.0.0
255.255.255.0
Solution :- 256-255=1 is difference
192.0.1.0 is first group
192.0.2.0 is second group and so on till now 255
192.0.255.0 is last group
now come to first group
192.0.1.0
we have discussed that 0 is not assign so that first valid & Last valid IP address
first valid ip is 192.0.1.1 192.0.1.2 upto last valid ip is 192.0.1.254
its broadcast address is 192.0.1.255 same process for every group
Example:- 192.0.0.0
255.255.255.128
Solution :- 255.255.255.128 here 128 is written as 10000000
so that subnet bit is 1 no of host is 7
now 256-128=128 difference
now 256 /128= 2 here 2 group is possible
let go to given ip address 192.0.0.0 to 192.0.0.128 (127 is loop back IP so that leave it) .it is first group
192.0.0.0 to 192.0.0.128 first group
192.0.0.129 to 192.0.0.255 Second Group lets go to first group
192.0.0.0 to 192.0.0.128
192.0.0.1 is first valid IP Address 192.0.0.2 is Second
Last valid IP address is 192.0.0.254 its broadcast IP Address is 192.0.0.128
lets go to Second group
192.0.0.129 to 192.0.0.255
First Valid IP is 192.0.0.130 & Last Valid IP is 192.0.0.254 broadcast IP is 192.0.0.255
Example:- 200.100.50.20
255.255.255.240
Solution :- 240 written as 11110000 so that no of network bit is 2 ka power 4=16 no of host is
2 ka power 4 -2= 14 now
256-240= 16 so that 16 is difference
16/256= 16 so that 16 group
200.100.50.0 to 200.100.50.15 first group
200.100.50.16 to 200.100.50.31 second group till now 16 group
let go to first group 200.100.50.0 to 200.100.50.15
first valid IP is 200.100.50.1 second Valid IP is 200.100.50.2
Last valid Ip is 200.100.20.14 its broadcast ip is 200.100.20.15
lets go to second group
200.100.50.16 to 200.100.50.31
First valid IP is 200.100.50.17
Last Valid IP is 200.100.50.30
Broadcast Ip is 200.100.50.31
Example:- 193.126.11.0
255.255.255.0
Solution:- no of network is 24
no of host is 2 ka power 8 -2 =254
difference is 256-255=1
group 193.126.0.0
193.126.1.0
193.126.2.0 to 193.126.11.0
lets go to 193.126.11.0 group
First valid IP is 193.126.11.1
Last Valid IP is 193.126.11.254
Broad cast IP is 193.126.11.255
lets go to 193.126.0.0 group
First valid IP is 193.126.0.1
Last Valid IP is 193.126.0.254
Broadcast IP is 193.126.0.255
Next IP is 193.126.12.0
Example: 192.0.0.0
255.255.255.224
Solution::- 224 is written as 11100000 so that
n/w bit =24
no of host =2 ka power 5 -2 = 32-2 =30
difference = 256-224=32
no of group = 256/32=8
192.0.0.0 to 192.0.0.31 first group
192.0.0.32 to 192.0.0.63 second group
192.0.0.64 to 192.0.0.95 third group
192.0.0.96 to 192.0.0.127 4th group
192.0.0.128 to 192.0.0.159 5th group
192.0.0.160 to 192.0.0.191 6th group
192.0.0.192 to 192.0.0.223 7th group
192.0.0.224 to 192.0.0.255 8th group
now lets go to first group 192.0.0.0 to 192.0.0.31
First Valid IP is 192.0.0.1 ,second 192.0.0.2
Last Valid IP is 192.0.0.30
Broadcast IP is 192.0.0.31
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