Must Remind (Keep in Mind)
128 | 1 0 0 0 0 0 0 0 (i.e. 1 Bit ON and 7 OFF)
192 | 1 1 0 0 0 0 0 0 (i.e. 2 Bits ON and 6 OFF)
224 | 1 1 1 0 0 0 0 0 (i.e. 3 Bits ON and 5 OFF)
240 | 1 1 1 1 0 0 0 0 (i.e. 4 Bits ON and 4 OFF)
248 | 1 1 1 1 1 0 0 0 (i.e. 5 Bits ON and 3 OFF)
252 | 1 1 1 1 1 1 0 0 (i.e. 6 Bits ON and 2 OFF)
254 | 1 1 1 1 1 1 1 0 (i.e. 7 Bits ON and 1 OFF)
255 | 1 1 1 1 1 1 1 1 (i.e. 8 Bits ON and 0 OFF)
PROBLEM 1.
172.16.0.0 (Network Address)
255.255.128.0 (/17)
Network = 2^1 = 2
Hosts = 2^15 - 2 = 32766 ( 7 bits in 3rd octet and 8 bits in 4th octet)
valid subnets = 256-128=128.
Block Size = 0 , 128
That subetting is performing in the tird octet , So subnet are really 0.0 and 128.0 These are the exact numbers we used with Class C we used that in the 3rd Octet then add 0 in the 4th Octet for the network address. (MUST READ)
Network1
172.16.0.0 (Network ID)
172.16.0.1
172.16.0.2
172.16.0.3
.
.
.
172.16.127.252
172.16.127.253
172.16.127.254
172.16.127.255 (Broadcast ID)
when the address 0.255 after that 3rd octet will be started from 1.0 then 1.1, 1.2, 1.3 similarly address reach 1.255 3rd octet will start From 2.0 Similarly all the address will allocate in a same way. (MUST READ)
NETWORK 2
172.16.128.0 (Network ID)
172.16.128.1
172.16.128.2
172.16.128.3
.
.
.
172.16.255.252
172.16.255.253
172.16.255.254
172.16.255.255 (Broadcast ID)
