Subnetting Class B Address Problem 1

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)