Wednesday, May 13, 2020

Subnetting Of University

Four Block:
256   128     64    32    16        18     4      2    1
Original Network : 192.32.16.0
A Block=136 Hosts
B Block=95 Hosts
C Block=51 Hosts
D Block=24 Hosts
For Block A:
For Block B:
28=256
27=128
Total IP Bits: 32
Total IP Bits: 32
So 32-8=24
So 32-7=25
IP= 192.32.16.0
IP= 192.32.17.0
Mask=255.255.255.0
Mask=255.255.255.128
  First Host Address: 192.32.16.1
  First Host Address: 192.32.17.1
Last Host Address: 192.32.16.254
Last Host Address: 192.32.17.126
Broadcast Address: 192.32.16.255
Broadcast Address: 192.32.17.127
For Block C:
For Block D:
26 =64
25=32
 Total IP Bits: 32
 Total IP Bits:32
So 32-6=26
So, 32-5=27
192.32.16.0/26
192.32.16.0/27
IP= 192.32.17.128
IP= 192.32.17.192
Mask=255.255.255.192
Mask=255.255.255.224
First Host Address: 192.32.17.129
 First Host Address: 192.32.17.193
Last Host Address: 192.32.17.190
 Last Host Address: 192.32.17.222
Broadcast Address: 192.32.17.191
 Broadcast Address:192.32.17.223














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