What is the last valid host on the subnetwork 192.168.153.228/30? explain

#1

What is the last valid host on the subnetwork 192.168.153.228/30? explain

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#2

last valid host on the sub-network 192.168.153.228/30 will be
192.168.153.230/30
as we know that it is a Class C IP so by default its network bit is /16
now during sub-netting we took 6 more bit as a network bit.
so if we convert 192.168.153.228 into binary it will be
11000000.10101000.10011001.11101000
so last two bit of the last octet is host bit 11101000
so in this last two bit we can change
and how many ip will we get?
ans: 2^(Host Bit)=2^H=2^2=4
so from above formula we can conclude that we can get 4 IP address
but as we know the first and last IP is reserved for network representation and broadcast address consecutively. so we subtract 2 from total IP block
so in this case (4-2) =2
so only 2 IP is usable in host.

  1. 192.168.153.229 (for host)

  2. 192.168.153.230 (for host)

              &
    

The reserved IP are

  1. 192.168.153.228 (represent network)
  2. 192.168.153.231 (Broadcast IP for this block)

so i think i tried my best to make you understand. so if you find any further problem please feel free to ask. thank you.

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