Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 1)
Your router has the following IP address on Ethernet0: Which of the following can be valid host IDs on the LAN interface attached to the router?
1 only
2 and 3 only
3 and 4 only
None of the above
Answer: Option
The router's IP address on the E0 interface is, which is This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.
34 comments Page 2 of 4.

Felix said:   8 years ago
The valid host range will be (172.16,2.1-

The network address :

The broadcast address: 172,16.3.255.

The network will ends then it will become which is still on the range of the network.

Davit said:   8 years ago

IP starts from 2.1 and ends with 3.254. So we can't use these numbers. It means we can use - I think so:x.

Neeraj said:   8 years ago NID. FIRST VALID HOST. LAST VALID HOST. BROAD CAST. it's not a valid address.

Between in these option valid address is

Akira said:   7 years ago
Can you please explain this from the scratch?

Nikhil said:   7 years ago
How did we get the subnet mask as

Nirmal said:   7 years ago

Because 7 bits are opened in the third octet.

Jot Singh said:   7 years ago
Guys What I think is.

So the subnets would be divided by a factor of 2.

Starting from a network address and broadcast address).

As we are considering E0 starting from _._.2.1.

The subnet it comes under is as n/w id) and the flow is as follows: NID. FIRST VALID HOST. *VALID ADDRESS(as it is not a n/w id). LAST VALID HOST. BROAD CAST.

According to which it should be none of these networks.
Hope it will help you.

Akshay said:   7 years ago
How you guys calculate? The router's IP address on the E0 interface is, which is How? Please tell me. I am new in subnets.

Anish said:   7 years ago
23 is a prefix length. So IPV4 has 32 bit of address.

32 - 23 = 9. Now IP address of ethernet 0 is

If we want to find out first address than we have to set 9 (32 - 23 = 9) rightmost bit as zero.

10101100.00111110.00000010.00000000 =

So, the first address is

last address -> put 9 rightmost bit as 1.

10101100.00111110.00000011.11111111 =

So the range = to said:   7 years ago
How can be is a host id? It is a network id.

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