# Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 1)
1.
Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?
1. 172.16.1.100
2. 172.16.1.198
3. 172.16.2.255
4. 172.16.3.0
1 only
2 and 3 only
3 and 4 only
None of the above
Explanation:
The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. 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.
Discussion:
34 comments Page 2 of 4.

Kapil said:   7 years ago
Can any one help me to understand subnetting?

Sundha said:   7 years ago
How it is. Can you explain me clearly?

Datta Ikhe said:   7 years ago

Given IP- 172.16.2.1/23.
First IP used as Network Id: 172.16.00000010.00000000(172.16.2.0).
First Valid Host IP: 172.16.00000011.00000000(172.16.3.0).
Broadcast IP: 172.16.3.255 & NOT 172.16.2.255.

So, 172.16.2.255 is a valid Host IP.

Shahnas said:   7 years ago
What means of /23?

Can we take it as subnet mask? and why we want to calculate 32-23=9?

Guys! can you tell me the in IP address and subnet mask of this?

Shipukangra@gmail.com said:   7 years ago
How can be 172.168.3.0 is a host id? It is a network id.

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 172.62.2.1/23.

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 = 172.62.2.1.

So, the first address is 172.62.2.1.

last address -> put 9 rightmost bit as 1.

10101100.00111110.00000011.11111111 = 172.62.3.255.

So the range = 172.62.2.1 to 172.62.3.255.

Akshay said:   7 years ago
How you guys calculate? The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. How? Please tell me. I am new in subnets.

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

Block size is 2(256-254:FROM LAST NON-ZERO OCTET IN SUBNET MASK).
So the subnets would be divided by a factor of 2.

Starting from

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

The subnet it comes under is as 127.16.2.0(being n/w id) and the flow is as follows:
172.16.2.0 NID.
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.0 *VALID ADDRESS(as it is not a n/w id).
172.16.3.1.
172.16.3.254 LAST VALID HOST.

According to which it should be none of these networks.