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?
- 172.16.1.100
- 172.16.1.198
- 172.16.2.255
- 172.16.3.0
Answer: Option
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:
36 comments Page 4 of 4.
Aswathy said:
1 decade ago
I think the valid host IDs are between 172.16.0.1 to 172.16.1.254.
And broadcast address is 172.16.1.255.
And broadcast address is 172.16.1.255.
RAJESH BADARIYA said:
1 decade ago
But as we always say that first and last ip is for networkk and host id than here why we are not considering 172.16.2.255 and 172.16.3.0.
Please tell since this network is divided into two bock then how?
Please tell since this network is divided into two bock then how?
Adhie said:
1 decade ago
the option C is broadcast address and the others is host address.
that given address is class B (from 1st octet 172).
So here is sub network addr:
172.16.0.0
172.16.2.0
...
172.16.254.0
that given address is class B (from 1st octet 172).
So here is sub network addr:
172.16.0.0
172.16.2.0
...
172.16.254.0
Vikash_jaiswal said:
1 decade ago
As given, router's IP address as 172.16.2.1/23 ;
it means last 9 bits (32-23) of binary form of 172.16.2.1 are responsible for deciding host ID.
When you make those last 9 bits 0 then u will get address as
172.16.2.0 and when you make those last 9 bits as 1 then you will get address as 172.16.3.255 ;
But 172.16.2.0 is used for network-ID (as 0 in last) and
172.16.3.255 is used for host-ID (as 255 in last).
So, rest of IPs will be vallid IPs for hosts.
it means last 9 bits (32-23) of binary form of 172.16.2.1 are responsible for deciding host ID.
When you make those last 9 bits 0 then u will get address as
172.16.2.0 and when you make those last 9 bits as 1 then you will get address as 172.16.3.255 ;
But 172.16.2.0 is used for network-ID (as 0 in last) and
172.16.3.255 is used for host-ID (as 255 in last).
So, rest of IPs will be vallid IPs for hosts.
Muhammad Rahman said:
1 decade ago
If the range is like
.
.
172.16.2.253
172.16.2.254
172.16.2.255
172.16.3.0
172.16.3.1
172.16.3.2
.
.
How come 172.16.3.0 is not a valid address ?
.
.
172.16.2.253
172.16.2.254
172.16.2.255
172.16.3.0
172.16.3.1
172.16.3.2
.
.
How come 172.16.3.0 is not a valid address ?
Salim said:
1 decade ago
How many network will divide by this subnet mask ?
How you know the number of host to each subnet will be from 2.1 to 3.254 ?
How you know the number of host to each subnet will be from 2.1 to 3.254 ?
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