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 2 of 4.
Chandni said:
7 years ago
Thanks for sharing this Subnetting problem, I was looking for idea for solving these kind of problems. Finally I got it.
(1)
Basir said:
8 years ago
I'm not getting this. Please anyone help it to me.
(1)
Ansar said:
8 years ago
According to me, the answer will be 172.16.2.255.
(1)
Neeraj said:
9 years ago
172.16.2.0 NID.
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.1.
172.16.3.255.
172.16.4.1.
172.16.4.254 LAST VALID HOST.
172.16.4.255 BROAD CAST.
172.16.3.0 it's not a valid address.
Between in these option valid address is 172.16.2.255
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.1.
172.16.3.255.
172.16.4.1.
172.16.4.254 LAST VALID HOST.
172.16.4.255 BROAD CAST.
172.16.3.0 it's not a valid address.
Between in these option valid address is 172.16.2.255
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 ?
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.
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
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?
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.
Aishwarya mishra said:
1 decade ago
The valid host id is from 172.16.2.0 to 172.16.3.254 [excluding 172.16.2.1(as this is itself routers id)].
And the broadcast add is 172.16.3.255 [here we are making all the last nine bits as 1.].
And the broadcast add is 172.16.3.255 [here we are making all the last nine bits as 1.].
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers