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
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:
34 comments Page 1 of 4.

Waseem said:   6 years ago
(Q).Given ip 172.16.2.1/23

(A). by default the given IP is class B(N.N.H.H), slash notation for Class B is /16 but from given it is /23, so the converted network bits are 7 (23-16).
formula for subnets is 2^n (note:- it indicates how many parts we need to divide a N/w)

Here n represents no. of converted N/w bits
n=7
from formula 2^7=128
Formula for Hosts is 2^h-2 (note:- it indicates how many IP's consists in a N/w)

Here h represents no. of host bits
-2 represents (one is N/W IP and another one is Broadcast IP, we can't use
these 2 IP's these are default)
h=32-23
=9
from formula 2^h
2^9=512
The given IP is defaultly class B(N.N.H.H) we should vary Host bits only,N/W bits are constant in every N/W.

So the range starts from 172.16.0.0.
As we got hosts from hosts formula we should divide our N/W with 512 hosts
so the range is as bellow
(1st subnet). 172.16.0.0 to 172.16.1.255
(2nd subnet). 172.16.2.0 to 172.16.3.255
(3rd subnet). 172.16.4.0 to 172.16.5.255

From given

The given ip 172.16.2.1 belongs to 2nd subnet range, we should neglect 1st and last IP's in this range those are 172.16.2.0 and 172.16.3.255(N/W ID and Broad cast ID).

Finally the valid IP's are 172.16.2.1 to 172.16.3.254
From given option's only option C belongs to above range.. So answer is 'C'.
(15)

Jot Singh said:   8 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
172.16.0.0-172.16.1.255(including a network address and broadcast address).

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.
172.16.3.255. BROAD CAST.

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

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.

Anish said:   8 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.

Babu said:   4 years ago
@All.

We have to understand host range is 1 to 254,0 and 255 is network & broadcast ID REMEMBER Only first and last bits Network & Broadcast ID.

So 172.16.2.1 to 172.16.2.255 then next change 172.16.3.0 to 172.16.3.255 Because we have 2^9-2=510 hosts.

That's why 3.254 is valid one So 2.55 is not Broadcast ID.

I hope you will understand.
(9)

Datta Ikhe said:   7 years ago
Answer C is correct because;

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.

Felix said:   9 years ago
The valid host range will be (172.16,2.1-172.16.3.254).

The network address : 172.16.2.0.

The broadcast address: 172,16.3.255.

The network 172.16.2.0 will ends 172.16.2.55 then it will become 172.16.3.0 which is still on the range of the network.

Neeraj said:   8 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

Davit said:   9 years ago
00000000.00000000.0000000|0.00000000.
76543210.76543210.7654321|0.76543210.
1.255.

IP starts from 2.1 and ends with 3.254. So we can't use these numbers. It means we can use 172.16.2.1 - 172.16.3.254. I think so:x.

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?


Post your comments here:

Your comments will be displayed after verification.