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:
34 comments Page 3 of 4.
Akira said:
7 years ago
Can you please explain this from the scratch?
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
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:
8 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.
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.
Felix said:
8 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.
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.
Gabe said:
8 years ago
Don't know if it helps but 172.16.1.100 and 172.16.1.198 are kinda off from the default gateway 172.16.2.1/23 notice that to there?
Varun tyagi said:
8 years ago
2 is the block size.
So 0, 2, 4, 6 is N.ID.
0.1, 0.2, 0.3 to 1.254 are host.
Similarly.
2.1, 2.2, 2.3, 2.4 to 2.254, 2.255, 3.1, 3.2 to 3.254 are host.
But 3.255 are B.id and 4.0 n.id.
So 0, 2, 4, 6 is N.ID.
0.1, 0.2, 0.3 to 1.254 are host.
Similarly.
2.1, 2.2, 2.3, 2.4 to 2.254, 2.255, 3.1, 3.2 to 3.254 are host.
But 3.255 are B.id and 4.0 n.id.
Santhosh said:
9 years 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.
And the broadcast add is 172.16.3.255.
Aishwarya mishra said:
9 years 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.].
Aswathy said:
9 years 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:
10 years 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?
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