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 3 of 4.
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 ?
Gabe said:
10 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?
Tonychrys said:
10 months ago
Well done for the detailed explanation @Waseem.
From h= 32- 23 = 9.
Here, How did you get that term 32? Please explain to me.
From h= 32- 23 = 9.
Here, How did you get that term 32? Please explain to me.
(1)
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 ?
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)
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.
Shipukangra@gmail.com said:
9 years ago
How can be 172.168.3.0 is a host id? It is a network id.
Nirmal said:
9 years ago
@Nikhil.
Because 7 bits are opened in the third octet.
Because 7 bits are opened in the third octet.
Malede said:
4 years ago
It is very helpful. Thanks everyone for explaining.
(3)
Basir said:
8 years ago
I'm not getting this. Please anyone help it to me.
(1)
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