Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 13)
13.
You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
Answer: Option
Explanation:
A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.
Discussion:
13 comments Page 2 of 2.
Haya said:
9 years ago
Why option A is not correct?
It is also a valid host. Because range is in between 172.16.16.1 to 172.16.19.254.
It is also a valid host. Because range is in between 172.16.16.1 to 172.16.19.254.
Pruthvi said:
1 decade ago
11111111.11111111.1 1 1 1 1 {1} 0 0.00000000/22
128 64 32 16 8 {4} 2 1
Here the host and network parts are separated at 4. So the block size is 4. This is a simple trick.
128 64 32 16 8 {4} 2 1
Here the host and network parts are separated at 4. So the block size is 4. This is a simple trick.
Mtk said:
1 decade ago
I couldn't understand how did the /22 mask could have block size of 4?
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