Subnetting Question confusion
Humper
Member Posts: 647
in CCNA & CCENT
I can do the subnetting questions out of Todd Lamles CCNA self-study guide fine...But I went to the subnettingquestions.com website and got this question.
Question: What is the first valid host on the subnetwork that the node 172.28.254.135 255.255.252.0 belongs to?
Answer: 172.28.252.1
I realize its a /22 with 6 subnet bits and 10host bits -- theres 64 subnets and 1024 hosts per subnet? Is that right? or is it total hosts?
I get confused by this question because I guess I can't go backwards. How do I find the first valid host from the information given to me in this question? I think I am missing something really simple, any help would be appreciated! Thanks
Question: What is the first valid host on the subnetwork that the node 172.28.254.135 255.255.252.0 belongs to?
Answer: 172.28.252.1
I realize its a /22 with 6 subnet bits and 10host bits -- theres 64 subnets and 1024 hosts per subnet? Is that right? or is it total hosts?
I get confused by this question because I guess I can't go backwards. How do I find the first valid host from the information given to me in this question? I think I am missing something really simple, any help would be appreciated! Thanks
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Comments
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gojericho0 Member Posts: 1,059 ■■■□□□□□□□Question: What is the first valid host on the subnetwork that the node 172.28.254.135 255.255.252.0 belongs to?
Answer: 172.28.252.1
I realize its a /22 with 6 subnet bits and 10host bits -- theres 64 subnets and 1024 hosts per subnet? Is that right? or is it total hosts?
172.28.254.135 /22 is a host in the range of the subnet:
172.28.252.0 - 172.28.255.255
So you can see there is 1024 hosts per subnet and the original class be address is originally being broken down into 64 sections.
The first section would be:
172.28.0.0 - 172.28.3.255 with the first host being 172.28.0.1
Does this make more sense?[/quote] -
EricO Member Posts: 94 ■■■□□□□□□□You should really do the logical anding of the subnet mask and the network address to find the answer.
11111111.11111111.11111100.00000000
x
10101100.00011100.11111111.10000111
=
10101100.00011100.11111100.00000000
or
172.28.252.0
and of course eliminating the zero would give the the first available host of
172.28.252.1