Subnetting question
I have a subnetting question, that needs to be cleared
Question is, How many subnets and hosts per subnet can you get from the network 172.17.0.0 255.255.255.0?
I answered 1 subnet and 254 hosts , but right answer came out as 256 subnets and 254 hosts
Which part is wrong here ?
Question is, How many subnets and hosts per subnet can you get from the network 172.17.0.0 255.255.255.0?
I answered 1 subnet and 254 hosts , but right answer came out as 256 subnets and 254 hosts
Which part is wrong here ?
Last Comment
ASKER
Oh i understand what you are saying dacselat, but here comes another doubt . actually B Class is usable IP range /12 , so shouldnt that be /12 to /24 = 12 bits of subnetting. 2^12 = 4096 subnets ?
For classless subnetting using CIDR, you are right .. it will have 254 hosts. However, using the old way of having Class A, B and C where the network 172.16.0.0 to 172.31.0.0 belonging to Class B which will indeed have 256 subnets with 254 hosts each.
It entirely depends on the equipment being used if it still sees the 172.16.0.0 to 172.31.0.0 networks as Class B.
It entirely depends on the equipment being used if it still sees the 172.16.0.0 to 172.31.0.0 networks as Class B.
But you're just subnetting 172.17.0.0/16, it's not the complete range of private addresses (class B: /12)
If you don't have the original mask, it's common to assume the default mask of the class.
If you don't have the original mask, it's common to assume the default mask of the class.
172.17.0.0 to 172.17.255.255 = 256 x (256-2) when recognized as Class B.
ASKER
Sorry , Still it isnt clear to me. Do you guys know any tips/tricks to easily findout/learn subnets,broadcast,1st and last host ? Its been a while for me...
Try this. Straight from a CCNA Boot Camp class.
Subnetting-procedure.doc
Subnetting-procedure.doc
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ASKER
philippe7v7,
you said it will 254 hosts each subnet ? how is that ? check what this ipsubnetter utility (ck459 provided, thanks dude) shows....
http://s118.photobucket.com/albums/o87/kavlins/ComputerErrors/?action=view¤t=1721700.jpg
Another question to all is , even though there is 256 subnets derived from 172.17.0.0 /24 , is it usable ? can i use it as 172.17.1.1 with 255.255.255.0 mask ?
you said it will 254 hosts each subnet ? how is that ? check what this ipsubnetter utility (ck459 provided, thanks dude) shows....
http://s118.photobucket.com/albums/o87/kavlins/ComputerErrors/?action=view¤t=1721700.jpg
Another question to all is , even though there is 256 subnets derived from 172.17.0.0 /24 , is it usable ? can i use it as 172.17.1.1 with 255.255.255.0 mask ?
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For each subnet you have 8 bits for hosts.
2^8=256 IP adresses, but you have to reduce 2 because the first address is reserved (Network Address) and the last too(broadcast address).
Then:
2^8 - 2 (reserved addresses) = 254 Ip addresses for hosts.
2^8=256 IP adresses, but you have to reduce 2 because the first address is reserved (Network Address) and the last too(broadcast address).
Then:
2^8 - 2 (reserved addresses) = 254 Ip addresses for hosts.
ASKER
If so , total usable hosts should be 256x256 = 65536 - 2 = 65534 ? not 254
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ASKER
ok guys now its clear, its due to CIDR (Classless Inter-Domain Routing ) method. i wasnt sure abt it...
Special Thanks to Tim and Dacselat for clearing it.
Special Thanks to Tim and Dacselat for clearing it.
It's tough to explain especially in a forum like this where we can't really graphically depict it. Hope we could all help kavlins. Take care.
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From /16 to /24 = 8 bits to subnetting:
2^8 = 256 subnets
For hosts:
32-24 = 8 bits (host) --------> 2^8 - 2 = 254 hosts/subnet