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Right your standard Class B address uses 16 bit for the network portion and 16 bits for the host portion so the Mask looks like this: 255.255.0.0 or 11111111.11111111.00000000.00000000.
Now you have borrowed 12 bits for the network portion and left 4 bits for the host so your mask now looks like 255.255.255.240 or 11111111.11111111.11111111.11110000 just like DVNT said.
In an 8 bit octet (byte):
1_____1___1____1___1___1__1__1 =
128___64__32___16__8___4__2__1
So 11110000 is the same as 128+64+32+16+0+0+0+0=240
In this configuration you have subnetted a class B into 4096 possible subnets with each subnet containing 16 hosts per subnet. 2 to the 12th=4096 and 2 to the 4th =16.
Let's say your class B is 172.16.0.0 then you're subnets would look like this:
_____Network Address____Host Range_____Broadcast
0)______172.16.0.0_______172.16.0.1-14___172.16..0.15
1)______172.16.0.16 _____172.16.0.17-30___172.16.0.31
2)______172.16.0.32______172.16.0.33-46___172.16.0.47
All would have Masks of 255.255.255.240
This would continue until
4096)_172.16.255.240___172.16.255.241-254__172.16.255.255
As you can see the first and last subnet are not useable because the first contains the network address of 172.16.0.0 and the last contains the broadcast address of 172.16.255.255. The same is true for each subnet, of the 16 host addresses only 14 are useable because the first host address is the network address for that subnet and the last host address is the broadcast address of that subnet. (these can become useable with some other networking techniques, but that is another lesson)
Clear as mud?
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Last edited by darrelld; 03-11-2004 at 08:47 PM.
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