Well 11 bits of subnetting is : 255.255.255.224
Class B subnet is always 255.255.x.x so 11 bits of subnetting means after
the first 2 255's you got 11 bits of subnetting. Wich is another 255 and
then 3 bits of the host adress wich comes to 224. So there you got your
255.255.255.224 subnet mask.
So your addresses can be : 172.16.3.33 to 172.16.3.62 !
Your first address starts with 32 .. but thats the network address so cant
be used .. so 33 is the first you can use. Your next subnet starts with 64
.. So 63 will be your last host you can use. But in this case 63 is your
broadcast address. So 62 is the last host you can use !
Hope this solves it for you :)
JT.
-----Original Message-----
From: Keith Woodworth [mailto:[EMAIL PROTECTED]]
Sent: dinsdag 10 oktober 2000 7:07
To: [EMAIL PROTECTED]
Subject: Subnet mask question..
Ive been at this for quite a while and the odd subnet question still gets
me.
Boson question:
IP address 172.16.3.57 w/ and 11-bit subnet mask. What are valid hosts?
I think ok class B, but I look at 11 bits as 255.224.0.0
(11111111.11100000.0.0) which does not go with the choices of answers I
had.
I got it wrong as the answer says an 11-bit mask is 255.255.255.224 when
using a class b address. Is the mask there not 27 bits? What am I missing
there? How do they get the above mask w/11 bits?
The valid hosts were:
172.16.3.33-172.16.3.62, which I think is valid for a 27 bit mask
Thanks for clue.
Keith
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