Hi!
I am still struggling wiht Tcp/ip
Can I ask a question?
Is 172.37.2.56/12 right formular?
I am asking this qestion because I thought that once 172.37.2.56/12 is class
B, it cannot have /12. Shouldn't it be allowed to use subnet mask starting
from /18 to /30?
This way, class A can have subnet mask starting from /10 to /30, and class C
can have sunbet mask starting from /26 to /30.
This is what I have understood.
Am I missing something?
I will appreciate your reply.
jeongwoo
------Original Message------
From: Ken Wahl <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED]
Sent: July 13, 2000 6:45:06 AM GMT
Subject: Re: 2 questions
Shane,
Here is an alternative for calculating your subnet addresses, valid host
range and broadcast address that I found easier to wrap my mind around.
Having determined that a /12 is a 255.240.0.0 subnet mask look to the
octet in the subnet mask that the subnet boundary falls on and subtract
that value from 256.
In this case 256-240=16. The subnet addresses will be multiples of 16 up
to but not including the value you subtracted from 256 (240). So the
possible subnets for a /12 or 255.240.0.0 mask will be
16, 32, 48, 64, 80, 96, 112, up to 224 and the valid host ranges will be
all those between the subnets except the last one which is the broadcast.
For example:
Subnet 172.16.0.0
Valid Hosts = 172.16.0.1 - 172.31.255.254
Broadcast = 172.31.255.255
Subnet 172.32.0.0
Valid Hosts = 172.32.0.1 - 172.47.255.254
Broadcast = 172.47.255.255
Subnet 172.48.0.0
Valid Hosts = 172.48.0.1 - 172.63.255.254
Broadcast = 172.63.255.255
See the pattern?
Your particular question asks about 172.37.2.56/12 .
37 is greater than 32 but less than 48 so it is in the 172.32.0.0 subnet
and the valid host range and broadcast address are those shown above for
that particular subnet.
Regards,
--
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On Wed, 12 Jul 2000, Lawrence Dwyer wrote:
> Shane,
>
> The /12 represents the first 12 bits of a possible 32 of masking.
>
> 11111111.11110000.00000000.00000000
>
> 255 128+64+32+16=240
>
> This mask is written as 255.240.0.0
>
> 37 is written as 00100101 (32+4+1)
>
> Since the 2nd octet is 11110000, the 1's represent matching #'s
("static")
> and the
> 0's represent any number.
>
> the 4 bits of 37 stay the same (0010) this gives you a range of
> 0010 0000 - 0010 1111 = 32-47
>
> You now have your network range
> 172.32.0.0 - 172.47.255.255
> 1st number is the network number, all 1's (the last) is your broadcast
> number.
>
> HTH,
> Larry
>
>
>
>
>
> Shane Stockman wrote:
>
> > (1) If I have an address of 172.37.2.56/12 how would I work out the IP
> > range , broadcast address ,subnet mask ,subnet address.?
> > (2) What are the different config reg settings and what would they do ?
> > e.g 0x2102 or 0X0101
> >
> > Thanks in advance
> > ________________________________________________________________________
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> >
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> --
> Lawrence Dwyer, MCSE CCNA
> Project Officer
> Telemedicine Advanced
> Technology Research Center
>
> (301) 619-7946
>
>
> ___________________________________
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