"That depends" on what you mean when you say "the # of subnets" with the mask
12. It can be 1 or several thousand depending on how far down you wish to move
the mask. There is only the given network until you change it. A /12 broken into
/24's would give you 4,096 (2^12).
>From your examples of /12 and /20 below, I think what you are missing is you
seem to only be counting those possible subnets at the next octet boundry. You
say that a /12 would give you 16 and /20 would give you 16 subnets with a 2^4.
True, but that is ONLY if you were to break the /12 to /16's and the /20 into
/24's (octet boundries). What if you wished to break /20 into /25's for smaller
offices or /23's for much larger ones? Then you would have 32 and 8
respectively. (I know I am not using the -2 part of the process because we are
discussing the binary form of possibles, not the textbook "usable" subnets).
But a /20 mask from your /12 example would give you 2^8 (256) subnets, not 16.
If you decided on a /21, then it would be 2^9 (512), a /22 would be 1024.
VLSM subnetting is just taking a current defined network range and moving the
bits of the mask farther down to use those numbers more efficiently.
Larry
jeongwoo park wrote:
> Hi Lawrence.
> I really appreciate your kind explanation.
> Please check if I understood correctly.
> When subnet mask 12 changes to 13, I guess, according to what I have known
> of, it is calculated like this: # of subnets with the mask 12 was 2^4 -2=14,
> and # of subnets with the mask 13 will be 2^5-2=30 subnets.
> Therefore, the increased # of subnet: 30-14=16 by increasing one more bit of
> mask.
> I think that with the subnet mask 12, I can have up to 14 subnets (2^4)-2
> <---(This part is still not clear, because it seems to say that as long as I
> have the subnet mask 12, whether it is class A, B, or C, it will always have
> 14 subnets. Is it correct?) How about this: 172.37.2.56/20<---This one also
> looks to me that 4 extra bits have been borrowed from 3rd octet so that it
> can have (2^4)-2=14 more subnets.)And how about this: 12.37.2.56/12<---This
> one is class A. It looks to me that it also will have up to 14 subnets
> simply it has /12. Am I missing something?
>
> Wasn't the original ip address 172.37.2.56/12 one that belonged to
> 170.32.0.0/12 subnet whose range is 170.32.0.0 to 170.47.255.255?
>
> Am I getting close to your point or am I uncontrollably misunderstanding?
>
> Again, I appreciate your help.
> I look forward to your reply.
> Thanks very much.
>
> Jeongwoo.
>
> ------Original Message------
> From: [EMAIL PROTECTED] (Lawrence Dwyer)
> To: jeongwoo park <[EMAIL PROTECTED]>, Groupstudy <[EMAIL PROTECTED]>
> Sent: July 13, 2000 12:55:23 AM GMT
> Subject: Re: 2 questions
>
> As long as you only have the /12, and only your assigned address range
> 172.32.0.0-172.47.255.255 then there will be no more subnets.
> Now you could move the bit masking to the right 1 bit and that would give
> you 2
> subnets with a /13..
> /14 will give you 4 subnets, /15 will give you 8 subnets and so on. Each
> move
> of the masking bit by one, changes the # by a power of 2. Think of the word.
> Sub Net. You are taking one large range of addresses and breaking it into
> smaller ranges by increasing the bits in the subnet mask. Remember, a subnet
> mask really only defines a range of addresses with a common network number
> expressed in binary. Variable Length Subnet Masking (VLSM) just means that
> you
> can move that masking back and forth. As you increase the bits in the
> subnet
> mask, you are decreasing the range of possible addresses accepted by that
> mask.
> They are Sub Networks of your original range.
> So
> if you were to move your masking bit 1 to the right from /12 to /13
> Then you are going from 11111111.11110000.0.0 (255.240.0.0)
> to 11111111.11111000.0.0 (255.248.0.0) and you would have 2 sub networks of
> your original network range.
> 172.32.0.0-172.39.255.255/13 and
> 172.40.0.0-172.47.255.255/13
>
> If you remember the binary from the 1st answer 37 was 0010 0101
> With 12 bits of masking, you have 4 bits in the 2nd octet, "locking in" the
> 0010
> With 13 bits and now 2 subnets you are "locking in" 00100 and 00101 in two
> seperate sub ranges
> bits 12 13th address range
> 0010 0 000-111 is 32-39
> 0010 1 000-111 is 40-47
>
> OK one more for clarification
> With a /14 you are now increasing you mask by 2 bits from the original,
> which
> gives you 4 subnetworks.
> bits 12 13th 14th add range
> 0010 0 0 00-11 is 32-35
> 0010 0 1 00-11 is 36-39
> 0010 1 0 00-11 is 40-43
> 0010 1 1 00-11 is 44-47
>
> There was a range of 8 numbers in the second octet per network at /13, but
> at
> /14 that range was cut in half to 4 while the # of networks doubled. At /15
> it
> would do so again; range of 2 (32-33, 34-35,....,46-47) with 8 networks.
> The network number is the 1st and the broadcast is the last in a particular
> range.
> The number of subnets is only limited by the mask you give it (up to a /30).
>
> I really hope it helps, subnet theory can be quite an "Ah Ha!" experience,
> once
> you understand them, you can go backwards and forwards very easily.
> :)
> Larry
>
> jeongwoo park wrote:
>
> > Hi Lawrence!
> > I have an additional question.
> > How many subnets will this address 172.37.2.56/12 have?
> > And what are they? What is the broadcast ip address in a particular
> subnet?
> > Could you tell me how you calculate?
> > I am a tcp/ip newbie.
> > I have read book about tcp/ip. But it confuse all the time.
> > I will appreciate your help.
> > Thanks in adv.
> >
> > ........................................................
> > iWon.com http://www.iwon.com why wouldn't you?
> > ........................................................
>
> ........................................................
> iWon.com http://www.iwon.com why wouldn't you?
> ........................................................
>
> ___________________________________
> UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html
> FAQ, list archives, and subscription info: http://www.groupstudy.com
> Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
--
Lawrence Dwyer, MCSE CCNA
Project Officer
Telemedicine Advanced
Technology Research Center
(301) 619-7946
___________________________________
UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html
FAQ, list archives, and subscription info: http://www.groupstudy.com
Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
