Finding what's after the slash
Joseph S D Yao
jsdy at cospo.osis.gov
Thu May 11 20:51:58 UTC 2000
On Thu, May 11, 2000 at 02:10:53PM -0300, augusto bott wrote:
> Hi Gurus!
>
> I've seen lots of messages, texts and pages around and was trying to figure
> out how one can know what to write after a slash ("/") on a subnet...
>
> like, for instance, 10.1.1.0/26 meaning ip's from 10.1.1.1 untill 10.1.1.50
> (is this statement correct?)
Let's find out.
IPv4 has 32 bits. 32 - 26 = 6. 2^6 == 64. So this subnet must have
64 members, STARTING ON A BOUNDARY DIVISIBLE BY 64. Well, 0 is
divisible by 64, so the members are 10.1.1.0 - 10.1.1.63.
BUT ... the first number is the network number (10.1.1.0) and the last
number is the broadcast address on this subnet (10.1.1.63). So, the
USABLE members are 10.1.1.1 - 10.1.1.62.
> i.e., for a net like 192.168.2.45 untill 192.168.2.221 it would be what?
> Thanks...
>
> Augusto Bott
> augusto.bott at via-rs.net
It would not be anything. .45 is not divisible by any power of two
[except 1], and so cannot be the start of any subnet. Further, there
are (221 - 45) + 1 = 177 members of the above set. The next larger
power of two is 256 (2^8). This means that the subnet that includes
all of the above addresses must be 192.168.2.0/24. This is the network
whose network number is 192.168.2.0, whose broadcast address is
192.168.2.255, and whose useful members are 192.168.2.1 - 192.168.2.254.
It's all about powers of two! The most able members of the networking
community are those who, at a very young age, realized the special
beauty of the sequence:
0 1
1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 256
...
All subnets must contain a number of members that is equal to a power
of two [including the network number and the broadcast address]. All
subnets must start at an even multiple of the number of members.
Comprendo? ;-)
--
Joe Yao jsdy at cospo.osis.gov - Joseph S. D. Yao
COSPO/OSIS Computer Support EMT-B
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