repost minus 256...

this has eclipsed all others for amount of hits, and rightly so...  i wrote and posted this on the internet back in 2004...

pre-req's

comfort with binary to decimal conversion

comfort with exponents of 2, ie. 2 to the x = ?

comfort with the three classful boundaries...

the thing about ccna and all it's variants, on up through ccie is that you can count on the numbers... if you are afraid of the numbers and their various permutations and calculations, then all of this will be difficult at best...

this site was created for the ccent/ccna candidate... anything more advanced than that is gravy...

if you do a search in the search bar for ccna, you will find over a hundred posts directly related to ccna... and don't forget the hex tutorials...

there is safety in numbers...

the minus 256 technique

this presupposes knowledge of binary and ip addressing conventions

rule 1. remember that the first octet only ever designates the class

of ip, ie. a b or c

rule 2. the first octet that contains a zero bit, is always the octet

where the action occurs, ie, 255.255.255.0, calculation happens in 4th

octet; or 255.255.0.0, calculation happens in third octet; or

255.255.248.0, calculation occurs in 3rd octet, and so on.

rule 3. see rule number 1. The first octet always tells you the class

of address no matter the octet where subnetting occurs.  Subnetting

calculation always happens in the octet of the ip address that the

subnet mask designates with its first instance of less than 255, or

more simply, the first instance of a zero bit.

therefore, given 172.16.10.10 mask of 255.255.248.0, we know that the

calculation will happen in the ip's 3rd octet.  The mask designates

that with 248.  it is imperative that this is understood.

another way of looking at it in the above example is; the octet in the

subnet mask with the first instance of less than 255, or the first

zero bit, is the multiplier.

rule 4.  when the multiplier (first zero bit octet or octet with first

instance of less than 255) is determined always subtract it from 256

to determine the ranges.

ie. 256-248=8, hence 8 is the multiplier.

using 172.16.10.10 with 255.255.248.0, it is determined that we have a

class b address 172, our calculation must happen in the 3rd octet, and

we must subtract 256 from 248 to get 8.

the rest is academic:

the multiplier ( has determined our first subnet range

8 16 24 32 40, etc

the first range (excepting the use of subnet zero) begins with 8 and

ends with 15, the second range begins with 16 and ends with 31, next

range begins with 32, and so on up to 255.

Important: there are 256 numbers total comprising the range 0-255,

including the zero.

in the ip 172.16.10.10 /21 (notice the use of bit count; this equals

248 as well.  to determine the number to subtract from 256 in  bit

count form, you need to add the bits...

1st octet 8 bits, second octet 8 bits, third octet 5 bits, hence 8 + 8

+ 5 =/21 or 248 or

172.16.10.10 /21 = 172.16.10.10 255.255.248.0

our calculation takes place in the octet designated by the first

instance of a zero, or in our example, /21 or 255.255.248.0.  we

determine that 10 is the number occupying the third octet in our

example, and our multiplier has determined the first possible subnet

is 8 (excepting subnet zero)

so, since 10 falls between 8 and 15 (16 begins the next subnet or

network), our valid range for the address has been determined.

8    16   32...

9    17

14   30

15   31

so our octet 3 number, which is 10 in the example, can only fall

between the range of 8 (the  network), 9 our first valid host, 14 our

last valid host, and 15 which is the broadcast address for the network

our number ten resides in.

if we changed our third octet number to 172.16.20.10 /21 or

255.255.248.0, we know that our calculation still takes place in the

3rd octet, but the number 20 falls between the network 16, the

broadcast 31, and within the valid range of hosts which is 17-30...

one more example:

192.168.100.100 255.255.255.192

the class of address is C

the action takes place in octet 4

subtract 192 from 256 which equals 64 and we can determine the

network, the broadcast and the valid range of hosts because 64 is our

multiplier

hence:

64     128    192

65     129    193

126    190    254

127    191    255

our number in octet 4 is 100, our number 100 falls between 64 (the

first network) and 128 (the second network).  the subnet address is

192.168.100.64

   the first valid host is 192.168.100.65

   the last valid host in the range is 192.168.100.126

   and the broadcast address is 192.168.100.127

it takes a little time and effort, and a lot of practice, but you'll

eventually be able to do it without paper.