Step by step method subnetting and working out subnets
- Thread starter mrapoc
- Start date
Rastalovich
New member
Sweet.
Now tell every1 else
Now tell every1 else
depends what info ur given/gotta work out
to work out number of hosts write out your binary values from 2^7 to 2^0 (128, 64, 32 down to 1)
These represent the octet you are working on (each section of an ip address is valued by the values of 1 or 0 in corresponding spaces in the binary 2^7 to 2^0 which you wrote out. For example, 192 would be 11000000 (128 + 64).
So to work out the subnet you would need for say, 26 hosts - you look at how many 0s you would need in the last section of the ip address not subnetted (class a, class b, class c) - class c would be for example 192.168.44 .whatever <whatever being hosts>
class b would be 192.168. whatever . whatever and so on
anyway, class c - 192.168.44 would be
11000000 . 10101000 . 00111100
then we look at the last octet, for 26 hosts the closest we can get whilst allowing it is 32. so we need 0s from 32 down to 1. ala 11000000
11000000 . 10101000 . 00111100 . 11000000
We can only use 30 of these due to one being assigned network address and the other broadcast.
then...
work out how much each subnet goes up by to help working out different network addresses
so we have 11000000 < the last bit "borrowed" (the last "1") is the value of 64 meaning each subnet goes up by 64 each time.
somewhere along those lines, however im tired and its probably fundamentally wrong somewhere so perhaps anyone else wants to do the favour of correcting me.
to work out number of hosts write out your binary values from 2^7 to 2^0 (128, 64, 32 down to 1)
These represent the octet you are working on (each section of an ip address is valued by the values of 1 or 0 in corresponding spaces in the binary 2^7 to 2^0 which you wrote out. For example, 192 would be 11000000 (128 + 64).
So to work out the subnet you would need for say, 26 hosts - you look at how many 0s you would need in the last section of the ip address not subnetted (class a, class b, class c) - class c would be for example 192.168.44 .whatever <whatever being hosts>
class b would be 192.168. whatever . whatever and so on
anyway, class c - 192.168.44 would be
11000000 . 10101000 . 00111100
then we look at the last octet, for 26 hosts the closest we can get whilst allowing it is 32. so we need 0s from 32 down to 1. ala 11000000
11000000 . 10101000 . 00111100 . 11000000
We can only use 30 of these due to one being assigned network address and the other broadcast.
then...
work out how much each subnet goes up by to help working out different network addresses
so we have 11000000 < the last bit "borrowed" (the last "1") is the value of 64 meaning each subnet goes up by 64 each time.
somewhere along those lines, however im tired and its probably fundamentally wrong somewhere so perhaps anyone else wants to do the favour of correcting me.
Rastalovich
New member
.. and what's the address usually given to a port as standard if nothing can be resolved ?
(nominally firewire ports etc)
(nominally firewire ports etc)
Rastalovich
New member
Simpler really, 169.254.x.x appears to be assigned to tcp/ip ports that have problems resolving.
The science behind why this happens is probably known by Jim/Nat/etc, all I know is that it happens
and I remember some1 touching on it during a server 2k3 course.
The science behind why this happens is probably known by Jim/Nat/etc, all I know is that it happens
and I remember some1 touching on it during a server 2k3 course.name='Rastalovich' said:Simpler really, 169.254.x.x appears to be assigned to tcp/ip ports that have problems resolving.
The science behind why this happens is probably known by Jim/Nat/etc, all I know is that it happens
Don't think there is much more behind it than simply 169.254.x.x and 255.255.0.0 being the defaults for the APIPA mechanism in all modern versions of windows. Although someone may know why it chooses to default to these, but while I can't see any logic to it I know some protocols still function over connections in that range.
