hail hail said:
I've got my semester 1 exam coming up after christmas, any ccna people help this sink in as my brain is turning to mush
132 64 32 16 8 4 2 1
working from the left side is subnets,right side hosts i think
so if i needed a network of 25 users i would count from right side 2/4/8/16/32, so my address table will go up in 32's
ie 0 (in here would be usable addresses) 31
32 63
64 127
128 255
256
am i on the right track as i'm not sure thats right
Also what does the /26 mean??
Sounds like you need to go over your coursework again matey

The /26 is how many 1s you have in the binary octets signifying the subnet
so for example /24 is
255.255.255.0
which means you have 24 1s
The subnet address translates from decimal to binary as
11111111.11111111.1111111.00000000
With calculating what hosts are usable you examine the octet after the last 255 octet. so for example /26
255.255.255.192
which is in binary
11111111.11111111.11111111.11000000
You take the 192, and subtract that from 256, this leaves you with 64, which means your subnet goes up in groups of 64.
Now, in the group of 64, the first number of each group (0,64,128,192) is the network address (and therefore not a valid IP address), and the LAST IP address in the range (63,127,191) is the broadcast address and also not a valid choice for an IP address.
So in this example you would have 4 subnets
1)
Subnet Address: x.x.x.0
Broadcast Address: x.x.x.63
Valid IPs: x.x.x.1 - x.x.x.62
2)
Subnet Address: x.x.x.64
Broadcast Address: x.x.x.127
Valid IPs: x.x.x.65 - x.x.x.126
3)
Subnet Address: x.x.x.128
Broadcast Address: x.x.x.191
Valid IPs: x.x.x.129 - x.x.x.190
4)
Subnet Address: x.x.x.192
Broadcast Address: x.x.x.255
Valid IPs: x.x.x.193 - x.x.x.254
I'm 99% sure all these numbers are right, but there are a lot of them, so I apologise if I got anything wrong
