Hi Lynkaran,
I’ll explain how to do the first example, see if you can solve them with my technique. Let’s start with 112.10.78.40/22:
First we need to figure out what the subnet mask since /22 doesn’t tell us much. You need to write this down in binary and convert it to decimal:
first 8 bits = 11111111 (255 in decimal)
next 8 bits = 11111111 (255 in decimal)
next 8 bits = 11111100 (252 in decimal)
next 8 bits = 00000000 (0 in decimal)
So now we know the subnet mask is 255.255.252.0
How many hosts do we have per subnet? There are 2 + 8 host bits so 10 host bits in total. You can use this formula for this:
210 - 2 = 1024.
You need to remove 2 since one address is the network address and the other one is the broadcast address.
What subnets do we have? There’s a quick way to calculate this. Take number 256 - subnet mask = size of subnet:
256 - 252 = 4
Now we can write down the subnets, I’ll start with 0:
#1 112.10.0.0/22
#2 112.10.4.0/22
#3 112.10.8.0/22
#4 112.10.12.0/22
#5 112.10.16.0/22
#6 112.10.20.0/22
#7 112.10.24.0/22
…and so on
You can see it increases with 4 every time, you now have your network addresses. What about the broadcast address? That’s the last address in each subnet.
For subnet #1 that will be 112.10.3.255 and for subnet #5 it’s 112.10.19.255.
See if you can do the other examples on your own with these steps. All you need is practice, practice, practice.
Hope this helps.
Rene