Subnet question

general9mm

Hi

Can you please tell me what is the simplest way to answer subnet question when ip addresses are give to you at random.

I can never get this sort of question right.

E.G. Q. This question is part of a sim, given the network of 192.168.199.0 /27, assign the first ip address of the thrid subnet to serial 0.

Can you please tell me how to calculate this. Do you just determine the host block size? Really not sure.

many thanks

bmauro

Everyone finds their own way that is easiest. What may work for me might now work as well for you.

Still - I approach this by finding the block size first. I look at a /27 as a /24 +3 bits. A /27 is equal to 224.

Then I take 256 - 224 which is 32. That tells me how large the subnets are.

0-31
32-63
64-95
96-127
etc., etc.

Once you quickly write down the subnets it's easy to see that the first IP address in the third subnet is going to be .65

Kaminsky

Easiest way is to work in binary in your head...

192.168.199.0 /27 gives the subnet mask of:

11111111.11111111.11111111.111 00000 = 255.255.255.224

256 Rule => 256-224 => 32 ip addresses per subnet.

Q: Assign the first ip address of the thrid subnet to serial 0.
As this is a class C it is easy to start working out

192.168.199.0 - 192.168.199.31 :SN=192.168.199.0 B/C = 192.168.199.31 Valid Hosts = 192.168.199.01 - 192.168.199.30

192.168.199.32 - 192.168.199.63 :SN=192.168.199.32 B/C = 192.168.199.63 Valid Hosts = 192.168.199.33 - 192.168.199.62

192.168.199.64 - 192.168.199.95 :SN=192.168.199.64 B/C = 192.168.199.95 Valid Hosts = 192.168.199.65 - 192.168.199.94

So, the first ip address in subnet 3 is 192.168.199.65
(This is assuming Subnet 0 is allowed. If not, Subnet 1 starts at 32-63)

I will leave you to assign it to serial 0

The real work in these questions is done in identifying the binary mask and then doing the 256 rule to get the subnet range.

You can do these in your head....

Check out this link: http://www.subnettingquestions.com/default_uk.asp for practice questions of this type.


Hope that helps.

general9mm

Hi

thanks for all the responses. I am assuming that the trick is to just pick the host bits, calculate the block size and that should provide the answer.

correct me if i am worng.

thanks again.

DirtySouth

I always picture the following numbers in my head:

128 64 32 16 8 4 2 1

If the subnet mask is /27 & its a class C network, then I know the network is using borrowing three bits....(27-24=3).

I then count three over...(128...64...32), so now I know that the subnets will be in multiples of 32. From there you just count in 32's.

0-31 (1st subnet)
32-63 (2nd. subnet)
64-95 (3rd. subnet)

We know that you cannot use .64 on an interface because it is a network address, so the first usable IP in that range is .65. This method may seem strange to some, but I'm able to calculate subnets in my head so that's cool with me.