I haven’t done the academy, but going off technical knowledge, they are probably wanting to know the network address in standard format, so 10.200.20.___. In a /27 network, you can have 8 subnets. They are wanting to know the 3rd network address from the 8 available.

Don’t worry about CIDR notation or anything, just the network address for THAT subnet.

For this one they want it in four subnets so there will be a range of addresses and I need to know how to lay them out as in 10.200.200.___ - ___ or 10.200.20.___ - 10.200.20.___ another way I tried is 10.200.20.-. If it wasn’t for the submission seaming to need to be exact I would be done with this question

For this one they want it in four subnets so there will be a range of addresses and I need to know how to lay them out as in 10.200.200.___ - ___ or 10.200.20.___ - 10.200.20.___ another way I tried is 10.200.20.-. If it wasn’t for the submission seaming to need to be exact I would be done with this question

Have you tried using /29 as the mask (I haven’t done this lab either, but if I was asked to divide 10.10.10.0/27 into four, it would be 10.10.10.0/29 etc.)

It’s pretty basic. Think of the range you get from a /27 network. 32 addresses, minus 2 for the network (0) and the broadcast (31), means possible ips are 10.10.10.1 to 10.10.10.30. You need to break this whole range into 4, so take the 32 total addresses and divide by 4. Now you have network address 10.10.10.0 to broadcast 10.10.10.7, then network address 10.10.10.8 to broadcast 10.10.10.15 then network 10.10.10.16 to broadcast 10.10.10.23, then network 10.10.10.24 to broadcast 10.10.10.31.

(Quote)
Have you tried using /29 as the mask (I haven’t done this lab either, but if I was asked to divide 10.10.10.0/27 into four, it would be 10.10.10.0/29 etc.)

Breaking the /27 into 4 subnets would result in 4 separate /29 ranges. 0/29, 8/29, 16/29 and 24/29.

Might be easier to just do the math in your head for such a small range.

So the reason this question was giving me a hard time was because I was not using the full address. You have to split the 32 addresses into 4 equal parts, so 8 addresses per part. I was starting at 10.200.20.8 instead of starting at 10.200.20.0 because this address is still a part of the address space.

(Its says I’m responding to Androwranni, but this is just a general message)