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.)
I’m on the same boat as you, I find out that you solved, but I’m some lost on this one lol, I really hate this questions that requires a exact entry
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)
10.200.20.16
If you see /27 that means it’s already a subnet. So you need subnet a subnet.
Using a table, subnet mask/27 has 32 hosts and 8 subnet. You need 4 new subnets - 8*4=32.
32 subnet has 8 hosts.
Your new subnets are
10.200.20.0
10.200.20.8
10.200.20.16
10.200.20.24
how do you know its already a subnet, i followed the method mentioned in the module and i get the 3rd subnet as 10.200.20.32. It divides 64 by 4 not 32, why is it different in this case?
Yeah, I didn’t get this kinda stuff either at first.
The question is in my opinion a bit to vague in its language for a fundamentals course.
This seeing as it kinda assumes that the person understands that a /27 “network” is already a subnet, and thats often not the case.
But to make it simple: any network that is not a /8, /16 or /24 network is a subnet.
Format is going to be 10.200.20.*
I recommend watching a few youtube videos on subnetting, getting a pen/paper and writing it out a few times. That’s what I did until I finally figured it out.
The IP addresses will not go past .31 as that is the broadcast address.