CIDR Question

DXU76

I have a practice exam question that I don't understand. Hoping someone can explain it to me. Thanks in advance.

A service provider has been assigned the 192.168.0.0/16 address. Which of the following addresses is a valid host address for the CIDR block 192.168.8.0/21?
A. 192.168.8.64 255.255.255.240
B. 192.168.9.13 255.255.252.0
C. 192.168.10.31 255.255.255.224
D. 192.168.16.111 255.255.255.0

I know that A. and C. are the wrong answers. A. is wrong because it is a network address. C. is wrong because it is a broadcast address.

So that leaves me with B. and D. answers. Both are host addresses. But B. is the right answer. My question is: Why is D. a wrong answer? 192.168.16.111 can exist 'legally' both in the 192.168.16.0/24 and the 192.168.16.0/21 subnets, right? Isn't that the rule of thumb in CIDR?

Thanks.

EdTheLad

You have been given a CIDR block of 192.168.8.0/21, this block covers the addresses in the range 192.168.8.1 -> 192.168.15.254
The address in D is outside your CIDR block.

DXU76

Ok I got it now, thanks.

hectorjhrdz

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my god this is just marvelous!!! i didn't remember that info


very helpful

thnks

ignign0kt

I'm having trouble understanding what the question is asking and how to solve it.
Could someone explain it and go through it for me? I also had this question on a practice test and was stumped.

Netstudent

well the block in the question is 192.168.8.0/21 . Don't even worry about 192.168.0.0/16.
All that is telling you is that you have 16 bits in octets 3 and 4 to play with. So looking at /21 you should know that the subnet mask is 255.255.248.0

Now look in the the 3rd Octet and subtract that from 256.

256-248 = 8

Therefore your subnets will be in increments of 8.
192.168.0.0 192.168.8.0 192.168.16.0 192.168.24.0

Those are some of your subnetworks.

The question asks What is a valid host in the 192.168.8.0 subnetwork or block.

Valid hosts in this subnetwork ranges from 192.168.8.1 - 192.168.15.254
Broadcast is 192.168.15.255

Next subnetwork would be 192.168.16.0

The answer D is 192.168.16.111...As you can see this is clearly out of the 192.168.8.0/21 block

I hope I explained that right. Sometimes what confuses me is the subnet masks in the answer choices. All of those different subnet masks kinda throw me off sometimes.

ignign0kt

Well I know why D wasn't the answer, but I don't get by the others aren't. And why do they each have different subnet masks? I thought all of them were going to be in the /21 mask.

Netstudent

Well looking at A you see that 192.168.8.0/21 has been further broken down. Think of it like a Block within a larger block. 192.168.8.64 255.255.255.240 is a smaller block within the larger block of 192.168.8.0 255.255.248.0

That kinda answers why they all have different subnet masks. I'm not an expert yet so my description might be kinda vague.

Now then, looking at 192.168.8.64 255.255.255.240...Go through your subnetting process.
256-240 = 16 Thus we have increments of 16 in the last Octet

You should automatically see that 64 is divisible by 16. Therefore 192.168.8.64 MUST be a subnetwork and not a valid host.

Do the same for answer C and you will see that it is a Broadcast address.

See even though the Correct answer B has a different subnet mask, it still is a valid host within the larger block of 192.168.8.0/21

This is the beauty of CIDR...nothing goes to waste if everything stays contiguous.

They don't all have to stay as /21..If they did that would be a massive waste..
You would then have 32 subnets with 2046 hosts per subnet. Do you think this company would need that many hosts per subnet? NO probably not. This is where VLSM comes into play. Anyways hope this helps.

ignign0kt

So they're basically all individual separate ip addresses on different networks, but 1 happens to fall in the 192.168.8.0/21 host range?
So to solve I just subnet all the addresses and find out which one lies between 192.168.8.1 and 192.168.15.255?

EDIT: Does VLSM have to do with determining the answer? Because I thought even if you doing VLSM, you need to borrow a minimum of 2 more bits. And /22 (the prefix of the right answer) is only 1 more than in the quesiton (21)



The explanation to the answer on the test is even more confusing..



What the heck does it mean the first 21 bits are the same? 3 of the answer choices have the same first 21 bits. I actually think I understand the part where it says all host bits are off or on. If the host bits are off in 192.168.8.64, it can't be a host address.

Netstudent

okay that explanation is rather confusing. But yes they are IP's for different blocks and yes we subnetted to find the valid host in between that range. Maybe I shouldn't have brought up VLSM, I was just using it to retiterate the fact that you don't have to use the same subnet mask.

ignign0kt

Well the thing is, I know what VLSM is, and I know how to create a VLSM from scratch... but I don't get how they came up with the addresses. I guess I just don't have enough experience with them to know them in depth. I thought in order to use VLSM, you had to borrow more bits, and then subnet a subnet address... but I don't get how the address 192.168.9.13 could come about from using VLSM. I guess I just need to read about it more and get more practice.
I need a break from studying... it's so frustrating lol