can someone please explain this CIDR problem..

First, I'd like to just understand what's going on here, and also what's a quick way to figure out this kind of problem while testing. I'm guessing you shouldn't spend any more than like a minute on this type of question while testing.

I understand the basics if CIDR in terms of figuring out which mask corresponds to which CIDR notation, and also determining the number of hosts based on the notation, and also given enough time which hosts are valid, but when you get in the VLSMs and figuring out valid hosts I'm lost. Here's an example from my transcender test.

if you've 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? First off, when they say CIDR block do they mean subnet ID? Is the address of 192.168.8.0 a network ID arrived at by the /21 (24

mask, where the IDs would be 0, 8, 16, 24, 32, etc?

If so, how can ip addresses with a different mask be a host address in that block?

the options are..

192.168.8.64 /28
192.168.9.13 /22
192.168.10.31 /27
192.168.16.111 /24

Does the fact that the /28, /27 and /24 addresses are all pushing the hosts over to the 4th octet give you an immediate clue that they won't be valid hosts for a /21?

I dunno...I so confused....I've read the stuff on the cisco sight and I understand the basics of applying the VLSM.

I already have the answer, so not looking for that...looking to understand what's going on and a quick way to figure this type of thing out.

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Comments

eurotrash

192.168.8.0/21

11111111.11111111.11111000.00000000
11000000.10101000.00001000.00000000

options (in order)

11111111.11111111.11111111.11110000
11000000.10101000.00001000.01000000

11111111.11111111.11111100.00000000
11000000.10101000.00001001.00001101

11111111.11111111.11111111.11100000
11000000.10101000.00001010.00011111

11111111.11111111.11111111.00000000
11000000.10101000.00010000.01101111

i think the answer is 192.168.9.13/22

methoselah

I think it goes like this...

192.168.8.0 /21 = 192.168.8.0 255.255.248.0
Block size = 256-248 = 8
Classless Networks using /21 mask : (0.0, 8.0, 16.0…)

Valid hosts for network 8.0 is (8.1- 15.254). 15.255 is broadcast.

Choices A, B, C fall between x.x.8.1 – x.x.15.254, and are valid so far.
Choice D is outside x.x.8.0 –x.x.15.255 ----> INVALID.

Take closer look at Choices A, B, C
********
A)
192.168.8.64 /28 = 192.168.8.64 255.255.255.240
Block size = 256 -240 = 16
Sub-networks : (0, 16, 32, 48, 64, 80,…)

192.168.8.64 is a sub-Network address

> INVALID

********

192.168.9.13 /22 = 192.168.9.13 255.255.252.0
Block size = 256 -252 = 4
Sub- networks : (0.0, 4.0, 8.0, 12.0,…)

192.168.9.13 /22 is between (x.x.8.0 – x.x.11.255)

> VALID

********
C)
192.168.10.31 /27 = 192.168.10.31 255.255.255.224
Block size = 256 -224 = 32
Sub- networks : (0, 32, 64…)

192.168.10.31 /27 is a broadcast address

> INVALID

*********
D)

(Different network)

pizzafart

Sleemie I think the question you got was worded poorly. Overall, it sounds like you're on the right track when it comes to subnetting. Some of the terminology gets thrown about loosely and that's probably what's confusing.

In this question, "CIDR block" could be considered:
A) A single broadcast domain designated for possible host addresses of 192.168.8.1 - 192.168.15.254. The broadcast address would be 192.168.15.255 and the Subnet ID would be 192.168.8.0.

If this was what the question intended then it is true that none of the answers are correct because the masks are bogus. You can't slap a host with a /22 mask onto a /21 network.

Because none of the answers work in this context, we have to make some assumptions:

The "Cider block" must simply be referring to that same range of addresses which could potentially be carved up into any number of smaller networks.

See, a "CIDR" block doesn't specifically mean one or the other. It's a loose term. It's typically meant as a range of addresses used to carve up into smaller networks, but, it doesn't have to be. It could just mean one big network. If I remember my BCSI exam, there aren't really many protocols that use the verbage "vlsm" and "cidr" as actual IOS commands. These terms are mostly conceptual and don't always map well to actual router functions.

Also, think about this:
A "subnet" can mean a few different things. Traditionally, it's a network that has a mask past the typical class A/B/C boundaries. Though, consider the following two networks in relation to eachother:

150.64.0.0 /10 and 150.72.0.0 /13.

You could say that the /13 network is a "subnet" of the /10 network if you wanted because all of the /13 addresses fall within the /10 range. This is reasonable terminology even though both networks still have masks that are to the left of the standard class B mask designation. It might not be "proper", though, I don't know what else to call it. A "CIDR block of a block"? That just sounds akward.

Also, it could help to know that a "subnet ID" can be called a "network ID", a "network address", a "network", a "subnet address", a "subnet number" and probably some others. What's really important is to define what the layer 2 broadcast domain is so that we can consider the appropriate IP broadcast address(s) and network number(s). Because the question doesn't offer enough details, the best we can do with this question to is assume:

If a particular answer has a mask that signifies a network (subnet, what ever you want to call it) of which all the addresses fall within the given 192.168.8.0 /21 "CIDR" range; that answer is correct, assuming that the answer follows proper subnetting rules otherwise.

Hopefully this helps.

I understand the basics if CIDR in terms of figuring out which mask corresponds to which CIDR notation, and also determining the number of hosts based on the notation, and also given enough time which hosts are valid, but when you get in the VLSMs and figuring out valid hosts I'm lost. Here's an example from my transcender test.

if you've 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? First off, when they say CIDR block do they mean subnet ID? Is the address of 192.168.8.0 a network ID arrived at by the /21 (24

EdTheLad

pizzafart wrote:

In this question, "CIDR block" could be considered:
A) A single broadcast domain designated for possible host addresses of 192.168.8.1 - 192.168.15.254. The broadcast address would be 192.168.15.255 and the Subnet ID would be 192.168.8.0.

If this was what the question intended then it is true that none of the answers are correct because the masks are bogus. You can't slap a host with a /22 mask onto a /21 network.

CIDR refers to blocks of addresses, so i wouldnt go thinking about hosts when playing around with CIDR.Think about blocks of networks.After you have the networks assigned think about the host addresses.
I see nothing wrong with the wording of the question.

"You can't slap a host with a /22 mask onto a /21 network"
You have misunderstood the question.
The /21 block is not a network address but a summary of many networks which include the /22 network.This /21 address is what the router will advertise as its summary to its neighbors.

pizzafart

Sleemie, everyone has a right to an opinion - mine is that Ed's response doesn't do anything to help your situation. I don't care to debate this so feel free to come to your own conclusions.

Allan