Subnetting question...

So I have my Network+ under my belt already. When preparing for it I went to subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online often to make sure I had subnetting down. I never had to know about how many hosts and subnets per network, so I just skipped them before.

So I was watching the CBT nuggets for the CCENT last night that had to do with subnetting, and wanted to pound away at making sure I have it down. I'm pretty confident in my ability, but plan on doing it non stop until I test in a few months to make sure I know it with doing as little written math as possible.

My question is: I got the following question and I don't quite understand it...

Question: How many subnets and hosts per subnet can you get from the network 172.27.0.0/24?

Answer: 256 subnets and 254 hosts

I can't wrap my head around how the heck it came to 256 subnets. What am I missing here? Before I clicked for the answer I said to myself "that's one network and 254 hosts".

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Comments

odysseyelite

Its a class B address. The default Subnet address would be 255.255.0.0

If you are watching CBT nuggets, the binary would be 11111111.11111111.11111111.00000000

They borrowed 8 bits FOR MORE NETWORKS to have a 255.255.255.0 address.

2^8 = 256 networks

Now count how many zero bits (hosts) you have (

2^8 - 2 (Always subtract 2 IP's for the network address and broadcast address) = 254 hosts

SdotLow

Errr...I guess I made a very poor assumption.

I assumed /24 implied a class C address. Thinking over the videos from last night I see I may have made a stupid mistake (I remember having to dip into the last octet on a Class

. When defining the class of an address, is the amount of 0's in each octet what makes it a class?

So for example..

170.0.0.0 = Class A

170.28.0.0 = Class B

170.94.198.0 = Class C

So 170.28.0.0 would never be considered a Class C address, because the last two octets have 0's?

Every address I've ever had to break down had the class provided already ;(

ehnde

SdotLow wrote: »

Errr...I guess I made a very poor assumption.

I assumed /24 implied a class C address. Thinking over the videos from last night I see I may have made a stupid mistake (I remember having to dip into the last octet on a Class . When defining the class of an address, is the amount of 0's in each octet what makes it a class?

So for example..

170.0.0.0 = Class A

170.28.0.0 = Class B

170.94.198.0 = Class C

So 170.28.0.0 would never be considered a Class C address, because the last two octets have 0's?

Every address I've ever had to break down had the class provided already ;(

No, you can establish if you are dealing with Class A, B, or C by the first octet. 0 - 127 in the first octet is class A. 128 - 191 is class B. 192 - 223 is class C, and above that is multicast and experimental address ranges.

Shanman

0 - 126 class A default mask /8

128 - 191 class B default mask /16

192 - 223 class C default mask /24

odysseyelite

No, you need to go back and remember you class IP ranges

172 is a class B address because of the number it starts with:

Class A 1.0.0.1 to 126.255.255.254
Class B 128.1.0.1 to 191.255.255.254
Class C 192.0.1.1 to 223.255.254.254

Class A default subnet 255.0.0.0
Class B default subnet 255.255.0.0
Class C default subnet 255.255.255.0

SdotLow

Wow. I so should have known that. Sorry for wasting everyones time ;(

Been spreading myself to thin in a lot of respects, this reaffirms that I need to not do that and just focus on one thing at a time.

Thanks for the responses and again, I apologize for the stupid question.

odysseyelite

SdotLow wrote: »

Wow. I so should have known that. Sorry for wasting everyones time ;(

Been spreading myself to thin in a lot of respects, this reaffirms that I need to not do that and just focus on one thing at a time.

Thanks for the responses and again, I apologize for the stupid question.

The only stupid questions are the ones not asked. Everyone has their moments.

Good luck on your studies.

h4ckn3t

I know this is old, but where the OP got confused is he was assuming Classless Inter domain Routing (CIDR) was being used. If that was the case then there would only be one network and 254 hosts.