quick question about something in Sybex CCNA 5th edition

Hi everyone!
I'm new here, so be easy on me. Just started my CCNA studies, it's coming along good...just very very slowly if you know what I mean. I am coming from almost no background in networking, but I really do enjoy this (even if I still suck at it now)!

I have a question regarding something I read in the Sybex CCNA 5th edition:

on pg 124 you will find something like this:

So, what private IP address should I use? (topic title)

That’s a really great question: Should you use Class A, Class B, or even Class C private
addressing when setting up your network? Let’s take Acme Corporation in San Francisco as
an example. This company is moving into a new building and needs a whole new network
(what a treat this is!). They have 14 departments, with about 70 users in each. You could probably
squeeze one or two Class C addresses to use, or maybe you could use a Class B, or even a
Class A, just for fun.
The rule of thumb in the consulting world is, when you’re setting up a corporate network—
regardless of how small it is—you should use a Class A network address because it gives
you the most flexibility and growth options. For example, if you used the 10.0.0.0 network
address with a /24 mask, then you’d have 65,536 networks, each with 254 hosts. Lots of
room for growth with that network!
But if you’re setting up a home network, you’d opt for a Class C address because it is the easiest
for people to understand and configure. Using the default Class C mask gives you one network
with 254 hosts—plenty for a home network.
With the Acme Corporation, a nice 10.1.x.0 with a /24 mask (the x is the subnet for each department)
makes this easy to design, install, and troubleshoot.

My question: can you please explain to me just how a Class A 10.0.0.0 network address with a /24 mask can yield 65,536 networks, each with 254 hosts? Is a Class A address not network.node.node.node? If so, then should it not be just the opposite? that is yielding lots and lots of host on one network?

Sorry for this boring question, I am sure I am just missing something simple here.

Thank you!

Jared

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Comments

GeorgeMcFly22

Normally network 10.0.0.0 use 255.0.0.0 or /8 as subnet-mask, or like you say network.node.node.node
This would give you 1 network with :
2power24(3x8bits) min 2 (network and broadcast ip) = 16777214 hosts

Thats why they burried 16 bits from the node portion and make it network portion (this is called subnetting)
This gives us 10.0.0.0 with 255.255.255.0 or /24 or network.network.network.node
This makes 2power16(bits burried) = 65536 networks
and 2power8(bits over) min 2 = 254 hosts

JNeko

I am over half way through the book now and reviewing subnetting again, it's going along well, except for this question:

11. Which of the following IP addresses fall into the CIDR block of 110.68.4.0/18? (Choose three.)
A. 110.68.8.32
B. 110.68.7.64
C. 110.67.6.255
D. 110.66.3.254
E. 110.65.5.128
F. 110.64.12.128

Answer: B, C, E.

Is this really the answer? I am afraid I don't understand how this could be the answer.

Can someone help me out? I have looked at this for like an hour already...><

Thank you!

jrmcent

11. Which of the following IP addresses fall into the CIDR block of 110.68.4.0/18? (Choose three.)
A. 110.68.8.32
B. 110.68.7.64
C. 110.67.6.255
D. 110.66.3.254
E. 110.65.5.128
F. 110.64.12.128

I am not positive, but I dont think those can be the answers. With a 255.255.192.0, the third octet is your interesting octet, increasing by 64.
So the first two subnets would be:
110.0.0.1 - 10.0.63.254 BA=10.0.63.255
110.0.64.1 - 10.0.127.254 BA=10.0.127.255

The 110.68.4.0 would be in the 110.68.0.0 subnet, which would be from
110.68.0.1 - 110.68.63.254 BA=110.68.63.255

I am not sure is thats the question that he is asking, but the only answer that falls in that subnet is 110.68.7.64 (B). A also falls in that range, so I am not sure why that wasn't one of the answers. I could be wrong, but I would think its a typo.

carveone

JNeko wrote:

I am over half way through the book now and reviewing subnetting again, it's going along well, except for this question:

11. Which of the following IP addresses fall into the CIDR block of 110.68.4.0/18? (Choose three.)
A. 110.68.8.32
B. 110.68.7.64
C. 110.67.6.255
D. 110.66.3.254
E. 110.65.5.128
F. 110.64.12.128

Take some of the questions you see in books with a really large grain of salt. Mistakes and typos are plentiful and can make your head spin if you take what you read as gospel. It's fairly clear that /18 was not what was meant given that:

110.68.4.0/18 gives you an increment of 64 in the third octet. This would give you a base network of 110.68.0.0 and a valid ip range of 110.68.0.1 to 110.68.63.255

I'm pretty much saying what jrmcent said! So A and B are the answers.

What the author meant was 110.68.4.0 with an increment of 4 in the third octet. So he obviously took 16 and added 2 (2^2 = 4) to make 18 like a twit. He should have taken 2 bits from the right hand side to make 255.255.252.0 or /22. So the question becomes:

Which of the following IP addresses fall into the CIDR block of 110.68.4.0/22.

Now it's B, C and E.

Conor.

carveone

Doh! I read B,C,D as 110.68.something. Never mind. Nothing fits together in any way in that question so I'd say that the question as stated is garbage. Ignore it and move on. Life's too short

mobri09

I just ran into that question as well. My coworker said it is wrong. Infact he looked it up and sybex corrected the question with another wrong answer.

JNeko

Awesome, thanks everyone! Nice to know my hour of just staring and trying to figure out why the answer was not A/B was with good reason. Man, I don't know about anyone else, but I felt pretty good with subnetting after reading the chapter in the beginning, but after moving on in the book, and then reviewing this once again, wow, even though I got most of the answers right it took me FOREVER to calculate them out in my head, I would be screwed if I took the CCNA ><

mobri09 wrote:

I just ran into that question as well. My coworker said it is wrong. Infact he looked it up and sybex corrected the question with another wrong answer.

heh heh

blackmage439

One rule of thumb: you won't get anywhere trying to calculate subnets in your head unless you have that drilled in your skull. Paper, pencil, and a calculator capable of converting decimal to binary are my best friends for getting subnets calculated quickly. All of those are given to you when you take the test (at least I have been told that), so don't worry about wasting time calculating those darn octets in your head. :P

Tazmanian

Just checked my book. It has been corrected (probably still dozens of typos left though).
The question should read .......fall into CIDR block 110.68.4.0/22 (not /18 ).
So that gives you a mask of 252 with a block size of four.
The subnet would then be 110.68.4.0 to 110.68.7.255. Obviosly you can't use the network or broadcast addresses but everything in between is good to go.
That makes B,C and E the right answers.
I think!!!!

Tazmanian

Re-read your post and the possible answers are totally wrong too.
Here's the right question and options.

11. Which of the following IP addresses fall into the CIDR block of 110.68.4.0/22? (Choose three.)
A. 110.68.8.32
B. 110.68.7.64
C. 110.68.6.255
D. 110.68.3.254
E. 110.68.5.128
F. 110.68.12.128

Answer: B, C, E.

Hope that makes more sense.

carveone

And Tazmanian just did what I did already

Anyway, to blackmage439: The only thing you are given in the exam is a whiteboard style marker and two sheets of gridded plastic to write on with the marker. You aren't allowed bring anything else into the exam.

It is quite useful to know what exactly is going on in subnetting and to remember a few key nets: /24 and /16 are obvious, but at least /30 should be known as it comes up so often. I used the initial time in the exam to write down:

mask  128  192  224  240  248  252  254  255
incr  128   64   32   16    8    4    2    1
      /25  /26  /27  /28  /29  /30  /31  /32
nets    2    4    8   16   32   64  128  256

      /17  /18  /19  /20  /21  /22  /23  /24

There's no substitute for not having to think in that exam room

blackmage439

Sorry, my mistake. I was told the test program can display a calculator for you to use during the test... I thought I heard that from an instructor. I heard about the markerboard of doom already tho.

Oh well.