Subnetting question

Have a question or want to start a discussion? Post it! No Registration Necessary.  Now with pictures!

Threaded View
Okay, I got my CCNA a couple months ago and am going toward my MCSE
now....however, I haven't dealt with subnetting since my course
material in Cisco so please forgive me .....

I have the McGraw/Hill MCSA/E 70-291 book and am covering subnetting.
I haven't dealt with subnetting in a while....here's my question
(straight from the book)

If you have an IP of 10.10.1.0 /23 how many subnets does this give
you?

Well, wouldn't you take 2 to the power of 7 giving you 128?  The book
says the correct answer is 32,766.

How in the world do you get 32,766 when the 3rd octet of the mask has
7 networking bits?

I'm either way off base or the book has had 3 wrong answers when it
comes to subnetting which I find hard to believe.  Please help me out
here.


Re: Subnetting question
I think I managed to answer my own question...someone confirm please.

Since 10.10.1.0 /23 is a Class A network, then I really only have 15
bits to play with (since the first octet is sorta locked b/c the
default netmask for Class A is 255.0.0.0)...therefore you'd take 2^15
and get 32768 (if you can't use subnet zero you'd get 32766)

Am I right? I know I am...but I need someone to lock it in stone for
me please :)


Re: Subnetting question
tash wrote:
Quoted text here. Click to load it

That is correct. Class A network means the first octek is the network
and next 3 are the host. So using a /23 gives you 15 bits for the subnet
address and 7 for the host. Since you're not using subnet zero you get...

Networks .. (2^15)-2 = 32766
Host ...... (2^7)-2  = 126

Terry

Re: Subnetting question
Quoted text here. Click to load it

oops, that should read ....

Host ........ (2^9)-2 = 510

;-)


Re: Subnetting question
Quoted text here. Click to load it

oops, that should read ....

Host ........ (2^9)-2 = 510

;-)


Re: Subnetting question

Quoted text here. Click to load it

BINGO



Site Timeline