Networking gurus - quick easy question

Soldato
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Yes, this is for an assignment - so I don't expect you to just give me the answer :) I thought I was happy with my solution, but now I'm second guessing myself...

With a subnet mask of 255.255.248.0 how many hosts can be created?

My answer is 2047, minus the 2 reserved, for a total of 2045.

If this isn't correct, can you give me some pointers?

And go easy on me....this is my first networking class :o

Thanks :cool:
 
(2^N) - 2

Where N is the number of available bits.

So, you've 8 free bits from the final octet and three from the third octet.

(2^11) - 2

2048 - 2

2046.

For the number of subnets it's pretty much the same equation:

2^N.
 
Last edited:
(2^N) - 2

Where N is the number of available bits.

So, you've 8 free bits from the final octet and three from the third octet.

(2^11) - 2

2048 - 2

2046.

For the number of subnets it's pretty much the same equation:

2^N.

Ah-ha, that was my original answer. I then confused myself by converting the 11 binary bits to decimal (which is 2047) and then subtracting the 2 :(

Stu - yeah I will. Although my career choice strictly isn't networking, I do have to take the network basics class as part of my course.

Thanks guys - off to hand this puppy in ;)
 
Ah-ha, that was my original answer. I then confused myself by converting the 11 binary bits to decimal (which is 2047) and then subtracting the 2 :(

Stu - yeah I will. Although my career choice strictly isn't networking, I do have to take the network basics class as part of my course.

Thanks guys - off to hand this puppy in ;)

It can be a little annoying getting your head round 0 also being counted.

So your 2047 actually was 2048 as 0 counts.
 
http://forums.overclockers.co.uk/forumdisplay.php?f=6

Also, get this sub-netting lark down. It's integral to your career choice. I know guys who have been in this game for 10 years and struggle to subnet in their head on the spot, quite embarrassing when they are designing huge networks.

Not any more really, IPv6 means it's somewhere close to impossible work out in your head, you need to memorise enough to be able to infer it really or use a calculator.

How many addresses in a /40? (2^88)-2. Yeah, good luck with that in your head...
 
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