I get the same results as Marko, but this may lay it out so you (and 
others) can see the development:

IP address = 32 bits
Network portion = 22 bits
Host portion = 10 bits
Total addresses for host portion = 2^10 = 1024

Start with 192.168.24.0/22
Focus on the 3rd octet (network_host): 000110_00

400 hosts requires 9 bits (2^8 = 256, 2^9 = 512)
and you will have some left in this block
divide the /22 into two blocks of 512 addresses each:
0001100_0 (.24/23) and 0001111_0 (.26/23)
use .24/23 for the 400-host network

200 hosts requires 8 bits (2^7 = 128, 2^8 = 256)
and there will be some left in this block, too
divide the .26/25 into 2 blocks of 256 addresses each:
00011110 (.26/24) and 00011111 (.27/24)
use .26/24 for the 200-host network

50 hosts requires 6 bits (2^5 = 32, 2^6 = 64)
and you will again have some leftovers
divide the .27/24 into 4 blocks of 64 addresses each
now looking at the 4th octet:
00_000000 (.0/26), 01_000000 (.64/26), 10_000000 (.128/26), and 
11_000000 (.192/26)
use the first two for the 50-host networks

and the rest is easy

My personal rule is to always start with the biggest blocks and work 
down from there.


HTH

Annlee

Steven Aiello wrote:

> I just started my routing class for my CCNP.  We are covering CIDR.  The 
> book is VEEEEEERY vague on how the bit patterns break down and are used.
> 
> 
> This was a problem posed in one of my CCNP labs
> 
> I have network number
> 
> 192.168.24.0 / 22
> 
> from this I need
> networks with
> 
> 400 hosts
> 200 hosts
> 50  hosts
> 50  hosts
> 2   hosts (for serial int - no ip un-numbered allowed )
> 2   hosts
> 2   hosts
> 
> Also no NATing
> 
> Thanks all I really could use the help
> 
> Steve
> **Please support GroupStudy by purchasing from the GroupStudy Store:
> http://shop.groupstudy.com
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