No offense, but this is CCNA material.  If you are going for your CCNP, then
you should already have your CCNA and know the answer.  But anyway...

If you need a network with 400 hosts, the smallest subnet would have a /23
mask.  So take the first part of your given network and assign it to that:

192.168.24.0/23 (192.168.24.0-192.168.25.255)

Then you need one with 200 hosts.  Well, that could fit within a /24 subnet,
so assign the next available to that:

192.168.26.0/24 (192.168.26.0-192.168.26.255)

Now you only have 192.168.27.0/24 left from the original 192.168.24.0/23
(which covered 192.168.24.0-192.168.27.255).  You need two 50's, so that
should fit within /26 subnets each.  Assign them:

192.168.27.0/26 (192.168.27.0-192.168.27.63)
192.168.27.64/26 (192.168.27.64-192.168.27.191)

Finally, you need three subnets that can have two hosts each, which would
fit within /30 subnets.  So assign:

192.168.27.192/30
192.168.27.196/30
192.168.27.200/30


Fred Reimer - CCNA


Eclipsys Corporation, 200 Ashford Center North, Atlanta, GA 30338
Phone: 404-847-5177  Cell: 770-490-3071  Pager: 888-260-2050


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-----Original Message-----
From: Steven Aiello [mailto:[EMAIL PROTECTED] 
Sent: Tuesday, September 09, 2003 8:02 AM
To: [EMAIL PROTECTED]
Subject: Please Help - CIDR - How the bits work [7:75050]

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
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