From: Andrew (arousch@xxxxxxxx)
Date: Mon Apr 23 2001 - 17:51:26 GMT-3
I'm may be a little confused by your question but here's a few things to
ponder...
Just do the simple binary math in your head. Think to yourself:
/24 = 1 network (in this case: 192.168.100.0)
/23 = 2 networks (in this case: 192.168.110.0->192.168.111.0)
/21 = 8 networks (in this case: 192.168.96.0->192.168.103.0)
The /21 is going to cover your 192.168.100.0/24 so you can toss the
/24. The /23 is not contiguous to the networks in the /21.
Practice the math behind it and you will find that these 'riddles' will no
longer mystify you. (Think Base2 math.)
-A
At 01:40 PM 4/23/01 -0700, Troy Edington wrote:
>I am having an argument with a fellow co-worker and I am wondering if my
>fellow peers on this list can help me with this. I think it is fairly
>simple
>
>You are given three network addresses
>
>192.168.100.0/24
>192.168.110.0/23
>192.168.96.0/21
>
>The question states that you need to make a supernet or summarize the
>following 3 network addresses as much as possible. Assume there are no
>other networks anywhere. But make sure you summarize it as much as
>possible.
>
>In getting a question like this I said ok,
>
>To summarize as much as possible I would say
>
>128.0.0.0/1 There is only a potential for 2 networks here.
>
>However, given the following two choices of summarization, which one would
>be considered more correct.
>
>192.168.96.0/20 or
>192.168.96.0/19
>
>Thanks
>
>Troy
>**Please read:http://www.groupstudy.com/list/posting.html
**Please read:http://www.groupstudy.com/list/posting.html
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