Phil,
May I give one small suggestion. Don't do it in decimal form. Use the
binary form. This is the safest way and the easiest way. (some people will
disagree) I will show you how to do it fast and easy.
I do this before every exam that subnetting/supernetting may come up.
On a piece of paper, put in two roll of numbers
192 224 240 248 252 254 255
128 64 32 16 8 4 2 1
Now lets do your question.
You need 10 + 1 = 11 IP address. Look at the second roll of numbers to see
where 11 fall between. To the left is 16. 16 -2 is 14 which is available
IP address for the mask of 240 (which is right on top of 16).
Now is the hard part. Under the second roll of number, put in the 0 and 1's
for the network.
192 224 240 248 252 254 255
128 64 32 16 8 4 2 1
0 0 0 0 x x x x
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 1
and so on.
By looking at the binary, you will be able to see all the valid range really
fast and correct.
Remember, all the IP address, subnet mask, wide card mask, supernetting are
binary. IF you do your math in binary, you don't have to worry about
conversion mistakes.
BTW. if you get a question on subnetting, remember that to use your first
range of 1-15, you need to have subnet 0 in the config.
I hope this help.
Albert
-----Original Message-----
From: Circusnuts [mailto:[EMAIL PROTECTED]]
Sent: Monday, September 04, 2000 5:01 AM
To: [EMAIL PROTECTED]
Subject: Quick Remedial Subnetting Theory (not hard I swear)...
I broke out the CCNA book, to practice readings for a couple of up & coming
interviews (this week). Here is what I have forgotten :-) OK- I can
calculate the mask (no issues here), but I cannot logically figure out how I
would start the host addressing. Case in point (bare with me, this all
looks soooo familiar it might just set back in while writing this :-)
192.16.12.0 is my given class C
I need 10 users (hosts- 11 counting the Ethernet Interface)
So this means I must take 2 to the 4th (16-2= 14 usable) on the mask side &
that leaves me with 2 to the 4th on the Host addressing side (same 14
usable).
How to I proceed with the first to last IP's available... i.e.
My nature tendency, it to look @ things this way...
192.16.12._ to _, does this mean I am now in the 16's if I recall.
192.16.12.1 thru 15, 192.16.12.16 thru 31, 192.16.12.32 thru 63, etc...
The book I have (full of typo's) starts with a 192.16.12.144 as the first
address...
Hope this is not too confusing...
Thanks All !!!
Phil
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