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 Topic: Subnetting Question |
 
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networxing
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Joined: 05 May 2006
Posts: 3
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| Posted: 05 May 2006 at 5:37am | IP Logged |
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Hi everyone.
Just have few questions about Subnetting. Iam Using the shortcut method(magic no method) described in the Cisco press ICND book. However Iam stuck and cannot figure out how to solve this problem. Here goes...
1) Given the IP address 140.1.1.1 and the mask 255.255.255.248, what are the assignable
IP addresses in this subnet?
I cant figure out which octet in the subnet mask is the so called "interesting" octet. The IP address given is a class B address, so is the interesting octet the 3rd octet? but the book at times says that the interesting octet is the octet with non 0's or 1's( as in other than 0 or 255)?
Another example to make myself clear---
2) IP 167.88.99.66 and the mask 255.255.255.192 or
even
3) 10.5.118.3 and the mask 255.255.255.0
In the last example even though there is no so called "difficult" octets, which octet do we use to find the "magic no". Although in the example the resulting magic no will be same i.e 1, no matter which octet i choose, but the end result would certainly be different.
In short, I cant figure out which octet is the interesting octet if subnetting has been done on more than 8 bits, specially on a class A and B IP address/subnet.
Can someone pls help me, if iam not clear pls let me know.
Thanks
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Wildcat_Dude
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Joined: 20 August 2003
Location: United States
Posts: 2066
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| Posted: 05 May 2006 at 6:37am | IP Logged |
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Personally, I think you're jumping the gun just a bit and you're in over your head a little. Go back to the start of subnetting and re-read and do the simpler exercises first as you're still not grasping it just yet.
Also check out my favorite subnetting website Ralph's Web Site Tutorials
Subnetting will get easier, but it takes a lot A LOT of practice to the point you'll be dreaming about subnetting questions and techniques and suddenly, a bright light will appear and you'll come outta the fog and see it all clearly 
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Wildcat_Dude
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| Posted: 05 May 2006 at 6:40am | IP Logged |
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Oh and if you don't have it, I also recommend Todd Lammle's 5th Edition by Sybex as I love his method of learning how to subnet quickly in your head (for me it worked).
Once you've reviewed the materials, look again at the question you asked us (or your book) and see can figure it out. Remember on the ccna exam, this is one of you questions you'll have to answer in 30 seconds or less (if you want to answer *all* the questions in the time alloted on the exam).
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networxing
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Joined: 05 May 2006
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| Posted: 05 May 2006 at 7:07am | IP Logged |
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"Remember on the ccna exam, this is one of you questions you'll have to answer in 30 seconds or less (if you want to answer *all* the questions in the time alloted on the exam)".
Exactly... thats why i want to learn the shortcut method..I know how to solve this question using the anding process...probably i should've mentioned that in my original post..:)
Does anyone know how to solve atleast the first question using the shortcut method using "magic no".
Thanks guys.
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Wildcat_Dude
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| Posted: 05 May 2006 at 8:00am | IP Logged |
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Think of the 256 rule
The 256 rule is a nice subnetting rule.
The 256 rule is simple, you subtract the subnet mask from 256 to get the block size. For example, 255.255.255.240 has a block size of 256-240, or 16. Meaning the first subnet after the zero subnet is 16, second subnet is 32, first subnets broadcast is 31, host ranges 17~30. That’s it, done.
While the 256 rule is simple to do, sometimes it can get tricky. How many hosts can you get on the network this host is on: 172.16.56.34/23. Well, first step, you need to convert /23 into a subnet mask. If you have the chart, this is simple, just find /23. Without the chart, you have to count. 8,16 and 24 are the “even” subnet masks, 255.0.0.0, 255.255.0.0, 255.255.255.0. Since we know 23 is more then 16, and less then 24, we know it’s 255.255.?.0 mask. What you have to do, is know the valid subnet mask numbers by heart, or be able to figure them out: 0,128,192,224,240,248,252,254,255. /8,/16 and /24 all give you a 0 subnet mask (and 255 on the one before). Add one, /9,/17,/25, you get 128 mask. and just keep going on your counting.
Since we know /24 is 255.255.255.0, you can take a short cut, and use one lower mask. /23 is 255.255.254.0. Now we take the 256 - 254, and we have a block size of 2. The first network is 172.16.0.0 and the second one is 172.16.2.0, up to 172.16.56.0, where our host is on.
You may be thinking, well, block size is 2, subtract the network address and the broadcast address, 2-2 = 0, so something is invalid. Well, it would be, if it wasn’t for the 8 host addresses to the left of it. So the address of the next network is 172.16.58.0, so take one step back, the broadcast address of our 56 network is 172.16.57.255, not 172.16.57.0 like you may have been thinking. Valid host ranges on out network is 172.16.56.1~172.16.57.254. Rather then counting them, you know that there’s 2 in the block size of the c part of the address, 56 and 57, and there’s 256 addresses in each of them, so 256*2 = 512, - the network and broadcast, 510 host addresses. Like I said, anything larger then 510 you may need scrap paper, but any masks of /24 or more, this 256 rule is great.
Edited by Wildcat_Dude on 05 May 2006 at 9:22am
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networxing
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Joined: 05 May 2006
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| Posted: 05 May 2006 at 8:40am | IP Logged |
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[QUOTE=Wildcat_Dude] Think of the 256 rule
The 256 rule is a nice subnetting rule.
Thanks very much for your detailed and informed reply. Much apprieciated!!!
ok applying the same 256 rule on this example...
140.1.1.1/29
29 = 255.255.255.248
therefore 256-248 = 8
Using "8" I get the subnet no as 140.1.1.8 (incorrect), obviously thats not what you get when you do the anding process. With anding you get 140.1.1.0 as subnet no and 140.1.1.7 as broadcast address(correct).
back to the 256 rule.....if i take the third octet then i get..
256-255 = 1
I get 140.1.1.0 as the subnet which is correct, but the broadcast will be 140.1.1.255 which is incorrect.
The only way i can get the correct broadcast is by doing
0+8(256-248)-1 in the last octet.
I figure that when the corresponding octet value in the IP address is less than the magic no then we simply replace it with zero and use the magic no to get the broadcast address,no of subnets etc. Is that correct?
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Wildcat_Dude
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| Posted: 05 May 2006 at 9:27am | IP Logged |
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What class address is 140.1.1.1? You already said its a class B address.
In this class, what octects indictate the network portion of this address?
Now think about it again. That's why I suggested you re-read at the beginning. I don't want you to get overwhelmed. I promise, the light will come on eventually
(and Todd Lammle's explanation is far better than mine is)
Edited by Wildcat_Dude on 05 May 2006 at 9:37am
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Detectorist
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| Posted: 05 May 2006 at 1:27pm | IP Logged |
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Hey Networxing:
Thanks for asking the question. I'm also studying for the CCNA and I think I have subnetting down pat.
Your basic network is 140.1.1.1
This is a Class B address.
Therefore the 140.1 is your Network part and the last 2 octets are the host part. When subnetting it's the host section that gets changed, borrowed, and added to.
140.1.1.1/29 mask is 255.255.255.248, which is correct.
Block size is '8', which is also correct.
Remember, since your 3rd octet is 255, it is incremented by exactly 1.
Your first subnet is 140.1.0.0
In order to find your second subnet simply increment the last octet by 8. So the second subnet would be 0.8 or 140.1.0.8 . With me so far?
The last subnet possible using 0 in the 3rd octet is 0.248, the first host for the last subnet then is 0.249, last host is 0.254, broadcast is 0.255. Every subnet gets 6 hosts.
the next subnet after the 0.254 is found by incrementing the 3rd octet by 1. Therefore it is 1.0, then 1.8, 1.16, 1.24....then 2.0, 2.8, 2.16......255.0, 255.8
With a class A address the first subnet would be 0.0.0 then 0.0.8 and so forth.
I hope I have helped you since many books don't really explain it.
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