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 Post subject: Subnetting Help
PostPosted: Fri Feb 17, 2012 5:32 pm 
Little Foot
Little Foot

Joined: Sat Nov 05, 2011 1:02 pm
Posts: 151
Hopefully someone can help me with subnetting. I found some good articles and i have the majority of it down on how to do it but i'm still rusty in some places.

I did learn subnetting about 5 years ago to complete my CCNA but haven't used it since then so now that i'm trying to finish up my MCSA i have to relearn subnetting for test 70-291. Any help would be appreciated as i'm going through subnettingquestions.com right now and a few trick me up.


1. How many subnets and hosts per subnet from 10.0.0.0 /20?
Answer = 4096 subs and 4094 hosts.

----how did they get that answer. I thought i was doing good because i figured out /20 was a 255.255.240.0 mask which means each subnet increments by 16 so i counted subnets like "10.0.0.0 (1), 10.0.16.0 (2), 10.0.32.0 (3), 10.0.48.0 (4)..." and so on by increments of 16 which gave me 17 subs and of course 254 hosts for each subnet (1-255). I have no idea how they gathered 4096 subs and 4094 hosts.

***I'm going through the site now so i'll probably be posting more questions that stump me. THANKS AGAIN****


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 Post subject: Re: Subnetting Help
PostPosted: Fri Feb 17, 2012 5:47 pm 
Little Foot
Little Foot

Joined: Sat Nov 05, 2011 1:02 pm
Posts: 151
Alright two more which got me puzzled.

-What is the valid host range from 172.31.1.134 /28?
Answer = 172.31.1.129 - 172.31.1.142
**I worked it out to be 172.31.1.128-172.31.1.144. I see my mistake as from number to number should be 14 hosts but remember this question as it will affect my reasoning for the next one pertaining to it being a Class B address yet the subnet 255.255.240.0 is only causing me to make adjustments to the last octet**

-What is the broadcast of 10.190.208.0 255.255.240.0?
Answer = 10.190.223.255
**I answered incorrectly of 10.190.208.254. What the hell? This is the exact same subnet (255.255.240.0) as the first question and the only difference is that this is a Class A address. Since the Class B address above had me only changing the last octet (in the d class range) so i assumed it was the same for this question as its the same subnet. Now it seems like if i have a Class A address i should always change the 3rd octet while Class B address's have me change the 4th octets. Makes no sense to me that its skipping an octet. I would assume by logic that if i have a Class A address such as this on (10.190.208.0) then i have free range to change numbers after the first octet (the 190.208.0) and if i had a Class B address then i could change the third octet over. In these two questions it turned out that a Class A can only change the third octet and on and Class B changes only 4th octet.

Please tell me how stupid i am and where i am going wrong? Am i overthinking it?


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 Post subject: Re: Subnetting Help
PostPosted: Mon Feb 20, 2012 8:07 pm 
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Masterchi wrote:
1. How many subnets and hosts per subnet from 10.0.0.0 /20?
Answer = 4096 subs and 4094 hosts.

----how did they get that answer. I thought i was doing good because i figured out /20 was a 255.255.240.0 mask which means each subnet increments by 16 so i counted subnets like "10.0.0.0 (1), 10.0.16.0 (2), 10.0.32.0 (3), 10.0.48.0 (4)..." and so on by increments of 16 which gave me 17 subs and of course 254 hosts for each subnet (1-255). I have no idea how they gathered 4096 subs and 4094 hosts.


10.0.0.0 is a Class A network. You are correct in counting out increments of 16, but you have to keep going throughout the entire 10 network (10.0.0.0 through 10.255.255.255). 10.0.0.0, 10.0.16.0... 10.1.0.0, 10.1.16.0 ... 10.2.0.0, 10.2.16.0... etc.

Now for hosts. Let's use subnet 0 as an example. Here we have 10.0.0.0 through 10.0.15.255 (one less than the start of the next subnet, 10.0.16.0). This gives us 4096 addresses (256 * 16), of which 4094 are usable. I'm guessing you were incorrectly counting from 10.0.0.0 through 10.0.0.255.

It's late and I'll come back for the second post tomorrow.


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 Post subject: Re: Subnetting Help
PostPosted: Wed Feb 22, 2012 5:37 am 
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Masterchi wrote:
Alright two more which got me puzzled.

-What is the valid host range from 172.31.1.134 /28?
Answer = 172.31.1.129 - 172.31.1.142
**I worked it out to be 172.31.1.128-172.31.1.144. I see my mistake as from number to number should be 14 hosts but remember this question as it will affect my reasoning for the next one pertaining to it being a Class B address yet the subnet 255.255.240.0 is only causing me to make adjustments to the last octet**


In this question, you don't really need to worry about classes. A /28 gives you a block of 16 addresses and that's all you really need to be concerned with here.


Masterchi wrote:
-What is the broadcast of 10.190.208.0 255.255.240.0?
Answer = 10.190.223.255
**I answered incorrectly of 10.190.208.254. What the hell? This is the exact same subnet (255.255.240.0) as the first question and the only difference is that this is a Class A address. Since the Class B address above had me only changing the last octet (in the d class range) so i assumed it was the same for this question as its the same subnet. Now it seems like if i have a Class A address i should always change the 3rd octet while Class B address's have me change the 4th octets. Makes no sense to me that its skipping an octet. I would assume by logic that if i have a Class A address such as this on (10.190.208.0) then i have free range to change numbers after the first octet (the 190.208.0) and if i had a Class B address then i could change the third octet over. In these two questions it turned out that a Class A can only change the third octet and on and Class B changes only 4th octet.

Please tell me how stupid i am and where i am going wrong? Am i overthinking it?


Again, forget about classes on this one, that's only going to add extra complexity you don't need to worry about here. A /20 gives you an increment of 16 in the third octet, so the range of this subnet will be 10.190.208.0 through 10.190.223.255 (or one address less than the beginning of the next subnet, 10.190.224.0 (208 + 16 = 224)). The broadcast address for this network is the last valid address, or 10.190.223.255.

Until you get all this back committed to memory, it helps to have a cheat sheet handy. Go here and grab the IPv4 Subnetting Cheat Sheet (I kept a lot of these handy when I was working on the CCNA). You can also go here to check your work.


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 Post subject: Re: Subnetting Help
PostPosted: Wed Feb 22, 2012 5:27 pm 
Little Foot
Little Foot

Joined: Sat Nov 05, 2011 1:02 pm
Posts: 151
Thanks for that, i think i'm slowly starting to get it now. Much appreciated.


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 Post subject: Re: Subnetting Help
PostPosted: Thu Feb 23, 2012 5:57 am 
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Masterchi wrote:
Thanks for that, i think i'm slowly starting to get it now. Much appreciated.


No problem. I like to hit a question or two like this on forums once in a while so I don't forget it myself since I don't use it very much otherwise. :)


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