subnetting on 291?
nel
Member Posts: 2,859 ■□□□□□□□□□
just a quick q?
do you have to know this like the back of your hand? also does anything like vlsm etc come into the exam or not?
cheers
do you have to know this like the back of your hand? also does anything like vlsm etc come into the exam or not?
cheers
Xbox Live: Bring It On
Bsc (hons) Network Computing - 1st Class
WIP: Msc advanced networking
Bsc (hons) Network Computing - 1st Class
WIP: Msc advanced networking
Comments
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doom969 Member Posts: 304Hi,
I got a few questions that needed subnetting on 291. Like, calculating the correct subnet or mask was a part in finding the good answer. I also got 2 sims were subnetting was involved, so you had to be pretty sure, or you'd miss the sim. I wouldnt say that you have to know it like the back of your hand, but you definitly have to be able to do it. As for Variable length subnets, I did not get it on the test, but its in the Ms-Press Book so i gues you could get it on the exam. Maybe I was just lucky.
GLuck to you
Doom969Doom969
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Scooty70 Member Posts: 2 ■□□□□□□□□□For me I had one question that was related to VLSM
If you intend on doing CCNA then yes know it like the back of your hand and be able to do it in your head. I saw a previous post asking why you would use it and it's something that you will come across in networking, especially in the ISP world and is very useful for consverving addres space. We have for example a /28 assigned by our ISP for things like mail.xyz and vpn.xyz etc
To work this out simply we know that a /24 (Class C) gives you one network and 254 hosts (with the .255 being the broadcast address and .0 being the network address)
In the above example we have a /28 which means we have borrowed 4 bits of the network address range to subnet
We add these 4 bits up
128 + 64 + 32 + 16 = 240
Remembering that if we add up all the bits in an octet it would equal 255 (128 + 64 + 32 + 16 + 8 + 4 + 2 + 1), but we can't forget we actually start at 0, so this gives us a number of 256 to work with (0 to 255 equals 256 numbers)
If we subtract 240 from 256 we end up with 16
Remembering we must have a broadcast and a network address this leaves us with 14 hosts
Using say 192.168.1.0 as an eample this means that
192.168.1.0 is the network address
192.168.1.1 - 192.168.1.14 are the available hosts
192.168.1.15 is the broadcast address
192.168.1.16 is the next network address
192.168.1.17 - 192.168.1.30 are the availale hosts
192.168.1.31 is the boradcast address
and so on
We can work out with all certainty that .0, .16, .32, .48, .64 etc are the networks (just keep adding 16)
Subtract 1 will always be the broadcast address and plus one will be the first available host
A question states that you need x hosts or x number of networks, what is the closest power of 2
Example is we need 20 hosts, closest power of 2 is 5, 2 times 2 five times equals 32, the same to the power of 4 is only 16. We need to then steal 5 bits from the network, add the last 5 bits in the octet up and subtract from 256 and this is the / VLSM you need, in this case a /27 or 255.255.255.224 (256 less 32 equals 224)
I hope that makes some sort of sense, just remember it's all to the power of 2
Scott -
nel Member Posts: 2,859 ■□□□□□□□□□cheers for the replies.
i know how to subnet and how to use vlsm but i cant do it in my head lol!
so i was just wondering how much MS push it.
cheers againXbox Live: Bring It On
Bsc (hons) Network Computing - 1st Class
WIP: Msc advanced networking