@DeusQain Hey Qain! would you be willing to help me study for my networking exam? I have midterms this week and would love the extra help from someone in the field. I am just getting into ipv4 ANDing and am having a bit of trouble with it, not the ANDing itself that is fairly easy. Also if you have any tips on remembering the OSI model I would be greatly appreciative.
Thanks for your time!
Dragon
I cant remember the last two remembering words in this sequence i was told completely but it went something like this
All
People
Seem
To
Need
Doughnut
Pizza
I dont know why doughnut pizza thats just what i remember xD theres also Data Processing :p
or
People Dont Need Those Stupid Packets Anyway
(edit)
Oh and if its not obvious thats
Physical layer 1
Data link layer 2
Network layer 3
Transport lays 4
Session layer 5
Presentation layer 6
Application layer 7
(end edit)
What is it about ANDing your not getting?
haha nice, it wasn't the anding itself, and I cant seem to put it to words. what we are doing is, looking at an IP address, given a number of hosts or networks needed, and then seeing how many networks/hosts can be assigned to that ip address
I have no problem with class C IP addresses its when I get into class B and A that I start getting a bit confused
So heres something I got from some network guy sometime somewhere in the past. Maybe this is useful?
(finds notebook)
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 64 32 16 8 4 2 1
-----------------------------------------------------
128 192 224 240 248 252 254 255
-----------------------------------------------------
A 1 2 3 4 5 6 7 8 255
B 9 10 11 12 13 14 15 16 255
C 17 18 19 20 21 22 23 24 255
D 25 26 27 28 29 30 31 32 255So that took ages to write out.
I was never told exactly how to use this but figured it out for the most part. Its basically a table that will allow you to quickly come up with how many hosts are in a network or make a network with a specific number of hosts.
So for example (really simple one) say you want a network with 2 addresses, say for connecting a single server to a router.
(ive no good way of highlighting the right parts)
you want 2 addresses so you look at the second line, you choose 4 (4 ip addresses , 2 usable);
The third line is the mask for that part of the subnet mask, the next section is basically your CIDR number. You know that with only two addresses your only going to change the last part of the mask so the rest will use 255 (on the side as a reminder of sorts).
The last line (D) you just go directly down and get to 30 which is your CIDR mask. the actualy subnet mask number for the fourth oct is line 3 (252)
So your mask for a 2 IP network is 255.255.255.252 or /30
--
Another example, say you get 10.0.0.0/12 how many addresses
/12 ends up being int he B range. It has a subnet mask of 255.240.0.0 and a new network starts after every 16 addresses in the second oct.
the first range would be 10.0.0.1 - 10.15.555.554
the second would be 10.16.0.1 - 10.31.255.254
10.32.0.1 - 10.46.255.254
(remembering that ive excluded the two unusable addresses in each range)
Not sure if thats helpful or answers what your asking, its been a little bit since ive done networking.
This is a really good thing to memorize, it is fairly simple and can save a lot of time that you might waste trying to figure out how many hosts a network has or what a strange subnet mask would be. My advice if they give you paper to work stuff out on write this down immediately and refer to it as often as possible, Also find the CCNA 1 exams online like THIS and go through as many of the exams as you can. They help alot.
That's exactly what I did, take a few minutes to write it down and it saves a lot of time later.
I wish I had advice that would be on par with these two.
Unfortunately, I process all that data in my head in Binary.
So, when people say, 192.168.25.23/24
I think
11000000.10101000.00011001.00010111/11111111.11111111.11111111.00000000
Network is 192.169.25.0 with a netmask 255.255.255.0 with a broadcast of 192.168.25.255
Leaves 254 host addresses
Now, if you went class A, 10.24.224.125/8
00001010.00011000.11100010.01111101/11111111.00000000.00000000.00000000
This tells me that the network is 10.0.0.0 with a netmask of 255.0.0.0 broadcast is 10.255.255.255
Host addresses are a lot. 16,777,214
Now if we want to start with VLSM, we get into the fun world.
say we had 10.255.23.18/20
To make this REALLY simple we just put them on top of each other. and bam!
00001010.11111111.00010111.00010010
11111111.11111111.11110000.00000000
00001010.11111111.00010000.00000000 <--- here is our network 10.255.16.0
00001010.11111111.00011111.11111111 <---Our broadcast address is 10.255.31.255
Which says our usable host space is 10.255.16.1<-->10.255.31.254 (4,096)
As for the OSI model, I'm down with the "People Don't Need Those Stupid Packets Anyway.
ahh guess that comes with years of practice still useful information though! however I do have a question about the /8 you added to the end of the IP is that the number of networks? currently doing a bit of HW :p
I chose e because it was closest to the 5 needed hosts/networks however something just doesn't seem to be adding up right, I am looking through my notes and must have missed that part in particular. @bladezz @Eden
The number 4,096, is the number of hosts in the network.
204.175.220.0 <--- Network number 1.
a. 11111111.11111111.11111111.10000000
This one gives you a single additional network. 204.175.220.0 and 204.175.220.128
b. 11111111.11111111.10000000.00000000
This one is a supernet, and it's network address would be 204.175.128.0 (your InterNIC IP, would be a host not a Net ID)
c. 11111111.11111111.11111111.11000000
This one will give you 3 additional networks 204.175.220.0 204.175.220.64 204.175.220.128 204.175.220.192
d. 11111111.11111111.11111111.11100000
This one gives you 7 additional networks:
204.175.220.0 204.175.220.32 204.175.220.64 204.175.220.96 204.175.220.128 204.175.220.160 204.175.220.192 204.175.220.224
e. 11111111.11111111.11111111.11110000
This one!! Gives you 15 additional networks
204.175.220.0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240
Since Qain never mentioned. The /8 is the (Classless Inter-Domain Routing) CIDR mask, its basically the shorthand for subnet masks.
A /8 is a standard class A subnet mask as in 255.0.0.0 or as Qain likes 11111111.00000000.00000000.00000000