Now I remember why I hated that class as a sophomore

Destonomos · Jan 7, 2009

Well I just got to the subnetting part in the CCNA book and I've been on pages 264-269 for the last 2 hours at work reading over subnetting again. Good lord I hate it. I remember taking an exam on this as a sophomore and almost failing. Now I remember why...

Gott · Jan 7, 2009

Is there anything specific to subnetting that you're not getting that we could possibly help with?

QwertyJuan · Jan 7, 2009

What do the 255's stand for??

Destonomos · Jan 7, 2009

Gott said:
Is there anything specific to subnetting that you're not getting that we could possible help with?

Thanks for wanting to help me but I think I have it down it just takes so long to do.

Let me make sure I got this right. Check what I write to make sure I have it down right.

Lets say I want to do buisness in 20 countries with 5 offices per country subnetting by office.

20x5 = 100 subnets

2^6 = 64 so that won't work
2^7 = 128 there we go

I want to use a class C address so we are going to start with 192.168.1.0

I also need to barrow 7 bits so my subnet mask is going to be:

255.255.255.254

7-2 = 5

2^5 = 32

so....
Subnet ID...........Broadcast
192.168.1.32 - 192.168.1.63
192.168.1.64 - 192.168.1.95
192.168.1.96 - 192.168.1.127
192.168.1.128 - 192.168.1.159
192.168.1.160 - 192.168.1.191
192.168.1.192 - 192.168.1.223

All addresses inside are Host Id's

Is that right?

XOR != OR · Jan 7, 2009

Looks like you are getting hosts and networks confused. From the problem description, it looks like you want 100 subnets. Which means you'll need 7 bits for the network, leaving 1 for the host. Given the assumption that we have a single class C to work with, it's impossible to complete the assignment.

First we need to know how many hosts per subnet, then we can build out the rest of the table.

ElvisG · Jan 7, 2009

XOR is right. You have to know how many host you will be needing first. Then you can find out how many subnets you will need, not the other way around.

Destonomos · Jan 7, 2009

ElvisG said:
XOR is right. You have to know how many host you will be needing first. Then you can find out how many subnets you will need, not the other way around.

yeah I think I'm starting to get it but not totally yet. I'm reading some how to's on some ccna forums now.

XOR != OR · Jan 7, 2009

Destonomos said:
yeah I think I'm starting to get it but not totally yet. I'm reading some how to's on some ccna forums now.

If it makes you feel any better, once you get it, you get it. I learned this at the start of last year and then didn't touch it for 8 months, but it all came back to me once I started doing it again.

atomiser · Jan 7, 2009

XOR != OR said:
If it makes you feel any better, once you get it, you get it. I learned this at the start of last year and then didn't touch it for 8 months, but it all came back to me once I started doing it again.

^^^ this!

it took me weeks of being shown different ways of doing it, and just practicing it over and over and over again...and then it just 'clicks'. i still struggle a bit doing it in my head, but i have the principles down so give me a bit of paper and i can have it done in no time!

i was helping out one of the guys at work, and hadn't touched it for a couple of months, and i too had it back within a few minutes.

persevere, it's worth it!

edit: and, obviously, just ask questions if you get stuck!

Destonomos · Jan 8, 2009

I'm at work again today but once I get some tickets done and get some free time I'll show another example to see if I'm "getting it" I think I get it now.

Destonomos · Jan 8, 2009

ok so let me try this again....

Lets say you have a class C address and you want 15 users per subnet.

2^3=8
2^4=16 but it won't work because you need 1 for broadcast and ID
2^5=32 there we go

so...

11111111.111111111.11100000.00000000
255.255.224.0 subnet address

since the last 1 ends on the 32 that is my hop count.
so with lets say an ip address of 192.168.1.0 here would be my ranges

ID...................Broadcast
192.168.1.0 - 192.168.1.31
192.168.1.32 - 192.168.1.63
192.168.1.64 - 192.168.1.95
192.168.1.96 - 192.168.1.127
192.168.1.128 - 192.168.1.159

and so on...

the - is all the addresses in between that are possible host IP's

I have two questions. Here they go.

So that answers if you have a certain number of host IP's but what if you want a certain number of host IP's AND a certain number of subnets? What then? Also if I get this correctly a classful set of IP's does the 255.255.255.0 for C and 255.255.0.0 for B and 255.0.0.0 for A addresses and the blocks of addresses for each A,B, and C.

In classless the IP ranges in classful don't matter? Like in my example I used a class B address (noted with the 192 at the beg.) but I wanted a class C address so I just stuck a 1 at the end of it.

Ockie · Jan 8, 2009

I hated subnetting when I was first introduced to it, now it's used daily. It's not uncommon for somone to ask me for a /28 say a /26 network.

My advice is to suck it up and really try and understand it because it's used more than most people realize.

Destonomos · Jan 8, 2009

Ockie said:
I hated subnetting when I was first introduced to it, now it's used daily. It's not uncommon for somone to ask me for a /28 say a /26 network.

My advice is to suck it up and really try and understand it because it's used more than most people realize.

um... ok? Not trying to be rude but that is what i'm trying to do. (see post before yours)

fantasticdan · Jan 8, 2009

Destonomos said:
ok so let me try this again....

Lets say you have a class C address and you want 15 users per subnet.

2^3=8
2^4=16 but it won't work because you need 1 for broadcast and ID
2^5=32 there we go

so...

11111111.111111111.11100000.00000000
255.255.224.0 subnet address

since the last 1 ends on the 32 that is my hop count.
so with lets say an ip address of 192.168.1.0 here would be my ranges

ID...................Broadcast
192.168.1.0 - 192.168.1.31
192.168.1.32 - 192.168.1.63
192.168.1.64 - 192.168.1.95
192.168.1.96 - 192.168.1.127
192.168.1.128 - 192.168.1.159

and so on...

the - is all the addresses in between that are possible host IP's

I have two questions. Here they go.

So that answers if you have a certain number of host IP's but what if you want a certain number of host IP's AND a certain number of subnets? What then? Also if I get this correctly a classful set of IP's does the 255.255.255.0 for C and 255.255.0.0 for B and 255.0.0.0 for A addresses and the blocks of addresses for each A,B, and C.

In classless the IP ranges in classful don't matter? Like in my example I used a class B address (noted with the 192 at the beg.) but I wanted a class C address so I just stuck a 1 at the end of it.

Hey Mate, you're getting a buit mixed up here.

11111111.11111111.11111111.11111111 = 255.255.255.255 = /32 (single static IP, only really used on Public)
11111111.11111111.11111111.11111110 = 255.255.255.254 = /31 (2 static IPs, again only used really on public because we have only a network and a broadcast)
11111111.11111111.11111111.11111100 = 255.255.255.252 = /30 (now were getting somewhere, 2 Host Ips)
11111111.11111111.11111111.11111000 = 255.255.255.248 = /29 (6 host Ips)
11111111.11111111.11111111.11110000 = 255.255.255.240 = /28 (14 host ips)
11111111.11111111.11111111.11100000 = 255.255.255.224 = /27 (30 host Ips)
11111111.11111111.11111111.11000000 = 255.255.255.192 = /26 (62 host Ips)
11111111.11111111.11111111.10000000 = 255.255.255.128 = /25 (126 host Ips)
11111111.11111111.11111111.00000000 = 255.255.255.0 = /24 (254 host ips)

11111111.11111111.11111110.00000000 = 255.255.254.0 = /23 (512 host Ips)
11111111.11111111.11111100.00000000 = 255.255.252.0 = /22 (1022 host Ips)
11111111.11111111.11111000.00000000 = 255.255.248.0 = /21 (2046 host Ips)

You're not likely to need more than that at the moment.

Your correct in that 192.168.1.0-192.168.1.31 is a class C. The network is 192.168.1.0 and the broadcast is 192.168.1.31. Only the 30 hosts .1-.31 are usable.

However unless your using VLSM you should drop the first and last subnets so 192.168.1.0-192.168.1.31 and 192.168.1.224-192.168.1.255 aren't used either.

Your 3 classless ranges are 10.0.0.0 for A, 172.16.0.0-172.31.255.255 (i'm not 100% sure on this one) for B and 192.168.0.0. for class C. All other IPs are Public (classful)

Motley · Jan 8, 2009

A real quick thing I always do, when determining how many hosts you need, and what subnet to apply is:

I need 30 hosts, ok take (256 - 30) + 1 (broadcast) +1 (network)

So... 256 - 32 = 224 = 255.255.255.224 subnet mask, you can do that in your head!!!

Ockie · Jan 9, 2009

Destonomos said:
um... ok? Not trying to be rude but that is what i'm trying to do. (see post before yours)

Tried a little morale boosting but I guess some people rather be rude about it

Destonomos · Jan 9, 2009

Ockie said:
Tried a little morale boosting but I guess some people rather be rude about it

You can't get mad. I said not trying to be rude.

Now I remember why I hated that class as a sophomore

Destonomos

[H]ard|Gawd

Gott

Supreme [H]ardness

QwertyJuan

[H]F Junkie

Destonomos

[H]ard|Gawd

XOR != OR

[H]F Junkie

ElvisG

Limp Gawd

Destonomos

[H]ard|Gawd

XOR != OR

[H]F Junkie

atomiser

Gawd

Destonomos

[H]ard|Gawd

Destonomos

[H]ard|Gawd

Ockie

* Self Proclaimed Storage King *

Destonomos

[H]ard|Gawd

fantasticdan

n00b

Motley

2[H]4U

Ockie

* Self Proclaimed Storage King *

Destonomos

[H]ard|Gawd

Now I remember why I hated that class as a sophomore

[H]ard|Gawd

Supreme [H]ardness

[H]F Junkie

[H]ard|Gawd

[H]F Junkie

Limp Gawd

[H]ard|Gawd

[H]F Junkie

Gawd

[H]ard|Gawd

[H]ard|Gawd

*** Self Proclaimed Storage King ***

[H]ard|Gawd

n00b

2[H]4U

*** Self Proclaimed Storage King ***

[H]ard|Gawd

* Self Proclaimed Storage King *

* Self Proclaimed Storage King *