Subnetting...Shortcuts 8

[bobbyforhire]

Ok so as i stated before i think i'm getting good on this..So good that i have decided to put all of my cisco equipment together and set it up right.

now, i will have a connection from my pix to my router. Just to make sure this is the correct standard for my subnetting.

Because I only need two hosts i have selected the following.

172.0.0.0 as my base. I will asign my pix 172.0.0.1 and my router 172.0.0.2 both using a subnet 255.255.255.252

Can anyone confirm that this is the correct subnet to use. Also to increase my time on the test, does anyone have any shortcuts that you use?


172.0.0.0
CLASS B
2 Host needed
128 64 32 16 8 4 2 1
i need 2 bits to get 2 hosts.

*Tip

HOSTS = 0'S (RIGHT TO LEFT)
NETWORK = 1'S (LEFT TO RIGHT)


255.255.0.0 = 11111111.11111111.00000000.00000000
turn that into 11111111.11111111.11111111.11111100

255 - 3 = 252
Lowest INC = 4

172.0.0.0 - 172.0.0.3 ( Giving me 1 and 2 to use)

 

[bobbyforhire]

Also besides just using good math, does this increase performance in any way? I know this sounds dumb but does it ever pay off to do it the right way?

 

[burtsbees]

Well, just remember that broadcast addresses are always odd numbers in the last octet, and network addresses are always even.
You can subnet any private range to make a point-to-point link. 172.0.0.0 is not a private IP address (RFC1918), so be careful if you're hooking this up to the internet.
And no, this does not increase performance.

Burt

 

[bobbyforhire]

Ah...forgot that 172 is the only one out of the three that you have to "chop" it up 172.16.x.x - 172.32.255.255

I'm not sure i understand what you mean by broadcast addresses are always odd numbers in the last octet and network addresses are always even.

thnx

 

[maczen]

Let's try this..

You are the admin for your company and have 3 LANS needing a minimum of 200 users per LAN. You also have three p-to-p WANS that you will need to address. You have been provided the address 172.16.32.0 /22. Conserve the addresses as much as possible.

This is the type of thing that you will want to be able to do with ease. Honestly, these are not very difficult. First, determine how to subnet. You need 3 LANS... To get that you will need to borrow 2 network bits (2n=4). Note that you need a min of 200 users per LAN and you will have 254 in this example. So here you are...

172.16.32.0 /24
Network, Subnet, Host
NNNNNNNN.NNNNNNNN.NNNNNNSS.HHHHHHHH
172.16.00100000.00000000 or 172.16.32.0 (Range 1-254)
172.16.00100001.00000000 or 172.16.33.0 (Ditto)
172.16.00100010.00000000 or 172.16.34.0 (Ditto)
172.16.00100011.00000000 or 172.16.35.0 (Will use for WANS)

WANS are Point-to-Point so we only need two usable addresses per subnet... We are beginning with 172.16.35.0 /24.. We need three /30 subnets...
SO...
172.16.00100011.00000000 or 172.16.35.0 /24 becomes
NNNNNNNN.NNNNNNNN.NNNNNNNN.SSSSSSHH
(We would actually be wasting an entire subnet on /30's here but that is fine for this example.. and we could actually address this way!)

Now we have...
172.16.35.0 /30
172.16.35.4 /30
172.16.35.8 /30
as our three point-to-point WAN subnets with many more available...

If you look at the binary you will see why...
172.16.35.0
172.16.00100011.00000000 .0
172.16.00100011.00000001 .1
172.16.00100011.00000010 .2
172.16.00100011.00000011 .3
---------------------------
172.16.00100011.00000100 .4 etc.

You can also check yourself quickly by subtracting...
256
-252
---
4

OK.. It is getting late on a Saturday night and I have just powered up my lab.. I am REALLY lucky that the Mrs. is in school as well and has to study too... LoL

I looked this over twice and believe it all to be accurate but you will definitely hear about it if I missed something.. LoL Hope this helps!!!

B Haines
CCNA, FOI

 

[maczen]

By the way.. you can have sixty-four /30's in the above example to address you various point to point links..

B Haines
CCNA, FOI

 

[burtsbees]

A simple example of the even and odd thing...
Cisco will ask which address is the broadcast address for subnet blablabla...let's say it's 192.168.1.0/28
The choices might be
A)192.168.1.0
B)192.168.1.255
C)192.168.1.16
D)192.168.1.14
E)192.168.1.15
F)192.168.1.1

Since they are asking for a broadcast address, we can eliminate choices A, C and D right off the top because the last octet is an even number. So you tell me...what's the answer?
Another short cut---let's do the same example...192.168.1.0/28
Every network address will be multiples of 16...16, 32, 48, 64, etc. So .0 is the first subnet, .16 the second, .32 the third, .48 the fourth, etc.
192.168.1.0
192.168.16.0
192.168.32.0
192.168.48.0
The broadcast for each previous subnet will be the number right before...so we have subnet 4---192.168.1.48---the broadcast for the subnet before it is 192.168.1.47, one before .48. So the broadcast for 192.168.1.48---take the next network address and subtract .1---192.168.1.64 minus 1=192.168.1.63---the broadcast for 192.168.1.48.
This works for ANY mask---let's take a /20...

10.10.0.0/20
Well, the same exact rules apply, except since the network boundary falls in the third subnet, then the rules apply to the third subnet, not the fourth.
10.10.0.0/20, the first subnet is 10.10.0.0, second is 10.10.16.0, the third is 10.10.32.0, etc. Broadcast for the second is 10.10.31.255.
Here's a guy that made a blogspot---a very good one, easy to follow, will help immensely on the CCNA...

http://subnettingmadeeasy.blogspot.com/

Burt

 

[SteveTheGeek]

just remember that broadcast addresses are always odd numbers in the last octet, and network addresses are always even."

Nice, simple, and very useful, thanks!!

And bobbyforhire, for the exam and also (in my experience) in practice it's handy to remember that a /30 (.252 mask) gives you two usable IP addresses and is generally used by ISPs at the demarc between their gear and yours.

 

[burtsbees]

Well, the same exact rules apply, except since the network boundary falls in the third subnet, then the rules apply to the third subnet, not the fourth."
Third OCTET, not subnet. Sorry, guys...

Burt

 

[maczen]

This is good stuff!

B Haines
CCNA, FOI

 

[bobbyforhire]

Ok, I just finished up the Subnetting and my time on each problem is around 2:30. If anyone out there can toss me a few test questions i would really appreciate it. Ill reply with my answer and show my work!

 

[bobbyforhire]

oh btw...who here is star happy?

 

[Cluebird]

Subnetting is just doubling and halving. When you move the bit boundary in a mask to the right, you double the number of networks but cut the size in half. When you move the bit boundary to the left, you halve the number of networks but double their size. Within an octet, the actual IP addresses are determined by the block size (how many host bits are available in the octet). Subnetting is one of the easiest topics if you don't get hung up on math and just look at patterns. The key to address space is the mask.

 

[Cluebird]

Try these...I will intentionally throw a few curves at you.

1. 208.23.23.0 Subnet for 15 networks with max hosts. What is the second usable host on the third usable network?

2. 212.44.53.0 Subnet for 15 hosts with max networks. What is the broadcast for the 5th usable subnet.

3. 230.55.55.0 What mask gives 4 networks with max hosts?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.192.0
D. None of the above

4. What type of address is this: 194.255.172.64/27?

5. What type of address is this? 172.56.33.255/23?

6. Is 197.23.15.160/28 usable for my

http://WWW server?

Why or why not?

Enjoy

 

[bobbyforhire]

Due to me being at work i can do 1 at a time for right now. Let me know if this one is right.


1. 208.23.23.0 Subnet for 15 networks with max hosts. What is the second usable host on the third usable network?

Answer:
208.23.23.34 (First usable is 208.23.23.33)

Work:
208.23.23.0
15 networks
class c

What is the Second usable host on the third usable network


15 = 4 bits

128 64 32 16 8 4 2 1


11111111.11111111.11111111.11110000

255
-15
-----
240

Lowest INC = 16

= 255.255.255.240

------------------------------------------------------------
208.23.23.0 - 208.23.23.15
208.23.23.16 - 208.23.23.31
208.23.23.32 - 208.23.23.47
208.23.23.48 - 208.23.23.48

 

[Cluebird]

Got it. As a reference, I did all four of these in less than 60 seconds.

 

[bobbyforhire]

I could be wrong but there are 6


Ok looking at number 2...I think it's a trick question you only want 15 hosts...i can get there in two networks. Am i right?

 

[maczen]

By saying "subnet for 15 hosts" and "max networks" he is saying that you will create as many subnets as possible so long as each can accommodate fifteen hosts.

212.44.53.0 /24
NNNNNNNN.NNNNNNNN.NNNNNNNN.SSSHHHHH

That gives you 8 subnets that can accommodate 30 hosts each...

NOTE: That x.x.x.SSSSHHHH Will NOT work as that would give you 16H - 2H due to one network address and one broadcast address = 14H where the requirement was 15...

Now... What are the subnet ranges for this design?
212.44.53.0 - x.x.x.x
x.x.x.x - x.x.x.x
x.x.x.x - x.x.x.x
x.x.x.x - x.x.x.x
x.x.x.x - x.x.x.x
x.x.x.x - x.x.x.x
x.x.x.x - x.x.x.x
x.x.x.x - x.x.x.x


B Haines
CCNA R&S, ETA FOI

 

[bobbyforhire]

see thats what had me confused was "max networks".

Ok so..

212.44.53.0 Subnet for 15 hosts with max networks. What is the broadcast for the 5th usable subnet.
15 hosts (even number add 1)
16 hosts
255.255.255.0
11.11.11.00000000
12481632
11.11.11.11000000
255.255.255.192
(im starting to see a pattern instead of 1's and 0's)

Now here is where i start getting iffy. In the event that you reach 255 you should move the next octet up one. The broadcast address is the last address and the network address is the first address in the network. let me know if i'm right on this one


212.44.53.0 - 212.44.53.31
212.44.53.32 - 212.44.53.63
212.44.53.64 - 212.44.53.127
212.44.53.128 - 212.44.53.255
212.44.54.0 - 212.44.54.31 <-5th Broadcast

 

[bobbyforhire]

oh, i meant odd number add one.