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TOPIC: Subnetting in 5 Mins.... really great stuff...

Subnetting in 5 Mins.... really great stuff... 7 years 5 months ago #30017

  • cscrtr
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Hi all dear friends,

I am not sure who has written this.. just found on net...when I saw few posts here regarding subnetting then really wanting to share this great stuff with you all.... Now I can easily do subnetting by using this method.....Enjoy...

***********************************

You think Subnetting is a beast?
You think you have to be Superbrain to understand it?

You are wrong!

Here the step-by-step course.
After reading and some self-training, you should be able to fix Subnetting-Questions in CCNA Exam
without any problems in a snatch.
Relax!

What is a Subnetmask?

With Subnetmasks, we can divide an IP-Address in network-part and in host-part.
A given IP-Network can be divided in smaller parts. Each of this smaller parts is called a "Subnet".

If we for example have the network

192.168.10.0 255.255.255.0
We have here ONE Class C - network, with 253 useable IPs for Client-PCs.


The useable IP Range of this network is

192.168.10.1 - 192.168.10.254

The very last IP of each Subnet is called Broadcast-Address.
This address is in that example 192.168.10.255 and its NOT useable for host-pcs.

If we want to divide this network in two parts, we must use subnetting.

With Subnetmask 255.255.255.128 we would divide the network in two parts.

192.168.10.1 - 192.168.10.127

192.168.10.128 - 192.168.10.255

Image File

img209.imageshack.us/img209/6452/subnettingdg5.jpg



So in this example, BEFORE we had one big Network.
With the change of the Subnetmask we did divide it in two smaller networks.

First with Subnetmask 255.255.255.0 we had THIS network:
192.168.10.0 >>> This is the "Network-IP" which is NOT useable for Host-PCs

192.168.10.1
192.168.10.2
192.168.10.3
192.168.10.4
192.168.10.5
...
...
...
192.168.10.253
192.168.10.254
192.168.10.255 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs

Now with Subnetmask 255.255.255.128 we have THIS two networks:

First Subnet:

192.168.10.0 >>> This is the "Network-IP" which is NOT useable for Host-PCs

192.168.10.1
192.168.10.2
192.168.10.3
192.168.10.4
192.168.10.5
...
...
...
192.168.10.125
192.168.10.126
192.168.10.127 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs

Second Subnet:

192.168.10.128 >>> This is the "Network-IP" which is NOT useable for Host-PCs

192.168.10.129
192.168.10.130
192.168.10.131
192.168.10..132
192.168.10.133
...
...
...
192.168.10.253
192.168.10.254
192.168.10.255 >>>This is the Broadcast-IP, which is NOT useable for Host-PCs

The Subnetmask defines how big the subnet is.
That means - how many Client-PCs will have place in that subnetwork.

A Subnetmask of 255.255.255.0 means in binary

11111111.11111111.11111111.00000000

So, what do we see?

4 Blocks, divided with a ".". Each of these blocks is also called "octett". Because - each Block has 8 bits.

To be able to do subnet-calculation, we first must understand binary calculation.

Lets take the first block.

The first "1" stands for a 128.

The second "1" stands for a 64.

The third "1" stands for a 32.

The fourth "1" stands for a 16.

The fifth "1" stands for a 8.

And so on. That means:

11111111=255

11110000=240

11100000=224

If we see something like "/24", that means that 24 bits are set to "1", from the left side.

Examples:

/16 = 255.255.0.0 = 11111111.11111111.00000000.00000000

/20 = 255.255.240.0 = 11111111.11111111.11110000

If we would take a subnetmask of 255.255.255.255 that would be

128+64+32+16+8+4+2+1.128+64+32+16+8+4+2+1.128+64+32+16+8+4+2+1.
128+64+32+16+8+4+2+1

and in binary it would be

11111111.11111111.11111111.11111111

Calculation of Subnetmask big enough for a specified number of Hosts

If they ask..

"create a subnet with minimum 10 host IPs"

than

1. calculate a power of two, that is minimum 10

2^3=8. That is not enough
2^4=16 That is higher than 10. Good.


2. Now put the LAST 4 Bits of your subnetmask to 0.

11111111.11111111.11111111.11110000

That is in decimal

255.255.255.240

With THIS Subnetmask, you have minimum 10 Host-Ips in the Subnet, without wasting to much IP-Addresses.

Other example

If they ask

Create a subnet with minimum 70 Host-IPs

1. Calculate a Power of 2 that is MINIMUM 70

2^6=64. Not enough.
2^7=128. Thats higher than 70. Good.

2. Put the LAST 7 Bits of your Subnetmask to 0.

11111111.11111111.11111111.10000000

That is in decimal

255.255.255.128

You have a Subnetmask, with more than 70 Host-IPs.


Calculation what is the Broadcast-IP of a Subnet

When they ask
"There is subnet 172.16.64.0/20. What is the BROADCAST ADDRESS of that Subnet, dude?"



1. Step

/20 meens 255.255.240.0

2. Step

Now analyze the Subnet Oktett to find out the "network-jumps"

240 means 11110000

The LAST of the 1s is under decimal 16. That are our "network jumps"
(128/64/32/16/8/4/2/1)
3. Step

Write down the network-jumps


172.16.64.0 - 172.16.79.255
+16 172.16.80.0 - 172.16.95.255
+16 172.16.96.0 - 172.16.111.255
+16 172.16.112.0 - 172.16.127.255

Because the NEXT Subnet in the example is 172.16.80.0, the broadcast must be 172.16.79.255, cause THAT is the IP BEFORE the next Subnet starts = the BroadcastAddress.





Other example of Broadcast-IP calculation:

If it would be 172.16.64.0 /26

Same procedure

/26 means 255.255.255.192


192 is binary 11000000
The LAST 1 stands under the 64. That are in that example our "net-jumps".

172.16.64.0 - 172.16.64.63 <<<in this example THIS is the broadcastaddress of first subnet
172.16.64.64 - 172.16.64.127
172.16.64.128 - 172.16.64.191



Calculating first and last possible IP for a Host

You have Network 192.168.20.32 /27
The very first IP is reserved for Default Gateway!
What is the first and last valid IP for a Host-PC?

1.Step

/27 is 255.255.255.224

2.Step

224 means 11100000

The LAST 1 is under the 32. That are our "network-jumps" in this example
(128/64/32/16/8/4/2/1)

Valid IPs in that subnet:
192.168.20.33 - 192.168.20.62
(192.168.20.63 is NOT useable, this is the very last IP and so the BROADCAST-IP).
192.168.20.64 <<<this is the network-IP of the NEXT Subnet!

So, because the very first IP is reserved for Default Gateway, our first Host-PC IP would be
192.168.20.34
The very last Host-PC IP would be
192.168.20.62

###################################################################

Here some examples of real exam questions, and step by step solutions:


Given that you have a class B IP address network range, which of the subnet masks
below will allow for 100 subnets with 500 usable host addresses per subnet?
A. 255.255.0.0
B. 255.255.224.0
C. 255.255.254.0
D. 255.255.255.0
E. 255.255.255.224

Solution:

Allways the same game... Like in 5 minute course..

Power of 2 that is minimum 500?
2^7=128
2^8=256
2^9=512 >>voila!

Now - put the last 9 Bits of your Subnetmask to "0"

11111111.11111111.11111110.00000000

That is in decimal
255.255.254.0


=====================================================================
If a host on a network has the address 172.16.45.14/30, what is the address of the
subnetwork to which this host belongs?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.18

Solution:

172.16.45.14/30

/30 means 11111111.11111111.11111111.11111100

The last of the ones stands under the "4". That is our increment or network jump.
172.16.45.0 - 172.16.45.3
172.16.45.4 - 172.16.45.7
172.16.45.8 - 172.16.45.11
172.16.45.12 - 172.16.45.15
172.16.45.16 - 172.16.45.19

As we see, the Ip is in the Range of 172.16.45.12 - 172.16.45.15.
So the network Address is 172.16.45.12

=================================================
QUESTION NO: 9
Which two of the addresses below are available for host addresses on the subnet
192.168.15.19/28? (Select two answer choices)
A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above

Solution:

/28 means 11111111.11111111.11111111.11110000

The last 1 stands under the 16. This is increment or network-jumps.

192.168.15.0 - 192.168.15.15
192.168.15.16 - 192.168.15.31
192.168.15.32 - 192.168.15.47

Only A and C are IPs in the right range.
Also E is in the right range. But - this is not useable for hosts, cause its broadcastaddress.
So answer is A and C.

################################################################


Calculation of Wildcard-Masks (Needed for Access Lists and OSPF Configuration)

You have Network 192.168.32.0 /28
Only THIS network should be denied of accessing a network or server.

1. Step
calculate the wildcard mask

/28 means 255.255.255.240

binary this is

11111111.11111111.11111111.11110000

For wildcard-mask only the ZEROS are interesting.

11110000 Make a addition of all the fields, that are set to zero

128/64/32/16/8/4/2/1
That is 8+4+2+1=15

So the wildcard-mask will be
0.0.0.15

access-list will be

access-list 1 deny 192.168.32.0 0.0.0.15
access-list 1 permit ip any any

now, we have to bind that access-list to a routerinterface. In the example, this is e0.

interface e0
ip access-group 1 out (or in!)
exit



PS.

Its good to write on a BIG paper the powers of 2

2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
2^10=1024
2^11=2048
2^12=4096

And write on that paper the numbers

128 192 224 240 248 252 254

Cause this are the Numbers, you will allways need in calculating Subnets.

Burn them in your mind! Hang the paper in front of your eyes to never forget them.
Then you will be able to calculate Subnets in your head in a half second!

Isnt live easy?


****************

cscrtr
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Re: Subnetting in 5 Mins.... really great stuff... 7 years 5 months ago #30030

  • KiLLaBeE
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Now can you show us how to subnet in 1 minute? I've heard that you need to be able to subnet in under 1 minute for the CCNA
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Re: Subnetting in 5 Mins.... really great stuff... 7 years 5 months ago #30065

  • donjuan
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good info bro... thanks for sharing... :D
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Re: Subnetting in 5 Mins.... really great stuff... 7 years 5 months ago #30066

  • KiLLaBeE
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cscrtr,
I was joking when I made that post.....you're post is actually good :-D
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