LEARN SUBNETTING BEAST IN 5 MINUTES!!!!!  

Sunday, August 31, 2008

The credit for this courses goes to MR.spacyfreak

He is currently pursing his ccsa and done with following
courses

CCNA / MCSE / MCSA / MCP / LPI
70-270 / 70-219 / 70-216 / 70-218 /
70-224 / 70-215 / 70-217 /
70-292 / LPI 101 / LPI 102
/ 640-801 / 642-502 / 642-551



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





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 <<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 <<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?


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