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-Question s 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
Reduced: 72% of original size [ 800 x 508 ] - Click to view full image
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 <<
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 <<
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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