I am pretty sure you know the class of address right??????
Remember this CLASS A RANGE 0-127
CLASS B 128-191
CLASS C 192-223
The rest is CLASS D AND CLASS E........
A.B.C.D, obviously this is the IP address format and each of the letters is one octet (8 bits)
The ranges given above is the first octet........(A)
So if you find an IP ADDRESS WITH 170.32.88.254 YOU SHOULD GUESS IT IS CLASS B TYPE. SIMILARLY 10.0.25.1 IS .... IS......
YOU SHOULD GUESSS (CLASS A).
After you have identified the class then remmember the CLASS A FOR FIRST OCTET represents Network....
CLASS B FIRST TWO OCTECTS REPRESENT Network
CLASS C FIRST THREE OCTECTS REPRESENT NETWORK.....
For now just remember this.... Class A first octet ,,,,,, network
Class B first two octects network...
CLass c frist three octectsts represent
networs
Class A : A:B:C:D
Class B : A:B:C:D
Class C : A:B:C:D
And any book will go with the same old mouth yawning stuff of IETF, ADRESS SPACE BLAH BLAH......
Simply to put it....Class A first 8 bits represent network so 24 bits represent Host
Class B FIRST 16 Bits represent N/W AND the next 16 bits hosts
Class c... i think you can guess
CLASS A
Suppose the IETF is awarding you America.0.0.0, so all the host within America are yours are all the host within India. SO THAT WILL BE LIKE 2 TO THE POWER 24 HOSTS
Class B
England.Newcastle.0.0 so all the host with New Castle are yours. 2 TO THE POWER 16 HOST
Class C
India.Hyderabad.Kalyan Nagar.0 So all the host within India, Hyderabad and Kalyan Nagar are yours. 2 TO THE POWER 8 HOSTS.
But the catch here is that you have fixed no of hosts and fixed number of network......, , you cant move back you cant move front
Subneting is basically giving you the flexibility in addressing no you can say
India.Hyderabad.Kalyan Nagar.Area 1 so you can have the flexiblity of giving these many number of bit of the 32 bits for networks and these many for hosts.... so you can see i need only 40 host in Area 1 that is how many bits is that...
******2^x => 40 that make is 64 (Binary number... blah blah) so you need 6 bits for the hosts
and how many bit for area 1 ,,, you would be a genius if you get this right
India. Hyderabad. Kalyan Nagar.| Area 1|Hosts
8 : 8 : 8: 2|6 bits
So area 1 will have ....... 2 bits because in an octet there will be 8 bits...
or with subnetting you could now say
England.New Castle.Hydrid Road.0
or you could take it one step further
England .New Castle. Hydribid Road. Road 1
it depends........ JUST ABOUT BALANCE OF POWER, INSTEAD OF GIVING ONE ORGANISATION ONE ADDRESS YOU GIVE PLENTLY OF PEOPLE PLENTY OF AUTHORITY.
Now just for now think of this way private addresses.. blah blah, now things like NAT PAT, GLOBALLY ROUTABLE ADDRESSES WILL COME , THIS IS JUST TO GET YOU STARTED WITH SUBNETTING...
NOTE ALSO 0.0.0.0 is a default route , loopbacks i mean you have plenty but just think of it like this now.. only okay...this is more of a numbers and mathematics class.......
PLEASE DO THIS EXERCISE
www.subnettingquestions.com
Eg. 192.168.13.50/26 annadu annuko,...... immediate after looking at the first octet you should guess which class of address it is CLASS C , so just like Arnold in Terminator you should see.....
that as
192.168.13.|50 /26 okay ,
now since they asked for /26 this means the first 26..... let me repeat ... the first 26 bits represent networks... that is from LEFT -> RIGHT, that is the number 192.168.13.0|Host represent the network will tell you why the 50 became the 0 in a minute....
***** Please dont fix that Class C should only start after Third octet, it can start even before /24 meaning i can also say 192.168.13.50 / 22 but that would be route summarization.. all that stuff but just cling on to this now.....plus a concept of VLSM... CIDR... but not to confuse you with that technical jargon*****
Now this /26 means 2 bits.... of the 4th octet....
means 126+64=192
256-192 = 64... always it is 256...
64 multiples upto ... 256 0,64,128,192,256....
so .50 belongs to 0-63 so the network address for area 1 is
192.168.13.0 so if you want o go to any host in area 1 it will be 192.168.13.0 and and the host address will be from 192.168.13.1- 192.168.13.62 i know i know left out the 63 it is all 1111 so broad cast.... so forget about it now ... i will not give more information than this today,
This is enough for you today trust me you will thank me for this!!!!!
