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the simplest and fastest way to subnet
raaki_88
Member Posts: 13 ■□□□□□□□□□
in CCNA & CCENT
i found these techniques in subnetting secrets and found them very effective hoping that these will help you too
firstly you need to have remember two tables ...
1.basic table of first host , last host table
2.the crucial one to all calculation table ...
here are the two
subnet
firsthost
lasthost
broadcast
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
128
192
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024
********************************************************************************
*********************************************
let me check with the formatter , if it comes correctly i will go ahead explaining the concept if not i will host it to image shack and explain it
regards
raaki
firstly you need to have remember two tables ...
1.basic table of first host , last host table
2.the crucial one to all calculation table ...
here are the two
subnet
firsthost
lasthost
broadcast
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
128
192
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024
********************************************************************************
*********************************************
let me check with the formatter , if it comes correctly i will go ahead explaining the concept if not i will host it to image shack and explain it
regards
raaki
#When You Aim For Perfection You Discover It As a Moving Target#
Comments

Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□this is the first example
class c
identify subnet , first and last host of the following ip or what ever it may be
here is the example 1
192.168.12.68 / 25 ***************** 25 bits so ...
firstly count in multiples of 8 till we get 24 .... as one octet has 8 bits right ...
so 8 + 8 + 8 + (1)
left ..
rest is the childs play... really .. now go to the table 2 and borrow or tick 1 bit horizontally and vertically so in this case
see the table two and come back you will find number 128 under subnets and 128 under bits ..
so subnet is 128 and bits 128 indicate that difference should be 128 >we will see this point later
so our subnet is
255.255.255.128
as we all know we need to start any subnet with 0 .. so start it
subnet
firsthost
lasthost
broadcast
192.168.12.0
192.168.12.1
192.168.12.126(broad1)
192.168.12.127(simple next subnet 1)
(see the difference of 128
which we got in bits )
192.168.12.128
hence in this way it can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets ******** *(1 bit vertically)
128 **** *(1 bit horizontally)
192
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024
hence in this way it can be determined
more examples to follow .. let me know if this makes sense to anyone
regards
raaki#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□another class c example
192.200.200.167 / 28
now again count till 24 ...now are left with another 4 bits ... rest is childs play as usual... tick 4 horizontally and 4 vertically in the second table...
after seeing the table and seeing the number under subnets it is 240 and under bits it is 16 .. so simple subnet mask ends with 240 and it has a diff of 16 for every
so our subnet mask is ... 255.255.255.240
subnet we write
so we will start with 0 as usual
subnet
firsthost
lasthost
broadcast
192.200.200.0
192.200.200.1
192.200.200.16
192.200.200.32
192.200.200.48
etc
etc with increments of 16
192.200.200.160
192.200.200.161
192.200.200.174
192.200.200.175
our need fills here
192.200.200.176
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
*
*
*
*
128
*
192
*
224
*
240
*
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□moving on to class b
150.200.155.23/18
18 bits .. so multiples of 8 ... now 24 will not work as 18 is less than 24 .. so next 8 multiple is 16 ... so ... we need two more bits from 16 to 18
so go to table two tick two horizontally and two vertically ... and you are done
so seeing the table ... under subnets it stopped at 192.. remember we had only 16 bits now not 24 .. so our mask would be
255.255.192.0
now see under the bits table ... we are stopped at 64 .. so for this we need to have increments of 64 ... as usual start with 0
subnet
firsthost
lasthost
broadcast
150.200.0.0
150.200.64.0
150.200.128.0
150.200.128.1
150.200.191.254
150.200.191.255(next subnet1)
our ip statisfies here
150.200.192.0
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
*
*
128
*
192
*
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□this is some what tricky .. please do clear the concepts in the above cases
191.20.56.65 / 25
as usual ...greater than 24 .. so we need to add one more bit to make it 25 .. so tick in the table two one start horizontally and vertically and rest is gone
so seeing it ... we got under subnets the value is 128 .. so the subnet mask is
255.255.255.128
now going on with next ..we got 128 under bits so the difference is 128..but now we need to have four octet subnets for every third octet one .. if this
confuses you see the below table
start with 0
since class b
x.x.0.0
subnet
firsthost
lasthost
broadcast
191.20.0.0
191.20.0.128
191.20.1.0
191.20.1.128
191.20.2.0
191.20.2.128
etc etc till our ip
191.20.56.0
191.20.56.1
191.20.56.126
191.20.56.127
this satisfies our ip
191.20.56.128
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
*
128
*
192
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□moving on to class a
10.210.204.70/12
multiples of 8 .... 12 is not greater than 16 .. so multiple of 8 satisfying it is ...8...so to get 12 we need to add 4 more bits ...
there fore ..tick 4 bits horizontally and 4 bits vertically and game over
seeing the table under subnets it stops at 240..so remember we have only one multiple of 8 .. so only one 255 in subnet mask..
therefore subnet mask is 255.240.0.0
now seeing under bits section...it has stopped at 16 so...multiples of 16 or difference of 16 ..we need to start with zero  for class a x.0.0.0
subnet
firsthost
lasthost
broadcast
10.0.0.0
10.16.0.0
10.32.0.0
etc till our ip
10.208.0.0
10.208.0.1
10.223.255.254
10.223.255.255
this satisfies our ip
10.224.0.0
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
*
*
*
*
128
*
192
*
224
*
240
*
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□last class a
20.100.55.3 / 26
multiple of 8 again .. it ends at 24...to get to 26 we need to have two bits .. so tick two bits vertically and horizontally in the table two ..
under subnets it stopped at 192
so our subnet mask is
255.255.255.192
and under bits it stopped at 64 .. so the subnets increment at rate of 64
lets start with zero subnet
subnet
firsthost
lasthost
broadcast
20.0.0.0
20.0.0.64
20.0.0.128
20.0.0.192
etc
etc till our ip
20.100.55.0
20.100.55.1
20.100.55.62
20.100.55.63
our ip falls here ..
20.100.55.64
20.100.55.128
20.100.55.192
hence in this way this can be determined
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
*
*
128
*
192
*
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
4
8
16
32
64
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□this is what i liked in this tutorial.. designing subnets was never been easy for me and also very much confusing .. thanks to the author iam now satisfied with my designing subnets knowledge
192.168.1.0
4 subnets required
each subnet needs at least 10 hosts
firstly for its a class c addr ..so for our subnet mask it should have three octets full ... as class c is 255.255.255.x
so
fill it out as 255.255.255.x
now we need 4 subnets ... so tick till we meet our subnet requirements in the table
now as marked two bits under subnets ...you also need to mark two bits under subnets in the upper colomn
hence in this way this can be determined
as it stopped at 192..our subnet mask is ends with 192..there fore our subnet mask is 255.255.255.192
now as you tick two bits under upper subnets coloum...there are 6 bits remaining unticked right ..(unticked 224 240 248 252 254 255 total 6)
tick them under hosts wink.gif
look the table and come back ... we are left at 64 ..as we need to subtract two for broadcast and subnets .. there would be 62
solve simply
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
128
*
192
*
224
240
248
252
254
255
powers of 2
subnets
hosts (2)
2
*
*
4
*
*
8
*
16
*
32
*
64
*
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□200.100.20.0
9 subnets
each host needs atleast 10 hosts
class c addr ..so subnet mask is 255.255.255.x
now we need 9 subnets .. tick in the subnets coloum till we meet the requirement of 9 subnets ...
so look at the table .. we have ticked 4 of them .. so tick 4 bits in the upper subnet table
it stopped at 240 so our subnet mask is 255.255.2555.240
now we are left with 4 empty bits (248 , 252, 254 , 255) tick them under hosts ... and subtract two we are done
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
128
*
192
*
224
*
240
*
248
252
254
255
powers of 2
subnets
hosts (2)
2
*
*
4
*
*
8
*
*
we have met the requirement of 9 subnets
16
*
*(162 = 14 hosts )
32
64
128
256
512
1024#When You Aim For Perfection You Discover It As a Moving Target# 
Optionsnel Member Posts: 2,859 ■□□□□□□□□□Thanks for sharing that info if it helps anyone learning it.Xbox Live: Bring It On
Bsc (hons) Network Computing  1st Class
WIP: Msc advanced networking 
Optionsraaki_88 Member Posts: 13 ■□□□□□□□□□this is the final one and interesting one
130.100.0.0
we need 30 subnets
each subnet with atleast 1000 hosts
class b ..
so subnet mask is 255.255.x.x
we need 30 right .. so tick in the subnet table until we meet the requirement
after looking at the table ..we ticked 5 of the subnet ones ..so even tick the 5 bits of upper subnets
it ends at 248 so our subnet mask is 255.255.248.0
now we have unticked three of them
(252 ,254 , 255 ***********important (3 of the unticked third octet and 8 of the remaining 4th octet so a total of 11..so tick 11 in the host bits portion and
subtract 2 to get the number of hosts)
the crucial one
bits
128
64
32
16
8
4
2
1
subnets
128
*
192
*
224
*
240
*
248
*
252
254
255
powers of 2
subnets
hosts (2)
2
*
*
4
*
*
8
*
*
16
*
*
32
*(we met 30 here)
*
64
*
128
*
256
*
512
*
1024
*
2048
* (total of 11 bits ticked 2048  2 = 2046 hosts / subnet)
hope this helped anyone ... any comments / suggestions are warmly welcomed ...
regards
rakesh#When You Aim For Perfection You Discover It As a Moving Target# 
Optionskelargo Member Posts: 5 ■□□□□□□□□□I like to count with my fingers. four fingers on each hand and start by counting backwards.
right pinky=1, 2, 4, 8 then my left hand 16, 32, 64, 128.
then to get the mask, add from lefttoright:
left pinky = 128, then 192, 224, 240 then right hand 248, 252, 254, 255
each finger is a bit; quick and dirty visual aid.

OptionsJavonR Member Posts: 245kelargo wrote:I like to count with my fingers. four fingers on each hand and start by counting backwards.
right pinky=1, 2, 4, 8 then my left hand 16, 32, 64, 128.
then to get the mask, add from lefttoright:
left pinky = 128, then 192, 224, 240 then right hand 248, 252, 254, 255
each finger is a bit; quick and dirty visual aid.
LOL, I do the same thing! 
Optionskevin31 Member Posts: 154Hi
thanks for this good tutorial but could you explain the power of 2 as I just cant get my head round that?
thanksLAB  4 X 2651XM's 1 X 2620 3 X 2950 1 X 2509 AS 1 X 3550 
Optionsbashtie Member Posts: 25 ■□□□□□□□□□kevin31 as you know computers are working with binary ... so its base 2
2^0
2^1
2^2
2^3
and so on ...
thats what is meant by the power of 2 thing
the rightmost bit is 2^0
the one left to this is 2^1
and so on ...
hope this helps 
Optionscraigandem Member Posts: 1 ■□□□□□□□□□kevin31 as you know computers are working with binary ... so its base 2
2^0
2^1
2^2
2^3
and so on ...
thats what is meant by the power of 2 thing
the rightmost bit is 2^0
the one left to this is 2^1
and so on ...
hope this helps 
Optionslol2dubs Registered Users Posts: 1 ■□□□□□□□□□I don't want to dig up the dead, but this thread helped me understand how to subnet immensely.
I'd like to add the note that I've memorized to allow me to subnet in under 10 seconds, as recommended per the Academic version of ICND1 for anyone who may come across this via google like I did. I put this together in Notepad++ in the Java language so it looked all nice and pretty, and attached an image for you guys.
128 64 32 16 8 4 2 1
128
192
224
240
248
252
254
255
8 16 24 32
^^^ Count multiples of 8 till you
reach one of these below your bit notation
 Take left over amount and tick
horizontally and vertically on chart above
So 192.168.12.68 /25 = 8+8+8+(1);
1 = H 128 V 128
H = range
V = mask
where h = horizontal and v = vertical
>Subnet = 192.168.12.0 FirstHost = 192.168.12.1
>LastHost = 192.168.12.126 Broadcast = 192.168.12.127
To determine which subnet an address belongs to:
Example 172.31.77.201/27
27  24(8+8+ = 3 = 32 & 224
201 divided by 32 = 6.2ish
32 * 6 = 192
>Subnet = 172.31.77.192 FirstHost = 172.31.77.193
>LastHost = 172.31.77.222 Broadcast = 172.31.77.223
2dubs