the simplest and fastest way to subnet

raaki_88raaki_88 Member Posts: 13 ■□□□□□□□□□
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
#When You Aim For Perfection You Discover It As a Moving Target#

Comments

  • raaki_88raaki_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(broad-1)
    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#
  • raaki_88raaki_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#
  • raaki_88raaki_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 subnet-1)

    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#
  • raaki_88raaki_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#
  • raaki_88raaki_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#
  • raaki_88raaki_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#
  • raaki_88raaki_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#
  • raaki_88raaki_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
    *
    *(16-2 = 14 hosts )

    32

    64

    128


    256


    512


    1024
    #When You Aim For Perfection You Discover It As a Moving Target#
  • nelnel 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
  • raaki_88raaki_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#
  • kelargokelargo 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 left-to-right:
    left pinky = 128, then 192, 224, 240 then right hand 248, 252, 254, 255

    each finger is a bit; quick and dirty visual aid.

    :D
  • JavonRJavonR Member Posts: 245
    kelargo 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 left-to-right:
    left pinky = 128, then 192, 224, 240 then right hand 248, 252, 254, 255

    each finger is a bit; quick and dirty visual aid.

    :D

    LOL, I do the same thing!
  • kevin31kevin31 Member Posts: 154
    Hi

    thanks for this good tutorial but could you explain the power of 2 as I just cant get my head round that?

    thanks
    LAB - 4 X 2651XM's 1 X 2620 3 X 2950 1 X 2509 AS 1 X 3550
  • bashtiebashtie 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
  • craigandemcraigandem Member Posts: 1 ■□□□□□□□□□
    bashtie wrote: »
    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
    What do you do that result of 2^y number? This is the part i don't under stand...
  • lol2dubslol2dubs 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+icon_cool.gif = 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
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