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shankar

Tip To Remember Subnetting For Cisco Exam

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Got this off another site but it's useful here also:

 

To remember the subnetting tables all you have to do is start with "4" and double it until you get to "16384" Write them downward on a sheet of paper and when you are done just subtract 2 from each number.

 

ie:

4 = 2

8 = 6

16 = 14

32 = 30

64 = 62

 

once you have done that all you need to do is reverse the order of all the numbers going back up the sheet:

 

subnets hosts

2 62

6 30

14 14

30 6

62 2

 

See how the numbers flip flop between each column? My example is for class C but it works for class B just the same.

 

Once you have the subnet/host numbers written out, just remember the following numbers .192, .224, .240, .248, .252

class C:

sub hosts

.192 /26 2 62

.224 /27 6 30

.240 /28 14 14

.248 /29 30 6

.252 /30 62 2

 

The numbers with a slash (ie /26) are just short hand ways of writing out subnets. They can be really confusing if you are trying to learn subnetting for the first time. Just rember that ip addresses are made up of 32 bit addresses, or /32. These 32 bit addresses are broken down into class A,B, and C. class B are from /18 to /30 and class C go from /26 to /30. The reason the numbers don't go up to /32 are because it goes against the rules of subnetting (according to Cisco), I don't have any other reason why.

You need to memorize this stuff!!! When I went to work I jotted notes all over my desk and tool boxes just so I would see it all the time.

 

Here is the class B example:

(1). start with 4 and double it till 16384: 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384.

 

(2). subtract 2 from each number: 2, 6, 14, 30, 62, 126, 254, 510, 1022, 2046, 4094, 8190, 16382.

 

(3). write them downward on a sheet of paper and then write them back up in reverse order:

2 16384

6 8190

14 4094

30 2046

62 1022

126 510

254 254

510 126

1022 62

2046 30

4094 14

8190 6

16382 2

 

(4) Finally you just have to add the net number to your list... Rember these numbers: .192.0 (/1) .224.0 (/19) .240.0 (/20) .248.0 (/21) .252.0 (/22) .254.0(/23) .255.0 (/24) .255.128 (/25) .255.192 (/26) .255.224 (/27) .255.240 (/28) .255.248 (/29) .255.252. (/30)

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While learning "192,224,240,248,252" by heart can be extremely useful in your day-to-day life as a network administrator (I know those numbers sure saved me a lot of time! :) ), my advice for the people who are new to subnetting is this: write it in binary!!!

 

It is extremely easy to "misplace" a bit if you try doing the calculations in your head (yes, it happens to me too - after many years of subnetting :) ). But if you know what each bit of the IP address and subnet mask is and you put those bits on paper, you cannot go wrong.

 

I know some of you will say that "it takes too much time". And I agree - in my daily life, I don't write the binary form of the address on paper (hey, I don't even write the address at all :) ). But when the result of those calculations is really important (such as when you're sitting your CCNA exam), the few seconds it takes you to write down the 8 or 16 bits on the paper can make the difference between a correct and a wrong answer to that question. (and after a little practice, writing down those few bits will definitely take you less time than writing columns upon columns of powers of 2).

 

I've been an Academy instructor for several years now, and this is the advice I give to all my students. I hope that it will benefit some of the people here, too.

 

Before I get to another subject, I will try to give you an example (and hopefully persuade you to do things the "old-fashioned" way :P ). Suppose you are given the IP address of 192.168.17.151, with a subnet mask of 255.255.255.224. And you are asked what the network address is.

 

Write down the address and subnet mask in binary (you can safely skip an octet if the corresponding octet in the subnet mask is 255):

 

192.168.017.10010111

255.255.255.11100000

 

Draw a line where the 1s of the subnet mask end, and the 0s begin:

 

192.168.017.100|10111

255.255.255.111|00000

 

The bits in the IP address to the right of that line are the host bits. Since we have both 1s and 0s there, we know that this is a host address. We just need to replace them with 0s to get the network address (or with 1s, if you want the broadcast):

 

192.168.017.100|00000

255.255.255.111|00000

 

So our network address is 192.168.17.128. Wasn't that easy? :)

 

 

The numbers with a slash (ie /26) are just short hand ways of writing out subnets. They can be really confusing if you are trying to learn subnetting for the first time. Just rember that ip addresses are made up of 32 bit addresses, or /32. These 32 bit addresses are broken down into class A,B, and C. class B are from /18 to /30 and class C go from /26 to /30.

 

 

I'm sorry, but this is plain wrong. First of all, the whole idea behind using CIDR (Clasless InterDomain Routing) and the /xx notation is to get rid of the old "classes" notion. Secondly, it's not the subnet mask that makes the difference between a class B and a class C address - it is the first bits of the address itself ("0" for class A, "10" for class B, "110" for class C). So you cannot say that "class B are from /18 to /30".

 

For more details on CIDR and why we do not need (or want!) classes anymore, check out the following link:

http://public.pacbell.net/dedicated/cidr.html

 

 

The reason the numbers don't go up to /32 are because it goes against the rules of subnetting (according to Cisco), I don't have any other reason why.

 

 

The reason is simple: the first and last addresses of every subnet are reserved. The first one (the one that has all the host bits equal to 0) is the network address, and the last one (all host bits = 1) is the broadcast address. If we tried to use a /31 subnet mask, we would get a subnet that has only the network and broadcast address - no room left for hosts! And we wouldn't have much use for such a subnet... :)

Edited by bogd
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Hi all,

 

Do any of you find find subnetting hard? Have you spent money on simple solutions which simply don't deliver? Are you confused by too many books showing different ways to subnet or colleagues expressing their opinions? Do you need to write out charts (as above) just to find your subnet?

 

I honestly believe I have come up with the easiest technique to learn to subnet and I guarantee that it does not involve charts, videos, writing out binary, or any other resources that are not allowed in the exam. This simply uses your head and some VERY BASIC arithmetic.

 

This technique has been praised highly both by how2pass and Cisco.com forums.

 

Please view the post at

http://www.subnettingmadeeasy.[websitebanned]

 

Cheers and happy subnetting!

Edited by Lord Flasheart
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Hi Guys,

 

 

If anyone have CCNA dump latest please send to me xxxxxxxxxx .since i am going to write the exam.

Edited by gordonccie
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finger%20method.jpg

 

class A 1st octet 0-127: 16m

class B 2nd octet 127-191: 65thou

class C 3rd octet 192-223: 254

class D 4th octet 224-239: Multicast

class E 240 -255 reserve for experiment

 

 

Decimal to Hexadecimal

9 - 9

10 - A

11 - B

12 - C

13 - D

14 - E

15 - F

 

How to get BINARY: using our formula

let say 10.125.4.1

 

10. 125 4. 1

0000 1010. 0111 1101. 0000 0100. 0000 0001

 

if you are using our formula in second level you can get the binary.

Using the fingering method in your two hands in one hands you have a for

finger exept the tumbnail. in for finger in left side you have first finger

128, second is 64, third is 32 fourth is 16. and in right finger you have

8,4,2,1 in first, second, third, and last. and look at your finger right now

its look like Binary righ!using this method.

 

example:

in the last given IP address in first octet is 10 so we are going to used

finger method. look at your finger two hands.in 10 in binary is 0000 1010.

so in finger is:

 

128 64 32 16 8 4 2 1 second level using finger method

0 0 0 0 1 0 1 0 in binary its look like this. because 8+2= 10

 

in your finger you need to close 0 in binary and 1 who is raise so

the answer is 10 GET IT!

 

more example: using 125

 

128 64 32 16 8 4 2 1

0 1 1 1 1 1 0 1 64+32+16+8+4+1 = 125

 

look at your finger again do the same thing 0 is close

in your finger and 1 is raise Got IT!

 

note: you better practice this finger method make it easy to you the subnetting

etc..always remember your finger....:) NO need to have a paper to calculate what

binary of this, awts how can you solve it if you dont have a finger just imagine

hehe. remember no need to have a paper in your exam or interview of solving problem

like this all you need to do is finger and your mind............ :P

 

 

Private IP address

A. 10.0.0.0 - 10.x.x.x

B. 172.16.0.0 - 172.31.x.x

C. 192.168.0.0 - 192.168.x.x

 

HOW TO GET A SUBNET

 

Using this formula you can subnet all ip that you need to subnet.

Or anything to get subnet and host. And specially route aggregation.

 

Some Rules:

 

1.Using those valid mask

2.The next level 128i, 64i, 32i etc. this formula is used to get the

first valid IP address and the last valid IP address.

3.In slash /1 - / 32 this is used to know which mask are you going to

use and which First valid IP. Take a look at the formula

 

Example: 192.168.4.0/22

 

If you look at the formula in /22 is class C right, and in line of /22

and its valid mask is 252 and the first valid IP is 4 and the last is 7

like this (192.168.7.254/22) next subnet is 192.168.8.0/22 it because

of this 4, 8, 12, 16, 22 etc. So the next valid IP after using 4-7 is 8

so that’s why the next valid IP for the next subnet is 192.168.8.0/22 to

192.168.11.254/255 got it’?

 

2nd Formula: how to get subnet and host

/class a,b, or c + /Bits = SUBNET

/32 - /Bits = HOST

 

172.198.0.0/16 = let say the boss ask you. give me 8 subnet for that IP address.

 

first using 2nd Formula in SUBNET in a first slash we are going

to use /16 because boss gives us class B.

Second we are going to use /BITS formula of this is:

 

.................|------------->>> answer: /BITS is 4 so /4

128i 64i 32i 16i 8i 4i 2i 1i

 

explaination : why we are using 8i to calculate how many bits that are

we going to use it because boss ask for 8 subnet count backward

8,7,6,5,4,3,2,1 so in valid ip the answer is 4BITS.

 

NOTE: if the given subnet is 7, 15, 31, 63, 127 you are goin

to add 1 Bits like this

 

let say 7 so 7 is...|-------------->> /4bits again

128i 64i 32i 16i 8i 4i 2i 1i

because i said you need to add 1 Bits in 7,15 etc.

 

 

let say 15 is |-------------->> /5bits

128i 64i 32i 16i 8i 4i 2i 1i

because i said you need to add 1 Bits in 7,15 etc.

 

But if look like this 62,30,29, 126 you do not need to add.

if you follow the rules you can make it..

 

let say 65 is |-------------------------->> /7bits

________128i 64i 32i 16i 8i 4i 2i 1i

because it is in the range of 64 remember 64 - 126, 32 - 62 and if its 63 you need to add 1 bits.

 

 

 

SECOND: now that we no which Bits we are going to use lets start it.

 

/16 + /4bits = /20 the in line of /20 again think..... the valid mask

is 240, and the valid IP is 16i right.you can make it, i know you are not an idiot....

 

what now..

 

answer:

Network: Subnet & mask Network # Valid Hosts Broadcast

Network A 172.198.16.0 /20 172.198.16.0 172.198.16-31.1-254 172.198.31.255

NETWORK B 172.198.32.0 /20 172.198.32.0 172.198.32-47.1-254 172.198.47.255

NETWORK C 172.198.48.0 /20 172.198.48.0 172.198.48-63.1-254 172.198.63.255

NETWORK D 172.198.64.0 /20 172.198.64.0 172.198.64-79.1-254 172.198.79.255

NETWORK E 172.198.80.0 /20 172.198.80.0 172.198.80-95.1-254 172.198.95.255

NETWORK F 172.198.96.0 /20 172.198.96.0 172.198.32-96.111-254 172.198.111.255

NETWROK G 172.198.112.0 /20 172.198.112.0 172.198.112-127.1-254 172.198.127.255

 

 

 

Using host it is almost the same the procedure the difference between the two is host are

using /32 - /bits unlike subnet /classes + /bits.... the bits that you are going to use is

like in subnet in 8 subnet 8i,4i,2i,1i answer for bits is /4 remember in subnet example..

 

Example of host..

 

192.168.21.0/24 for 62 host remember in previous example...

 

 

...........|----------------------->>> /6 so answer: /32 - /6 = /26

128i 64i 32i 16i 8i 4i 2i 1i

remember in host always use /32 to minus /bits..

 

 

answer: 192.168.21.64 /26 first valid ip mask 255.255.255.192

to

192.168.21.127 /26 last valid ip

 

 

 

how to know route aggregation: aggregation is the whole or total....

 

letsay you have 16.0.0.0 to 19.0.0.0/8 if you count it 4,8,12,16,20 it

is in the range of 16 right first valid ip so we are using 4i to aggregate

the ip answer is: 252.0.0.0 in mask...

 

more sample : 172.168.24.0 to 172.168.32.0/24 in 8,16,24,32 in the range of 24 right

answer is 8i for 255.255.248.0

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Also go through my tutorial...hmmmmmmm

 

 

hxxp://www.sadikhov.com/forum/index.php?showtopic=162652

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This subnetting thing is made overly difficult by conversion between number bases, all you need to know is a little bit of simple mental arithmatic.

 

Lets start!! Forget about the address for the moment as all of the magic is in the subnetmask :) Also DO NOT FORGET IP SUBNET ZERO

 

TWO RULES YOU NEED TO EMBED IN YOUR HEAD ARE:

 

The range of address's you have available starts at 0 and ends with 255 and that = 256 address's

 

Networks always start with an even number and end with an odd number, if your calculation tells you otherwise you have done something wrong!!

 

Two masks that we will use for an example are:

 

255.255.255.192

255.255.255.224

 

We know that the largest value in any octet will be 255, so lets do a little mental calculation!

 

255.255.255.255 - 255.255.255.192 = 63 Instantly we now know by using this subnet mask we have and our address range is 0-63 where 0 will be the net and 63 the broadcast, all other address's 1-62 are host.

 

So based on that and knowing that our first network begins with zero we can work out our subnets

 

Net 1 = 0-63

Net 2 = 64-127

Net 3 = 128-191

Net 4 = 192-255

 

255.255.255.255 - 255.255.255.224 = 31 Instantly we now know by using this subnet mask we have and our address range is 0-31 where 0 will be the net and 31 the broadcast, all other address's 1-30 are host.

 

So based on that and knowing that our first network begins with zero we can work out our subnets

 

Net 1 = 0-31

Net 2 = 32-63

Net 3 = 64-95

Net 4 = 96-127

Net 5 = 128-159

Net 6 = 160-191

Net 7 = 192-223

Net 8 = 224-255

 

You can recheck your subnet just by doing 8 x 32 = 256 "Remember the available address range starts at zero.

 

Try these yourself and see how quick and accurate you can be without using subnet calculators or overly complex number base conversions.

 

Enjoy

 

Mark

 

255.255.255.240

255.255.255.248

255.255.255.252

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Sir Mark,

 

 

can you please give other examples on your process of subnetting. please include solutions for ip adresses

 

thank you

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Marks process is very nice but it requires YOU to sit down and try it out and understand it.

 

Give it a try and then come back with your attempts at pass or fail. We will then assist.

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HI

 

I want to subnet 192.168.20.0/24 for 2 groups, one has 70 hosts and consist of one router one switch. Can I use VLSM to save on hosts or will end up providing more than required?

 

I'm sorry but this is all new to me and I can only logically see me using a 192.168.20.1/25 split offering 126 hosts with 56 left hanging.

 

B RGDS

Bammer

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Hi

 

Split it down into smaller subnets and add more sub-if's if it's a router on stick scenario, or if you have a L3 switch create SVI's

 

Maybe something like

 

192.168.20.0/26 = 0-63 gives you 61 usable address's after your GW address

192.168.20.0/28 = 64-79 gives you 13 usable address's after your GW address

 

Then just subnet the remainder

 

That only uses around 38% of your available address range and you can just summarise at the demarc to your network.

 

Mark

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To add to the discussion above. Here is a nice way of summarizing the information for memory on the exam:

 

Bits Borrowed 1 2 3 4 5 6 7 8

Bit Value 128 64 32 16 8 4 2 1

Subnet Mask 128 192 224 240 248 252 254 255

# of Subnets 0 2 6 14 30 62 126 254

 

1. Understand the basics of subnetting & assigning ip addresses

2. Memorize the table above (very easy if you have step #1 down)

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I created an easy to replicate, very powerful chart for test purposes. Obviously, knowing the fundamentals of IPv4 is critical to your success as an engineer. For testing I found having the chart that's not hard to create in 1 minute which will insure that 100% of my answers to subnet related questions are CORRECT is priceless. I will admit learning to use this chart may seem hard, but trust me - once you see how it is designed you will laugh :)

 

I have CCNA, and I will tell you CCNA candidates that the #1 thing that you find on almost every question is a subnet problem. If you cannot reverse engineer IP's in networks then you cannot troubleshoot ANYTHING

 

Here's the link: http://www.johnpatricklockie.com/rmmylife/?p=163

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Wow thats a lot of information to remember, all you need to know is your two times tables and you can subnet anything

 

Mark

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I agree with Mark...

 

Long discussions of subnetting is in my humble opinion why most people don't understand the topic very well.

 

2 to the 2nd power - 2 works miracles...

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HI

 

I want to subnet 192.168.20.0/24 for 2 groups, one has 70 hosts and consist of one router one switch. Can I use VLSM to save on hosts or will end up providing more than required?

 

I'm sorry but this is all new to me and I can only logically see me using a 192.168.20.1/25 split offering 126 hosts with 56 left hanging.

 

B RGDS

Bammer

 

You got the concept. Most of Cisco's subnetting questions would be like this. The 56 other IPs are indeed left hanging. Compare that if you were to dedicate a /24 for only 70 hosts. A /25 is smaller than a /24.

 

I'd agree with the others - in IPv4 subnetting, everything's just about 2^x.

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can i ask something? still about subnetting.

 

starting IP would be 150.150.10.0

 

different host requirements:

 

IP Phone = 144

Voice with Data Users: 15 rooms / 24 ports each room

 

 

how would i divide the IP so that i would eliminate network waste due to improper way of subnetting.

 

 

thank you....

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can i ask something? still about subnetting.

 

starting IP would be 150.150.10.0

subnet mask would be 255.255.0.0

 

different requirements:

 

1st network = 150 host

succeeding networks: 30(subnets) X 40(host) =======(1 network = 40 host)

 

 

 

how would i divide the IP so that i would eliminate network waste due to improper way of subnetting.

 

 

thank you....

 

 

repost due to unclear instructions above.

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For Class C address Subnetting (192-223 as first octet in IP address) only i.e 255.255.255.?

Maximum Bits that can be borrowed is 7 (you cannot borrow all It will be a host)

FORMULA

N=no of bits borrowed

Prefix=32-N

Subnet=256-2^N

Boundaries of Subnets=256-Subnet

 

e.g

N=2 Most cases its WAN

Prefix=32-2=30 (or/30)

Subnet=256-2^2=256-4=252 (255.255.255.252)

 

Boundaries of Subnets=256-252=4 (ie. 0-3,4-7,8-11 etc)

-Remember The 3,7,11 always come as odd numbers

 

You can check with any class C address

 

I will be posting soon for Class A and Class B

 

:)

With Regards

Edited by Pawan_Bhattarai
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I am poor in Math but I considered this logic tip during my review. I really appreciate this advise. High five!

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Oh man I wish I'd found this forum sooner. It looks amazing here, so much to see for ccna's like me.

 

I saw this thread and thought I might add something I found a few days ago. This UK CCIE guy has put a 'shortcuts' PDF together and is selling it for about $5 and it's full of the sort of thing on here but for $5 I guess it might be nice to see something a bit more 'together' and I've asked him loads of questions and he always gets back to me...just an idea for the forum and certainly not an advert. I bought it and I figured it was great. Anyway he was at www.defaultrouteuk.com so it's up to you if you look.

 

I liked the way he put it together to help people like us doing our CCNA. He's got online exams to run through so you really get it when you've finished. Oh man he's got some wicked tips and tricks too to make subnetting totally easy. Some of those tips are already in this thread actually but I loved the exponent of 2 stuff this ccie goes through, just made it all really easy. I'm sure I passed my exam because of this guy's tips...finished with 5 minutes to go. You don't need pass4sure.

 

Anyway - great forum here!

 

can i ask something? still about subnetting.

 

starting IP would be 150.150.10.0

subnet mask would be 255.255.0.0

 

different requirements:

 

1st network = 150 host

succeeding networks: 30(subnets) X 40(host) =======(1 network = 40 host)

 

 

 

how would i divide the IP so that i would eliminate network waste due to improper way of subnetting.

 

 

thank you....

 

 

repost due to unclear instructions above.

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Class A Subnets

In Class A, only the first octet is used as Network identifier and rest of three octets are used to be assigned to Hosts (i.e. 16777214 Hosts per Network). To make more subnet in Class A, bits from Host part are borrowed and the subnet mask is changed accordingly.

For example, if one MSB (Most Significant Bit) is borrowed from host bits of second octet and added to Network address, it creates two Subnets (21=2) with (223-2) 8388606 Hosts per Subnet.

The Subnet mask is changed accordingly to reflect subnetting. Given below is a list of all possible combination of Class A subnets:

Class A Subnets 

In case of subnetting too, the very first and last IP address of every subnet is used for Subnet Number and Subnet Broadcast IP address respectively. Because these two IP addresses cannot be assigned to hosts, sub-netting cannot be implemented by using more than 30 bits as Network Bits, which provides less than two hosts per subnet.

Class B Subnets

By default, using Classful Networking, 14 bits are used as Network bits providing (214) 16384 Networks and (216-2) 65534 Hosts. Class B IP Addresses can be subnetted the same way as Class A addresses, by borrowing bits from Host bits. Below is given all possible combination of Class B subnetting:

Class B Subnets

Class C Subnets

Class C IP addresses are normally assigned to a very small size network because it can only have 254 hosts in a network. Given below is a list of all possible combination of subnetted Class B IP address:

Class C Subnets

 

Regards

Christian

Network Admin

 

Edited by Darby Weaver
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