Subnetting Basics

8/22/2019 Subnetting Basics

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IP ADDRESSING

EASY STEPS TO SUBNET


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Objectives

Understand what an IP address is.

Understand why we need to subnet.

Be able to subnet

Be able to interpret CIDR notation


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IP Address

An IP address is a 32-bit sequence of 1s and 0s. To make the IP address easier to use, the address is

usually written as four decimal numbers separated byperiods.

This way of writing the address is called the dotteddecimal format.


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IP Private Addresses

No two machines that connect to a public network can have

the same IP address because public IP addresses are globaland standardized

Private IP addresses are a solution to the problem of the

exhaustion of public IP addresses. Addresses that fall withinthese ranges are not routed on the Internet backbone:

Connecting a network using private addresses to the

Internet requires the usage of NAT


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Logical Addressing

At the network layer, we use logical, hierarchicaladdressing.

With Internet Protocol (IP), this address is a 32-bit

addressing scheme divided into four octets. Do you remember the classes 1st octets value?

Class A: 1 - 126

Class B: 128 - 191

Class C: 192 - 223 Class D: 224 - 239 (multicasting)

Class E: 240 - 255 (experimental)


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Network vs. Host

N H H H

Class A: 27 = 126 networks; 224 > 16 million hosts

N N H H

Class B : 214 = 16,384 networks; 216 > 65,534 hosts

N N N H

Class C : 221 > 2 million networks; 28 = 254 hosts


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Why Subnet?

Remember: we are usually dealing with abroadcast topology.

Can you imagine what the network trafficoverhead would be like on a network with 254hosts trying to discover each others MACaddresses?

Subnetting allows us to segment LANs intological broadcast domains called subnets,thereby improving network performance.


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Four Subnetting Steps

To correctly subnet a given network address intosubnet addresses, ask yourself the followingquestions:

1. How many bits do I need to borrow?

2. Whats the subnet mask?

3. Whats the magic number or multiplier?

4. What are the first three subnetwork addresses?

Lets look at each of these questions in detail


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1. How many bits to borrow?

First, you need to know how many bits you haveto work with.

Second, you must know either how manysubnets you need or how many hosts per subnetyou need.

Finally, you need to figure out the number ofbits to borrow.


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How many bits do I have to work with?

Depends on the class of your network address.

Class C: 8 host bits

Class B: 16 host bits

Class A: 24 host bits

Remember: you must borrow at least 2 bits for

subnets and leave at least 2 bits for host addresses. 2 bits borrowed allows 2

2- 2 = 2 subnets


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How many subnets or hosts do I need?

A simple formula:

Host Bits = Bits Borrowed + Bits Left

HB = BB + BL

I need x subnets: x22BB

I need x hosts: x22BL

Remember: we need to subtract two toprovide for the subnetwork andbroadcast addresses.


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Class C Example: 210.93.45.0

Design goals specify at least 5 subnets so how manybits do we borrow?

How many bits in the host portion do we have to workwith (HB)?

Whats the BB in our HB = BB + BL formula? (8 =

BB + BL) 2 to the what power will give us at least 5 subnets?

23

- 2 = 6 subnets


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How many bits are left for hosts?

HB = BB + BL

8 = 3 + BLBL = 5

So how many hosts can we assign to each

subnet?2

5- 2 = 30 hosts


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Class B Example: 185.75.0.0

Design goals specify no more than 126 hosts persubnet, so how many bits do we need to leave (BL)?

How many bits in the host portion do we have to workwith (HB)?

Whats the BL in our HB = BB + BL formula? (16 =BB + BL)

2 to the what power will insure no more than 126 hostsper subnet and give us the most subnets?

27

- 2 = 126 hosts


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How many bits are left for subnets?

HB = BB + BL

16 = BB + 7BB = 9

So how many subnets can we have?

29

- 2 = 510 subnets


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2. Whats the subnet mask?

We determine the subnet mask by adding up the decimal valueof the bits we borrowed.

In the previous Class C example, we borrowed 3 bits. Below isthe host octet showing the bits we borrowed and their decimal

values.

128 64 32 16 8 4 2 1

1 1 1

We add up the decimal value of these bits and get 224.Thats the last non-zero octet of our subnet mask.

So our subnet mask is 255.255.255.224


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3. Whats the magic number?

To find the magic number or the multiplier

we will use to determine the subnetworkaddresses, we subtract the last non-zero octet

from 256.

In our Class C example, our subnet mask was255.255.255.224. 224 is our last non-zero octet.

Our magic number is 256 - 224 = 32


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Last Non-Zero Octet

Memorize this table. You should be able to: Quickly calculate the last non-zero octet when given the

number of bits borrowed.

Determine the number of bits borrowed given the last non-

zero octet. Determine the amount of bits left over for hosts and the

number of host addresses available.

Bits

Borrowed

Non-Zero

Octet Hosts

2 192 62

3 224 30

4 240 14

5 248 66 252 2


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4. What are the subnets?

We now take our magic number and use it as amultiplier.

Our Class C address was 210.93.45.0.

We borrowed bits in the fourth octet, so thats whereour multiplier occurs 1st subnet: 210.93.45.32

2nd subnet: 210.93.45.64

3rd subnet: 210.93.45.96 We keep adding 32 in the fourth octet to get all six

available subnet addresses.


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Host & Broadcast Addresses

Now you can see why we subtract 2 when determiningthe number of host address.

Lets look at our 1st subnet: 210.93.45.32

What is the total range of addresses up to our nextsubnet, 210.93.45.64? 210.93.45.32 to 210.93.45.63 or 32 addresses

.32 cannot be assigned to a host. Why?

.63 cannot be assigned to a host. Why? So our host addresses are .33 - .62 or 30 host addresses-

-just like we figured out earlier.


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CIDR


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CIDR Notation

Classless Interdomain Routing is a method ofrepresenting an IP address and its subnet mask with aprefix.

For example: 192.168.50.0/27 What do you think the 27 tells you?

27 is the number of 1 bits in the subnet mask. Therefore,255.255.255.224

Also, you know 192 is a Class C, so we borrowed 3 bits!! Finally, you know the magic number is 256 - 224 = 32, so the

first useable subnet address is 197.168.50.32!!

Lets see the power of CIDR notation.


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202.151.37.0/26

Subnet mask? 255.255.255.192

Bits borrowed?

Class C so 2 bits borrowed Magic Number?

256 - 192 = 64

First useable subnet address?

202.151.37.64 Third useable subnet address?

64 + 64 + 64 = 192, so 202.151.37.192


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198.53.67.0/30

Subnet mask? 255.255.255.252

Bits borrowed?

Class C so 6 bits borrowed Magic Number?

256 - 252 = 4

Third useable subnet address?

4 + 4 + 4 = 12, so 198.53.67.12 Second subnets broadcast address?

4 + 4 + 4 - 1 = 11, so 198.53.67.11


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200.39.89.0/28

What kind of address is 200.39.89.0?

Class C, so 4 bits borrowed

Last non-zero octet is 240

Magic number is 256 - 240 = 16

32 is a multiple of 16 so 200.39.89.32 is a subnetaddress--the second subnet address!!

Whats the broadcast address of 200.39.89.32? 32 + 16 -1 = 47, so 200.39.89.47


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194.53.45.0/29

What kind of address is 194.53.45.26? Class C, so 5 bits borrowed

Last non-zero octet is 248

Magic number is 256 - 248 = 8

Subnets are .8, .16, .24, .32, ect.

So 194.53.45.26 belongs to the third subnet address(194.53.45.24) and is a host address.

What broadcast address would this host use tocommunicate with other devices on the same subnet? It belongs to .24 and the next is .32, so 1 less is .31

(194.53.45.31)


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No Worksheet Needed!

After some practice, you should never need asubnetting worksheet again.

The only information you need is the IP address andthe CIDR notation.

For example, the address 221.39.50/26 You can quickly determine that the first subnet address

is 221.39.50.64. How? Class C, 2 bits borrowed

256 - 192 = 64, so 221.39.50.64 For the rest of the addresses, just do multiples of 64

(.64, .128, .192).


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The Key!!

MEMORIZE THIS TABLE!!!

Bits

Borrowed

Non-Zero

Octet

1 128

2 192

3 224

4 2406 252

7 254

8 255


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Practice On Your Own

Below are some practice problems. Take out a sheetof paper and calculate... Bits borrowed

Last non-zero octet

Second subnet address and broadcast address

1. 192.168.15.0/26

2. 220.75.32.0/30

3. 200.39.79.0/294. 195.50.120.0/27

5. 202.139.67.0/28


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Address Class

Bits

Borrowed

Last Non-

Zero Octet

Magic

Number

2nd Subnet's

Address

2nd Subnet's

Broadcast

192.168.15.0/26 C 2 192 64 192.168.15.128 192.168.15.191

220.75.32.0/30 C 6 252 4 220.75.32.8 220.75.32.11

200.39.79.0/29 C 5 248 8 200.39.79.16 220.39.79.23195.50.120.0/27 C 3 224 32 195.50.120.64 195.50.120.95

202.139.67.0/28 C 4 240 16 202.139.67.32 202.139.67.47

132.59.0.0/19 B 3 224 32 132.59.64.0 132.59.95.255

64.0.0.0/16 A 8 255 1 64.2.0.0 64.2.255.255

Challenge:

Dont Cheat Yourself!!

Work them out before you check your answers. Clickthe back button if youre not done. Otherwise, clickanywhere else in the screen to see the answers.

Date post:	08-Aug-2018
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Subnetting Basics

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