32 Bit Number 4 Octets

8 Bits

Two pieces of information:1. Network Address2. Host Address

IP Clas

s

First Octet

Networks

Hosts

A 1-126 126 16,777,214

B 128-191 16,384 65,534

C 192-223 2,097,152

254

255.0.0.08 network

bits24 host bits

255.255.0.016 network

bits16 host bits

255.255.255.0

24 network bits

8 host bits

Class A Class B Class C

Benefits Reduced network traffic

Optimized network performance Simplified management

Security purposes Facilitated spanning of large geographical distances

8th digit 7th digit 6th digit 5th digit 4th digit 3rd digit 2nd digit 1st digit

128 64 32 16 8 4 2 1

16th digit 15th digit 14th digit 13th digit 12th digit 11th digit 10th digit 9th digit

32,768 16,384 8192 4096 2048 1024 512 256

32 Bit Binary Definitive ChartLet’s look at a couple of charts that will help us see how we count in binary, and how we apply bits

to figure out sub-netting. As you can see, each digit doubles in value as it travels left.

32nd digit 31st digit 30th digit 29th digit 28th digit 27th digit 26th digit 25th digit

2,147,483,648 1,073,741,824 536,870,912

268,435,456

134,217, 728

67,108,864 33,554,432 16,777,216

24th digit 23rd digit 22nd digit 21st digit 20th digit 19th digit 18th digit 17th digit

8,388,608 4,194,304 2,097,152 1,048,576 524,288 262,144 131,072 65,536

Borrowed

Bits

128 64 32 16 8 4 2 1 Value LSB

1 1 0 0 0 0 0 0 0 128 128

2 1 1 0 0 0 0 0 0 192 64

3 1 1 1 0 0 0 0 0 224 32

4 1 1 1 1 0 0 0 0 240 16

5 1 1 1 1 1 0 0 0 248 8

6 1 1 1 1 1 1 0 0 252 4

7 1 1 1 1 1 1 1 0 254 2

8 1 1 1 1 1 1 1 1 255 1

Applying Borrowed Bits to Subnet Masks

1) Determine class of network. 1-126=Class A 128-91=Class B 192-223=Class C

2) Establish the default subnet mask and number of host bits. Class A=255.0.0.0 Class B=255.255.0.0 Class C=255.255.255.0

3) Determine the number of bits to borrow or retain and how many remaining. Define subnet mask. Universal calculator (2n-2>/X) X=#of networks and n=represents # bits in subnet mask

4) Determine the number of subnets and hosts.5) Establish the least significant subnet bit.

Your subnets will increment by the least significant bit (lsb).

6) List the first and last subnet in the series, as well as the first and last host on each subnet.

209.168.19.0 Create 4 networks with largest possible number of host ID’s

1. Determine class: “C”2. Default Subnet mask and host bits: 255.255.255. 0 / 8 host bits3. Determine number of bits to borrow to create 4 networks. Define custom

subnet mask.

128 64 32 16 8 4 2 127 26 25 24 23 22 21 20

Universal Calculator: (2n-2>/X) X=#of networks and n=# bits in subnet mask.

Solve for borrowed bits Custom subnet mask 2n-2>/X x.x.x.11100000 23-2>/4 128+64+32= 224Borrow 3 bits 255.255.255.224

4. Determine number of subnets and hosts:

Solve for subnets: 2borrowed -2 23= 8-2 = 6 subnets Solve for hosts: 2retained – 2 25= 32-2 = 30 hosts

128 64 32 16 8 4 2 127 26 25 24 23 22 21 20

5. Establish the least significant subnet bit: x.x.x.11100000

The least significant bit is 32, therefore our subnets should increment by 32.

6. List the first and last subnet in the series, as well as the first and last host on each subnet. Let’s look at how that looks in Binary.

128 64 32 16 8 4 2 10 0 1 0 0 0 0 0

Bit # 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

With a Class “C” Network we have 8 host bits:

In our example, we have borrowed 3 bits to create additional networks so, our bits has to be divided as follows:

List out the first and last subnet, and first and last host on each subnet: This is where understanding binary, and how to list a series is important.Remember that the first in a series is all zeros and a one, and the last in a series is all ones and a zero.

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

0 0 1 0 0 0 0 0

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

0 0 1 0 0 0 0 1

Let’s look at how we list out our first subnet, remember the first in a series is all zeros and a one.

Our first subnet in binary is “32” because we borrowed 3 bits, which also is our least significant bit. Our first subnet is 209.168.19.32.

Let’s look at how we find the first host on the first subnet, remember the rule?

Our first host on our first subnet is “33” (32+1) because that is the value of all of the bits in this octet. Our first host on the first subnet is 209.168.19.33.

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

0 0 1 1 1 1 1 0

The last host on our first subnet is “62” (32+16+8+4+2). The last host on the first subnet is 209.168.19.62.

Our first subnet looks like this 209.168.19.32First host 209.168.19.33 Last host 209.168.19.62

How would you figure out the last host on the first subnet? The last in a series is all ones and a zero.

Let’s look at how we would determine the last subnet, remember the last in a series is all ones and a zero. The last subnet is 209.168.19.192

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0

Tip: Use the custom subnet mask minus the LSB (224 – 32 = 192)

The first host on the last subnet?

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 1

The first host on the last subnet is 209.168.19.193.

The last host on the last subnet?

Network Bits Host Bits

Bit# 8 7 6 5 4 3 2 1

Value 128 64 32 16 8 4 2 1

1 1 0 1 1 1 1 0

The last host on the last subnet is 209.168.19.222.

The last subnet is 209.168.19.192 First host 209.168.19.193 Last host 209.168.19.222

1st subnet 209.168.19.32 First host 209.168.19.33 Last host 209.168.19.622nd subnet 209.168.19.64 First host 209.168.19.65 Last host 209.168.19.943rd subnet 209.168.19.96 First host 209.168.19.97 Last host 209.168.19.1264th subnet 209.168.19.128 First host 209.168.19.129 Last host

209.168.19.1585th subnet 209.168.19.160 First host 209.168.19.161 Last host

209.168.19.1906th subnet 209.168.19.192 First host 209.168.19.193 Last host

209.168.19.222