Lab – Calculating IPv4 Subnets
Objectives
Part 1: Determine IPv4 Address
Subnetting
Part 2: Calculate IPv4 Address
Subnetting
Background / Scenario
The ability to work with IPv4 subnets and
determine network and host information based on a given IP address and subnet
mask is critical to understanding how IPv4 networks operate. The first part is
designed to reinforce how to compute network IP address information from a
given IP address and subnet mask. When given an IP address and subnet mask, you
will be able to determine other information about the subnet.
Required Resources
·
1 PC (Windows 7 or 8 with
Internet access)
·
Optional: IPv4 address
calculator
Part 1:
Determine IPv4 Address
Subnetting
In Part 1, you will determine the network
and broadcast addresses, as well as the number of hosts, given an IPv4 address
and subnet mask.
REVIEW: To determine the network address, perform binary ANDing on the
IPv4 address using the subnet mask provided. The result will be the network
address. Hint: If the subnet mask has decimal value 255 in an octet, the result
will ALWAYS be the original value of that octet. If the subnet mask has decimal
value 0 in an octet, the result will ALWAYS be 0 for that octet.
Example:
IP Address 192.168.10.10
Subnet
Mask 255.255.255.0
==========
Result (Network) 192.168.10.0
Knowing this, you may only have to
perform binary ANDing on an octet that does not have 255 or 0 in its subnet
mask portion.
Example:
IP Address 172.30.239.145
Subnet
Mask 255.255.192.0
Analyzing this example, you can see that
you only have to perform binary ANDing on the third octet. The first two octets
will result in 172.30 due to the subnet mask. The fourth octet will result in 0
due to the subnet mask.
IP Address 172.30.239.145
Subnet
Mask 255.255.192.0
==========
Result (Network) 172.30.?.0
Perform binary ANDing on the third octet.
Decimal Binary
239 11101111
192 11000000
=======
Result 192 11000000
Analyzing this example again produces the
following result:
IP
Address 172.30.239.145
Subnet
Mask 255.255.192.0
==========
Result (Network) 172.30.192.0
Continuing with this example, determining
the number of hosts per network can be calculated by analyzing the subnet mask.
The subnet mask will be represented in dotted decimal format, such as
255.255.192.0, or in network prefix format, such as /18. An IPv4 address always
has 32 bits. Subtracting the number of bits used for the network portion (as
represented by the subnet mask) gives you the number of bits used for hosts.
Using our example above, the subnet mask
255.255.192.0 is equivalent to /18 in prefix notation. Subtracting 18 network
bits from 32 bits results in 14 bits left for the host portion. From there, it
is a simple calculation:
2(number of host bits) -
2 = Number of hosts
214 = 16,384 – 2 =
16,382 hosts
Determine the network and broadcast
addresses and number of host bits and hosts for the given IPv4 addresses and prefixes
in the following table.
|
IPv4 Address/Prefix
|
Network Address
|
Broadcast Address
|
Total Number of Host Bits
|
Total Number of Hosts
|
|
192.168.100.25/28
|
192.168.100.16
|
192.168.100.31
|
4
|
2^4-2=14
|
|
172.30.10.130/30
|
172.30.10.128
|
172.30.10.131
|
2
|
2^2-2=2
|
|
10.1.113.75/19
|
10.1.96.0
|
10.1.127.255
|
13
|
2^13-2=8190
|
|
198.133.219.250/24
|
198.133.219.0
|
198.133.219.255
|
8
|
2^8-2=254
|
|
128.107.14.191/22
|
128.107.12.0
|
128.107.15.255
|
10
|
2^10-2=1022
|
|
172.16.104.99/27
|
172.16.104.96
|
172.16.104.127
|
5
|
2^5-2=30
|
Part 2:
Calculate IPv4 Address
Subnetting
When given an IPv4 address, the original
subnet mask and the new subnet mask, you will be able to determine:
·
Network address of this subnet
·
Broadcast address of this
subnet
·
Range of host addresses of this
subnet
·
Number of subnets created
·
Number of hosts per subnet
The following
example shows a sample problem along with the solution for solving this problem:
|
Given:
|
|
|
Host
IP Address:
|
172.16.77.120
|
|
Original
Subnet Mask
|
255.255.0.0
|
|
New
Subnet Mask:
|
255.255.240.0
|
|
Find:
|
|
|
Number
of Subnet Bits
|
4
|
|
Number
of Subnets Created
|
16
|
|
Number
of Host Bits per Subnet
|
12
|
|
Number
of Hosts per Subnet
|
4,094
|
|
Network
Address of this Subnet
|
172.16.64.0
|
|
IPv4
Address of First Host on this Subnet
|
172.16.64.1
|
|
IPv4
Address of Last Host on this Subnet
|
172.16.79.254
|
|
IPv4
Broadcast Address on this Subnet
|
172.16.79.255
|
Let’s analyze how this table was completed.
The original subnet mask was 255.255.0.0
or /16. The new subnet mask is 255.255.240.0 or /20. The resulting difference
is 4 bits. Because 4 bits were borrowed, we can determine that 16 subnets were
created because 24 = 16.
The new mask of 255.255.240.0 or /20
leaves 12 bits for hosts. With 12 bits left for hosts, we use the following
formula: 212 = 4,096 – 2 = 4,094 hosts per subnet.
Binary ANDing will help you determine the
subnet for this problem, which results in the network 172.16.64.0.
Finally, you need to determine the first
host, last host, and broadcast address for each subnet. One method to determine
the host range is to use binary math for the host portion of the address. In
our example, the last 12 bits of the address is the host portion. The first
host would have all significant bits set to zero and the least significant bit
set to 1. The last host would have all significant bits set to 1 and the least
significant bit set to 0. In this example, the host portion of the address
resides in the 3rd and 4th octets.
|
Description
|
1st Octet
|
2nd Octet
|
3rd Octet
|
4th Octet
|
Description
|
|
Network/Host
|
nnnnnnnn
|
nnnnnnnn
|
nnnnhhhh
|
hhhhhhhh
|
Subnet Mask
|
|
Binary
|
10101100
|
00010000
|
01000000
|
00000001
|
First Host
|
|
Decimal
|
172
|
16
|
64
|
1
|
First Host
|
|
Binary
|
10101100
|
00010000
|
01001111
|
11111110
|
Last Host
|
|
Decimal
|
172
|
16
|
79
|
254
|
Last Host
|
|
Binary
|
10101100
|
00010000
|
01001111
|
11111111
|
Broadcast
|
|
Decimal
|
172
|
16
|
79
|
255
|
Broadcast
|
Step 1:
Fill out the tables below with appropriate
answers given the IPv4 address, original subnet mask, and new subnet mask.
a.
Problem 1:
|
Given:
|
|
|
Host
IP Address:
|
192.168.200.139
|
|
Original
Subnet Mask
|
255.255.255.0
|
|
New
Subnet Mask:
|
255.255.255.224
|
|
Find:
|
|
|
Number
of Subnet Bits
|
3
|
|
Number
of Subnets Created
|
2^3=8
|
|
Number
of Host Bits per Subnet
|
5
|
|
Number
of Hosts per Subnet
|
2^5-2=30
|
|
Network
Address of this Subnet
|
192.168.200.128
|
|
IPv4
Address of First Host on this Subnet
|
192.168.200.129
|
|
IPv4
Address of Last Host on this Subnet
|
192.168.200.158
|
|
IPv4
Broadcast Address on this Subnet
|
192.168.200.159
|
b.
Problem 2:
|
Given:
|
|
|
Host
IP Address:
|
10.101.99.228
|
|
Original
Subnet Mask
|
255.0.0.0
|
|
New
Subnet Mask:
|
255.255.128.0
|
|
Find:
|
|
|
Number
of Subnet Bits
|
9
|
|
Number
of Subnets Created
|
2^9=512
|
|
Number
of Host Bits per Subnet
|
15
|
|
Number
of Hosts per Subnet
|
2^15-2=32766
|
|
Network
Address of this Subnet
|
10.101.0.0
|
|
IPv4
Address of First Host on this Subnet
|
10.101.0.1
|
|
IPv4
Address of Last Host on this Subnet
|
10.101.127.254
|
|
IPv4
Broadcast Address on this Subnet
|
10.101.127.255
|
c.
Problem 3:
|
Given:
|
|
|
Host
IP Address:
|
172.22.32.12
|
|
Original
Subnet Mask
|
255.255.0.0
|
|
New
Subnet Mask:
|
255.255.224.0
|
|
Find:
|
|
|
Number
of Subnet Bits
|
3
|
|
Number
of Subnets Created
|
2^3=8
|
|
Number
of Host Bits per Subnet
|
13
|
|
Number
of Hosts per Subnet
|
2^13-2=8190
|
|
Network
Address of this Subnet
|
172.22.32.0
|
|
IPv4
Address of First Host on this Subnet
|
172.22.32.1
|
|
IPv4
Address of Last Host on this Subnet
|
172.22.63.254
|
|
IPv4
Broadcast Address on this Subnet
|
172.22.63.255
|
d.
Problem 4:
|
Given:
|
|
|
Host
IP Address:
|
192.168.1.245
|
|
Original
Subnet Mask
|
255.255.255.0
|
|
New
Subnet Mask:
|
255.255.255.252
|
|
Find:
|
|
|
Number
of Subnet Bits
|
6
|
|
Number
of Subnets Created
|
2^6=64
|
|
Number
of Host Bits per Subnet
|
2
|
|
Number
of Hosts per Subnet
|
2^2-2=2
|
|
Network
Address of this Subnet
|
192.168.1.244
|
|
IPv4
Address of First Host on this Subnet
|
192.168.1.245
|
|
IPv4
Address of Last Host on this Subnet
|
192.168.1.246
|
|
IPv4
Broadcast Address on this Subnet
|
192.168.1.247
|
e.
Problem 5:
|
Given:
|
|
|
Host
IP Address:
|
128.107.0.55
|
|
Original
Subnet Mask
|
255.255.0.0
|
|
New
Subnet Mask:
|
255.255.255.0
|
|
Find:
|
|
|
Number
of Subnet Bits
|
8
|
|
Number
of Subnets Created
|
2^8=256
|
|
Number
of Host Bits per Subnet
|
8
|
|
Number
of Hosts per Subnet
|
2^8-2=254
|
|
Network
Address of this Subnet
|
128.107.0.0
|
|
IPv4
Address of First Host on this Subnet
|
128.107.0.1
|
|
IPv4
Address of Last Host on this Subnet
|
128.107.0.254
|
|
IPv4
Broadcast Address on this Subnet
|
128.107.0.255
|
f.
Problem 6:
|
Given:
|
|
|
Host
IP Address:
|
192.135.250.180
|
|
Original
Subnet Mask
|
255.255.255.0
|
|
New
Subnet Mask:
|
255.255.255.248
|
|
Find:
|
|
|
Number
of Subnet Bits
|
5
|
|
Number
of Subnets Created
|
2^5=32
|
|
Number
of Host Bits per Subnet
|
3
|
|
Number
of Hosts per Subnet
|
2^3-2=6
|
|
Network
Address of this Subnet
|
192.135.250.176
|
|
IPv4
Address of First Host on this Subnet
|
192.135.250.177
|
|
IPv4
Address of Last Host on this Subnet
|
192.135.250.182
|
|
IPv4
Broadcast Address on this Subnet
|
192.135.250.183
|
Reflection
Why is the subnet mask so important when analyzing
an IPv4 address?
__________The subnet mask determines everything about the address.
Merely looking at an IPv4 address tells you nothing. You need the subnet mask
to fill in all the important pieces of information. _____________________________________________________________________________
http://erdoganakbiyik.blogspot.com/
https://www.youtube.com/channel/UCDsUnmBfVdEPkcC8FlzPKcg
No comments:
Post a Comment