Subnet Tutorials: September 2013

7.9.13

Variable Length Subnet Mask - VLSM

Variable Length Subnet Mask-VLSM

1.With Variable Length Subnet Mask (VLSM), a network administrator can use a long mask on networks with few hosts, and a short mask on subnets with many hosts.
2.A VLSM is a means of allocating IP addressing resources to subnets according to their individual need rather than some general network-wide rule.
3.Implementing VLSM is often referred to as "subnetting a subnet", and can be used to maximize addressing efficiency.
4.VLSM allows an organization to use more than one subnet mask within the same network address space.
5.When using VLSM to subnet an address, not all of the subnets have to be the same size.
A different subnet mask may be applied to some of the subnets to further subnet the address.
6.VLSM allows to divide big networks into smaller, different-sized slices. This enables you to make maximum use of your valuable IP address space.
7.In order to take advantage of VLSM, the proper routing protocol must be selected.

IP Subnet zero

In the old days, IP Subnet-zero was not default in routers. i.e. Some older technologies don't allow the use of this first subnet.
The first IP is used as subnet and last IP as the broadcast.
It has nothing to do with the network and broadcast addresses.

Example: 1

Let's take the 10.0.0.0/8 network, and use  3 bits for subnetting.
That is 255.0.0.0 will be 255.11100000.00000000.00000000
                         = 255.224.0.0
 
Subnets = 2^3 - 2 = 254.
Valid subnets = 256 – 224 = 32. 0, 32, 64, 96, 128, 160, 192, 224. 
Block Size = 2^5 = 32.
So, by using the first 3 bits, we can create 8 subnets,
they would be:
10.0.0.0/11    first subnet
10.32.0.0/11
10.64.0.0/11
10.96.0.0/11
10.128.0.0/11
10.160.0.0/11
10.192.0.0/11
10.224.0.0/11  last subnet

In binary, the first 3 bits of the 2nd octet for these networks
would look like this:
000   This is the first subnet 
001
010
011
100
101
110
111   This is the last subnet in the range

If the subnet 0 is not allowed, it actually means that we can't use subnet 000 and subnet 111 in this example. (10.0.0.0/11 and 10.224.0.0/11).
The calculation for subnets, without subnet 0, is 2^(power of borrowed bits) – 2 = Amount of Subnets.

Example: 2

When we do VLSM, say we break 192.168.1.0/24 to four subnets
192.168.1.0/26 – can’t be used as 192.168.1.0/24 is already named
192.168.1.64/26
192.168.1.128/26
192.168.1.192/26 – can’t be used as it contains 192.168.1.255/24
which is already used as broadcast.

IP subnet-zero, allows for the first and last subnet in a
subnetted network to be used.

IP Subnet zero

Route Summarization Practice

Subnet Route Summarization Practice Questions

In large internetworks, hundreds, or even thousands of network addresses can exist. It is often problem for routers to maintain these volumes of routes in their routing tables. Route summarization can reduce the number of routes that a router must maintain, because it is a method of representing a series of network numbers in a single summary address. Or in other words, summarization is a method of finding a common network address for all the networks taking part in the summary process.

In Summarization we take multiple small network and logically group them into 1 single big network.

Summarization gives one single network ID to route if some networks are contagious.

A summary route is announced by the summarizing router as long as at least one specific route in its routing table matches the summary route.

When doing summary, forget about the original mask or CIDR notation. Just convert to binary, check where they are separated, count common bits, no matter whether they are 0s or 1s and that is your new summary mask.

For example, in the figure above, router R1 can either send five routing update entries or
summarize of the four addresses into a single network number. If router R1 summarizes the information into a single network number entry, the following things happen:
1. Bandwidth is saved on the link between routers R1 and R2.
2. Router R2 needs to maintain only one route and therefore saves memory.
3. Router R2 also saves CPU resources, because it evaluates packets against fewer entries in its routing table.

Example 1:

Summarize the network given below:
172.16.0.0/24
172.16.1.0/24
172.16.2.0/24
172.16.3.0/24

Solution

The binary equivalents of the shown addresses are as follows:
3rd octet is the significant one, let's convert 3rd octet to binary.
172.16.0.0 = 172.16.000000 00.00000000
172.16.1.0 = 172.16.000000 01.00000000
172.16.2.0 = 172.16.000000 10.00000000
172.16.3.0 = 172.16.000000 11.00000000
Here the first 16 bits (172.16) are fixed means the third portion is varying from 0 to 7.
The last 2 bits of the 3rd octet makes the difference here, first 6 bits of the 3rd octet is identical.
The remaining 6 ones (111111) are added to our fixed 16 bit network portion, 16+6 = 22, or 8+8+6=22, these are the identical bits starting from the leftmost.
Now 22 bits are fixed represents the network portion and the remaining is 12 bits are our host portion.
So the summary address is 172.16.0.0/22.

Example 2:

Summarize the network given below:
172.16.4.0/24
172.16.5.0/24
172.16.6.0/24
172.16.128.0/24

Solution

The binary equivalents of the shown addresses are as follows:
3rd octet is the significant one, let's convert 3rd octet to binary.
172.16.4.0 = 10101100.00010000.00000100.00000000
172.16.5.0 = 10101100.00010000.00000101.00000000
172.16.6.0 = 10101100.00010000.00000110.00000000
172.16.128.0 = 10101100.00010000.10000000.00000000
Here the first 16 bits (172.16) are fixed means the third portion is varying from 0 to 7.
Because of the last subnet (172.16.128.0/24), a good summarization is not possible
with these subnets. Now 16 bits are fixed represents the network portion and the remaining is 16 bits are our host portion.
The summary address is 172.16.0.0/16.
We must drop back to the classful summarization of 172.16.0.0/16. Because discontiguous network addressing can
destroy our network summarization efficiency.

Example 3:

Summarize the network given below:
172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24

Solution

172.1.000001 00.00000000
172.1.000001 00.10000000
172.1.000001 01.00000000
172.1.000001 10.00000000
172.1.000001 11.00000000
---------------------------------
172.1.00000100.00000000 - Our summary network.
mask is: 8 bits+8bits+6 bits= /22
or in binary: 11111111.11111111.11111100.00000000=/22
So the summary address is 172.1.4.0/22.

Example 4:

Summarize the network given below:
192.168.1.0/24
192.168.2.0/24
192.168.3.0/24
192.168.4.0/24

Solution

Check for the common bits in all the octets….
192.168.1.0 ---- 11000000.10101000.00000 001.00000000
192.168.2.0 ---- 11000000.10101000.00000 011.00000000
192.168.3.0 ---- 11000000.10101000.00000 011.00000000
192.168.4.0 ---- 11000000.10101000.00000 100.00000000
-----------------------------------------------
192.168.00000000.00000000 - Our summary network.
mask is: 8 bits+8 bits+5 bits(common bits)= /21
or in binary: 11111111.11111111.11111000.00000000=/21
So the summary address is 192.168.0.0/21.

Example 5:

Summarize the network given below:
192.168.0.0/24
192.168.1.0/24
192.168.2.0/24
192.168.3.0/24
192.168.4.0/24
192.168.5.0/24
192.168.6.0/24
192.168.7.0/24

Solution

3rd octet is the significant one, lets convert 3rd octet to binary.

192.168.0.0 = 192.168. 00000 000 .0
192.168.1.0 = 192.168. 00000 001 .0
192.168.2.0 = 192.168. 00000 010 .0
192.168.3.0 = 192.168. 00000 011 .0
192.168.4.0 = 192.168. 00000 100 .0
192.168.5.0 = 192.168. 00000 101 .0
192.168.6.0 = 192.168. 00000 110 .0
192.168.7.0 = 192.168. 00000 111 .0
-----------------------------------------------
192.168.00000000.00000000 - Our summary network.
mask is: 8 bits+8bits+5 bits= /21
or in binary: 11111111.11111111.11111000.00000000=/21
So the summary address is 192.168.0.0/21.

Example 6:

Summarize the network given below:
192.168.112.0/24
192.168.113.0/24
192.168.114.0/24
192.168.115.0/24
192.168.116.0/24
192.168.117.0/24
192.168.118.0/24
192.168.119.0/24

Solution

192.168.112.0 = 192.168.01110 000.00000000
192.168.113.0 = 192.168.01110 001.00000000
192.168.114.0 = 192.168.01110 010.00000000
192.168.115.0 = 192.168.01110 011.00000000
192.168.116.0 = 192.168.01110 100.00000000
192.168.117.0 = 192.168.01110 101.00000000
192.168.118.0 = 192.168.01110 110.00000000
192.168.119.0 = 192.168.01110 111.00000000
Based on these binary values, you can see that the first 21 bits on each one of these IP addresses are the same.
Here the first 16 bits (192.168) are fixed means the third portion is varying from 0 to 7.
The last 3 bits of the 3rd octet makes the difference here, first 5 bits of the 3rd octet is identical.
It means we need three bits in third portion that is 11111(000) as we take these 000 for our use.
The remaining 5 ones (11111) are added to our fixed 16 bit network portion, 16+5 = 21. i.e. as each octet is 8 bits, the mask will be 8+8+5=21, these are the identical bits starting from the leftmost.
Now 21 bits are fixed represents the network portion and the remaining is 11 bits are our host portion.
So the summary address is 192.168.112.0/21.

Exercise

Exercise 1:

10.1.1.0/24
10.1.2.0/24
10.1.3.0/24
10.1.4.0/24

Answer:

 Summary: 10.1.0.0/21

Exercise 2:

10.10.1.0/27
10.10.1.32/28
10.10.1.48/26
10.10.1.64/25
10.10.1.128/25

Answer:

Summary: 10.10.1.0/24

Exercise 3:

10.10.1.0/27
10.10.1.32/28
10.10.1.48/28
10.10.1.64/26
10.10.1.128/29
10.10.1.136/29
10.10.1.144/28

Answer:

Summary: 10.10.1.0/24

5.9.13

Broadcast Valid Host Range Class C

Subnet, Broadcast, Valid Host Range:Class C

Example 1:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.0 255.255.255.128(/25).

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.128
                  = 255. 255. 255. 10000000
Subnets = 2^1 = 2.
Hosts = 2^7 – 2 = 126.
Block Size = 2^7 = 128.
Valid subnets = 256 – 128 = 128, means we have a block size of
128 in the forth octet.
Subnets are 0, 128.
Address 0 falls in the 0 subnet.
Subnet = 192. 168. 10. 0
The subnets would be 192.168.10.0, 192.168.10.128.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.126	192.168.10.127
192.168.10.128	192.168.10.129-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.0.
The broadcast address is: 192.168.10.127.
The valid host range: 192.168.10.1-192.168.10.126.

Example 2:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.0 255.255.255.192(/26).

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.192
                  = 255. 255. 255. 11000000
Subnets = 2^2 = 4.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, means we have a block size of
64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 0 falls in the 0 subnet.
Subnet = 192. 168. 10. 0
The subnets would be 192.168.10.0, 192.168.10.64, 192.168.10.128,
192.168.10.192.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.62	192.168.10.63
192.168.10.64	192.168.10.65-192.168.10.126	192.168.10.127
192.168.10.128	192.168.10.129-192.168.10.190	192.168.10.191
192.168.10.192	192.168.10.193-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.0.
The broadcast address is: 192.168.10.63.
The valid host range: 192.168.10.1-192.168.10.62.

Example 3:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.0 255.255.255.224(/27).

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.224
                  = 255. 255. 255. 11100000
Subnets = 2^3 = 8.
Hosts = 2^5 – 2 = 30.
Block Size =2^5 = 32.
Valid subnets = 256 – 224 = 32, means we have a block size of
32 in the forth octet.
Subnets are 0, 32, 64, 96, 128, 160, 192, 224.
Address 0 falls in the 0 subnet.
Subnet = 192. 168. 10. 0
The subnets would be 192.168.10.0, 192.168.10.32, 192.168.10.64,
192.168.10.96, 192.168.10.128, 192.168.10.160, 192.168.10.192,
192.168.10.224.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.30	192.168.10.31
192.168.10.32	192.168.10.33-192.168.10.62	192.168.10.63
192.168.10.64	192.168.10.65-192.168.10.94	192.168.10.95
192.168.10.96	192.168.10.97-192.168.10.126	192.168.10.127
192.168.10.128	192.168.10.129-192.168.10.158	192.168.10.159
192.168.10.160	192.168.10.161-192.168.10.190	192.168.10.191
192.168.10.192	192.168.10.193-192.168.10.222	192.168.10.223
192.168.10.224	192.168.10.225-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.0.
The broadcast address is: 192.168.10.31.
The valid host range: 192.168.10.1-192.168.10.30.

Example 4:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.248.0(/21).

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.240
                  = 255. 255. 255. 11110000
Subnets = 2^4 = 16.
Hosts = 2^4 – 2 = 14.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, means we have a block size of
16 in the forth octet.
Subnets are 0, 16, 32, 48, 64, 80, … , 240.
Address 0 falls in the 0 subnet.
Subnet = 192. 168. 10. 0
The subnets would be 192.168.10.0, 192.168.10.16, 192.168.10.32,
192.168.10.48, 192.168.10.64, 192.168.10.80, … , 192.168.10.240.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.30	192.168.10.31
192.168.10.32	192.168.10.33-192.168.10.46	192.168.10.47
192.168.10.48	192.168.10.49-192.168.10.62	192.168.10.63
192.168.10.64	192.168.10.65-192.168.10.78	192.168.10.79
192.168.10.80	192.168.10.81-192.168.10.94	192.168.10.95
.	.	.
192.168.10.240	192.168.10.241-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.0.
The broadcast address is: 192.168.10.15.
The valid host range: 192.168.10.1-192.168.10.14.

Example 5:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.0 255.255.255.248(/29).

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
                  = 255. 255. 255. 11111000
Subnets = 2^5 = 32.
Hosts = 2^3 – 2 = 8.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a block size of
8 in the forth octet.
Subnets are 0, 8, 16, 24, 32, 40, … , 248.
Address 0 falls in the 0 subnet.
Subnet = 192. 168. 10. 0
The subnets would be 192.168.10.0, 192.168.10.8, 192.168.10.16,
192.168.10.24, 192.168.10.32, 192.168.10.40, … , 192.168.10.248.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.6	192.168.10.7
192.168.10.8	192.168.10.9-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.22	192.168.10.23
192.168.10.24	192.168.10.25-192.168.10.30	192.168.10.31
192.168.10.32	192.168.10.33-192.168.10.38	192.168.10.39
192.168.10.40	192.168.10.41-192.168.10.46	192.168.10.47
.	.	.
192.168.10.240	192.168.10.241-192.168.10.246	192.168.10.247
192.168.10.248	192.168.10.249-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.0.
The broadcast address is: 192.168.10.7.
The valid host range: 192.168.10.1-192.168.10.6.

Example 6:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.0 255.255.255.252(/30).

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.252
                  = 255. 255. 255. 11111100
Subnets = 2^6 = 64.
Hosts = 2^2 – 2 = 2.
Block Size = 2^2 = 4.
Valid subnets = 256 – 252 = 4, means we have a block size of
4 in the forth octet.
Subnets are 0, 4, 8, 12, 16, … , 248, 252.
Address 0 falls in the 0 subnet.
Subnet = 192. 168. 10. 0
The subnets would be 192.168.10.0, 192.168.10.4, 192.168.10.8,
192.168.10.12, 192.168.10.16, … , 192.168.10.248, 192.168.10.252.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.2	192.168.10.3
192.168.10.4	192.168.10.5-192.168.10.6	192.168.10.7
192.168.10.8	192.168.10.9-192.168.10.10	192.168.10.11
192.168.10.12	192.168.10.13-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.18	192.168.10.19
.	.	.
192.168.10.248	192.168.10.249-192.168.10.250	192.168.10.251
192.168.10.252	192.168.10.253-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.0.
The broadcast address is:192.168.10.3.
The valid host range: 192.168.10.1-192.168.10.2.

Example 7:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.68 255.255.255.192.

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.192
                  = 255. 255. 255. 11000000
Subnets = 2^2 = 4.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 4.
Valid subnets = 256 – 192 = 64, means we have a block size of
64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 68 falls in the 64 subnet.
Subnet = 192. 168. 10. 64
The subnets would be 192.168.10.0, 192.168.10.64, 192.168.10.128,
192.168.10.192.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.62	192.168.10.63
192.168.10.64	192.168.10.65-192.168.10.126	192.168.10.127
192.168.10.128	192.168.10.129-192.168.10.190	192.168.10.191
192.168.10.192	192.168.10.193-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.64.
The broadcast address is: 192.168.10.127.
The valid host range: 192.168.10.65-192.168.10.126.

Example 8:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.65 255.255.255.224.

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.224
                  = 255. 255. 255. 11100000
Subnets = 2^3 = 8.
Hosts = 2^5 – 2 = 30.
Block Size = 2^5 = 32.
Valid subnets = 256 – 224 = 32, means we have a block size of
32 in the forth octet.
Subnets are 0, 32, 64, 96, … , 224.
Address 65 falls in the 64 subnet.
Subnet = 192. 168. 10. 64
The subnets would be 192.168.10.0, 192.168.10.32, 192.168.10.64,
192.168.10.96, … ,192.168.10.224.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.30	192.168.10.31
192.168.10.32	192.168.10.33-192.168.10.62	192.168.10.63
192.168.10.64	192.168.10.65-192.168.10.94	192.168.10.95
192.168.10.96	192.168.10.97-192.168.10.126	192.168.10.127
.	.	.
192.168.10.224	192.168.10.225-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.64.
The broadcast address is: 192.168.10.95.
The valid host range: 192.168.10.65-192.168.10.94.

Example 9:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.38 255.255.255.240.

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.240
                  = 255. 255. 255. 11110000
Subnets = 2^ = 16.
Hosts = 2^4 – 2 = 14.
Block Size = 2^5 = 32.
Valid subnets = 256 – 224 = 16, means we have a block size of
16 in the forth octet.
Subnets are 0, 16, 32, 48, … , 240.
Address 38 falls in the 32 subnet.
Subnet = 192. 168. 10. 32
The subnets would be 192.168.10.0, 192.168.10.16, 192.168.10.32,
192.168.10.48, … ,192.168.10.240.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.30	192.168.10.31
192.168.10.32	192.168.10.33-192.168.10.46	192.168.10.47
192.168.10.48	192.168.10.49-192.168.10.62	192.168.10.63
.	.	.
192.168.10.240	192.168.10.241-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.32.
The broadcast address is: 192.168.10.47.
The valid host range: 192.168.10.33-192.168.10.46.

Example 10:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.26 255.255.255.248.

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
                  = 255. 255. 255. 11111000
Subnets = 2^5 = 32.
Hosts = 2^3 – 2 = 6.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a block size of
8 in the forth octet.
Subnets are 0, 8, 16, 24, 32, … , 248.
Address 26 falls in the 24 subnet.
Subnet = 192. 168. 10. 24
The subnets would be 192.168.10.0, 192.168.10.8, 192.168.10.16,
192.168.10.24, … ,192.168.10.248.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.6	192.168.10.7
192.168.10.8	192.168.10.9-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.22	192.168.10.23
192.168.10.24	192.168.10.25-192.168.10.30	192.168.10.31
.	.	.
192.168.10.248	192.168.10.249-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.24.
The broadcast address is: 192.168.10.31.
The valid host range: 192.168.10.25-192.168.10.30.

Example 11:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.13 255.255.255.252.

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.252
                  = 255. 255. 255. 11111100
Subnets = 2^6 = 64.
Hosts = 2^2 – 2 = 2.
Block Size = 2^2 = 4.
Valid subnets = 256 – 252 = 4, means we have a block size of
4 in the forth octet.
Subnets are 0, 4, 8, 12, 16, … , 252.
Address 13 falls in the 12 subnet.
Subnet = 192. 168. 10. 12
The subnets would be 192.168.10.0, 192.168.10.8, 192.168.10.12,
192.168.10.16, … ,192.168.10.252.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.2	192.168.10.3
192.168.10.4	192.168.10.5-192.168.10.6	192.168.10.7
192.168.10.8	192.168.10.9-192.168.10.10	192.168.10.11
192.168.10.12	192.168.10.13-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.18	192.168.10.19
.	.	.
.	.	.
192.168.10.252	192.168.10.253-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.12.
The broadcast address is: 192.168.10.15.
The valid host range: 192.168.10.13-192.168.10.14.

Example 12:

Calculate Subnet, Broadcast, Valid Host Range of 192.168.10.99 255.255.255.248.

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
                  = 255. 255. 255. 11111000
Subnets = 2^5 = 32.
Hosts = 2^3 – 2 = 6.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a block size of
8 in the forth octet.
Subnets are 0, 8, 16, 24, 32, … , 88, 96, 248.
Address 99 falls in the 96 subnet.
Subnet = 192. 168. 10. 96
The subnets would be 192.168.10.0, 192.168.10.8, 192.168.10.16,
192.168.10.24, 192.168.10.24,
…, 192.168.10.248.

Network	Assignable Host	Broadcast
192.168.10.0	192.168.10.1-192.168.10.6	192.168.10.7
192.168.10.8	192.168.10.9-192.168.10.14	192.168.10.15
192.168.10.16	192.168.10.17-192.168.10.22	192.168.10.23
192.168.10.24	192.168.10.25-192.168.10.30	192.168.10.31
192.168.10.32	192.168.10.33-192.168.10.38	192.168.10.39
.	.	.
.	.	.
192.168.10.88	192.168.10.89-192.168.10.94	192.168.10.95
192.168.10.96	192.168.10.97-192.168.10.102	192.168.10.103
.	.	.
.	.	.
192.168.10.248	192.168.10.249-192.168.10.254	192.168.10.255

The subnet is: 192.168.10.96.
The broadcast address is: 192.168.10.103.
The valid host range: 192.168.10.97-192.168.10.102.

Example 13:

192.168.100.17 255.255.255.248(/29)
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
                  = 255. 255. 255. 11111000
Subnets = 2^5 = 32.
Hosts = 2^3 – 2 = 6.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a block size of
8 in the forth octet.
Subnets are 0, 8, 16, 24, 32, … , 88, 96, 248.
Address 17 falls in the 16 subnet.
Subnet = 192. 168. 100. 16
The subnets would be 192.168.100.0, 192.168.100.8, 192.168.100.16,
192.168.100.24, …, 192.168.100.248.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.6	192.168.100.7
192.168.100.8	192.168.100.9-192.168.100.14	192.168.100.15
192.168.100.16	192.168.100.17-192.168.100.22	192.168.100.23
192.168.100.24	192.168.100.25-192.168.100.30	192.168.100.31
.	.	.
.	.	.
192.168.100.248	192.168.100.249-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.16.
The broadcast address is: 192.168.100.23.
The valid host range: 192.168.100.17-192.168.100.22.

Example 14:

192.168.100.25 255.255.255.252
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.252
                  = 255. 255. 255. 11111100
Subnets = 2^6 = 64.
Hosts = 2^2 – 2 = 2.
Block Size = 2^2 = 4.
Valid subnets = 256 – 252 = 4, means we have a block size of
4 in the forth octet.
Subnets are 0, 4, 8, 12, 16, 20, 24, 28, … 252.
Address 25 falls in the 24 subnet.
Subnet = 192. 168. 100. 24
The subnets would be 192.168.100.0, 192.168.100.4, 192.168.100.8,
192.168.100.12, 192.168.100.16, 192.168.100.20, 192.168.100.24,
192.168.100.28 ,…, 192.168.100.252.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.2	192.168.100.3
192.168.100.4	192.168.100.5-192.168.100.6	192.168.100.7
192.168.100.8	192.168.100.9-192.168.100.10	192.168.100.11
192.168.100.12	192.168.100.13-192.168.100.14	192.168.100.15
192.168.100.16	192.168.100.17-192.168.100.18	192.168.100.19
192.168.100.20	192.168.100.21-192.168.100.22	192.168.100.23
192.168.100.24	192.168.100.25-192.168.100.26	192.168.100.27
192.168.100.28	192.168.100.29-192.168.100.30	192.168.100.31
.	.	.
.	.	.
192.168.100.252	192.168.100.253-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.24.
The broadcast address is: 192.168.100.27.
The valid host range: 192.168.100.25-192.168.100.26.

Example 15:

192.168.100.37 255.255.255.240(/28)
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.240
                  = 255. 255. 255. 11110000
Subnets = 2^4 = 16.
Hosts = 2^4 – 2 = 14.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, means we have a block size of
16 in the forth octet.
Subnets are 0, 16, 32, 48, … 240.
Address 37 falls in the 32 subnet.
Subnet = 192. 168. 100. 32
The subnets would be 192.168.100.0, 192.168.100.16,
192.168.100.32, 192.168.100.48, …, 192.168.100.240.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.14	192.168.100.15
192.168.100.16	192.168.100.17-192.168.100.30	192.168.100.31
192.168.100.32	192.168.100.33-192.168.100.46	192.168.100.47
192.168.100.48	192.168.100.49-192.168.100.62	192.168.100.63
.	.	.
.	.	.
192.168.100.240	192.168.100.241-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.32.
The broadcast address is: 192.168.100.47.
The valid host range: 192.168.100.33-192.168.100.46.

Example 16:

192.168.100.37 255.255.255.248(/29)
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
                  = 255. 255. 255. 11111000
Subnets = 2^5 = 32.
Hosts = 2^3 – 2 = 6.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a block size of
8 in the forth octet.
Subnets are 0, 8, 16, 24, 32, 40, 48, … 248.
Address 37 falls in the 32 subnet.
Subnet = 192. 168. 100. 32
The subnets would be 192.168.100.0, 192.168.100.8, 192.168.100.16,
192.168.100.24, 192.168.100.32, 192.168.100.40, 192.168.100.48, … ,
192.168.100.248.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.6	192.168.100.7
192.168.100.8	192.168.100.9-192.168.100.14	192.168.100.15
192.168.100.16	192.168.100.17-192.168.100.22	192.168.100.23
192.168.100.24	192.168.100.25-192.168.100.30	192.168.100.31
92.168.100.32	192.168.100.33-192.168.100.38	192.168.100.39
192.168.100.40	192.168.100.41-192.168.100.46	192.168.100.47
192.168.100.48	192.168.100.49-192.168.100.54	192.168.100.55
.	.	.
192.168.100.248	192.168.100.249-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.32.
The broadcast address is: 192.168.100.39.
The valid host range: 192.168.100.33-192.168.100.38.

Example 17:

192.168.100.66 255.255.255.224(/27)
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.224
                  = 255. 255. 255. 11100000
Subnets = 2^3 = 8.
Hosts = 2^5 – 2 = 30.
Block Size = 2^5 = 32.
Valid subnets = 256 – 224 = 32, means we have a block size of
32 in the forth octet.
Subnets are 0, 32, 64, 96, … 224.
Address 66 falls in the 64 subnet.
Subnet = 192. 168. 100. 32
The subnets would be 192.168.100.0, 192.168.100.32,192.168.100.64,
192.168.100.96, …, 192.168.100.224.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.30	192.168.100.31
192.168.100.32	192.168.100.33-192.168.100.62	192.168.100.63
192.168.100.64	192.168.100.65-192.168.100.94	192.168.100.95
192.168.100.96	192.168.100.97-192.168.100.126	192.168.100.127
.	.	.
.	.	.
192.168.100.224	192.168.100.225-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.64.
The broadcast address is: 192.168.100.95.
The valid host range: 192.168.100.65-192.168.100.94.

Example 18:

192.168.100.99 255.255.255.128(/25)
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.128
                  = 255. 255. 255. 10000000
Subnets = 2^1 = 2.
Hosts = 2^7 – 2 = 126.
Block Size = 2^7 = 128.
Valid subnets = 256 – 128 = 128, means we have a block size of
128 in the forth octet.
Subnets are 0, 128.
Address 99 falls in the 0 subnet.
Subnet = 192. 168. 100. 0
The subnets would be 192.168.100.0, 192.168.100.128.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.126	192.168.100.127
192.168.100.128	192.168.100.129-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.0.
The broadcast address is: 192.168.100.127.
The valid host range: 192.168.100.1-192.168.100.126.

Example 19:

192.168.100.99 255.255.255.192(/26)
What is the valid subnet?
What is the broadcast address?
What is the valid host range?

Solution

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.192
                  = 255. 255. 255. 11000000
Subnets = 2^2 =4.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, means we have a block size of
64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 99 falls in the 64 subnet.
Subnet = 192. 168. 100. 64
The subnets would be 192.168.100.0, 192.168.100.64,
192.168.100.128, 192.168.100.192.

Network	Assignable Host	Broadcast
192.168.100.0	192.168.100.1-192.168.100.62	192.168.100.63
192.168.100.64	192.168.100.65-192.168.100.126	192.168.100.127
192.168.100.128	192.168.100.129-192.168.100.190	192.168.100.191
192.168.100.192	192.168.100.193-192.168.100.254	192.168.100.255

The subnet is: 192.168.100.64.
The broadcast address is: 192.168.100.127.
The valid host range: 192.168.100.65-192.168.100.126.

Broadcast Valid Host Range Class B

Subnet, Broadcast, Valid Host Range:Class B

Example 1:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.128.0(/17).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.128.0
                  = 255. 255. 10000000. 00000000
Subnets = 2^1 = 2.
Hosts = 2^15 – 2 = 32766.
Block Size = 2^7 = 128.
Valid subnets = 256 – 128 = 128, means we have a
block size of 128 in the third octet.
Subnets are 0, 128.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.128.0.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.127.254	172.16.127.255
172.16.8.0	172.16.128.1-172.16.255.254	172.16.255.255

The subnet is: 172.16.0.0.
The broadcast address is:172.16.127.255.
The valid host range: 172.16.0.1-172.16.127.254.

Example 2:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.192.0(/18).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.192.0(/19)
                  = 255. 255. 11000000. 00000000
Subnets = 2^2 = 4.
Hosts = 2^14 – 2 = 16382.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, means we have a
block size of 64 in the third octet.
Subnets are 0, 64, 128, 192.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.64.0, 172.16.128.0,
172.16.192.0.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.63.254	172.16.63.255
172.16.64.0	172.16.64.1-172.16.127.254	172.16.127.255
172.16.128.0	172.16.128.1-172.16.191.254	172.16.191.255
172.16.192.0	172.16.192.1-172.16.255.254	172.16.255.255

The subnet is: 172.16.0.0.
The broadcast address is:172.16.63.255.
The valid host range: 172.16.0.1-172.16.63.254.

Example 3:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.240.0(/20).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.240.0
                  = 255. 255. 11110000. 00000000
Subnets = 2^4 = 16.
Hosts = 2^12 – 2 = 4094.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, means we have a
block size of 16 in the third octet.
Subnets are 0, 16, 32, 48, … , 240.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.16.0, 172.16.32.0,
172.16.48.0, … , 172.16.240.0.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.15.254	172.16.15.255
172.16.16.0	172.16.16.1-172.16.31.254	172.16.31.255
172.16.32.0	172.16.32.1-172.16.47.254	172.16.47.255
172.16.48.0	172.16.48.1-172.16.63.254	172.16.63.255
.
.
172.16.240.0	172.16.240.1-172.16.254.255	172.16.255.255

The subnet is: 172.16.0.0.
The broadcast address is: 172.16.15.255.
The valid host range: 172.16.0.1-172.16.15.254.

Example 4:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.248.0(/21).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.248.0
                  = 255. 255. 11111000. 00000000
Subnets = 2^5 = 32.
Hosts = 2^11 – 2 = 2048.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a
block size of 8 in the third octet.
Subnets are 0, 8, 16,24, … , 248.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.8.0, 172.16.16.0,
172.16.24.0, … , 172.16.248.0.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.7.254	172.16.7.255
172.16.8.0	172.16.8.1-172.16.15.254	172.16.15.255

172.16.16.0	172.16.16.1-172.16.23.254	172.16.23.255
172.16.24.0	172.16.24.1-172.16.31.254	172.16.31.255
.
.
.
172.16.240.0	172.16.240.1-172.16.247.254	172.16.247.255
172.16.248.0	172.16.248.1-172.16.255.254	172.16.255.255

The subnet is: 172.16.0.0.
The broadcast address is: 172.16.7.255.
The valid host range: 172.16.0.1-172.16.7.254.

Example 5:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.254.0(/23).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.254.0
                  = 255. 255. 11111110. 00000000
Subnets = 2^7 = 128.
Hosts = 2^9 – 2 = 510.
Block Size = 2^1 = 2.
Valid subnets = 256 – 254 = 2, means we have a
block size of 2 in the third octet.
Subnets are 0, 2, 4, 6, … , 254.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.2.0, 172.16.4.0,
172.16.6.0, … , 172.16.254.0.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.1.254	172.16.1.255
172.16.2.0	172.16.2.1-172.16.3.254	172.16.3.255
172.16.4.0	172.16.4.1-172.16.5.254	172.16.5.255
172.16.6.0	172.16.6.1-172.16.7.254	172.16.7.255
.
.
172.16.254.0	172.16.254.1-172.16.255.254	172.16.255.255

The subnet is: 172.16.0.0.
The broadcast address is: 172.16.1.255.
The valid host range: 172.16.0.1-172.16.1.254.

Example 6:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.255.0(/24).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.0
                  = 255. 255. 11111111. 00000000
Subnets = 2^8 = 256.
Hosts = 2^8 – 2 = 254.
Block Size = 2^0 = 1.
Valid subnets = 256 – 255 = 1, means we have a
block size of 1 in the third octet.
Subnets are 0, 1, 2, 3, … , 255.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.1.0, 172.16.2.0,
172.16.3.0, … , 172.16.255.0.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.0.254	172.16.0.255
172.16.1.0	172.16.1.1-172.16.1.254	172.16.1.255
172.16.2.0	172.16.2.1-172.16.2.254	172.16.2.255
172.16.3.0	172.16.3.1-172.16.3.254	172.16.3.255
.	.	.
172.16.254.0	172.16.254.1-172.16.253.254	172.16.253.255
172.16.255.0	172.16.255.1-172.16.254.254	172.16.254.255

The subnet is: 172.16.0.0.
The broadcast address is: 172.16.0.255.
The valid host range: 172.16.0.1-172.16.0.254.

Example 7:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.255.128(/25).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.128
                  = 255. 255. 11111111. 10000000
Subnets = 2^9 = 512.
Hosts = 2^7 – 2 = 126.
Block Size = 2^7 = 128.
Valid subnets = 256 – 128 = 128, means we have a block size of
128 in the forth octet.
Subnets are 0, 128.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.0.128.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.0.126	172.16.0.127
172.16.0.128	172.16.0.129-172.16.0.254	172.16.0.255

The subnet is: 172.16.0.0.
The broadcast address is: 172.16.0.127.
The valid host range: 172.16.0.1-172.16.0.126.

Example 8:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.255.192(/26).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.192
                  = 255. 255. 11111111. 11000000
Subnets = 2^10 = 1024.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, means we have a block size of
64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 16 falls in the 0 subnet.
Subnet = 172. 16. 0. 0
The subnets would be 172.16.0.0, 172.16.0.64, 172.16.0.128,
172.16.0.192.

Network	Assignable Host	Broadcast
172.16.0.0	172.16.0.1-172.16.0.62	172.16.0.63
172.16.0.64	172.16.0.65-172.16.0.126	172.16.0.127
172.16.0.128	172.16.0.129-172.16.0.190	172.16.0.191
172.16.0.192	172.16.0.193-172.16.0.254	172.16.0.255

The subnet is: 172.16.0.0.
The broadcast address is: 172.16.0.63.
The valid host range: 172.16.0.1-172.16.0.62.

Example 9:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.5 255.255.255.128(/25).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.128(/25)
                  = 255. 255. 11111111. 10000000
10.5    = 00001010. 00000101
255.128 = 11111111. 10000000
------------------------------
    And = 00001010. 00000000 = 10.0
Subnet = 172. 16. 10. 0
Subnets = 2^9 = 512.
Hosts = 2^7 – 2 = 126.
Block Size = 2^7 = 128.
Valid subnets = 256 – 128 = 128, means we have a block size of
128 in the forth octet.
Subnets are 0, 128.
Address 5 falls in the 0 subnet.
Subnet = 172. 16. 10. 0
The subnets would be 172.16.10.0, 172.16.10.128.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.126	172.16.10.127
172.16.10.128	172.16.10.129-172.16.10.245	172.16.10.255

The subnet is: 172.16.10.0.
The broadcast address is:172.16.10.127.
The valid host range: 172.16.10.1-172.16.10.126.

Example 10:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.13 255.255.255.252.

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.252
                  = 255. 255. 11111111. 11111100

Subnets = 2^14 = 16384.
Hosts = 2^2 – 2 = 2.
Block Size = 2^5 = 32.
Valid subnets = 256 – 252 = 4, means we have a block size of
4 in the forth octet.
Subnets are 0, 4, 8, 12, 16,… , 252.
Address 13 falls in the 12 subnet.
Subnet = 172. 16. 10. 12
The subnets would be 172.16.10.0, 172.16.10.4, 172.16.10.8,
172.16.10.12, 172.16.10.16 … , 172.16.10.252.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.2	172.16.10.3
172.16.10.4	172.16.10.5-172.16.10.6	172.16.10.7
172.16.10.8	172.16.10.9-172.16.10.10	172.16.10.11
172.16.10.12	172.16.10.13-172.16.10.14	172.16.10.15
172.16.10.16	172.16.10.17-172.16.10.18	172.16.10.19
.
172.16.10.252	172.16.10.253-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.12.
The broadcast address is:172.16.10.15.
The valid host range: 172.16.10.13-172.16.10.14.

Example 11:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.17 255.255.255.252(/30).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.252(/30)
                  = 255. 255. 11111111. 11111100
    172.16.10.17  = 10101100. 00010000. 00001010. 00011110
255.255.255.252   = 11111111. 11111111. 11111111. 11110000
-----------------------------------------------------------
              And = 10101100. 00010000. 00001010. 00010000
                  = 172. 16. 10. 16 = Subnet
Subnets = 2^14 = 16384.
Hosts = 2^2 – 2 = 2.
Block Size = 2^2 = 4.
Valid subnets = 256 – 252 = 4, means we have a block size of 4
in the forth octet.
Subnets are 0, 4, 8, 12, 16, 20, … , 252.
Address 17 falls in the 16 subnet.
Subnet = 172. 16. 10. 16
The subnets would be 172.16.10.0, 172.16.10.4, 172.16.10.8,
172.16.10.12, 172.16.10.16, … , 172.16.10.252.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.2	172.16.10.3
172.16.10.4	172.16.10.5-172.16.10.6	172.16.10.7
172.16.10.8	172.16.10.9-172.16.10.10	172.16.10.11
172.16.10.12	172.16.10.13-172.16.10.14	172.16.10.15
172.16.10.16	172.16.10.17-172.16.10.18	172.16.10.19
.
.
172.16.10.252	172.16.10.253-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.16.
The broadcast address is:172.16.10.19.
The valid host range: 172.16.10.17-172.16.10.18.

Example 12:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.66 255.255.255.224(/27).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.224(/27)
                  = 255. 255. 11111111. 11100000
Subnets = 2^3 = 8.
Hosts = 2^5 – 2 = 30.
Block Size = 2^5 = 32.
Valid subnets = 256 – 224 = 32, means we have a block size of
32 in the forth octet.
Subnets are 0, 32, 64, 96, … , 224.
Address 66 falls in the 64 subnet.
Subnet = 172. 16. 10. 64
The subnets would be 172.16.10.0, 172.16.10.32, 172.16.10.64,
172.16.10.96, … , 172.16.10.224.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.30	172.16.10.31
172.16.10.32	172.16.10.33-172.16.10.62	172.16.10.63
172.16.10.64	172.16.10.65-172.16.10.94	172.16.10.95
172.16.10.96	172.16.10.97-172.16.10.126	172.16.10.127
.
.
172.16.10.224	172.16.10.225-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.64.
The broadcast address is:172.16.10.95.
The valid host range: 172.16.10.65-172.16.10.94.

Example 13:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.90 255.255.255.192.

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.192
                  = 255. 255. 11111111. 11000000
Subnets = 2^2 = 4.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, means we have a block size of
64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 90 falls in the 64 subnet.
Subnet = 172. 16. 10. 64
The subnets would be 172.16.10.0, 172.16.10.64, 172.16.10.128,
172.16.10.192.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.62	172.16.10.63
172.16.10.64	172.16.10.65-172.16.10.126	172.16.10.127
172.16.10.128	172.16.10.129-172.16.10.190	172.16.10.191
172.16.10.192	172.16.10.193-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.64.
The broadcast address is:172.16.10.127.
The valid host range: 172.16.10.65-172.16.10.126.

Example 14:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.33 255.255.255.224(/27).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.224(/27)
                  = 255. 255. 11111111. 11100000
Subnets = 2^11 = 2048.
Hosts = 2^5 – 2 = 30.
Block Size = 2^5 = 32.
Valid subnets = 256 – 224 = 32, means we have a block size of
32 in the forth octet.
Subnets are 0, 32, 64, 96, … , 224.
Address 33 falls in the 32 subnet.
Subnet = 172. 16. 10. 32
The subnets would be 172.16.10.0, 172.16.10.32, 172.16.10.64,
172.16.10.96, … , 172.16.10.224.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.30	172.16.10.31
172.16.10.32	172.16.10.33-172.16.10.62	172.16.10.63
172.16.10.64	172.16.10.65-172.16.10.94	172.16.10.95
172.16.10.96	172.16.10.97-172.16.10.126	172.16.10.127
.
.
172.16.10.224	172.16.10.225-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.32.
The broadcast address is:172.16.10.63.
The valid host range: 172.16.10.33-172.16.10.62.

Example 15:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.33 255.255.255.240(/28).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.240(/28)
                  = 255. 255. 11111111. 11110000
Subnets = 2^12 = 4096.
Hosts = 2^4 – 2 = 14.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, means we have a block size of
16 in the forth octet.
Subnets are 0, 16, 32, 48, … , 240.
Address 33 falls in the 32 subnet.
Subnet = 172. 16. 10. 32
The subnets would be 172.16.10.0, 172.16.10.16, 172.16.10.32,
172.16.10.48, … , 172.16.10.240.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.14	172.16.10.15
172.16.10.16	172.16.10.17-172.16.10.30	172.16.10.31
172.16.10.32	172.16.10.33-172.16.10.46	172.16.10.47
172.16.10.48	172.16.10.49-172.16.10.62	172.16.10.63
.
.
172.16.10.240	172.16.10.241-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.32.
The broadcast address is:172.16.10.47.
The valid host range: 172.16.10.33-172.16.10.46.

Example 16:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.33 255.255.255.248.

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.248
                  = 255. 255. 11111111. 11111000
Subnets = 2^13 = 8192.
Hosts = 2^3 – 2 = 6.
Block Size = 2^3 = 8.
Valid subnets = 256 – 248 = 8, means we have a block size of
8 in the forth octet.
Subnets are 0, 8, 16, 24, 32, 40, … , 248.
Address 33 falls in the 32 subnet.
Subnet = 172. 16. 10. 32
The subnets would be 172.16.10.0, 172.16.10.8, 172.16.10.16,
172.16.10.24, 172.16.10.32, 172.16.10.40, … , 172.16.10.248.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.6	172.16.10.7
172.16.10.8	172.16.10.9-172.16.10.14	172.16.10.15
172.16.10.16	172.16.10.17-172.16.10.22	172.16.10.23
172.16.10.24	172.16.10.25-172.16.10.31	172.16.10.32
172.16.10.32	172.16.10.33-172.16.10.38	172.16.10.39
.
172.16.10.248	172.16.10.249-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.32.
The broadcast address is: 172.16.10.39.
The valid host range: 172.16.10.33-172.16.10.38.

Example 17:

Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.65 255.255.255.192(/26).

Solution

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.192(/26)
                  = 255. 255. 11111111. 11000000
Subnets = 2^10 = 1024.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, means we have a block size of
64 in the forth octet.
Subnets are 0, 64, 128, 192.
Address 65 falls in the 64 subnet.
Subnet = 172. 16. 10. 64
The subnets would be 172.16.10.0, 172.16.10.64, 172.16.10.128,
172.16.10.192.

Network	Assignable Host	Broadcast
172.16.10.0	172.16.10.1-172.16.10.62	172.16.10.63
172.16.10.64	172.16.10.65-172.16.10.126	172.16.10.127
172.16.10.128	172.16.10.129-172.16.10.190	172.16.10.191
172.16.10.192	172.16.10.193-172.16.10.254	172.16.10.255

The subnet is: 172.16.10.64.
The broadcast address is:172.16.10.127.
The valid host range: 172.16.10.65-172.16.10.126.

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