Subnet Tutorials: Broadcast Valid Host Range Class A

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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.

Adscend

5.9.13

Broadcast Valid Host Range Class C

Subnet, Broadcast, Valid Host Range:Class C

Broadcast Valid Host Range Class B

Subnet, Broadcast, Valid Host Range:Class B