Subnet, Broadcast, Valid Host Range:Class A
Class A
Example: 1
Calculating Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.0.0 (/16) subnet.
Example: 2
Calculating Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.255.192(/26) subnet.
Example: 3
Calculating Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.240.0(/20) subnet.
Example: 4
If your host ID is 10.20.80.30/20, what is your subnet, broadcast address, and valid host range?
Example: 5
Write the subnet, broadcast address and valid host range of IP 10.10.10.5/20.
Example: 6
10.10.10.10/20, is a host assignable IP? If yes , determine the subnet IP and broadcast of that address.
Example: 7
Is 10.1.3.65/23 a host assignable IP? If yes , determine the subnet IP and broadcast of that address.
Example 1:
Calculate Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.0.0 (/16) subnet.Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.0.0
= 255. 11111111. 00000000. 00000000
Subnets = 2^8 = 256.
Hosts = 2^16 – 2 = 65,534.
Block Size = 2^0 = 1.
Valid subnets = 256 – 255 = 1, which means we have a block size
of 1 in the second octet.
Subnets are 0, 1, 2, 3, … , 255.
Address 0 falls in the 0 subnet.
Subnet = 10. 0. 0. 0
The subnets would be 10.0.0.0, 10.1.0.0, 10.2.0.0, 10.3.0.0, etc.,
up to 10.255.0.0.
The subnets would be 10.0.0.0, 10.1.0.0, 10.2.0.0, 10.3.0.0, etc., up to 10.255.0.0.| Network | Assignable Host | Broadcast |
|---|---|---|
| 10.0.0.0 |
10.0.0.1-10.0.255.254
|
10.0.255.255
|
10.1.0.0
|
10.1.0.1-10.1.255.254
|
10.1.255.255
|
10.2.0.0
|
10.2.0.1-10.2.255.254
|
10.2.255.255
|
10.3.0.0
|
10.3.0.1-10.3.255.254
|
10.3.255.255
|
.
|
||
.
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||
.
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||
10.254.0.0
|
10.254.0.1-10.254.255.254
|
10.254.255.255
|
10.255.0.0
|
10.255.0.1-10.255.255.254
|
10.255.255.255
|
The broadcast address is:10.0.255.255.
The valid host range: 10.0.0.1-10.0.255.254.
Example 2:
Calculate Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.255.192(/26) subnet.Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.255.192
= 255. 11111111. 11111111. 11000000
Subnets = 2^18 = 262144.
Hosts = 2^6 – 2 = 62.
Block Size = 2^6 = 64.
Valid subnets = 256 – 192 = 64, which 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 = 10. 0. 0. 0
The subnets would be 10.0.0.0, 10.0.0.64, 10.0.0.128, 10.0.0.192.
| Network | Assignable Host | Broadcast |
|---|---|---|
| 10.0.0.0 |
10.0.0.1-10.0.0.62
|
10.0.0.63
|
10.0.0.64
|
10.0.0.65-10.0.0.126
|
10.0.0.127
|
10.0.0.128
|
10.0.0.129-10.0.0.190
|
10.0.0.191
|
10.0.0.192
|
10.0.0.193-10.0.0.254
|
10.0.0.255
|
The broadcast address is: 10.0.0.63.
The valid host range: 10.0.0.1-10.0.0.62.
Example 3:
Calculating Subnet, Broadcast, Valid Host Range of 10.0.0.0 255.255.240.0(/20) subnet.Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.240.0
= 255. 11111111. 11110000. 00000000
Subnets = 2^12 = 4094.
Hosts = 2^12 – 2 = 4096.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, which means we have a
block size of 16 in the third octet.
Subnets are 0, 16, 32, 48, … , 240.
Address 0 falls in the 0 subnet.
Subnet = 10. 0. 0. 0
The subnets would be 10.0.0.0, 10.0.16.0, 10.0.32.0,
10.0.48.0, … , 10.0.240.0.
| Network | Assignable Host | Broadcast |
|---|---|---|
| 10.0.0.0 |
10.0.0.1-10.0.15.254
|
10.0.15.255
|
10.0.16.0
|
10.0.16.1-10.0.31.254
|
10.0.31.255
|
10.0.32.0
|
10.0.32.1-10.0.47.254
|
10.0.47.255
|
10.0.48.0
|
10.0.48.1-10.0.63.254
|
10.0.63.255
|
.
. . |
||
10.0.240.0
|
10.0.240.1-10.0.255.254
|
10.0.255.255
|
The broadcast address is:10.0.15.255.
The valid host range: 10.0.0.1-10.0.15.254.
Example 4:
If your host ID is 10.20.80.30/20, what is your subnet, broadcast address, and valid host range?Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask, /20 = 255.255.240.0
= 255. 11111111. 11110000. 00000000
Subnets = 2^12 = 4096.
Hosts = 2^12 – 2 = 4094.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, which means we have a
block size of 16 in the third octet.
Subnets are 0, 16, 32, 48, 64, 80, 96, …, 240.
Address 80 falls in the 80 subnet.
Subnet = 10. 20. 80. 0
The subnets would be 10.20.0.0, 10.20.16.0, 10.20.48.0,
10.20.64.0, … , 10.20. 240 .0.
| Network | Assignable Host | Broadcast |
|---|---|---|
10.20.0.0 |
10.20.0.1-10.20.15.254
|
10.20.15.255
|
10.20.16.0
|
10.20.16.1-10.20.15.254
|
10.20.31.255
|
10.20.32.0
|
10.20.32.1-10.20.15.254
|
10.20.47.255
|
10.20.48.0
|
10.20.48.1-10.20.15.254
|
10.20.63.255
|
10.20.64.0
|
10.20.64.1-10.20.79.254
|
10.20.79.255
|
10.20.80.0
|
10.20.80.1-10.20.95.254
|
10.20.95.255
|
10.20.96.0
|
10.20.96.1-10.20.211.254
|
10.20.111.255
|
The broadcast address is:10.20.95.255.
The valid host range: 10.20.80.1-10.20.95.254.
Example 5:
Write the subnet, broadcast address and valid host range of IP 10.10.10.5/20.Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.240.0
= 255. 11111111. 11110000. 00000000
Subnets = 2^12 = 4096.
Hosts = 2^12 – 2 = 4094.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, which means we have a
block size of 16 in the third octet.
Subnets are 0, 16, 32, 48, … , 240.
Address 10 falls in the 0 subnet.
Subnet = 10. 10. 0. 0
The subnets would be 10.10.0.0, 10.10.16.0, 10.10.32.0,
10.10.48.0, … , 10.10.240.0.
| Network | Assignable Host | Broadcast |
|---|---|---|
10.10.0.0 |
10.10.0.1-10.10.15.254
|
10.10.15.255
|
10.10.16.0
|
10.10.16.1-10.10.31.254
|
10.10.31.255
|
10.10.32.0
|
10.10.32.1-10.10.47.254
|
10.10.47.255
|
10.10.48.0
|
10.10.48.1-10.10.63.254
|
10.10.63.255
|
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10.10.240.0
|
10.10.240.1-10.10.255.254
|
10.10.255.255
|
The broadcast address is: 10.10.15.255.
The valid host range: 10.10.0.1-10.10.15.254.
Example 6:
10.10.10.10/20, is a host assignable IP? If yes , determine the subnet IP and broadcast of that address.Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.240.0
= 255. 11111111. 11110000. 00000000
Subnets = 2^12 = 4096.
Hosts = 2^12 – 2 = 4094.
Block Size = 2^4 = 16.
Valid subnets = 256 – 240 = 16, which means we have a
block size of 16 in the third octet.
Subnets are 0, 16, 32, 48, … , 240.
Address 10 falls in the 0 subnet.
Subnet = 10. 10. 0. 0
The subnets would be 10.10.0.0, 10.10.16.0, 10.10.32.0,
10.10.48.0, … , 10.10.240.0.
| Network | Assignable Host | Broadcast |
|---|---|---|
| 10.10.0.0 |
10.10.0.1-10.10.15.254
|
10.10.15.255
|
10.10.16.0
|
10.10.16.1-10.10.31.254
|
10.10.31.255
|
10.10.32.0
|
10.10.32.1-10.10.47.254
|
10.10.47.255
|
10.10.48.0
|
10.10.48.1-10.10.63.254
|
10.10.63.255
|
.
|
.
|
.
|
10.10.240.0
|
10.10.240.1-10.10.255.254
|
10.10.255.255
|
The broadcast address is: 10.10.15.255.
The valid host range: 10.10.0.1-10.10.15.254.
Example 7:
Is 10.1.3.65/23 a host assignable IP? If yes , determine the subnet IP and broadcast of that address.Solution:
Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.254.0
= 2555. 11111111. 11111110. 00000000
Subnets = 2^15 = 32768.
Hosts = 2^9 – 2 = 510.
Block Size = 2^4 = 16.
Valid subnets = 256 – 254 = 2, which means we have a
block size of 2 in the third octet.
Subnets are 0, 2, 4, 6, 8, … , 254.
Address 3 falls in the 2 subnet.
Subnet = 10. 1. 2. 0
The subnets would be 10.1.0.0, 10.1.2.0, 10.1.4.0, 10.1.6.0,
10.1.8.0… , 10.1.254.0.
| Network | Assignable Host | Broadcast |
|---|---|---|
| 10.1.0.0 |
10.1.0.1-10.1.1.254
|
10.1.1.255
|
10.1.2.0
|
10.1.2.1-10.1.3.254
|
10.1.3.255
|
10.1.4.0
|
10.1.4.1-10.1.5.254
|
10.1.5.255
|
10.1.6.0
|
10.1.6.1-10.1.7.254
|
10.1.7.255
|
10.1.8.0
|
10.1.8.1-10.1.9.254
|
10.1.9.255
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10.1.254.0
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10.1.254.1-10.1.255.254
|
10.1.255.255
|
The subnet is: 10.1.2.0
The broadcast address is:10.1.3.255
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