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).
Example: 2
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.192.0(/18).
Example: 3
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.240.0(/20).
Example: 4
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.248.0(/21).
Example: 5
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.254.0(/23).
Example: 6
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.255.0(/24).
Example: 7
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.255.128(/25).
Example: 8
Calculate Subnet, Broadcast, Valid Host Range of 172.16.0.0 255.255.255.192(/26).
Example: 9
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.5 255.255.255.128(/25).
Example: 10
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.13 255.255.255.252.
Example: 11
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.17 255.255.255.252(/30).
Example: 12
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.66 255.255.255.224(/27).
Example: 13
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.90 255.255.255.192.
Example: 14
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.33 255.255.255.224(/27).
Example: 15
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.33 255.255.255.240(/28).
Example: 16
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.33 255.255.255.248.
Example: 17
Calculate Subnet, Broadcast, Valid Host Range of 172.16.10.65 255.255.255.192(/26).
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 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 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 |
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172.16.240.0 | 172.16.240.1-172.16.254.255 | 172.16.255.255 |
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 |
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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 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 |
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172.16.254.0 | 172.16.254.1-172.16.255.254 | 172.16.255.255 |
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 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 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 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 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 |
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172.16.10.252 | 172.16.10.253-172.16.10.254 | 172.16.10.255 |
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 |
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172.16.10.252 | 172.16.10.253-172.16.10.254 | 172.16.10.255 |
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 |
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172.16.10.224 | 172.16.10.225-172.16.10.254 | 172.16.10.255 |
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 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 |
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172.16.10.224 | 172.16.10.225-172.16.10.254 | 172.16.10.255 |
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 |
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172.16.10.240 | 172.16.10.241-172.16.10.254 | 172.16.10.255 |
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 |
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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 |
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172.16.10.248 | 172.16.10.249-172.16.10.254 | 172.16.10.255 |
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 broadcast address is:172.16.10.127.
The valid host range: 172.16.10.65-172.16.10.126.