Subnet Tutorials: Last Valid Host Address

Last Valid Host Address

Class A

Example 1:

What is the last valid host on the subnetwork 10.5.208.0/20?

Solution:

  Class  A addresses use a default mask of 255.0.0.0.
  Given  subnet mask = 255.255.240.0
  Block size = 256-240 = 16, which  means we have a block size of
  16 in the third octet.
  208 is divisible by 16.
  That is 10.5.208.0/20 is a network address.
  Next subnet address = 10.5.224.0/20.
  Broadcast address is 10.5.223.255/20 of 10.5.208.0/20  network.
  Valid host range = 10.5.208.1-10.5.223.254
  The last valid IP = 10.5.223.254

Example 2:

What is the last valid host on the subnetwork 10.73.0.0/20?

Solution:

  Class  A addresses use a default mask of 255.0.0.0.
  Given  subnet mask = 255.255.240.0
  Block size = 256-240 = 16, which  means we have a block size of
  16 in the third octet.
  0 is divisible by 16.
  That is 10.73.0.0/20 is a  network address.
  Next subnet address = 10.73.16.0/20.
  Broadcast address is 10.73.15.255/20 of 10.73.0.0/20 network.
  Valid host range = 10.73.0.1-10.73.15.254 
  The last valid IP = 10.73.15.254

Example 3:

What is the last valid host on the subnetwork 10.174.16.0 255.255.240.0?

Solution:

  Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.240.0
Block size = 256-240 = 16, which means we have a block size
of 16 in the third octet.
16 is divisible by 16.
That is 10.174.16.0/20 is a network address.
Next subnet address = 10.174.32.0/20.
Broadcast address is 10.174.31.255/20 of 10.174.16.0/20 network.
Valid host range = 10.174.16.1-10.174.31.254
The last valid IP = 10.174.31.254

Example 4:

What is the last valid host on the subnetwork 10.236.240.0/20?

Solution:

Class A addresses use a default mask of 255.0.0.0.
Given subnet mask = 255.255.240.0 
Block size = 256-240 = 16, which means we have a block size of 
16 in the third octet.
240 is divisible by 16.
That is 10.236.240.0/20is a network address.
Next subnet address = 10.236.256.0/20.
Broadcast address is 10.236.255.255/20 of 10.236.240.0/20 network.
Valid host range = 10.236.240.1-10.236.255.254 
The last valid IP = 10.236.255.254

Class B

Example 5:

What is the last valid host on the subnetwork 172.19.24.0 255.255.254.0?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.254.0 
Block size = 256-254 = 2, which means we have a block size of 
2 in the third octet.
24 is divisible by 2.
That is 172.19.24.0/23 is a network address.
Next subnet address = 172.19.26.0/23.
Broadcast address is 172.19.25.255/23 of 172.19.24.0/23 network.
Valid host range = 172.19.24.1-172.19.25.254 
The last valid IP = 172.19.25.254

Example 6:

What is the last valid host on the subnetwork 172.19.130.0 255.255.254.0?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.254.0
Block size = 256-254 = 2, which means we have a block size of
2 in the third octet.
130 is divisible by 2.
That is 172.19.130.0/23 is a network address.
Next subnet address = 172.19.132.0/23.
Broadcast address is 172.19.131.255/23 of 172.19.130.0/23 network.
Valid host range = 172.19.130.1-172.19.131.254
The last valid IP = 172.19.131.254

Example 7:

What is the last valid host on the subnetwork 172.24.4.0/22?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.252.0
Block size = 256-252 = 4, which means we have a block size of
4 in the third octet.
4 is divisible by 4.
That is 172.24.4.0/22is a network address.
Next subnet address = 172.24.8.0/22.
Broadcast address is 172.24.7.255/22 of 172.24.4.0/22network.
Valid host range = 172. 24. 4. 1 to 172. 24. 7. 254
The last valid IP = 172. 24. 7. 254

Example 8:

What is the last valid host on the subnetwork 172.25.182.0/23?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.254.0
Block size = 256-254 = 2, which means we have a block size
of 2 in the third octet.
182 is divisible by 2.
That is 172.25.182.0/23is a network address.
Next subnet address = 172.25.184.0/23.
Broadcast address is 172.25.183.255/23 of 172.25.182.0/23 network.
Valid host range = 172.25.182.1-172.25.183.254
The last valid IP = 172.25.183.254

Example 9:

What is the last valid host on the subnetwork 172.29.142.0/23?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.254.0
Block size = 256-254 = 2, which means we have a block size of 2 in the third octet.
142 is divisible by 2.
That is 172.29.142.0/23 is a network address.
Next subnet address = 172.29.144.0/23.
Broadcast address is 172.29.143.255/23 of 172.29.142.0/23 network.
Valid host range = 172.29.142.1-172.29.143.254
The last valid IP = 172.29.143.254

Example 10:

What is the last valid host on the subnetwork 172.16.148.0/26?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.192
Block size = 256-192 = 64, which means we have a block size of 16 in the forth octet.
0 is divisible by 64.
That is 172.16.148.0/26 is a network address.
Next subnet address = 172.16.148.64/26.
Broadcast address is 172.16.148.63/26 of 172.16.148.0/26 network.
Valid host range = 172. 16. 148. 1 to 172. 16. 148. 62
The last valid IP = 172. 16. 148. 62

Example 11:

What is the last valid host on the subnetwork 172.18.1.128 255.255.255.240?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.240
Block size = 256-240 = 16, which means we have a block size of 16 in the forth octet.
128 is divisible by 16.
That is 172.18.1.128/28 is a network address.
Next subnet address = 172.18.1.144/28.
Broadcast address is 172.18.1.143/28 of 172.18.1.128/28 network.
Valid host range = 172. 18. 1. 129 to 172. 18. 1. 142
The last valid IP = 172. 18. 1. 142

Example 12:

What is the last valid host on the subnetwork 172.19.222.112 255.255.255.240?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.240
Block size = 256-240 = 16, which means we have a block size of 16 in the forth octet.
112 is divisible by 16.
That is 172.19.222.112/28 is a network address.
Next subnet address = 172.19.222.128/28.
Broadcast address is 172.19.222.127/28 of 172.19.222.112/28 network.
Valid host range = 172.19.222.113-172.19.222.126
The last valid IP = 172.19.222.126

Example 13:

What is the last valid host on the subnetwork 172.22.148.128/26?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.192
Block size = 256-192 = 64, which means we have a block size
of 16 in the forth octet.
182 is divisible by 64.
That is 172.22.148.128/26 is a network address.
Next subnet address = 172.22.148.192/26.
Broadcast address is 172.22.148.191/26 of 172.22.148.128/26 network.
Valid host range = 172.22.148.129-172.22.148.190
The last valid IP = 172.22.148.190

Example 14:

What is the last valid host on the subnetwork 172.25.127.64 255.255.255.224?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.224
Block size = 256-224 = 32, which means we have a block size of 32 in the forth octet.
64 is divisible by 32.
That is 172.25.127.64/27 is a network address.
Next subnet address = 172.25.127.96/27.
Broadcast address is 172.25.127.95/27 of 172.25.127.64/27 network.
Valid host range = 172. 25. 127. 65 to 172. 25. 127. 94
The last valid IP = 172. 25. 127. 94

Example 15:

What is the last valid host on the subnetwork 172.28.50.64 255.255.255.192?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.192
Block size = 256-192 = 64, which means we have a block size of 64 in the forth octet.
64 is divisible by 64.
That is 172.28.50.64/26is a network address.
Next subnet address = 172.28.50.128/26.
Broadcast address is 172.28.50.127/26 of 172.28.50.64/26network.
Valid host range = 172. 28. 50. 65 to 172. 28. 50. 126
The last valid IP = 172. 28. 50. 126

Example 16:

What is the last valid host on the subnetwork 172.29.254.128 255.255.255.128?

Solution:

Class B addresses use a default mask of 255.255.0.0.
Given subnet mask = 255.255.255.128
Block size = 256-128 = 128, which means we have a block size of 16 in the forth octet.
182 is divisible by 128.
That is 172.29.254.128/25is a network address.
Next subnet address = 172.29.254.256/25.
Broadcast address is 172.29.254.255/25 of 172.29.254.128/25 network.
Valid host range = 172. 29. 254. 129 to 172. 29. 254. 254
The last valid IP = 172. 29. 254. 254

Class C

Example 17:

What is the last valid host on the subnetwork 192.168.27.184 255.255.255.248?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
Block size = 256-248 = 8, which means we have a block size of 8 in the forth octet.
184 is divisible by 8.
That is 192.168.27.184/29 is a network address.
Next subnet address = 192.168.27.192/29.
Broadcast address is 192.168.27.191/29 of 192.168.27.184/29 network.
Valid host range = 192. 168. 27. 185 to 192. 168. 27. 190
The last valid IP = 192. 168. 27. 190

Example 18:

What is the last valid host on the subnetwork 192.168.37.128/27?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.224
Block size = 256-224 = 32, which means we have a block size of 32 in the forth octet.
128 is divisible by 32.
That is 192.168.37.128/27 is a network address.
Next subnet address = 192.168.37.160/27.
Broadcast address is 192.168.37.159/27 of 192.168.37.128/27 network.
Valid host range = 192. 168. 37. 129 to 192. 168. 37. 158
The last valid IP = 192. 168. 37. 158

Example 19:

What is the last valid host on the subnetwork 192.168.40.160 255.255.255.224?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.224
Block size = 256-224 = 32, which means we have a block size of 32 in the forth octet.
160 is divisible by 32.
That is 192.168.40.160/27 is a network address.
Next subnet address = 192.168.40.192/27.
Broadcast address is 192.168.40.191/27 of 192.168.40.160/27 network.
Valid host range = 192.168.40.161-192.168.40.190
The last valid IP = 192.168.40.190

Example 20:

What is the last valid host on the subnetwork 192.168.62.240 255.255.255.248?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
Block size = 256-248 = 8, which means we have a block size of
8 in the forth octet.
240 is divisible by 8.
That is 192.168.62.240/29 is a network address.
Next subnet address = 192.168.62.248/29.
Broadcast address is 192.168.62.247/29 of 192.168.62.240/29
network.
Valid host range = 192. 168. 62. 241 to 192. 168. 62. 246
The last valid IP = 192. 168. 62. 246

Example 21:

What is the last valid host on the subnetwork 192.168.74.224 255.255.255.240?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.240
Block size = 256-240 = 16, which means we have a block size of
16 in the forth octet.
224 is divisible by 16.
That is 192.168.74.224/28 is a network address.
Next subnet address = 192.168.74.240/28.
Broadcast address is 192.168.74.239/28 of 192.168.74.224/28
network.
Valid host range = 192.168.74.225/28 -192.168.74.238
The last valid IP = 192.168.74.238

Example 22:

What is the last valid host on the subnetwork 192.168.95.64/29?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.248
Block size = 256-248 = 8, which means we have a block size of
8 in the forth octet.
64 is divisible by 8.
That is 192.168.95.64/29 is a network address.
Next subnet address = 192.168.95.72/29.
Broadcast address is 192.168.95.71/29 of 192.168.95.64/29
network.
Valid host range = 192. 168. 95. 65 to 192. 168. 95. 70
The last valid IP = 192. 168. 95. 70

Example 23:

What is the last valid host on the subnetwork 192.168.115.128 255.255.255.192?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.192
Block size = 256-192 = 64, which means we have a block size of
64 in the forth octet.
128 is divisible by 64.
That is 192.168.115.128/26 is a network address.
Next subnet address = 192.168.115.192/26.
Broadcast address is 192.168.115.191/26 of 192.168.115.128/26
network.
Valid host range = 192. 168. 115. 129 to 192. 168. 115. 190
The last valid IP = 192. 168. 115. 190

Example 24:

What is the last valid host on the subnetwork192.168.129.80/28?

Solution:

Class C addresses use a default mask of 255.255.255.0.
Given subnet mask = 255.255.255.240
Block size = 256-240 = 16, which means we have a block size of
16 in the forth octet.
80 is divisible by 16.
That is 192.168.129.80/28 is a network address.
Next subnet address = 192.168.129.96/28.
Broadcast address is 192.168.129.95/28 of 192.168.129.80/28
network.
Valid host range = 192.168.129.81-192.168.129.94
The last valid IP = 192.168.129.94

Exercise

Exercise 1:

What is the last valid host on the subnetwork 172.27.117.192 255.255.255.224?

Answer:

The last valid IP = 172.27.117.222

Exercise 2:

What is the last valid host on the subnetwork 172.28.215.96 255.255.255.224?

Answer:

The last valid IP = 172. 28. 215. 126

Exercise 3:

What is the last valid host on the subnetwork 172.29.55.128 255.255.255.192?

Answer:

The last valid IP = 172.29.55.190

Exercise 4:

What is the last valid host on the subnetwork 172.31.183.128 255.255.255.192?

Answer:

The last valid IP = 172. 31. 183. 190

Exercise 5:

What is the last valid host on the subnetwork 192.168.179.96 255.255.255.224?

Answer:

The last valid IP = 192.168.179.126

Exercise 6:

What is the last valid host on the subnetwork 192.168.249.0 255.255.255.128?

Answer:

The last valid IP = 192. 168. 249. 126

Subnet Tutorials

Adscend

4.9.13

Last Valid Host Address

Last Valid Host Address

Class A

Class B

Class C

Exercise

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