Subnet Practice Examples

Class C Addresses

Practice example 1:

Network address: 192.168.10.0
Subnet mask: 255.255.255.224
1. How many subnets?
Answer... Hide Ans: 6
2. How many hosts?
Answer... Hide Ans: 30
3. What are the valid subnets?
Answer... Hide Ans: 32, 64, 96, 128, 160, 192
4. Fill in the table below.

Subnet 1 Subnet 2 Subnet 3 Subnet 4 Subnet 5 Subnet 6 Meaning
Answer... Hide 32 Answer... Hide 64 Answer... Hide 96 Answer... Hide 128 Answer... Hide 160 Answer... Hide 192 The subnet address
Answer... Hide 33 Answer... Hide 65 Answer... Hide 97 Answer... Hide 129 Answer... Hide 161 Answer... Hide 193 The first valid host
Answer... Hide 62 Answer... Hide 94 Answer... Hide 126 Answer... Hide 158 Answer... Hide 190 Answer... Hide 222 Our last valid host
Answer... Hide 63 Answer... Hide 95 Answer... Hide 127 Answer... Hide 159 Answer... Hide 191 Answer... Hide 223 The broadcast address

Practice example 2:

Network address: 192.168.10.0 Subnet mask: 255.255.255.240 1. How many subnets?
Answer... Hide Ans: 14
2. How many hosts?
Answer... Hide Ans: 14
3. What are the valid subnets?
Answer... Hide Ans: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224
4. Fill in the table below.

Subnet 16 32 48 64 80 96 112 128 144 160 176 192 208 224
First Host . . . . . . . . . . . . . .
Last Host
Broadcast


Answer... Hide

Subnet 16 32 48 64 80 96 112 128 144 160 176 192 208 224
First Host 17 33 49 65 81 97 113 129 145 161 177 193 209 225
Last Host 30 46 62 78 94 110 126 142 158 174 190 206 222 238
Broadcast 31 47 63 79 95 111 127 143 159 175 191 207 223 239

Practice example 3:

Network address: 192.168.10.0 Subnet mask: 255.255.255.248 1. How many subnets?
Answer... Hide Ans: 30
2. How many hosts?
Answer... Hide Ans: 6
3. What are the valid subnets?
Answer... Hide Ans: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240
4. Fill in the table below (only the first three and last three subnets).

Subnet 8 16 24 224 232 240
First Host            
Last Host            
Broadcast            


Answer... Hide

Subnet 8 16 24 224 232 240
First Host 9 17 25 225 233 241
Last Host 14 22 30 230 238 246
Broadcast 15 23 31 231 239 247

Practice example 4:

Network address: 192.168.10.0 Subnet mask: 255.255.255.252 1. How many subnets?
Answer... Hide Ans: 62
2. How many hosts?
Answer... Hide Ans: 2
3. What are the valid subnets?
Answer... Hide Ans: 4, 8, 12 etc., all the way to 248
4. Fill in the table below (only the first three and last three subnets).

Subnet 4 8 12 240 244 248
First Host          
Last Host          
Broadcast          


Answer... Hide

Subnet 4 8 12 240 244 248
First Host 5 9 13 241 245 249
Last Host 6 10 14 242 246 250
Broadcast 7 11 15 243 247 251

Practice example 5:

Write the subnet, broadcast address and valid host range for the following:
1. 172.16.10.5 255.255.255.128
Answer... Hide Ans: Subnet is 172.16.10.0, broadcast is 172.16.10.127 and valid host range is 172.16.10.1 to 126. You need to ask yourself, “Is the subnet bit in the fourth octet on or off?” If the host address has a value of less than 128 in the fourth octet, then the subnet bit must be off. If the value of the fourth octet is higher than 128, then the subnet bit must be on. In this case, the host address is 10.5 and the bit in the fourth octet must be off. The subnet must be 172.16.10.0.


2. 172.16.10.33 255.255.255.224
Answer... Hide Ans: Subnet is 172.16.10.32, broadcast is 172.16.10.63 and valid host range is 172.16.10.33 to 10.62. 256-224=32. 32+32=64. The subnet is 10.32 and the next subnet is 10.64, so the broadcast address must be 10.63.


3. 172.16.10.65 255.255.255.192
Answer... Hide Ans: Subnet is 172.16.10.64, broadcast is 172.16.10.127 and valid host range is 172.16.10.65 to 172.16.10.126. 256-192=64. 64+64=128, so the network address must be 172.16.10.64, with a broadcast of 172.16.10.127.


4. 172.16.10.17 255.255.255.252
Answer... Hide Ans: Network is 172.16.10.16, broadcast is 172.16.10.19 and valid host range is 172.16.10.17 to 18. 256-252=4. 4+4=8, plus 4=12, plus 4=16, plus 4=20. So the subnet is 172.16.10.16 and the broadcast must be 10.19.


5. 172.16.10.33 255.255.255.240
Answer... Hide Ans: Network is 172.16.10.32, broadcast is 172.16.10.47 and valid host range is 172.16.10.33 to 46. 256-240=16. 16+16=32, plus 16=48. The subnet is 172.16.10.32 and the broadcast is 172.1610.47.


6. 192.168.100.25 255.255.255.252
Answer... Hide Ans: Subnet is 192.168.100.24, broadcast is 192.168.100.27 and valid host range is 192.168.100.25 to 26. 256-252=4. 4+4=8, plus 4=12, plus 4=16, plus 4=20, plus 4=24, plus 4=28. The subnet is 100.24 and the broadcast is 100.27.


7. 192.168.100.17, with 4 bits of subnetting
Answer... Hide Ans: Subnet is 192.168.100.16, broadcast is 192.168.100.31 and valid host range is 192.168.100.17 to 30. 256-240=16. 16+16=32. The subnet is, then, 100.16 and the broadcast is 100.31 because 32 is the next subnet.


8. 192.168.100.66, with 3 bits of subnetting
Answer... Hide Ans: Subnet is 192.168.100.64, broadcast is 192.168.100.95 and valid host range is 192.168.100.65 to 94. 256-224=32. 32+32=64, plus 32=96. The subnet is 100.64 and the broadcast is 100.95.


9. 192.168.100.17 255.255.255.248
Answer... Hide Ans: Subnet is 192.168.100.16, broadcast is 192.168.100.23 and valid host range is 192.168.100.17 to 22. 256-248=8. 8+8=16, plus 8=24. The subnet is 16 and the broadcast is 23.


10. 10.10.10.5 255.255.255.252
Answer... Hide Ans: Subnet is 10.10.10.4, broadcast is 10.10.10.7 and valid host range is 10.10.10.5 to 6. 256-252=4. 4+4=8.

-- JimSkon - 2012-11-07

Topic revision: r2 - 2012-12-06 - JimSkon
 
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