Simple ways of subnetting class A network addresses
Having treated how to subnet class B and Class C addresses, I did mention that class A was next in line. Starting from class C, then to class B was a deliberate move since it is easier to understand C simply because it has less host bits (8) than the rest.
With class A you have a total of 24 host bits to manipulate. In this demonstration, I will share with us how to subnet class A addresses with three examples. The examples have been carefully selected to address issues students face when subnetting class A addresses. Lets get going.
Range of class A address: 027; usable is 126. What this means is that if the values in the first octet of an IP fall between 0 and 127, then that IP is a class A address, although you can only use from 1 to 126.
Example 1: Given 10.0.0.0/10, determine the following:
(i) The number of subnets that the given IP and subnetmask will produce
(ii) The number of hosts per subnet generated in (i) above
(iii) With the aid of a table, list the subnets, first, last and broadcast IPs in each subnet.
/10=255.192.0.0
Answers:
(i) Number of subnets =2^x where x equals the number of bits borrowed. If the default subnetmask for class A is /8 which is 255.0.0.0 and we are given /10 in this question, it therefore means we borrowed 2 bits. Substituting 2 into the formula, we have 2^2=4. We have 4 subnets.
(ii) Number of hosts per subnets=2^y2, where y equals the number of off bits. To get our off bit is simple. 3210=22. IPv4 is a 32bit address but we are give /10 in this question, therefore, we simply subtract 10 from 32 to get 22. Now substituting 22 into the formula, we get 2^222=4,194,302. we will have 4,194,302 per each of the four subnets.
(iii)We need the block size to be able to list out subnets. Block size is 256192=64. Because the subnet bit falls in the second octet, we will list our subnets with the change happening in the second octet.
Example 2: Given 10.0.0.0/17, determine the following:
(i) The number of subnets that the given IP and subnetmask will produce
(ii) The number of hosts per subnet generated in (i) above
(iii) With the aid of a table, list the subnets, first, last and broadcast IPs in each subnet.
/17=255.255.128.0
Answers:
(i) Number of subnets=2^9=512
(ii) Number of hosts per subnet=2^152=32,766
(iii)
First eight, using the last three octet( that 0.0.0 instead of 10.0.0.0, just because of space)
Last eight, using the last three octets (0.0.0 instead of 10.0.0.0, just because of space)
Example 3: Given 10.0.0.0/24, determine the follow:
(i) The number of subnets that the given IP and subnetmask will produce
(ii) The number of hosts per subnet generated in (i) above
(iii) With the aid of a table, list the subnets, first, last and broadcast IPs in each subnet.
/24=255.255.255.0
Answers:
(i) Number of subnets=2^16=65,536
(ii) Number of hosts=2^82=254
(iii) Our valid subnets, hosts and broadcasts are as given in the tables below:
First eight subnets, using the last three octets:
Last eight subnets, using the last three octets:
With class A you have a total of 24 host bits to manipulate. In this demonstration, I will share with us how to subnet class A addresses with three examples. The examples have been carefully selected to address issues students face when subnetting class A addresses. Lets get going.
Range of class A address: 027; usable is 126. What this means is that if the values in the first octet of an IP fall between 0 and 127, then that IP is a class A address, although you can only use from 1 to 126.
Example 1: Given 10.0.0.0/10, determine the following:
(i) The number of subnets that the given IP and subnetmask will produce
(ii) The number of hosts per subnet generated in (i) above
(iii) With the aid of a table, list the subnets, first, last and broadcast IPs in each subnet.
/10=255.192.0.0
Answers:
(i) Number of subnets =2^x where x equals the number of bits borrowed. If the default subnetmask for class A is /8 which is 255.0.0.0 and we are given /10 in this question, it therefore means we borrowed 2 bits. Substituting 2 into the formula, we have 2^2=4. We have 4 subnets.
(ii) Number of hosts per subnets=2^y2, where y equals the number of off bits. To get our off bit is simple. 3210=22. IPv4 is a 32bit address but we are give /10 in this question, therefore, we simply subtract 10 from 32 to get 22. Now substituting 22 into the formula, we get 2^222=4,194,302. we will have 4,194,302 per each of the four subnets.
(iii)We need the block size to be able to list out subnets. Block size is 256192=64. Because the subnet bit falls in the second octet, we will list our subnets with the change happening in the second octet.
Subnets

10.0.0.0

10.64.0.0

10.128.0.0

10.192.0.0

First IP

10.0.0.1

10.64.0.1

10.128.0.1

10.192.0.1

Last IP

10.63.255.254

10.127.255.254

10.191.255.254

10.255.255.254

Broadcast

10.63.255.255

10.127.255.255

10.191.255.255

10.255.255.255

Example 2: Given 10.0.0.0/17, determine the following:
(i) The number of subnets that the given IP and subnetmask will produce
(ii) The number of hosts per subnet generated in (i) above
(iii) With the aid of a table, list the subnets, first, last and broadcast IPs in each subnet.
/17=255.255.128.0
Answers:
(i) Number of subnets=2^9=512
(ii) Number of hosts per subnet=2^152=32,766
(iii)
First eight, using the last three octet( that 0.0.0 instead of 10.0.0.0, just because of space)
subnets

0.0.0

0.128.0

1.0.0

1.128.0

2.0.0

2.128.0

3.0.0

3.128.0

First IP

0.0.1

0.128.1

1.0.1

1.128.1

2.0.1

2.128.1

3.0.1

3.128.1

Last IP

0.127.254

0.255.254

1.127.254

1.255.254

2.127.254

2.255.254

3.127.254

3.255.254

Broadcast

0.127.255

0.255.255

1.127.255

1.255.255

2.127.255

2.255.2555

3.127.255

3.255.255

Last eight, using the last three octets (0.0.0 instead of 10.0.0.0, just because of space)
Subnet

252.0.0

252.128.0

253.0.0

253.128.0

254.0.0

254.128.0

255.0.0

255.128.0

First IP

252.0.1

252.128.1

253.0.1

253.128.1

25454.0.1

254.128.1

255.0.1

255.128.1

Last IP

252.127.254

252.255.254

253.127.254

253.255.254

254.127.254

254.255.254

255.127.254

255.255.254

BC

252.127.255

252.255.255

253.127.255

253.255.255

254.127.255

254.255.2555

255.127.255

255.255.255

Example 3: Given 10.0.0.0/24, determine the follow:
(i) The number of subnets that the given IP and subnetmask will produce
(ii) The number of hosts per subnet generated in (i) above
(iii) With the aid of a table, list the subnets, first, last and broadcast IPs in each subnet.
/24=255.255.255.0
Answers:
(i) Number of subnets=2^16=65,536
(ii) Number of hosts=2^82=254
(iii) Our valid subnets, hosts and broadcasts are as given in the tables below:
First eight subnets, using the last three octets:
Subnet

0.0.0

0.1.0

0.2.0

0.3.0

0.4.0

0.5.0

0.6.0

0.7.0

First IP

0.0.1

0.1.1

0.2.1

0.3.1

0.4.1

0.5.1

0.6.1

0.7.1

Last IP

0.0.254

0.1.254

0.2.254

0.3.254

0.4.254

0.5.254

0.6.254

0.7.254

BC

0.0.255

0.1.255

0.2.255

0.3.255

0.4.255

0.5.255

0.6.255

0.7.255

Last eight subnets, using the last three octets:
Sub

255.248.0

255.249.0

255.250.0

255.251.0

255.252.0

255.253.0

255.254.0

255.255.0

First IP

255.248.1

255.249.1

255.250.1

255.251.1

255.252.1

255.253.1

255.254.1

255.255.1

Last IP

255.248.254

255.249.254

255.250.254

255.251.254

255.252.254

255.253.254

255.254.254

255.255.254

BC

255.248.255

255.249.255

255.250.255

255.251.255

255.252.255

255.253.255

255.254.255

255.255.255

Great! Informative post
ReplyDelete