Cisco CCNA: subnetting class A addresses, from start to finish.

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:
0-27; usable is 1-26. 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^y-2, where y equals the number of off bits. To get our off bit is
simple. 32-10=22. IPv4 is a 32-bit 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 2^22-2=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 256-192=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=2^15-2=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=2^8-2=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

 

All questions regarding this should be dropped in the comment box. Thank you.

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Timigate

I believe that Africa will advance technologically if knowledge is shared among the living than out of greed, taken to the grave.

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