Simple Ways Of Subnetting Course Of Didactics A Network Addresses

Having treated how to subnet class B and Class C addresses, I did lift that degree Influenza A virus subtype H5N1 was adjacent inward line. Starting from degree C, too then to degree B was a deliberate motility since it is easier to sympathize C merely because it has less host bits (8) than the rest.
With degree Influenza A virus subtype H5N1 you lot convey a amount of 24 host bits to manipulate. In this demonstration, I volition part amongst us how to subnet degree Influenza A virus subtype H5N1 addresses amongst 3 examples. The examples convey been carefully selected to address issues students human face upwards when subnetting degree Influenza A virus subtype H5N1 addresses. Lets teach going.





Range of degree Influenza A virus subtype H5N1 address: 0-27; usable is 1-26. What this agency is that if the values inward the commencement octet of an IP autumn betwixt 0 too 127, too then that IP is a degree Influenza A virus subtype H5N1 address, although you lot tin hand notice alone operate from 1 to 126.

Example 1: Given 10.0.0.0/10, create upwards one's heed the following:

(i) The number of subnets that the given IP too subnetmask volition hit

(ii) The number of hosts per subnet generated inward (i) inward a higher house

(iii) With the assistance of a table, listing the subnets, first, end too broadcast IPs inward 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 degree Influenza A virus subtype H5N1 is /8 which is 255.0.0.0 too nosotros are given /10 inward this question, it hence agency nosotros borrowed ii bits. Substituting ii into the formula, nosotros convey 2^2=4. We convey 4 subnets.

(ii) Number of hosts per subnets=2^y-2, where y equals the number of off bits. To teach our off flake is simple. 32-10=22. IPv4 is a 32-bit address but nosotros are hand /10 inward this question, therefore, nosotros merely subtract ten from 32 to teach 22. Now substituting 22 into the formula, nosotros teach 2^22-2=4,194,302. nosotros volition convey 4,194,302 per each of the 4 subnets.

(iii)We demand the block size to live able to listing out subnets. Block size is 256-192=64. Because the subnet flake falls inward the mo octet, nosotros volition listing our subnets amongst the alter happening inward the mo 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, create upwards one's heed the following:

(i) The number of subnets that the given IP too subnetmask volition produce
(ii) The number of hosts per subnet generated inward (i) above
(iii) With the assistance of a table, listing the subnets, first, end too broadcast IPs inward 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 end 3 octet( that 0.0.0 instead of 10.0.0.0, exactly 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 end 3 octets (0.0.0 instead of 10.0.0.0, exactly 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, create upwards one's heed the follow:

(i) The number of subnets that the given IP too subnetmask volition produce
(ii) The number of hosts per subnet generated inward (i) above
(iii) With the assistance of a table, listing the subnets, first, end too broadcast IPs inward 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 too broadcasts are equally given inward the tables below:

First 8 subnets, using the end 3 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 8 subnets, using the end 3 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

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