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Subnetting -addressing classes
15 years 8 months ago #29987
by G33KKitty
Subnetting -addressing classes was created by G33KKitty
Hey guys..this is my first post*exciting*well it would be if i didnt need help!
i was wondering if one of you could possibly help me please?
i have a university assignment due in and this is the question i am stuck on
Your company is in the process of requesting an IP address from ARIN. As part of your initial information gathering, you know that you will need at least 10 different subnets with at least 250 hosts each. Which IP address class will you request from ARIN? Justify your answer.
Would any of you be able to help me please?i would be ever so grateful.
Than kyou:)kittyx
i was wondering if one of you could possibly help me please?
i have a university assignment due in and this is the question i am stuck on
Your company is in the process of requesting an IP address from ARIN. As part of your initial information gathering, you know that you will need at least 10 different subnets with at least 250 hosts each. Which IP address class will you request from ARIN? Justify your answer.
Would any of you be able to help me please?i would be ever so grateful.
Than kyou:)kittyx
15 years 8 months ago #29989
by skylimit
"...you are never too old to learn" anon
Replied by skylimit on topic Re: Subnetting -addressing classes
I'll give this a shot.
Using the class B address as an example. 172.16.1.0/16 (default mask). A netmask of 255.255.240.0 (/20) will give you 16 subnets (2^4) i.e. including the network (zero) subnet and broadcast subnet. The number of hosts on the /20 subnet will be 2^n - 2 = 2^12 - 2 = 4094. Yea, it way more than 256 hosts but that should be fine. If you went for say a class c address with a /28 mask you'll get 16 subnets but less than 256 hosts which would not satisfy your need so I think a class B address will be what you want so you can then subnet it...
Binary conversion:
255.255.240.0 = 11111111.11111111.11110000.00000000
The bits in bold are the subnet bits so 240 is obtained by adding the binary valuse 128+64+32+16 (left to right)
So a 172.16.1.0/20 will be ideal...
Sorry not the best explanation. I guess someone will come up with a better explanation.
PS. the subnetting tutorial on firewall.cx is also very useful.
Using the class B address as an example. 172.16.1.0/16 (default mask). A netmask of 255.255.240.0 (/20) will give you 16 subnets (2^4) i.e. including the network (zero) subnet and broadcast subnet. The number of hosts on the /20 subnet will be 2^n - 2 = 2^12 - 2 = 4094. Yea, it way more than 256 hosts but that should be fine. If you went for say a class c address with a /28 mask you'll get 16 subnets but less than 256 hosts which would not satisfy your need so I think a class B address will be what you want so you can then subnet it...
Binary conversion:
255.255.240.0 = 11111111.11111111.11110000.00000000
The bits in bold are the subnet bits so 240 is obtained by adding the binary valuse 128+64+32+16 (left to right)
So a 172.16.1.0/20 will be ideal...
Sorry not the best explanation. I guess someone will come up with a better explanation.
PS. the subnetting tutorial on firewall.cx is also very useful.
"...you are never too old to learn" anon
- broadcaststorm
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15 years 7 months ago #30275
by broadcaststorm
Replied by broadcaststorm on topic Re: Subnetting -addressing classes
If you only need 250 hosts as a max, then a /24 (regular class C address) would do, and you can just ask ARIN for how ever many class C network addresses as you need. You have a requirement for 10 networks, so you can just buy 10 Class C addresses. Future expansion then simply requires buying more class C numbers.
If you need future expansion without the extra potential cost of buying more addresses, then a /23 class B address will give you up to 510 hosts (more than double your original requirement) and 128 networks, which is plenty more than 10, so you have room for expansion of both hosts and networks, as well as the freedom to VLSM and still retain useful network sizes.
If cost is an issue, you may find that a /23 address range is a lot cheaper than a /20 range from ARIN or your ISP. If the cost is the same then buy as large an address range as is available, but IME, it usually does come down to money. However if the assignment is to get as close to 250 hosts as possible regardless of any other considerations, then a single class C address per network address range as required will do.
If you need future expansion without the extra potential cost of buying more addresses, then a /23 class B address will give you up to 510 hosts (more than double your original requirement) and 128 networks, which is plenty more than 10, so you have room for expansion of both hosts and networks, as well as the freedom to VLSM and still retain useful network sizes.
If cost is an issue, you may find that a /23 address range is a lot cheaper than a /20 range from ARIN or your ISP. If the cost is the same then buy as large an address range as is available, but IME, it usually does come down to money. However if the assignment is to get as close to 250 hosts as possible regardless of any other considerations, then a single class C address per network address range as required will do.
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