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ipv6 subnetting

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  #1  
Old 20-11-2009
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Join Date: Nov 2009
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ipv6 subnetting

Need some help understanding why A is the correct answer...

Question states you need to configure IPV6 addresses on all computers in network. A global address prefix is assigned as 3FFA:FF2B:4D:B000::/41. You have to assign a subnet to four departments. Which subnetted address prefix will you assign to the fourth department?

A. 3FFA:FF2B:4D:C800::/43
B. 3FFA:FF2B:4D:B400::/43
C. 3FFA:FF2B:4D:C000::/43
D. 3FFA:FF2B:4D:F000::/45
E. None of the above.

The correct answer they give is A ,but having difficulty understanding. Looking at http://technet.microsoft.com/en-us/l...6997.aspx#ECAA
I determined F = 0xB000, n = 4, s = 2^s = 4 s=2, f = ?, i = ? (looks like they increment by 1536(decimal) 0x600(hex) but not understanding how they get that increment?
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  #2  
Old 2 Weeks Ago
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Join Date: Mar 2010
Posts: 1
Re: ipv6 subnetting

Not sure if you ever received a reply to this question but for what its worth here it goes:
The question actually is wrong. I believe this may be a typing error or something simillar.
The question really shoud state the following:

Question states you need to configure IPV6 addresses on all computers in network. A global address prefix is assigned as 3FFA:FF2B:4D:B000::/51. You have to assign a subnet to four departments. Which subnetted address prefix will you assign to the fourth department?

A. 3FFA:FF2B:4D:C800::/53
B. 3FFA:FF2B:4D:B400::/53
C. 3FFA:FF2B:4D:C000::/53
D. 3FFA:FF2B:4D:F000::/55
E. None of the above.

Review the question using these figures and with reference to the Microsoft posting you mentioned earlier you would find that:

F = B000 (the hexadecimal value of the subnet ID being subnetted otherwise known as the starting subnet)
n = 4 (number of subnets required)
s = 2 (the number of bits you are using for subnetting)
m = 51 (the prefix length of the address prefix being subnetted)
f = m - 48 OR 51 - 48 = 3

i = 2^16-(f+s)

i is the incremental value between each successive subnet ID.


Hence i = 2^16 - (3 + 2) or 2^11 = 2048 (800 Hex)

Hence first subnet is F = B000
Second subnet = B000 + 800 = B800
Third Subnet = B800 + 800 = C000
Fourth Subnet = C000 + 800 = C800

The net subnet prefix of course if the existing prefix (51) plus the number of bits required to add 4 new subnets which was 2 OR
51 + 2 = 53

This is how they arrived at the answer A. For those who are need more information on the formular used, pls go to
http://technet.microsoft.com/en-us/l...6997.aspx#ECAA

There really is a lot of misinformation on IPV6 subnetting out there and its already a challenging topic on its own without this added complication of typo's and a whole load of guys who are copying the wrong answers without a clue of how this works and passing on the answers as their own!!!
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