From: Edouard Zorrilla (ezorrilla@tsf.com.pe)
Date: Mon Mar 16 2009 - 09:30:15 ART
2^128 - 2^32 = 2^32(2^96-1) so in the limit 2^96-1 ~ 2^96
So the answer should be 2^32(2^96-1) ~ 2^32(2^96) ~ 2^128 so the answer
should be A.
Jut math I guess,
Regards
----- Original Message -----
From: "Tharak Abraham" <tharakabraham@gmail.com>
To: "slo" <nebulite@gmail.com>
Cc: "Cisco certification" <ccielab@groupstudy.com>
Sent: Friday, March 13, 2009 8:15 AM
Subject: OT:Re: IPV6 vs IPv4 addresses
> Sorry for the minor err in option b.
> Q.How many more IP addresses are available with IPv6 than IPv4 ?
>
> a: 2^128
> b: 2^96
> c. 2^64
> d. 2^32
>
> @Slo : as long as IPv6 and IPv4 addresses doesn't overlap with each other,
> then why not 2^128?
> In this sense, even if i extinguish 100% of IPv4 why cant i have
> 2^128 more addresses in IPv6 !
>
>
>
> On Fri, Mar 13, 2009 at 1:59 PM, slo <nebulite@gmail.com> wrote:
>
>> IPv6 = 2^128
>> IPv4 = 2^32
>>
>> IPv6/IPv4 = 2^(128-32) = 2^96
>>
>> If you want know How many more than IPv4 , I think is 2^(96-1) = 2^95
>>
>> So, a, b, c, d, all are wrong.
>>
>> I will choice C 2^96
>>
>> -------
>> slo
>>
>> Best Regards!
>> ------------------------------------------------------------
>> Shen Lei CCIE#13947
>> Software Business Department
>> Jiangsu Fujitsu Telecommunications Technology Co.,Ltd. (JFTT) No.118,
>> Dengwei Rd.,Suzhou, P.R. China
>> Postcode: 215011
>> Mail :shenlei@cn.fujitsu.com
>> ------------------------------------------------------------
>>
>>
>>
>>
>> -----Original Message-----
>> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
>> Tharak Abraham
>> Sent: Friday, March 13, 2009 8:32 PM
>> To: Cisco certification
>> Subject: OT : IPV6 vs IPv4 addresses
>>
>> One of my friend got a question as follows in a tech exam.
>>
>> Looks easy but mind the word "more"
>>
>> Q.How many more IP addresses are available with IPv6 than IPv4 ?
>>
>> a: 2^128
>> b: 2^92
>> c. 2^64
>> d. 2^32
>>
>>
>> Best Regards,
>> Tharak Abraham
>>
>>
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>
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