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    <div class="moz-cite-prefix">On 01/22/2013 08:37 AM, august forsakov
      wrote:<br>
    </div>
    <blockquote
cite="mid:CAHT2OAFPb6J1Cu-mMWRFeNzSOSJF3JYJ8mzXfBtGgi4=u7vPtA@mail.gmail.com"
      type="cite">
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        <blockquote class="gmail_quote">
          <div lang="EN-AU">
            <div>
              <p class="MsoNormal">
                <span>IPv6 addresses are assigned to organizations in
                  much larger blocks as compared to IPv4 address
                  assignments—the recommended allocation is a</span><span> </span><span>/48</span><span> </span><span>block

                  which contains 2<sup>80</sup>addresses, being 2<sup>48</sup></span><span> </span><span>or
                  about</span><span> </span><span>2.8×10<sup>14</sup></span><span> </span><span>times

                  larger than the entire IPv4 address space of 2<sup>32</sup></span><span> </span><span>addresses
                  and about</span><span> </span><span>7.2×10<sup>16</sup></span><span> </span><span>times

                  larger than the</span><span> </span><span>/8</span><span> </span><span>blocks

                  of IPv4 addresses, which are the largest allocations
                  of IPv4 addresses. The total pool, however, is
                  sufficient for the foreseeable future, because there
                  are 2<sup>128</sup></span><span> </span><span>or about</span><span> </span><span>3.4×10<sup>38</sup></span><span> </span><span>(340</span><span> </span><span><a
                    moz-do-not-send="true"
                    href="http://en.wikipedia.org/wiki/10%5E12"
                    title="10^12" target="_blank"><span>trillion</span></a></span><span> </span><span>trillion

                  trillion) unique IPv6 addresses.</span></p>
              <p class="MsoNormal">
                <span>Each RIR can divide each of its multiple</span><span> </span><span>/23</span><span> </span><span>blocks

                  into 512</span><span> </span><span>/32</span><span> </span><span>blocks,

                  typically one for each ISP; an ISP can divide its</span><span> </span><span>/32</span><span> </span><span>block

                  into</span><span> </span><span>65536</span><span> </span><span>/48</span><span> </span><span>blocks,

                  typically one for each customer;<sup><a
                      moz-do-not-send="true"
                      href="http://en.wikipedia.org/wiki/IPv6_address#cite_note-16"
                      target="_blank"><span>[16]</span></a></sup></span><span> </span><span>customers

                  can create</span><span> </span><span>65536</span><span> </span><span>/64</span><span> </span><span>networks

                  from their assigned</span><span> </span><span>/48</span><span> </span><span>block,

                  each having 2<sup>64</sup></span><span> </span><span>addresses.
                  In contrast, the entire IPv4 address space has only 2<sup>32</sup></span><span> </span><span>(about4.3×10<sup>9</sup>)
                  addresses.</span></p>
              <p class="MsoNormal">
                <span>...</span><span></span></p>
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          </div>
        </blockquote>
        <div><br>
        </div>
        <div>However many trillion, quadrillion or quintillion ip
          addresses are possible with IPv6 there will always be a
          shortage at some point...</div>
        <div>Any technological renewal wave will show that to be true,
          so we _will_ run out of IPv6...</div>
        <div>Maybe we'll want to assign each gene an ip address as will
          as its associated body cell... hahaha</div>
      </div>
    </blockquote>
    <br>
    I think quoting the number of addresses in IPv6 ranges rather than
    number of subnets is misleading at best and dangerous at worst.<br>
    <br>
    No one is going to use <span>2<sup>64</sup> MAC addresses in a /64. 
      They're going to use somewhere between 2<sup>1</sup></span> and
    maybe 2<sup>24</sup><span> at the outside in the very largest L2
      data centre deployments.  (Assuming they find some way of dealing
      with broad-/multicast issues and are using something like MAC
      address rewriting [1].)<br>
      <br>
      We should be talking in terms of numbers of /64 subnets.  Using
      the figures above, /23 == 512 ISPs, each of which has 65536
      customers (/48), each of which has 65536 subnets (/64).  That
      corresponds exactly with the number of /24 subnets available to
      organisations using the 10.0.0.0/8 block internally under IPv4
      today (with the obvious advantage of not needing NAT).<br>
      <br>
      When i see those sort of numbers i'm glad Jeff Doyle is suggesting
      that each building gets a /48 instead of each customer - i can see
      it not being enough even within a short space of time, given
      current industry talk about in-car networks (CANs?) and the
      "Internet of things".<br>
      <br>
      Regards,<br>
      Paul<br>
      <br>
      [1] </span><a class="moz-txt-link-freetext" href="http://www.cs.duke.edu/courses/fall10/cps296.2/lectures/16RoutingTopology.pdf">http://www.cs.duke.edu/courses/fall10/cps296.2/lectures/16RoutingTopology.pdf</a><br>
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