<?xml version="1.0" encoding="UTF-8"?>
<rss xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#" xmlns:taxo="http://purl.org/rss/1.0/modules/taxonomy/" version="2.0">
  <channel>
    <title>topic Re: 3 Node Simplivity cluster usable capacity in HPE SimpliVity</title>
    <link>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172483#M3499</link>
    <description>&lt;P&gt;Here is the general formula I found to use when sizing for the SimpliVity DVP capacity efficiency:&lt;/P&gt;&lt;P&gt;( Production data + (production data * percent expected growth/100) ) / 2.25 efficiency&lt;/P&gt;&lt;P&gt;For example a customer with 125 TB of data and expected growth of 25%:&lt;/P&gt;&lt;P&gt;( 125 + ( 125 * 25/100 ) ) / 2.25 = ~69.44 TB&lt;BR /&gt;&lt;BR /&gt;If a node failed, I would remove that amount of capacity from the equation-&lt;BR /&gt;Here is a good web page that may help you with usable capacity.&lt;BR /&gt;&lt;A href="http://www.vhersey.com/2016/05/16/simplivity-effective-storage-capacity/" target="_blank"&gt;http://www.vhersey.com/2016/05/16/simplivity-effective-storage-capacity/&lt;/A&gt;&lt;/P&gt;&lt;P&gt;&amp;nbsp;&lt;/P&gt;</description>
    <pubDate>Thu, 18 Aug 2022 14:11:59 GMT</pubDate>
    <dc:creator>MikeSeden</dc:creator>
    <dc:date>2022-08-18T14:11:59Z</dc:date>
    <item>
      <title>3 Node Simplivity cluster usable capacity</title>
      <link>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172478#M3498</link>
      <description>&lt;P&gt;Hello everyone,&lt;/P&gt;&lt;P&gt;I have a question about Simplivity usable space in 3 node clusters.&lt;/P&gt;&lt;P&gt;As I understand if in 2 node cluster reported physical space is 59TB, that means the actual usable space is ~29TB (50% goes HA), similar to RAID1&lt;/P&gt;&lt;P&gt;In 3 node case, does RAID5 level of usable space and availability apply? Meaning only 1 node can fail and usable space would 2/3 of total physical space and 1/3 to HA?&lt;/P&gt;</description>
      <pubDate>Mon, 22 Aug 2022 03:48:17 GMT</pubDate>
      <guid>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172478#M3498</guid>
      <dc:creator>steez</dc:creator>
      <dc:date>2022-08-22T03:48:17Z</dc:date>
    </item>
    <item>
      <title>Re: 3 Node Simplivity cluster usable capacity</title>
      <link>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172483#M3499</link>
      <description>&lt;P&gt;Here is the general formula I found to use when sizing for the SimpliVity DVP capacity efficiency:&lt;/P&gt;&lt;P&gt;( Production data + (production data * percent expected growth/100) ) / 2.25 efficiency&lt;/P&gt;&lt;P&gt;For example a customer with 125 TB of data and expected growth of 25%:&lt;/P&gt;&lt;P&gt;( 125 + ( 125 * 25/100 ) ) / 2.25 = ~69.44 TB&lt;BR /&gt;&lt;BR /&gt;If a node failed, I would remove that amount of capacity from the equation-&lt;BR /&gt;Here is a good web page that may help you with usable capacity.&lt;BR /&gt;&lt;A href="http://www.vhersey.com/2016/05/16/simplivity-effective-storage-capacity/" target="_blank"&gt;http://www.vhersey.com/2016/05/16/simplivity-effective-storage-capacity/&lt;/A&gt;&lt;/P&gt;&lt;P&gt;&amp;nbsp;&lt;/P&gt;</description>
      <pubDate>Thu, 18 Aug 2022 14:11:59 GMT</pubDate>
      <guid>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172483#M3499</guid>
      <dc:creator>MikeSeden</dc:creator>
      <dc:date>2022-08-18T14:11:59Z</dc:date>
    </item>
    <item>
      <title>Re: 3 Node Simplivity cluster usable capacity</title>
      <link>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172498#M3500</link>
      <description>&lt;P&gt;Hi Steez&lt;/P&gt;&lt;P&gt;Hope you are doing well.&lt;/P&gt;&lt;P&gt;We always keep two full copies of hives that is why It will be a total capacity / 2&lt;BR /&gt;OR&amp;nbsp;&lt;BR /&gt;Usable capacity per cluster =[ #nodes * (Usable capacity per node)] /2&lt;/P&gt;&lt;P&gt;Hope this helps.!!&lt;/P&gt;&lt;P&gt;Regards&lt;BR /&gt;Mahesh.&lt;/P&gt;</description>
      <pubDate>Thu, 18 Aug 2022 17:06:37 GMT</pubDate>
      <guid>https://community.hpe.com/t5/hpe-simplivity/3-node-simplivity-cluster-usable-capacity/m-p/7172498#M3500</guid>
      <dc:creator>Mahesh202</dc:creator>
      <dc:date>2022-08-18T17:06:37Z</dc:date>
    </item>
  </channel>
</rss>

