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    <title>topic Re: Procedure to locate/idnetify an iCAP processor in Operating System - HP-UX</title>
    <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789248#M607271</link>
    <description>iCAP CPUs are bound to complexes, not to cell boards, so if you took a cell board out of one complex and put it in another then you would be moving ACTIVE CPUs from that complex to another.&lt;BR /&gt;&lt;BR /&gt;So for example if I had:&lt;BR /&gt;&lt;BR /&gt;System A - a rx8640 with 4 cell boards/32 cores and iCAP reported 24 Active/8 iCAP cores&lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 3 cell boards/24 cores and no iCAP&lt;BR /&gt;&lt;BR /&gt;and I took a cell board out of system A and put it in system B, I would end up with:&lt;BR /&gt;&lt;BR /&gt;System A - a rx8640 with 3 cell boards/24 cores and iCAP reported 16 Active/ 8 iCAP cores&lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 3 cell boards/32 cores and no iCAP&lt;BR /&gt;&lt;BR /&gt;See section A.5 on p189 of the same manual.&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
    <pubDate>Thu, 19 May 2011 09:53:59 GMT</pubDate>
    <dc:creator>Duncan Edmonstone</dc:creator>
    <dc:date>2011-05-19T09:53:59Z</dc:date>
    <item>
      <title>Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789243#M607266</link>
      <description>Hi Experts,&lt;BR /&gt;There are three iCAP CPUs in a superdome partition. How can I locate/idnetify which are iCAP CPUs (cell number, CPU slot number).</description>
      <pubDate>Wed, 18 May 2011 08:42:56 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789243#M607266</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-18T08:42:56Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789244#M607267</link>
      <description>There is nothing "physically" different about an iCAP processor versus an active processor. All that happens is that at boot time a system with iCAP codewords applied will read data on the number of iCAP CPUs out of NVRAM on the system, and not use that many CPUs. This is left to the OS (and/or vPar software) to determine.&lt;BR /&gt;&lt;BR /&gt;Why do you care which CPU is on and which is off?&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Thu, 19 May 2011 05:52:14 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789244#M607267</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-19T05:52:14Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789245#M607268</link>
      <description>Thanks Duncan.&lt;BR /&gt;So if I move a CPU or a full cell board with 4 CPUs from an iCAP enabled partition to another partition, what will be the status of CPU there? (active or inactive)?</description>
      <pubDate>Thu, 19 May 2011 08:11:53 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789245#M607268</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-19T08:11:53Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789246#M607269</link>
      <description>This is discussed in the iCAP manual:&lt;BR /&gt;&lt;BR /&gt;&lt;A href="http://bizsupport2.austin.hp.com/bc/docs/support/SupportManual/c01942677/c01942677.pdf" target="_blank"&gt;http://bizsupport2.austin.hp.com/bc/docs/support/SupportManual/c01942677/c01942677.pdf&lt;/A&gt;&lt;BR /&gt;&lt;BR /&gt;See section 4.11 on p67&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Thu, 19 May 2011 08:26:02 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789246#M607269</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-19T08:26:02Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789247#M607270</link>
      <description>Hi Thanks,&lt;BR /&gt;This document discuss about the same complex with more than one partition.&lt;BR /&gt;If I move the CPU or Cell board from one complex (iCAP enabled)to other complex (non iCAP). What would be the status of CPUs in this 2nd complex?</description>
      <pubDate>Thu, 19 May 2011 09:16:11 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789247#M607270</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-19T09:16:11Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789248#M607271</link>
      <description>iCAP CPUs are bound to complexes, not to cell boards, so if you took a cell board out of one complex and put it in another then you would be moving ACTIVE CPUs from that complex to another.&lt;BR /&gt;&lt;BR /&gt;So for example if I had:&lt;BR /&gt;&lt;BR /&gt;System A - a rx8640 with 4 cell boards/32 cores and iCAP reported 24 Active/8 iCAP cores&lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 3 cell boards/24 cores and no iCAP&lt;BR /&gt;&lt;BR /&gt;and I took a cell board out of system A and put it in system B, I would end up with:&lt;BR /&gt;&lt;BR /&gt;System A - a rx8640 with 3 cell boards/24 cores and iCAP reported 16 Active/ 8 iCAP cores&lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 3 cell boards/32 cores and no iCAP&lt;BR /&gt;&lt;BR /&gt;See section A.5 on p189 of the same manual.&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Thu, 19 May 2011 09:53:59 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789248#M607271</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-19T09:53:59Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789249#M607272</link>
      <description>Great!&lt;BR /&gt;So after moving the cell board, the system B is &lt;BR /&gt;&amp;gt;&amp;gt;System B - a rx8640 with 3 cell boards/32 cores and no iCAP&lt;BR /&gt;it would be&lt;BR /&gt;System B - a rx8640 with 4 cell boards/32 cores and no iCAP&lt;BR /&gt;??&lt;BR /&gt;And what about all the 32 cores in system B, active?&lt;BR /&gt;&lt;BR /&gt;Best Regards,&lt;BR /&gt;Abid Iqbal</description>
      <pubDate>Thu, 19 May 2011 10:18:51 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789249#M607272</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-19T10:18:51Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789250#M607273</link>
      <description>Hi,&lt;BR /&gt;Also your opinion will be of great value to me.&lt;BR /&gt;Below is a new situation.&lt;BR /&gt;System A - a rx8640 with 3 cell boards/24 cores and iCAP reported 18 Active/6 iCAP cores&lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 2 cell boards/12 cores (six on each cell) and no iCAP&lt;BR /&gt;&lt;BR /&gt;and I took a cell board out of system A and put it in system B, It would end up with:&lt;BR /&gt;&lt;BR /&gt;System A - a rx8640 with 2 cell boards/16 cores and iCAP reported 16 Active/ 0 iCAP cores&lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 3 cell boards/20 cores and no iCAP ( all active cores)&lt;BR /&gt; Is it true??&lt;BR /&gt;Regards,&lt;BR /&gt;Abid Iqbal</description>
      <pubDate>Thu, 19 May 2011 10:34:00 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789250#M607273</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-19T10:34:00Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789251#M607274</link>
      <description>Answering the first question:&lt;BR /&gt;&lt;BR /&gt;&amp;gt; And what about all the 32 cores in system B, active?&lt;BR /&gt;&lt;BR /&gt;Yes, all active because iCAP (inactive) cores have stayed with iCAP system (system A)&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Thu, 19 May 2011 10:58:48 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789251#M607274</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-19T10:58:48Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789252#M607275</link>
      <description>&amp;gt; Below is a new situation. &lt;BR /&gt;&amp;gt; System A - a rx8640 with 3 cell boards/24 cores and iCAP reported 18 Active/6 iCAP cores &lt;BR /&gt;&amp;gt; &lt;BR /&gt;&amp;gt; System B - a rx8640 with 2 cell boards/12 cores (six on each cell) and no iCAP &lt;BR /&gt;&amp;gt; &lt;BR /&gt;&amp;gt; and I took a cell board out of system A and put it in system B, It would end up with: &lt;BR /&gt;&amp;gt; &lt;BR /&gt;&amp;gt; System A - a rx8640 with 2 cell boards/16 cores and iCAP reported 16 Active/ 0 iCAP cores &lt;BR /&gt;&amp;gt; &lt;BR /&gt;&amp;gt; System B - a rx8640 with 3 cell boards/20 cores and no iCAP ( all active cores&lt;BR /&gt;&lt;BR /&gt;&lt;BR /&gt;No, we don't give away activations! In the above you went from a total across the 2 systems with a total of 30 active / 6 iCAP to 36 active / 0 iCAP! You certainly can't do that. The number of inactive cores on system A has to remain the same if you did this you would end up with:&lt;BR /&gt;&lt;BR /&gt;System A - a rx8640 with 2 cell boards/16 cores and iCAP reported 10 Active/6 iCAP cores &lt;BR /&gt;&lt;BR /&gt;System B - a rx8640 with 3 cell boards/20 cores (six on the 2 existing cells, 8 on the new one) and no iCAP &lt;BR /&gt;&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Thu, 19 May 2011 11:04:59 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789252#M607275</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-19T11:04:59Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789253#M607276</link>
      <description>But here the situation on system A is different.&lt;BR /&gt;Find the attached zip file containing&lt;BR /&gt;1. icapstatus_before_removing_cell.txt&lt;BR /&gt;"total cores 24, active 18 inactive 6"&lt;BR /&gt;2. icapstatus_after_removing_cell.txt&lt;BR /&gt;"total cores 16, active 16 inactive 0"&lt;BR /&gt;from system A.&lt;BR /&gt;&lt;BR /&gt;This means all the CPUs on system A are active.&lt;BR /&gt;And now if I add the removed cell to System B&lt;BR /&gt;the number of active cell would increase by 8 cores and no inactive core.&lt;BR /&gt;Need your expert opinion.&lt;BR /&gt;</description>
      <pubDate>Thu, 19 May 2011 11:29:48 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789253#M607276</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-19T11:29:48Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789254#M607277</link>
      <description>OK, notice this in the icapstatus_after_removing_cell:&lt;BR /&gt;&lt;BR /&gt;Number of cores without usage rights:                                        6&lt;BR /&gt;Number of inactive cores:                                                    0&lt;BR /&gt;Number of cores that must be deactivated (insufficient usage rights):        6&lt;BR /&gt;&lt;BR /&gt;So iCAP thinks you should deactivate 6 cores... I'm not sure why this hasn't been enforced, but I would expect it to be at least at the next occasion that you reboot the partition. When did you take out the cell board? What version of iCAP are you run nning? &lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Thu, 19 May 2011 11:56:12 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789254#M607277</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-19T11:56:12Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789255#M607278</link>
      <description>Hi,&lt;BR /&gt;You mean to say, at next reboot, the six cores will be marked iCAP(inactive)?&lt;BR /&gt;&lt;BR /&gt;I removed the cell a day before and did not reboot the partition yet.&lt;BR /&gt;&lt;BR /&gt;Below is the version&lt;BR /&gt;iCAP-Mgr A.04.00.07.33  iCAP Manager SMH &lt;BR /&gt;iCOD B.11.31.09.00.00.140 Instant Capacity</description>
      <pubDate>Fri, 20 May 2011 03:17:08 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789255#M607278</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-20T03:17:08Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789256#M607279</link>
      <description>Abid,&lt;BR /&gt;&lt;BR /&gt;Yes, I mean precisely that - this complex is currently "out of compliance", and the iCAP software will correct that at next reboot (or at least it should).&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Fri, 20 May 2011 04:54:23 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789256#M607279</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-20T04:54:23Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789257#M607280</link>
      <description>Duncan,&lt;BR /&gt;It is confusing to me.&lt;BR /&gt;I have usage rights for 18 active core and no usage rights for 6 iCAP cores.(total 24 cores)&lt;BR /&gt;Now I removed 8 cores and left with 16 cores, as per my usage rights, I can use these 16 cores as active, which is 2 cores less than 18 cores. &lt;BR /&gt;If I add more cores to this partition and this becomes greater than 18, then the cores next to 18 would become inactive.&lt;BR /&gt;Please correct if I am wrong.&lt;BR /&gt;Regards,&lt;BR /&gt;Abid Iqbal</description>
      <pubDate>Fri, 20 May 2011 05:15:05 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789257#M607280</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-20T05:15:05Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789258#M607281</link>
      <description>Abid,&lt;BR /&gt;&lt;BR /&gt;No - You had Usage rights for 18 cores and iCAP for 6 cores.&lt;BR /&gt;&lt;BR /&gt;You took out a cell board which had 8 active cores. You can't "move" inactive cores out of a complex, iCAP stays with the complex.&lt;BR /&gt;&lt;BR /&gt;So you now have usage rights for 10 active cores and 6 iCAP cores. &lt;BR /&gt;&lt;BR /&gt;You can keep asking the same question, but the answer will remain the same - you can't get more active cores in total across your environments by just moving cell boards to other systems!&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Fri, 20 May 2011 06:41:54 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789258#M607281</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-20T06:41:54Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789259#M607282</link>
      <description>Hi,&lt;BR /&gt;I don't agree with this statement.&lt;BR /&gt;Coz this contradicts the document provided by you.&lt;BR /&gt;Page no. 67-68&lt;BR /&gt;Figure: 4-8, 4-9, 4-10&lt;BR /&gt;</description>
      <pubDate>Mon, 23 May 2011 03:39:46 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789259#M607282</guid>
      <dc:creator>Abid Iqbal</dc:creator>
      <dc:date>2011-05-23T03:39:46Z</dc:date>
    </item>
    <item>
      <title>Re: Procedure to locate/idnetify an iCAP processor</title>
      <link>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789260#M607283</link>
      <description>really?&lt;BR /&gt;&lt;BR /&gt;I think you need to read that section again more carefully... that section gives details of what happens when you take a cell out of a partition, but *it remains in the iCAP system*.&lt;BR /&gt;&lt;BR /&gt;I think I also pointed you at section A.5 on p189 which describes the situation when you actually take the cell board out of the system, specifically:&lt;BR /&gt;&lt;BR /&gt;"However, a complex can also get out of compliance if a cell is removed from the complex. For&lt;BR /&gt;example, if a cell contains inactive cores that are contributing to compliance, and the cell is&lt;BR /&gt;removed, there will be fewer inactive cores in the complex. This can result in the complex being&lt;BR /&gt;out of compliance, and temporary capacity might begin to be debited."&lt;BR /&gt;&lt;BR /&gt;Keep trying to change the rules if you like!&lt;BR /&gt;&lt;BR /&gt;HTH&lt;BR /&gt;&lt;BR /&gt;Duncan</description>
      <pubDate>Mon, 23 May 2011 15:15:52 GMT</pubDate>
      <guid>https://community.hpe.com/t5/operating-system-hp-ux/procedure-to-locate-idnetify-an-icap-processor/m-p/4789260#M607283</guid>
      <dc:creator>Duncan Edmonstone</dc:creator>
      <dc:date>2011-05-23T15:15:52Z</dc:date>
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