An administrator has a virtual machine configured with the following settings:
ESXi version: 5.1
CPU: vCPUs 6
Memory: 48GB
Hardware version: 7
VMware Tools: Installed
Which two actions must the administrator take in order to utilize vNUMA? (Choose two.)
A.
Upgrade the ESXi host to vSphere 5.5 or later.
B.
Upgrade to Virtual Hardware version 8.
C.
Configure numa.vcpu.min to 5
D.
Configure numa.vcpu.min to 6
Explanation:
Explanation/Reference:
B. is right
Virtual NUMA, which requires virtual hardware version 8 or later, can in some cases provide significant performance benefits for wide virtual machines
http://www.vmware.com/content/dam/digitalmarketing/vmware/en/pdf/techpaper/vmware-perfbest-practices-vsphere6-0-white-paper.pdf
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I think D is correct.
6 vcpu’s, and not 5.
uma.vcpu.min
Minimum number of “virtual CPUs” in a virtual machine that are required in order to generate a virtual NUMA topology.
http://pubs.vmware.com/vsphere-60/index.jsp#com.vmware.vsphere.resmgmt.doc/GUID-3E956FB5-8ACB-42C3-B068-664989C3FF44.html?resultof=%2522%256e%2575%256d%2561%252e%2576%2563%2570%2575%252e%256d%2569%256e%2522%2520
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B & C
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B/D
You can enable vNUMA for virtual machines with eight or fewer vCPUs by adding to the .vmx file the line:
numa.vcpu.min = X
(where X is the number of vCPUs in the virtual machine).
http://www.vmware.com/content/dam/digitalmarketing/vmware/en/pdf/techpaper/vmware-perfbest-practices-vsphere6-0-white-paper.pdf
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I’m onboard with B & D. I’m not sure that it would ever make sense to make the minimum an odd number given that NUMA is meant for balancing which can only be done with an even number of objects. Plus the same VM is already 6 CPUs so why set the minimum to anything less than that?
https://kb.vmware.com/selfservice/microsites/search.do?language=en_US&cmd=displayKC&externalId=1026063
https://support.rackspace.com/how-to/numa-vnuma-and-cpu-scheduling/
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As per the link to the VMware-perfbest-practices-vsphere6-0-white-paper.pdf provide by our friend above going by the name of “google”:
You can enable vNUMA for virtual machines with eight or fewer vCPUs by adding to the .vmx file the
line:
numa.vcpu.min = X
(where X is the number of vCPUs in the virtual machine).
I would therefor go for b and d and not C!
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Technically it will work on 5. It’s one of those questions where both answers are correct. One answer is the logical option. The other option is the minimum required.
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I disagree on the logical and minimum required.
Minimum required is 6 (because VM has 6 vCPU’s) 5 is less then minimum required.
same for logical answer, makes the most sense to choose 6 over 5 in the numa.vcpu.min as 6 is required.
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As per the URL above by Bee197 from the vSphere 6 Doc Center “Virtual NUMA Controls” article, the default value for numa.vcpu.min is 9 (nine), which seems to suggest it’s ok to have odd number of vCPUs in general, but again – it makes little sense…
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I tend to agree with what “None” said above: both C and D are correct. The default value for this parameter (nine) means that a virtual NUMA topology will be generated for any VM that has nine and more vcpus. By setting it to 5 or 6 (or even 3 and 4) would allow for the same affect as far as this 6 vcpu VM is concerned.
So, if the choice was between 5 and everything more than 6, it would definitely be ‘C’ (5).
This leads me to conclude it’s yet another “oops” by VMwareEdu folks…
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It’s B & C on account of the Virtual NUMA control settings doc provided by Bee above. The setting is a Base-0 counter i.e. 9 being the default is actually 10 vCPU’s in the VM.
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@Clamps: Not correct I would say.
p.48/49: http://www.vmware.com/content/dam/digitalmarketing/vmware/en/pdf/techpaper/vmware-perfbest-practices-vsphere6-0-white-paper.pdf
By default, vNUMA is enabled only for virtual machines with more than eight vCPUs
This means for 9 or more vCPUs (numa.vcpu.min is set to 9 by default)
You can enable vNUMA for virtual machines with eight or fewer vCPUs by adding to the .vmx file the line: numa.vcpu.min = X (where X is the number of vCPUs in the virtual machine).
So answer definitely should be D and B.
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