Greetings
I need to make a CPU ugrade on ESXI 7.0.2 server, in order to switch from a cpu without Hypertrading 4 cores (E5-2609) to one with Hypertrading and 8 cores (E5-2670).
The problem is that I cannot migrate the VMs, as I have a Starter Kit license, if I turn off the server and replace the CPU, leaving the VMs in the server, can I run into any problems?
thank you
Assuming that you made sure that the new CPU is supported in the host without further modification, you should have no issues with powering the VM back on after the replacement.
Please ensure that none of the VMs is in suspended state!
Regardless of the hardware modification, you should always ensure that you have up-to-date backups of your VMs.
André
Not exactly an upgrade as it appears that you are downgrading from Ivy Bridge to Sandy Bridge. You lose certain features (example: RDRAND instruction is one you lose, that is the Secure Key in the comparison). E5-2670 v2 would have 10c/20t.
https://ark.intel.com/content/www/us/en/ark/compare.html?productIds=75275,64595
The effect isn't so much with vSphere but what the VMs and the guest OS and applications running inside the VMs would no longer see to be available.
Just an example, I am not saying that it will happen but quite possible:
Secure Key means the RDRAND instruction. That is used for generating a random number. Random numbers are important and widely used in cryptography. So a guest OS and/or application(s) running inside the VM might previously using RDRAND (because it was present) now have to fall back to use some other algorithm to generate random number. As it is no longer a single instruction (RDRAND) but a series of instructions, it could take longer and/or consume more CPU cycles whenever random number needs to be generated. Worst case, the quality of the randomness of the number might be affected. For example, if there is an application in a VM that generates one time passwords (OTPs), the quality/strength of the OTP might suffer without RDRAND (depending on the quality of the fallback algorithm).
ok thank you for your answer.