The perfect setup

[Udbytossen]

Hi Forum. 
After beeing running with VMware for a long time - I'd a breakdown this morning - that makes me wonder on how the perfect setup is 

My Issue is regarding the disk that VMware ESXi is installed on. 
My Setup was ( my thoughts ) pretty secure since - my OS was installed on 2x32GB Sandisk USB setup in RAID 1:
But today the whole hyper freezes - and only option was to reboot - after rebooting -error description was No Image information on both drives. 

So I'm wondering if there is anyway to a more secure setup ( This is the secod time - i've tried that VMware ESXi is crashed) 
So how do other people do this - to secure the server the best way 

Is this instead of running RAID1 - use them as standalone disk where the disk are cloned - So if disk 1 fails - i'll try boot on disk 2 
Am I just unlucky that this is happening? My thoughts are if something goes wrong on the 1 of the USB - then it'll replicate the error  from disk1 to disk2 - so offcorse it'll fail 

But then if I use the Big Storage desvice ( RAID10 ) and it'll crash - I'm would allso loose all VM's which I didn't do this time 

So what would be the right and perfect way to run VMware OS on - to secure it from crashing again 

[e_espinel]

Hello.
RAID 1 is the safest and should not happen what happened to you. The first thing would be to check the basics:
Firmware levels in all hardware, if they are outdated you should upgrade (this is recommended)
Have the latest patches for the version of VMware vSphere that is installed on the host.
Verify that all host hardware, including USB Sandisk, is supported by VMware (compatibility matrix)

 VMware Compatibility Guide

https://www.vmware.com/resources/compatibility/search.php

Best of luck.

[bryanvaneeden]

Hi @Udbytossen,

I agree with @Udbytossen. The RAID 1 configuration for this situation is perfect, one disk/usb/sd card can fail this way and the hypervisor will continue running. I've been doing this for years and it works fine. What happend to you was probably a glitch because that is not expected behaviour.