I have had this question a couple of times already, how many hosts do I need per site when the Primary FTT is set to 1 and the Secondary FTT is set to 1 and RAID-5 is used as the Failure Tolerance Method? The answer is straight forward, you have a local RAID-5 set locally in each site. RAID-5 is a 3+1 configuration, meaning 3 data blocks and 1 parity block. As such each site will need 4 hosts at a minimum. So if the requirement is PFTT=1 and SFTT=1 with the Failure Tolerance Method (FTM) set to RAID-5 then the vSAN Stretched Clustering configuration will be: 4+4+1. Note, that also when you use RAID-1 you will need at minimum 3 hosts per site. This because locally you will have 2 “data” components and 1 witness component.
From a capacity stance, if you have a 100GB VM and do PFTT=1, SFTT=1 and FTM set to RAID-1 then you have a local RAID-1 set in each site. Which means 100GB requires 200GB in each location. So 200% required local capacity, 400% for the total cluster. Using the below table you can easily see the overhead. Note that RAID-5 and RAID-6 are only available when using all-flash.
I created a quick table to help those going through this exercise. I did not include “FTT=3” as this in practice is not used too often in stretched configurations.
|Description||PFTT||SFTT||FTM||Hosts per site||Stretched Config||Single site capacity||Total cluster capacity|
|Standard Stretched across locations with local protection||1||1||RAID-1||3||3+3+1||200% of VM||400% of VM|
|Standard Stretched across locations with local RAID-5||1||1||RAID-5||4||4+4+1||133% of VM||266% of VM|
|Standard Stretched across locations with local RAID-6||1||2||RAID-6||6||6+6+1||150% of VM||300% of VM|
|Standard Stretched across locations no local protection||1||0||RAID-1||1||1+1+1||100% of VM||200% of VM|
|Not stretched, only local RAID-1||0||1||RAID-1||3||n/a||200% of VM||n/a|
|Not stretched, only local RAID-5||0||1||RAID-5||4||n/a||133% of VM||n/a|
|Not stretched, only local RAID-6||0||2||RAID-6||6||n/a||150% of VM||n/a|
Hope this helps!
Wojciech Marusiak says
Phil H says
Do you have any feedback on how a 2+2+1 stretched cluster would fit into this? You have a 1+1+1 but not 2+2+1.
Duncan Epping says
The article provides the minimum node count for a certain policy. With 2+2+1 you can do:
PFTT = 1
SFTT = 0
As for SFTT=1 you need 3+3+1 as the table indicates