Over the last couple of weeks I have been talking to customers a lot about VSAN 6.2 and how to design / size their environment correctly. Since the birth of VSAN we have always spoken about 10% cache to capacity ratio to ensure performance is where it needs to be. When I say 10% cache to capacity ratio, I should actually say the following:
The general recommendation for sizing flash capacity for Virtual SAN is to use 10% of the anticipated consumed storage capacity before the NumberOfFailuresToTolerate is considered.
Reality is though that what most customers did was they looked at their total capacity, cut it in half (FTT=1) and then said “we will take 10%” of that. So a 10TB VSAN Datastore would require “10% of 5TB” in terms of cache. This is fast way of indeed calculating what your caching requirements are… That is, if ALL of your virtual machines have the same “availability” requirements. Because even in 6.1 and prior the outcome would change if you had VMs which required FTT=2 or FTT=3 or even FTT=0. (Although I would not recommend FTT=0.)
With VSAN 6.2 this is amplified even more. Why? Well as you hopefully read, VSAN 6.2 introduces space efficiency functionality (for all-flash) like deduplication, compression, RAID-5 or RAID-6 over the network. The following diagram depicts what that looks like. In this case we show RAID-6 with 4 data blocks and 2 parity blocks, which is capable of tolerating 2 failures anywhere in the cluster of 6 hosts.
If you look at the above, and take that old “FTT=1” or “FTT=2” diagram in mind you quickly realize that the effective capacity per datastore is not as easy to calculate as it was in the past. Lets take a look at a simple example to show the impact which using certain data services can have on your design / sizing.
- 1000 VMs with on average 50GB disk space required
- 1000 * 50GB = 50TB
Lets take a look at both FTT=1 and FTT=2 with and without Raid-5/6 enabled. The calculations are pretty simple. Note that “FTT” stands for “Failures to Tolerate” and FTM stands for “Failure Tolerance Method”.
|1||Raid-1||1000 VMs * 50GB * 2 (overhead)||100TB|
|1||Raid-5/6||1000 VMs * 50GB * 1.33 (overhead)||66.5TB|
|2||Raid-1||1000 VMs * 50GB * 3 (overhead)||150TB|
|2||Raid-5/6||1000 VMs * 50GB * 1.5 (overhead)||75TB|
Now if you look at the table, you will see there is a big difference between the capacity requirements for FTT=2 using RAID-1 and FTT=2 using RAID-5/6 even between the FTT=1 variations the difference is significant. You can imagine that when you base your required cache capacity simply on the required disk capacity that the math will be off. Assuming that the amount of hot data in all cases is 10% the difference could be substantial. However, when you base your cache requirements on the initial 10% of “1000* 50GB” the result never changes!
And in this case I haven’t even taken deduplication and compression in to account, you can imagine that with a data reduction of 2x using VSAN compression and deduplication that the math will change again for the caching tier, well that is if you do it wrong and calculate it based on the actual capacity layer… To summarize, when you do your VSAN design and sizing, make sure to always base it on the Virtual Machine size, it is the safest and definitely the easiest way to do the math!