Creating a RAID-5 using Storage Spaces in Windows 11

Question

Hi, first time posting here. I tried to search but I couldn't find and I don't know if this is something very specific or not.

I want to try the new Storage Spaces in Windows 11. Some years ago I bought an external enclosure with 5-bay for hard drives and also provide RAID 5 functionality out-of-the-box. It works pretty good and no issue. This time I want to mount another RAID 5 but using the built-in Storage Spaces available in Windows 11.

My previous RAID5 was using 5 HDD with 8 TB each, so the total available was 32 TB (29.1 in reality).

My new RAID5 will be with 6 HDD with 8 TB each, that's suppose to be 40 TB from what I understand (or ~35 GB in reality).

I started creating an Storage Pool in Windows 11 and then associated all the 6 storage drives

Then my options from what I understand are

• Resiliency type: Parity
• Total pool capacity: 43.6 TB
• Available pool capacity: 43.6 TB
• Size (maximum): 29.0 TB
• Including resiliency: 43.5 TB

Supposedly the Size is auto-calculated depending of Total pool and Resiliency type, but my question is why is so low? The size will be less than my current RAID5 with 5 disks, it should be more with 6 disks.

If I change the size to the correct value I get

• Size (maximum): 36.3 TB
• Including resiliency: 54.4 TB

I don't know if I am doing something wrong or I don't fully understand this. If I create the Storage with the default options effectively it creates a new Disk but with less size than my current Storage. Any help will be appreciated. Thanks in advance.

Answer 1

RAID Definition - What is a RAID array? (techterms.com)

Answer 2

Hi Horace

I already know how RAID-5 works, in fact, that's the reason why I'm asking. Because like I mentioned in the beginning, with 6 disk with 8 TB each, RAID-5 should be 40 TB, and even considering that the disk aren't 8TB exactly, is not even near to 29.0 TB as showed in the screenshot. It should be near 33~35 in reality

Answer 3

the overhead associated with storing the parity information is the equivalent to one full disk.

Example, a RAID 5 array contains three 1 TB hard disks, then the array's usable capacity will be 2 TB, not 3 TB. Each drive loses some space to the indexing.

there's more and more overhead as you increase drives... the overhead increases exponentially, unless I misread your question...

Answer 4

Exactly... with 6 disk, the parity should only be 1 disk and the available capacity should be the other 5 disks, the math doesn't sum up correctly and that's why I'm confused...

6 Disks * 8 TB = 48 TB

RAID5 (6-1) * 8 TB = 40 TB

considering that the disks are not really 8 TB, using 7.3 TB as the real capacity the math still doesn't sum up

6 Disks * 7.3 TB = 43.8 TB (equivalent to the total pool capacity)

RAID5 (6-1) * 7.3 TB = 36.5 TB

but why according to the storage space is 29.0 !?? Is a lot less

Answer 5

ok... So, keep in mind...
All drives hold data and parity information for all the "other drives"

Raid 3 and 4 only hold info on a single drive, 5 is a tad different. I seem to remember an equation to explain this lost area..

Take your total capacity and times that by .92 (this is roughly the area that the OS will occupy, I may be a "slight" percentage off here)
so, 48TB x .92 = 44.16 This would be "roughly" the lost storage area due to the OS overhead

(pretty close to what you're looking for, no?