Exam 70-740 topic 1 question 273 discussion

Summary: Deployment Cache drives Capacity drives Cache behavior (default) NVMe + SSD + HDD NVMe SSD + HDD Read + Write for HDD, Write-only for SSD https://docs.microsoft.com/en-us/windows-server/storage/storage-spaces/understand-the-cache

I had this on exam,had 100% on storage part. I`ve selected 18tb + write

Cache drives do not contribute usable storage capacity. All data stored in the cache is also stored elsewhere, or will be once it de-stages. This means the total raw storage capacity of your deployment is the sum of your capacity drives only. https://docs.microsoft.com/en-us/windows-server/storage/storage-spaces/understand-the-cache

S2D will create two way mirror so the available disk space should be 18/2 = 9

Got this q on exam (11AUG2020) The correct answer is 18 write

Yes, 18TB and Write only cache for the SSDs

I think the 1st answer is 18TB. The link Sabarazzzdev proposed says as follow. "Cache drives do not contribute usable storage capacity. All data stored in the cache is also stored elsewhere, or will be once it de-stages." In this paragraph "usable storage capacity" means S2D virtual disks. "elsewhere" isn't included in S2D virtual disks. Because of these the link concluded as follow. "This means the total raw storage capacity of your deployment is the sum of your capacity drives only."

Check the summary paragraph in this page https://docs.microsoft.com/en-us/windows-server/storage/storage-spaces/understand-the-cache

The answer is 18TB and Write for SSD

The first half of the question wants to know the total amount of disk space available for S2D virtual disks and 22.5 TB is the total amount available (xziomal9's post has the formula). The question isn't asking how S2D carves up the space and what space is available for long-term storage.

https://s2dcalc.blob.core.windows.net/www/index.html This results in 6.0 TB of physical capacity and 1.5 TB of cache per node So 18 TB is mostly correct "If devices of multiple types (NVMe, SSD, or HDD) are present, the fastest are automatically used as cache. Each faster device binds dynamically to several slower devices, to accelerate I/O for consistent low latency. These devices do not contribute to the overall capacity, since everything stored in cache is also stored elsewhere." "This deployment has 18.0 TB total physical capacity, excluding cache. We recommend setting aside the equivalent of one capacity drive per server, up to 4 drives, for in-place recovery. This guarantees an immediate, in-place, parallel repair can succeed after the failure of any drive, even before it is replaced. In this case, that's 3 drives x 1.0 TB each." .... This is only recommendation "Total 18 TB : (5 TB Usable , 10 TB Resiliency , 3 TB Reserve )"

Capacity >> SSD + HDD = 18Tb Cache >> Read + Write for HDD, Write-only for SSD

Just to set anyone mind at ease who still isn't convinced it's 18tb... "In deployments with multiple types of drives, Storage Spaces Direct automatically uses all drives of the "fastest" type for caching. The remaining drives are used for capacity. Which type is "fastest" is determined according to the following hierarchy, NVME, SSD and then HDD." https://docs.microsoft.com/en-us/windows-server/storage/storage-spaces/understand-the-cache#cache-drives-are-selected-automatically

I've just used an s2d calc based on this and the answer was 18tb of physical capacity and 4.5tb of cache.

18 ot 22.5?

Got this question in exam

The answer should be 18T ,Read + Write