## Monday, October 21, 2019

### How many storage devices does a workload require?

I read an interesting paper that was motivated by the inability of existing storage engines to utilize all of the read IO provided by new and fast storage devices. That is excellent motivation but the story has more nuance.

A storage device provides capacity, endurance and reads. For now read means read operations and I ignore read throughput and latency. For a given device and workload none of the dimensions (capacity, endurance, reads) might be saturated. When saturation occurs it is usually limited to one of the dimensions. In that case if you want to reduce the cost of storage then change something to be more efficient in the saturated dimension. When capacity (endurance, read IO) is the bottleneck then reducing space (write, read) amplification is an answer.

It is hard to saturate a storage device in all dimensions so I am cautious when insufficient utilization in any dimension is cited as a problem. Too much saturation leads to lousy quality of service. Besides, workloads rarely have constant workloads -- web-scale varies by time of day.

When endurance or capacity are the bottleneck then it can be OK to not use all of the read IO provided by a new storage device. A simple model for this is:
• Storage device supplies r units of read IO, w units of write endurance and s units of capacity.
• The workload demands R units of read IO, W units of write endurance and S units of capacity.
• max(R/r, W/w, S/s) storage devices are required when amplification is ignored
• max(R*ar/r, W*aw/w, S*as/s) storage devices are required when amplification is not ignored where ar, aw and as are read, write and space amplification
• To reduce the number of storage devices focus on the saturated dimension and tune or change the index structure
An example

For (R=10000, W=1000, S=100, r=10, w=10, s=10, ar=4, aw=10, as=2) then max(R*ar/r, W*aw/w, S*as/s) = max(10000*4/10, 1000*10/10, 100*2/10) = max(4000, 1000, 20) = 4000 and read IO is the bottleneck. But change S from 100 to 100000 and this becomes max(4000, 1000, 20000) = 20000 and capacity is the bottleneck.