Persistent Storage

Storage Devices

• persistent/nonvolatile, retains data after power down
  • Hard Drive (HDD) / Spinning Disk
    • large capacity at low cost, block level access, not byte addressable
    • physical motion needed to read and write, milliseconds access latency
  • Solid State Drive (SSD)
  • large capacity at intermediate cost (~3x HDD), block level access, not byte addressable
  • no physical moving parts, microsecond access latency

Hard Drive

• what does a spinning disk look like?


• head: moves above the platter (3nm), reads data from and writes data to disk sectors
• sector: unit of reads and writes on disk, 512 bytes, contains error correcting code
• track: length varies across disk, outer tracks have more sectors
• separated by unused guard regions to reduce likelihood of neighboring corruption
• only outer half of radius is used, most sectors are in the outer half
• what happens on a disk read?
1. a read request is sent to disk controller
2. find the right platter and surface, arm moves to the right track
3. head reads while the disk spins (desired sector will spin under the head)
4. transfer data read back to the host
• access latency
• total time = seek time + rotation time + transfer time
• seek time: time to move disk arm over the desired track (1-20ms)
• rotation time: time for the desired sector to rotate under the disk head (based on RPM, 4-15ms)
  • e.g. 7200 RPM = 120 RPS = 0.12 rotation per ms = 8.3 ms per rotation
  • reasonable to assume it takes half a rotation to get to the desired sector, so 8.3/2 = 4ms
• transfer time: time to transfer data onto/off the disk (based on disk bandwidth, often < 4us per sector)
• e.g. 120 MiB/s bandwidth = 125829120 B/s
• sector = 512 bytes, 512 / 125829120 * 1000 = 0.004 ms per sector
• sequential vs random access
• how many seeks would we need to do if we read 10 consecutive sectors on the same track?
• how many seeks would we need to do if we read 10 random sectors on different tracks?
• IOPS (I/O operation per second)
• # of I/O requests / total latency
• access patterns of the I/O request matters
• IOPS of 10 random reads (one sector per read)?
• IOPS of 10 sequential reads (one sector per read)?

Solid State Drive

• no moving parts, NAND-based flash

• components
  • page: unit of read and write, 2-4 KB, not the same as VM pages!
  • block: unit of erasure, 1-8 MB, span hundreds of pages
  • can only write to an empty page, if no empty page, must erase an entire block to create empty pages first
• operations
1. read (a page): can read any page, fast sequential and random access, tens of microseconds
2. program (a page): program a page in an erased block by setting certain bits to 0 to write data, tens of microseconds
3. erase (a block): erase a block by setting all bits in the block to 1, slow (a few ms)
• what happens to existing data in the block?
• once a block is erased, it's ready to be programmed
• reliability
• a block becomes unusable after a certain number(10-100K) of program/erase operations
• repeated writes to the same page is bad for endurance
• wear leveling: try to spread writes across the blocks as evenly as possible
• SSD needs a way to flexibly remap logical blocks to actual block & page
• Flash Translation Layer (FTL)
  • allows client (OS) to perform read & write operation on logical block/page
  • translates read/write requests for logical block/page to read, erase, program operations on actual block/page
  • allows SSD to do internal management (wear leveling, garbage collection, see OSTEP: SSD)
• latency
• sequential access still faster than random access, but much closer than hard drive
• total time = access latency + transfer time + (erasure time if applicable)
  • e.g. transfer rate of 500MiB/s = 524288000 B/s
  • transfer 4KiB of data = 4096 bytes, 4096 / 524288000 * 1000000 = 7.8us
  • total time for reading 4KiB of data = 10us (access latency) + 7.8 (transfer time)
• IOPS
• # of I/O operations / total time