HW7
released: 11/23, 21:00

> Answer the following questions.
> Submit your answers to Canvas assignments. There is an entry for this homework.
> 
> 
> 1. Disk performance
> 
> Here is some numbers of a disk, CS5600-Disk:
> 
>   --Spindle Speed: 6000 RPM
>     (note: this number is per minute.
>      So, on average, how long does CS5600-Disk
>      rotate to a certain sector?)
> 
>   --Avg Seek Time, read/write: 10ms / 12 ms
>     (note: this is the time moving heads to the right track.)
> 
>   --Transfer rate: 64 MB/s
>     (note: this is the rate of reading/writing sequential data.
>      Assume 1MB=10^6B here.)
> 
> When CS5600-Disk reads/writes a sector (512 Bytes), it needs to
> (1) move the head to the right track,
> (2) wait rotating to the right sector,
> and (3) read/write the data.
> 
> Questions:
> 
> (a) How long would it take to do 500 sector reads, spread out
> randomly over the disk (and serviced in FIFO order)?
> [Write down your calculation and
> write the final result in seconds with one decimal place accuracy.]

Answer:

  total time = seek + rotation + transfer

  for one sector:

    seek = 10ms      [seek time is given]

    rotation = (60 s/1min *  1min/6000 rotations * 1000 ms/s) / 2 = 5ms

                     [*on average*, it takes the disk to rotate half a circle
                      to read a sector]

    transfer = 512 Bytes * 1s/64MB * 1MB/10^6 bytes = 8 * 10^-6 s = 0.008 ms

                     [the time to read 512B from the disk]

  total time = (10+5+0.008) * 500 = 7.5s


> 
> 
> (b) How long would it take to do 500 sector writes, spread out
> randomly over the disk (and serviced in FIFO order)?
> [Write down your calculation and
> write the final result in seconds with one decimal place accuracy.]
> 

Answer:

  for one sector:
    seek = 12ms
    rotation = (60 s/1min *  1min/6000 rotations * 1000 ms/s) / 2 = 5ms
    transfer = 512 Bytes * 1s/64MB * 1MB/10^6 bytes = 8 * 10^-6 s = 0.008 ms

  total time = (12+5+0.008) * 500 = 8.5s

> 
> (c) How long would it take to do 500 sector reads, SEQUENTIALLY on the
> CS5600-Disk? (FIFO order once more)
> [hint: notice that some actions only appear once in this case.]
> [Write down your calculation and
> write the final result in milliseconds with one decimal place accuracy.]
> 

Answer:

  seek = 10ms           [this happens once]
  rotation = 5ms        [this happens once]
  transfer = 512 Bytes * 500 / 64 M/s = 4ms
      [note: this transfer time implicitly includes the rotating time]

  total time = 10ms + 5ms + 4ms = 19ms

> 
> 
> 2. SSD
> 
> CS5600-SSD is an SSD that has 1 flash bank, 4 blocks (in the same bank),
> and 8 pages (2 in each block).
> CS5600-SSD uses the log-structured FTL we learn in class.
> 
> The current state of the CS5600-SSD is as follows:
> 
>          +---------------------------------------+
>  blocks  | block 0 | block 1 | block 2 | block 3 |  [update 11/22: was "block 2"; a typo]
>          +---------+---------+---------+---------+
>  pages   | P1 | P2 | P3 | P4 | P5 | P6 | P7 | P8 |  [update 11/23: was "P5 P6"; typos]
>          +----+----+----+----+----+----+----+----+
>  data    | A  | B  | C  | A' |    |    |    |    |
>          +----+----+----+----+----+----+----+----+
> 
>  mapping: A' => P4, B => P2, C => P3
> 
> A, B, C are three logical pages.
> Whenever they get updated, we add an apostrophe ("'") to their names,
> for example, after an update to page A, A becomes A'.
> In other words, (A, A', A'', A''',...) is a series of snapshots to page A.
> 
> Questions:
> 
> (a) draw the CS5600-SSD status (including the mapping) after running
> 
>   write(B')
>   write(C')
>   write(A'')
>   write(B'')
> 

Answer:
         +---------------------------------------+
 blocks  | block 0 | block 1 | block 2 | block 3 |
         +---------+---------+---------+---------+
 pages   | P1 | P2 | P3 | P4 | P5 | P6 | P7 | P8 |
         +----+----+----+----+----+----+----+----+
 data    | A  | B  | C  | A' | B' | C' | A''| B''|
         +----+----+----+----+----+----+----+----+

 mapping: A'' => P7, B'' => P8, C'=> P6

[note:
   "mapping: A'' => P5, B'' => P6, C'=> P6"
   is a correct answer as well because "P7"/"P8" was "P5"/"P6", which are typos.
]

> 
> 
> (b) (following the last question) now CS5600-SSD runs a round of garbage
> collection which recycles all blocks that do not contain valid pages.
> Draw the status after the garbage collection.
> 

Answer:
         +---------------------------------------+
 blocks  | block 0 | block 1 | block 2 | block 3 |
         +---------+---------+---------+---------+
 pages   | P1 | P2 | P3 | P4 | P5 | P6 | P7 | P8 |
         +----+----+----+----+----+----+----+----+
 data    |    |    |    |    | B' | C' | A''| B''|
         +----+----+----+----+----+----+----+----+

 mapping: A'' => P7, B'' => P8, C'=> P6