released: 04/07, 9:00
due: 04/13, 23:59

> 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: 12000 RPM
>     note: 
>      ** this speed number is per minute.
>      ** on average, it takes the disk to rotate half a circle to reach a sector.
>      ** So, on average, how long does CS5600-Disk rotate to a sector?
>         [this isn't a question you need to answer but one you should think of.]
> 
>   --Avg Seek Time, read/write: 5ms / 6 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)?  (2 points)
> [Write down your calculation and
> write the final result in seconds with two decimal place accuracy.]
> 

[
Answer:

  total time = seek + rotation + transfer

  for one sector:

    seek = 5ms      [seek time is given]

    rotation = (60 s/1min *  1min/12000 rotations * 1000 ms/s) / 2 = 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 = (5+2.5+0.008) * 500 = 3.75s
]


> 
> 
> (b) How long would it take to do 500 sector writes, spread out
> randomly over the disk (and serviced in FIFO order)?  (2 points)
> [Write down your calculation and
> write the final result in seconds with two decimal place accuracy.]
> 


[
Answer:

  for one sector:
    seek = 6ms
    rotation = (60 s/1min *  1min/12000 rotations * 1000 ms/s) / 2 = 2.5ms
    transfer = 512 Bytes * 1s/64MB * 1MB/10^6 bytes = 8 * 10^-6 s = 0.008 ms

  total time = (6+2.5+0.008) * 500 = 4.25s
]



> 
> 
> (c) How long would it take to do 500 sector reads, SEQUENTIALLY on the
> CS5600-Disk? (FIFO order once more) (2 points)
> [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 = 5ms           [this happens once]
  rotation = 2.5ms        [this happens once]
  transfer = 512 Bytes * 500 / 64 M/s = 4ms
      [note: this transfer time implicitly includes the rotating time]

  total time = 5ms + 2.5ms + 4ms = 11.5ms
]


> 
> 
> 
> 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 |
>          +---------+---------+---------+---------+
>  pages   | P1 | P2 | P3 | P4 | P5 | P6 | P7 | P8 |
>          +----+----+----+----+----+----+----+----+
>  data    | D  | B  | C  | A' |    |    |    |    |
>          +----+----+----+----+----+----+----+----+
> 
>  mapping: A' => P4, B => P2, C => P3, D => P1
> 
> -- A, B, C, and D are four logical pages.
> -- Whenever a page gets 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.
>    But they refer to the same logical page (from a program's point of view).
> 
> Questions:
> 
> (2 points)
> (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    | D  | B  | C  | A' | B' | C' | A''| B''|
           +----+----+----+----+----+----+----+----+

   mapping: A'' => P7, B'' => P8, C'=> P6, D => P1
]
> 
> 
> (2 points)
> (b) (following the last question) now CS5600-SSD runs a round of garbage
> collection which recycles all blocks that do not contain valid pages.
>   [updated 04/10: note that this is a naive GC; it **only** "recycles all
>    blocks that do not contain valid pages"]
> Draw the SSD 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    | D  | B  |    |    | B' | C' | A''| B''|
	         +----+----+----+----+----+----+----+----+
	
	 mapping: A'' => P7, B'' => P8, C'=> P6, D => P1
]