> released: 03/30, 12:00
> due: 04/05, 23:59
> 
> Answer the following questions.
> Submit your answers to Canvas assignments.
> There is an entry for this homework.
> 
> 
> 1. MMIO and WeensyOS (4 points)
> 
> Below is a snippet of code from "lib.cc".
> This is the new version piece of code in the final panel of handout week11b
> (https://naizhengtan.github.io/23spring/notes/handout_w11b.pdf).
> 
> Read it and answer the following questions.
> 
>    1  inline void console_printer::putc(unsigned char c, int color) {
>    2      while (cell_ >= console + CONSOLE_ROWS * CONSOLE_COLUMNS) {
>    3          scroll();
>    4      }
>    5      if (c == '\n') {
>    6          int pos = (cell_ - console) % 80;
>    7          while (pos != 80) {
>    8              *cell_++ = ' ' | color;
>    9              ++pos;
>   10          }
>   11      } else {
>   12          *cell_++ = c | color;
>   13      }
>   14  }
> 
> 
> 1.a If we want to replace all empty spaces after a new line with '_', how
>     should we modify the code?
>     That is, if printing "abc\n123" which is used to look like:
>         """
>         abc
>         123
>         """
>     now should be:
>         """
>         abc_______________
>         123
>         """
> 
>     (2 points) Write down the line numbers and your modifications to
>     make this happen.
> 

   [answer:
      line 8:  *cell_++ = '_' | color;
   ]

> 
> 
> 
> 1.b Then, we want to fix the color of the characters to be yellow and we will
>     ignore the input argument "color" (line 1), meaning whatever input "color" is,
>     the color of the characters will be yellow.
> 
>   [ A crash course of 16-color coding
>     ----
> 
>     In WeensyOS, the input "color" will look like this:
> 
>       0x0F00
>         ^^^^
>         ||++-> these two should be zeros for color.
>         |+-> this hex digit means the color of the character.
>         +-> this hex digit means the color of the background.
> 
>     WeensyOS uses a 16-color system.
>     (why 16? a digit of hex number can only represent 16 numbers, 0x0 to 0xf.)
>     And here are the colors and their decimal code:
> 
>              0    black
>              1    blue
>              2    green
>              3    cyan
>              4    red
>              5    magenta
>              6    brown
>              7    light gray
>              8    dark gray
>              9    bright blue
>             10    bright green
>             11    bright cyan
>             12    bright red
>             13    bright magenta
>             14    bright yellow
>             15    bright white
>     ]
> 
>     (2 points) Write down the line numbers and your modifications to always
>     print yellow characters.
> 
>     [Aside, why the main body of the console doesn't change to yellow? because
>     main body's print does not use the above code. See k-memviewer.cc for how
>     WeensyOS prints the main body.]
> 

     [answer:
          line 12:    *cell_++ = c | 0x0E00;
     ]

> 
> 
> 
> 2. Disk performance (6 points)
> 
> CS5600 staff designs a disk, CS5600-Disk:
> 
>   --Spindle Speed: 6000 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 is not a question you need to answer but one you should think of.]
> 
>   --Avg Seek Time, read/write: 5ms / 10 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. Again, this is 10^6B, not 2^20B.)
> 
> When CS5600-Disk reads/writes a sector (512 Bytes), it needs to
> (1) move the head to the right track (seeking),
> (2) wait rotating to the right sector (rotation),
> and (3) read/write the data (data transfer).
> This means, the time finishing a job (like reading one sector)
> equals to the sum of time spending on the above three phases.
> 
> Questions:
> 
> 2.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 milliseconds.
> 

[answer:

  For reading one sector:

    (1) seeking time = 5ms [given]

    (2) avg rotation time = 60s / 6000 / 2 = 5 ms

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

    (3) transfer time = 512B / (64MB/s) = 0.008 ms

   total time = #sectors * (seeking time + rotation time + data transfer time)
              = 500 * (5+5+0.008) ms
              = 5004 ms
]

> 
> (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 milliseconds.
> 

[answer:

  For writing one sector:

    (1) seeking time = 10ms [given]
    (2) avg rotation time = 60s / 6000 / 2 = 5 ms
    (3) transfer time = 512B / (64MB/s) = 0.008 ms

   total time = #sectors * (seeking time + rotation time + data transfer time)
              = 500 * (10+5+0.008) ms
              = 7504 ms
]

> 
> 
> (c) How long would it take to do 500 sector reads, SEQUENTIALLY on the
> CS5600-Disk? (FIFO order once more) (2 points)
>   [notice that seek and rotation will appear only once in sequential reads.]
> Write down your calculation and write the final result in milliseconds.


[answer:

  (1) seeking time = 5ms           [this happens once]
  (2) avg rotation time = 5ms      [this happens once]
  (3) data transfer time = 512 Bytes * 500 / 64 MB/s = 4ms
         [note: this transfer time implicitly includes the rotating time]

  total time = 5ms + 5ms + 4ms = 14ms
]