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• CD-R Read Speed Experiments

  21 mai 2011, par Multimedia Mike — Science Projects, Sega Dreamcast

  I want to know how fast I can really read data from a CD-R. Pursuant to my previous musings on this subject, I was informed that it is inadequate to profile reading just any file from a CD-R since data might be read faster or slower depending on whether the data is closer to the inside or the outside of the disc.

  Conclusion / Executive Summary
  It is 100% true that reading data from the outside of a CD-R is faster than reading data from the inside. Read on if you care to know the details of how I arrived at this conclusion, and to find out just how much speed advantage there is to reading from the outside rather than the inside.

  Science Project Outline

  • Create some sample CD-Rs with various properties
  • Get a variety of optical drives
  • Write a custom program that profiles the read speed

  Creating The Test Media
  It’s my understanding that not all CD-Rs are created equal. Fortunately, I have 3 spindles of media handy : Some plain-looking Memorex discs, some rather flamboyant Maxell discs, and those 80mm TDK discs :

  
  
  

  My approach for burning is to create a single file to be burned into a standard ISO-9660 filesystem. The size of the file will be the advertised length of the CD-R minus 1 megabyte for overhead— so, 699 MB for the 120mm discs, 209 MB for the 80mm disc. The file will contain a repeating sequence of 0..0xFF bytes.

  Profiling
  I don’t want to leave this to the vagaries of any filesystem handling layer so I will conduct this experiment at the sector level. Profiling program outline :

  • Read the CD-ROM TOC and get the number of sectors that comprise the data track
  • Profile reading the first 20 MB of sectors
  • Profile reading 20 MB of sectors in the middle of the track
  • Profile reading the last 20 MB of sectors

  Unfortunately, I couldn’t figure out the raw sector reading on modern Linux incarnations (which is annoying since I remember it being pretty straightforward years ago). So I left it to the filesystem after all. New algorithm :

  • Open the single, large file on the CD-R and query the file length
  • Profile reading the first 20 MB of data, 512 kbytes at a time
  • Profile reading 20 MB of sectors in the middle of the track (starting from filesize / 2 - 10 MB), 512 kbytes at a time
  • Profile reading the last 20 MB of sectors (starting from filesize - 20MB), 512 kbytes at a time

  Empirical Data
  I tested the program in Linux using an LG Slim external multi-drive (seen at the top of the pile in this post) and one of my Sega Dreamcast units. I gathered the median value of 3 runs for each area (inner, middle, and outer). I also conducted a buffer flush in between Linux runs (as root : 'sync; echo 3 > /proc/sys/vm/drop_caches').

  LG Slim external multi-drive (reading from inner, middle, and outer areas in kbytes/sec) :

  • TDK-80mm : 721, 897, 1048
  • Memorex-120mm : 1601, 2805, 3623
  • Maxell-120mm : 1660, 2806, 3624

  So the 120mm discs can range from about 10.5X all the way up to a full 24X on this drive. For whatever reason, the 80mm disc fares a bit worse — even at the inner track — with a range of 4.8X - 7X.

  Sega Dreamcast (reading from inner, middle, and outer areas in kbytes/sec) :

  • TDK-80mm : 502, 632, 749
  • Memorex-120mm : 499, 889, 1143
  • Maxell-120mm : 500, 890, 1156

  It’s interesting that the 80mm disc performed comparably to the 120mm discs in the Dreamcast, in contrast to the LG Slim drive. Also, the results are consistent with my previous profiling experiments, which largely only touched the inner area. The read speeds range from 3.3X - 7.7X. The middle of a 120mm disc reads at about 6X.

  Implications
  A few thoughts regarding these results :

  • Since the very definition of 1X is the minimum speed necessary to stream data from an audio CD, then presumably, original 1X CD-ROM drives would have needed to be capable of reading 1X from the inner area. I wonder what the max read speed at the outer edges was ? It’s unlikely I would be able to get a 1X drive working easily in this day and age since the earliest CD-ROM drives required custom controllers.
  • I think 24X is the max rated read speed for CD-Rs, at least for this drive. This implies that the marketing literature only cites the best possible numbers. I guess this is no surprise, similar to how monitors and TVs have always been measured by their diagonal dimension.
  • Given this data, how do you engineer an ISO-9660 filesystem image so that the timing-sensitive multimedia files live on the outermost track ? In the Dreamcast case, if you can guarantee your FMV files will live somewhere between the middle and the end of the disc, you should be able to count on a bitrate of at least 900 kbytes/sec.

  Source Code
  Here is the program I wrote for profiling. Note that the filename is hardcoded (#define FILENAME). Compiling for Linux is a simple 'gcc -Wall profile-cdr.c -o profile-cdr'. Compiling for Dreamcast is performed in the standard KallistiOS manner (people skilled in the art already know what they need to know) ; the only variation is to compile with the '-D_arch_dreamcast' flag, which the default KOS environment adds anyway.

  PLAIN TEXT

  C :

  1. #ifdef _arch_dreamcast
  2.   #include <kos .h>
  3.  
  4.   /* map I/O functions to their KOS equivalents */
  5.   #define open fs_open
  6.   #define lseek fs_seek
  7.   #define read fs_read
  8.   #define close fs_close
  9.  
  10.   #define FILENAME "/cd/bigfile"
  11. #else
  12.   #include <stdio .h>
  13.   #include <sys /types.h>
  14.   #include </sys><sys /stat.h>
  15.   #include </sys><sys /time.h>
  16.   #include <fcntl .h>
  17.   #include <unistd .h>
  18.  
  19.   #define FILENAME "/media/Full disc/bigfile"
  20. #endif
  21.  
  22. /* Get a current absolute millisecond count ; it doesn’t have to be in
  23. * reference to anything special. */
  24. unsigned int get_current_milliseconds()
  25. {
  26. #ifdef _arch_dreamcast
  27.   return timer_ms_gettime64() ;
  28. #else
  29.   struct timeval tv ;
  30.   gettimeofday(&tv, NULL) ;
  31.   return tv.tv_sec * 1000 + tv.tv_usec / 1000 ;
  32. #endif
  33. }
  34.  
  35. #define READ_SIZE (20 * 1024 * 1024)
  36. #define READ_BUFFER_SIZE (512 * 1024)
  37.  
  38. int main()
  39. {
  40.   int i, j ;
  41.   int fd ;
  42.   char read_buffer[READ_BUFFER_SIZE] ;
  43.   off_t filesize ;
  44.   unsigned int start_time, end_time ;
  45.  
  46.   fd = open(FILENAME, O_RDONLY) ;
  47.   if (fd == -1)
  48.   {
  49.     printf("could not open %s\n", FILENAME) ;
  50.     return 1 ;
  51.   }
  52.   filesize = lseek(fd, 0, SEEK_END) ;
  53.  
  54.   for (i = 0 ; i <3 ; i++)
  55.   {
  56.     if (i == 0)
  57.     {
  58.       printf("reading inner 20 MB...\n") ;
  59.       lseek(fd, 0, SEEK_SET) ;
  60.     }
  61.     else if (i == 1)
  62.     {
  63.       printf("reading middle 20 MB...\n") ;
  64.       lseek(fd, (filesize / 2) - (READ_SIZE / 2), SEEK_SET) ;
  65.     }
  66.     else
  67.     {
  68.       printf("reading outer 20 MB...\n") ;
  69.       lseek(fd, filesize - READ_SIZE, SEEK_SET) ;
  70.     }
  71.     /* read 20 MB ; 40 chunks of 1/2 MB */
  72.     start_time = get_current_milliseconds() ;
  73.     for (j = 0 ; j <(READ_SIZE / READ_BUFFER_SIZE) ; j++)
  74.       if (read(fd, read_buffer, READ_BUFFER_SIZE) != READ_BUFFER_SIZE)
  75.       {
  76.         printf("read error\n") ;
  77.         break ;
  78.       }
  79.     end_time = get_current_milliseconds() ;
  80.     printf("%d - %d = %d ms => %d kbytes/sec\n",
  81.       end_time, start_time, end_time - start_time,
  82.       READ_SIZE / (end_time - start_time)) ;
  83.   }
  84.  
  85.   close(fd) ;
  86.  
  87.   return 0 ;
  88. }

• Anomalie #1876 : Non prise en compte des guillemets dans le surlignage

  14 juin 2011, par cedric -

  voir aussi #2025

• Understanding the VP8 Token Tree

  7 juin 2010, par Multimedia Mike — VP8

  I got tripped up on another part of the VP8 decoding process today. So I drew a picture to help myself understand it. Then I went back and read David Conrad’s comment on my last post regarding my difficulty understanding the VP8 spec and saw that he ran into the same problem. Since we both experienced the same hindrance in trying to sort out this matter, I thought I may as well publish the picture I drew.

  VP8 defines various trees for decoding different syntax elements. There is one tree for decoding the tokens and it is expressed in the VP8 spec as such :

  PLAIN TEXT

  C :

  1. const tree_index coef_tree [2 * (num_dct_tokens - 1)] =
  2. {
  3.  -dct_eob, 2,        /* eob = "0"  */
  4.   -DCT_0, 4,        /* 0  = "10" */
  5.   -DCT_1, 6,        /* 1  = "110" */
  6.    8, 12,
  7.    -DCT_2, 10,      /* 2  = "11100" */
  8.     -DCT_3, -DCT_4,    /* 3  = "111010", 4 = "111011" */
  9.    14, 16,
  10.     -dct_cat1, -dct_cat2, /* cat1 = "111100", cat2 = "111101" */
  11.    18, 20,
  12.     -dct_cat3, -dct_cat4, /* cat3 = "1111100", cat4 = "1111101" */
  13.     -dct_cat5, -dct_cat6 /* cat4 = "1111110", cat4 = "1111111" */
  14. } ;

  Here is what the table looks like when you make a tree out of it (click for full size image) :

  
  
  

  The catch is that it makes no sense for an end-of-block (EOB) token to follow a 0 token since EOB already indicates that the remainder of the coefficients should be 0 anyway. Thus, the spec states that, "decoding of certain DCT coefficients may skip the first branch, whose preceding coefficient is a DCT_0." I confess, I didn’t understand what "skip the first branch" meant until I drew the tree.

  
  
  

  For those wondering why it might be sub-optimal (clarity-wise) for a spec to simply regurgitate vast chunks of C code, this makes a decent case. As you can see, the spec makes certain assumptions about how a binary tree should be organized in a static array (node n points to elements n*2 and n*2+1 as its branches ; leaves are either negative or 0). This is the second method I have seen ; another piece of code (not the VP8 spec) had the nodes in the first half of the array and pointed to leaves in the second half. There must be other arrangements.