How to calculate average seek time

hard drive

I'm reading Megatron 747 example from the book, "Database System Complete Book 2nd edition" (Example 13.2 pg.565-566). They give the following information:

  • 8 platters = 16 surfaces
  • 2^16 = 65536 tracks per surface
  • 256 sectors per track
  • 4096 bytes per sector

When they are calculating the average seek time, they do the following calculation:

1 + (65536/3)/4000 = 6.46 millisecond.

I understand that 65536 is the number of tracks as given and 1/3 is for averaging the distance, but I can't figure out why they added 1 and where they get the 4000.

Can anyone please help me out??

Best Answer

You don't even need to know anything about the topic (in fact, what you seem to know just gets you off track, this is about seek time only). Just look for all occurrences of 4000 in the description (there's just one), and a related occurrence of 1 millisecond in the same paragraph.

Quoting Database System - The Complete Book (2nd edition), emphasis mine:

To move the head assembly between cylinders takes one millisecond to start and stop, plus one additional millisecond for every 4000 cylinders traveled. Thus, the heads move one track in 1.00025 milliseconds and move from the innermost to the outermost track, a distance of 65,536 tracks, in about 17.38 milliseconds.

Rephrasing the calculation: It takes 17.38 ms to seek from innermost to outermost cylinder. Removing the 1 ms for start/stop leaves 16.38 ms for actual movement. One third of that (moving the average distance according to the author) is 5.46. Add the 1 ms for start/stop back in, you have your answer.

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