Posted by hussein on August 13, 1999 at 10:24:30:
In Reply to: 9.2 and 9.4 posted by John on August 12, 1999 at 18:39:32:
Interleaving factor = 3
Time for one revolution = 1/7200 rpm = 8.3 ms
Number of tracks = 2100
Sectors/track = 63
Bytes/sector = 512
Sectors/cluster = 8
Assume that each cluster must fit within one track.
Then, Clusters/track = 7
Hence, Max Bytes/track = 7*8*512 = 28K
Therefore, on four tracks we can store 112K.
The last 16K (last 4 clusters) would fit on the 5th track, but because of the interleaving factor of 3, we need at least two revolutions of the disk to read in this data.
Assume that the first seek must traverse 1/3 of the tracks on average.
Assume that a new seek (with startup time) is required for each track read operation.
Assume that rotational delay is needed after every seek operation.
Assume on average that rotational delay is half the time taken for one revolution.
Time
= first track time + 3*(time for middle tracks) + time for last track
= (2100/3*0.08+3+3.5*8.3)+3*(0.08+3+3.5*8.3)+(0.08+3+2.5*8.3)
= 208.27 ms