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A disk is advertised with a seek time of 3 ms, 512 bytes per sector and 128 sectors per track.  The disk rotates at 5, 200 rpm.

i.    Determine the average rotational delay for the disk.

ii.    Determine the time required to read a 4 Mbyte file.  You are to assume that the file occupies sectors on adjacent tracks.
asked in CO & Architecture by (7 points) | 98 views

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5200 rotation --------> 60 seconds

  1     rotation  --------> $\frac{60}{5200}$ = 11.53 msec

i) Avg rotational delay = $\frac{11.53}{2}$ = 5.76 msec

1 track data = 128 × 512 = 65536 Byte

now,

1 track data wiil be fetched in 1 rotation

65536 byte -------> 11.53 msec

  4 Mbyte    -------> $\frac{11.53}{65536}$ × $2^{22}$  = 737.92 msec

Number of tracks required to store 4 Mbyte = $\frac{2^{22}}{65536}$ = 64

Since data are stored in adjacent track, seek time and rotation time will be considered for every track change

ii) Time to read 4 Mbyte = ((Seek time + Avg rotation time) × 64)  + Time to transfer 4 Mbyte

                                            = ((3 msec + 5.76 msec) × 64) + 737.92 msec

                                            = 560.64 msec + 737.92 msec

                                            = 1298.56 msec
answered by Junior (751 points)


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