A 2.5 inch diameter disk has 8 platters with each platter having two data recording surfaces each platter on disk has 4084 tracks each track has 400 sectors and one sector can store 1 MB of data .Calculate the storage capacity of the disk in bytes if this disk has a seek time of 2millisecond and rotates at the speed of 6000 rpm find the access time for the disk make suitable assumption if any
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Surface m=8X2=16
Track (t)=4048
Sector per track (p)=400
Byte per sector (s)=1MB X 1024 X 102 bytes
Capacity =mtps
=16 X 4048 X 400 X 1MB X 1024 X 1024 bytes
Access time= seesk time + latency time
Latency time = (1/(RPM/60))*0.5
=(1/(6000/60))*0.5
=(1/100)*0.5*1000 ms
=5ms
Access time =2ms +5ms=7ms
hope i answered your question.
Track (t)=4048
Sector per track (p)=400
Byte per sector (s)=1MB X 1024 X 102 bytes
Capacity =mtps
=16 X 4048 X 400 X 1MB X 1024 X 1024 bytes
Access time= seesk time + latency time
Latency time = (1/(RPM/60))*0.5
=(1/(6000/60))*0.5
=(1/100)*0.5*1000 ms
=5ms
Access time =2ms +5ms=7ms
hope i answered your question.
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