4KB sector, 7200RPM, 2ms average seek time, 80MB/s transferrate, 0.4ms controller overhead, average waiting time in queue is 1second. What is the estimated time to access this sector on thisHard Drive Disk? (The result should be given in ms)
Answer
Time to access sector = seek time + average rotational time+controller overhead + sector read time+ waiting time in queue
Seek time = 2ms
average rotational time = 7200/60=120 rev per second or 8.3 revper ms, average is half of that 8.3/2 =4.15 ms
Controller overhead= 0.4ms
Sector read time= 4*2^10/(80*2^20) seconds
=0.048 ms
Waiting time in queue = 1sec=1000ms
Time to access sector= 2+4.15+0.4+0.048+1000
= 1006.598ms