3. Suppose a disk drive in your system has a rotation speed of 7,200 rpm. and se

Question

3. Suppose a disk drive in your system has a rotation speed of 7,200 rpm. and seek times of 0.68 ms track-to-track, or 21 ms full ‘stroke’ (inner edge to outer edge). (a) What is the average latency to access this drive? (b) If the sustained read rate is 42 Mbits/sec, how much more time after the average latency (you calculated above) will it take to read 750 Kbytes (assuming all those bytes are in sequential sectors on the same track)? How long to write that same amount of data if the sustained write speed is 32 Mbits/sec?

Explanation / Answer

Please find my Answer:

(a) formula for average latency

((1/spindle speed/60)) *0,5*1000

rpm= 7200

((1/7200/60))*0.5*1000

so 60/7200*0.5*1000

0.00833*0.5*1000

4.16ms

(b)sustained read rate after average latency

FORMULA

Seek time + Average rotational latency delay + Sector read time + Overhead controller

0.68ms+4.16ms+(42/1000)+21ms=25.88ms

Seek time + Average rotational latency delay + Sector write time + Overhead controller

0.68ms+4.16ms+(32/1000)+21ms=25.872ms

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