What would be the average I/O time required to transfer a 6KB block of data from

Question

What would be the average I/O time required to transfer a 6KB block of data from a hard disk drive to the memory if the drive spins at 5400 RPM? The delay involved in moving the disk’s read-write head assembly to the right track is 12 milliseconds on an average. The host and disk controllers need a negotiating time of 0.7 millisecond. The maximum theoretical per second transfer capability of the disk is 800 Mb of data. How would your calculations change if the maximum theoretical per second transfer capability of the disk is 800 MB of data?

Explanation / Answer

The average I/O time to transfer 6KB of block from hard disk is as follows:

1.Average Rotational delay = Time taken for 0.5 Rotations
                           = (60*0.5*1000)/5400 = 5.55ms

2.Average Seek Time = 12ms

3.Overhead time = 0.7 ms

4. Transfer time for 6KB with 800Mb per second = (6000*8)/(800 * 10^6) = 0.06 ms
   Transfer time for 6KB with 800MB per second = (6000*8)/(800 * 8 * 10^6) = 0.007 ms

   Average IO time (800 Mb/sec) =   Average Rotational delay + Average Seek Time + Overhead time + Transfer time
                   =   5.55 + 12 + 0.7 + 0.06 = 18.31 ms

   Average IO time (800 MB/sec) =   Average Rotational delay + Average Seek Time + Overhead time + Transfer time
                   =   5.55 + 12 + 0.7 + 0.007 = 18.257 ms

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.