A disk drive has a speed of 12000 RPM, an average seek time of 5 ms, and a trans

Question

A disk drive has a speed of 12000 RPM, an average seek time of 5 ms, and a transfer rate of 140 MB/sec. How much time does it take to perform 50 random reads of 10 KB? (In other words, the 50 reads are from random spots on the drive, but each individual read is performed sequentially.) Give your answer in units of seconds. Given the same information as in c) above, what is the I/O rate for the 50 reads? Give your answer in MB/sec.

Answer is: (Please check)

Disk Access Time = seek time + rotational delay + transfer time

                             = 5 + (0.5*60*1000/12000) + 10*1024*50*1000/140*2^20

                            = 5 + 2.5 + 3.48

                           = 10.98 ms

Given the same information as in c) above, what is the I/O rate for the 50 reads? Give your answer in MB/sec.

Explanation / Answer

Disk Access Time calculated is absolutely correct and it is 10.98 ms.

Now for calculating I/O rate we will do the below

Given 50 reads is of 10 KB

I/O rate will be 10 KB / 10.98 ms

which is 10 KB / 10.98 * 10-3 s

which is 10 MB / 10.98 s

which is 0.91 MB / s

= 0.91 MB / s

Hence the I/O rate is 0.91 MB /s

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