Question

A magnetic disk has an average seek time of 8 ms. The transferrate is 50 MB/sec. The

disk rotates at 10,000 rpm and the controller overhead is 0.3msec. Find the average time to read or write 1024 bytes.

Answer 1

Answer:

11.32milisec.

Explanation:

Average seek time = 8ms

transfer rate = 50MB/sec

disk rotate = 10000 rpm

controller overhead = 0.3 msec

Average time = ?

Average time is the addition of the seek time, transfer time, rotation time and ontroller overhead

$Average\,time = Seek\,time+Rotation\,time+Transfer\,time+Controller\,overhead$

now, find the rotation time,

Rotation time is the time at which half rotation is made.

$Rotation\,time=\frac{0.5}{rotation\,per\,second}$

convert rotation per minute to rotation per second.

$rps=\frac{10000}{60}$

substitute in the above formula.

$Rotation\,time=\frac{0.5}{\frac{10000}{60} }=0.003sec$

or 3milisec

now, find the transfer time by formula:

$Transfer\,time=\frac{Number\,of \,read\,or\,write\,bytes}{transfer\,rate}$

so, read or write bytes is 1024

$[tex]Transfer\,time=\frac{1024}{50*10^{6} }=0.02milisec$[/tex]

The average time is:

$[tex]Average\,time= 8+3+0.02+0.3=11.32milisec$[/tex]