MadeEasy Subject Test: Operating System - Disks

Question

Consider a typical disk that rotates at 30000 rotations per minute(RPM) and has a transfer rate of 40Mbps.If the average seek time of the disk is twice the average rotational delay and the controller's transfer time is 8 times the disk transfer time,the average time to read or write a 256 byte sector of the disk is __________msec.(upto 4 decimal places)

My queries-:

should the time taken by the controller be taken into account when calculating the average time?

I have calculated it in the following way-:

time taken=seek time+rotational latency+transfer time+controller time

=2+1+0.0512+0.4096

=3.4608

Comments

Just solved same question :)
https://gateoverflow.in/8251/gate2015-2_49

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in the link Arjun sir didn't consider transfer time ...so it should be added or not??

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added

Answer 1

Yes you are right, found one reference from geeks
ref: http://cse.unl.edu/~jiang/cse430/Lecture%20Notes/reference-ppt-slides/Disk_Storage_Systems_2.ppt

avg. rotational latency $= \frac{1}{2} \times \frac{60}{30000}= 1\ msec$

avg. seek time $= 2\ msec$

Transfer time $= \frac{256 \times 8}{40 \times 10^6}=0.0512\ msec$

Controller overhead $=8(0.0512)=0.4096\ msec$

avg $T_{read/write}= 2+1+0.0512+0.4096=3.4608\ msec$

So, the average time to read or write a 256 byte sector of the disk is 3.4608 msec