Percentage Improvement

Question

A magnetic disk has following specs:

# 0f tracks=1024

#of sectors/track=512

#of Bytes/sector =512

Disk Rotation speed=7200rpm

a)What is the percentage of improvement(approx) in average access time ,if disk speed is doubled?

a)16% b)18% c)19% d)None

Comments

is it d)?

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No

with old speed total access time is ~ 13 ms

with new speed ~ 11 ms

% improvement = 2/11 * 100 ~ around 18 must be the answer

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explain how u get 13ms and 11ms values of access time

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Can anyone explain in detail

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I think data transfer rate must be given. Are you missing anythng in the question?

Answer 1

seek time is not required

now consider only RL AND TRANSFER TIME.

RPM=7200

7200R=60 SEC

1R=60/7200=8.33MS

TO COVER 1 TRACK 8.33MS SINCE HOW MUCH DATA TO BE TRANSFERED IS NOT MENTIONED WE CONSIDER THIS AS THE DATA TRANSFER TIME

SO TOTAL TIME=8.33+(8.33/2)=12.49MS

NOW SPEED IS DOUBLED MEANS RPM=7200*2=14400RPM

1R=4.16MS

TOTAL TIME=4.16+(4.16/2)=6.24

IMPROVE=  (OLD-NEW)/OLD=(12.49-6.24)/6.24=APPROX 50%