seek time

Question

the seek time of a disk is 30ms.it rotates at the rate of 30 rotations/second.the capacity of each track is 300 words.the access time is (approximately)

 

guys can any one solve this plzzzz...

Comments

The following question is given in the booklet of M*d* Easy. But, the answer given is 50 ms. But checked in different websites found the posted answer as 46.78 ms. So the answer should be that.

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So in GATE exam Hall can you go and check every websites?

Answer 1

Access time is defined as time to access one word from memory

So, it may be the answer

1 rot = 1/30 sec

In 1 rotation ---> 100 words

For 1 word ---> 1/300 rotations

So, 1 word transfer time = 1/300*1/30 sec

So, avg access time= 30 ms(seek time)+ 1/60 sec (avg rotational latency) + 1/ 9000 sec (transfer time)

= 30+16.6667+.1111 ms

= 46.78 ms(approx)

Comments

Transfer time = track size/ data transfer rate  = 300/ (1/30) ....am i doing wrong ?

@srestha

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@admin  Correction needed: 1 rotation -→ 300 words(not 100 words)