TEST SERIES question

Question

Given Ts : Transfer time.

b: Number of bytes to be transfered.

N: Number of bytes on a track.

r: Rotational speed.

Which of the following expression gives the total avg access time?

a. Ts+(r/2)+(rb/N)

b. Ts+(1/2r)+(rb/N)

c. Ts+(1/2r)+(N/rb)

d. Ts+(1/2r)+(b/rN)

Answer 1

Avg access time=seek time+rotational latency+ transfer

Rotational latency=1/2 rotational time

=1/2r

Transfer time =b/Nr

 So ans is d

Comments

where is the seek time given, and avg rotational latency is rotation time/2 na????

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this is the printing mistake . the answer will be d .