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An application loads $100$ libraries at startup. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as $10$ ms. Rotational speed of disk is $6000$ rpm. If all $100$ libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected.)

1. $0.50 \ s$
2. $1.50 \ s$
3. $1.25 \ s$
4. $1.00 \ s$
edited | 3.4k views

$\text{Disk access time} =\text{Seek time}+\text{Rotational latency}+\text{Transfer time}$ (given that transfer time is neglected)
$\text{Seek time}=10 \text{ ms}$
$\text{Rotational speed}=6000 \text{ rpm}$

• $60\;s \to 6000 \text{ rotations}$
• $1 \text{ rotation}\to 60/6000 \;s$
• $\text{Rotational latency} =1/2 \times 60/6000 \;s =5 \text{ ms}$

Total time to transfer one library $=10+5=15 \text{ ms}$
$\therefore$ Total time to transfer $100$ libraries $=100 \times 15 \text{ ms}=1.5 s$

Correct Answer: $B$

edited
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tational latency---1/2 *60/6000 sec=5msec

please tell me, why are u dividing by 2 ?

+14
thats defined in the formula. It is the average rotational latency and since min. rotation needed being 0 (when access is to current location) and maximum being 1 full rotation, the average latency is taken as the time for 1/2 rotation.
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@arjun sir
the average rotational latency is typically based on the empirical relation that the average latency in milliseconds for such a drive is one-half the rotational period.
but how will we come to know here we need to take average rotational latency?

+1
When do we need to divide the seek time by 2? I had divided both by 2, and got 1sec as ans. Kind of confused here
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Why we are not taking average of seek time?
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It is not average seek time
It is seek time
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Formula :-

Average rotation latency = (1/2)× rotation time

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