6.2.2 Describe access time in terms of latency (rotational delay), seek time and transfer time.
(No teaching notes for this one.)
JSR Notes:
The first thing to mention here is that the way you do what is stated above is to add up the three values; which is pretty logical, but don’t just define the three things in a question like this, add them up as well.
Basically, to access data from a hard drive with a read/write head on an arm, the arm has to rotate, and then it has to wait for the particular sector to get around to it. Keep in mind that the disks are always spinning from the time you turn the computer on. So upon rotation to the correct track, the hard drive just has to wait – certainly not very long, but consider that a millisecond is a millisecond, and that’s actually a lot compared to the CPU that is working thousands of times faster than this.
So which is which. Well, in terms of the read/write head itself, it is only “taking the initiative” as it were, to seek, when it rotates. So it is the rotation of the read/write arm which is the seek time; and it has a definite, measurable time, from the outer to inner-most track, and this is a characteristic of the particular drive. The latency, then, is the time beyond this that the read/write head needs to wait as the disk spins around. Certainly, it could get lucky, and the particular track in question could be there almost immediately, or it could have to wait for an almost full rotation. In terms of the specs of the hard drive, that would be the revolutions per minute – 7200 RPM is standard.
If you can remember that latency is also rotational delay, you won’t get it wrong.
But don’t forget in the overall calculation, the time it takes to transfer the signal is also important, and is also a characteristic of a particular hard drive.
You’d be looking at an overall access time in the order of 10 milliseconds:
For example:
Seek: 6 ms
Latency: 4.17 ms
Transfer Time: 0.17 ms
10.34 ms