Hard Drive Seek Time using FIFO, SSF, SCAN A hard disk spins at 6000 rpm (revolutions per minute), and it takes 100 μs (on average) for the head to traverse one track. Consider the following sequence of disk track requests: 27, 129, 110, 186, 147, 41, 10, 64, 120. Assume that initially the head is at track 30 and is moving in the direction of decreasing track numbers. Compute the time it takes to serve the requests using (1) FIFO, (2) SSF (Shortest-Seek First), (3) SCAN.

I understand the difference between the different algorithms and I generally get how to calculate it. 

My questions is, when I calculate the Average rotational time, i.e. (60/6000) / 2 = 5ms, do I need to multiply this by the number of tracks moved?

My FIFO calculation as an example:

(3+102+19+76+39+106+31+54+56) * (5ms + 0.1ms) = 2478.6ms

I add up the number of tracks moved and multiply this by the average rotational time + the time it takes to traverse each track.

Is this approach correct?

Thank you
@greybeard why he divided by 2
on average we will need to wait for half a rotation of the disk for the correct sector to come under the head. Thus, on average, the rotational latency is half the time it takes the disk to make a complete revolution.

do I need to multiply this by the number of tracks moved?
No. Because average rotational time is the time to move head from one sector to any other sector in same track. This movement is done by disk rotation.

In your question we will consider any sector in given track. So when track changes from 30 to 27, it takes 3 track movement and 1 average rotation so time 3*0.1+5

In general, total time = seek time + rotating latency

Seek time : time to change track

Edited

(3+102+19+76+39+106+31+54+56)*0.1 + number_of_time_movement_occured * rotational_latency
= 486 * 0.1 + 9 * 5
= 48.6 + 45
= 93.6