Question

Consider a hard disk with a rotational speed of 15000 rpm. The time to move the
read/write head from a track to its adjacent track is 1 millisecond. Initially, the head
is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes.
It is necessary to transfer data from 10 randomly located sectors in each of the
following tracks in the order: 5, 12 and 7.
The total time for the data transfer (in milliseconds) from the hard disk is _________.
(rounded off to one decimal place)

Hint:

First find seek time by summing track jumps 0 to 5 to 12 to 7 at 1 ms per track. Then find rotation time from 15000 rpm (4 ms per revolution), and note that each of the 10 random sectors per track needs an average half-rotation wait plus one sector's transfer time (4/400 ms). Multiply per-sector cost by 10, then by 3 tracks, and add the seek time.

Answer 1

Correct Answer: 77.3

Given data: rotational speed = 15000 rpm, seek time between adjacent tracks = 1 millisecond, sectors per track = 400, sector size = 1024 bytes. The head starts at track 0 and must read 10 randomly located sectors from each of the tracks in the order 5, 12, then 7.

Step 1: Time for one full rotation

Rotational speed = 15000 rpm = 15000/60 = 250 rotations per second.

\[ T_{rotation} = \frac{1}{250}\text{ s} = 4\text{ ms} \]

Step 2: Seek time between tracks

The head moves in the order 0 to 5 to 12 to 7. The number of tracks crossed at each move is:

\[ |5-0| + |12-5| + |7-12| = 5 + 7 + 5 = 17 \]

Since each adjacent track move takes 1 ms:

\[ T_{seek} = 17 \times 1 = 17\text{ ms} \]

Step 3: Average rotational latency for one sector access

Because the 10 sectors needed on each track are randomly located and not accessed in any physically sorted order, each individual sector request must, on average, wait for half a rotation before the required sector arrives under the head:

\[ T_{latency} = \frac{T_{rotation}}{2} = \frac{4}{2} = 2\text{ ms} \]

Step 4: Transfer time for one sector

With 400 sectors spread over one rotation, the time for one sector to pass under the head (its transfer time) is:

\[ T_{transfer} = \frac{T_{rotation}}{400} = \frac{4}{400} = 0.01\text{ ms} \]

Step 5: Time to service one random sector

\[ T_{sector} = T_{latency} + T_{transfer} = 2 + 0.01 = 2.01\text{ ms} \]

Step 6: Time to service 10 random sectors on one track

\[ 10 \times 2.01 = 20.1\text{ ms} \]

Step 7: Time to service 10 sectors on each of the 3 tracks (5, 12, 7)

\[ 3 \times 20.1 = 60.3\text{ ms} \]

Step 8: Total transfer time

\[ T_{total} = T_{seek} + 60.3 = 17 + 60.3 = 77.3\text{ ms} \]

Final Answer:

\[ \boxed{77.3\text{ ms}} \]