Question

Assume that we are using a hard disk with the following characteristics:

 Capacity 30 GB

 Average seek time 6 msec

 Spindle speed 15,000 rpm

 Bytes per sector 512

 Sectors per cluster 8

 Sectors per track 400.

Assume that we have a file of 20, 000 records.

Each record has 64 bytes. Answer the following questions:

A. How much is the average rotational delay?

B. How many records can be stored in one sector?

C. How many clusters are needed for the file?

D. How much is the time for reading one track, including the seek time, the average rotational delay, and the transfer time?

E. Assuming contiguous storage, namely, the records are stored in clusters from the same track as long as possible, how much is the total time for reading the whole file?

F. Assuming the minimal transfer unit is one cluster, how much is the time for reading one cluster, including the seek time, the average rotational delay, and the transfer time?

G. Assuming random storage of the records and every record is stored in a different cluster, reading a cluster will include the seek time, the average rotational delay, and the transfer time for that cluster, how much is the total time for reading the whole file?

Answer 1

As per Chegg answering guidelines, I'm answering the first 4 parts:-

A) 15000 RPM.

For 15000 rotations time needed = 60 seconds = 60000 ms

Therefore, for 1 rotation time needed = 60000/15000 = 4 ms

Therefore, average rotational latency = (0.5)*4 = 2ms

B) Bytes per sector = 512 bytes

Record size = 64 bytes

Therefore, number of records that can be accomadated in one sector = 512/64 = 8 records

C) Total number of records in the file = 20000

size of each record = 64 bytes

Therefore size of file = 64*20000 = 1280000 bytes

Number of sectors in each cluster = 8

Size of each sector = 512

Therefore, size of each cluster = 512*8 = 4096 bytes

Therefore, number of clusters needed to accomodate the file = 1280000/4096 = [312.5] = 313

D) Average seek time = 6 ms

Average rotational latency = 2 ms [From A]

Transfer time:

rotation speed as calculated in part A) = 4 ms

Number of bytes in a track = 400*512 = 2048 bytes

Therefore, transfer rate = 2048/4 = 512 bytes/ms

Therefore time to transfer bytes of 1 track = 512/512 ms = 1 ms

Therefore, time to read one track = average rotational latency + average seek time + transfer time = 2 + 6 + 1 = 9ms