Chapter 7, Problem 31E

a.

Explanation of Solution

Given:

Given that the disk drive has the following features:

Number of surfaces = 6

Tracks per surface = 953

Sectors per track = 256

Bytes per sector = 512 bytes

Seek time = 6.5ms

Rotational speed = 5400rpm

To find: Capacity of the drive

Solution:

The capacity of the drive is manipulated by multiplying number of surfaces, tracks per surface, sectors per track and bytes per sector.

$\begin{array}{l}\text{Capacity of the drive = number of surfaces × tracks per surface × sectors per track × bytes per sector}\\ \text{Capacity of thex}\end{array}$

b.

Explanation of Solution

Given:

Given that the disk drive has the following features:

Number of surfaces = 6

Tracks per surface = 953

Sectors per track = 256

Bytes per sector = 512 bytes

Seek time = 6.5ms

Rotational speed = 5400rpm

To find: Access time

Solution:

The summation of rotational delay and seek time is called access time.

Rotational delay = $\frac{\text{1}}{\text{(}\frac{\text{RPM}}{\text{60}}\text{)}}\text{×0}$

c.

Explanation of Solution

Faster disk:

“Yes”, this disk faster than the disk that was specified in Exercise 30 because, the access time “12.66ms” in “Exercise 30” is slower than the access time “12...