Given: Given that the disk drive has the following features: Number of surfaces = 4 Tracks per surface = 1024 Sectors per track = 128 Bytes per sector = 512 bytes Seek time = 5ms Rotational speed = 5000rpm 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. Capacity of the drive = number of surfaces × tracks per surface × sectors per track × bytes per sector Capacity&

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Chapter 7, Problem 28E

a)

Explanation of Solution

Given:

Given that the disk drive has the following features:

Number of surfaces = 4

Tracks per surface = 1024

Sectors per track = 128

Bytes per sector = 512 bytes

Seek time = 5ms

Rotational speed = 5000rpm

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.

Capacity of the drive = number of surfaces × tracks per surface × sectors per track × bytes per sectorCapacity&

b)

Explanation of Solution

Given:

Given that the disk drive has the following features:

Number of surfaces = 4

Tracks per surface = 1024

Sectors per track = 128

Bytes per sector = 512 bytes

Seek time = 5ms

Rotational speed = 5000rpm

To find: Access time

Solution:

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

Rotational delay = 1(RPM60)×0

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