Answers to Self-Review Questions - Java Software Solutions: Foundations of Program Design - page 736

a. 3 KB 3 * 1,024 bytes 3,072 bytes = approximately 3 thou-

sand bytes
b. 2 MB

2 * 1,048,576 bytes 2,097,152 bytes = approximately

2.1 million bytes
c. 4 GB

4 * 1,073,741,824 bytes 4,294,967,296 bytes = approxi-

mately 4.3 billion bytes

SR 1.8 There are eight bits in a byte. Therefore:

a. 8 bytes 8 * 8 bits 64 bits

b. 2 KB 2 * 1,024 bytes 2,048 bytes 2,048 * 8 bits = 16,384

bits
c. 4 MB 4 * 1,048,576 bytes 4,194,304 bytes 4,194,304 * 8

bits 33,554,432 bits

SR 1.9 Under the stated conditions, one hour of music would require

288,000,000 bytes (40,000 2). Dividing this number by

the number of bytes in a megabyte (1,048,576 bytes) gives approxi-
mately 275 MB. Note that a typical audio CD has a capacity of about
650 MB and can store about 70 minutes of music. This coincides with
an actual sampling rate of 41,000 measurements per second, two bytes
of storage space per measurement, and the need to store two streams of
music to produce a stereo effect.

60 60

SR 1.10 The two primary hardware components are main memory and the CPU.

Main memory holds the currently active programs and data. The CPU
retrieves individual program instructions from main memory, one at a
time, and executes them.

SR 1.11 A memory address is a number that uniquely identifies a particular

memory location in which a value is stored.

SR 1.12 Main memory is volatile, which means the information that is stored

in it will be lost if the power supply to the computer is turned off.
Secondary memory devices are nonvolatile; therefore, the information
that is stored on them is retained even if the power goes off.

SR 1.13 The word that best matches is

a. peripheral b. controller c. modem d. main or RAM
e. secondary or ROM f. RAM g. CPU

1.3 Networks
SR 1.14 A file server is a network computer that is dedicated to storing and pro-

viding programs and data that are needed by many network users.