Essentials of Computer Organization and Architecture
Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123036
Author: Linda Null
Publisher: Jones & Bartlett Learning
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Chapter 5, Problem 27E

a.

Explanation of Solution

Explanation:

The size of the words present in memory unit is 24 bits or 3 Bytes.

The number of operations present in the instruction set is 150.

To accommodate the 150 operations we require Ceiling(log2

b.

Explanation of Solution

Explanation:

The size of the words present in memory unit is 24 bits or 3 Bytes.

The number of operations present in the instruction set is 150.

To accommodate the 150 operations we require Ceiling(log2(150))=8bits

c.

Explanation of Solution

Explanation:

The size of the words present in memory unit is 24 bits or 3 Bytes.

The number of operations present in the instruction set is 150.

To accommodate the 150 operations we require Ceiling(log2(150))=8bits.

Each instruction is identified by using its opcode and hence, the number of bits required by the opcode is 8 bits

d.

Explanation of Solution

Explanation:

The size of the words present in memory unit is 24 bits or 3 Bytes.

If N is the length of the binary string then, the largest unsigned integer that is stored in the string can be 2N1.

Substitute, “3 Bytes or 24 bits” for “N” in the above formula,

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