Chapter 5, Problem 28E

a.

Explanation of Solution

Explanation:

The number of words present in the memory unit is 256K.

The size of the words present in memory unit is 32 bits or 4 Bytes.

Each instruction includes opcode, mode, register and memory address.

To accommodate the 7 modes, the number of bits required is ${\text{Ceiling(l}}_{}$

b.

Explanation of Solution

Explanation:

The number of words present in the memory unit is 256K.

The size of the words present in memory unit is 32 bits or 4 Bytes.

Each instruction includes opcode, mode, register and memory address.

To accommodate the 60 registers, the number of bits required is ${\text{Ceiling(log}}_{}$

c.

Explanation of Solution

Explanation:

The size of the words present in memory unit is 32 bits or 4 Bytes.

Each instruction includes opcode, mode, register and memory address.

The memory is word addressable and there are 256K words present in the memory unit.

To accommodate each word, the number of bits required can be calculated as below ${\text{Ceiling(log}}_{}$

d.

Explanation of Solution

Explanation:

The size of the words present in memory unit is 32 bits or 4 Bytes.

The memory is word addressable and includes 256K words.

Each instruction includes opcode, mode, register and memory address.

The total length of an instruction is 32 bits whereas mode, register and memory address includes 3 bits, 6 bits and 18 bits respectively and forms 27 bits as whole...