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 ${\text{Ceiling(log}}_{\text{2}}$

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 ${\text{Ceiling(log}}_{\text{2}}\text{(150))}\text{=}\text{8}\text{bits}$

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 ${\text{Ceiling(log}}_{\text{2}}\text{(150))}\text{=}\text{8}\text{bits}$.

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 ${\text{2}}^{\text{N}}-1$.

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

$\begin{array}{c}\text{Unsigned}\end{array}$