Chapter 4, Problem 14E

a)

Explanation of Solution

Given:

A computer has a memory unit with 32 bits per word

Number of operations in the instruction is 110.

Number of required bits for the opcode:

Convert the number of operations of instruction set into binary form:

b)

Explanation of Solution

Number of required bits to specify the register:

The number of registers is 8 in memory. So, the user needs to determine the binary representation of the number. The conversion is as follows:

$\begin{array}{c}{\text{n = ceiling(log}}_2\end{array}$

c)

Explanation of Solution

Number of bits left in the address part of the instruction:

From given memory unit, it has “32 bits” per word and number of required bits for opcode is “7 bits” and register is “3 bits”.

The number of bits left for the address part is calculated as follows:

d)

Explanation of Solution

Maximum allowable size for memory:

The formula for finding number of words in memory is ${\text{2}}^{\text{N}}$.

Here, “N” is the number of address bits.  The number of address bits is “22” which is fetch from above subpart

e)

Explanation of Solution

Largest unsigned binary number in one word of memory:

The formula for calculating the largest unsigned binary number is $\text{Number of bits per word - 1}$.

Note: The number of address bits is “(2^{length of word in bits} )-1”